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Learning Outcomes
• Describe the properties of a suspension.
• Describe the properties of a colloid.
• Distinguish among suspensions. colloids, and solutions.
Suspensions
Take a glass of water and throw in a handful of sand or dirt. Stir it and stir it and stir it. Have you made a solution? Sand and dirt do not dissolve in water, and, though it may look homogenous for a few moments, the sand or dirt gradually sinks to the bottom of the glass (see figure below). Some medications are delivered as suspensions and must be mixed well before the doses measured to make sure the patient is receiving the correct amount of medication.
A suspension is a heterogeneous mixture in which some of the particles settle out of the mixture upon standing. The particles in a suspension are far larger than those of a solution, so gravity is able to pull them down out of the dispersion medium (water). The diameter for the dispersed particles in a suspension, such as the sand in the suspension described above, is typically at least 1000 times greater than those in a solution. Unlike a solution, the dispersed particles can be separated from the dispersion medium by filtering. Suspensions are considered heterogeneous because the different substances in the mixture will not remain uniformly distributed if they are not actively being mixed.
Colloids
A colloid is a heterogeneous mixture in which the dispersed particles are intermediate in size between those of a solution and a suspension. The particles are spread evenly throughout the dispersion medium, which can be a solid, liquid, or gas. Because the dispersed particles of a colloid are not as large as those of a suspension, they do not settle out upon standing. The table below summarizes the properties and distinctions between solutions, colloids, and suspensions.
Table $1$: Properties of Solutions, Colloids, and Suspensions
Solution Colloids Suspensions
Homogeneous Heterogeneous Heterogeneous
Particle size: $0.01$-$1 \: \text{nm}$; atoms, ions or molecules Particle size: $1$-$1000 \: \text{nm}$, dispersed; large molecules or aggregates Particle size: over $1000 \: \text{nm}$, suspended: large particles or aggregates
Do not separate on standing Do not separate on standing Particles settle out
Cannot be separated by filtration Cannot be separated by filtration Can be separated by filtration
Do not scatter light Scatter light (Tyndall effect) May either scatter light or be opaque
Colloids are unlike solutions because their dispersed particles are much larger than those of a solution. The dispersed particles of a colloid cannot be separated by filtration, but they scatter light, a phenomenon called the Tyndall effect.
Tyndall Effect
Colloids are often confused with true homogenous solutions because the individual dispersed particles of a colloid cannot be seen. When light is passed through a true solution, the dissolved particles are too small to deflect the light. However, the dispersed particles of a colloid, being larger, do deflect light (see figure below). The Tyndall effect is the scattering of visible light by colloidal particles. You have undoubtedly "seen" a light beam as it passes through fog, smoke, or a scattering of dust particles suspended in air. All three are examples of colloids. Suspensions may scatter light, but if the number of suspended particles is sufficiently large, the suspension may simply be opaque, and the light scattering will not occur.
Examples of Colloids
Listed in the table below are examples of colloidal systems, most of which are very familiar. Some of these are shown below (see figure below). The dispersed phase describes the particles, while the dispersion medium is the material in which the particles are distributed.
Table $2$: Classes of Colloids
Class of Colloid Dispersed Phase Dispersion Medium Examples
Sol and gel solid liquid paint, jellies, blood, gelatin, mud
Solid aerosol solid gas smoke, dust in air
Solid emulsion liquid solid cheese, butter
Liquid emulsion liquid liquid milk, mayonnaise
Liquid aerosol liquid gas fog, mist, clouds, aerosol spray
Foam gas solid marshmallow
Foam gas liquid whipped cream, shaving cream
Emulsions
Butter and mayonnaise are examples of a class of colloids called emulsions. An emulsion is a colloidal dispersion of a liquid in either a liquid or a solid. A stable emulsion requires an emulsifying agent to be present. Mayonnaise is made in part of oil and vinegar. Since oil is nonpolar, and vinegar is a polar aqueous solution, the two do not mix and would quickly separate into layers. However, the addition of egg yolk causes the mixture to become stable and not separate. Egg yolk is capable of interacting with both the polar vinegar and the nonpolar oil. The egg yolk is called the emulsifying agent. Soap acts as an emulsifying agent because one end of a soap molecule is polar, and the other end is nonpolar. This allows the grease to be removed from your hands or your clothing by washing with soapy water.
Supplemental Resources
• Types of Mixtures - Solutions, Suspensions, Colloids: www.edinformatics.com/math_sc.../mixtures.html
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/07%3A_Solids_Liquids_and_Gases/7.06%3A_Colloids_and_Suspensions.txt |
Learning Outcomes
• List examples of solutions made from different solute-solvent combinations.
• Distinguish between saturated and unsaturated solutions.
• Explain the effects of temperature on the solubility of solids and gases.
• Use a solubility curve to determine the solubilities of substances at various temperatures.
• Explain the effect of pressure on the solubility of gases.
Previously, we looked at the primary characteristics of a solution and how water is able to dissolve solid solutes. There are many examples of solutions that do not involve water at all, or solutions that involve solutes that are not solids. The table below summarizes the possible combinations of solute-solvent states, along with examples of each.
Table $1$: Solute-Solvent Combinations
Solute State Solvent State Example
liquid gas water in air
gas gas oxygen in nitrogen (gas mixture)
solid liquid salt in water
liquid liquid alcohol in water
gas liquid carbon dioxide in water
solid solid zinc in copper (brass alloy)
liquid solid mercury in silver and tin (dental amalgam)
Air is a homogeneous mixture of many different gases and, therefore, qualifies as a solution. Solid-solid solutions, such as brass, bronze, and sterling silver, are called alloys. Fish depend on oxygen gas that is dissolved in the water found in oceans, lakes and rivers (see figure below). While solid-liquid and aqueous solutions comprise the majority of solutions encountered in the chemistry laboratory, it is important to be aware of the other possibilities.
Rate of Dissolving
We know that the dissolving of a solid by water depends upon the collisions that occur between the solvent molecules and the particles in the solid crystal. Anything that can be done to increase the frequency of those collisions and/or to give those collisions more energy will increase the rate of dissolving. Imagine that you were trying to dissolve some sugar in a glassful of tea. A packet of granulated sugar would dissolve faster than a cube of sugar. The rate of dissolving would be increased by stirring, or agitating the solution. Finally, the sugar would dissolve faster in hot tea than it would in cold tea.
Surface Area
The rate at which a solute dissolves depends upon the size of the solute particles. Dissolving is a surface phenomenon, since it depends on solvent molecules colliding with the outer surface of the solute. A given quantity of solute dissolves faster when it is ground into small particles than if it is in the form of a large chunk, because more surface area is exposed. The packet of granulated sugar exposes far more surface area to the solvent and dissolves more quickly than the sugar cube.
Agitation of the Solution
Dissolving sugar in water will occur more quickly if the water is stirred. The stirring allows fresh solvent molecules to continually be in contact with the solute. If it is not stirred, then the water right at the surface of the solute becomes saturated with dissolved sugar molecules, meaning that it is more difficult for additional solute to dissolve. The sugar cube would eventually dissolve because random motions of the water molecules would bring enough fresh solvent into contact with the sugar, but the process would take much longer. It is important to realize that neither stirring nor breaking up a solute affect the overall amount of solute that dissolves. It only affects the rate of dissolving.
Temperature
Heating up the solvent gives the molecules more kinetic energy. The more rapid motion means that the solid or liquid solvent molecules collide with the solute with greater frequency, and the collisions occur with more force. Both factors increase the rate at which the solid or liquid solute dissolves. As we will see in the next section, a temperature change not only affects the rate of dissolving, but it also affects the amount of solute that can be dissolved.
Types of Solutions
Table salt $\left( \ce{NaCl} \right)$ readily dissolves in water. Suppose that you have a beaker of water to which you add some salt, stirring until it dissolves. Then you add more, and that dissolves as well. If you keep adding more and more salt, eventually you will reach a point at which no more of the salt will dissolve, no matter how long or how vigorously you stir it. Why? On the molecular level, we know that the action of the water causes the individual ions to break apart from the salt crystal and enter the solution, where they remain hydrated by water molecules. What also happens is that some of the dissolved ions collide back again with the crystal and remain there. Recrystallization is the process of dissolved solute returning to the solid state. At some point, the rate at which the solid salt is dissolving becomes equal to the rate at which the dissolved solute is recrystallizing. When that point is reached, the total amount of dissolved salt remains unchanged.
When the solution equilibrium point is reached and no more solute will dissolve, the solution is said to be saturated. A saturated solution is a solution that contains the maximum amount of solute that is capable of being dissolved. At $20^\text{o} \text{C}$, the maximum amount of $\ce{NaCl}$ that will dissolve in $100. \: \text{g}$ of water is $36.0 \: \text{g}$. If any more $\ce{NaCl}$ is added past that point, it will not dissolve because the solution is saturated. What if more water is added to the solution instead? Now, more $\ce{NaCl}$ would be capable of dissolving, since there is additional solvent present. An unsaturated solution is a solution that contains less than the maximum amount of solute that is capable of being dissolved. The figure below illustrates the above process, and shows the distinction between unsaturated and saturated.
How can you tell if a solution is saturated or unsaturated? If more solute is added and it does not dissolve, then the original solution was saturated. If the added solute dissolves, then the original solution was unsaturated. A solution that has extra undissolved solute at the bottom of the container, must be saturated (see figure below).
Solubility Values
The solubility of a substance is the amount of that substance that is required to form a saturated solution in a given amount of solvent at a specified temperature. Solubility is often measured in grams of solute per $100 \: \text{g}$ of solvent. The solubility of sodium chloride in water is $36.0 \: \text{g}$ per $100 \: \text{g}$ of water at $20^\text{o} \text{C}$. The temperature must be specified because solubility varies with temperature. For gases, the pressure must also be specified. Solubility is specific for a particular solvent. In other words, the solubility of sodium chloride would be different in another solvent. For the purposes of this text, the solubility of a substance will refer to aqueous solubility unless otherwise specified. Solubilities for different solutes have a very wide variation, as can be seen by the data presented in the table below.
Table $2$: Solubility of Solutes at Different Temperatures ($\text{g}$ solute in $100 \: \text{g} \: \ce{H_2O}$)
Substance $0^\text{o} \text{C}$ $20^\text{o} \text{C}$ $40^\text{o} \text{C}$ $60^\text{o} \text{C}$ $80^\text{o} \text{C}$ $100^\text{o} \text{C}$
$\ce{AgNO_3}$ 122 216 311 440 585 733
$\ce{Ba(OH)_2}$ 1.67 3.89 8.22 20.94 101.4 __
$\ce{C_{12}H_{22}O_{11}}$ 179 204 238 287 362 487
$\ce{Ca(OH)_2}$ 0.189 0.173 0.141 0.121 __ 0.07
$\ce{KCl}$ 28.0 34.2 40.1 45.8 51.3 56.3
$\ce{KI}$ 128 144 162 176 192 206
$\ce{KNO_3}$ 13.9 31.6 61.3 106 167 245
$\ce{LiCl}$ 69.2 83.5 89.8 98.4 112 128
$\ce{NaCl}$ 35.7 35.9 36.4 37.1 38.0 39.2
$\ce{NaNO_3}$ 73 87.6 102 122 148 180
$\ce{CO_2}$ $\left( 1 \: \text{atm} \right)$ 0.335 0.169 0.0973 0.058 __ __
$\ce{O_2}$ $\left( 1 \: \text{atm} \right)$ 0.00694 0.00537 0.00308 0.00227 0.00138 0.00
Factors Affecting Solubility
The solubility of a solid or a liquid solute in a solvent is affected by the temperature, while the solubility of a gaseous solute is affected by both the temperature and the pressure of the gas. We will examine the effects of temperature and pressure separately.
Temperature
The solubility of the majority of solid substances increases as the temperature increases. However, the effect is difficult to predict and varies widely from one solute to another. The temperature dependence of solubility can be visualized with the help of a solubility curve, which is a graph of the solubility vs. temperature. Examine the solubility curves shown (see figure below).
Notice how the temperature dependence of $\ce{NaCl}$ is fairly flat, meaning that an increase in temperature has relatively little effect on the solubility of $\ce{NaCl}$. The curve for $\ce{KNO_3}$, on the other hand, is very steep; an increase in temperature dramatically increases the solubility of $\ce{KNO_3}$.
Several substances listed on the graph - $\ce{HCl}$, $\ce{NH_3}$, and $\ce{SO_2}$ - have solubilities that decrease as the temperature increases. These substances are all gases over the indicated temperature range when at standard pressure. When a solvent with a gas dissolved in it is heated, the kinetic energy of both the solvent and solute increases. As the kinetic energy of the gaseous solute increases, its molecules have a greater tendency to escape the attraction of the solvent molecules and return back to the gas phase. As a result, the solubility of a gas decreases as the temperature increases. This has some profound environmental consequences. Industrial plants situated near bodies of water often use that water as a coolant, returning the warmer water back to the lake or river. This increases the overall temperature of the water, which lowers the quantity of dissolved oxygen, affecting the survival of fish and other organisms.
Solubility curves can be used to determine if a given solution is saturated or unsaturated. Suppose that $80 \: \text{g}$ of $\ce{KNO_3}$ is added to $100 \: \text{g}$ of water at $30^\text{o} \text{C}$. According to the solubility curve, approximately $48 \: \text{g}$ of $\ce{KNO_3}$ will dissolve at $30^\text{o} \text{C}$. This means that the solution will be saturated, since $48 \: \text{g}$ is less than $80 \: \text{g}$. We can also determine that there will be $80 - 48 = 32 \: \text{g}$ of undissolved $\ce{KNO_3}$ remaining at the bottom of the container. Now, suppose that this saturated solution is heated to $60^\text{o} \text{C}$. According to the curve, the solubility of $\ce{KNO_3}$ at $60^\text{o} \text{C}$ is about $107 \ \text{g}$ the solution is now unsaturated, since it still contains only the original $80 \: \text{g}$ of solute, all of which is now dissolved. Then, suppose the solution is cooled all the way down to $0^\text{o} \text{C}$. The solubility at $0^\text{o} \text{C}$ is about $14 \: \text{g}$, meaning that $80 - 14 = 66 \: \text{g}$ of the $\ce{KNO_3}$ will recrystallize.
Some solutes, such as sodium acetate, do not recrystallize easily. Suppose an exactly saturated solution of sodium acetate is prepared at $50^\text{o} \text{C}$. As it cools back to room temperature, crystals do not immediately appear in the solution, even though the solubility of sodium acetate is lower at room temperature. A supersaturated solution is a solution that contains more than the maximum amount of solute that is capable of being dissolved at a given temperature. The recrystallization of the excess dissolved solute is a supersaturated solution can be initiated by the addition of a tiny crystal of solute, called a seed crystal. The seed crystal provides a nucleation site on which the excess dissolved crystals can begin to grow. Recrystallization from a supersaturated solution is typically very fast.
Pressure
Pressure has very little effect on the solubility of solids or liquids, but it has a significant effect on the solubility of gases. Gas solubility increases as the partial pressure of a gas above the liquid increases. Suppose a certain volume of water is in a closed container with the space above it occupied by carbon dioxide gas at standard pressure. Some of the $\ce{CO_2}$ molecules come into contact with the surface of the water and dissolve into the liquid. Now suppose that more $\ce{CO_2}$ is added to the space above the container, causing a pressure increase. More $\ce{CO_2}$ molecules are now in contact with the water, so more of them dissolve. Thus the solubility increases as the pressure increases. As with a solid, the $\ce{CO_2}$ that is undissolved reaches an equilibrium with the dissolved $\ce{CO_2}$, represented by the following equation.
$\ce{CO_2} \left( g \right) \rightleftharpoons \ce{CO_2} \left( aq \right)$
At equilibrium, the rate of gaseous $\ce{CO_2}$ dissolving is equal to the rate of dissolved $\ce{CO_2}$ coming out of the solution.
When carbonated beverages are packaged, they are done so under high $\ce{CO_2}$ pressure so that a large amount of carbon dioxide dissolves in the liquid. When the bottle is opened, equilibrium is disrupted because the $\ce{CO_2}$ pressure above the liquid decreases. Immediately, bubbles of $\ce{CO_2}$ rapidly exit the solution and escape out of the top of the open bottle. The amount of dissolved $\ce{CO_2}$ decreases. If the bottle is left open for an extended period of time, the beverage becomes "flat" as more and more $\ce{CO_2}$ comes out of the liquid.
The relationship of gas solubility to pressure is described by Henry's law, named after English chemist William Henry (1774 - 1836). Henry's law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/07%3A_Solids_Liquids_and_Gases/7.07%3A_Solubility.txt |
These are homework exercises to accompany Chapter 7 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Answers are available below the solutions.
Questions
(click here for solutions)
Q7.1.1
In which state(s) of matter are atoms
1. closest together?
2. farthest apart?
3. filling the container?
4. fixed in position relative to one another?
5. moving past one another?
6. taking on the shape of the container?
Q7.1.2
Which of the following statements are true? Correct any false statements.
1. All substances exist as a liquid at room temperature and pressure.
2. Water changes from liquid to solid at 32°C.
3. All substances can exist as solids, liquids, or gases.
Q7.1.3
Which state of matter is most compressible?
Q7.1.4
Use online resources to find the boiling point of ethanol and dimethyl ether. Which one is higher? Why?
Q7.1.5
Describe the relationship between boiling point and altitude.
Q7.1.6
Where is the boiling point of ethanol the highest? The lowest?
1. Lexington, KY
2. New Orleans, LA
3. Salt Lake City, UT
4. Same at all locations.
(click here for solutions)
7.2.1
What phase change is described by each term? Is the process endothermic or exothermic?
1. sublimation
2. vaporization
3. fusion
4. deposition
Q7.2.2
List two phase changes that consume energy.
Q7.2.3
List two phase changes that release energy.
Refer to Table 7.2.1 for enthalpy values.
Q7.2.4
What is the enthalpy of fusion, vaporization, freezing, and condensation for each substance?
1. oxygen, O2
2. ethane, C2H6
3. carbon tetrachloride, CCl4
4. lead, Pb
Q7.2.5
How much energy is needed to vaporize 1.4 moles of ammonia (NH3)?
Q7.2.6
How much energy is needed to melt 3.0 moles of ice (H2O)?
Q7.2.7
What is the change in energy when 2.0 moles of ethanol is condensed?
Q7.2.8
What is the change in energy when 2.2 moles of oxygen is condensed?
Q7.2.9
Using the molar mass of water, convert the molar heats of fusion and vaporization for water from units of kJ/mol to kJ/g.
Q7.2.10
Calculate the quantity of heat that is absorbed or released during each process.
1. 655 g of water vapor condenses at 100°C
2. 8.20 kg of water is frozen
3. 40.0 mL of ethanol is vaporized. The density of ethanol is 0.789 g/mL.
4. 25.0 mL of ethanol condenses. The density of ethanol is 0.789 g/mL.
Q7.2.11
Various systems are each supplied with 9.25 kJ of heat. Calculate the mass of each substance that will undergo the indicated process with this input of heat.
1. melt ice at 0°C
2. vaporize water at 100°C
3. vaporize ethanol at 351 K
Q7.2.12
15.5 kJ of energy is released from each change. What mass of substance is involved?
1. condensation of NH3
2. freezing water
3. condensation of ethanol
Q7.2.13
What is $\Delta H_{vap}$ for benzene (C6H6) if 7.88 kJ of energy is needed to vaporize 20.0 g of benzene?
(click here for solutions)
7.3.1
How are gases different from liquids and solids in terms of the distance between the particles?
Q7.3.2
Under what conditions do gases exhibit the most ideal behavior?
Q7.3.3
Which of the following are behaviors of a gas that can be explained by the kinetic-molecular theory?
a. Gases are compressible.
b. Gases exert pressure.
c. All particles of a gas sample move at the same speed.
d. Gas particles can exchange kinetic energy when they collide.
e. Gas particles move in a curved-line path.
Q7.3.4
What is an elastic collision?
Q7.3.5
Perform the indicated conversions for the following pressure measurements.
1. 1.721 atm to mmHg
2. 559 torr to kPa
3. 91.1 kPa to atm
4. 2320 mmHg to atm
Q7.3.6
1. A typical barometric pressure in Redding, California, is about 755 mmHg. Calculate this pressure in atm and kPa.
2. A typical barometric pressure in Denver, Colorado, is 615 mmHg. What is this pressure in atmospheres and kilopascals?
Q7.3.7
How does the average kinetic energy of an air sample near a campfire compare to the average kinetic energy of a sample of air that is far away from it?
(click here for solutions)
7.4.1
Complete the missing temperature values in the table.
oC oF K
25
99
32
0
300
65
Q7.4.2
What units must temperature be in for gas law calculations?
Q7.4.3
Based on R = 0.08206 $\frac{L\dot atm}{mol \dot K}$, what units should be used in ideal gas law calculations?
Q7.4.4
A 1.00 mol sample of gas is at 300 K and 4.11 atm. What is the volume of the gas under these conditions?
Q7.4.5
What is the pressure in a 2.5 L container with 2.5 moles of gas at 293 K?
Q7.4.6
How many moles of carbon monoxide, CO, are in an 11.2-L sample at 744 torr at 55 °C?
Q7.4.7
A weather balloon contains 8.80 moles of helium at a pressure of 0.992 atm and a temperature of 25 °C at ground level. What is the volume of the balloon under these conditions?
Q7.4.8
A balloon inflated with three breaths of air has a volume of 1.7 L. At the same temperature and pressure, what is the volume of the balloon if five more same-sized breaths are added to the balloon?
Q7.4.9
The volume of an automobile air bag was 66.8 L when inflated at 25 °C with 77.8 g of nitrogen (N2) gas. What was the pressure in the bag in kPa?
Q7.4.10
How many moles of gaseous boron trifluoride, BF3, are contained in a 4.3410-L bulb at 788.0 K if the pressure is 1.220 atm? How many grams of BF3?
Q7.4.11
How is the combined gas law is simplified for each set of conditions?
1. constant V and n
2. constant n
3. constant P and V
4. constant T and n
5. constant V and T
6. constant P and n
7. constant T
Q7.4.12
A nitrogen sample has a pressure of 0.56 atm with a volume of 2.0 L. What is the final pressure if the volume is compressed to a volume of 0.75 L? Assume constant moles and temperature.
Q7.4.13
A 2.50-L volume of hydrogen measured at –196 °C is warmed to 100 °C. Calculate the volume of the gas at the higher temperature, assuming no change in pressure.
Q7.4.14
A high altitude balloon is filled with 1.41 × 104 L of hydrogen at a temperature of 21 °C and a pressure of 745 torr. What is the volume of the balloon at a height of 20 km, where the temperature is –48 °C and the pressure is 63.1 torr?
Q7.4.15
A cylinder of medical oxygen has a volume of 35.4 L, and contains O2 at a pressure of 151 atm and a temperature of 25 °C. What volume of O2 does this correspond to at normal body conditions, that is, 1 atm and 37 °C?
Q7.4.16
A 0.50 L container of helium expands to 1.50 L. By what factor does the pressure change? Assume constant moles and temperature.
Q7.4.17
A sample of oxygen gas has an initial pressure and volume of 1.0 L and 1.0 atm. What is the final pressure if the volume is compressed to 0.50 L? Assume constant moles and temperature.
Q7.4.18
A sample of gas has a volume of 2.75 L at a temperature of 100 K. What is the volume of the gas when the temperature increases to 200 K? Assume constant pressure and moles.
Q7.4.19
What is the final volume of a gas that was originally at 0.75 L at 25°C and a final temperature of 50°C? Assume constant pressure and moles.
Q7.4.20
A sample of nitrogen is at 45°C with a volume of 2.5 L. What is the final temperature in °C if the volume is compressed to 1.4 L? Assume constant pressure and moles.
Q7.4.21
A 2.00 mole sample of gas is in a 3.50 L container. What happens to the volume when an additional 0.75 moles of gas is added? Assume pressure and temperature are constant.
Q7.4.22
A 1.85 mole sample of helium has a volume of 2.00 L. Additional helium is added at constant pressure and temperature until the volume is 3.25 L. What is the total moles of helium present in the sample? What mass of helium was added?
Q7.4.23
If the temperature of a fixed amount of a gas is doubled at constant volume, what happens to the pressure?
Q7.4.24
If the volume of a fixed amount of a gas is tripled at constant temperature, what happens to the pressure?
(click here for solutions)
7.5.1
Describe the solution, solvent, and solute.
Q7.5.2
How do solutions differ from compounds? Are solutions heterogeneous or homogeneous mixtures?
Q7.5.3
When KNO3 is dissolved in water, the resulting solution is significantly colder than the water was originally. Is the dissolution of KNO3 an endothermic or an exothermic process?
Q7.5.4
What are the differences between strong, weak and non-electrolytes.
Q7.5.5
Write dissociation equations for the following strong electrolytes.
1. NaCl(s)
2. CoCl3(s)
3. Li2S(s)
4. MgBr2(s)
5. CaF2(s)
Q7.5.6
Based on the given information, identify each as a strong, weak, or non-electrolyte.
1. C6H12O6(s) $\rightarrow$ C6H12O6(aq)
2. NaCl is added to water and the conductivity increases dramatically.
3. 1.5 moles of HCl are added to a container of water. The resulting solution has 1.5 moles of H+ ions and 1.5 moles of Cl$^{\-}$ ions.
4. Acetic acid (CH3COOH) partially dissociates in water.
5. An HCN solution contains 0.50 moles of HCN molecules and 0.05 moles of H+ ions and 0.05 moles of Cl$^{\-}$ ions.
6. Acetone is added to water and the conductivity does not change.
(click here for solutions)
7.6.1
How can you distinguish between a suspension and a solution?
Q7.6.2
How big are the particles in a colloid compared to those of a suspension and a solution?
Q7.6.3
What is the Tyndall effect? Why don’t solutions demonstrate the Tyndall effect?
Q7.6.4
Explain the difference between the dispersed phase and the dispersing medium of a colloid.
Q7.6.5
Identify each of the following descriptions or examples as being representative of a solution, suspension, or colloid. More than one answer may apply.
a. dispersed particles can be filtered out
b. heterogeneous
c. particles are not visible to the unaided eye
d. paint
e. lemonade with no pulp
f. particle size larger than 1 nm
g. milk
h. particles do not settle upon standing
i. fog
(click here for solutions)
7.7.1
Describe the difference between saturated and unsaturated solutions.
Q7.7.2
What are two things that you could do to change an unsaturated solution into a saturated solution?
Q7.7.3
A given solution is clear and colorless. A single crystal of solute is added to the solution. Describe what happens in each of the following situations.
1. The original solution was saturated.
2. The original solution was unsaturated.
Q7.7.4
List the original states (solid, liquid, or gas) of the solute and solvent that are combined to make each of the following solutions.
1. an alloy
2. salt water
3. carbonated water
4. oil in gasoline
Q7.7.5
Answer the following using the solubility curve diagram.
1. How many grams of NH4Cl are required to make a saturated solution in 100 g of water at 70°C?
2. How many grams of NH4Cl could be dissolved in 200 g of water at 70°C?
3. At what temperature is a solution of 50 grams of KNO3 dissolved in 100 grams of water a saturated solution?
4. Which two substances in the above graph have the same solubility at 85°C?
5. How many grams of NaNO3 can be dissolved in 100 grams of water to make a saturated solution at 25°C?
6. How much KI can be dissolved in 5 grams of water at 20°C to make a saturated solution?
Q7.7.6
An exactly saturated solution of KClO3 is prepared at 90°C using 100 grams of water. If the solution is cooled to 20°C, how many grams of KClO3 will recrystallize (i.e. come out of solution)?
Q7.7.7
Indicate whether the following solutions are unsaturated or saturated.
1. 22 grams of KClO3 is dissolved in 100 g of water at 50°C.
2. 60 grams of KNO3 is dissolved in 100 g of water at 50°C.
3. 50 grams of NaCl is dissolved in 100 g of water at 50°C.
Q.7.7.8
Under which set of conditions is the solubility of a gas in a liquid the greatest?
1. low temperature and low pressure
2. low temperature and high pressure
3. high temperature and low pressure
4. high temperature and high pressure
Answers
7.1: States of Matter
Q7.1.1
1. solid
2. gas
3. gas
4. solid
5. liquid and gas
6. liquid and gas
Q7.1.2
Which of the following statements are true? Correct any false statements.
1. All substances exist as a liquid at room temperature and pressure. at some temperature and pressure.
2. Water changes from liquid to solid at 32°C °F.
3. True (although some states are rarely seen for some substances).
Q7.1.3
gas
Q7.1.4
ethanol 78°C; dimethyl ether $-$24°C
Ethanol has stronger intermolecular forces due to having hydrogen bonding which is not seen in dimethyl ether. The stronger the intermolecular forces, the higher the boiling point.
Q7.1.5
As the altitude increases, the boiling point decreases
Q7.1.6
1. Lexington, KY (altitude = 978 feet)
2. New Orleans, LA (altitude = 2 feet) - HIGHEST
3. Salt Lake City, UT (altitude = 4226 feet) - LOWEST
7.2: Heat and Changes of State
Q7.2.1
1. solid to gas; endothermic
2. liquid to gas; endothermic
3. solid to liquid; endothermic
4. gas to solid; exothermic
Q7.2.2
Any two of fusion, vaporization, or sublimation.
Q7.2.3
Any two of freezing, condensation, deposition.
Q7.2.4
Substance $\Delta H_{fus}$ (kJ/mol) $\Delta H_{vap}$ (kJ/mol) $\Delta H_{freezing}$ (kJ/mol) $\Delta H_{condensation}$ (kJ/mol)
oxygen, O2 0.44 6.82 $-$0.44 $-$6.82
ethane, C2H6 2.85 14.72 $-$2.85 $-$14.72
carbon tetrachloride, CCl4 2.67 30.0 $-$2.67 $-$30.0
lead, Pb 4.77 178 $-$4.77 $-$178
Q7.2.5
$1.4\;mol \; \text{NH}_{3}\left ( \frac{23.35 \;kJ}{mol} \right )=33 \; kJ$
Q7.2.6
$3.0\;mol \; \text{H}_2\text{O}\left ( \frac{6.01 \;kJ}{mol} \right )=18 \; kJ$
Q7.2.7
$2.0\;mol \; \text{CH}_3\text{CH}_2\text{OH}\left ( \frac{-38.56\; kJ}{mol} \right )=-77 \; kJ$
Q7.2.8
$2.2\;mol \; \text{O}_2\left ( \frac{-6.82\; kJ}{mol} \right )=-15 \; kJ$
Q7.2.9
$\frac{6.01\; kJ}{mol} \left ( \frac{1\;mol}{18.02\;g} \right )=\frac{0.334\;kJ}{g}$
$\frac{40.7\; kJ}{mol} \left ( \frac{1\;mol}{18.02\;g} \right )=\frac{2.26\;kJ}{g}$
Q7.2.10
1. $655\;g\; \text{H}_2\text{O}\left ( \frac{1\;mol}{18.02\;g} \right ) \left (\frac{-40.7\; kJ}{mol} \right )=-1.48\times10^3\;kJ$
2. $8.20\;kg\; \text{H}_2\text{O}\left ( \frac{1000\;g}{1\;kg} \right ) \left ( \frac{1\;mol}{18.02\;g} \right ) \left (\frac{-6.01\; kJ}{mol} \right )=-2.73\times10^3\;kJ$
3. $40.0\;mL\; \text{CH}_3\text{CH}_2\text{OH}\left ( \frac{0.789\;g}{1\;mL} \right ) \left ( \frac{1\;mol}{46.07\;g} \right ) \left (\frac{38.56\; kJ}{mol} \right )=26.4\;kJ$
4. $25.0\;mL\; \text{CH}_3\text{CH}_2\text{OH}\left ( \frac{0.789\;g}{1\;mL} \right ) \left ( \frac{1\;mol}{46.07\;g} \right ) \left (\frac{-38.56\; kJ}{mol} \right )=-16.5\;kJ$
Q7.2.11
1. $9.25\;kJ \left ( \frac{mol}{6.01\;kJ} \right )\left ( \frac{18.02\;g}{mol} \right )=27.7\;g\;\text{H}_2\text{O}$
2. $9.25\;kJ \left ( \frac{mol}{40.7\;kJ} \right )\left ( \frac{18.02\;g}{mol} \right )=4.10\;g\;\text{H}_2\text{O}$
3. $9.25\;kJ \left ( \frac{mol}{38.56\;kJ} \right )\left ( \frac{46.07\;g}{mol} \right )=11.1\;g\;\text{CH}_3\text{CH}_2\text{OH}$
Q7.2.12
1. $-15.5\;kJ \left ( \frac{mol}{-23.35\;kJ} \right )\left ( \frac{17.03\;g}{mol} \right )=11.3\;g\;\text{NH}_3$
2. $-15.5\;kJ \left ( \frac{mol}{-6.01\;kJ} \right )\left ( \frac{18.02\;g}{mol} \right )=46.5\;g\;\text{H}_2\text{O}$
3. $-15.5\;kJ \left ( \frac{mol}{-38.56\;kJ} \right )\left ( \frac{46.07\;g}{mol} \right )=18.5\;g\;\text{CH}_3\text{CH}_2\text{OH}$
Q7.2.13
Find the moles of benzene.
$20.0\;g\;\text{C}_6\text{H}_6\left ( \frac{1\;mol}{78.11\;g} \right )=0.256\;mol\;\text{C}_6\text{H}_6$
Combine the energy with the moles to calculate the enthalpy of vaporization.
$\Delta H_{vap}=\frac{7.88\;kJ}{0.256\;mol}=\frac{30.8\;kJ}{mol}$
7.3: Kinetic-Molecular Theory
Q7.3.1
Gas particles are much farther from one another than liquid or solid particles.
Q7.3.2
Gases have the most ideal behavior at high temperatures (molecules moving more quickly than at low temperatures so less time to interact) and at low pressure (molecules are farther apart from one another than at high pressure).
Q7.3.3
1. Molecules are very far apart from one another and are compressible.
2. Gases are in constant random motion so they collide with the walls of the container.
3. False. Molecules of the same substance are moving at a range of speeds.
4. Collisions are elastic. Energy is exchanged but not lost when two particles coll
5. False. Particles move in a straight line.
Q7.3.4
A collision in which no energy is lost.
Q7.3.5
1. $1.721\;atm\left ( \frac{760\;mmHg}{1\;atm} \right )=1308\;mmHg$
2. $559\;torr\left ( \frac{101.3\;kPa}{760\;torr} \right )=74.5\;kPa$
3. $91.1\;kPa\left ( \frac{1\;atm}{101.3\;kPa} \right )=0.899\;atm$
4. $2320\;mmHg\left ( \frac{1\;atm}{760\;mmHg} \right )=3.05\;atm$
Q7.3.6
1. $755\;mmHg\left ( \frac{1\;atm}{760\;mmHg} \right )=0.993\;atm$ $755\;mmHg\left ( \frac{101.3\;kpa}{760\;mmHg} \right )=101\;kPa$
2. $615\;mmHg\left ( \frac{1\;atm}{760\;mmHg} \right )=0.809\;atm$ $615\;mmHg\left ( \frac{101.3\;kpa}{760\;mmHg} \right )=82.0\;kPa$
Q7.3.7
Closer to the fire, it is warmer and the kinetic energy of the particles (and therefore the average speed) will be greater.
7.4:The Ideal Gas Equation
Q7.4.1
oC oF K
25 77 298
37 99 310
32 90 305
$-$273 $-$459 0
27 80 300
18 65 291
Q7.4.2
Kelvin
Q7.4.3
P (atm), V (L), n (mol), T (K)
Q7.4.4
A 1.00 mol sample of gas is at 300 K and 4.11 atm. What is the volume of the gas under these conditions?
$PV=nRT$
$\left ( 4.11\; atm \right )V=\left ( 1.00\;mol \right )\left ( 0.08206\frac{L\cdot atm}{mol \cdot K} \right )\left ( 300\;K \right )$
$V=5.99\;L$
Q7.4.5
$PV=nRT$
$P\left ( 2.5\;L \right )=\left ( 2.5\;mol \right )\left ( 0.08206\frac{L\cdot atm}{mol \cdot K} \right )\left ( 293\;K \right )$
$P=24 \;atm$
Q7.4.6
$744\;torr \left (\frac{1\;atm}{760\;mmHg} \right ) =0.979\; atm$
T = 55°C + 273.15 = 328 K
$PV=nRT$
$\left ( 0.979\;atm \right )\left ( 11.2\;L \right )=n \left ( 0.08206\frac{L\cdot atm}{mol \cdot K} \right )\left ( 328\;K \right )$
$n=0.407 \;mol$
Q7.4.7
T = 25°C + 273.15 = 298 K
$PV=nRT$
$\left ( 0.992\;atm \right )V=\left (8.80\; mol \right ) \left ( 0.08206\frac{L\cdot atm}{mol \cdot K} \right )\left ( 298\;K \right )$
$V=217 \;L$
Q7.4.8
x = volume of one breath of air
$3x=1.7\;L$
$x=0.57\;L$
Balloon will have a total of 8 breaths of air (3 original plus 5 additional)
$V = 8x=8(0.57\;L) = 4.6 L$
Q7.4.9
$77.8\;g\;N_2 \left (\frac{1\;mol}{28.02\;g} \right )=2.78\;mol\;N_2$
T = 25°C + 273.15 = 298 K
$PV=nRT$
$P \left (66.8\;L \right )=\left (2.78\; mol \right ) \left ( 0.08206\frac{L\cdot atm}{mol \cdot K} \right )\left ( 298\;K \right )$
$P=1.02 \;atm$
$1.02\;atm \left (\frac{101.3\;kPa}{1\;atm} \right ) = 103\;kPa$
Q7.4.10
$PV=nRT$
$\left (1.220\;atm \right ) \left (4.3410\;L \right )=mol \left ( 0.08206\frac{L\cdot atm}{mol \cdot K} \right )\left ( 788.0\;K \right )$
$n=0.08190 \;mol\; BF_3$
$0.08190\;mol \left (\frac{67.82\;g}{mol} \right )=5.554\;g\;BF_3$
Q7.4.11
1. $\frac{P_i}{T_i}=\frac{P_f}{T_f}$
2. $\frac{P_iV_i}{T_i}=\frac{P_fV_f}{T_f}$
3. $\frac{1}{n_iT_i}=\frac{1}{n_fT_f}$ or $n_iT_i=n_fT_f$
4. $P_iV_i=P_fV_f$
5. $\frac{P_i}{n_i}=\frac{P_f}{n_f}$
6. $\frac{V_i}{T_i}=\frac{V_f}{T_f}$
7. $\frac{P_iV_i}{n_i}=\frac{P_fV_f}{n_f}$
Q7.4.12
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_f}{n_fT_f}$
$P_iV_i=P_fV_f$
$0.56\;atm\cdot2.0L=P_f\cdot0.75\;L$
$P_f=1.5\;atm$
Q7.4.13
$K=-196^{\circ}C+273.15=77\;K$
$K=100^{\circ}C+273.15=373\;K$
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_f}{n_fT_f}$
$\frac{V_i}{T_i}=\frac{V_f}{T_f}$
$\frac{2.50\;L}{77\;K}=\frac{V_f}{373\;K}$
$V_f=12\;L$
Q7.4.14
$K=21^{\circ}C+273.15=294\;K$
$K=-48^{\circ}C+273.15=225\;K$
$745\;torr\left(\frac{1\;atm}{760\;torr}\right)=0.980\;atm$
$63.1\;torr\left(\frac{1\;atm}{760\;torr}\right)=0.0830\;atm$
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_f}{n_fT_f}$
$\frac{P_iV_i}{T_i}=\frac{P_fV_f}{T_f}$
$\frac{0.980\;atm\cdot1.4\times10^4\;L}{294\;K}=\frac{P_i\cdot0.0830\;atm}{225\:}$
$P_f=1.27\times10^5\;atm$
Q7.4.15
$K=25^{\circ}C+273.15=298\;K$
$K=-37^{\circ}C+273.15=310\;K$
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_f}{n_fT_f}$
$\frac{P_iV_i}{T_i}=\frac{P_fV_f}{T_f}$
$\frac{151\;atm\cdot35.4\;L}{298\;K}=\frac{1\;atm\cdot V_f}{310\:}$
$V_f=5.56\times10^3\;L$
Q7.4.16
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_f}{n_fT_f}$
$P_iV_i=P_fV_f$
Set the initial pressure = x to calculate the factor of change in terms of x.
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_f}{n_fT_f}$
$P_iV_i=P_fV_f$
$x\cdot0.50\;L =P_f\cdot1.50\;L$
$P_f=\frac{1}{3}x$
The final pressure is one third of the original pressure.
Q7.4.17
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_f}{n_fT_f}$
$P_iV_i=P_fV_f$
$1.0\;atm\cdot1.0\;L =P_f\cdot0.50\;L$
$P_f=2.0\;atm$
Q7.4.18
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_f}{n_fT_f}$
$\frac{V_i}{T_i}=\frac{V_f}{T_f}$
$\frac{2.75\;L}{100\;K}=\frac{V_f}{200\;K}$
$V_f=5.50\;L$
Q7.4.19
$K=25^{\circ}C+273.15=298\;K$
$K=-50^{\circ}C+273.15=323\;K$
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_f}{n_fT_f}$
$\frac{V_i}{T_i}=\frac{V_f}{T_f}$
$\frac{0.75\;L}{298\;K}=\frac{V_f}{323\;K}$
$V_f=0.813\;L$
Q7.4.20
$K=45^{\circ}C+273.15=318\;K$
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_f}{n_fT_f}$
$\frac{V_i}{T_i}=\frac{V_f}{T_f}$
$\frac{2.5\;L}{318\;K}=\frac{1.4\;L}{T_f}$
$T_f=178\;K$
$^{\circ}C=K-273.15$
$^{\circ}C=178-273.15$
$^{\circ}C=-95\;K$
Q7.4.21
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_f}{n_fT_f}$
$\frac{V_i}{n_i}=\frac{V_f}{n_f}$
$\frac{3.50\;L}{2.00\;mol}=\frac{V_f}{2.75\;mol}$
$V_f=4.81\;L$
Note the final moles is 2.75 because the problem says that 0.75 moles of gas is added to the original amount of 2.00 moles.
Q7.4.22
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_f}{n_fT_f}$
$\frac{V_i}{n_i}=\frac{V_f}{n_f}$
$\frac{2.00\;L}{1.85\;mol}=\frac{3.25\;L}{n_f}$
$n_f=3.01\;mol$
$\text{moles added} = 3.01\;mol-1.85\;mol$
$\text{moles added} = 1.16\;mol$
$1.16\;mol\;He\left(\frac{4.003\;g}{mol}\right)=4.64\;g\; \text{He}$
Q7.4.23
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_f}{n_fT_f}$
$\frac{P_i}{T_i}=\frac{P_f}{T_f}$
The temperature is doubled so $T_f=2\cdot T_i$
Let $P_i=x$ to see the factor the pressure changes.
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_f}{n_fT_f}$
$\frac{P_i}{T_i}=\frac{P_f}{T_f}$
$\frac{x}{T_i}=\frac{P_f}{2\cdot T_i}$
$x=\frac{P_f}{2}$
$P_f=2x$
The final pressure is twice the initial pressure.
Q7.4.24
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_f}{n_fT_f}$
$P_iV_i=P_fV_f$
The volume is tripled so $V_f=3\cdot V_i$
Let $P_i=x$ to see the factor the pressure changes.
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_f}{n_fT_f}$
$P_iV_i=P_fV_f$
$x\cdot V_i=P_f\cdot 3V_i$
$x=3P_f$
$P_f=\frac{1}{3}x$
The final pressure is one-third of the initial pressure.
7.5: Aqueous Solutions
Q7.5.1
The solute is present in the smaller amount, the solvent is present in the larger amount, and the solution is the combination of the solute and solvent.
Q7.5.2
Solutions are a homoogeneous mixture of two or more compounds.
Q7.5.3
Endothermic because heat was needed to dissolve the KNO3. Heat present in the solution was consumed by the dissolution process.
Q7.5.4
Strong electrolytes completely dissociate into ions in aqueous solution and are conductors of electricity. Weak electrolytes partially dissociate into ions in aqueous solutions are are weak conductors of electricity. Non-electrolytes do not dissociate into ions in aqueous solution and are poor conductors of electricity.
Q7.5.5
1. NaCl(s) → Na+(aq) + Cl(aq)
2. CoCl3(s) → Co3+(aq) + 3Cl(aq)
3. Li2S(s) → 2Li+(aq) + S2–(aq)
4. MgBr2(s) → Mg2+(aq) + 2Br(aq)
5. CaF2(s) → Ca2+(aq) + 2F(aq)
Q7.5.6
Based on the given information, identify each as a strong, weak, or non-electrolyte.
1. non-electrolyte
2. strong electrolyte
3. strong electrolyte
4. weak electrolyte
5. weak electrolyte
6. non-electrolyte
7.6: Colloids and Suspensions
Q7.6.1
A suspension can be separated from the solvent by filtration while a solution cannot because particles settle out of suspensions but not solutions.
Q7.6.2
Particles in a solution are less than 1 nanometer, colloids have particles from 1-1000 nm, and suspensions have particles over 1000 nm.
Q7.6.3
The Tyndall effect is the scattering of visible light by particles. The particles in colloids are large enough to scatter light while the particles in solutions are too small to scatter light. Solutions are transparent (we can see through them) because the particles are so small.
Q7.6.4
The dispersed phase is present in the smaller amount and the dispersing medium is present in a larger amount.
Q7.6.5
1. suspension
2. colloids and suspensions
3. solution
4. colloid
5. solution
6. colloids and suspensions
7. colloid
8. solutions and colloids
9. colloids
7.7: Solubility
Q7.7.1
A saturated solution has the maximum amount of solute dissolved. An unsaturated solution does not have the maximum amount dissolved; additional solute can be added and will dissolve.
Q7.7.2
1. Addition of solute to the solution until no more dissolves.
2. Removal of solvent such as through evaporation.
Q7.7.3
1. The added solute will not dissolve.
2. The added solute will dissolve.
Q7.7.4
1. The solute and solvent are both solids.
2. The solute is a solid and the solvent is a liquid.
3. The solute is a gas and the solvent is a liquid.
4. The solute and solvent are both liquids.
Q7.7.5
1. 60 g NH4Cl
2. 120 g NH4Cl
3. 31oC
4. HCl and KClO3
5. 90 g
6. 7 g
Q7.7.6
At 90°C, 50 g of KClO3 will dissolve in 100 g of water for a saturated solution. At 20°C, only 10 g of KClO3 is dissolved in 100 g of water for a saturated solution. 40 grams of KClO3 will precipitate out of solution.
Q7.7.7
1. saturated
2. unsaturated
3. saturated (with additional undissolved solute)
Q.7.7.8
The solubility of a gas in a liquid is the greatest at low temperature and high pressure. | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/07%3A_Solids_Liquids_and_Gases/7.08%3A_Solutions_%28Exercises%29.txt |
• 8.1: Concentrations of Solutions
There are several ways to express the amount of solute present in a solution. The concentration of a solution is a measure of the amount of solute that has been dissolved in a given amount of solvent or solution. A concentrated solution is one that has a relatively large amount of dissolved solute. A dilute solution is one that has a relatively small amount of dissolved solute. However, these terms are relative, and we need to be able to express concentration in a more exact, quantitative manner
• 8.2: Chemical Equilibrium
Chemical equilibrium can be attained whether the reaction begins with all reactants and no products, all products and no reactants, or some of both. It may be tempting to think that once equilibrium has been reached, the reaction stops. Chemical equilibrium is a dynamic process. The forward and reverse reactions continue to occur even after equilibrium has been reached. Because the rates of the reactions are the same, there is no change in the relative concentrations of reactants and products.
• 8.4: Osmosis and Diffusion
Fish cells, like all cells, have semipermeable membranes. Eventually, the concentration of "stuff" on either side of them will even out. A fish that lives in salt water will have somewhat salty water inside itself. Put it in freshwater, and the freshwater will, through osmosis, enter the fish, causing its cells to swell, and the fish will die. What will happen to a freshwater fish in the ocean?
• 8.5: Acid-Base Definitions
The Swedish chemist Svante Arrhenius (1859 - 1927) was the first to propose a theory to explain the observed behavior of acids and bases. Because of their ability to conduct a current, he knew that both acids and bases contained ions in solution. An Arrhenius acid is a compound which ionizes to yield hydrogen ions (H+) in aqueous solution. An Arrhenius base is a compound which ionizes to yield hydroxide ions (OH−) in aqueous solution.
• 8.6: The pH Concept
Expressing the acidity of a solution by using the molarity of the hydrogen ion is cumbersome because the quantities are generally very small. Danish scientist Søren Sørensen (1868 - 1939) proposed an easier system for indicating the concentration of H+ called the pH scale. The letters pH stand for the power of the hydrogen ion. The pH of a solution is the negative logarithm of the hydrogen-ion concentration.
• 8.7: Properties of Solutions (Exercises)
These are homework exercises to accompany Chapter 8 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Solutions are available below the questions.
• 8.3: Le Chatelier's Principle
The description of how a system responds to a stress to equilibrium has become known as Le Châtelier's principle: When a chemical system that is at equilibrium is disturbed by a stress, the system will respond in order to relieve the stress. Stresses to a chemical system involve changes in the concentrations of reactants or products, changes in the temperature of the system, or changes in the pressure of the system.
08: Properties of Solutions
Learning Outcomes
• Define concentration.
• Use the terms concentrated and dilute to describe the relative concentration of a solution.
• Calculate the molarity of a solution.
• Calculate percentage concentration (m/m, v/v, m/v).
• Describe a solution whose concentration is in $\text{ppm}$ or $\text{ppb}$.
• Use concentration units in calculations.
• Determine equivalents for an ion.
• Complete calculations relating equivalents to moles, volumes, or mass.
• Complete dilution calculations.
There are several ways to express the amount of solute present in a solution. The concentration of a solution is a measure of the amount of solute that has been dissolved in a given amount of solvent or solution. A concentrated solution is one that has a relatively large amount of dissolved solute. A dilute solution is one that has a relatively small amount of dissolved solute. However, these terms are relative, and we need to be able to express concentration in a more exact, quantitative manner. Still, concentrated and dilute are useful as terms to compare one solution to another (see figure below). Also, be aware that the terms "concentrate" and "dilute" can be used as verbs. If you were to heat a solution, causing the solvent to evaporate, you would be concentrating it, because the ratio of solute to solvent would be increasing. If you were to add more water to an aqueous solution, you would be diluting it because the ratio of solute to solvent would be decreasing.
Percent Concentration
One way to describe the concentration of a solution is by the percent of the solution that is composed of the solute. This percentage can be determined in one of three ways: (1) the mass of the solute divided by the mass of solution, (2) the volume of the solute divided by the volume of the solution, or (3) the mass of the solute divided by the volume of the solution. Because these methods generally result in slightly different vales, it is important to always indicate how a given percentage was calculated.
Mass Percent
When the solute in a solution is a solid, a convenient way to express the concentration is a mass percent (mass/mass), which is the grams of solute per $100 \: \text{g}$ of solution.
$\text{Percent by mass} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100\%$
Suppose that a solution was prepared by dissolving $25.0 \: \text{g}$ of sugar into $100 \: \text{g}$ of water. The percent by mass would be calculated as follows:
$\text{Percent by mass} = \frac{25 \: \text{g sugar}}{125 \: \text{g solution}} \times 100\% = 20\% \: \text{sugar}$
Sometimes, you may want to make a particular amount of solution with a certain percent by mass and will need to calculate what mass of the solute is needed. For example, let's say you need to make $3.00 \times 10^3 \: \text{g}$ of a sodium chloride solution that is $5.00\%$ by mass. You can rearrange and solve for the mass of solute.
\begin{align} \% \: \text{by mass} &= \frac{\text{mass of solute}}{\text{mass of solution}} \times 100\% \ 5.00\% &= \frac{\text{mass of solute}}{3.00 \times 10^3 \: \text{g solution}} \times 100\% \ \text{mass of solute} &= 150. \: \text{g} \end{align}
You would need to weigh out $150 \: \text{g}$ of $\ce{NaCl}$ and add it to $2850 \: \text{g}$ of water. Notice that it was necessary to subtract the mass of the $\ce{NaCl}$ $\left( 150 \: \text{g} \right)$ from the mass of solution $\left( 3.00 \times 10^3 \: \text{g} \right)$ to calculate the mass of the water that would need to be added.
Volume Percent
The percentage of solute in a solution can more easily be determined by volume when the solute and solvent are both liquids. The volume of the solute divided by the volume of the solution expressed as a percent, yields the percent by volume (volume/volume) of the solution. If a solution is made by taking $40. \: \text{mL}$ of ethanol and adding enough water to make $240. \: \text{mL}$ of solution, the percent by volume is:
\begin{align} \text{Percent by volume} &= \frac{\text{volume of solute}}{\text{volume of solution}} \times 100\% \ &= \frac{40 \: \text{mL ethanol}}{240 \: \text{mL solution}} \times 100\% \ &= 16.7\% \: \text{ethanol} \end{align}
Frequently, ingredient labels on food products and medicines have amounts listed as percentages (see figure below).
It should be noted that, unlike in the case of mass, you cannot simply add together the volumes of solute and solvent to get the final solution volume. When adding a solute and solvent together, mass is conserved, but volume is not. In the example above, a solution was made by starting with $40 \: \text{mL}$ of ethanol and adding enough water to make $240 \: \text{mL}$ of solution. Simply mixing $40 \: \text{mL}$ of ethanol and $200 \: \text{mL}$ of water would not give you the same result, as the final volume would probably not be exactly $240 \: \text{mL}$.
The mass-volume percent is also used in some cases and is calculated in a similar way to the previous two percentages. The mass/volume percent is calculated by dividing the mass of the solute by the volume of the solution and expressing the result as a percent.
For example, if a solution is prepared from $10 \: \ce{NaCl}$ in enough water to make a $150 \: \text{mL}$ solution, the mass-volume concentration is
\begin{align} \text{Mass-volume concentration} & \frac{\text{mass solute}}{\text{volume solution}} \times 100\% \ &= \frac{10 \: \text{g} \: \ce{NaCl}}{150 \: \text{mL solution}} \times 100\% \ &= 6.7\% \end{align}
Parts per Million and Parts per Billion
Two other concentration units are parts per million and parts per billion. These units are used for very small concentrations of solute such as the amount of lead in drinking water. Understanding these two units is much easier if you consider a percentage as parts per hundred. Remember that $85\%$ is the equivalent of 85 out of a hundred. A solution that is $15 \: \text{ppm}$ is 15 parts solute per 1 million parts solution. A $22 \: \text{ppb}$ solution is 22 parts solute per billion parts solution. While there are several ways of expressing two units of $\text{ppm}$ and $\text{ppb}$, we will treat them as $\text{mg}$ or $\mu \text{g}$ of solutes per $\text{L}$ solution, respectively.
For example, $32 \: \text{ppm}$ could be written as $\frac{32 \: \text{mg solute}}{1 \: \text{L solution}}$ while $59 \: \text{ppb}$ can be written as $\frac{59 \: \mu \text{g solute}}{1 \: \text{L solution}}$.
Molarity
Chemists primarily need the concentration of solutions to be expressed in a way that accounts for the number of particles present that could react according to a particular chemical equation. Since percentage measurements are based on either mass or volume, they are generally not useful for chemical reactions. A concentration unit based on moles is preferable. The molarity $\left( \text{M} \right)$ of a solution is the number of moles of solute dissolved in one liter of solution. To calculate the molarity of a solution, you divide the moles of solute by the volume of the solution expressed in liters.
$\text{Molarity} \: \left( \text{M} \right) = \frac{\text{moles of solute}}{\text{liters of solution}} = \frac{\text{mol}}{\text{L}}$
Note that the volume is in liters of solution and not liters of solvent. When a molarity is reported, the unit is the symbol $\text{M}$, which is read as "molar". For example, a solution labeled as $1.5 \: \text{M} \: \ce{NH_3}$ is a "1.5 molar solution of ammonia".
Example $1$
A solution is prepared by dissolving $42.23 \: \text{g}$ of $\ce{NH_4Cl}$ into enough water to make $500.0 \: \text{mL}$ of solution. Calculate its molarity.
Solution
Step 1: List the known quantities and plan the problem.
Known
• Mass of $\ce{NH_4Cl} = 42.23 \: \text{g}$
• Molar mass of $\ce{NH_4Cl} = 53.50 \: \text{g/mol}$
• Volume of solution $= 500.0 \: \text{mL} = 0.5000 \: \text{L}$
Unknown
• Molarity $= ? \: \text{M}$
The mass of the ammonium chloride is first converted to moles. Then, the molarity is calculated by dividing by liters. Note that the given volume has been converted to liters.
Step 2: Solve.
$42.23 \: \text{g} \: \ce{NH_4Cl} \times \frac{1 \: \text{mol} \: \ce{NH_4Cl}}{53.50 \: \text{g} \: \ce{NH_4Cl}} = 0.7893 \: \text{mol} \: \ce{NH_4Cl}$
$\frac{0.7893 \: \text{mol} \: \ce{NH_4Cl}}{0.5000 \: \text{L}} = 1.579 \: \text{M}$
Step 3: Think about your result.
The molarity is $1.579 \: \text{M}$, meaning that a liter of the solution would contain 1.579 moles of $\ce{NH_4Cl}$. Having four significant figures is appropriate.
Dilutions
When additional water is added to an aqueous solution, the concentration of that solution decreases. This is because the number of moles of the solute does not change, but the total volume of the solution increases. We can set up an equality between the moles of the solute before the dilution (1) and the moles of the solute after the dilution (2).
$\text{mol}_1 = \text{mol}_2$
Since the moles of solute in a solution is equal to the molarity multiplied by the volume in liters, we can set those equal.
$M_1 \times L_1 = M_2 \times L_2$
Finally, because the two sides of the equation are set equal to one another, the volume can be in any units we choose, as long as that unit is the same on both sides. Our equation for calculating the molarity of a diluted solution becomes:
$M_1 \times V_1 = M_2 \times V_2$
Additionally, the concentration can be in any other unit as long as $M_1$ and $M_2$ are in the same unit.
Suppose that you have $100. \: \text{mL}$ of a $2.0 \: \text{M}$ solution of $\ce{HCl}$. You dilute the solution by adding enough water to make the solution volume $500. \: \text{mL}$. The new molarity can easily be calculated by using the above equation and solving for $M_2$.
$M_2 = \frac{M_1 \times V_1}{V_2} = \frac{2.0 \: \text{M} \times 100. \: \text{mL}}{500. \: \text{mL}} = 0.40 \: \text{M} \: \ce{HCl}$
The solution has been diluted by a factor of five, since the new volume is five times as great as the original volume. Consequently, the molarity is one-fifth of its original value. Another common dilution problem involves deciding how much a highly concentrated solution is required to make a desired quantity of solution with a lower concentration. The highly concentrated solution is typically referred to as the stock solution.
Example $2$
Nitric acid $\left( \ce{HNO_3} \right)$ is a powerful and corrosive acid. When ordered from a chemical supply company, its molarity is $16 \: \text{M}$. How much of the stock solution of nitric acid needs to be used to make $8.00 \: \text{L}$ of a $0.50 \: \text{M}$ solution?
Solution
Step 1: List the known quantities and plan the problem.
Known
• Stock $\ce{HNO_3} \: \left( M_1 \right) = 16 \: \text{M}$
• $V_2 = 8.00 \: \text{L}$
• $M_2 = 0.50 \: \text{M}$
Unknown
• Volume of stock $\ce{HNO_3} \: \left( V_1 \right) = ? \: \text{L}$
The unknown in the equation is $V_1$, the necessary volume of the concentrated stock solution.
Step 2: Solve.
$V_1 = \frac{M_2 \times V_2}{V_1} = \frac{0.50 \: \text{M} \times 8.00 \: \text{L}}{16 \: \text{M}} = 0.25 \: \text{L} = 250 \: \text{mL}$
Step 3: Think about your result.
$250 \: \text{mL}$ of the stock $\ce{HNO_3}$ solution needs to be diluted with water to a final volume of $8.00 \: \text{L}$. The dilution from $16 \: \text{M}$ to $0.5 \: \text{M}$ is a factor of 32.
Equivalents
Concentration is important in healthcare because it is used in so many ways. It's also critical to use units with any values to ensure the correct dosage of medications or report levels of substances in blood, to name just two.
Another way of looking at concentration such as in IV solutions and blood is in terms of equivalents. One equivalent is equal to one mole of charge in an ion. The value of the equivalents is always positive regardless of the charge. For example, $\ce{Na^+}$ and $\ce{Cl^-}$ both have 1 equivalent per mole.
$\begin{array}{ll} \textbf{Ion} & \textbf{Equivalents} \ \ce{Na^+} & 1 \ \ce{Mg^{2+}} & 2 \ \ce{Al^{3+}} & 3 \ \ce{Cl^-} & 1 \ \ce{NO_3^-} & 1 \ \ce{SO_4^{2-}} & 2 \end{array}$
Equivalents are used because the concentration of the charges is important than the identity of the solutes. For example, a standard IV solution does not contain the same solutes as blood but the concentration of charges is the same.
Sometimes, the concentration is lower in which case milliequivalents $\left( \text{mEq} \right)$ is a more appropriate unit. Just like metric prefixes used with base units, milli is used to modify equivalents so $1 \: \text{Eq} = 1000 \: \text{mEq}$.
Example $3$
How many equivalents of $\ce{Ca^{2+}}$ are present in a solution that contains 3.5 moles of $\ce{Ca^{2+}}$?
Solution
Use the relationship between moles and equivalents of $\ce{Ca^{2+}}$ to find the answer.
$3.5 \: \text{mol} \cdot \frac{2 \: \text{Eq}}{1 \: \text{mol} \: \ce{Ca^{2+}}} = 7.0 \: \text{Eq} \: \ce{Ca^{2+}}$
Example $4$
A patient received $1.50 \: \text{L}$ of saline solution which has a concentration of $154 \: \text{mEq/L} \: \ce{Na^+}$. What mass of sodium did the patient receive?
Solution
Use dimensional analysis to set up the problem based on the values given in the problem, the relationship for $\ce{Na^+}$ and equivalents and the molar mass of sodium. Note that if this problem had a different ion with a different charge, that would need to be accounted for in the calculation.
$1.50 \: \text{L} \cdot \frac{154 \: \text{mEq}}{1 \: \text{L}} \cdot \frac{1 \: \text{Eq}}{1000 \: \text{mEq}} \cdot \frac{1 \: \text{mol} \: \ce{Na^+}}{1 \: \text{Eq}} \cdot \frac{22.99 \: \text{g}}{1 \: \text{mol} \: \ce{Na^+}} = 5.31 \: \text{g} \: \ce{Na^+}$
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/08%3A_Properties_of_Solutions/8.01%3A_Concentrations_of_Solutions.txt |
Learning Outcomes
• Explain chemical equilibrium.
• Write expression for calculating $K$.
• Calculate and compare Q and K values.
• Predict relative amounts of reactants and products based on equilibrium constant $K$.
Hydrogen and iodine gases react to form hydrogen iodide according to the following reaction:
$\ce{H_2} \left( g \right) + \ce{I_2} \left( g \right) \rightleftharpoons 2 \ce{HI} \left( g \right)$
\begin{align} &\text{Forward reaction:} \: \: \ce{H_2} \left( g \right) + \ce{I_2} \left( g \right) \rightarrow 2 \ce{HI} \left( g \right) \ &\text{Reverse reaction:} \: \: 2 \ce{HI} \left( g \right) \rightarrow \ce{H_2} \left( g \right) + \ce{I_2} \left( g \right) \end{align}
Initially, only the forward reaction occurs because no $\ce{HI}$ is present. As soon as some $\ce{HI}$ has formed, it begins to decompose back into $\ce{H_2}$ and $\ce{I_2}$. Gradually, the rate of the forward reaction decreases while the rate of the reverse reaction increases. Eventually the rate of combination of $\ce{H_2}$ and $\ce{I_2}$ to produce $\ce{HI}$ becomes equal to the rate of decomposition of $\ce{HI}$ into $\ce{H_2}$ and $\ce{I_2}$. When the rates of the forward and reverse reactions have become equal to one another, the reaction has achieved a state of balance. Chemical equilibrium is the state of a system in which the rate of the forward reaction is equal to the rate of the reverse reaction.
Chemical equilibrium can be attained whether the reaction begins with all reactants and no products, all products and no reactants, or some of both. The figure below shows changes in concentration of $\ce{H_2}$, $\ce{I_2}$, and $\ce{HI}$ for two different reactions. In the reaction depicted by the graph on the left (A), the reaction begins with only $\ce{H_2}$ and $\ce{I_2}$ present. There is no $\ce{HI}$ initially. As the reaction proceeds towards equilibrium, the concentrations of the $\ce{H_2}$ and $\ce{I_2}$ gradually decrease, while the concentration of the $\ce{HI}$ gradually increases. When the curve levels out and the concentrations all become constant, equilibrium has been reached. At equilibrium, concentrations of all substances are constant.
In reaction B, the process begins with only $\ce{HI}$ and no $\ce{H_2}$ or $\ce{I_2}$. In this case, the concentration of $\ce{HI}$ gradually decreases while the concentrations of $\ce{H_2}$ and $\ce{I_2}$ gradually increase until equilibrium is again reached. Notice that in both cases, the relative position of equilibrium is the same, as shown by the relative concentrations of reactants and products. The concentration of $\ce{HI}$ at equilibrium is significantly higher than the concentrations of $\ce{H_2}$ and $\ce{I_2}$. This is true whether the reaction began with all reactants or all products. The position of equilibrium is a property of the particular reversible reaction and does not depend upon how equilibrium was achieved.
Conditions for Equilibrium and Types of Equilibrium
It may be tempting to think that once equilibrium has been reached, the reaction stops. Chemical equilibrium is a dynamic process. The forward and reverse reactions continue to occur even after equilibrium has been reached. However, because the rates of the reactions are the same, there is no change in the relative concentrations of reactants and products for a reaction that is at equilibrium. The conditions and properties of a system at equilibrium are summarized below.
1. The system must be closed, meaning no substances can enter or leave the system.
2. Equilibrium is a dynamic process. Even though we don't necessarily see the reactions, both forward and reverse are taking place.
3. The rates of the forward and reverse reactions must be equal.
4. The amount of reactants and products do not have to be equal. However, after equilibrium is attained, the amounts of reactants and products will be constant.
The description of equilibrium in this concept refers primarily to equilibrium between reactants and products in a chemical reaction. Other types of equilibrium include phase equilibrium and solution equilibrium. A phase equilibrium occurs when a substance is in equilibrium between two states. For example, a stoppered flask of water attains equilibrium when the rate of evaporation is equal to the rate of condensation. A solution equilibrium occurs when a solid substance is in a saturated solution. At this point, the rate of dissolution is equal to the rate of recrystallization. Although these are all different types of transformations, most of the rules regarding equilibrium apply to any situation in which a process occurs reversibly.
Red blood cells transport oxygen to the tissues so they can function. In the absence of oxygen, cells cannot carry out their biochemical responsibilities. Oxygen moves to the cells attached to hemoglobin, a protein found in the red cells. In cases of carbon monoxide poisoning, $\ce{CO}$ binds much more strongly to the hemoglobin, blockin oxygen attachment and lowering the amount of oxygen reaching the cells. Treatment involves the patient breathing pure oxygen to displace the carbon monoxide. The equilibrium reaction shown below illustrates the shift toward the right when excess oxygen is added to the system:
$\ce{Hb(CO)_4} \left( aq \right) + 4 \ce{O_2} \left( g \right) \rightleftharpoons \ce{Hb(O_2)_4} \left( aq \right) + 4 \ce{CO} \left( g \right)$
Equilibrium Constant
Consider the hypothetical reversible reaction in which reactants $\ce{A}$ and $\ce{B}$ react to form products $\ce{C}$ and $\ce{D}$. This equilibrium can be shown below, where the lowercase letters represent the coefficients of each substance.
$a \ce{A} + b \ce{B} \rightleftharpoons c \ce{C} + d \ce{D}$
As we have established, the rates of the forward and reverse reactions are the same at equilibrium, and so the concentrations of all of the substances are constant. Since that is the case, it stands to reason that a ratio of the concentration for any given reaction at equilibrium maintains a constant value. The equilibrium constant $\left( K_\text{eq} \right)$ is the ratio of the mathematical product of the products of a reaction to the mathematical product of the concentrations of the reactants of the reaction. Each concentration is raised to the power of its coefficient in the balanced chemical equation. For the general reaction above, the equilibrium constant expression is written as follows:
$K_\text{eq} = \frac{\left[ \ce{C} \right]^c \left[ \ce{D} \right]^d}{\left[ \ce{A} \right]^a \left[ \ce{B} \right]^b}$
The concentrations of each substance, indicated by the square brackets around the formula, are measured in molarity units $\left( \text{mol/L} \right)$.
The value of the equilibrium constant for any reaction is only determined by experiment. As detailed in the above section, the position of equilibrium for a given reaction does not depend on the starting concentrations and so the value of the equilibrium constant is truly constant. It does, however, depend on the temperature of the reaction. This is because equilibrium is defined as a condition resulting from the rates of forward and reverse reactions being equal. If the temperature changes, the corresponding change in those reaction rates will alter the equilibrium constant. For any reaction in which a $K_\text{eq}$ is given, the temperature should be specified.
When $K_\text{eq}$ is greater than 1, the numerator is larger than the denominator so the products are favored, meaning the concentration of its products are greater than that of the reactants.
If $K_\text{eq}$ is less than 1, then the reactants are favored because the denominator (reactants) is larger than the numerator (products).
When $K_\text{eq}$ is equal to 1, then the concentration of reactants and products are approximately equal.
Reaction Quotient
The reaction quotient, $Q$, is used when questioning if we are at equilibrium. The calculation for $Q$ is exactly the same as for $K$ but we can only use $K$ when we know we are at equilibrium. Comparing $Q$ and $K$ allows the direction of the reaction to be predicted.
• $Q$ = $K$ equilibrium
• $Q$ < $K$ reaction proceeds to the right to form more products and decrease amount of reactants so value of $Q$ will increase
• $Q$ > $K$ reaction proceeds to the left to form more reactants and decrease amount of products so value of $Q$ will decrease
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/08%3A_Properties_of_Solutions/8.02%3A_Chemical_Equilibrium.txt |
Learning Outcomes
• Define osmosis and diffusion.
• Distinguish among hypotonic, hypertonic, and isotonic solutions.
• Describe a semipermeable membrane.
• Predict behavior of blood cells in different solution types.
• Describe flow of solvent molecules across a membrane.
• Identify the polar and nonpolar regions of a cell membrane.
• Explain the components present in a phospholipid.
Fish cells, like all cells, have semipermeable membranes. Eventually, the concentration of "stuff" on either side of them will even out. A fish that lives in salt water will have somewhat salty water inside itself. Put it in freshwater, and the freshwater will, through osmosis, enter the fish, causing its cells to swell, and the fish will die. What will happen to a freshwater fish in the ocean?
Osmosis
Imagine you have a cup that has $100 \: \text{mL}$ water, and you add $15 \: \text{g}$ of table sugar to the water. The sugar dissolves and the mixture that is now in the cup is made up of a solute (the sugar) that is dissolved in the solvent (the water). The mixture of a solute in a solvent is called a solution.
Imagine now that you have a second cup with $100 \: \text{mL}$ of water, and you add $45 \: \text{g}$ of table sugar to the water. Just like the first cup, the sugar is the solute, and the water is the solvent. But now you have two mixtures of different solute concentrations. In comparing two solutions of unequal solute concentration, the solution with the higher solute concentration is hypertonic, and the solution with the lower solute concentration is hypotonic. Solutions of equal solute concentration are isotonic. The first sugar solution is hypotonic to the second solution. The second sugar solution is hypertonic to the first.
You now add the two solutions to a beaker that has been divided by a semipermeable membrane, with pores that are too small for the sugar molecules to pass through, but are big enough for the water molecules to pass through. The hypertonic solution is one one side of the membrane and the hypotonic solution on the other. The hypertonic solution has a lower water concentration than the hypotonic solution, so a concentration gradient of water now exists across the membrane. Water molecules will move from the side of higher water concentration to the side of lower concentration until both solutions are isotonic. At this point, equilibrium is reached.
Red blood cells behave the same way (see figure below). When red blood cells are in a hypertonic (higher concentration) solution, water flows out of the cell faster than it comes in. This results in crenation (shriveling) of the blood cell. On the other extreme, a red blood cell that is hypotonic (lower concentration outside the cell) will result in more water flowing into the cell than out. This results in swelling of the cell and potential hemolysis (bursting) of the cell. In an isotonic solution, the flow of water in and out of the cell is happening at the same rate.
Osmosis is the diffusion of water molecules across a semipermeable membrane from an area of lower concentration solution (i.e., higher concentration of water) to an area of higher concentration solution (i.e., lower concentration of water). Water moves into and out of cells by osmosis.
• If a cell is in a hypertonic solution, the solution has a lower water concentration than the cell cytosol, and water moves out of the cell until both solutions are isotonic.
• Cells placed in a hypotonic solution will take in water across their membranes until both the external solution and the cytosol are isotonic.
A red blood cell will swell and undergo hemolysis (burst) when placed in a hypotonic solution. When placed in a hypertonic solution, a red blood cell will lose water and undergo crenation (shrivel). Animal cells tend to do best in an isotonic environment, where the flow of water in and out of the cell is occurring at equal rates.
Diffusion
Passive transport is a way that small molecules or ions move across the cell membrane without input of energy by the cell. The three main kinds of passive transport are diffusion (or simple diffusion), osmosis, and facilitated diffusion. Simple diffusion and osmosis do not involve transport proteins. Facilitated diffusion requires the assistance of proteins.
Diffusion is the movement of molecules from an area of high concentration of the molecules to an area with a lower concentration. For cell transport, diffusion is the movement of small molecules across the cell membrane. The difference in the concentrations of the molecules in the two areas is called the concentration gradient. The kinetic energy of the molecules results in random motion, causing diffusion. In simple diffusion, this process proceeds without the aid of a transport protein. It is the random motion of the molecules that causes them to move from an area of high concentration to an area with a lower concentration.
Diffusion will continue until the concentration gradient has been eliminated. Since diffusion moves materials from an area of higher concentration to the lower, it is described as moving solutes "down the concentration gradient". The end result is an equal concentration, or equilibrium, of molecules on both sides of the membrane. At equilibrium, movement of molecules does not stop. At equilibrium, there is equal movement of materials in both directions.
Not everything can make it into your cells. Your cells have a plasma membrane that helps to guard your cells from unwanted intruders.
The Plasma Membrane and Cytosol
If the outside environment of a cell is water-based, and the inside of the cell is also mostly water, something has to make sure the cell stays intact in this environment. What would happen if a cell dissolved in water, like sugar does? Obviously, the cell could not survive in such an environment. So something must protect the cell and allow it to survive in its water-based environment. All cells have a barrier around them that separates them from the environment and from other cells. This barrier is called the plasma membrane, or cell membrane.
The Plasma Membrane
The plasma membrane (see figure below) is made of a double layer of special lipids, known as phospholipids. The phospholipid is a lipid molecule with a hydrophilic ("water-loving") head and two hydrophobic ("water-hating") tails. Because of the hydrophilic and hydrophobic nature of the phospholipid, the molecule must be arranged in a specific pattern as only certain parts of the molecule can physically be in contact with water. Remember that there is water outside the cell, and the cytoplasm inside the cell is mostly water as well. So the phospholipids are arranged in a double layer (a bilayer) to keep the cell separate from its environment. Lipids do not mix with water (recall that oil is a lipid), so the phospholipid bilayer of the cell membrane acts as a barrier, keeping water out of the cell, and keeping the cytoplasm inside the cell. The cell membrane allows the cell to stay structurally intact in its water-based environment.
The function of the plasma membrane is to control what goes in and out of the cell. Some molecules can go through the cell membrane to enter and leave the cell, but some cannot. The cell is therefore not completely permeable. "Permeable" means that anything can cross a barrier. An open door is completely permeable to anything that wants to enter or exit through the door. The plasma membrane is semipermeable, meaning that some things can enter the cell, and some things cannot.
Molecules that cannot easily pass through the bilayer include ions and small hydrophilic molecules, such as glucose, and macromolecules, including proteins and RNA. Examples of molecules that can easily diffuse across the plasma membrane include carbon dioxide and oxygen gas. These molecules diffuse freely in and out of the cell, along their concentration gradient. Though water is a polar molecule, it can also diffuse through the plasma membrane.
Cytosol
The inside of all cells also contain a jelly-like substance called cytosol. Cytosol is composed of water and other molecules, including enzymes, which are proteins that speed up the cell's chemical reactions. Everything in the cell sits in the cytosol, like fruit in a Jell-o mold. The term cytoplasm refers to the cytosol and all of the organelles, the specialized compartments of the cell. The cytoplasm does not include the nucleus. As a prokaryotic cell does not have a nucleus, the DNA is in the cytoplasm.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/08%3A_Properties_of_Solutions/8.04%3A_Osmosis_and_Diffusion.txt |
Learning Outcomes
• Describe the properties of acids and bases.
• Define an acid and a base according to the Arrhenius theory.
• Define an acid and a base according to the Brønsted-Lowry theory.
• Identify the conjugate acid-base pairs in a Brønsted-Lowry acid-base reaction.
Acid-Base Properties
Acids are very common in some of the foods that we eat. Citrus fruits such as oranges and lemons contain citric acid and ascorbic acid, which is better known as vitamin C (see figure below). Carbonated sodas contain phosphoric acid. Vinegar contains acetic acid. Your own stomach utilizes hydrochloric acid to digest food. Bases are less common as foods, but they are nonetheless present in many household products (see figure below). Many cleaners contain ammonia, a base. Sodium hydroxide is found in drain cleaner. Antacids, which combat excess stomach acid, are comprise of bases such as magnesium hydroxide or sodium hydrogen carbonate.
Acids
Acids are a distinct class of compounds because of the properties of their aqueous solutions. Those properties are outlined below.
1. Aqueous solutions of acids are electrolytes, meaning that they conduct an electrical current. Some acids are strong electrolytes because they ionize completely in water. Other acids are weak electrolytes which partially ionize when dissolved in water.
2. Acids have a sour taste. Lemons, vinegar, and sour candies all contain acids.
3. Acids change the color of certain acid-base indicators .Two common indicators are litmus and phenolphthalein. Litmus turns red in the presence of an acid, while phenolphthalein is colorless.
4. Acids react with some metals to yield hydrogen gas.
5. Acids react with bases to produce a salt and water. When equal moles of an acid and a base are combined, the acid is neutralized by the base. Water and an ionic compound called a salt are produced.
Bases
Bases have properties that mostly contrast with those of acids.
1. Aqueous solutions of bases are also electrolytes. Bases can be either strong or weak, just as acids can.
2. Bases often have a bitter taste and are found in foods less frequently than acids. Many bases, like soaps, are slippery to the touch.
3. Bases also change the color of indicators. Litmus turns blue in the presence of a base (see figure below), while phenolphthalein turns pink.
4. Bases do not react with metals in the way that acids do.
5. Bases react with acids to produce a salt and water.
Arrhenius Acids and Bases
The Swedish chemist Svante Arrhenius (1859 - 1927) was the first to propose a theory to explain the observed behavior of acids and bases. Because of their ability to conduct a current, he knew that both acids and bases contained ions in solution. An Arrhenius acid is a compound which ionizes to yield hydrogen ions $\left( \ce{H^+} \right)$ in aqueous solution. An Arrhenius base is a compound which ionizes to yield hydroxide ions $\left( \ce{OH^-} \right)$ in aqueous solution.
Arrhenius Acids
Acids are molecular compounds with ionizable hydrogen atoms. Only hydrogen atoms that are part of a highly polar covalent bond are ionizable. The hydrogen atom is attracted to the lone pair of electrons in a water molecule when $\ce{HCl}$ is dissolved in water. The result is that the $\ce{H-Cl}$ bond breaks, with both bonding electrons remaining with the $\ce{Cl}$, forming a chloride ion. The $\ce{H^+}$ ion attaches to the water molecule, forming a polyatomic ion called the hydronium ion. The hydronium ion $\left( \ce{H_3O^+} \right)$ can be thought of as a water molecule with an attached hydrogen ion.
Equations showing the ionization of an acid in water are frequently simplified by omitting the water molecule.
$\ce{HCl} \left( g \right) \rightarrow \ce{H^+} \left( aq \right) + \ce{Cl^-} \left( aq \right)$
This is merely a simplification of the previous equation, but it is commonly used. Any hydrogen ions in an aqueous solution will be attached to water molecules as hydronium ions, $\ce{H_3O^+}$, even if it is written as $\ce{H^+}$.
Not all hydrogen atoms in molecular compounds are ionizable. In methane $\left( \ce{CH_4} \right)$, the hydrogen atoms are covalently bonded to carbon in bonds that are only slightly polar. The hydrogen atoms are not capable of ionizing, and methane has no acidic properties. Acetic acid $\left( \ce{CH_3COOH} \right)$ (see figure below) belongs to a class of acids called carboxylic acids. There are four hydrogen atoms in the molecule, but only the one hydrogen that is bonded to an oxygen atom is ionizable.
The table below lists some of the more common acids.
Table $1$: Common Acids
Acid Name Formula
hydrochloric acid $\ce{HCl}$
nitric acid $\ce{HNO_3}$
sulfuric acid $\ce{H_2SO_4}$
phosphoric acid $\ce{H_3PO_4}$
acetic acid $\ce{CH_3COOH}$
hypochlorous acid $\ce{HClO}$
A monoprotic acid is an acid that contains only one ionizable hydrogen. Hydrochloric acid and acetic acid are monoprotic acids. A polyprotic acid is an acid that contains multiple ionizable hydrogens. Most common polyprotic acids are either diprotic (such as $\ce{H_2SO_4}$), or triprotic (such as $\ce{H_3PO_4}$).
Arrhenius Bases
Bases are ionic compounds which yield the hydroxide ion $\left( \ce{OH^-} \right)$ upon dissociating in water. The table below lists several of the more common bases.
Table $2$: Common Bases
Base Name Formula
Sodium hydroxide $\ce{NaOH}$
Potassium hydroxide $\ce{KOH}$
Magnesium hydroxide $\ce{Mg(OH)_2}$
Calcium hydroxide $\ce{Ca(OH)_2}$
All of the bases listed in the table are solids at room temperature. Upon dissolving in water, each dissociates into a metal cation and the hydroxide ion.
$\ce{NaOH} \left( s \right) \overset{\ce{H_2O}}{\rightarrow} \ce{Na^+} \left( aq \right) + \ce{OH^-} \left( aq \right)$
Sodium hydroxide is a very caustic substance also known as lye. Lye is used as a rigorous cleaner and is an ingredient in the manufacture of soaps. Care must be taken with strong bases like sodium hydroxide, as exposure can lead to severe burns (see figure below).
Brønsted-Lowry Acids and Bases
The Arrhenius definition of acids and bases is somewhat limited. There are some compounds whose properties suggest that they are either acidic or basic, but which do not qualify according to the Arrhenius definition. An example is ammonia $\left( \ce{NH_3} \right)$. An aqueous solution of ammonia turns litmus blue, reacts with acids, and displays various other properties that are common for bases. However, it does not contain the hydroxide ion. In 1923, a broader definition of acids and bases was independently proposed by the Danish chemist Johannes Brønsted (1879 - 1947) and the English chemist Thomas Lowry (1874 - 1936). A Brønsted-Lowry acid is a molecule or ion that donates a hydrogen ion in a reaction. A Brønsted-Lowry base is a molecule or ion that accepts a hydrogen ion in a reaction. Because the most common isotope of hydrogen consists of a single proton and a single electron, a hydrogen ion (in which the single electron has been removed) is commonly referred to as a proton. As a result, acids and bases are often called proton donors and proton acceptors, respectively, according to the Brønsted-Lowry definition. All substances that are categorized as acids and bases under the Arrhenius definition are also define as such under the Brønsted-Lowry definition. The new definition, however, includes some substances that are left out according to the Arrhenius definition.
Brønsted-Lowry Acid-Base Reactions
An acid-base reaction according the Brønsted-Lowry definition is a transfer of a proton from one molecule or ion to another. When ammonia is dissolved in water, it undergoes the following reversible reaction.
$\begin{array}{ccccccc} \ce{NH_3} \left( aq \right) & + & \ce{H_2O} \left( l \right) & \rightleftharpoons & \ce{NH_4^+} \left( aq \right) & + & \ce{OH^-} \left( aq \right) \ \text{base} & & \text{acid} & & \text{acid} & & \text{base} \end{array}$
In this reaction, the water molecule is donating a proton to the ammonia molecule. The resulting products are the ammonium ion and the hydroxide ion. The water is acting as a Brønsted-Lowry acid, while the ammonia is acting as a Brønsted-Lowry base. The hydroxide ion that is produced causes the solution to be basic.
We can also consider the reverse reaction in the above equation. In that reaction, the ammonium ion donates a proton to the hydroxide ion. The ammonium ion is a Brønsted-Lowry acid, while the hydroxide ion is a Brønsted-Lowry base. Most Brønsted-Lowry acid-base reactions can be analyzed in this way. There is one acid and one base as reactants, and one acid and one base as products.
In the above reaction, water acted as an acid, which may seem a bit unexpected. Water can also act as a base in a Brønsted-Lowry acid-base reaction, as long as it reacts with a substance that is a better proton donor. Shown below is the reaction of water with the hydrogen sulfate ion.
$\begin{array}{ccccccc} \ce{HSO_4^-} \left( aq \right) & + & \ce{H_2O} \left( l \right) & \rightleftharpoons & \ce{H_3O^+} \left( aq \right) & + & \ce{SO_4^{2-}} \left( aq \right) \ \text{acid} & & \text{base} & & \text{acid} & & \text{base} \end{array}$
Water is capable of being either an acid or a base, a characteristic called amphoterism. An amphoteric substance is one that is capable of acting as either an acid or a base by donating or accepting hydrogen ions.
Conjugate Acids and Bases
When a substance that is acting as a Brønsted-Lowry acid donates its proton, it becomes a base in the reverse reaction. In the reaction above, the hydrogen sulfate ion $\left( \ce{HSO_4^-} \right)$ donates a proton to water and becomes a sulfate ion $\left( \ce{SO_4^{2-}} \right)$. The $\ce{HSO_4^-}$ and the $\ce{SO_4^{2-}}$ are linked to one another by the presence or absence of the $\ce{H^+}$ ion. A conjugate acid-base pair is a pair of substances related by the loss or gain of a single hydrogen ion. A conjugate acid is the particle produced when a base accepts a proton. The hydrogen sulfate ion is the conjugate acid of the sulfate ion. A conjugate base is the particle produced when an acid donates a proton. The sulfate ion is the conjugate base of the hydrogen sulfate ion.
A typical Brønsted-Lowry acid-base reaction contains two conjugate acid-base pairs as shown below.
$\ce{HNO_2} \left( aq \right) \ce{PO_4^{3-}} \left( aq \right) \rightleftharpoons \ce{NO_2^-} \left( aq \right) + \ce{HPO_4^{2-}} \left( aq \right)$
One conjugate acid-base pair is $\ce{HNO_2}/\ce{NO_2^-}$, while the other pair is $\ce{HPO_4^{2-}}/\ce{PO_4^{3-}}$.
The buffer systems in blood have conjugate acid-base pairs which help maintain the correct acid and base concentration in blood.
Table $3$: Acid-Base Definitions
Type Acid Base
Arrhenius $\ce{H^+}$ ions in solution $\ce{OH^-}$ ions in solution
Brønsted-Lowry $\ce{H^+}$ donor $\ce{H^+}$ acceptor
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/08%3A_Properties_of_Solutions/8.05%3A_Acid-Base_Definitions.txt |
Learning Outcomes
• Calculate between pH/$\left[ \ce{H^+} \right]$ and pOH/$\left[ \ce{OH^-} \right]$
• Convert among pH, pOH, hydrogen-ion concentration, and hydroxide-ion concentration for a given solution.
Self-Ionization of Water
Water is a molecular compound, so you may not necessarily expect it to break apart into ions. However, sensitive experiments show that water is actually a very weak electrolyte. When two molecules of water collide, there can be a transfer of a hydrogen ion from one molecule to the other. The products are a positively charged hydronium ion and a negatively charged hydroxide ion.
$\ce{H_2O} \left( l \right) + \ce{H_2O} \left( l \right) \rightleftharpoons \ce{H_3O^+} \left( aq \right) + \ce{OH^-} \left( aq \right)$
The self-ionization of water is the process in which water ionizes to hydronium ions and hydroxide ions. As with other aqueous acid-base reactions, the process is often simplified to show the ionization of just one water molecule into a hydrogen ion and a hydroxide ion.
$\ce{H_2O} \left( l \right) \rightleftharpoons \ce{H^+} \left( aq \right) + \ce{OH^-} \left( aq \right)$
Either equation is adequate, though the first is more accurate, since hydrogen ions in aqueous solution will always be attached to water molecules. Further discussion of acids and acid ionizations in this book will primarily show hydrogen ions in aqueous solution as $\ce{H^+}$, but keep in mind that this is just a commonly used abbreviation for the more accurate hydronium $\left( \ce{H_3O^+} \right)$ structure.
In pure water, the concentrations of hydrogen and hydroxide ions are equal to one another. Any aqueous solution in which $\left[ \ce{H^+} \right] = \left[ \ce{OH^-} \right]$ is said to be neutral.
For any neutral solution at $25^\text{o} \text{C}$, each of these ions has a concentration of $1.0 \times 10^{-7} \: \text{M}$.
An acidic solution is a solution in which the concentration of hydrogen ions is greater than the concentration of hydroxide ions. For example, hydrogen chloride ionizes to produce $\ce{H^+}$ and $\ce{Cl^-}$ ions upon dissolving in water.
$\ce{HCl} \left( g \right) \rightarrow \ce{H^+} \left( aq \right) + \ce{Cl^-} \left( aq \right)$
This increases the concentration of $\ce{H^+}$ ions in the solution.
A basic solution is a solution in which the concentration of hydroxide ions is greater than the concentration of hydrogen ions. Solid potassium hydroxide dissociates in water to yield potassium ions and hydroxide ions.
$\ce{KOH} \left( s \right) \rightarrow \ce{K^+} \left( aq \right) + \ce{OH^-} \left( aq \right)$
The pH Scale
Expressing the acidity of a solution by using the molarity of the hydrogen ion is cumbersome because the quantities are generally very small. Danish scientist Søren Sørensen (1868 - 1939) proposed an easier system for indicating the concentration of $\ce{H^+}$ called the pH scale. The letters pH stand for the power of the hydrogen ion. The pH of a solution is the negative logarithm of the hydrogen-ion concentration.
$\text{pH} = -\text{log} \: \left[ \ce{H^+} \right]$
In pure water or a neutral solution $\left[ \ce{H^+} \right] = 1.0 \times 10^{-7} \: \text{M}$. Substituting this value into the pH expression:
$\text{pH} = -\text{log} \: \left[ 1.0 \times 10^{-7} \right] = -\left( -7.00 \right) = 7.00$
The pH of pure water or any neutral solution is thus 7.00. Due to the somewhat less intuitive rules for dealing with significant figures in the context of logarithms, only the numbers ot the right of the decimal point in the pH value are the significant figures. Since $1.0 \times 10^{-7}$ has two significant figures, the pH can be reported as 7.00.
A logarithmic scale condenses the range of acidity to numbers that are easy to use. For example, a solution in which $\left[ \ce{H^+} \right] = 1.0 \times 10^{-4} \: \text{M}$ has a hydrogen-ion concentration that is 1000 times higher than in pure water. The pH of such a solution is 4.00, a difference of 3 pH units. Notice that when $\left[ \ce{H^+} \right]$ is written in scientific notation and the coefficient is 1, the pH is simply the exponent with the sign changed. The pH of a solution in which $\left[ \ce{H^+} \right] = 1.0 \times 10^{-2} \: \text{M}$ is 2.0 and the pH of a solution in which $\left[ \ce{H^+} \right] = 1.0 \times 10^{-10} \: \text{M}$ is 10.0. If the coefficient is not equal to 1, a calculator must be used to find the pH. For example, the pH of a solution in which $\left[ \ce{H^+} \right] = 2.3 \times 10^{-5} \: \text{M}$ can be found as shown below.
$\text{pH} = -\text{log} \: \left[ 2.3 \times 10^{-5} \right] = 4.64$
As we saw earlier, a solution in which $\left[ \ce{H^+} \right]$ is higher than $1 \times 10^{-7} \: \text{M}$ acidic, while a solution in which $\left[ \ce{H^+} \right]$ is lower than $1.0 \times 10^{-7} \: \text{M}$ is basic. Consequently, solutions with pH values of less than 7 are acidic, while solutions with pH values higher than 7 are basic. The figure below illustrates this relationship, along with some examples of the pH for various solutions.
The pH scale is generally presented as running from 0 to 14, though it is possible to have a pH of less than 0 or greater than 14. For example, a highly concentrated $3.0 \: \text{M}$ of $\ce{HCl}$ has a negative pH.
$\text{pH} = -\text{log} \left( 3.0 \right) = -0.48$
When the pH of a solution is known, the concentration of the hydrogen ion can be calculated. The inverse of the logarithm (or antilog) is the $10^x$ key on a calculator.
$\left[ \ce{H^+} \right] = 10^{-\text{pH}}$
For example, suppose that you have a solution with a pH of 9.14. $\left[ \ce{H^+} \right]$ c an be found as follows:
$\left[ \ce{H^+} \right] = 10^{-\text{pH}} = 10^{-9.14} = 7.24 \times 10^{-10} \: \text{M}$
The pOH Concept
As with the hydrogen-ion concentration, the concentration of the hydroxide ion can be expressed logarithmically by the pOH. The pOH of a solution is the negative logarithm of the hydroxide-ion concentration.
$\text{pOH} = -\text{log} \: \left[ \ce{OH^-} \right]$
The pH of a solution can be related to the pOH. Consider a solution with pH $=$ 4.0. The pOH of the solution would be 10. This example illustrates the following relationship. This is a result of the concentration of the $\ce{H^+}$ and $\ce{OH^-}$ ions present in pure water which we will not discuss further.
$\text{pH} + \text{pOH} = 14$
The pOH scale is similar to the pH scale, in that a pOH of 7 is indicative of a neutral solution. A basic solution has a pOH of less than 7, while an acidic solution has a pOH of greater than 7. The pOH is convenient to use when finding the hydroxide ion concentration from a solution with a known pH.
Example $1$
Find the hydroxide concentration of a solution with a pH of 4.42.
Solution
Step 1: List the known values and plan the problem.
Known
• pH $=$ 4.42
• pH $+$ pOH $=$14
Unknown
• $\left[ \ce{OH^-} \right]$
First, the pOH is calculated, followed by the $\left[ \ce{OH^-} \right]$
Step 2: Solve.
$\text{pOH} = 14 - \text{pH} = 14 - 4.42 = 9.58$
$\left[ \ce{OH^-} \right] = 10^{-\text{pOH}} = 10^{-9.58} = 2.6 \times 10^{-10} \: \text{M}$
Step 3: Think about your result.
The pH is that of an acidic solution, and the resulting hydroxide-ion concentration is less than $1 \times 10^{-7} \: \text{M}$. The answer has two significant figures because the given pH has two decimal places.
The diagram below shows all of the interrelationships between $\left[ \ce{H^+} \right]$, $\left[ \ce{OH^-} \right]$, pH, and pOH.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/08%3A_Properties_of_Solutions/8.06%3A_The_pH_Concept.txt |
These are homework exercises to accompany Chapter 8 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Answers are below the questions.
Questions
(click here for solutions)
Q8.1.1
How is a concentrated solution different from a dilute solution?
Q8.1.2
What is the molarity of a solution prepared with 0.75 moles NaCl in 250. mL solution?
Q8.1.3
What is the mass percent of an aqueous solution prepared by dissolving 12.0 g of solute into 40.0 g of water?
Q8.1.4
What is the volume percent of a solution prepared by adding enough water to 200. mL of acetone to make a total volume of 1.60 L?
Q8.1.5
What mass of glucose is in 250.0 mL of solution that is 5.00% by mass? Assume the density of the solution is 1.00 g/mL.
Q8.1.6
For a solution that contains 25.0 g of NaCl in 300.0 mL of water, find each of the following. The density of the water is 1.00 g/mL. Assume the NaCl does not contribute to the volume of the solution
1. mass percent
2. mass/volume percent
Q8.1.7
For a solution that contains 15.0 mL of methanol 125 mL of ethanol, find each of the following. The density of methanol is 0.792 g/mL and the density of ethanol is 0.789 g/mL.
1. mass percent
2. mass/volume percent
3. volume percent
Q8.1.8
A saline solution has a mass percent concentration of 10.5%. What mass of NaCl is present in 150.0 mL of the solution? Assume the density of the solution is 1.00 g/mL.
Q8.1.9
Calculate the molarity for each solution.
1. 87.2 g of Na2SO4 in enough water to make 500. mL of solution
2. 61.8 g of NH3 in enough water to make 7.00 L of solution
3. 100. mL of ethanol (C2H5OH) in 500. mL of solution (The density of ethanol is 0.789 g/mL.)
Q8.1.10
How many moles of KF are contained in 180.0 mL of a 0.250 M solution?
Q8.1.11
Calculate how many grams of each solute would be required in order to make the given solution.
1. 3.40 L of a 0.780 M solution of iron(III) chloride, FeCl3
2. 60.0 mL of a 4.10 M solution of calcium acetate, Ca(CH3COO)2
Q8.1.12
What volume of a 0.500 M solution of NaI could be prepared with 113 g of solid NaI?
Q8.1.13
Calculate the molarity of the solutions prepared from the following dilutions.
1. 125 mL of 2.00 M HCl is diluted to a volume of 4.00 L.
2. 1.85 mL of 6.30 M AgNO3 is diluted to a volume of 5.00 mL.
Q8.1.14
What volume of 12 M HCl is required to prepare 6.00 L of a 0.300 M solution?
Q8.1.15
What mass of lead is present in 50.0 mL of solution with a lead concentration of 12 ppm?
Q8.1.16
What mass of mercury is present in 175 mL of solution with a mercury concentration of 25 ppb?
Q8.1.17
What is the concentration, in units of ppm, for a solution that contains 34 g of iron in 365 mL of water?
Q8.1.18
How many equivalents are there in 2.0 moles of the ion of each element below?
1. magnesium
2. aluminum
3. sulfur
4. bromine (Br)
5. cesium (Cs)
6. barium (Ba)
Q8.1.19
How many equivalents are present in 2.50 moles of ions for each of the elements in the previous question?
Q8.1.20
How many moles of Ca2+ are given to a patient if they receive 250.0 mL of a solution with a concentration of 132 mEq/L?
Q8.1.21
How many grams of K+ are given to a patient if they receive 500.0 mL of a solution with a concentration of 98 mEq/L?
Q8.1.22
A solution contains 128 mEq/L of Sr2+. What volume of solution is needed to have a total mass of 3.93 g of strontium ions?
(click here for solutions)
Q8.2.1
What is chemical equilibrium?
Q8.2.2
If the reaction H2 + I2 ⇌ 2HI is at equilibrium, do the concentrations of HI, H2, and I2 have to be equal?
Q8.2.3
Do the concentrations at equilibrium depend upon how the equilibrium was reached?
Q8.2.4
What does the equilibrium constant tell us?
Q8.2.5
What does it mean if the Keq is > 1?
Q8.2.6
What does it mean if the Keq is < 1?
Q8.2.7
Does the equilibrium state depend on the starting concentrations?
(click here for solutions)
Q8.3.1
Define Le Chatelier’s principle.
Q8.3.2
List the three factors types of changes that can disturb the equilibrium of a system.
Q8.3.3
How will each change affect the reaction?
PCl5(g) + heat ⇌ PCl3(g) + Cl2(g)
1. Addition of PCl5
2. Addition of Cl2
3. Removal of PCl3
4. Increasing temperature
5. Decreasing temperature
6. Decreasing volume
Q8.3.4
How will each change affect the reaction?
HNO2(aq) ⇌ H+(aq) + NO2(aq)
1. Removal of HNO2
2. Addition of HCl (i.e. adding more H+)
3. Increasing volume
4. Decreasing volume
5. Removal of NO2
6. Addition of OH (which will react with and remove H+)
Q8.3.5
How will each change affect the reaction?
CO2(g) + C(s) ⇌ 2CO(g) $\Delta \text{H}=172.5\; kJ$
1. Addition of CO2
2. Removal of CO2
3. Increasing temperature
4. Decreasing temperature
5. Increasing volume
6. Addition of CO
Q8.3.6
How will each change affect the reaction?
H2(g) + I2(g) ⇌ 2HI(g) $\Delta \text{H} = -9.48\; kJ$
1. Addition of H2
2. Removal of H2
3. Increasing temperature
4. Decreasing temperature
5. Increasing volume
6. Decreasing volume
(click here for solutions)
Q8.4.1
What are some of the features of a semipermeable membrane?
Q8.4.2
Two solutions are separated by a semipermeable membrane. Solution A contains 25.0 g of NaCl in 100.0 mL of water and solution B contains 35.0 g of NaCl in 100.0 mL of water.
1. Which one has a higher concentration?
2. Which way will water molecules flow?
3. Which volume will increase?
4. Which volume will decrease?
5. What will happen to the concentration of solution A?
6. What will happen to the concentration of solution B?
Q8.4.3
What do the prefixes hyper, hypo, and iso mean?
Q8.4.4
Cells are placed in a solution and the cells then undergo hemolysis. What can be said about the relative concentrations of solute in the cell and the solution?
Q8.4.5
Describe the relative concentrations inside and outside a red blood cell when crenation occurs.
Q8.4.6
A saltwater fish is placed in a freshwater tank. What will happen to the fish? Describe the flow of water molecules to explain the outcome.
Q8.4.7
What makes up the "head" region of a phospholipid? Is it hydrophobic or hyrdrophillic?
Q8.4.8
What makes up the "tail" region of a phospholipid? Is it hydrophobic or hyrdrophillic?
(click here for solutions)
Q8.5.1
Which statement below is true? Explain.
1. All Arrhenius bases are also Brønsted-Lowry bases.
2. All Brønsted-Lowry bases are also Arrhenius bases.
Q8.5.2
Classify each of the following as an acid, base, or neither.
1. LiOH
2. HClO4
3. CH3COOH
4. Sr(OH)2
5. CH4
6. CH4OH
Q8.5.3
What does it mean to say that a substance is amphoteric?
Q8.5.4
Identify each reactant in the following reactions as an acid or a base according to the Brønsted-Lowry theory.
1. HIO3(aq) + H2O(l) ⇌ IO3(aq) + H3O+(aq)
2. F(aq) + HClO(aq) ⇌ HF(aq) + ClO(aq)
3. H2PO4(aq) + OH(aq) ⇌ HPO42(aq) + H2O(l)
4. CO32(aq) + H2O(l) ⇌ HCO3(aq) + OH(aq)
Q8.5.5
Referring to question 4, identify the conjugate acid-base pairs in each reaction.
Q8.5.6
Write the formula of each acid’s conjugate base.
1. HNO3
2. HSO3
3. H3AsO4
4. HCOOH
5. HPO42
6. H2S
7. HS
8. HCO3
9. H2CO3
10. H3PO4
11. NaHSO4
Q8.5.7
Write the formula of each base’s conjugate acid.
1. BrO3
2. NH3
3. CH3COO
4. HCO3
5. CN
6. HPO42
7. HS
8. SO42
9. CO32
10. HCO3
11. PH3
Q8.5.8
Explain why the hydrogen phosphate ion (HPO42) is amphoteric.
(click here for solutions)
Q8.6.1
Describe the process by which water self-ionizes, and explain why pure water is considered to be neutral.
Q8.6.2
Indicate whether solutions with the following pH values are acidic, basic, or neutral.
a. pH = 9.4
b. pH = 7.0
c. pH = 5.0
Q8.6.3
How can the pOH of a solution be determined if its pH is known? (Hint: Write a mathematical expression.)
Q8.6.4
Find pH and pOH of each solution.
1. [H+] = 2.3 × 10−4 M
2. [H+] = 8.7 × 10−10 M
3. [OH] = 1.9 × 10−9 M
4. [OH] = 0.60 M
Q8.6.5
Find pH and pOH of each solution.
1. [H+] = 1.0 × 10−5 M
2. [H+] = 2.8 × 10−11 M
3. [OH] = 1.0 × 10−2 M
4. [OH] = 4.4 × 10−9 M
Q8.6.6
Determine [H+] and [OH] in aqueous solutions with the following pH or pOH values.
1. pH = 1.87
2. pH = 11.15
3. pH = 0.95
4. pOH = 6.21
5. pOH = 13.42
6. pOH = 7.03
Q8.6.7
You have prepared 1.00 L of a solution with a pH of 5.00. What is the pH of the solution if 0.100 L of additional water is added to it? (Hint: Calculate the moles of H+ ions present in the solution.)
Q8.6.8
How much water would need to be added to the original solution in question 8 in order to bring the pH to 6.00?
Answers
8.1: Concentrations of Solutions
Q8.1.1
Concentrated solutions have more solute per unit of solvent or solution.
Q8.1.2
$M=\frac{mol\;solute}{L\;soln}$
$M=\frac{0.75\;mol}{0.250\;L}$
$M=3.0\;M$
Q8.1.3
$mass\; \%=\frac{g\;solute}{g\;soln}\times100$
$mass\; \%=\frac{12.0\;g}{40.0\;g+12.0\;g}\times100$
$mass\; \%=23.1\;\%$
Remember, the mass of the solution includes both the solute and solvent.
Q8.1.4
$volume\; \%=\frac{L\;solute}{L\;soln}\times100$
$volume\; \%=\frac{0.200\;L}{1.60\;L}\times100$
$volume\; \%=12.5\;\%$
Volumes can also be used in mL (or any other unit) as long as both volumes are in the same unit.
Q8.1.5
Write the concentration in "expanded form" which shows the relationship to then be used in dimensional analysis.
$5.00\; \%\;m/m=\frac{5.00\;g\;glucose}{100\;g\;solution}$
$250.0\;mL\;soln\left (\frac{1.00\;g\;soln}{mL\;soln}\right )\left (\frac{5.00\;g\;glucose}{100\;g\;soln}\right )=12.5\;g\;glucose$
Q8.1.6
1. $\%\;m\diagup m=\frac{mass\;solute}{mass\;solution}\times 100$
$\%\;m\diagup m=\frac{25.0\;g}{25.0\;g+300\;g}\times 100$
$\%\;m\diagup m=7.69\;\%$
2. The NaCl does not contribute to the volume of the solution so only the volume of the water is used for the volume fo the solution. Given the density is 1.00 g/mL, the volume of the solution is 300.0 mL.
$\%\;m\diagup v=\frac{mass\;solute}{volume\;solution}\times 100$
$\%\;m\diagup v=\frac{25.0\;g}{25.0\;mL+300\;mL}\times 100$
$\%\;m\diagup v=8.33\;\%$
Q8.1.7
The parts of this problem require both the volume and mass of solute and solvent. The volume of the solute and solvent are given so first, find the mass of the solute and solvent so all the values are present before we start calculating the concentrations.
$15.0\;mL\;\text{methanol}\left (\frac{0.792\;g}{mL\;\text{methanol}}\right)=11.9\;g\;\text{methanol}$
$125.0\;mL\;\text{ethanol}\left (\frac{0.789\;g}{mL\;\text{ethanol}}\right)=98.6\;g\;\text{ethanol}$
1. $\%\;m\diagup m=\frac{mass\;solute}{mass\;solution}\times 100$
$\%\;m\diagup m=\frac{11.9\;g\;\text{methanol}}{11.9\;g+98.6\;g}\times 100$
$\%\;m\diagup m=10.8\;\%$
2. $\%\;m\diagup v=\frac{mass\;solute}{volume\;solution}\times 100$
$\%\;m\diagup v=\frac{11.9\;g\;\text{methanol}}{15.0\;mL+125\;mL}\times 100$
$\%\;m\diagup v=8.50\;\%$
3. $\%\;v\diagup v=\frac{volume\;solute}{volume\;solution}\times 100$
$\%\;v\diagup v=\frac{15.0\;mL\;\text{methanol}}{15.0\;mL+125\;mL}\times 100$
$\%\;v\diagup v=10.7\;%$
Q8.1.8
Write the concentration in "expanded form" which shows the relationship to then be used in dimensional analysis.
$10.5\;\%\;m\diagup m=\frac{10.5\;g\;\text{NaCl}}{100\;g\;soln}$
$150.0\;mL\;soln\left (\frac{1.00\;g\;soln}{mL\;soln}\right )\left (\frac{10.5\;g\;\text{NaCl}}{100\;g\;soln}\right )=15.8\;g\;\text{NaCl}$
Q8.1.9
1. $M=\frac{0.614\;mol\;\text{Na}_2\text{SO}_4}{0.500\;L\;soln}=1.23\;M$
2. $M=\frac{3.63\;mol\;\text{NH}_3}{7.00\;L\;soln}=0.519\;M$
3. $M=\frac{1.71\;mol\;\text{EtOH}}{0.500\;L\;soln}=3.43\;M$
Q8.1.10
$0.250\;M=\frac{0.250\;mol\;\text{KF}}{1\;L\;soln}$
$180.0\;mL\left(\frac{10^{-3}\;L}{1\;mL}\right)\left(\frac{0.250\;mol\;\text{KF}}{1\;L\;soln}\right)=0.0450\;mol\;\text{KF}$
Q8.1.11
1. $3.40\;L\left(\frac{0.780\;mol}{1\;L}\right)\left(\frac{162.2\;g}{mol}\right)=430.\;g\;\text{FeCl}_3$
2. $60.0\;mL\left(\frac{10^{-3}\;L}{1\;mL}\right)\left(\frac{4.10\;mol}{1\;L}\right)\left(\frac{158.17\;g}{mol}\right)=38.9\;g\;\text{Ca(CH}_3\text{COO)}_2$
Q8.1.12
$0.500\;M=\frac{0.500\;mol\;\text{NaI}}{1\;L\;soln}$
$113\;g\;\text{NaI}\left(\frac{1\;mol}{149.89\;g}\right)\left(\frac{1\;L}{0.500\;mol}\right)=1.51\;L\;soln$
Q8.1.13
1. $C_1V_1=C_2V_2$
$2.00\;M\cdot0.125\;L=C_2\cdot4.00\;L$
$C_2=0.0625\;M$
2. $C_1V_1=C_2V_2$
$6.30\;M\cdot1.85\;mL=C_2\cdot5.00\;mL$
$C_2=2.33\;M$
Q8.1.14
$C_1V_1=C_2V_2$
$0.300\;M\cdot6.00\;L=12\;M\cdotV_2$
$V_2=0.15\;L$
Q8.1.15
$12\;ppm\;\text{Pb}=\frac{12\;mg\;\text{Pb}}{1\;L\;soln}$
$50.0\;mL\left(\frac{10^{-3}\;L}{1\;mL}\right)\left(\frac{12\;mg} {1\;L}\right)=0.60\;mg\;\text{Pb}$
Q8.1.16
$25\;ppb\;\text{Hg}=\frac{25\;\mu g\;\text{Hg}}{1\;L\;soln}$
$175\;mL\left(\frac{10^{-3}\;L}{1\;mL}\right)\left(\frac{25\;\mu g} {1\;L}\right)=4.4\;\mu g\;\text{Hg}$
Q8.1.17
$ppm=\frac{mg}{L}$
$34\;g\;\text{Fe}\left(\frac{1\;mg}{10^{-3}\;g}\right)=3.4\times10^4\;mg\;\text{Fe}$
$365\;mL\left(\frac{10^{-3}\;L}{1\;mL}\right)=0.365\;L$
$ppm=\frac{3.4\times10^4\;mg\;\text{Fe}}{0.365\;L}=9.3\times10^4\;ppm\;\text{Fe}$
Q8.1.18
1. $2.0\;mol\;\text{Mg}^{2+}\left(\frac{2\;Eq}{1\;mol}\right)=4.0\;Eq$
2. $2.0\;mol\;\text{Al}^{3+}\left(\frac{3\;Eq}{1\;mol}\right)=6.0\;Eq$
3. $2.0\;mol\;\text{S}^{2-}\left(\frac{2\;Eq}{1\;mol}\right)=4.0\;Eq$
4. $2.0\;mol\;\text{Br}^-\left(\frac{1\;Eq}{1\;mol}\right)=2.0\;Eq$
5. $2.0\;mol\;\text{Cs}^{+}\left(\frac{1\;Eq}{1\;mol}\right)=2.0\;Eq$
6. $2.0\;mol\;\text{Ba}^{2+}\left(\frac{2\;Eq}{1\;mol}\right)=4.0\;Eq$
Q8.1.19
1. $2.50\;mol\;\text{Mg}^{2+}\left(\frac{2\;Eq}{1\;mol}\right)=5.00\;Eq$
2. $2.50\;mol\;\text{Al}^{3+}\left(\frac{3\;Eq}{1\;mol}\right)=7.50\;Eq$
3. $2.50\;mol\;\text{S}^{2-}\left(\frac{2\;Eq}{1\;mol}\right)=5.00\;Eq$
4. $2.50\;mol\;\text{Br}^-\left(\frac{1\;Eq}{1\;mol}\right)=2.50\;Eq$
5. $2.50\;mol\;\text{Cs}^{+}\left(\frac{1\;Eq}{1\;mol}\right)=2.50\;Eq$
6. $2.50\;mol\;\text{Ba}^{2+}\left(\frac{2\;Eq}{1\;mol}\right)=5.00\;Eq$
Q8.1.20
$250.0\;mL\left(\frac{10^{-3}\;L}{1\;mL}\right)\left(\frac{132\;mEq}{L}\right)\left(\frac{10^{-3}\;Eq}{1\;mEq}\right)\left(\frac{1\;mol\;\text{Ca}^{2+}}{2\;Eq}\right)=0.0165\;mol\;\text{Ca}^{2+}$
Q8.1.21
$500.0\;mL\left(\frac{10^{-3}\;L}{1\;mL}\right)\left(\frac{98\;mEq}{L}\right)\left(\frac{10^{-3}\;Eq}{1\;mEq}\right)\left(\frac{1\;mol\;\text{K}^{+}}{1\;Eq}\right)\left(\frac{39.10\;g}{mol}\right)=1.92\;g\;\text{K}^{+}$
Q8.1.22
$3.93\;g\;\text{Sr}^{2+}\left(\frac{1\;mol}{87.62\;g}\right)\left(\frac{2\;Eq}{1\;mol}\right)\left(\frac{1\;mEq}{10^{-3}\;Eq}\right)\left(\frac{1\;L}{128\;mEq}\right)=0.701\;L\;soln$
8.2: Chemical Equilibrium
Q8.2.1
The rate of the forward reaction equals the rate of the reverse reaction.
Q8.2.2
No, the concentrations are constant but the concentrations do not have to be equal.
Q8.2.3
No.
Q8.2.4
The ratio of products and reactants when the system is at equilibrium.
Q8.2.5
More products than reactants are present at equilibrium.
Q8.2.6
More reactants than products present at equilibrium.
Q8.2.7
No. The equilibrium ratio does not depend on the initial concentrations.
8.3: Le Chatelier's Principle
Q8.3.1
Le Chatelier’s principle states that a system at equilibrium is disturbed, it will respond in a way to minimize te disturbance.
Q8.3.2
temperature, change in amount of substance, change in pressure through change in volume
Q8.3.3
1. shift right
2. shift left
3. shift right
4. shift right
5. shift left
6. shift left
Q8.3.4
1. shift left
2. shift left
3. no effect
4. no effect
5. shift right
6. shift right
Q8.3.5
1. shift right
2. shift left
3. shift right
4. shift left
5. shift right
6. shift left
Q8.3.6
1. shift right
2. shift left
3. shift left
4. shift right
5. no effect
6. no effect
8.4: Osmosis and Diffusion
Q8.4.1
A semipermeable membrane allows some substances to pass through but not others.
Q8.4.2
Two solutions are separated by a semipermeable membrane. Solution A contains 25.0 g of NaCl in 100.0 mL of water and solution B contains 35.0 g of NaCl in 100.0 mL of water.
1. Solution B
2. A $\rightarrow$ B
3. B
4. A
5. increase
6. decrease
Q8.4.3
hyper - higher
hypo - lower
iso - same
Q8.4.4
Cells contain fluid with higher concentration than solution outside the cell.
Q8.4.5
Cells contain fluid with a lower concentration than the solution outside the cell.
Q8.4.6
Water molecules will flow from the tank water into the fish because the fish has a higher concentration of salt. If the fish absorbs too much water, it will die.
Q8.4.7
The "head" region is a phosphate group and it is hydrophillic.
Q8.4.8
The "tail" is a hydrocarbon tail and it is hydrophobic.
8.5: Acid-Base Definitions
Q8.5.1
1. TRUE
2. FALSE - Bronsted-Lowry acid-base definitions are broader.
Q8.5.2
Classify each of the following as an acid, base, or neither.
1. base (contains metal and -OH group)
2. acid (formula starts with H and isn't water)
3. acid (contains -COOH which is carboxylic acid functional group)
4. base (contains metal and -OH group)
5. neither
6. neither (-OH group has to be with metal)
Q8.5.3
Amphoteric substances can act as an acid or base.
Q8.5.4
Identify each reactant in the following reactions as an acid or a base according to the Brønsted-Lowry theory.
1. HIO3(aq) - acid; H2O(l) - base; IO3(aq) - base ; H3O+(aq) - acid
2. F(aq) - base; HClO(aq) - acid; HF(aq) - acid; ClO(aq) - base
3. H2PO4(aq) - acid; OH(aq) - base; HPO42−(aq) - base; H2O(l) - acid
4. CO32−(aq) - base; H2O(l) - acid; HCO3(aq) - acid; OH(aq) - base
Q8.5.5
1. HIO3/IO3 and H3O+/H2O
2. HF/Fand HClO/ClO
3. H2PO4/HPO42− and H2O/OH
4. HCO3/CO32− and H2O/OH
Q8.5.6
1. NO3
2. SO32−
3. H2AsO4
4. HCOO(the H that is removed comes from the carboxylic acid functional group)
5. PO43−
6. HS
7. S2−
8. CO32
9. HCO3
10. H2PO4
11. Na2SOor NaSO4
Q8.5.7
Write the formula of each base’s conjugate acid.
1. HBrO3
2. NH4+
3. CH3COOH
4. H2CO3
5. HCN
6. H2PO4
7. H2S
8. HSO4
9. HCO3
10. H2CO3
11. PH4+
Q8.5.8
HPO42− can act as a base and accept a proton to form H2PO4and it can act as an acid and donate a proton to form PO43−.
8.6: The pH Concept
Q8.6.1
H2O(l) H+(aq) + OH(aq)
It's neutral because there are equal amounts of H+ and OH.
Q8.6.2
Indicate whether solutions with the following pH values are acidic, basic, or neutral.
a. basic
b. neutral
c. acidic
Q8.6.3
pH + pOH = 14
Q8.6.4
1. pH = 3.64; pOH = 10.36
2. pH = 9.06; pOH = 4.94
3. pOH = 8.72; pH = 5.28
4. pOH = 0.22; pH = 13.78
Q8.6.5
1. pH = 5.00; pOH = 9.00
2. pH = 10.55; pOH = 4.94
3. pOH = 8.72; pH = 5.28
4. pOH = 0.22; pH = 13.78
Q8.6.6
1. $[\text{H}^+]=1.3\times 10^{-2}\;M, [\text{OH}^-]=7.4\times10^{-13}\;M$
2. $[\text{H}^+]=7.1\times 10^{-12}\;M, [\text{OH}^-]=1.4\times10^{-3}\;M$
3. $[\text{H}^+]=0.11\;M, [\text{OH}^-]=8.9\times10^{-14}\;M$
4. $[\text{OH}^-]=6.2\times10^{-7}\;M, [\text{H}^+]=1.6\times 10^{-8}\;M$
5. $[\text{OH}^-]=3.8\times10^{-14}\;M, [\text{H}^+]=0.26\;M$
6. $[\text{OH}^-]=9.3\times10^{-8}\;M, [\text{H}^+]=1.1\times 10^{-7}\;M$
Q8.6.7
Given pH = 5.00, we know $[\text{H}^+]=1.0\times 10^{-5}\;M$ which means $M=\frac{1.0\times10^{-5}\;mol\;\text{H}^+}{1.00\;L}$.
If 0.100 L of water is added to 1.00 L, then the volume changes to 1.10 L but the moles of H+ does not change. The molarity can be calclulated with the same number of moles and the new volume.
$M=\frac{1.0\times10^{-5}\;mol\;\text{H}^+}{1.00+0.100\;L}$
$M=9.1\times10^{-6}\;M$
$pH = 5.04$
Q8.6.8
How much water would need to be added to the original solution in question 8 in order to bring the pH to 6.00?
To get to pH = 6.00, we need $[\text{H}^+]=1.0\times 10^{-6}\;M$.
Use the dilution formula to calculate the total volume of solution.
$C_1V_1=C_2V_2$
$1.0\times10^{-5}\;M\cdot 1.00\;L=1.0\times10^{-6}\;M\cdot V_2$$V_2=10.\;L$
The total volume is 10. L so 9 L needs to be added to the original 1 L solution. | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/08%3A_Properties_of_Solutions/8.07%3A_Properties_of_Solutions_%28Exercises%29.txt |
Learning Outcomes
• Define Le Chatelier's principle.
• Predict how the change in amounts of substances, temperature, or pressure will affect amounts of reactants and products present at equilibrium.
Le Chatelier's Principle
Chemical equilibrium was studied by the French chemist Henri Le Chatelier (1850 - 1936) and his description of how a system responds to a stress to equilibrium has become known as Le Chatelier's principle: When a chemical system that is at equilibrium is disturbed by a stress, the system will respond in order to relieve the stress. Stresses to a chemical system involve changes in the concentrations of reactants or products, changes in the temperature of the system, or changes in the pressure of the system. We will discuss each of these stresses separately. The change to the equilibrium position in every case is either a favoring of the forward reaction or a favoring of the reverse reaction. When the forward reaction is favored, the concentrations of products increase, while the concentrations of reactants decrease. When the reverse reaction is favored, the concentrations of the products decrease, while the concentrations of reactants increase.
$\begin{array}{lll} \textbf{Original Equilibrium} & \textbf{Favored Reaction} & \textbf{Result} \ \ce{A} \rightleftharpoons \ce{B} & \text{Forward:} \: \ce{A} \rightarrow \ce{B} & \left[ \ce{A} \right] \: \text{decreases}; \: \left[ \ce{B} \right] \: \text{increases} \ \ce{A} \rightleftharpoons \ce{B} & \text{Reverse:} \: \ce{A} \leftarrow \ce{B} & \left[ \ce{A} \right] \: \text{increases}; \: \left[ \ce{B} \right] \: \text{decreases} \end{array}$
Effect of Concentration
A change in concentration of one of the substances in an equilibrium system typically involves either the addition or the removal of one of the reactants or products. Consider the Haber-Bosch process for the industrial production of ammonia from nitrogen and hydrogen gases.
$\ce{N_2} \left( g \right) + 3 \ce{H_2} \left( g \right) \rightleftharpoons 2 \ce{NH_3} \left( g \right)$
If the concentration of one substance in a system is increased, the system will respond by favoring the reaction that removes that substance. When more $\ce{N_2}$ is added, the forward reaction will be favored because the forward reaction uses up $\ce{N_2}$ and converts it to $\ce{NH_3}$. The forward reaction speeds up temporarily as a result of the addition of a reactant. The position of equilibrium shifts as more $\ce{NH_3}$ is produced. The concentration of $\ce{NH_3}$ increases, while the concentrations of $\ce{N_2}$ and $\ce{H_2}$ decrease. After some time passes, equilibrium is reestablished with new concentrations of all three substance. As can be seen in the figure below, if more $\ce{N_2}$ is added, a new equilibrium is achieved by the system. The new concentration of $\ce{NH_3}$ is higher because of the favoring of the forward reaction. The new concentration of the $\ce{H_2}$ is lower .The concentration of $\ce{N_2}$ is higher than in the original equilibrium, but went down slightly following the addition of the $\ce{N_2}$ that disturbed the original equilibrium. By responding in this way, the value of the equilibrium constant for the reaction, $K_\text{eq}$, does not change as a result of the stress to the system.
In other words, the amount of each substance is different but the ratio of the amount of each remains the same.
If more $\ce{NH_3}$ were added, the reverse reaction would be favored. This "favoring" of a reaction means temporarily speeding up the reaction in that direction until equilibrium is reestablished. Recall that once equilibrium is reestablished, the rates of the forward and reverse reactions are again equal. The addition of $\ce{NH_3}$ would result in increased formation of the reactants, $\ce{N_2}$ and $\ce{H_2}$.
An equilibrium can also be disrupted by the removal of one of the substances. If the concentration of a substance is decreased, the system will respond by favoring the reaction that replaces that substance. In the industrial Haber-Bosch process, $\ce{NH_3}$ is removed from the equilibrium system as the reaction proceeds. As a result, the forward reaction is favored so that more $\ce{NH_3}$ is produced. The concentrations of $\ce{N_2}$ and $\ce{H_2}$ decrease. Continued removal of $\ce{NH_3}$ will eventually force the reaction to go to completion until all of the reactants are used up. If either $\ce{N_2}$ or $\ce{H_2}$ were removed from the equilibrium system, the reverse reaction would be favored and the concentration of $\ce{NH_3}$ would decrease.
The effect of changes in concentration on an equilibrium system according to Le Chatelier's principle is summarized in the table below.
Table $1$
Stress Response
addition of reactant forward reaction favored
addition of product reverse reaction favored
removal of reactant reverse reaction favored
removal of product forward reaction favored
Effect of Temperature
Increasing or decreasing the temperature of a system at equilibrium is also a stress to the system. The equation for the Haber-Bosch process is written again below, as a thermochemical equation (i.e. it contains information about the energy gained or lost when the reaction occurs).
$\ce{N_2} \left( g \right) + 3 \ce{H_2} \left( g \right) \rightleftharpoons 2 \ce{NH_3} \left( g \right) + 91 \: \text{kJ}$
The forward reaction is the exothermic direction: the formation of $\ce{NH_3}$ releases heat which is why that is shown as a product. The reverse reaction is the endothermic direction: as $\ce{NH_3}$ decomposes to $\ce{N_2}$ and $\ce{H_2}$, heat is absorbed. An increase in the temperature for this is like adding a product because heat is being released by the reaction. If we add a product then the reaction proceeds towards the formation of more reactants. Reducing the temperature for this system would be similar to removing a product which would favor the formation of more products. The amount of $\ce{NH_3}$ will increase and the amount of $\ce{N_2}$ and $\ce{H_2}$ will decrease.
For changes in concentration, the system responds in such a way that the value of the equilibrium constant, $K_\text{eq}$ is unchanged. However, a change in temperature shifts the equilibrium and the $K_\text{eq}$ value either increases or decreases. As discussed in the previous section, values of $K_\text{eq}$ are dependent on the temperature. When the temperature of the system for the Haber-Bosch process is increased, the resultant shift in equilibrium towards the reactants means that the $K_\text{eq}$ value decreases. When the temperature is decreased, the shift in equilibrium towards the products means that the $K_\text{eq}$ value increases.
Le Chatelier's principle as related to temperature changes can be illustrated easily be the reaction in which dinitrogen tetroxide is in equilibrium with nitrogen dioxide.
$\ce{N_2O_4} \left( g \right) + \text{heat} \rightleftharpoons 2 \ce{NO_2} \left( g \right)$
Dinitrogen tetroxide $\left( \ce{N_2O_4} \right)$ is colorless, while nitrogen dioxide $\left( \ce{NO_2} \right)$ is dark brown in color. When $\ce{N_2O_4}$ breaks down into $\ce{NO_2}$, heat is absorbed (endothermic) according to the forward reaction above. Therefore, an increase in temperature (adding heat) of the system will favor the forward reaction. Conversely, a decrease in temperature (removing heat) will favor the reverse reaction.
Effect of Pressure
Changing the pressure of an equilibrium system in which gases are involved is also a stress to the system. A change in the pressure on a liquid or a solid has a negligible effect. We will return again the equilibrium for the Haber-Bosch process. Imagine the gases are contained in a closed system in which the volume of the system is controlled by an adjustable piston as shown in the figure below.
On the far left, the reaction system contains primarily $\ce{N_2}$ and $\ce{H_2}$, with only one molecule of $\ce{NH_3}$ present. As the piston is pushed inwards, the pressure of the system increases according to Boyle's law. This is a stress to the equilibrium. In the middle image, the same number of molecules is now confined in a smaller space and so the pressure has increased. According to Le Chatelier's principle, the system responds in order to relieve the stress. In the image on the right, the forward reaction has been favored and more $\ce{NH_3}$ is produced. The overall result is a decrease in the number of gas molecules in the entire system. This in turn decreases the pressure and provides a relief to the original stress of a pressure increase. An increase in pressure on an equilibrium system favors the reaction which products fewer total moles of gas. In this case, it is the forward reaction that is favored.
A decrease in pressure on the above system could be achieved by pulling the piston outward, increasing the container volume. The equilibrium would respond by favoring the reverse reaction in which $\ce{NH_3}$ decomposes to $\ce{N_2}$ and $\ce{H_2}$. This is because the overall number of gas molecules would increase and so would the pressure. A decrease in pressure on an equilibrium system favors the reaction which produces more total moles of gas. This is summarized in the table below.
Table $2$
Stress Response
pressure increase reaction produces fewer gas molecules
pressure decrease reaction produces more gas molecules
Like changes in concentration, the $K_\text{eq}$ value for a given reaction is unchanged by a change in pressure. The amounts of each substance will change but the ratio will not. It is important to remember when analyzing the effect of a pressure change on equilibrium that only gases are affected. If a certain reaction involves liquids or solids, they should be ignored. For example, calcium carbonate decomposes according to the equilibrium reaction:
$\ce{CaCO_3} \left( s \right) \rightleftharpoons \ce{CaO} \left( s \right) + \ce{O_2} \left( g \right)$
Oxygen is the only gas in the system. An increase in the pressure of the system slows the rate of decomposition of $\ce{CaCO_3}$ because the reverse reaction is favored. When a system contains equal moles of gas on both sides of the equation, pressure has no effect on the equilibrium position, as in the formation of $\ce{HCl}$ from $\ce{H_2}$ and $\ce{Cl_2}$.
$\ce{H_2} \left( g \right) + \ce{Cl_2} \left( g \right) \rightleftharpoons 2 \ce{HCl} \left( g \right)$
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/08%3A_Properties_of_Solutions/803%3A_Le_Chateliers_Principle.txt |
• 9.1: Acid and Base Strength
Acids can be classified as strong or weak based on the extent to which they produce H3O+ when dissolved in water.
• 9.2: Buffers
A buffer is a solution of a weak acid or a base and its salt. Both components must be present for the system to act as a buffer to resist changes in pH. The salt is the conjugate of the weak acid or of the weak base.
• 9.3: Equilibrium Applications (Exercises)
These are homework exercises to accompany Chapter 9 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Solutions are available below the questions.
09: Equilibrium Applications
Learning Outcomes
• Define weak acids and bases.
• Write an equation representing the behavior of a weak acid.
• Explain differences between strong and weak acids and strong and weak bases.
• List the 6 strong acids.
• Calculate p$K_\text{a}$ and p$K_\text{b}$values.
• Rank acids in order of strength based on their $K_\text{a}$ and p$K_\text{a}$values.
• Rank bases in order of strength based on their $K_\text{b}$ and p$K_\text{b}$values.
So far, we have primarily been defining acids by their ability to donate an $\ce{H^+}$ ion and bases by their ability to accept an $\ce{H^+}$ ion. However, acids and bases vary in their relative ability to undergo these processes. Which was mentioned when we talked about weak electrolytes.
In general, acids can be classified as strong or weak based on the extent to which they produce $\ce{H_3O^+}$ when dissolved in water. For a generic acid, we can write the following equilibrium reaction:
$\ce{HA} \left( aq \right) + \ce{H_2O} \left( l \right) \rightleftharpoons \ce{H_3O^+} \left( aq \right) + \ce{A^-} \left( aq \right)$
Using the usual shorthand notation, this equation can also be written as follows:
$\ce{HA} \left( aq \right) \rightleftharpoons \ce{H^+} \left( aq \right) + \ce{A^-} \left( aq \right)$
This type of equilibrium, in which a proton is being transferred to water, is often indicated by writing the equilibrium constant as $K_\text{a}$. The relative position of this equilibrium for a given acid determines whether it will be considered strong or weak. When dissolved in water, a strong acid will completely transfer its proton to the solvent. In terms of the equilibrium above, the products will be heavily favored $\left( K_\text{a} \gg 1 \right)$. In fact, the products are so heavily favored that the reverse reaction is often not even considered, and the proton transfer is written as unidirectional. For example, the strong acid $\ce{HCl}$ can dissociated in water according to the following reaction:
$\ce{HCl} \left( aq \right) + \ce{H_2O} \left( l \right) \rightarrow \ce{H_3O^+} \left( aq \right) + \ce{Cl^-} \left( aq \right)$
which is sometimes written as
$\ce{HCl} \left( aq \right) \rightarrow \ce{H^+} \left( aq \right) + \ce{Cl^-} \left( aq \right)$
to simplify the equation by eliminating the water in the equation because the "$aq$" indicates that water is present. At equilibrium, essentially no intact $\ce{HCl}$ molecules are still present in solution.
In contrast, the equilibrium for a weak acid favors the reactants. A particularly common type of weak acid is an organic molecule that contains a carboxyl group $\ce{COOH}$. For example, acetic acid (the acidic component of vinegar) has the formula $\ce{CH_3COOH}$. Its dissociation equation can be written as follows:
$\ce{CH_3COOH} \left( aq \right) + \ce{H_2O} \left( l \right) \rightleftharpoons \ce{H_3O^+} \left( aq \right) + \ce{CH_3COO^-} \left( aq \right)$
sometimes written as $\ce{CH_3COOH} \left( aq \right) \rightleftharpoons \ce{H^+} \left( aq \right) + \ce{CH_3COO^-} \left( aq \right)$. Because we are dealing with a weak acid, $K_\text{a}$ for this equilibrium is much less than 1. At equilibrium, most of the acetic acid molecules are still intact, and only a small percentage have transferred their protons to the solvent. The $K_\text{a}$ values for some weak acids are listed in the table below. All weak acids are not equally
Table $1$
Acid Name Structure $K_\text{a}$
hydrofluoric acid $\ce{H-F}$ $7.1 \times 10^{-4}$
nitrous acid $\ce{O=N-O-H}$ $4.5 \times 10^{-4}$
formic acid $\ce{HCOOH}$ $1.7 \times 10^{-4}$
acetic acid $\ce{CH_3COOH}$ $1.8 \times 10^{-5}$
hydrocyanic acid $\ce{H-CN}$ $4.9 \times 10^{-10}$
The nature of p$K_\text{a}$ is also used to indicate the strength of an acid. p$K_\text{a}$ is determined much like pH by taking the negative logarithm of $K_\text{a}$. As with pH, it is used to make values easier to manage. As an acid's strength increases, its $K_\text{a}$ value increases and its p$K_\text{a}$ value decreases as shown in the table above. Most acids that you will encounter in general chemistry courses are weak acids. There are six common strong acids (see table below). If you recognize these six then you can assume any other acids are weak.
Acids
• Hydrochloric acid, $\ce{HCl}$
• Hydrobromic acid, $\ce{HBr}$
• Hydroiodic acid, $\ce{HI}$
• Perchloric acid, $\ce{HClO_4}$
• Nitric acid, $\ce{HNO_3}$
• Sulfuric acid, $\ce{H_2SO_4}$
Strong vs. Weak Bases
Analogous to the acid dissociation reaction from the previous section, we can write the reaction between a generic base and water as follows:
$\ce{B} \left( aq \right) + \ce{H_2O} \left( l \right) \rightleftharpoons \ce{BH^+} \left( aq \right) + \ce{OH^-} \left( aq \right)$
The equilibrium constant for a reaction in which a base is deprotonating water (taking water's hydrogen atom) is often given the symbol $K_\text{b}$. Strong bases and weak bases can then be defined based on the position of this equilibrium. A weak base would have a very small $K_\text{b}$ value (much less than 1), indicating that most molecules of the base do not remove a proton from water. Conversely, a strong base would have a $K_\text{b}$ value greater than or equal to 1.
Nitrogen-containing compounds are a common type of weak base. The lone pair on the nitrogen atom can accept a proton from water as follows:
$\ce{NH_3} \left( aq \right) + \ce{H_2O} \left( l \right) \rightleftharpoons \ce{NH_4^+} \left( aq \right) + \ce{OH^-} \left( aq \right)$
The equilibrium constant for this reaction is quite low, so most of the $\ce{NH_3}$ molecules will not remove a proton from water. $K_\text{b}$ and p$K_\text{b}$ values for a few weak bases are listed in the table below.
Table $3$
Base $K_\text{b}$
ethylamine $\left( \ce{CH_3CH_2NH_2} \right)$ $5.6 \times 10^{-4}$
methylamine $\left( \ce{CH_3NH_2} \right)$ $4.4 \times 10^{-4}$
ammonia $\left( \ce{NH_3} \right)$ $1.8 \times 10^{-5}$
The only strong bases that are commonly used in general chemistry courses are ionic compounds composed of metal cations and hydroxide anions, such as $\ce{NaOH}$, $\ce{KOH}$, or $\ce{Ba(OH)_2}$.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/09%3A_Equilibrium_Applications/9.01%3A_Acid_and_Base_Strength.txt |
Learning Outcomes
• Define buffer.
• Define buffer capacity.
• Describe how a buffer controls pH.
• Identify the components of a buffer solution.
Diabetes mellitus is a disorder of glucose metabolism in which insulin production by the pancreas is impaired. Since insulin helps glucose enter the cells, a decrease of this hormone means that glucose cannot be used in its normal fashion. When this happens, the body begins to break down fats, producing a decrease in blood pH. Chemical systems in the body can balance this pH shift for a while, but excessive acid production can create serious problems if not corrected by administering insulin to restore normal glucose use.
Buffers
If only $1.0 \: \text{mL}$ of $0.10 \: \text{M}$ hydrochloric acid is added to $1.0 \: \text{L}$ of pure water the pH drops drastically from 7.0 to 4.0. This is a 1000-fold increase in the acidity of the solution. For many purposes, it is desirable to have a solution which is capable of resisting such large changes in pH when relatively small amounts of acid or base are added to them. Such a solution is called a buffer. A buffer is a solution of a weak acid or a base and its salt. Both components must be present for the system to act as a buffer to resist changes in pH. The salt is the conjugate of the weak acid or of the weak base. It can be shown as the ion or with the counter ion. $\left( \ce{COO^-} \: \text{or} \: \ce{COONa} \right)$
Some common buffer systems are listed in the table below. Note that the two components of the buffer system differ by only one hydrogen ion $\left( \ce{H^+} \right)$.
Table $1$: Some Common Buffers
Buffer system Buffer components pH of buffer (equal molarities of both components)
Acetic acid/acetate ion $\ce{CH_3COOH}/\ce{CH_3COO^-}$ 4.74
Carbonic acid/hydrogen carbonate ion $\ce{H_2CO_3}/\ce{HCO_3^-}$ 6.38
Dihydrogen phosphate ion/hydrogen phosphate ion $\ce{H_2PO_4^-}/\ce{HPO_4^{2-}}$ 7.21
Ammonia/ammonium ion $\ce{NH_3}/\ce{NH_4^+}$ 9.25
One example of a buffer is a solution made of acetic acid (the weak acid) and sodium acetate (a conjugate of the acid). The pH of a buffer consisting of $0.50 \: \text{M} \: \ce{CH_3COOH}$ and $0.50 \: \text{M} \: \ce{CH_3COONa}$ is 4.74. If $10.0 \: \text{mL}$ of $1.0 \: \text{M} \: \ce{HCl}$ is added to $1.0 \: \text{L}$ of the buffer, the pH only decreases to 4.73. This ability to "soak up" the additional hydrogen ions from the $\ce{HCl}$ that was added is due to the reaction below.
$\ce{CH_3COO^-} \left( aq \right) + \ce{H^+} \left( aq \right) \rightleftharpoons \ce{CH_3COOH} \left( aq \right)$
Since both the acetate ion and the acetic acid were already present in the buffer, the only thing that changes is the ratio of one to the other. Small changes in that ratio have only very minor effects on the pH.
If $10.0 \: \text{mL}$ of $1.0 \: \text{M} \: \ce{NaOH}$ were added to another $1.0 \: \text{L}$ of the same buffer, the pH would only increase to 4.76. In this case, the buffer takes up the additional hydroxide ions.
$\ce{CH_3COOH} \left( aq \right) + \ce{OH^-} \left( aq \right) \rightleftharpoons \ce{CH_3COO^-} \left( aq \right) + \ce{H_2O} \left( l \right)$
Again the ratio of acetate ion to acetic acid changes only slightly, this time causing a very small increase in the pH. All buffers follow Le Chatelier's principle and respond to a stress on the system by responding to minimize the stress.
It is possible to add so much acid or base to a buffer that its ability to resist a significant change in pH is overwhelmed. The buffer capacity is the amount of acid or base that can be added to a buffer solution before a large change in pH occurs. The buffer capacity is exceeded when the number of moles of $\ce{H^+}$ or $\ce{OH^-}$ that are added to the buffer exceeds the number of moles of the buffer components.
Human blood contains a buffer of carbonic acid $\left( \ce{H_2CO_3} \right)$ and bicarbonate anion $\left( \ce{HCO_3^-} \right)$ in order to maintain blood pH between 7.35 and 7.45, as a value higher than 7.8 (alkalosis) or lower than 6.8 (acidosis) can lead to death. In this buffer, hydronium and bicarbonate anion are in equilibrium with carbonic acid. The bicarbonate neutralizes excess acids in the blood while the carbonic acid neutralizes excess bases.
Furthermore, carbonic acid can decompose into $\ce{CO_2}$ gas and water, resulting in a second equilibrium system between carbonic acid and water. Because $\ce{CO_2}$ is an important component of the blood buffer, its regulation in the body, as well as that of $\ce{O_2}$, is extremely important. The effect of this can be important when the human body is subjected to strenuous conditions.
$\ce{CO_2} + \ce{H_2O} \rightleftharpoons \ce{H_2CO_3} \rightleftharpoons \ce{HCO_3^-} + \ce{H^+}$
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/09%3A_Equilibrium_Applications/9.02%3A_Buffers.txt |
These are homework exercises to accompany Chapter 9 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Answers are below the questions.
Questions
(click here for solutions)
Q9.1.1
Describe the difference between a strong acid and a weak acid.
Q9.1.2
Describe the difference between a strong base and a weak base.
Q9.1.3
Identify each of the following as a strong acid, weak acid, strong base, or weak base.
1. HCl
2. NaOH
3. KOH
4. HNO2
5. HNO3
6. HF
7. NH3
8. Ba(OH)2
9. CH3CH2COOH
Q9.1.4
Write an equation representing the behavior of each substance in question 3. Pay attention to the type of arrow used in the equation.
Q9.1.5
A solution is prepared by dissolving 15.0 grams of NaOH in enough water to make 500.0 mL of solution. Calculate the pH of the solution.
Q9.1.6
A solution is prepared by dissolving 22.0 grams of HCl in enough water to make 300.0 mL of solution. Calculate the pH of the solution.
(click here for solutions)
Q9.2.1
What is a buffer?
Q9.2.2
What is the purpose of a buffer?
Q9.2.3
Determine whether or not each of these pairs can act as a buffer.
1. HCl/Cl
2. HF/F
3. H2SO4/HSO4
4. HSO4/SO42
5. H2O/NaOH
6. HNO2/KNO2
7. HCl/NaOH
Q9.2.4
Write the formula of the conjugate base needed to form a buffer with each of the following weak acids.
1. HClO3
2. H2PO4
3. CH3COOH
Q9.2.5
Write the formula of the conjugate acid needed to form a buffer with each of the following weak bases.
1. NH3
2. CH3NH2
Q9.2.6
Describe buffer capacity.
Answers
9.1: Acid and Base Strength
Q9.1.1
A strong acid completely dissociates into ions and a weak acid doesn't completely dissociate into ions.
Q9.1.2
A strong base is a base (metal with an -OH) group that dissociates completely into ions. A weak base is a proton acceptor but not ll of the molecules will accept a proton.
Q9.1.3
Identify each of the following as a strong acid, weak acid, strong base, or weak base.
1. HCl is a strong acid. See list of 6 strong acids.
2. NaOH is a strong base. It has a metal with an -OH group.
3. KOH is a strong base. It has a metal and an -OH group.
4. HNO2 is a weak acid. The formula starts with H but it's not water so it's an acid. Recognize it is weak because it is not on the list of 6 strong acids.
5. HNO3 is a strong acid. The formula starts with H but it's not water so it's an acid. See the list of 6 strong acids.
6. HF is a weak acid. the formula starts with H but it's not water so it's an acid. Recognize it is weak becauce it is not on the list of 6 strong acids.
7. NH3 is a weak base. Amines are weak bases.
8. Ba(OH)2 is a strong base. It has a metal with an -OH group.
9. CH3CH2COOH is a weak base. It has a carboxylic acid functional group so it's an acid. Carboxylic acids are all weak (also, not on the list of 6 strong acids).
Q9.1.4
Write an equation representing the behavior of each substance in question 3. Pay attention to the type of arrow used in the equation.
1. HCl(aq) $\rightarrow$ H+(aq) + Cl(aq)
2. NaOH(aq) $\rightarrow$ Na+(aq) + OH(aq)
3. KOH(aq) $\rightarrow$ K+(aq) + OH(aq)
4. HNO2(aq) $\rightleftharpoons$ H+(aq) + NO2(aq)
5. HNO3(aq) $\rightarrow$ H+(aq) + NO3(aq)
6. HF(aq) $\rightleftharpoons$ H+(aq) + F(aq)
7. NH3(aq) + H2O(l)$\rightleftharpoons$ NH4+(aq) + OH(aq)
8. Ba(OH)2(aq) $\rightarrow$ Ba2+(aq) + 2OH(aq)
9. CH3CH2COOH(aq) $\rightleftharpoons$ H+(aq) + CH3CH2COO(aq)
Q9.1.5
NaOH is a strong base and completely dissociates (see reaction). Since it completely dissociates, the concentration of NaOH equals the concentration of OH. We need to calculate the concentration of NaOH.
NaOH(aq) $\rightarrow$ Na+(aq) + OH(aq)
$15.0\;g\;\text{NaOH}\left(\frac{1\;mol}{40.00\;g}\right)=0.375\;mol\;\text{NaOH}$
$M=\frac{mol\;solute}{L\;soln}=\frac{0.375\;mol}{0.500\;L}=0.750\;M\;NaOH$
[NaOH] = [OH] = 0.750 M
Use [OH] to find pOH.
$p\text{OH} = -log[\text{OH}^-]=-log[0.750]=0.125$
Now, convert from pOH to pH. Note that 14 is an exact number in this context so it does not affect significant figures.
$p\text{H}+p\text{OH}=14$
$p\text{H}=14-p\text{OH}$
$p\text{H}=14-0.125$
$p\text{H}=13.875$
Q9.1.6
HCl is a strong acid and completely dissociates (see reaction). Since it completely dissociates, the concentration of HCl equals the concentration of H+. We need to calculate the concentration of HCl.
HCl(aq) $\rightarrow$ H+(aq) + Cl(aq)
$22.6\;g\;\text{HCl}\left(\frac{1\;mol}{36.46\;g}\right)=0.603\;mol\;\text{HCl}$
$M=\frac{mol\;solute}{L\;soln}=\frac{0.603\;mol}{0.300\;L}=2.01\;M\;HCl$
[HCl] = [H+] = 2.01 M
$p\text{H} = -log[\text{H}^+]=-log[2.01]=-0.303$ (pH can be less than zero if it is a strong acid with a concentration greater than 1 M)
9.2: Buffers
Q9.2.1
A buffer is a weak acid and its conjugate base (or a weak base and its conjugate acid) that helps maintain the pH of a solution.
Q9.2.2
The purpose of a buffer is to resist change of pH in a solution.
Q9.2.3
1. HCl/Cl cannot because HCl is a strong acid.
2. HF/Fcan because HF is a weak acid and F is its conjugate base. It will be added to mixture as a salt (ionic compound) of F (i.e. NaF, KF, etc)
3. H2SO4/HSO4cannot because H2SO4 is a strong acid.
4. HSO4/SO42can because HSO4 is a weak acid and SO42 is its conjugate base. Both compounds will be added to the mixture as salts (i.e. NaHSO4 and Na2SO4).
5. H2O/NaOH cannot becase NaOH is a strong base and NaOH and H2O are not a conjugate acid-base pair.
6. HNO2/KNO2 can because HNO2 is a weak acid and KNO2 contains its conjugate base. Note that KNO2 is a strong electrolyte so it dissociates into K+ and NO2. HNO2 and NO2 form a conjugate acid-base pair.
7. HCl/NaOH cannot because HCl is a strong acid, NaOH is a strong base, and they do not form a conjugate acid-base pair.
Q9.2.4
Write the formula of the conjugate base needed to form a buffer with each of the following weak acids.
1. ClO3(or a salt such as NaClO3)
2. HPO42(or a salt such as Na2HPO4)
3. CH3COO(or a salt such as CH3COONa)
Q9.2.5
Write the formula of the conjugate acid needed to form a buffer with each of the following weak acids.
1. NH4+ (or a salt such as NH4Cl )
2. CH3NH3+ (or a salt such as CH3NH3Cl)
Q9.2.6
Buffer capacity is the amount of acid or base that can be added to a buffer solution before it can no longer resist significant changes in the pH of the solution. Adding small amounts of acid or base will change the pH of a buffer by a small amount and the buffer continues to be effective. If larger amounts of acid or base are added, the buffer capacity is reached and the solution can no longer resist changes in pH. | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/09%3A_Equilibrium_Applications/9.03%3A_Equilibrium_Applications_%28Exercises%29.txt |
• 10.1: Nuclear Radiation
Nuclear reactions are very different from chemical reactions. In chemical reactions, atoms become more stable by participating in a transfer of electrons or by sharing electrons with other atoms. In nuclear reactions, it is the nucleus of the atom that gains stability by undergoing a change of some kind. Some elements have no stable isotopes, which means that any atom of that element is radioactive. For some other elements, only certain isotopes are radioactive.
• 10.2: Fission and Fusion
Nuclear fission is a process in which a very heavy nucleus splits into smaller nuclei of intermediate mass. Because the smaller nuclei are more stable, the fission process releases tremendous amounts of energy. Nuclear fusion is a process in which light-mass nuclei combine to form a heavier and more stable nucleus. Fusion produces even more energy than fission. In the sun and other stars, four hydrogen nuclei combine at extremely high temperatures & pressures to produce a helium nucleus.
• 10.3: Half-Life
The rate of radioactive decay is often characterized by the half-life of a radioisotope. Half-life (t1/2) is the time required for one half of the nuclei in a sample of radioactive material to decay. After each half-life has passed, one half of the radioactive nuclei will have transformed into a new nuclide.
• 10.4: Physical and Chemical Changes
A physical change is a change to a sample of matter in which some properties of the material change, but the identity of the matter does not. In contrast, a chemical property describes the ability of a substance to undergo a specific chemical change.
• 10.5: Chemical Equations
Chemical reactions are occurring all around you. Plants use sunlight to drive their photosynthetic process and produce energy. Cars and other vehicles burn gasoline in order to power their engines. Batteries use electrochemical reactions to produce energy and power many everyday devices. Many chemical reactions are going on inside you as well, especially during the digestion of food.
• 10.6: Nuclear and Chemical Reactions (Exercises)
These are homework exercises to accompany Chapter 10 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Solutions are available below the questions.
10: Nuclear and Chemical Reactions
Learning Outcomes
• Define radioactivity.
• Describe a radioisotope.
• Explain how radioactivity involves a change in the nucleus of a radioisotope.
• Explain the characteristics of the forms of radiation.
• Describe and write equations for the primary types of radioactive decay.
Marie Curie (1867 - 1934) was a Polish scientist who pioneered research into nuclear radiation (Figure $1$). She was awarded the Nobel Prize in physics in 1903 along with her husband Pierre and Antoine Henri Becquerel for their work on radioactivity. She was awarded a second Nobel Prize in 1911, this time in chemistry, for her continued research on radioactive elements. In this lesson, you will learn about radioactivity, the reasons why certain elements and isotopes are radioactive, and the most common types of radioactive decay processes.
Radioactivity
Radioactivity was discovered quite by accident. In 1896, Henri Becquerel was studying the effect of certain uranium salts on photographic film plates. He believed that the salts had an effect on the film only when they had been exposed to sunlight. He accidentally found that uranium salts that had not been exposed to sunlight still had an effect on the photographic plates. The Curies, associates of Becquerel at the time, showed that the uranium was emitting a type of ray that interacted with the film. Marie Curie called this radioactivity. Radioactivity is the spontaneous breakdown of an atom's nucleus by the emission of particles and/or radiation. Radiation is the emission of energy through space in the form of particles and/or waves.
Nuclear reactions are very different from chemical reactions. In chemical reactions, atoms become more stable by participating in a transfer of electrons or by sharing electrons with other atoms. In nuclear reactions, it is the nucleus of the atom that gains stability by undergoing a change of some kind. Some elements have no stable isotopes, which means that any atom of that element is radioactive. For some other elements, only certain isotopes are radioactive. A radioisotope is an isotope of an element that is unstable and undergoes radioactive decay. The energies that are released in nuclear reactions are many orders of magnitude greater than the energies involved in chemical reactions. Unlike chemical reactions, nuclear reactions are not noticeably affected by changes in environmental conditions, such as temperature or pressure.
The discovery of radioactivity and its effects on the nuclei of elements disproved Dalton's assumption that atoms are indivisible. A nuclide is a term for an atom with a specific number of protons and neutrons in its nucleus. As we will see, when nuclides of one type emit radiation, they are changed into different nuclides. Radioactive decay is spontaneous and does not required an input of energy to occur. The stability of a particular nuclide depends on the composition of its nucleus, including the number of protons, the number of neutrons, and the proton-to-neutron ratio.
The Band of Stability
Carbon-12, with six protons and six neutrons, is a stable nucleus, meaning that it does not spontaneously emit radioactivity. Carbon-14, with six protons and eight neutrons, is unstable and naturally radioactive. Among atoms with lower atomic numbers, the ideal ratio of neutrons to protons is approximately 1:1. As the atomic number increases, the stable neutron-proton ratio gradually increases to about 1.5:1 for the heaviest known elements. For example, lead-206 is a stable nucleus that contains 124 neutrons and 82 protons, a ratio of 1.51 to 1.
This observation is shown in the figure below. The band of stability is the range of stable nuclei on a graph that plots the number of neutrons in a nuclide against the number of protons. Known stable nuclides are shown with individual blue dots, while the 1:1 and 1.5:1 ratios are shown with a solid red line and a green line, respectively.
It should be noted that just because a nucleus is "unstable" (able to undergo spontaneous radioactive decay) does not mean that it will rapidly decompose. For example, uranium-238 is unstable because it spontaneously decays over time, but if a sample of uranium-238 is allowed to sit for 1000 years, only $0.0000155\%$ of the sample will have decayed. However, other unstable nuclei, such as berkelium-243, will be almost completely gone (>$99.9999\%$ decayed) in less than a day.
Radioactive Decay
Unstable nuclei spontaneously emit radiation in the form of particles and energy. This generally changes the number of protons and/or neutrons in the nucleus, resulting in a more stable nuclide. A nuclear reaction is a reaction that affects the nucleus of an atom. One type of a nuclear reaction is radioactive decay, a reaction in which a nucleus spontaneously disintegrates into a slightly lighter nucleus, accompanied by the emission of particles, energy, or both. An example is shown below, in which the nucleus of a polonium atom radioactively decays into a lead nucleus.
$\ce{^{210}_{84}Po} \rightarrow \ce{^{206}_{82}Pb} + \ce{^4_2He}$
Note that in a balanced nuclear equation, the sum of the atomic numbers (subscripts) and the sum of the mass numbers (superscripts) must be equal on both sides of the equation. Recall the notation system for isotopes, which shows both the atomic number and mass number along with the chemical symbol.
Because the number of protons changes as a result of this nuclear reaction, the identity of the element changes. Transmutation is a change in the identity of a nucleus as a result of a change in the number of protons. There are several different types of naturally occurring radioactive decay, and we will examine each separately.
Alpha Decay
An alpha particle $\left( \alpha \right)$ is a helium nucleus with two protons and two neutrons. Alpha particles are emitted during some types of radioactive decay. The net charge of an alpha particle is $2+$, and its mass is approximately $4 \: \text{amu}$. The symbol for an alpha particle in a nuclear equation is usually $\ce{^4_2He}$, though sometimes $\alpha$ is used. Alpha decay typically occurs for very heavy nuclei in which the nuclei are unstable due to large numbers of nucleons. For nuclei that undergo alpha decay, their stability is increased by the subtraction of two protons and two neutrons. For example, uranium-238 decays into thorium-234 by the emission of an alpha particle (see figure below).
Beta Decay
Nuclei above the band of stability are unstable because their neutron to proton ratio is too high. To decrease that ratio, a neutron in the nucleus is capable of turning into a proton and an electron. The electron is immediately ejected at a high speed from the nucleus. A beta particle $\left( \beta \right)$ is a high-speed electron emitted from the nucleus of an atom during some kinds of radioactive decay (see figure below). The symbol for a beta particle in an equation is either $\beta$ or $\ce{^0_{-1}e}$. Carbon-14 undergoes beta decay, transmutating into a nitrogen-14 nucleus.
$\ce{^{14}_6C} \rightarrow \ce{^{14}_7N} + \ce{^0_{-1}e}$
Note that beta decay increases the atomic number by one, but the mass number remains the same.
Positron Emission
Nuclei below the band of stability are unstable because their neutron to proton ratio is too low. One way to increase that ratio is for a proton in the nucleus to turn into a neutron and another particle called a positron. A positron is a particle with the same mass as an electron, but with a positive charge. Like the beta particle, a positron is immediately ejected from the nucleus upon its formation. The symbol for a positron in an equation is $\ce{^0_{+1}e}$. For example, potassium-38 emits a positron, becoming argon-38.
$\ce{^{38}_{19}K} \rightarrow \ce{^{38}_{18}Ar} + \ce{^0_1e}$
Positron emission decreases the atomic number by one, but the mass number remains the same.
Electron Capture
An alternate way for a nuclide to increase its neutron to proton ratio is by a phenomenon called electron capture. In electron capture, an electron from an inner orbital is captured by the nucleus of the atom and combined with a proton to form a neutron. For example, silver-106 undergoes electron capture to become palladium-106.
$\ce{^{106}_{47}Ag} + \ce{^0_{-1}e} \rightarrow \ce{^{106}_{46}Pd}$
Note that the overall result of electron capture is identical to positron emission. The atomic number decreases by one while the mass number remains the same.
Gamma Ray Emission
Gamma rays $\left( \gamma \right)$ are very high energy electromagnetic waves emitted from a nucleus. Gamma rays are emitted by a nucleus when nuclear particles undergo transitions between nuclear energy levels. This is analogous to the electromagnetic radiation emitted when excited electrons drop from higher to lower energy levels; the only difference is that nuclear transitions release much more energetic radiation. Gamma ray emission often accompanies the decay of a nuclide by other means.
$\ce{^{230}_{90}Th} \rightarrow \ce{^{226}_{88}Ra} + \ce{^4_2He} + \gamma$
The emission of gamma radiation has no effect on the atomic number or mass number of the products, but it reduces their energy.
Summary of Nuclear Radiation
The table below summarizes the main types of nuclear radiation, including charge, mass, symbol, and penetrating power. Penetrating power refers to the relative ability of the radiation to pass through common materials. Radiation with high penetrating power is potentially more dangerous because it can pass through skin and do cellular damage.
Table $1$ Summary of types of nuclear radiation.
Type Symbol Mass number Charge Penetration Power Shielding
Alpha particle $\ce{^4_2He}$ or $\alpha$ 4 $2+$ Low Paper, skin
Beta particle $\ce{^0_{-1}e}$ or $\beta$ 0 $1-$ Moderate Heavy cloth, plastic
Positron $\ce{^0_1e}$ or $\beta^+$ 0 $1+$ Moderate Heavy cloth, plastic
Gamma ray $\gamma$ or $^0_0\gamma$ 0 0 High Lead, concrete
Neutron $\ce{^1_0n}$ 1 0 High Water, lead
Supplemental Resources
• Atomic structure review: www.sciencegeek.net/Chemistry...t1Numbers2.htm
• Balancing Nuclear equations: www.sciencegeek.net/Chemistry...rEquations.htm
• Nuclear decay: www.sciencegeek.net/Chemistry...cleardecay.htm
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/10%3A_Nuclear_and_Chemical_Reactions/10.01%3A_Nuclear_Radiation.txt |
Learning Outcomes
• Define fission.
• Describe a nuclear chain reaction and how it is applied in both a fission bomb and in a nuclear power plant.
• Define fusion.
Nuclear Fission
The most stable nuclei are of intermediate mass. To become more stable, the heaviest nuclei are capable of splitting into smaller fragments. Nuclear fission is a process in which a very heavy nucleus (mass > 200) splits into smaller nuclei of intermediate mass. Because the smaller nuclei are more stable, the fission process releases tremendous amounts of energy. Nuclear fission may occur spontaneously or may occur as a result of bombardment. When uranium-235 is hit with a slow-moving neutron, it absorbs it and temporarily becomes the very unstable uranium-236. This nucleus splits into two medium-mass nuclei while also emitting more neutrons. The mass of the products is less than the mass of the reactants, with the lost mass being converted to energy.
Nuclear Chain Reactions
Because the fission process produces more neutrons, a chain reaction can result. A chain reaction is a reaction in which the material that starts the reaction is also one of the products and can start another reaction. Illustrated below is a nuclear chain reaction for the fission of uranium-235.
The original uranium-235 nucleus absorbs a neutron, splits into a krypton-92 nucleus and a barium-141 nucleus, and releases three more neutrons upon splitting.
$\ce{^{235}_{92}U} + \ce{^1_0n} \rightarrow \ce{^{92}_{36}Kr} + \ce{^{141}_{56}Ba} + 3 \ce{^1_0n}$
Those three neutrons are then able to cause the fission of three more uranium-235 nuclei, each of which release more neutrons, and so on. The chain reaction continues until all of the uranium-235 nuclei have been split, or until the released neutrons escape the sample without striking any more nuclei. If the size of the original sample of uranium-235 is sufficiently small, too many neutrons escape without striking other nuclei, and the chain reaction quickly ceases. The critical mass is the minimum amount of fissionable material needed to sustain a chain reaction. Atomic bombs and nuclear reactors are two ways to harness the large energy released during nuclear fission.
Atomic Bombs - Uncontrolled Nuclear Reactions
In an atomic bomb, or fission bomb, the nuclear chain reaction is designed to be uncontrolled, releasing huge amounts of energy in a short amount of time. A critical mass of fissionable plutonium is contained within the bomb, but not at a sufficient density. Conventional explosives are used to compress the plutonium, causing it to go critical and trigger a nuclear explosion.
Nuclear Power Plants- Controlled Nuclear Reactions
A nuclear power plant (see figure below) uses a controlled fission reaction to produce large amounts of heat. The heat is then used to generate electrical energy.
Uranium-235, the usual fissionable material in a nuclear reactor, is first packaged into fuel rods. In order to keep the chain reaction from processing unchecked, moveable control rods are placed in between the fuel rods. Control rods limit the amount of available neutrons by absorbing some of them and preventing the reaction from proceeding too rapidly. Common control rod materials include alloys with various amounts of silver, indium, cadmium, or boron. A moderator is a material that slows down high-speed neutrons. This is beneficial because slow-moving neutrons are more efficient at splitting nuclei. Water is often used as a moderator. The heat released by the fission reaction is absorbed by constantly circulating coolant water. The coolant water releases its heat to a steam generator, which turns a turbine and generates electricity. The core of the reactor is surrounded by a containment structure that absorbs radiation.
Nuclear Fusion
The lightest nuclei are also not as stable as nuclei of intermediate mass. Nuclear fusion is a process in which light-mass nuclei combine to form a heavier and more stable nucleus. Fusion produces even more energy than fission. In the sun and other stars, four hydrogen nuclei combine at extremely high temperatures and pressures to produce a helium nucleus. The concurrent loss of mass is converted into extraordinary amounts of energy (see figure below).
Fusion is even more appealing than fission as an energy source because no radioactive waste is produced and the only reactant needed is hydrogen. However, fusion reactions only occur at very high temperatures - in excess of $40,000,000^\text{o} \text{C}$. No known materials can withstand such temperatures, so there is currently no feasible way to harness nuclear fusion for energy production, although research is ongoing.
Uses of Radiation
As we saw earlier, different types of radiation vary in their abilities to penetrate through matter. Alpha particles have very low penetrating ability and are stopped by skin and clothing. Beta particles have a penetrating ability that is about 100 times that of alpha particles. Gamma rays have very high penetrating ability, and great care must be taken to avoid overexposure to gamma rays.
Exposure and Detection
Radiation emitted by radioisotopes is called ionizing radiation. Ionizing radiation is radiation that has enough energy to knock electrons off the atoms of a bombarded substance and produce ions. The roentgen is a unit that measures nuclear radiation and is equal to the amount of radiation that produces $2 \times 10^9$ ion pairs when it passes through $1 \: \text{cm}^3$ of air. The primary concern is that ionizing radiation can do damage to living tissues. Radiation damage is measured in rems, which stands for roentgen equivalent man . A rem is the amount of ionizing radiation that does as much damage to human tissue as is done by 1 roentgen of high-voltage x-rays. Tissue damage from ionizing radiation can cause genetic mutations due to interactions between the radiation and DNA, which can lead to cancer.
You are constantly being bombarded with background radiation from space and from geologic sources that vary depending on where you live. Average exposure is estimated to be about 0.1 rem per year. The maximum permissible does of radiation exposure for people in the general population is 0.5 rem per year. Some people are naturally at higher risk because of their occupations, so reliable instruments to detect radiation exposure have been developed. A Geiger counter is a device that uses a gas-filled metal tube to detect radiation (see figure below). When the gas is exposed to ionizing radiation, it conducts a current, and the Geiger counter registers this as audible clicks. The frequency of the clicks corresponds to the intensity of the radiation.
A scintillation counter is a device that uses a phosphor-coated surface to detect radiation by the emission of bright bursts of light. Workers who are at risk of exposure to radiation wear small portable film badges. A film badge consists of several layers of photographic film that can measure the amount of radiation to which the wearer has been exposed. Film badges are removed and analyzed at periodic intervals to ensure that the person does not become overexposed to radiation on a cumulative bases.
Medicine and Agriculture
Radioactive nuclides, such as cobalt-60, are frequently used in medicine to treat certain types of cancers. The faster growing cancer cells are exposed to the radiation and are more susceptible to damage than healthy cells. Thus, the cells in the cancerous area are killed by the exposure to high-energy radiation. Radiation treatment is risky because some healthy cells are also killed, and cells at the center of a cancerous tumor can become resistant to the radiation.
Radioactive tracers are radioactive atoms that are incorporated into substances so that the movement of these substances can be tracked by a radiation detector. Tracers are used in the diagnosis of cancer and other diseases. For example, iodine-131 is used to detect problems with a person's thyroid. A patient first ingests a small amount of iodine-131. About two hours later, the iodine uptake by the thyroid is determined by a radiation scan of the patient's throat. In a similar way, technetium-99 is used to detect brain tumors and liver disorders, and phosphorus-32 is used to detect skin cancer.
Radioactive tracers can be used in agriculture to test the effectiveness of various fertilizers. The fertilizer is enriched with a radioisotope, and the uptake of the fertilizer by the plant can be monitored by measuring the emitted radiation levels. Nuclear radiation is also used to prolong the shelf life of produce by killing bacteria and insects that would otherwise cause the food to spoil faster.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/10%3A_Nuclear_and_Chemical_Reactions/10.02%3A_Fission_and_Fusion.txt |
Learning Outcomes
• Define half-life as it relates to radioactive nuclides and solve half-life problems.
• Describe the general process by which radioactive dating is used to determine the age of various objects.
• Calculate the time for a sample to decay.
• Complete dosage calculations based on nuclide activity.
The rate of radioactive decay is often characterized by the half-life of a radioisotope. Half-life $\left( t_{1/2} \right)$ is the time required for one half of the nuclei in a sample of radioactive material to decay. After each half-life has passed, one half of the radioactive nuclei will have transformed into a new nuclide (see table below). The rate of decay and the half-life do not depend on the original size of the sample. They also do not depend upon environmental factors such as temperature and pressure.
Table $1$
Number of Half-Lives Passed Fraction Remaining Percentage Remaining Mass remaining starting with $80 \: \text{g}$
1 1/2 50 $40 \: \text{g}$
2 1/4 25 $20 \: \text{g}$
3 1/8 12.5 $10 \: \text{g}$
4 1/16 6.25 $5.0 \: \text{g}$
5 1/32 3.125 $2.5 \: \text{g}$
As an example, iodine-131 is a radioisotope with a half-life of 8 days. It decays by beta particle emission into xenon-131.
$\ce{^{131}_{53}I} \rightarrow \ce{^{131}_{54}Xe} + \ce{^0_{-1}e}$
After eight days have passed, half of the atoms of any sample of iodine-131 will have decayed, and the sample will now be $50\%$ iodine-131 and $50\%$ xenon-131. After another eight days pass (a total of 16 days or 2 half-lives), the sample will be $25\%$ iodine-131 and $75\%$ xenon-131. This continues until the entire sample of iodine-131. has completely decayed (see figure below).
Half-lives have a very wide range, from billions of years to fractions of a second. Listed below (see table below) are the half-lives of some common and important radioisotopes. Those with half-lives on the scale of hours or days are the ones most suitable for use in medical treatment.
Table $2$
Nuclide Half-Life $\left( t_{1/2} \right)$ Decay Mode
Carbon-14 5730 years $\beta^-$
Cobalt-60 5.27 years $\beta^-$
Francium-220 27.5 seconds $\alpha$
Hydrogen-3 12.26 years $\beta^-$
Iodine-131 8.07 days $\beta^-$
Nitrogen-16 7.2 seconds $\beta^-$
Phosphorus-32 14.3 days $\beta^-$
Plutonium-239 24,100 years $\alpha$
Potassium-40 $1.28 \times 10^9$ years $\beta^-$ and $\ce{e^-}$ capture
Radium-226 1600 years $\alpha$
Radon-222 3.82 days $\alpha$
Strontium-90 28.1 days $\beta^-$
Technetium-99 $2.13 \times 10^5$ years $\beta^-$
Thorium-234 24.1 days $\beta^-$
Uranium-235 $7.04 \times 10^8$ years $\alpha$
Uranium-238 $4.47 \times 10^9$ years $\alpha$
The following example illustrates how to use the half-life of a sample to determine the amount of radioisotope that remains after a certain period of time has passed.
Example $1$: Strontium-90
Strontium-90 has a half-life of 28.1 days. If you start with a $5.00 \: \text{mg}$ sample of the isotope, how much remains after 140.5 days have passed?
Solution
Step 1: List the known values and plan the problem.
Known
• Original mass $= 5.00 \: \text{mg}$
• $t_{1/2} =$ 28.1 days
• Time elapsed $=$ 140.5 days
Unknown
• Final mass of $\ce{Sr}$-90 $= ? \: \text{mg}$
First, find the number of half-lives that have passed by dividing the time elapsed by the half-life. Then, reduce the amount of $\ce{Sr}$-90 by half, once for each half-life.
Step 2: Solve.
Number of half-lives, days, mass
0 half-lives, 0 days, $5.00 \: \text{mg}$
1 half-life, 28.1 days, $2.50 \: \text{mg}$
2 half-lives, 56.2 days, $1.25 \: \text{mg}$
3 half-lives, 84.3 days, $0.613 \: \text{mg}$
4 half-lives, 112.4 days, $0.313 \: \text{mg}$
5 half-lives, 140.5 days, $0.156 \: \text{mg}$
Step 3: Think about your result.
According to the data above, the passage of 5 half-lives means $0.156 \: \text{mg}$ of the original $\ce{Sr}$-90 remains. The remaining $4.844 \: \text{mg}$ has decayed by beta particle emission to yttrium-90.
Radioactive Dating
Radioactive dating is a process by which the approximate age of an object is determined through the use of certain radioactive nuclides. For example, carbon-14 has a half-life of 5,730 years and is used to measure the age of organic material. The ratio of carbon-14 to carbon-12 in living things remains constant while the organism is alive because fresh carbon-14 is entering the organism whenever it consumes nutrients. When the organism dies, this consumption stops, and no new carbon-14 is added to the organism. As time goes by, the ratio of carbon-14 to carbon-12 in the organism gradually declines, because carbon-14 radioactively decays while carbon-12 is stable. Analysis of this ratio allows archaeologists to estimate the age of organisms that were alive many thousands of years ago. Carbon dating is effective until about 50,000 years. The ages of many rocks and minerals are far greater than the ages of fossils. Uranium-containing minerals that have been analyzed in a similar way have allowed scientists to determine that the Earth is over 4 billion years old.
Decay Series
In many instances, the decay of an unstable radioactive nuclide simply produces another radioactive nuclide. It may take several successive steps to reach a nuclide that is stable. A decay series is a sequence of successive radioactive decays that proceeds until a stable nuclide is reached. The terms reactant and product are generally not used for nuclear reactions. Instead, the terms parent and daugher nuclide are used to to refer to the starting and ending isotopes in a decay process. The figure below shows the decay series for uranium-238.
In the first step, uranium-238 decays by alpha emission to thorium-234 with a half-life of $4.5 \times 10^9$ years. This decreases its atomic number by two. The thorium-234 rapidly decays by beta emission to protactinium-234 ($t_{1/2} =$ 24.1 days). The atomic number increases by one. This continues for many more steps until eventually the series ends with the formation of the stable isotope lead-206.
Artificial Transmutation
As we have seen, transmutation occurs when atoms of one element spontaneously decay and are converted to atoms of another element. Artificial transmutation is the bombardment of stable nuclei with charged or uncharged particles in order to cause a nuclear reaction. The bombarding particles can be protons, neutrons, alpha particles, or larger atoms. Ernest Rutherford performed some of the earliest bombardments, including the bombardment of nitrogen gas with alpha particles to produce the unstable fluorine-18 isotope.
$\ce{^{14}_7N} + \ce{^4_2He} \rightarrow \ce{^{18}_9F}$
Fluorine-18 quickly decays to the stable nuclide oxygen-17 by releasing a proton.
$\ce{^{18}_9F} \rightarrow \ce{^{17}_8O} + \ce{^1_1H}$
When beryllium-9 is bombarded with alpha particles, carbon-12 is produced with the release of a neutron.
$\ce{^9_4Be} + \ce{^4_2He} \rightarrow \ce{^{12}_6C} + \ce{^1_0n}$
Transuranium Elements
Many, many radioisotopes that do not occur naturally have been generated by artificial transmutation. The elements technetium and promethium have been produced, since these elements no longer occur in nature. All of their isotopes are radioactive and have half-lives short enough that any amount of the elements that once existed have long since disappeared through natural decay. The transuranium elements are elements with atomic numbers greater than 92. All isotopes of these elements are radioactive and none occur naturally.
Dosing
Half-life calculations can be based on mass, percent remaining, or dose. Regardless of which one, the concept is still the same. Understanding the radioactivity and half-life of a sample is important for calculating the correct dose for a patient and determining the levels and duration of radioactive emission from a patient after treatment is received.
Frequently, dosages for radioactive isotopes are given the activity in volume. For example, the concentration of $\ce{I}$-137 is given as $50 \: \mu \text{Ci/mL}$ (microCurie per milliliter). This relationship can be used to calculate the volume needed for a particular dose. For example, a patient needs $125 \: \mu \text{Ci}$ of $\ce{I}$1-51. What volume of a $50 \mu \text{Ci}$ per $10 \: \text{mL}$ solution should be given?
$125 \: \mu \text{Ci} \left( \frac{10 \: \text{mL}}{50 \: \mu \text{Ci}} \right) = 25 \: \text{mL}$
Example $2$
A patient is given $\ce{I}$-131 to treat thyroid cancer. The patient receives $5.50 \: \text{mL}$ of a solution containing $50 \: \text{mCi}$ (milliCurie) in $2 \: \text{mL}$ (assume concentration is an exact number). What does (in $\text{mCi}$) is given to the patient? What will be the activity in $\text{mCi}$ after 24.21 days given that the half-life of $\ce{I}$-131 is 8.07 days?
Solution
The first part of the problem is to find the dose given to the patient. We are given the volume and the concentration (in units of radioactivity over volume).
$5.50 \: \text{mL solution} \left( \frac{50 \: \text{mCi}}{2 \: \text{mL}} \right) = 138 \: \text{mCi}$
The patient is given a does of $138 \: \text{mCi}$.
Now, we need to find the activity after 24.21 days have passed. After one half-life (8.07 days), the sample will have half as much activity $\left( 138/2 = 69.0 \: \text{mCi} \right)$. After two half-lives (total of 16.14 days), the sample will have half as much activity as after the first half-life $\left( 69.0/2 = 34.5 \: \text{mCi} \right)$. After three half-lives (total of 24.21 days), the sample will have half as much activity as after the second half-life $\left( 34.5/2 = 17.3 \: \text{mCi} \right)$. Therefore, after 24.21 days (3 half-lives), the radioactivity will be $17.3 \: \text{mCi}$.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/10%3A_Nuclear_and_Chemical_Reactions/10.03%3A_Half-Life.txt |
Learning Outcomes
• Distinguish between physical and chemical changes.
• Give examples of physical and chemical changes.
Physical Changes
As an ice cube melts, its shape changes as it acquires the ability to flow. However, its composition does not change. Melting is an example of a physical change. A physical change is a change to a sample of matter in which some properties of the material change, but the identity of the matter does not. Physical changes can further be classified as reversible or irreversible. The melted ice cube may be refrozen, so melting is a reversible physical change. Physical changes that involve a change of state are all reversible. Other changes of state include vaporization (liquid to gas), freezing (liquid to solid), and condensation (gas to liquid). Dissolving is also a reversible physical change. When salt is dissolved into water, the salt is said to have entered the aqueous state. The salt may be regained by boiling off the water, leaving the salt behind.
When a piece of wood is ground into sawdust, that change is irreversible since the sawdust could not be reconstituted into the same piece of wood that it was before. Cutting the grass or pulverizing a rock would be other irreversible physical changes. Firewood also represents an irreversible physical change since the pieces cannot be put back together to form the tree.
Chemical Changes
When exposed to air, an object made of iron will eventually begin to rust (see figure below).
As the rust forms on the surface of the iron, it flakes off to expose more iron, which will continue to rust. Rust is clearly a substance that is different from iron. Rusting is an example of a chemical change.
A chemical property describes the ability of a substance to undergo a specific chemical change. A chemical property of iron is that it is capable of combining with oxygen to form iron oxide, the chemical name of rust. A more general term for rusting and other similar processes is corrosion. Other terms that are commonly used in descriptions of chemical changes are burn, rot, explode, and ferment. Chemical properties are very useful as a way of identifying substances. However, unlike physical properties, chemical properties can only be observed as the substance is in the process of being changed into a different substance.
A chemical change is also called a chemical reaction. A chemical reaction is a process that occurs when one or more substances are changed into one or more new substances. Zinc $\left( \ce{Zn} \right)$ is a silver-gray element that can be ground into a powder. If zinc is mixed at room temperature with powdered sulfur $\left( \ce{S} \right)$, a bright yellow element, the result will simply be a mixture of zinc and sulfur. No chemical reaction occurs. However, if energy is provided to the mixture in the form of heat, the zinc will chemically react with the sulfur to form the compound zinc sulfide $\left( \ce{ZnS} \right)$. Pictured below are the substances involved in this reaction.
The reaction between zinc and sulfur can be depicted in something called a chemical equation. In words, we could write the reaction as:
$\text{zinc} + \text{sulfur} \rightarrow \text{zinc sulfide}$
A more convenient way to express a chemical reaction is to use the symbols and formulas of the substances involved:
$\ce{Zn} + \ce{S} \rightarrow \ce{ZnS}$
The substance(s) to the left of the arrow in a chemical equation are called reactants. A reactant is a substance that is present at the start of a chemical reaction. The substance(s) to the right of the arrow are called products. A product is a substance that is present at the end of a chemical reaction. In the equation above, zinc and sulfur are the reactants that chemically combine to form zinc sulfide as a product.
Recognizing Chemical Reactions
How can you tell if a chemical reaction is taking place? Certain visual clues indicate that a chemical reaction is likely (but not necessarily) occurring, including the following examples:
1. A change of color occurs during the reaction.
2. A gas is produced during the reaction.
3. A solid product, called a precipitate, is produced in the reaction.
4. A visible transfer of energy occurs in the form of light as a result of the reaction.
When zinc reacts with hydrochloric acid, the reaction bubbles vigorously as hydrogen gas is produced (see figure below). The production of a gas is also an indication that a chemical reaction may be occurring.
When a colorless solution of lead (II) nitrate is added to a colorless solution of potassium iodide, a yellow solid called a precipitate is instantly produced (see figure below). A precipitate is a solid product that forms from a reaction and settles out of a liquid mixture. The formation of a precipitate may also indicate the occurrence of a chemical reaction.
$\ce{Pb(NO_3)_2} \left( aq \right) + 2 \ce{KI} \left( aq \right) \rightarrow \ce{PbI_2} \left( s \right) + 2 \ce{KNO_3} \left( aq \right)$
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/10%3A_Nuclear_and_Chemical_Reactions/10.04%3A_Physical_and_Chemical_Changes.txt |
Learning Outcomes
• Describe chemical reactions using word equations.
• Write equations for chemical reactions.
• Use coefficients to balance chemical equations.
Chemical reactions are occurring all around you. Plants use sunlight to drive their photosynthetic process and produce energy. Cars and other vehicles burn gasoline in order to power their engines. Batteries use electrochemical reactions to produce energy and power many everyday devices. Many chemical reactions are going on inside you as well, especially during the digestion of food.
In math class, you have written and solved many mathematical equations. Chemists keep track of chemical reactions by writing equations as well. In any chemical reaction one or more substances, called reactants, are converted into one or more new substance, called products. The general form of the equation for such a process looks like this.
$\text{Reactants} \rightarrow \text{Products}$
Unlike in a math equation, a chemical equation does not use an equal sign. Instead the arrow is called a yield sign and so the equation is described as "reactants yield products".
Word Equations
You can describe a chemical reaction by writing a word equation. When silver metal is exposed to sulfur it reacts to form silver sulfide. Silver sulfide is commonly known as tarnish and turns the surface of silver objects dark and streaky black (see figure below). The sulfur that contributes to tarnish can come from traces of sulfur in the air or from food such as eggs. The word equation for the process is:
$\text{Silver} + \text{sulfur} \rightarrow \text{Silver sulfide}$
The silver and the sulfur are the reactants in the equation, while the silver sulfide is the product.
Another common chemical reaction is the burning of methane gas. Methane is the major component of natural gas and is commonly burned on a gas stove or in a Bunsen burner (see figure below). Burning is a chemical reaction in which some type of fuel is reacted with oxygen gas. The products of the reaction in the burning of methane as well as other fuels are carbon dioxide and water. The word equation for this reaction is:
$\text{Methane} + \text{oxygen} \rightarrow \text{carbon dioxide} + \text{water}$
Chemical Equations
Word equations are time-consuming to write and will not prove to be convenient for many of the things that chemists need to do with equations. A chemical equation is a representation of a chemical reaction that displays the reactants and products with chemical formulas. The chemical equation for the reaction of methane with oxygen is shown:
$\ce{CH_4} + \ce{O_2} \rightarrow \ce{CO_2} + \ce{H_2O}$
The equation above, called a skeleton equation, is an equation that shows only the formulas of the reactants and products with nothing to indicate the relative amounts. The first step in writing an accurate chemical equation is to write the skeleton equation, making sure that the formulas of all substances involved are written correctly. All reactants are written to the left of the yield arrow, separated from one another by a plus sign. Likewise, products are written to the right of the yield arrow, also separated with a plus sign.
It is often important to know the physical states of the reactants and products taking part in a reaction. To do this, put the appropriate symbol in parentheses after each formula: $\left( s \right)$ for solid, $\left( l \right)$, for liquid, $\left( g \right)$ for gas, and $\left( aq \right)$ for an aqueous (water-based) solution. At room temperature, the components of the previous reaction are in the following states:
$\ce{CH_4} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + \ce{H_2O} \left( l \right)$
The table below shows a listing of symbols used in chemical equations.
Table $1$: Symbols Used in Chemical Equations
Symbol Description
$+$ used to separate multiple reactants or products
$\rightarrow$ yield sign; separates reactants from products
$\rightleftharpoons$ replaces the yield sign for reversible reactions that reach equilibrium
$\left( s \right)$ reactant or product in the solid state
$\left( l \right)$ reactant or product in the liquid state
$\left( g \right)$ reactant or product in the gas state
$\left( aq \right)$ reactant or product in an aqueous solution (dissolved in water)
$\overset{\ce{Pt}}{\rightarrow}$ formula written above the arrow is used as a catalyst in the reaction
$\overset{\Delta}{\rightarrow}$ triangle indicates that the reaction is being heated
Balancing Chemical Equations
Suppose you were to write a word equation for building the ideal ham sandwich (see figure below). Perhaps you might come up with this:
$\text{Ham} + \text{cheese} + \text{tomato} + \text{pickles} + \text{bread} \rightarrow \text{ham sandwich}$
The reactants are the "parts" or ingredients of the ham sandwich while the sandwich itself is the product. There is something missing from your equation, however. There is no indication how many of each "reactant" is required to make the "product". For one thing, you would certainly need two slices of bread to make a conventional sandwich.
Let's say that the perfect ham sandwich $\left( \ce{HS} \right)$ is composed of 2 slices of ham $\left( \ce{H} \right)$, a slice of cheese $\left( \ce{C} \right)$, 1 slice of tomato $\left( \ce{T} \right)$, 5 pickles $\left( \text{P} \right)$, and 2 slices of bread $\left( \ce{B} \right)$. Accounting for the numbers of each reactant, as well as substituting symbols for words, your equation would become:
$2 \ce{H} + \ce{C} + \ce{T} + 5 \ce{P} + 2 \ce{B} \rightarrow \ce{HS}$
This now shows the correct quantities of the reactants. As one final improvement, we will change the "formula" of the product. Since the final sandwich contains all the reactants that went into it, its formula should reflect that.
$2 \ce{H} + \ce{C} + \ce{T} + 5 \ce{P} + 2 \ce{B} \rightarrow \ce{H_2CTP_5B_2}$
The subscript after each symbol in the product stands for the number of that particular reactant found on the reactant side of the equation: 2 for $\ce{H}$, 1 for $\ce{C}$, etc.
Since the equation now shows equal numbers of each sandwich part on both sides of the equation, we say that the equation is balanced. Chemical equations must also be balanced in a similar way. A balanced equation is a chemical equation in which mass is conserved and there are equal numbers of atoms of each element on both sides of the equation.
We can write a chemical equation for the reaction of carbon with hydrogen gas to form methane $\left( \ce{CH_4} \right)$.
$\begin{array}{ccccc} \ce{C} \left( s \right) & + & \ce{H_2} \left( g \right) & \rightarrow & \ce{CH_4} \left( g \right) \ 1 \: \ce{C} \: \text{atom} & & 2 \: \ce{H} \: \text{atoms} & & 1 \: \ce{C} \: \text{atom}, \: 4 \: \ce{H} \: \text{atoms} \end{array}$
In order to write a correct equation, you must first write the correct skeleton equation with the correct chemical formulas. Recall that hydrogen is a diatomic molecule and so is written as $\ce{H_2}$. When we count the number of atoms of both elements, shown under the equation, we see that the equation is not balanced. There are only 2 atoms of hydrogen on the reactant side of the equation, while there are 4 atoms of hydrogen on the product side. This violates the law of conservation of mass, which states that mass must be conserved in any chemical reaction or physical process. Another common way to express the law of conservation of mass is that matter cannot be created or destroyed.
John Dalton's atomic theory stated that chemical reactions are separations, combinations, or rearrangements of atoms. Atoms themselves cannot be created or destroyed. Dalton's theory explains the law of conservation of mass and the process of balancing an equation ensures that the law is followed. We can balance the above equation by adding a coefficient of 2 in front of the formula for hydrogen.
$\ce{C} \left( s \right) + 2 \ce{H_2} \left( g \right) \rightarrow \ce{CH_4} \left( g \right)$
A coefficient is a small whole number placed in front of a formula in an equation in order to balance it. The 2 in front of the $\ce{H_2}$ means that there are a total of $2 \times 2 = 4$ atoms of hydrogen as reactants. Visually, the reaction looks like:
In the balanced equation, there is one atom of carbon and four atoms of hydrogen on both sides of the arrow. Below are guidelines for writing and balancing chemical equations.
1. Determine the correct chemical formulas for each reactant and product.
2. Write the skeleton equation by placing the reactant(s) on the left side of the yield sign $\left( \rightarrow \right)$ and the product(s) on the right side. If there is more than one reactant or product, separate with plus signs.
3. Count the number of atoms of each element that appears as a reactant and as a product. If a polyatomic ion is unchanged on both sides of the equation, count it as a unit.
4. Balance each element one at a time by placing coefficients in front of the formula. No coefficient is written for a 1. It is best to begin by balancing elements that only appear in one formula on each side of the equation. You can only balance equations by using coefficients, NEVER change the subscripts in a chemical formula.
5. Check each atom or polyatomic ion to be sure that they are equal on both sides of the equation.
6. Make sure that all coefficients are in the lowest possible ratio. If necessary, reduce to the lowest ratio.
Example $1$
Aqueous solutions of lead (II) nitrate and sodium chloride are mixed. The products of the reaction are an aqueous solution of sodium nitrate and a solid precipitate of lead (II) chloride. Write the chemical equation for this reaction.
Solution
Step 1: Plan the problem.
Follow the steps for writing and balancing a chemical equation.
Step 2: Solve.
Write the skeleton equation with the correct formulas.
$\ce{Pb(NO_3)_2} \left( aq \right) + \ce{NaCl} \left( aq \right) \rightarrow \ce{NaNO_3} \left( aq \right) + \ce{PbCl_2} \left( s \right)$
Count the number of each atom or polyatomic ion on both sides of the equation (see table below).
The nitrate ions and the chlorine atoms are unbalanced. Start by placing a 2 in front of the $\ce{NaCl}$. This increases the reactant counts to 2 $\ce{Na}$ atoms and 2 $\ce{Cl}$ atoms. Then place a 2 in front of the $\ce{NaNO_3}$. The result is:
$\ce{Pb(NO_3)_2} \left( aq \right) + 2 \ce{NaCl} \left( aq \right) \rightarrow 2 \ce{NaNO_3} \left( aq \right) + \ce{PbCl_2} \left( s \right)$
The new count for each atom and polyatomic ion becomes (see table below).
Step 3: Think about your result.
The equation is now balanced since there are equal numbers of atoms of each element on both sides of the equation.
Count the number of each atom or polyatomic ion on both sides of the equation.
Table $2$: Unbalanced Reactants/Products
Reactants Products
1 $\ce{Pb}$ atom 1 $\ce{Pb}$ atom
2 $\ce{NO_3^-}$ 1 $\ce{NO_3^-}$
2 $\ce{Na}$ atoms 2 $\ce{Na}$ atoms
2 $\ce{Cl}$ atoms 1 $\ce{Cl}$ atom
The new count for each atom and polyatomic ion becomes:
Table $3$: Balanced Reactants/Products
Reactants Products
1 $\ce{Pb}$ atom 1 $\ce{Pb}$ atom
2 $\ce{NO_3^-}$ 2 $\ce{NO_3^-}$
2 $\ce{Na}$ atoms 2 $\ce{Na}$ atoms
2 $\ce{Cl}$ atoms 2 $\ce{Cl}$ atoms
Some equations provide a challenge to balancing when one or more of the elements can't be balanced simply by using one coefficient. Aluminum reacts with oxygen gas to form aluminum oxide according to the equation:
$\ce{Al} \left( s \right) + \ce{O_2} \left( g \right) \rightarrow \ce{Al_2O_3} \left( s \right)$
Since there are two oxygen atoms on the reactant side and 3 oxygen atoms on the product side, no single whole-number coefficient will balance the oxygen atoms. Find the lowest common multiple of 2 and 3, which is 6. Placing a 3 in front of the $\ce{O_2}$ and a 2 in front of the $\ce{Al_2O_3}$ will result in 6 oxygen atoms on both sides. Finish by balancing the aluminum with a 4.
$4 \ce{Al} \left( s \right) + 3 \ce{O_2} \left( g \right) \rightarrow 2 \ce{Al_2O_3} \left( s \right)$
The equation is balanced with 4 $\ce{Al}$ atoms and 6 $\ce{O}$ atoms on each side.
Finally, we will return to the equation from earlier where methane was reacted with oxygen to form carbon dioxide and water.
$\ce{CH_4} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + \ce{H_2O} \left( l \right)$
The element oxygen appears in two different places on the product side of the equation, so you should not start by trying to balance the oxygen. Instead, balance the carbon and the hydrogen first. The carbon is already balanced, but the hydrogen is balanced by placing a 2 in front of the water.
$\ce{CH_4} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + 2 \ce{H_2O} \left( l \right)$
Now count the total number of oxygen atoms on the product side: two from the $\ce{CO_2}$ and two from the 2 $\ce{H_2O}$ to give a total of four. Place a 2 in front of the $\ce{O_2}$.
$\ce{CH_4} \left( g \right) + 2 \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + 2 \ce{H_2O} \left( l \right)$
Balancing difficult equations can be a trial-and-error process and is a skill that requires practice. If you find that one particular strategy with a tough equation isn't working, start over and balance a different element first. Persistence will lead you to the correct balanced equation.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/10%3A_Nuclear_and_Chemical_Reactions/10.05%3A_Chemical_Equations.txt |
These are homework exercises to accompany Chapter 10 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Answers are below the questions.
Questions
(click here for solutions)
Q10.1.1
Write the symbol for the isotope described.
1. 12 protons, 12 electrons, 13 neutrons
2. 17 protons, 17 electrons, 20 neutrons
3. 53 protons, 53 electrons, 78 neutrons
4. 92 protons, 92 electrons, 146 neutrons
Q10.1.2
Determine the number of protons, neutrons, and electrons in each isotope.
1. $\ce{^{195}_{77}Ir}$
2. $\ce{^{209}_{82}Pb}$
3. $\ce{^{211}_{84}Po}$
4. $\ce{^{237}_{93}Np}$
Q10.1.3
Fill in the missing numbers in each equation.
1. $\ce{^{196}_{82}Pb}$ + $\ce{^0_{-1}e}$ → $_{\text{__}}^{\text{__}}\text{Tl}$
2. $\ce{^{28}_{15}P}$ → $_{\text{__}}^{\text{__}}\text{Si}$ + $\ce{^0_1e}$
3. $\ce{^{226}_{88}Ra}$ → $_{\text{__}}^{\text{__}}\text{Rn}$ + $\ce{^4_2 \alpha}$
4. $\ce{^{73}_{30}Zn}$ → $_{\text{__}}^{\text{__}}\text{Ga}$ + $\ce{^0_{-1}e}$
Q10.1.4
Fill in the blanks for each of the nuclear reactions below. State the type of decay in each case.
1. $\ce{^{198}_{79}Au}$ → _______ + $\ce{^0_{-1}e}$
2. $ce{^{57}_{27}Co}$ + $\ce{^0_{-1}e}$ → _______
3. $\ce{^{230}_{92}U}$ → _______ + $\ce{^4_2He}$
4. $\ce{^{128}_{56}Ba}$ → _______ + $\ce{^0_{1}e}$
5. $\ce{^{131}_{53}I}$ → $\ce{^{131}_{54}Xe}$ + _______
6. $\ce{^{239}_{94}Pu}$ → $\ce{^{235}_{92}U}$ + _______
Q10.1.5
Write balanced nuclear reactions for each of the following.
1. Francium-220 undergoes alpha decay.
2. Arsenic-76 undergoes beta decay.
3. Uranium-231 captures an electron.
4. Promethium-143 emits a positron.
(click here for solutions)
Q10.2.1
Describe the main difference between fission and fusion.
Q10.2.2
What is the difference between the fission reactions used in nuclear power plants and nuclear weapons?
Q10.2.3
How do the doses of radioisotopes used in diagnostic procedures and therapeutic treatment compare to one another?
(click here for solutions)
Q10.3.1
What percent of a sample remains after one half-life? Three half-lives?
Q10.3.2
The half-life of polonium-218 is 3.0 min. How much of a 0.540 mg sample would remain after 9.0 minutes have passed?
Q10.3.3
The half-life of hydrogen-3, commonly known as tritium, is 12.26 years. If 4.48 mg of tritium has decayed to 0.280 mg, how much time has passed?
Q10.3.4
The half-life of protactinium-234 is 6.69 hours. If a 0.812 mg sample of Pa-239 decays for 40.14 hours, what mass of the isotope remains?
Q10.3.5
2.86 g of a certain radioisotope decays to 0.358 g over a period of 22.8 minutes. What is the half-life of the radioisotope?
Q10.3.6
Use Table 10.3.2 above to determine the time it takes for 100. mg of carbon-14 to decay to 6.25 mg.
Q10.3.7
A radioisotope decays from 55.9 g to 6.99 g over a period of 72.5 hours. What is the half-life of the isotope?
Q10.3.8
A sample of a radioisotope with a half-life of 9.0 hours has an activity of 25.4 mCi after 36 hours. What was the original activity of the sample?
Q10.3.9
What volume of a radioisotope should be given if a patient needs 125 mCi of a solution which contains 45 mCi in 5.0 mL?
Q10.3.10
Sodium-24 is used to treat leukemia. A 36-kg patient is prescribed 145 μCi/kg and it is supplied to the hospital in a vial containing 250 μCi/mL. What volume should be given to the patient?
Q10.3.11
Using information from the previous question and knowing the half-life of Na-24 is 15 hours, calculate the total dose in μCi given to the patient. How long will it take for the radioactivity to be approximately 80 μCi?
Q10.3.12
Lead-212 is one of the radioisotopes used in the treatment of breast cancer. A patient needs a 15 μCi dose and it is supplied as a solution with a concentration of 2.5 μCi/mL. What volume does the patient need? Given the half-life of lead is 10.6 hours, what will be the radioactivity of the sample after approximately four days?
(click here for solutions)
Q10.4.1
Identify each of the following as a physical or chemical change.
1. melting ice
2. boiling water
3. cooking eggs
4. dissolving salt in water
5. burning match
6. metal reacting with HCl
7. mixing NaCl and KCl
8. decomposition of hydrogen peroxide
Q10.4.2
Give two signs that indicate a chemical change is occurring.
Q10.4.3
What doesn't change when a substance undergoes a physical change?
(click here for solutions)
Q10.5.1
Identify the reactants and products in each chemical reaction.
1. In photosynthesis, carbon dioxide and water react to form glucose and oxygen.
2. Magnesium oxide forms when magnesium is exposed to oxygen gas.
Q10.5.2
Write grammatically correct sentences that completely describe the chemical reactions shown in each equation. You may need to look up the names of elements or compounds.
1. 2H2O2(l) → 2H2O(l) + O2(g)
2. CuCO3(s) → CuO(s) + CO2(g)
3. 2Cs(s) + 2H2O(l) → 2CsOH(aq) + H2(g)
Q10.5.3
How many atoms of each element are represented by the following combinations of coefficients and chemical formulas?
1. 5Br2
2. 2NH3
3. 4(NH4)2SO4
4. 2CH3COOH
5. 3Fe(NO3)3
6. 2K3PO4
Q10.5.4
Balance the following equations.
1. Zn(s) + HCl(aq) → ZnCl2(aq) + H2(g)
2. Li(s) + N2(g) → Li3N(s)
3. Ca(OH)2 + HBr → CaBr2 + H2O
4. C4H10 + O2 → CO2 + H2O
5. NH3 + CuO → Cu + N2 + H2O
Q10.5.5
Balance the following equations.
1. Fe(s) + Cl2(g) → FeCl3(g)
2. C4H10O + O2 → CO2 + H2O
3. As + NaOH → Na3AsO3 + H2
4. SiO2 + HF → SiF4 + H2O
5. N2 + O2 + H2O → HNO3
Answers
10.1: Nuclear Radiation
Q10.1.1
Write the symbol for the isotope described.
1. $_{12}^{25}\text{Mg}$
2. $_{17}^{37}\text{Cl}$
3. $_{53}^{131}\text{I}$
4. $_{92}^{238}\text{U}$
Q10.1.2
1. 77 protons, 77 electrons, 118 neutrons
2. 82 protons, 82 electrons, 127 neutrons
3. 84 protons, 84 electrons, 127 neutrons
4. 93 protons, 93 electrons, 144 neutrons
Q10.1.3
1. $\ce{^{196}_{82}Pb}$ + $\ce{^0_{-1}e}$ →$_{81}^{196}\text{Tl}$
2. $\ce{^{28}_{15}P}$ → $_{14}^{28}\text{Si}$ + $\ce{^0_1e}$
3. $\ce{^{226}_{88}Ra}$ → $_{86}^{222}\text{Rn}$ + $\ce{^4_2 \alpha}$
4. $\ce{^{73}_{30}Zn}$ → $_{31}^{73}\text{Ga}$ + $\ce{^0_{-1}e}$
Q10.1.4
1. $\ce{^{198}_{79}Au}$ → $_{80}^{198}\text{Hg}$ + $\ce{^0_{-1}e}$, beta
2. $\ce{^{57}_{27}Co}$ + $\ce{^0_{-1}e}$ → $_{26}^{57}\text{Fe}$, electron capture
3. $\ce{^{230}_{92}U}$ → $_{90}^{226}\text{Th}$ + $\ce{^4_2He}$, alpha
4. $\ce{^{128}_{56}Ba}$ → $_{55}^{128}\text{Cs}$ + $\ce{^0_{1}e}$, positron
5. $\ce{^{131}_{53}I}$ → $\ce{^{131}_{54}Xe}$ + $_{-1}^{0}e$, beta
6. $\ce{^{239}_{94}Pu}$ → $\ce{^{235}_{92}U}$ + $_{2}^{4}\alpha$ (or can show as $_{2}^{4}\text{He}$), alpha
Q10.1.5
1. $_{87}^{220}\text{Fr}\;\rightarrow\;_{2}^{4}\text{He}\;+\;_{85}^{216}\text{At}$
2. $_{33}^{76}\text{As}\;\rightarrow\;_{-1}^{0}e\;+\;_{34}^{76}\text{Se}$
3. $_{92}^{231}\text{U}\;+\;_{-1}^{0}e\;\rightarrow\;_{91}^{231}\text{Pa}$
4. $_{61}^{143}\text{Pm}\;\rightarrow\;_{1}^{0}e\;+\;_{60}^{143}\text{Nd}$
10.2: Fission and Fusion
Q10.2.1
During fission, big nuclei split into smaller nuclei. During fusion, nuclei combine to form large nuclei.
Q10.2.2
Fission in nuclear power plants is controlled through limiting the availability of neutrons. Nuclear weapons are uncontrolled once the process initiates.
Q10.2.3
Diagnostic amounts are much smaller than therapeutic amounts.
10.3: Half-Life
Q10.3.1
1 half-life: 50%
3 half-lives: 12.5%
Q10.3.2
Time Half-lives Amount
0 minutes 0.540 mg
3 minutes 1 0.270 mg
6 minutes 2 0.135 mg
9 minutes 3 0.0675 mg
Q10.3.3
Amount Half-lives Time
4.48 mg 0 years
2.24 mg 1 12.26 years
1.12 mg 2 24.52 years
0.560 mg 3 36.78 years
0.280 mg 4 49.04 years
Q10.3.4
Time Half-lives Amount
0 hours 0.812 mg
6.69 hours 1 0.406 mg
13.38 hours 2 0.203 mg
20.07 hours 3 0.102 mg
26.76 hours 4 0.0508 mg
33.45 hours 5 0.0254 mg
40.14 hours 6 0.0127 mg
Q10.3.5
Amount Half-lives
2.86 g
1.43 g 1
0.715 g 2
0.358 g 3
It takes three half-lives to go from 2.86 g to 0.358 g in a total time of 22.8 minutes.
$22.8\;min\;\div\;3\;=7.60 \;min$
One half-life is 7.60 minutes.
Q10.3.6
Amount Half-lives Time
100. mg 0 years
50.0 mg 1 5730 years
25.0 mg 2 11460 years
12.5 mg 3 17190 years
6.25 mg 4 22920 years
Q10.3.7
Amount Half-lives
55.9 g
28.0 g 1
14.0 g 2
6.99 g 3
It takes three half-lives to go from 55.9 g to 6.99 g in a total time of 72.5 hours.
$72.5\;hr\;\div\;3\;=24.2 \;hr$
One half-life is 24.2 hours.
Q10.3.8
Fill in the time and half-lives from top to bottom. Start at the bottom of the amount column to fill it in because we know where we end up but not where we started.
Time Half-lives Activity
0 hours 406 mCi
9.0 hours 1 203 mCi
18 hours 2 102 mCi
27 hours 3 50.8 mCi
36 hours 4 25.4 mCi $\leftarrow$ START HERE
Q10.3.9
$125\;mCi\left(\frac{5.0\;mL}{45\;mCi}\right)=14\;mL$
Q10.3.10
Sodium-24 is used to treat leukemia. A 36-kg patient is prescribed 145 μCi/kg and it is supplied to the hospital in a vial containing 250 μCi/mL. What volume should be given to the patient?
$36\;kg\left(\frac{145\;\mu Ci}{kg}\right)\left(\frac{1\;mL}{250\;\mu Ci}\right)=21\;mL$
Q10.3.11
$21\;mL\left(\frac{250\;\mu Ci}{mL}\right)=5250\;\mu Ci$ is the total dose received
Amount Half-lives Time
5250 μCi 0 hours
2625 μCi 1 15 hours
1313 μCi 2 30 hours
656 μCi 3 45 hours
328 μCi 4 60 hours
164 μCi 5 75 hours
82 μCi 6 90 hours
Q10.3.12
Lead-212 is one of the radioisotopes used in the treatment of breast cancer. A patient needs a 15 μCi dose and it is supplied as a solution with a concentration of 2.5 μCi/mL. What volume does the patient need? Given the half-life of lead is 10.6 hours, what will be the radioactivity of the sample after approximately four days?
Volume given: $15\;\mu Ci\left(\frac{1\;mL}{2.5\;\mu Ci}\right)=6.0\;mL$
Elapsed time in hours: $4\;days\left(\frac{24\;hr}{day}\right)=96\;hr$
Time Half-lives Activity
0 hours 15 μCi
10.6 hours 1 7.5 μCi
21.2 hours 2 3.8 μCi
31.8 hours 3 1.9 μCi
42.4 hours 4 0.94 μCi
53.0 hours 5 0.47 μCi
63.6. hours 6 0.23 μCi
74.2 hours 7 0.12 μCi
84.8 hours 8 0.059 μCi
95.6 hours 9 0.029 μCi
10.4: Physical and Chemical Changes
Q10.4.1
1. physical
2. physical
3. chemical
4. physical
5. chemical
6. chemical
7. physical
8. chemical
Q10.4.2
Any two from change in color, formation of gas (i.e. bubbles), formation of precipitate, odor, change in temperature.
Q10.4.3
chemical composition (i.e. chemical formula is the same)
10.5: Chemical Equations
Q10.5.1
1. reactants: carbon dioxide and water; products: glucose and oxygen
2. reactants: magnesium and oxygen; product: magnesium oxide
Q10.5.2
Descriptions may vary.
1. Two moles of liquid hydrogen peroxide decomposes to form two moles of liquid water and one mole of gaseous hydrogen.
2. One mole of solid copper(II) carbonate decomposes to form one mole each of solid copper(II) oxide and gaseous carbon dioxide.
3. Two moles of solid cesium react with 2 moles of liquid water to form 2 moles of aqueous cesium hydroxide and 1 mole of gaseous hydrogen.
Q10.5.3
1. 10 Br
2. 2 N, 6 H
3. 8 N, 32 H, 4 S, 16 O
4. 4 C, 8 H, 4 O
5. 3 Fe, 9 N, 27 O
6. 6 K, 2 P, 8 O
Q10.5.4
1. Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)
2. 6 Li(s) + N2(g) → 2 Li3N(s)
3. Ca(OH)2 + 2 HBr → CaBr2 + 2 H2O
4. 2 C4H10 + 13 O28 CO2 + 10 H2O
5. 2 NH3 + 3 CuO → 3 Cu + N2 + 3 H2O
Q10.5.5
1. 2 Fe(s) + 3 Cl2(g) → 2 FeCl3(g)
2. C4H10O + 6 O24 CO2 + 5 H2O
3. 2 As + 6 NaOH → 2 Na3AsO3 + 3 H2
4. SiO2 + 4 HF → SiF4 + 2 H2O
5. 2 N2 + 5 O2 + 2 H2O → 4 HNO3 | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/10%3A_Nuclear_and_Chemical_Reactions/10.06%3A_Nuclear_and_Chemical_Reactions_%28Exercises%29.txt |
• 11.1: Oxidation Numbers
An oxidation number is a positive or negative number that is assigned to an atom to indicate its degree of oxidation or reduction. The term oxidation state is often used interchangeably with oxidation number. A partial electron transfer is a shift in the electron density near an atom as a result of a change in the other atoms to which it is covalently bonded. That charge shift is based on the relative electronegativities of the atoms involved in the bond.
• 11.2: The Nature of Oxidation and Reduction
Oxygen is an element that has been known for centuries. In its pure elemental form, oxygen is highly reactive, and it readily makes compounds with most other elements. It is also the most abundant element by mass in the Earth's crust. The class of reactions called oxidation and reduction were originally defined with respect to the element oxygen.
• 11.3: Types of Inorganic Reactions
Many chemical reactions can be classified as one of five basic types. Having a thorough understanding of these types of reactions will be useful for predicting the products of an unknown reaction. The five basic types of chemical reactions are combination, decomposition, single-replacement, double-replacement, and combustion. Analyzing the reactants and products of a given reaction will allow you to place it into one of these categories. Some reactions will fit into more than one category.
• 11.4: Entropy and Enthalpy
Previously, you learned that chemical reactions either absorb or release energy as they occur. The change in energy is one factor that allows chemists to predict whether a certain reaction will occur. In this lesson, you will learn about a second driving force for chemical reactions called entropy.
• 11.5: Spontaneous Reactions and Free Energy
The change in enthalpy and change in entropy of a reaction are the driving forces behind all chemical reactions. In this lesson, we will examine a new function called free energy, which combines enthalpy and entropy and can be used to determine whether or not a given reaction will occur spontaneously.
• 11.6: Rates of Reactions
Chemical kinetics is the study of the rates of chemical reactions. In this lesson, you will learn how to express the rate of a chemical reaction and about various factors that influence reaction rates.
• 11.7: Properties of Reactions (Exercises)
These are homework exercises to accompany Chapter 11 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Solutions are available below the questions.
11: Properties of Reactions
Learning Outcomes
• Assign oxidation numbers to free elements or elements in a compound or ion.
Oxidation Numbers
An oxidation number is a positive or negative number that is assigned to an atom to indicate its degree of oxidation or reduction. The term oxidation state is often used interchangeably with oxidation number. A partial electron transfer is a shift in the electron density near an atom as a result of a change in the other atoms to which it is covalently bonded. That charge shift is based on the relative electronegativities of the atoms involved in the bond.
Overall, the oxidation number of an atom in a molecule is the charge that the atom would have if all polar covalent and ionic bonds resulted in a complete transfer of electrons from the less electronegative atom to the more electronegative one. Oxidation numbers can be assigned using the set of rules outlined below.
1. The oxidation number of an atom in a neutral free element is zero. A free element is considered to be any element in an uncombined state, whether monatomic or polyatomic. For example, the oxidation number of each atom in $\ce{Fe}$, $\ce{Li}$, $\ce{N_2}$, $\ce{Ar}$, and $\ce{P_4}$ is zero.
2. The oxidation number of a monatomic (composed of one atom) ion is the same as the charge of the ion. For example, the oxidation numbers of $\ce{K^+}$, $\ce{Se^{2-}}$, and $\ce{Au^{3+}}$ are $+1$, $-2$, and $+3$, respectively.
3. The oxidation number of oxygen in most compounds is $-2$.
4. The oxidation number of hydrogen in most compounds is $+1$.
5. The oxidation number of fluorine in all compounds is $-1$. Other halogens usually have an oxidation number of $-1$ in binary compounds, but can have variable oxidation numbers depending on the bonding environment.
6. In a neutral molecule, the sum of the oxidation numbers of all atoms is zero. For example, in $\ce{H_2O}$, the oxidation numbers of $\ce{H}$ and $\ce{O}$ are $+1$ and $-2$, respectively. Because there are two hydrogen atoms in the formula, the sum of all the oxidation numbers in $\ce{H_2O}$ is $2 \left( +1 \right) + 1 \left( -2 \right) = 0$.
7. In a polyatomic ion, the sum of the oxidation numbers of all atoms is equal to the overall charge on the ion. For example, in $\ce{SO_4^{2-}}$, the oxidation numbers of $\ce{S}$ and $\ce{O}$ are $+6$ and $-2$, respectively. The sum of all oxidation numbers in the sulfate ion would be $1 \left( +6 \right) + 4 \left( -2 \right) = -2$, which is the charge of the ion.
An examination of the rules for assigning oxidation numbers reveals that there are many elements for which there are no specific rules, such as nitrogen, sulfur, and chlorine. These elements, as well as some others, can have variable oxidation numbers depending on the other atoms to which they are covalently bonded in a molecular compound. It is useful to analyze a few molecules in order to see the strategy to follow in assigning oxidation numbers to other atoms.
Oxidation numbers for the atoms in a binary ionic compound are easy to assign because they are equal to the charge of the ion (rule 2). In $\ce{FeCl_3}$, the oxidation number of iron is $+3$, while the oxidation number of chlorine is $-1$. In $\ce{Ca_3P_2}$, the calcium is $+2$, while the phosphorus is $-3$. This is because an ionic compound is in the form of a crystal lattice that is actually composed of these ions.
Assigning oxidation numbers for molecular compounds is trickier. The key is to remember rule 6: that the sum of all the oxidation numbers for any neutral species must be zero. Make sure to account for any subscripts which appear in the formula. As an example, consider the compound nitric acid, $\ce{HNO_3}$. According to rule 4, the oxidation number of hydrogen is $+1$. According to rule 3, the oxidation number of oxygen is $-2$. There is no rule regarding nitrogen, but its oxidation number can be calculated as follows.
$1 \left( +1 \right) + x + 3 \left( -2 \right) = 0, \: \text{where} \: x \: \text{is the oxidation number of nitrogen}$
$\text{Solving:} \: x = 0 - 1 - \left( -6 \right) = +5$
The oxidation number of the nitrogen atom in $\ce{HNO_3}$ is $+5$. Often when assigning oxidation numbers, it is convenient to write it above the symbol within the formula.
$\overset{+1}{\ce{H}} \overset{+5}{\ce{N}} \overset{-2}{\ce{O_3}}$
You may wonder if there are any limits on the value of oxidation numbers. The key point to consider is the octet rule. Since nitrogen has 5 valence electrons, the most that it can "lose" while forming bonds in a molecule is 5, so its highest possible oxidation number is $+5$. Alternatively, it could gain up to 3 electrons, and so its lowest (most negative) possible oxidation number is $-3$. Similarly, chlorine can have oxidation numbers ranging from $-1$ to $+7$.
Now consider the ionic compound sodium thiosulfate, $\ce{Na_2S_2O_3}$ (Figure $1$). It contains the thiosulfate polyatomic ion, $\ce{S_2O_3^{2-}}$. The sodium is not part of the covalently bonded polyatomic ion, and so its oxidation number is the same as it would be in a binary ionic compound, $+1$. The sulfur is the atom whose oxidation number is not covered by one of the rules. The oxidation number of sulfur is assigned the variable $x$ in the following calculation. Remember the sum of the oxidation numbers of all the elements must equal zero because $\ce{Na_2S_2O_3}$ is a neutral compound.
\begin{align}2 \left( +1 \right) + 2 \left( x \right) + 3 \left( -2 \right) &= 0 \[5pt] 2 + 2x -6 &=0 \[5pt] -4 + 2x &= 0 \[5pt] 2x &= +4 \[5pt] x &= +2 \end{align}
Sulfur has an oxidation number of $+2$ in $\ce{Na_2S_2O_3}$. Notice how the subscript of 2 for the $\ce{S}$ atom had to be accounted for by dividing the result of the subtraction by 2. When assigning oxidation numbers, you do so for each individual atom. In the above example, the oxidation number of sulfur could also have been determined by looking at just the thiosulfate ion, $\ce{S_2O_3^{2-}}$.
\begin{align} 2 \left( x \right) + 3 \left( -2 \right) &= -2 \[5pt] 2x - 6 &= -2 \[5pt] 2x &= +4 \[5pt] x &= +2 \end{align}
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/11%3A_Properties_of_Reactions/11.01%3A_Oxidation_Numbers.txt |
Learning Outcomes
• Define oxidation and reduction in terms of a gain or loss of oxygen or hydrogen atoms.
• Define oxidation and reduction in terms of a gain or loss of electrons.
• Identify the substances involved in oxidation and reduction in a reaction.
• Identify the oxidizing and reducing agents in a redox reaction.
• Write the oxidation or reduction sequence for a molecule (alcohol, aldehyde/ketone, carboxylic acid)
Oxygen is an element that has been known for centuries. In its pure elemental form, oxygen is highly reactive, and it readily makes compounds with most other elements. It is also the most abundant element by mass in the Earth's crust. The class of reactions called oxidation and reduction were originally defined with respect to the element oxygen.
Oxygen's Role in Reactions
Many elements simply combine with oxygen to form the oxide of that element. Heating magnesium in air allows it to combine with oxygen to form magnesium oxide.
$2 \ce{Mg} \left( s \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{MgO} \left( s \right)$
Many compounds react with oxygen as well, often in very exothermic processes that are generally referred to as combustion reactions. For example, when methane burns, carbon dioxide and water are produced.
$\ce{CH_4} \left( g \right) + 2 \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + 2 \ce{H_2O} \left( g \right)$
Carbon dioxide is an oxide of carbon, while water is an oxide of hydrogen. Early scientists viewed oxidation as a process in which a substance was reacted with oxygen to produce one or more oxides. In the previous examples, magnesium and methane are being oxidized.
Oxidation does not necessarily require heating. Iron that is exposed to air and water slowly oxidizes in a process commonly known as rusting. Bleaches contain various compounds such as sodium hypochlorite $\left( \ce{NaClO} \right)$ that can oxidize stains by the transfer of oxygen atoms, making the molecules in the stains more water-soluble and therefore easier to rinse off. Hydrogen peroxide $\left( \ce{H_2O_2} \right)$ releases oxygen as it spontaneously decomposes. It also acts as a bleach and is ued as an antiseptic that kills bacteria by oxidizing them.
The opposite of oxidation is called reduction. Since oxidation was originally defined as the addition of oxygen, reduction was therefore the removal of oxygen from a substance. Many naturally occurring metal ores are present as oxides. The pure metals can be extracted by reduction. For example, pure iron is obtained from iron (III) oxide by reacting it with carbon at high temperatures.
$2 \ce{Fe_2O_3} \left( s \right) + 3 \ce{C} \left( s \right) \rightarrow 4 \ce{Fe} \left( s \right) + 3 \ce{CO_2} \left( g \right)$
The removal of oxygen from the $\ce{Fe_2O_3}$ molecule means that it is being reduced to $\ce{Fe}$. Note that an oxidation process is also occurring simultaneously in this reaction; the carbon reactant is being oxidized to $\ce{CO_2}$. This is an important concept. In looking at organic compounds, we can describe oxidation and reduction in terms of the loss or gain of oxygen or the loss or gain of hydrogen.
If we look at the oxidation of ethanol ($\ce{CH_3CH_2OH}$, an alcohol) to form ethanal ($\ce{CH_3CH_2CHO}$, an aldehyde), we see that the number of bonds to oxygen has increased and the number of hydrogen atoms has decreased from six to four. Either or both of these indicate that an oxidation has occurred.
For a reduction, we look for the removal of oxygen or bonds to oxygen or the addition or hydrogen atoms. In the following reaction, we see ethanoic acid ($\ce{CH_3COOH}$, a carboxylic acid) being reduced to ethanal ($\ce{CH_3CH_2CHO}$, an aldehyde). Note that the number of oxygen atoms is reduced from two to one. There is no change in the number of hydrogen atoms, but both changes do not need to happen for evidence of oxidation or reduction.
If we see evidence of an oxidation or reduction, we know the other must have happened as well. Oxidation and reduction must happen together. Neither can happen alone in a reaction.
The Electron in Redox Reactions
The definitions of oxidation and reduction were eventually broadened to include similar types of reactions that do not necessarily involve oxygen. Oxygen is more electronegative than any element except for fluorine. Therefore, when oxygen is bonded to any element other than fluorine, electrons from the other atom are shifted away from that atom and toward the oxygen atom. An oxidation-reduction reaction (sometimes abbreviated as a redox reaction) is a reaction that involves the full or partial transfer of electrons from one reactant to another. Oxidation involves a full or partial loss of electrons, while reduction involves a full or partial gain of electrons.
An easy saying to remember the definitions of oxidation and reduction is "LEO the lion says GER" (see figure below)! LEO stands for Losing Electrons is Oxidation, while GER stands for Gaining Electrons is Reduction.
Redox Reactions and Ionic Compounds
In the course of a chemical reaction between a metal and a nonmetal, electrons are transferred from the metal atoms to the nonmetal atoms. For example, when zinc metal is heated in the presence of sulfur, the compound zinc sulfide is produced (see figure below). Two valence electrons from each zinc atom are transferred to each sulfur atom.
Since zinc is losing electrons during this reaction, it is being oxidized. The sulfur is gaining electrons, so it is being reduced. Each of these processes can be shown in a separate equation called a half-reaction. A half-reaction is an equation that shows either the oxidation or the reduction reaction that occurs during a redox reaction.
\begin{align} &\text{Oxidation:} \: \: \ce{Zn} \rightarrow \ce{Zn^{2+}} + 2 \ce{e^-} \ &\text{Reduction:} \: \: \ce{S} + 2 \ce{e^-} \rightarrow \ce{S^{2-}} \end{align}
It is important to remember that the two half-reactions occur simultaneously. The resulting ions that are formed are then attracted to one another, forming an ionic bond.
Redox reactions are also commonly described in terms of oxidizing and reducing agents. In the reaction above, the zinc is being oxidized by losing electrons. However, an isolated ionization would be a very high-energy process, so there must be another substance present to gain those lost electrons. In this case, the electrons are gained by sulfur. In other words, the sulfur is causing the zinc to be oxidized. Consequently, sulfur is referred to as the oxidizing agent. Conversely, the zinc causes the sulfur to gain electrons and become reduced, so zinc is the reducing agent. An oxidizing agent is a substance that causes oxidation by accepting electrons, and a reducing agent is a substance that causes reduction by losing electrons. Said another way, the oxidizing agent is the substance that is reduced, while the reducing agent is the substance that is oxidized.
Example $1$
When chlorine gas is bubbled into a solution of sodium bromide, a reaction occurs that produces aqueous sodium chloride and elemental bromine.
1. Determine what is being oxidized and what is being reduced.
2. Identify the oxidizing and reducing agents.
Solution
First, look at each element in the reaction and assign oxidation numbers to each element.
$\ce{Cl_2} \left( g \right) + 2 \ce{NaBr} \left( aq \right) \rightarrow 2 \ce{NaCl} \left( aq \right) + \ce{Br_2} \left( l \right)$
Chlorine in $\ce{Cl_2}$ and bromine in $\ce{Br_2}$ both have oxidation numbers of 0. $\ce{NaBr}$ and $\ce{NaCl}$ are ionic compounds so their ionic charges will be their oxidation numbers.
$\overset{0}{\ce{Cl_2}} \left( g \right) + 2 \overset{+1}{\ce{Na}} \overset{-1}{\ce{Br}} \left( aq \right) \rightarrow 2 \overset{+1}{\ce{Na}} \overset{-1}{\ce{Cl}} \left( aq \right) + \overset{0}{\ce{Br_2}} \left( l \right)$
a. The oxidation state of chlorine goes from 0 to $-1$ so it is gaining an electron and being reduced. The oxidation state of bromine goes from $-1$ to 0 so it is losing an electron and being oxidized. The oxidation state of sodium does not change so it is not involved in either the oxidation or the reduction.
b. The $\ce{Cl_2}$ is being reduced and is the oxidizing agent. The $\ce{Br^-}$ is being oxidized and is the reducing agent.
Redox Reactions and Molecular Compounds
The loss or gain of electrons is easy to see in a reaction in which ions are formed. However, in many reactions, electrons are not transferred completely. Recall that in a molecular compound, electrons are shared between atoms in a type of bond called a covalent bond. Even though electrons are not completely transferred between atoms, it is still common for reactions involving molecular compounds to be classified as redox reactions.
For example, when hydrogen gas is reacted with oxygen gas, water is formed as the product.
$2 \ce{H_2} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{H_2O} \left( l \right)$
We assign oxidation numbers to determine which species are gaining and losing electrons. Remember oxidation numbers tell us what the charge would be if the electrons were transferred completely even though they are shared, not transferred, in covalent bonds. For the synthesis of water, the oxidation numbers of the free elements (hydrogen and oxygen) are zero. For water, oxygen has a $-2$ oxidation number while hydrogen has a $+1$.
$2 \overset{0}{\ce{H_2}} \left( g \right) + \overset{0}{\ce{O_2}} \left( g \right) \rightarrow 2 \overset{+1}{\ce{H_2}} \overset{-2}{\ce{O}} \left( l \right)$
The oxidation state from hydrogen goes from 0 to $+1$, indicating oxidation. The oxidation state of oxygen goes from 0 to $-2$, indicating reduction. Therefore hydrogen is oxidized and is the reducing agent. Oxygen is reduced and is the oxidizing agent.
Gain or Loss of Hydrogen and Oxygen
Oxidation can also be defined as the addition of oxygen to a molecule or the removal of hydrogen. Reduction is therefore the addition of hydrogen or the removal of oxygen. Several classes of organic compounds are related to one another by oxidation and reduction reactions. Alkanes, alkenes, and alkynes represent different levels of oxidation of a hydrocarbon. When an alkane is heated in the presence of an appropriate catalyst, it can be oxidized to the corresponding alkene in a reaction called a dehydrogenation reaction. Two hydrogen atoms are removed in the process. The alkene can be further oxidized to an alkyne by the removal of two more hydrogen atoms. This is also considered an oxidation according to the modern definition, because the oxidation number of each carbon atom goes from $-3$ to $-2$ to $-1$.
$\text{oxidation:} \: \: \ce{CH_3CH_3} \overset{\ce{-H_2}}{\rightarrow} \ce{CH_2=CH_2} \overset{\ce{-H_2}}{\rightarrow} \ce{CH \equiv CH}$
The reactions are reversible, so an alkyne can be reduced first to an alkene and then to an alkane. These are addition reactions, as seen in the previous section.
$\text{reduction:} \: \: \ce{CH \equiv CH} \overset{\ce{+H_2}}{\rightarrow} \ce{CH_2=CH_2} \overset{\ce{+H_2}}{\rightarrow} \ce{CH_3CH_3}$
The alkane is the most reduced form of a hydrocarbon, while the alkyne is the most oxidized form.
The changes in electrons, oxidation number, number of oxygen atoms, and number of hydrogen atoms is summarized below for oxidation and reduction reactions.
Oxidation reactions in organic chemistry often involve the addition of oxygen to a compound (or an increase in the number of bonds to oxygen), which changes the functional group that is present. The following sequence shows how methane can be oxidized first to methanol (alcohol), then to methanal (aldehyde), then to methanoic acid (carboxylic acid), and finally to carbon dioxide. Each step in the process is either a gain of oxygen or a loss of hydrogen. Each step also represents energy, which explains why the complete combustion of alkanes to carbon dioxide is an extremely exothermic reaction.
$\begin{array}{ccccccccc} \ce{CH_4} & \overset{\text{gain of oxygen}}{\rightarrow} & \ce{CH_3OH} & \overset{\text{loss of hydrogen}}{\rightarrow} & \ce{CH_2O} & \overset{\text{gain of oxygen}}{\rightarrow} & \ce{HCOOH} & \overset{\text{loss of hydrogen}}{\rightarrow} & \ce{CO_2} \ \text{alkane} & & \text{alcohol} & & \text{aldehyde} & & \text{carboxylic acid} & & \text{carbon dioxide} \end{array}$
The opposite process is the reduction of a carboxylic acid to an alkane which involves the loss of oxygen and the gain of hydrogen.
$\begin{array}{ccccccc} \ce{CH_4} & \overset{\text{loss of oxygen}}{\leftarrow} & \ce{CH_3OH} & \overset{\text{gain of hydrogen}}{\leftarrow} & \ce{CH_2O} & \overset{\text{loss of oxygen}}{\leftarrow} & \ce{HCOOH} \ \text{alkane} & & \text{alcohol} & & \text{aldehyde} & & \text{carboxylic acid} \end{array}$
The oxidation of an alcohol can produce either an aldehyde or a ketone. A primary alcohol oxidized to form an aldehyde.
$\ce{CH_3CH_2OH} \underset{\ce{H^+}}{\overset{\ce{Cr_2O_7^{2-}}}{\rightarrow}} \ce{CH_3CHO}$
A secondary alcohol is oxidized to a ketone rather than an aldehyde. The oxidation of the simplest secondary alcohol, 2-propanol, yields propanone.
$\ce{CH_3CHOHCH_3} \underset{\ce{H^+}}{\overset{\ce{Cr_2O_7^{2-}}}{\rightarrow}} \ce{CH_3COCH_3}$
Tertiary alcohols cannot be oxidized in this way because the carbon to which the hydroxyl group is attached does not have another hydrogen atom attached to it and cannot undergo oxidation.
When a primary alcohol is oxidized to an aldehyde in the presence of water, the reaction can be difficult to stop because the aldehyde can be further oxidized to the corresponding carboxylic acid. For example, the oxidation of ethanal produces ethanoic (acetic) acid. Ethanol-containing beverages such as wine are susceptible to such oxidation if kept for long periods of time after having been opened and exposed to the air. Wine that has become oxidized will have an unpleasant vinegary taste due to the production of acetic acid. Unlike aldehydes, ketones are resistant to further oxidation because the carbonyl group is in the middle of the carbon chain, so the ketone cannot be converted to a carboxylic acid.
$\ce{CH_3CHO} \underset{\ce{H^+}}{\overset{\ce{Cr_2O_7^{2-}}}{\rightarrow}} \ce{CH_3COOH}$
Table $1$
Process Changes in electrons (always) Change in oxidation number (always) Change in oxygen (some reactions) Change in hydrogen (some reactions)
Oxidation lose increase gain lose
Reduction gain decrease lose gain
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/11%3A_Properties_of_Reactions/11.02%3A_The_Nature_of_Oxidation_and_Reduction.txt |
Learning Outcomes
• Classify a reaction as combination, decomposition, single-replacement, double-replacement, or combustion.
• Determine whether a reaction is an oxidation-reduction reaction.
• Predict the products and balance a combustion reaction.
Many chemical reactions can be classified as one of five basic types. Having a thorough understanding of these types of reactions will be useful for predicting the products of an unknown reaction. The five basic types of chemical reactions are combination, decomposition, single-replacement, double-replacement, and combustion. Analyzing the reactants and products of a given reaction will allow you to place it into one of these categories. Some reactions will fit into more than one category. Identifying a reaction as one of these types does not preclude it from also being an oxidation-reduction (redox) reaction. For example, many combination reactions can also be classified as a redox reaction.
Combination Reactions
A combination reaction, also known as a synthesis reaction, is a reaction in which two or more substances combine to form a single new substance. Combination reactions can also be called synthesis reactions .The general form of a combination reaction is:
$\ce{A} + \ce{B} \rightarrow \ce{AB}$
One example of a combination reaction is two elements combining to form a compound. Solid sodium metal reacts with chlorine gas to product solid sodium chloride.
$2 \ce{Na} \left( s \right) + \ce{Cl_2} \left( g \right) \rightarrow 2 \ce{NaCl} \left( s \right)$
This reaction can also be classified as a redox reaction due to the changes in oxidation states. Sodium goes from a 0 to +1 oxidation state while chlorine goes from a 0 to a $-1$ oxidation state.
Notice that in order to write and balance the equation correctly, it is important to remember the seven elements that exist in nature as diatomic molecules ($\ce{H_2}$, $\ce{N_2}$, $\ce{O_2}$, $\ce{F_2}$, $\ce{Cl_2}$, $\ce{Br_2}$, and $\ce{I_2}$).
One sort of combination reaction that occurs frequently is the reaction of an element with oxygen to form an oxide. Metals and nonmetals both react readily with oxygen under most conditions. Magnesium reacts rapidly and dramatically when ignited, combining with oxygen from the air to produce a fine powder of magnesium oxide.
$2 \ce{Mg} \left( s \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{MgO} \left( s \right)$
Decomposition Reactions
A decomposition reaction is a reaction in which a compound breaks down into two or more simpler substances. The general form of a decomposition reaction is:
$\ce{AB} \rightarrow \ce{A} + \ce{B}$
Most decomposition reactions require an input of energy in the form of heat, light, or electricity.
Binary compounds are compounds composed of just two elements. The simplest kind of decomposition reaction is when a binary compound decomposes into its elements. Mercury (II) oxide, a red solid, decomposes when heated to produce mercury and oxygen gas.
$2 \ce{HgO} \left( s \right) \rightarrow 2 \ce{Hg} \left( l \right) + \ce{O_2} \left( g \right)$
A reaction is also considered to be a decomposition reaction even when one or more of the products is still a compound. A metal carbonate decomposes into a metal oxide and carbon dioxide gas. For example, calcium carbonate decomposes into calcium oxide and carbon dioxide.
$\ce{CaCO_3} \left( s \right) \rightarrow \ce{CaO} \left( s \right) + \ce{CO_2} \left( g \right)$
Metal hydroxides decompose on heating to yield metal oxides and water. Sodium hydroxide decomposes to produce sodium oxide and water.
$2 \ce{NaOH} \left( s \right) \rightarrow \ce{Na_2O} \left( s \right) + \ce{H_2O} \left( g \right)$
Single-Replacement Reactions
A single-replacement reaction, also known as single-displacement or substitution reaction, is a reaction in which one element replaces a similar element in a compound. The general form of a single-replacement reaction is:
$\ce{A} + \ce{BC} \rightarrow \ce{AC} + \ce{B}$
In this general reaction, element $\ce{A}$ is a metal and replaces element $\ce{B}$, also a metal, in the compound. When the element that is doing the replacing is a nonmetal, it must replace another nonmetal in a compound, and the general equation becomes:
$\ce{Y} + \ce{XZ} \rightarrow \ce{XY} + \ce{Z}$
$\ce{Y}$ is a nonmetal and replaces the nonmetal $\ce{Z}$ in the compound with $\ce{X}$.
Magnesium is a more reactive metal than copper. When a strip of magnesium metal is placed in an aqueous solution of copper (II) nitrate, it replaces the copper. The products of the reaction are aqueous magnesium nitrate and solid copper metal.
$\ce{Mg} \left( s \right) + \ce{Cu(NO_3)_2} \left( aq \right) \rightarrow \ce{Mg(NO_3)_2} \left( aq \right) + \ce{Cu} \left( s \right)$
Many metals react easily with acids, and, when they do so, one of the products of the reaction is hydrogen gas. Zinc reacts with hydrochloric acid to produce aqueous zinc chloride and hydrogen (see figure below).
$\ce{Zn} \left( s \right) + 2 \ce{HCl} \left( aq \right) \rightarrow \ce{ZnCl_2} \left( aq \right) + \ce{H_2} \left( g \right)$
Double-Replacement Reactions
A double-replacement reaction, also known as double-displacement, is a reaction in which the positive and negative ions of two ionic compounds exchange places to form two new compounds. The general form of a double-replacement reaction is:
$\ce{AB} + \ce{CD} \rightarrow \ce{AD} + \ce{CB}$
In this reaction, $\ce{A}$ and $\ce{C}$ are positively-charged cations, while $\ce{B}$ and $\ce{D}$ are negatively-charged anions. Double-replacement reactions generally occur between substances in aqueous solution. In order for a reaction to occur, one of the products is usually a solid precipitate, a gas, or a molecular compound such as water.
A precipitate forms in a double-replacement reaction when the cations from one of the reactants combine with the anions from the other reactant to form an insoluble ionic compound. When aqueous solutions of potassium iodide and lead (II) nitrate are mixed, the following reaction occurs.
$2 \ce{KI} \left( aq \right) + \ce{Pb(NO_3)_2} \left( aq \right) \rightarrow 2 \ce{KNO_3} \left( aq \right) + \ce{PbI_2} \left( s \right)$
Combustion Reactions
A combustion reaction is a reaction in which a substance reacts with oxygen gas, releasing energy in the form of light and heat. Combustion reactions must involve $\ce{O_2}$ as one reactant. The combustion of hydrogen gas produces water vapor (see figure below).
$2 \ce{H_2} \left( g \right) + \ce{O_2} \left( g \right) + 2 \ce{H_2O} \left( g \right)$
Notice that this reaction also qualifies as a combination reaction.
Many combustion reactions occur with a hydrocarbon, a compound made up solely of carbon and hydrogen. The products of the combustion of hydrocarbons are always carbon dioxide and water. Many hydrocarbons are used as fuel because their combustion releases very large amount of heat energy. Propane $\left( \ce{C_3H_8} \right)$ is a gaseous hydrocarbon that is commonly used as the fuel source in gas grills.
$\ce{C_3H_8} \left( g \right) + 5 \ce{O_2} \left( g \right) \rightarrow 3 \ce{CO_2} \left( g \right) + 4 \ce{H_2O} \left( g \right)$
Example $1$
Ethanol can be used as a fuel source in an alcohol lamp. The formula for ethanol is $\ce{C_2H_5OH}$. Write the balanced equation for the combustion of ethanol.
Solution
Step 1: Plan the problem.
Ethanol and oxygen are the reactants. As with a hydrocarbon, the products of the combustion of an alcohol are carbon dioxide and water.
Step 2: Solve.
Write the skeleton equations: $\ce{C_2H_5OH} \left( l \right) + \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + \ce{H_2O} \left( g \right)$
Balance the equation.
$\ce{C_2H_5OH} \left( l \right) + 3 \ce{O_2} \left( g \right) \rightarrow 2 \ce{CO_2} \left( g \right) + 3 \ce{H_2O} \left( g \right)$
Step 3: Think about your result.
Combustion reactions must have oxygen as a reactant. Note that the water that is produced is in the gas state rather that the liquid state because of the high temperatures that accompany a combustion reaction.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/11%3A_Properties_of_Reactions/11.03%3A_Types_of_Inorganic_Reactions.txt |
Learning Outcomes
• Recall the meaning of exothermic and endothermic.
• Define entropy.
• Predict whether entropy change for a reaction is increasing or decreasing.
Previously, you learned that chemical reactions either absorb or release energy as they occur. The change in energy is one factor that allows chemists to predict whether a certain reaction will occur. In this lesson, you will learn about a second driving force for chemical reactions called entropy.
Enthalpy as a Driving Forces
The vast majority of naturally occurring reactions are exothermic. In an exothermic reaction, the reactants have a relatively high quantity of energy compared to the products. As the reaction proceeds, energy is released into the surroundings. Low energy can be thought of as providing a greater degree of stability to a chemical system. Since the energy of the system decreases during an exothermic reaction, the products of the system are more stable than the reactants. We can say that an exothermic reaction is an energetically favorable reaction.
If the drive toward lower energy were the only consideration for whether a reaction is able to occur, we would expect that endothermic reactions could never occur spontaneously. In an endothermic reaction, energy is absorbed during the reaction, and the products thus have a larger quantity of energy than the reactants. This means that the products are less stable than the reactants. Therefore, the reaction would not occur without some outside influence such as persistent heating. However, endothermic reactions do occur spontaneously, or naturally. There must be another driving force besides enthalpy change which helps promote spontaneous chemical reaction.
Entropy as a Driving Force
A very simple endothermic process is that of a melting ice cube. Energy is transferred from the room to the ice cube, causing it to change from the solid to the liquid state.
$\ce{H_2O} \left( s \right) + 6.01 \: \text{kJ} \rightarrow \ce{H_2O} \left( l \right)$
The solid state of water, ice, is highly ordered because its molecules are fixed in place. The melting process frees the water molecules from their hydrogen-bonded network and allows them a greater degree of movement. Water is more disordered than ice. The change from the solid to the liquid state of any substance corresponds to an increase in the disorder of the system.
There is a tendency in nature for systems to proceed toward a state of greater disorder or randomness. Entropy is a measure of the degree of randomness or disorder of a system. Entropy is an easy concept to understand when thinking about everyday situations. When the pieces of a jigsaw puzzle are dumped from the box, the pieces naturally hit the table in a very random state. In order to put the puzzle together, a great deal of work must be dome in order to overcome the natural entropy of the pieces. The entropy of a room that has been recently cleaned and organized is low. As time goes by, it likely will become more disordered, and thus its entropy will increase (see figure below). The natural tendency of a system is for its entropy to increase.
Chemical reactions also tend to proceed in such a way as to increase the total entropy of the system. How can you tell if a certain reaction shows an increase or a decrease in entropy? The states of the reactants and produces provide certain clues. The general cases below illustrate entropy at the molecular level.
1. For a given substance, the entropy of the liquid state is greater than the entropy of the solid state. Likewise, the entropy of the gas is greater than the entropy of the liquid. Therefore, entropy increases in processes in which solid or liquid reactants form gaseous products. Entropy also increases when solid reactants form liquid products.
2. Entropy increases when a substance is broken up into multiple parts. The process of dissolving increases entropy because the solute particles become separated from one another when a solution is formed.
3. Entropy increases as temperature increases. An increase in temperature means that the particles of the substance have greater kinetic energy. The faster moving particles have more disorder than particles that are moving more slowly at a lower temperature.
4. Entropy generally increases in reactions in which the total number of product molecules is greater than the total number of reactant molecules. An exception to this rule is when nongaseous products are formed from gaseous reactants.
The examples below will serve to illustrate how the entropy change in a reaction can be predicted.
$\ce{Cl_2} \left( g \right) \rightarrow \ce{Cl_2} \left( l \right)$
The entropy is decreasing because a gas is becoming a liquid.
$\ce{CaCO_3} \left( s \right) \rightarrow \ce{CaO} \left( s \right) + \ce{CO_2} \left( g \right)$
The entropy is increasing because a gas is being produced, and the number of molecules is increasing.
$\ce{N_2} \left( g \right) + 3 \ce{H_2} \left( g \right) \rightarrow 2 \ce{NH_3} \left( g \right)$
The entropy is decreasing because four total reactant molecules are forming two total product molecules. All are gases.
$\ce{AgNO_3} \left( aq \right) + \ce{NaCl} \left( aq \right) \rightarrow \ce{NaNO_3} \left( aq \right) + \ce{AgCl} \left( s \right)$
The entropy is decreasing because a solid is formed from aqueous reactants.
$\ce{H_2} \left( g \right) + \ce{Cl_2} \left( g \right) \rightarrow 2 \ce{HCl} \left( g \right)$
The entropy change is unknown (but likely not zero) because there are equal numbers of molecules on both sides of the equation, and all are gases.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/11%3A_Properties_of_Reactions/11.04%3A_Entropy_and_Enthalpy.txt |
Learning Outcomes
• Describe the meaning of a spontaneous reaction in terms of enthalpy and entropy changes.
• Define free energy.
• Determine the spontaneity of a reaction based on the value of its change in free energy at high and low temperatures.
The change in enthalpy and change in entropy of a reaction are the driving forces behind all chemical reactions. In this lesson, we will examine a new function called free energy, which combines enthalpy and entropy and can be used to determine whether or not a given reaction will occur spontaneously.
Spontaneous Reactions
A spontaneous reaction is a reaction that favors the formation of products at the conditions under which the reaction is occurring. A roaring bonfire (see figure below) is an example of a spontaneous reaction. A fire is exothermic, which means a decrease in the energy of the system as energy is released to the surroundings as heat. The products of a fire are composed mostly of gases such as carbon dioxide and water vapor, so the entropy of the system increases during most combustion reactions. This combination of a decrease in energy and an increase in entropy means that combustion reactions occur spontaneously.
A nonspontaneous reaction is a reaction that does not favor the formation of products at the given set of conditions. In order for a reaction to be nonspontaneous, one or both of the driving forces must favor the reactants over the products. In other words, the reaction is endothermic, is accompanied by a decrease in entropy, or both. Out atmosphere is composed primarily of a mixture of nitrogen and oxygen gases. One could write an equation showing these gases undergoing a chemical reaction to form nitrogen monoxide.
$\ce{N_2} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{NO} \left( g \right)$
Fortunately, this reaction is nonspontaneous at normal temperatures and pressures. It is a highly endothermic reaction with a slightly positive entropy change $\left( \Delta S \right)$. However, nitrogen monoxide is capable of being produced at very high temperatures, and this reaction has been observed to occur as a result of lightning strikes.
One must be careful not to confuse the term spontaneous with the notion that a reaction occurs rapidly. A spontaneous reaction is one in which product formation is favored, even if the reaction is extremely slow. You do not have to worry about a piece of paper on your desk suddenly bursting into flames, although its combustion is a spontaneous reaction. What is missing is the required activation energy to get the reaction started. If the paper were to be heated to a high enough temperature, it would begin to burn, at which point the reaction would proceed spontaneously until completion.
In a reversible reaction, one reaction direction may be favored over the other. Carbonic acid is present in carbonated beverages. It decomposes spontaneously to carbon dioxide and water according to the following reaction.
$\ce{H_2CO_3} \left( aq \right) \rightleftharpoons \ce{CO_2} \left( g \right) + \ce{H_2O} \left( l \right)$
If you were to start with pure carbonic acid in water and allow the system to come to equilibrium, more than $99\%$ of the carbonic acid would be converted into carbon dioxide and water. The forward reaction is spontaneous because the products of the forward reaction are favored at equilibrium. In the reverse reaction, carbon dioxide and water are the reactants, and carbonic acid is the product. When carbon dioxide is bubbled into water (see figure below), less than $1\%$ is converted to carbonic acid when the reaction reaches equilibrium. The reverse of the above reaction is not spontaneous. This illustrates another important point about spontaneity. Just because a reaction is not spontaneous does not mean that it does not occur at all. Rather, it means that the reactants will be favored over the products at equilibrium, even though some products may indeed form.
Gibbs Free Energy
Many chemical reactions and physical processes release energy that can be used to do other things. When the fuel in a car is burned, some of the released energy is used to power the vehicle. Free energy is energy that is available to do work. Spontaneous reactions release free energy as they proceed. Recall that the determining factors for spontaneity of a reaction are the enthalpy and entropy changes that occur for the system. The free energy change of a reaction is a mathematical combination of the enthalpy change and the entropy change.
$\Delta G^\text{o} = \Delta H^\text{o} - T \Delta S^\text{o}$
The symbol for free energy is $G$, in honor of American scientist Josiah Gibbs (1839 - 1903), who made many contributions to thermodynamics. The change in Gibbs free energy is equal to the change in enthalpy minus the mathematical product of the change in entropy multiplied by the Kelvin temperature. Each thermodynamic quantity in the equation is for substances in their standard states, as indicated by the $^\text{o}$ superscripts.
A spontaneous reaction is one that releases free energy, and so the sign of $\Delta G$ must be negative. Since both $\Delta H$ and $\Delta S$ can be either positive or negative, depending on the characteristics of the particular reaction, there are four different possible combinations. The outcomes for $\Delta G$ based on the signs of $\Delta H$ and $\Delta S$ are outlined in the table below. Recall that $- \Delta \text{H}$ indicates that the reaction is exothermic and a $+ \Delta \text{H}$ means the reaction is endothermic. For entropy, $+ \Delta \text{S}$ means the entropy is increasing and the system is becoming more disordered. A $- \Delta \text{S}$ means that entropy is decreasing and the system is becoming less disordered (more ordered).
Table $1$: Enthalpy, Entropy, and Free Energy Changes.
$\Delta H$ $\Delta S$ $\Delta G$
negative positive always negative
positive positive negative at higher temperatures, positive at lower temperatures
negative negative negative at lower temperatures, positive at higher temperatures
positive negative always positive
Keep in mind that the temperature in the Gibbs free energy equation is the Kelvin temperature, so it can only have a positive value. When $\Delta H$ is negative and $\Delta S$ is positive, the sign of $\Delta G$ will always be negative, and the reaction will be spontaneous at all temperatures. This corresponds to both driving forces being in favor of product formation. When $\Delta H$ is positive and $\Delta S$ is negative, the sign of $\Delta G$ will always be positive, and the reaction can never be spontaneous. This corresponds to both driving forces working against product formation.
When one driving force favors the reaction, but the other does not, it is the temperature that determines the sign of $\Delta G$. Consider first an endothermic reaction (positive $\Delta H$) that also displays an increase in entropy (positive $\Delta S$). It is the entropy term that favors the reaction. Therefore, as the temperature increases, the $T \Delta S$ term in the Gibbs free energy equation will begin to predominate and $\Delta G$ will become negative. A common example of a process which falls into this category is the melting of ice (see figure below). At a relatively low temperature (below $273 \: \text{K}$), the melting is not spontaneous because the positive $\Delta H$ term "outweighs" the $T \Delta S$ term. When the temperature rises above $273 \: \text{K}$, the process becomes spontaneous because the larger $T$ value has tipped the sign of $\Delta G$ over to being negative.
When the reaction is exothermic (negative $\Delta H$) but undergoes a decrease in entropy (negative $\Delta S$), it is the enthalpy term which favors the reaction. In this case, a spontaneous reaction is dependent upon the $T \Delta S$ term being small relative to the $\Delta H$ term, so that $\Delta G$ is negative. The freezing of water is an example of this type of process. It is spontaneous only at a relatively low temperature. Above $273. \: \text{K}$, the larger $T \Delta S$ value causes the sign of $\Delta G$ to be positive, and freezing does not occur.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/11%3A_Properties_of_Reactions/11.05%3A_Spontaneous_Reactions_and_Free_Energy.txt |
Learning Outcomes
• Define reaction rate.
• Define catalyst and explain its behavior in a chemical reaction.
• Describe how temperatures, concentration or pressure of reactant, and a catalyst affect the reaction rate.
• Explain the concept of activation energy.
• Label a diagram with reactants, products, enthalpy of forward and reverse reactions, activation energy of forward and reverse reactions, and activated complex.
• Calculate the enthalpy and activation energy with data on a reaction diagram.
Chemical kinetics is the study of the rates of chemical reactions. In this lesson, you will learn how to express the rate of a chemical reaction and about various factors that influence reaction rates.
Expressing Reaction Rate
Chemical reactions vary widely in the speeds with which they occur. Some reactions occur very quickly. It a lighted match is brought into contact with lighter fluid or another flammable liquid, it erupts into flame instantly and burns fast. Other reactions occur very slowly. A container of milk in the refrigerator will be good to drink for weeks before it begins to turn sour. Millions of years were required for dead plants under Earth's surface to accumulate and eventually turn into fossil fuels such as coal and oil.
Chemists need to be concerned with the rates at which chemical reactions occur. Rate is another word for speed. If a sprinter takes 11.0 seconds $\left( s \right)$ to run a 100 meter (\left( m \right)\) dash, his rate or speed is given by the distance traveled divided by the time (see figure below).
$\text{speed} = \frac{\text{distance}}{\text{time}} = \frac{100 \: \text{m}}{11.0 \: \text{s}} = 9.09 \: \text{m/s}$
The sprinter's average running rate for the race is $9.09 \: \text{m/s}$. We say that it is his average rate because he did not run at that speed for the entire race. At the very beginning of the race, while coming from a standstill, his rate must be slower until he is able to get up to his top speed. His top speed must the be greater than $9.09 \: \text{m/s}$ so that taken over the entire race, the average ends up at $9.09 \: \text{m/s}$.
Chemical reactions can't be measured in units of meters per second, as that would not make any sense. A reaction rate is the change in concentration of a reactant or product with time. Suppose that a simple reaction were to take place in which a 1.00 molar $\left( \text{M} \right)$ aqueous solution of substance $\ce{A}$ was converted to substance $\ce{B}$.
$\ce{A} \left( aq \right) \rightarrow \ce{B} \left( aq \right)$
Suppose that after 20.0 seconds, the concentration of $\ce{A}$ had dropped from $1.00 \: \text{M}$ to $0.72 \: \text{M}$ as it was being converted to substance $\ce{B}$. We can express the rate of this reaction as the change in concentration of $\ce{A}$ divided by the time.
$\text{rate} = -\frac{\Delta \left[ \ce{A} \right]}{\Delta t} = -\frac{\left[ \ce{A} \right]_\text{final} - \left[ \ce{A} \right]_\text{initial}}{\Delta t}$
A bracket around a symbol or formula means the concentration in molarity of that substance. The change in concentration of $\ce{A}$ is its final concentration minus its initial concentration. Because the concentration of $\ce{A}$ is decreasing over time, the negative sign is used. Thus, the rate for the reaction is positive, and the units are molarity per second or $\text{M/s}$.
$\text{rate} = -\frac{0.72 \: \text{M} - 1.00 \: \text{M}}{20.0 \: \text{s}} = -\frac{-0.28 \: \text{M}}{20.0 \: \text{s}} = 0.041 \: \text{M/s}$
Over the first 20.0 seconds of this reaction, the molarity of $\ce{A}$ decreases by an average rate of $0.041 \: \text{M}$ every second. In summary, the rate of a chemical reaction is measured by the change in concentration over time for a reactant or product. The unit of measurement for a reaction rate is molarity per second $\left( \text{M/s} \right)$.
Collision Theory
The behavior of the reactant atoms, molecules, or ions is responsible for the rates of a given chemical reaction. Collision theory is a set of principles based around the idea that reactant particles form products when they collide with one another, but only when those collisions have enough kinetic energy and the correct orientation to cause a reaction. Particles that lack the necessary kinetic energy may collide, but the particles will simply bounce off one another unchanged. The figure below illustrates the difference. In the first collision, the particles bounce off one another, and no rearrangement of atoms has occurred. The second collision occurs with greater kinetic energy, and so the bond between the two red atoms breaks. One red atom bonds with the other molecule as one product, while the single red atom is the other produce. The first collision is called an ineffective collision, while the second collision is called an effective collision.
Supplying reactant particles with energy causes the bonds between the atoms to vibrate with a greater frequency. This increase in vibrational energy makes a chemical bond more likely to break and a chemical reaction more likely to occur when those particles collide with other particles. Additionally, more energetic particles have more forceful collisions, which also increases the likelihood that a rearrangement of atoms will take place. The activation energy for a reaction is the minimum energy that colliding particles must have in order to undergo a reaction. Some reactions occur readily at room temperature because most of the reacting particles already have the requisite activation energy at that temperature. Other reactions only occur when heated because the particles do not have enough energy to react unless more is provided by an external source of heat.
Potential Energy Diagrams
Then energy changes that occur during a chemical reaction can be shown in a diagram called a potential energy diagram, sometimes called a reaction progress curve. A potential energy diagram shows the change in the potential energy of a system as reactants are converted into products. The figure below shows basic potential energy diagrams for an endothermic (left) and an exothermic (right) reaction. Recall that the enthalpy change $\left( \Delta H \right)$ is positive for an endothermic reaction and negative for an exothermic reaction. This can be seen in the potential energy diagrams. The total potential energy of the system increases for the endothermic reaction as the system absorbs energy from the surroundings. The total potential energy of the system decreases for the exothermic reaction as the system releases energy to the surroundings.
The activation energy for a reaction is illustrated in the potential energy diagram by the height of the hill between the reactants and the products. For this reason, the activation energy of a reaction is sometimes referred to as the activation energy barrier. Reacting particles must have enough energy so that when they collide, they can overcome this barrier (see figure below).
As discussed earlier, reactant particles sometimes collide with one another and yet remain unchanged by the collision. Other times, the collision leads to the formation of products. The state of the particles that is in between the reactants and products is called the activated complex. An activated complex is an unstable arrangement of atoms that exists momentarily at the peak of the activation energy barrier. Because of its high energy, the activated complex exists only for an extremely short period of time (about $10^{-13} \: \text{s}$). The activated complex is equally likely to either reform the original reactants or go on to form the products. The figure below shows the formation of a possible activated complex between colliding hydrogen and oxygen molecules. Because of their unstable nature and brief existence, very little is known about the exact structures of most activated complexes.
Factors Affecting Reaction Rates
By their nature, some reactions occur very quickly, while others are very slow. However, certain changes in the reaction conditions can have an effect on the rate of a given chemical reaction. Collision theory can be utilized to explain these rate effects.
Concentration
Increasing the concentration of one or more of the reacting substances generally increases the reaction rate. When more particles are present in a given amount of space, a greater number of collisions will naturally occur between those particles. Since the rate of a reaction is dependent on the frequency of collisions between the reactants, the rate increases as the concentration increases.
Pressure
When the pressure of a gas is increased, its particles are forced closer together, decreasing the amount of empty space between them. Therefore, an increase in the pressure of a gas is also an increase in the concentration of the gas. For gaseous reactions, an increase in pressure increases the rate of reaction for the same reasons as described above for an increase in concentration. Higher gas pressure leads to a greater frequency of collisions between reacting particles.
Surface Area
A large log placed in a fire will burn relatively slowly. If the same mass of wood were added to the fire in the form of small twigs, they would burn much more quickly. This is because the twigs provide a greater surface are than the log does. An increase in the surface area of a reactant increases the rate of a reaction. Surface area is larger when a given amount of a solid is present as smaller particles. A powdered reactant has a greater surface area than the same reactant as a solid chunk. In order to increase the surface area of a substance, it may be ground into smaller particles or dissolved into a liquid. In solution, the dissolved particles are separated from each other and will react more quickly with other reactants. The figure below shows the unfortunate result of high surface area in an unwanted combustion reaction. Small particles of grain dust are very susceptible to rapid reactions with oxygen, which can result in violent explosions and quick-burning fires.
Temperature
Raising the temperature of a chemical reaction results in a higher reaction rate. When the reactant particles are heated, they move faster and faster, resulting in a greater frequency of collisions. An even more important effect of the temperature increase is that the collisions occur with a greater force, which means the reactants are more likely to surmount the activation energy barrier and go on to form products. Increasing the temperature of a reaction increases not only the frequency of collisions, but also the percentage of those collisions that are effective, resulting in an increased reaction rate.
Paper is certainly a highly combustible material, but paper does not burn at room temperature because the activation energy for the reaction is too high. The vast majority of collisions between oxygen molecules and the paper are ineffective. However, when the paper is heated by the flame from a match, it reaches a point where the molecules now have enough energy to react. The reaction is very exothermic, so the heat released by the initial reaction will provide enough energy to allow the reaction to continue, even if the match is removed. The paper continues to burn rapidly until it is gone.
Catalysts
The rates of some chemical reactions can be increased dramatically by introducing certain other substances into the reaction mixture. Hydrogen peroxide is used as a disinfectant for scrapes and cuts, and it can be found in many medicine cabinets as a $3\%$ aqueous solution. Hydrogen peroxide naturally decomposes to produce water and oxygen gas, but the reaction is very slow. A bottle of hydrogen peroxide will last for several years before it needs to be replaced. However, the addition of just a small amount of manganese (IV) oxide to hydrogen peroxide will cause it to decompose completely in just a matter of minutes. A catalyst is a substance that increases the rate of a chemical reaction without being used up in the reaction. It accomplishes this task by providing an alternate reaction pathway that has a lower activation energy barrier. After the reaction occurs, a catalyst returns to its original state, so catalysts can be used over and over again. Because it is neither a reactant nor a product, a catalyst is shown in a chemical equation by being written above the yield arrow.
$2 \ce{H_2O_2} \left( aq \right) \overset{\ce{MnO_2}}{\rightarrow} 2 \ce{H_2O} \left( l \right) + \ce{O_2} \left( g \right)$
A catalyst works by changing the mechanism of the reaction, which can be though of as the specific set of smaller steps by which the reactants become products. The important point is that the use of a catalyst lowers the overall activation energy of the reaction (see figure below). With a lower activation energy barrier, a greater percentage of reactant molecules are able to have effective collisions, and the reaction rate increases.
Catalysts are extremely important parts of many chemical reactions. Enzymes in your body act as nature's catalysts, allowing important biochemical reactions to occur at reasonable rates. Chemical companies constantly search for new and better catalysts to make reactions go faster and thus make the company more profitable.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/11%3A_Properties_of_Reactions/11.06%3A_Rates_of_Reactions.txt |
These are homework exercises to accompany Chapter 11 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health.
Questions
(click here for solutions)
Q11.1.1
What is a free element, and what is the oxidation number for atoms that exist as a free element?
Q11.1.2
What is the highest oxidation number that sulfur can have? The lowest? Consider the number of valence electrons it has.
Q11.1.3
Determine the oxidation numbers of each of the atoms in the following.
1. KMnO4
2. OCl2
3. H2C2O4
4. Li3PO4
5. NaClO
6. Br2
7. ClF3
8. CaCl2
9. K2O
Q11.1.4
Determine the oxidation number of each of the atoms in the following ions.
1. NO2
2. NO3
3. Cr2O72
4. BrO3
5. ClO3
6. BO33
7. CO32
8. NH4+
9. CrO42
10. S2O32
(click here for solutions)
Q11.2.1
Explain why oxidation and reduction always occur together in a reaction.
Q11.2.2
What happens to the oxidizing agent in a redox reaction?
Q11.2.3
What happens to the reducing agent in a redox reaction?
Q11.2.4
Identify each process as an oxidation or a reduction.
1. Rb → Rb+
2. Te → Te2
3. 2H+ → H2
4. P3– → P
5. 2Cl → Cl2
6. Sn4+ → Sn2+
7. Br2 → Br
8. Fe2+ → Fe3+
Q11.2.5
For each equation, 1) identify the oxidation numbers of each element, 2) determine if it is a redox reaction or not, and for redox reactions, 3) identify the species being oxidized and the species being reduced, and 4) identify the oxidizing and reducing agents.
1. 2KClO3(s) → 2KCl(s) + 3O2(g)
2. H2(g) + CuO(s) → Cu(s) + H2O(l)
3. 2Al(s) + 3Mg(NO3)2(aq) → 2Al(NO3)3(aq) + 3Mg(s)
4. 2HNO3(aq) + 6HI(aq) → 2NO(g) + 3I2(s) + 4H2O(l)
5. AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
6. 2FeCl3(aq) + H2S(g) → 2FeCl2(aq) + 2HCl(aq) + S(s)
Q11.2.6
Identify the change occurring with respect to the gain or loss of hydrogen or oxygen.
a.
b.
c.
d.
e.
(click here for solutions)
Q11.3.1
Classify the following reactions as combination, decomposition, single replacement, double replacement, or combustion. For each, indicate if it is a redox reaction or not.
1. Cd(s) + H2SO4(aq) → CdSO4(aq) + H2(g)
2. 2Fe(s) + 3Cl2(g) → 2FeCl3(s)
3. C7H8(l) + 9O2(g) → 7CO2(g) + 4H2O(g)
4. 2NH4NO3(s) → 2N2(g) + O2(g) + 4H2O(g)
5. 2CoCl3(aq) + 3Pb(NO3)2(aq) → 2Co(NO3)3(aq) + 3PbCl2(s)
Q11.3.2
Balance the following equations and classify the following reactions as combination, decomposition, single replacement, double replacement, or combustion. For each, indicate if it is a redox reaction or not.
1. Na + Cl2 → NaCl
2. Na3PO4 + KOH → NaOH + K3PO4
3. P4 + O2 → P2O3
4. N2 + H2 → NH3
5. Al + HCl → H2 + AlCl3
6. H2O2 → H2O + O2
7. NH3 + CuO → Cu + N2 + H2O
8. NH4NO3 → N2O + H2O
Q11.3.3
What do the products of combustion reactions have in common?
Q11.3.4
Write and balance combustion reactions for the following compounds.
1. methane (CH4)
2. propane (C3H8)
3. octane (C8H18)
4. ethanol (CH3CH2OH)
5. sucrose (C12H22O11)
(click here for solutions)
Q11.4.1
In a certain reaction, the energy of the reactants is less than the energy of the products (reaction consumes energy). Is the reaction endothermic or exothermic?
Q11.4.2
What are the two driving forces for all chemical reactions and physical processes?
Q11.4.3
Does entropy determine the spontaneity of a reaction? Does enthalpy determine the spontaneity of a reaction?
Q11.4.4
How does an increase in temperature affect the entropy of a system?
Q11.4.5
Which system has the greater entropy in each of the following?
1. solid sodium chloride or a sodium chloride solution
2. bromine liquid or bromine vapor
3. 25 g of water at 80°C or 25 g of water at 50°C
4. liquid mercury or solid mercury
Q11.4.6
How does the entropy of a system change for each of the following processes?
1. A solid melts.
2. A liquid freezes.
3. A liquid boils.
4. A vapor condenses to a liquid.
5. Sugar dissolves in water.
6. A solid sublimes.
(click here for solutions)
Q11.5.1
What is true about the relative amounts of reactants and products at the end of a spontaneous reaction?
Q11.5.2
Can a proposed reaction be spontaneous and yet still not be observed to occur? Explain.
Q11.5.3
The forward reaction is spontaneous for a particular reversible reaction. What can you conclude about the spontaneity of the reverse reaction?
Q11.5.4
Explain how free energy is used to determine whether or not a reaction is spontaneous.
Q11.5.5
Under what conditions of enthalpy and entropy change is a reaction always spontaneous? Under what conditions is a reaction never spontaneous?
Q11.5.6
If the enthalpy change is unfavorable (endothermic), but the entropy change is favorable (increasing), would a high temperature or a low temperature be more likely to lead to a spontaneous reaction?
Q11.5.7
If the enthalpy change is favorable and the entropy change is favorable, would the reaction be spontaneous at high temperatures, low temperatures, or all temperatures?
(click here for solutions)
Q11.6.1
In what unit is the rate of a chemical reaction typically expressed?
Q11.6.2
A 2.50 M solution undergoes a chemical reaction. After 3.00 minutes, the concentration of the solution is 2.15 M. What is the rate of change in M/s?
Q11.6.3
Substance A disappears at a rate of 0.0250 M/s. If the initial concentration is 4.00 M, what is the concentration after one minute?
Q11.6.4
The concentration of product B increases from 0 to 1.75 M in 45 seconds. What is the rate of formation of B?
Q11.6.5
The concentration of product B increases from 0.50 M to 1.25 M in 2.5 seconds. What is the rate of formation of B?
Q11.6.6
Reactant B goes from 2.25 M to 1.50 M in 0.85 seconds. What is the rate of change of B?
Q11.6.7
Does every collision between reacting particles lead to the formation of products? Explain.
Q11.6.8
What two conditions must be met in order for a collision to be effective?
Q11.6.9
Explain why the activation energy of a reaction is sometimes referred to as a barrier.
Q11.6.10
Why is it difficult to study activated complexes?
Q11.6.11
Explain how reaction rates can be affected by
1. changes in concentration.
2. changes in pressure.
3. increased surface area.
4. changes in temperatures.
Q11.6.12
What is the effect of a catalyst on the rate of a reaction?
Q11.6.13
Explain how the presence of a catalyst affects the activation energy of a reaction.
Q11.6.14
Zinc metal reacts with hydrochloric acid. Which one would result in the highest rate of reaction?
1. A solid piece of zinc in 1 M HCl
2. A solid piece of zinc in 3 M HCl
3. Zinc powder in 1 M HCl
4. Zinc powder in 3 M HCl
Q11.6.15
Use the potential energy diagram below to answer the following questions.
1. What is the potential energy of the reactants?
2. What is the potential energy of the products?
3. What is the heat of reaction (ΔH = Eproducts − Ereactants)?
4. What is the potential energy of the activated complex?
5. What is the activation energy for the reaction?
6. Is the reaction endothermic or exothermic?
7. Which of the values in a-e above would be changed by the use of a catalyst in the reaction?
8. What is the activation energy of the reverse reaction?
9. What is the heat of reaction (ΔH = Eproducts − Ereactants) of the reverse reaction?
Q11.6.16
Use the potential energy diagram below to answer the following questions.
1. What is the potential energy of the reactants?
2. What is the potential energy of the products?
3. What is the heat of reaction (ΔH = Eproducts − Ereactants)?
4. What is the potential energy of the activated complex?
5. What is the activation energy for the reaction?
6. Is the reaction endothermic or exothermic?
7. Which of the values in a-e above would be changed by the use of a catalyst in the reaction?
8. What is the activation energy of the reverse reaction?
9. What is the heat of reaction (ΔH = Eproducts − Ereactants) of the reverse reaction?
Answers
11.1: Oxidation Numbers
Q11.1.1
Any element by itself, either monatomic or diatomic, without a charge. The oxidation number of any free element is zero.
Q11.1.2
Sulfur can have an oxidation number up to +6 or as low as $-2$.
Q11.1.3
1. $\overset{+1}{\text{K}}\overset{+7}{\text{Mn}}\overset{-2}{\text{O}_4}$
• Rule 2: K forms an ion with a 1+ charge so its oxidation number is +1.
• Rule 3: O has an oxidation number of $-2$.
• Rule 6 for Mn: $1+x+4(-2)=0\ 1+x+-8=0\ x-7=0\ x=7$
2. $\overset{-2}{\text{O}}\overset{+1}{\text{Cl}_2}$
• Rule 3: O has an oxidation number of $-2$.
• Rule 6 for Cl: $-2+2x=0\ 2x=-2\ x=1$
3. $\overset{+1}{\text{H}_2}\overset{+3}{\text{C}_2}\overset{-2}{\text{O}_4}$
• Rule 3: O has an oxidation number of $-2$.
• Rule 4: H has an oxidation number of \{+1\).
• Rule 6 for C: $2(+1)+2x+4(-2)=0\ 2+2x-8=0\ 2x-6=0\ 2x=6\x=3$
4. $\overset{+1}{\text{Li}_3}\overset{+5}{\text{P}}\overset{-2}{\text{O}_4}$
• Rule 2: Li forms an ion with a 1+ charge so its oxidation number is +1.
• Rule 3: O has an oxidation number of $-2$.
• Rule 6 for P: $3(+1)+x+4(-2)=0\ 3+x+-8=0\ x-5=0\ x=5$
5. $\overset{+1}{\text{Na}}\overset{+1}{\text{Cl}}\overset{-2}{\text{O}}$
• Rule 2: Na forms an ion with a 1+ charge so its oxidation number is +1.
• Rule 3: O has an oxidation number of $-2$.
• Rule 6 for Cl: $1+x+-2=0\ x-1=0\ x=1$
6. $\overset{0}{\text{Br}_2}$
• Rule 1: Oxidation number of a free element is zero.
7. $\overset{+3}{\text{Cl}} \overset{-1}{\text{F}_3}$
• Rule 5: F has an oxidation number of $-1$.
• Rule 6 for Cl: $x+3(-1)=0\ x-3=0\ x=3$
8. $\overset{+2}{\text{Ca}}\overset{-1}{\text{Cl}_2}$
• Rule 2: Ca forms an ion with a 2+ charge so its oxidation number is +2.
• Rule 6 for Cl: $2+2(x)=0\ 2x=-2\ x=-1$
9. $\overset{+1}{\text{K}_2}\overset{-2}{\text{O}}$
• Rule 2: K forms an ion with a 1+ charge so its oxidation number is +1.
• Rule 3: O has an oxidation number of $-2$.
Q11.1.4
Determine the oxidation number of each of the atoms in the following ions.
1. $\overset{+3}{\text{N}}\overset{-2}{\text{O}^-_2}$
• Rule 2: O has an oxidation number of $-2$.
• Rule 7 for N: $x+2(-2)=-1\ x-4=-1\ x=3$
2. $\overset{+5}{\text{N}}\overset{-2}{\text{O}^-_3}$
• Rule 2: O has an oxidation number of $-2$.
• Rule 7 for N: $x+3(-2)=-1\ x-6=-1\ x=5$
3. $\overset{+6}{\text{Cr}_2}\overset{-2}{\text{O}^{2-}_7}$
• Rule 2: O has an oxidation number of $-2$.
• Rule 7 for Cr: $2x+7(-2)=-2\ 2x-14=-2\ 2x=12\ x=6$
4. $\overset{+5}{\text{Br}}\overset{-2}{\text{O}^-_3}$
• Rule 2: O has an oxidation number of $-2$.
• Rule 7 for Br: $x+3(-2)=-1\ x-6=-1\ x=5$
5. $\overset{+5}{\text{Cl}}\overset{-2}{\text{O}^-_3}$
• Rule 2: O has an oxidation number of $-2$.
• Rule 7 for Cl: $x+3(-2)=-1\ x-6=-1\ x=5$
6. $\overset{+3}{\text{B}}\overset{-2}{\text{O}^-_3}$
• Rule 2: O has an oxidation number of $-2$.
• Rule 7 for B: $x+3(-2)=-3\ x-6=-3\ x=3$
7. $\overset{+4}{\text{C}}\overset{-2}{\text{O}^{2-}_3}$
• Rule 2: O has an oxidation number of $-2$.
• Rule 7 for C: $x+3(-2)=-2\ x-6=-2\ x=4$
8. $\overset{-3}{\text{N}}\overset{+1}{\text{H}^+_4}$
• Rule 2: H has an oxidation number of $+1$.
• Rule 7 for N: $x+4(+1)=+1\ x+4=1\ x=-3$
9. $\overset{+6}{\text{Cr}}\overset{-2}{\text{O}^{2-}_4}$
• Rule 2: O has an oxidation number of $-2$.
• Rule 7 for Cr: $x+4(-2)=-2\ x-8=-2\ x=6$
10. $\overset{+2}{\text{S}_2}\overset{-2}{\text{O}^{2-}_3}$
• Rule 2: O has an oxidation number of $-2$.
• Rule 7 for S: $2x+3(-2)=-2\ 2x-6=-2\ 2x=4\ x=2$
11.2: The Nature of Oxidation and Reduction
Q11.2.1
Oxidation and reduction are opposite processes. Electrons are lost by one substance and gained by another so they have a transfer of electrons.
Q11.2.2
The oxidizing agent is reduced.
Q11.2.3
The reducing agent is oxidized.
Q11.2.4
1. oxidation
2. reduction
3. reduction
4. oxidation
5. oxidation
6. reduction
7. reduction
8. oxidation
Q11.2.5
1. 2KClO3(s) → 2KCl(s) + 3O2(g)
1. $\overset{+1}{2\text{K}}\overset{+5}{\text{Cl}}\overset{-2}{\text{O}_3}\rightarrow \overset{+1}{2\text{K}}\overset{-1}{\text{Cl}}~+~\overset{0}{3\text{O}_2}$
2. redox
3. oxygen is being oxidized, chlorine is being reduced
4. KClO3 is the oxidizing and reducing agent (specifically, the oxygen is the reducing agent and the chlorine is the oxidizing agent)
2. H2(g) + CuO(s) → Cu(s) + H2O(l)
1. $\overset{0}{\text{H}_2}~+~\overset{+2}{\text{Cu}}\overset{-2}{\text{O}}\rightarrow \overset{0}{\text{Cu}}~+~\overset{+1}{\text{H}_2}\overset{-2}{\text{O}}$
2. redox
3. hydrogen is being oxidized, copper is being reduced
4. CuO is the oxidizing agent, H2 is the reducing agent
3. 2Al(s) + 3Mg(NO3)2(aq) → 2Al(NO3)3(aq) + 3Mg(s)
1. $overset{0}{2\text{Al}}~+~\overset{+2}{3\text{Mg}}\overset{+5} {\text{(N}}\overset{-2}{\text{O}_3\text{)}_2} \rightarrow \overset{+3}{2\text{Al}}\overset{+5} {\text{(N}}\overset{-2}{\text{O}_3\text{)}_3}~+~\overset{0}{3\text{Mg}}$
2. redox
3. aluminum is being oxidized, magnesium is being reduced
4. Mg(NO3)2 is the oxidizing agent, Al is the reducing agent
4. 2HNO3(aq) + 6HI(aq) → 2NO(g) + 3I2(s) + 4H2O(l)
1. $\overset{+1}{2\text{H}}\overset{+5}{\text{N}}\overset{-2}{\text{O}_3}~+~\overset{+1}{6\text{H}}\overset{-1}{\text{I}} \rightarrow \overset{+2}{2\text{N}}\overset{-2}{\text{O}}~+~\overset{0} {3\text{I}_2}~+~\overset{+1}{4\text{H}_2}\overset{-2}{\text{O}}$
2. redox
3. iodine is being oxidized, nitrogen is being reduced
4. HNO3 is the oxidizing agent, HI is the reducing agent
5. AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
1. $\overset{+1}{\text{Ag}}\overset{+5}{\text{N}}\overset{-2}{\text{O}_3}~+~\overset{+1}{\text{Na}}\overset{-1}{\text{Cl}} \rightarrow \overset{+1}{\text{Ag}}\overset{-1}{\text{Cl}}~+~\overset{+1}{\text{Na}}\overset{+5}{\text{N}}\overset{-2}{\text{O}_3}$
2. not redox
6. 2FeCl3(aq) + H2S(g) → 2FeCl2(aq) + 2HCl(aq) + S(s)
1. $\overset{+3}{2\text{Fe}}\overset{-1}{\text{Cl}_3}~+~\overset{+1}{\text{H}_2}\overset{-2}{\text{S}} \rightarrow \overset{+2}{2\text{Fe}}\overset{-1}{\text{Cl}_2}~+~\overset{+1}{2\text{H}}\overset{-1}{\text{Cl}}~+~\overset{0}{\text{S}}$
2. redox
3. sulfur is being oxidized, iron is being reduced
4. FeCl3 is the oxidizing agent, H2S is the reducing agent
Q11.2.6
1. Oxidation due to loss of hydrogen.
2. Reduction due to loss of oxygen.
3. Reduction due to loss of oxygen or gain of hydrogen.
4. Reduction due to gain of hydrogen.
5. Reduction due to gain of hydrogen.
11.3: Types of Chemical Reactions
Q11.3.1
1. single replacement (single displacement), redox
2. combination (synthesis), redox
3. combustion, redox
4. decomposition, redox
5. double replacement (double displacement), not redox
Q11.3.2
1. 2Na + Cl2 → 2NaCl, combination (synthesis), redox
2. Na3PO4 + 3KOH → 3NaOH + K3PO4, double replacement (double displacement), not redox
3. P4 + 3O2 → 2P2O3, combination (synthesis), redox
4. N2 + 3H2 → 2NH3, combination (synthesis), redox
5. 2Al + 6HCl → 3H2 + 2AlCl3, single replacement (single displacement), redox
6. 2H2O2 → 2H2O + O2, decomposition, redox
7. 2NH3 + 3CuO → 3Cu + N2 + 3H2O, decomposition, redox
8. NH4NO3 → N2O + 2H2O, decomposition, not redox
Q11.3.3
Combustion produces CO2 and H2O.
Q11.3.4
1. CH4 + 2O2 → CO2 + 2H2O
2. C3H8 + 5O2 → 3CO2 + 4H2O
3. 2C8H18 + 25O2 → 16CO2 + 18H2O
4. CH3CH2OH + 3O2 → 2CO2 + 3H2O
5. C12H22O11 + 12O2 → 12CO2 + 11H2O
11.4: Entropy
Q11.4.1
endothermic
Q11.4.2
enthalpy and entropy
Q11.4.3
Neither entropy nor enthalpy determine the spontaneity of a reaction but both contribute to determining the spontaneity.
Q11.4.4
Entropy increases with temperature.
Q11.4.5
Which system has the greater entropy in each of the following?
1. solution
2. vapor
3. 80°C
4. liquid
Q11.4.6
How does the entropy of a system change for each of the following processes?
1. increases
2. decreases
3. increasese
4. decreases
5. increases
6. increases
11.5: Spontaneous Reactions and Free Energy
Q11.5.1
There are more products than reactants.
Q11.5.2
Yes, because it is a very slow reaction.
Q11.5.3
The reverse reaction is not spontaneous.
Q11.5.4
Free energy ($\Delta G$) is negative for spontaneous reactions and positive for non-spontaneous reactions.
Q11.5.5
$\Delta G$ is negative (spontaneous) at all temperatures when $\Delta H$ is negative and $\Delta S$ is positive.
$\Delta G$ is positive (non-spontaneous) at all temperatures when $\Delta H$ is positive and $\Delta S$ is negative.
Q11.5.6
high temperature
Q11.5.7
all temperatures
11.6: Rates of Reactions
Q11.6.1
molarity per second, $\frac{M}{s}$ (may be written as $Ms^{-1}$)
Q11.6.2
$rate=\frac{\Delta [A]}{\Delta t}\=\frac{2.15-2.50\;M}{180. \;s}\ =\frac{-0.35\;M}{180.\;s}\ =-0.0019\;\frac{M}{s}$
Q11.6.3
$60\;s\left(\frac{0.0250\;M}{s}\right)=1.50\;M\;\text{lost}\ initial + change =final\ 4.00\;M+(-1.50\;M)=2.50\;M$
Q11.6.4
$rate=\frac{\Delta [B]}{\Delta t}\=\frac{1.75-0\;M}{45 \;s}\ =0.039\;\frac{M}{s}$
Q11.6.5
$rate=\frac{\Delta [B]}{\Delta t}\=\frac{1.25-0.50\;M}{2.5 \;s}\ =\frac{0.75\;M}{2.5\;s}\ =0.30\;\frac{M}{s}$
Q11.6.6
$rate=\frac{\Delta [B]}{\Delta t}\=\frac{1.50-2.25\;M}{0.85 \;s}\ =\frac{-0.75\;M}{0.85\;s}\ =-0.88\;\frac{M}{s}$
Q11.6.7
No, because collisions have to occur at the correct orientation with sufficient energy.
Q11.6.8
Correct orientation and sufficient energy.
Q11.6.9
Because it is the energy requirement that must be met before the reaction can occur. When there is not enough energy, the reaction is blocked from occurring.
Q11.6.10
They are very short-lived.
Q11.6.11
1. Increasing concentration increases the probability of an effective collision so the reaction rate will increase. The reverse is true for a decrease in concentration.
2. Change in pressure of a substance involved in the reaction will have the same effect as a change in concentration.
3. Increasing the surface area creates more places for reactants to interact which will increase the probability of an effective collision so the reaction rate will increase.
4. Increasing the temperature will increase the energy of collisions so a greater number of collisions will have sufficient energy to overcome the activation barrier which will result in an increase in the reaction rate. Decreasing the temperature will have the reverse effect.
Q11.6.12
Catalysts increase the rate of a reaction by lowering the activation energy.
Q11.6.13
Catalysts provide an alternate mechanism or "path" for the reaction to occur. This new mechanism has a lower activation energy so more collisions have enough energy to overcome the barrier which will increase the reaction rate.
Q11.6.14
d. The powder has a higher surface area than a solid piece of zinc and the higher concentration will result in a higher reaction rate.
Q11.6.15
1. 20 kJ/mol
2. 50 kJ/mol
3. 30 kJ/mol
4. 100 kJ/mol
5. 80 kJ/mol
6. endothermic
7. d (energy of the activated complex) and e (activation energy)
8. 50 kJ/mol
9. $-$30 kJ/mol
Q11.6.16
1. 45 kJ/mol
2. 15 kJ/mol
3. $-$30 kJ/mol
4. 92 kJ/mol
5. 47 kJ/mol
6. exothermic
7. d (energy of the activated complex) and e (activation energy)
8. 77 kJ/mol
9. 30 kJ/mol | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/11%3A_Properties_of_Reactions/11.07%3A_Properties_of_Reactions_%28Exercises%29.txt |
Organic reactions are chemical reactions involving organic compounds. The basic organic chemistry reaction types are addition reactions, elimination reactions, substitution reactions, pericyclic reactions, rearrangement reactions, photochemical reactions and redox reactions.
• 12.1: Organic Reactions
Organic reactions require the breaking of strong covalent bonds, which takes a considerable input of energy. In order for relatively stable organic molecules to react at a reasonable rate, they often must be modified with the use of highly reactive materials or in the presence of a catalyst. In this lesson, you will learn about several general categories of organic reactions.
• 12.2: Organic Reactions (Exercises)
These are homework exercises to accompany Chapter 12 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Solutions are available below the questions.
12: Organic Reactions
Learning Outcomes
• Identify and describe substitution, elimination, addition, hydrolysis, and condensation reactions.
• Predict products of each reaction type.
• Predict products of hydration reactions based on Markovnikov's rule.
• Explain why condensation reactions are also called dehydration reactions.
• Define "polymer".
Organic reactions require the breaking of strong covalent bonds, which takes a considerable input of energy. In order for relatively stable organic molecules to react at a reasonable rate, they often must be modified with the use of highly reactive materials or in the presence of a catalyst. In this lesson, you will learn about several general categories of organic reactions.
Substitution Reactions
A substitution reaction, which is the same as a single replacement reaction in inorganic reactions, is a reaction in which one or more atoms in a molecule are replaced with another atom or group of atoms. Alkyl halides are formed by the substitution of a halogen atom for a hydrogen atom. When methane reacts with chlorine gas, ultraviolet light can act as a catalyst for the reaction.
$\ce{CH_4} \left( g \right) + \ce{Cl_2} \left( g \right) \overset{\text{UV light}}{\rightarrow} \ce{CH_3Cl} \left( g \right) + \ce{HCl} \left( g \right)$
The reaction produces chloromethane and hydrogen chloride. When the mixture is allowed to react for longer periods of time, further substitution reactions may occur.
$\ce{CH_3Cl} \left( g \right) + \ce{Cl_2} \left( g \right) \overset{\text{UV light}}{\rightarrow} \ce{CH_2Cl_2} \left( g \right) + \ce{HCl} \left( g \right)$
The product above is dichloromethane. Further substitution products trichloromethane and tetrachloromethane, commonly called carbon tetrachloride. A mixture of products occurs in the reaction, with the relative amounts dependent upon the time that the reaction is allowed to proceed. Chlorofluorocarbons are produced by reacting chloroalkanes with $\ce{HF}$, because the fluorine atom makes a stronger bond to the carbon atom than chlorine does.
$\ce{CCl_4} \left( g \right) + \ce{HF} \left( g \right) \overset{\ce{SbF_5}}{\rightarrow} \ce{CCl_3F} \left( g \right) + \ce{HCl} \left( g \right)$
The fluorine atom substitutes for a chlorine atom in the reaction.
Elimination Reactions
An elimination reaction involves the removal of adjacent atoms from a molecule. This results in the formation of a multiple bond and the release of a small molecule, so they are called elimination reactions. They have the general form
A typical example is the conversion of ethyl chloride to ethylene:
$CH_3CH_2Cl \rightarrow CH_2=CH_2 + HCl$
Much of the approximately 26 million tons of ethylene produced per year in the United States is used to synthesize plastics, such as polyethylene. In the above reaction, the A–B molecule eliminated is HCl, whose components are eliminated as H+ from the carbon atom on the left and Cl from the carbon on the right. When an acid is produced, as occurs here, the reaction is generally carried out in the presence of a base (such as NaOH) to neutralize the acid. Other elimination reactions will produce H2, X2 (where X = halogen), or H2O. These reactions are often referred to by more descriptive terms such as dehydrogenation (removing hydrogen) or dechlorination (removing chlorine).
Addition Reactions
An addition reaction is a reaction in which an atom or molecule is added to an unsaturated molecule, making a single product. An addition reaction can often be thought of as adding a molecule across the double bond of an alkene or across the triple bond of an alkyne. Knowing that "ation" means to add, the specific names of these reactions, such as hydrogenation, hydration, or chlorination, should make sense. Note that hydrogenation (adding H2) and hydration (adding H2O) are very different processes.
One type of addition reaction is called hydrogenation. Hydrogenation is a reaction that occurs when molecular hydrogen is added to an alkene to produce an alkane or hydrogen is added to an alkyne to produce an alkene or alkane. The reaction is typically performed with the use of a transition metal catalyst. For example, ethene reacts with hydrogen to form ethane.
$\ce{CH_2=CH_2} \left( g \right) + \ce{H_2} \left( g \right) \overset{\ce{Pt}}{\rightarrow} \ce{CH_3CH_3} \left( g \right)$
Note that the hydrogenation reaction is also a redox reaction. Ethene is reduced, because the oxidation numbers of the carbon atoms change from $-2$ to $-3$ as a result of the reaction.
Vegetable oils consist of long carbon chains with carboxyl groups on the end; these molecules are referred to as fatty acids. The carbon chains of the fatty acids in vegetable oils are unsaturated, usually containing multiple double bonds. When hydrogen gas is blown through a sample of the oil, hydrogen atoms add across the double bonds. This conversion changes the substance from a liquid oil into a solid fat. The "hydrogenated" on a food product is an indication that oil (liquid) has been converted into fat (solid) by this process. Margarine is manufactured from unsaturated vegetable oil in this way by hydrogenating some of the double bonds making it a "partially hydrogenated vegetable oil".
Addition reactions are also useful ways to introduce a new functional group into an organic molecule. Alkyl halides can be produced from an alkene by the addition of either the elemental halogen or the hydrogen halide. A monosubstituted alkyl halide can be produced by the addition of a hydrogen halide to an alkene. Shown below is the formation of chloroethane.
$\ce{CH_2=CH_2} \left( g \right) + \ce{HCl} \left( g \right) \rightarrow \ce{CH_3CH_2Cl} \left( g \right)$
When the reactant is the diatomic halogen, the product is a disubstituted alkyl halide as in the addition of bromine to ethene.
$\ce{CH_2=CH_2} \left( g \right) + \ce{Br_2} \left( l \right) \rightarrow \ce{CH_2BrCH_2Br} \left( g \right)$
The addition of bromine to an unknown organic compound can be used as a test for unsaturation in the compound. Bromine has a distinctive brownish-orange color, while most bromoalkanes are colorless. When bromine is slowly added to a solution of the compound, the orange color will fade if it undergoes an addition reaction to produce an alkyl halide. If the orange color remains, then the original compound was already saturated, and no reaction occurred.
A hydration reaction is a reaction in which water is added to an alkene. Hydration reactions can take place when the alkene and water are heated to near $100^\text{o} \text{C}$ in the presence of a strong acid, which acts as a catalyst. Shown below is the hydration of ethene to produce ethanol.
$\ce{CH_2=CH_2} \left( g \right) + \ce{H_2O} \left( l \right) \rightarrow \ce{CH_3CH_2OH} \left( g \right)$
Unlike addition reactions involving $\ce{H_2}$ or $\ce{Br_2}$, the addition of water can have two possible products because an $\ce{-H}$ and an $\ce{-OH}$ are being added to the carbons in the double bond. Markovnikov's rule helps predict the main product in a hydration reaction. The rule states that the hydrogen atom from water will add to the carbon that originally had more hydrogen atoms. For example, look at the hydration of 1-butene.
Note that the first carbon in 1-butene started with two hydrogen atoms and the second carbon started with one hydrogen. Therefore, the hydrogen from water adds to the first carbon and the $\ce{-OH}$ group adds to the second carbon. When there are equal numbers of hydrogen atoms on both carbons in a double bond then the two products will form in approximately equal amounts. For example, the hydration of 2-pentene results in two products. In the first product, the $\ce{-OH}$ group is on the third carbon and in the second product, the $\ce{-OH}$ group is on the second carbon. While these two molecules will have similar properties, there will be differences.
Condensation Reactions
A condensation reaction is a reaction in which two molecules combine to form a single molecule. A small molecule, often water, is usually removed during a condensation reaction. Amino acids are important biological molecules that have an amine functional group on one end of the molecule and a carboxylic acid functional group on the other end. When two amino acids combine in a condensation reaction, a covalent bond forms between the amine nitrogen of one amino acid and the carboxyl carbon of the second amino acid. A molecule of water is then removed as a second product (see figure below).
This reaction forms a molecule called a dipeptide, and the resulting carbon-nitrogen covalent bond is often called a peptide bond. When repeated numerous times, a long molecule called a protein is eventually produced.
An esterification is a condensation reaction in which an ester is formed from an alcohol and a carboxylic acid. Esterification is a subcategory of condensation reactions because a water molecule is produced in the reaction. The reaction can be catalyzed by a strong acid, usually sulfuric acid. When the carboxylic acid, butanoic acid, is heated with an excess of methanol and a few drops of sulfuric acid, the ester methyl butanoate is produced. Methyl butanoate has the scent of pineapples. The reaction is shown below with both molecular and structural formulas. Esterification reactions are reversible.
$\ce{CH_3CH_2CH_2COOH} + \ce{HOCH_3} \overset{\ce{H_2SO_4}}{\rightarrow} \ce{CH_3CH_2CH_2COOCH_3} + \ce{H_2O}$
Saponification describes the alkaline hydrolysis reaction of an ester. The term saponification originally described the hydrolysis of long-chain esters called fatty acid esters to produce soap molecules, which are the salts of fatty acids. One such soap molecule is sodium stearate, formed from the hydrolysis of ethyl stearate.
$\begin{array}{ccccccc} \ce{C_{17}H_{35}COOC_2H_5} & + & \ce{NaOH} & \rightarrow & \ce{C_{17}H_{35}COO^-Na^+} & + & \ce{C_2H_5OH} \ \text{ethyl stearate} & & & & \text{sodium stearate (soap)} & & \end{array}$
Hydrolysis Reactions
Hydrolysis is the reverse of condensation. "Hydro" indicates that water is involved and "lysis" means to break apart. In a hydrolysis reaction, water is added and the molecule breaks apart, usually at a $\ce{C-O-C}$ linkage.
Polymerization
Polymers are very different than the other kinds of organic molecules that you have seen so far. Whereas other compounds are of relatively low molar mass, polymers are giant molecules of very high molar mass. Polymers are the primary components of all sorts of plastics and related compounds. A polymer is a large molecule formed of many smaller molecules covalently bonded to one another in a repeating pattern. The small molecules that make up the polymer are called monomers. Polymers are generally formed by either addition or condensation reactions. Teflon (see figure below) is a non-reactive, non-stick coating used on cookware as well as in containers and pipes for reactive or corrosive chemicals.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/12%3A_Organic_Reactions/12.01%3A_Organic_Reactions.txt |
These are homework exercises to accompany Chapter 12 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Answers are below the questions.
Questions
(click here for solutions)
Q12.1.1
Explain the difference between hydrolysis and condensation reactions.
Q12.1.2
What is being added in a hydrogenation reaction? Hydration? Bromination?
Q12.1.3
What is the product of the oxidation of a primary alcohol? A secondary alcohol?
Q12.1.4
Why is a condensation reaction also called a dehydration reaction?
Q12.1.5
Draw the structure for the primary product of each reaction..
1. CH3CH=CH2 + Br2
2. CH3CH2CH=CH2 + H2
3. CH3CH=CHCH3 + H2O →
4. CH3CH2CH=CH2 + H2O →
5. CH3CH2CH=CHCH3 + H2O →
Q12.1.6
Draw the structure of the compound that would be formed from the single oxidation of each compound.
a.
b.
c.
Q12.1.7
Draw the structure of the compound that would be formed from the complete oxidation of each compound.
a.
b.
c.
Q12.1.8
Draw the structure of the compound that would be formed from a single reduction of each compound.
a.
b.
c.
Q12.1.9
Draw the product(s) for the hydrolysis of the given reactant. The red bond in the reactant is the one that is broken.
Q12.1.10
Draw the product(s) for the condensation of the given reactants. The layout of the molecules indicates where the bond will form between them. If there are three molecules, two condensations will occur to produce one final product.
Q12.1.11
Compare the reactants and products in part d and e of 12.1.10. Are the reactants the same or different? Are the products the same or different? Why?
Answers
12.1: Organic Reactions
Q12.1.1
Hydrolysis is a molecule breaking apart after the addition of water. Condensation is two molecules coming together and producing H2O.
Q12.1.2
hydrogenation = adding hydrogen (H2)
hydration = adding water
bromination = adding bromine (Br2)
Q12.1.3
A primary alcohol oxidizes to an aldehyde and then to a carboxylic acid. A secondary alcohol oxidizes to a ketone.
Q12.1.4
Because water is released during a condensation reaction.
Q12.1.5
a.
Each bromine atom is added to one of the carbons in the double bond.
b.
CH3CH2CH2CH3
Each hydrogen atom is added to one of the carbons in the double bonds.
c.
Water is added as H and OH. The carbons in the double bond each have the same number of hydrogen atoms so Markovnikov's rule does not apply. The molecule is symmetrical so adding the OH group to either carbon results in the same molecule.
d.
Water is added as H and OH. The carbon atoms in the double bond have different number of hydrogen atoms so Markovnikov's rule does apply. The H is added to the carbon with more hydrogen atoms and the OH is added to the carbon with fewer hydrogen atoms.
e.
and
Water is added as H and OH. The carbon atoms in the double bond have the same number of hydrogen atoms on each so Markovnikov's rule does not apply. There are two possible products because adding the OH to the carbon on the left side of the double bond results in a different molecule than adding the OH to the carbon on the right side of the double bond. This differs from c because the molecule is not symmetrical.
Q12.1.6
a.
b.
c.
Q12.1.7
a.
b.
c.
Q12.1.8
a.
b.
c.
Q12.1.9
a.
b.
c.
d.
Q12.1.10
a.
b.
c.
d.
e.
Q12.1.11
The reactants are the same, but the products are different because they are combined in a different order. | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/12%3A_Organic_Reactions/12.02%3A_Organic_Reactions_%28Exercises%29.txt |
Amino acids are molecules containing an amine group(NH2), a carboxylic acid group(R-C=O-OH) and a side-chain( usually denoted as R) that varies between different amino acids. They are particularly important in biochemistry, where the term usually refers to alpha-amino acids. Proteins are biochemical compounds consisting of one or more polypeptides typically folded into a globular or fibrous form in a biologically functional way.
• 13.1: Amino Acids
An amino acid is a compound that contains both an amine group and a carboxyl group in the same molecule. While any number of amino acids can possibly be imagined, biochemists generally reserve the term for a group of 20 amino acids which are formed and used by living organisms.
• 13.2: Peptides
A peptide is a combination of amino acids in which the amine group of one amino acid has undergone a reaction with the carboxyl group of another amino acid. The reaction is a condensation reaction, forming an amide group.
• 13.3: Protein Structure
A polypeptide is a sequence of amino acids between ten and one hundred in length. A protein is a peptide that is greater than one hundred amino acids in length. The three-dimensional structure of a protein is very critical to its function. This structure can be broken down into four levels.
• 13.4: Amino Acids and Proteins (Exercises)
These are homework exercises to accompany Chapter 13 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Solutions are available below the questions.
13: Amino Acids and Proteins
Learning Outcomes
• Identify structural components of an amino acid.
• Define zwitterion and isoelectric point.
• Determine the charge on an amino acid when it is not at the isoelectric point.
• Label amino acids as polar and nonpolar and as acidic, basic, or neutral.
Athletics are very competitive these days at all levels, from school sports to the pros. Everybody is looking for that edge that will make them faster, stronger, more physically fit. One approach taken by many athletes is the use of amino acid supplements. The theory is that the increase in amino acids in the diet will lead to increased protein for muscles. However, the only real benefit comes to the people who make and sell the pills. Studies have not showed any advantage obtained by the athletes themselves. You're much better off just maintaining a healthy diet.
Amino Acids
An amino acid is a compound that contains both an amine group $\left( \ce{-NH_2} \right)$ and a carboxyl group $\left( \ce{-COOH} \right)$ in the same molecule. While any number of amino acids can possibly be imagined, biochemists generally reserve the term for a group of 20 amino acids which are formed and used by living organisms. The figure below shows the general structure of an amino acid. Either structure is considered correct for an amino acid.
The amine and carboxyl groups of an amino acid are both covalently bonded to a central carbon atom. That carbon atom is also bonded to a hydrogen atom and an $\ce{R}$ group. It is this $\ce{R}$ group which varies from one amino acid to another and is called the amino acid side chain.
The nature of the side chains accounts for the variability in physical and chemical properties of the different amino acids. Each amino acid is grouped based on the properties of the side chain. The groups are designated as polar (hydroxylic, sulfur-containing, amidic) , nonpolar (aliphatic and aromatic), acidic, or basic.
In addition to the full name of the amino acid, there are also one-letter and three-letter abbreviations for each. These abbreviations are especially helpful when listing the amino acids in a protein (a chain of many amino acids that will be discussed later).
Rules for classifying amino acids
The following rules (along with two exceptions) can help you classify amino acids as nonpolar, polar acidic (sometimes called acidic), polar basic (sometimes called basic), or polar neutral. We will look at two exceptions but note that the transition from nonpolar to polar neutral is a gradual transition (like the colors of a rainbow) so you may see variations in how amino acids are classified if you look at other sources.
1. Nonpolar amino acids (there are 9) contain aliphatic (hydrocarbon) chains or aromatic rings.
2. Polar acidic amino acids (2) contain a carboxylic acid (or carboxylate) group in the side chain (R group). This is in addition to the one in the backbone of the amino acid.
3. Polar basic amino acids (3) contain an amine (may be neutral or charged) group in the side chain (R group). This is in addition to the one in the backbone of the amino acid.
4. Polar neutral amino acids (6) contain a hydroxyl (-OH), sulfur, or amide in the R group).
There are two important exceptions to the above rules.
1. Tyrosine has an aromatic group and an -OH group and is considered polar neutral.
2. Methionine contains a sulfur but as a part of carbon chain. Sulfur has the same electronegativity as carbon, so it is considered nonpolar.
Zwitterion
Amino acids are typically drawn either with no charges or with a plus and minus charge (see figure 13.1.1). When an amino acid contains both a plus and a minus charge in the "backbone", it is called a zwitterion and has an overall neutral charge. The zwitterion of an amino acid exists at a pH equal to the isoelectric point. Each amino acid has its own pI value based on the properties of the amino acid. At pH values above or below the isoelectric point, the molecule will have a net charge which depends on its pI value as well as the pH of the solution in which the amino acid is found.
pH < pI
When pH is less than pI, there is an excess amount of $\ce{H^+}$ in solution. The excess $\ce{H^+}$ is attracted to the negatively charged carboxylate ion resulting in its protonation. The carbohydrate ion is protonated, making it neutral, leaving only a positive charge on the amine group. Overall, the amino acid will have a charge of $+1$.
pH > pI
When pH is greater than pI, there is an excess amount of $\ce{OH^-}$ in solution. The excess $\ce{OH^-}$ is attracted to the positively charged amine group resulting in the removal of an $\ce{H^+}$ ion to form (\ce{H_2O}\). The amine group has a neutral charge leaving only a negative charge on the carboxylate group. Overall, the amino acid will have a charge of $-1$.
Example $1$
1. Identify the amino acid pictured below.
2. Find the pI value for the amino acid.
3. Determine how the amino acid will exist at pH = 3.52
4. Determine how the amino acid will exist at pH = 9.34
5. Determine how the amino acid will exist at pH = 5.02
Solution:
a. Look at the side chain to identify the amino acid. The side chain contains $\ce{-CH_2SH}$ which matches the structure of cysteine.
b. The pI values for amino acids are found in the table of amino acids. For cysteine, pI = 5.02.
c. At pH = 3.52, the $\ce{H^+}$ concentration is high (low pH = more acidic = more $\ce{H^+}$). Therefore the $\ce{H^+}$ will add to the carboxylate ion and neutralize the negative charge. The amino acid will have a positive charge on the amine group left and will have an overall charge of $+1$.
d. At pH = 9.34, the $\ce{OH^-}$ concentration is high (high pH = more basic = less $\ce{H^+}$ = more $\ce{OH^-}$). Therefore the $\ce{OH^-}$ will be attracted to the positively charged amine group and will "steal" an $\ce{H^+}$ from it. As a result, the only remaining charge will be on the carboxylate ion so the amino acid will have a $-1$ charge.
e. At pH = 5.02, the pH = pI so the amino acid will exist as the zwitterion with both the positive and negative charges as shown above.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/13%3A_Amino_Acids_and_Proteins/13.01%3A_Amino_Acids.txt |
Learning Outcomes
• Define polypeptide.
• Identify amide bond.
• Predict product of condensation of amino acids.
• Name polypeptides given the abbreviation of the amino acids.
Peptide cells in our bodies have an intricate mechanism for the manufacture of proteins. Humans have to use other techniques in order to synthesize the same proteins in a lab. The chemistry of peptide synthesis is complicated. Both active groups on an amino acid can react and the amino acid sequence must be a specific one in order for the protein to function. Robert Merrifield developed the first synthetic approach for making proteins in the lab, a manual approach which was lengthy and tedious (and, he won the Nobel Prize in Chemistry in 1984 for his work). Today, however, automated systems can crank out a peptide in a very short period of time.
Peptides
A peptide is a combination of amino acids in which the amine group of one amino acid has undergone a reaction with the carboxyl group of another amino acid. The reaction is a condensation reaction, forming an amide group $\left( \ce{CO-N} \right)$, shown below.
A peptide bond is the amide bond that occurs between the amine nitrogen of one amino acid and the carboxyl carbon of another amino acid. The resulting molecule is called a dipeptide. Notice that the particular side chains of each amino acid are irrelevant since the $\ce{R}$ groups are not involved in the peptide bond.
The dipeptide has a free amine group on one end of the molecule (known as the $\ce{N}$-terminus) and a free carboxyl group on the other end (known as the $\ce{C}$-terminus). Each is capable of extending the chain through the formation of another peptide bond. The particular sequence of amino acids in a longer chain is called an amino acid sequence. By convention, the amino acid sequence is listed in the order such that the free amine group is on the left end of the molecule and the free carboxyl group is on the right end of the molecule. For example, suppose that a sequence of the amino acids glycine, tryptophan, and alanine is formed with the free amine group as part of the glycine and the free carboxyl group as part of the alanine. The amino acid sequence can be easily written using the abbreviations as Gly-Trp-Ala. This is a different sequence from Ala-Trp-Gly because the free amine and carboxyl groups would be on different amino acids in that case.
Example $1$
Draw the polypeptide Asp-Val-Ser.
Solution
1. Identify the structures of each of the three given amino acids and draw them in the same order as given in the name.
2. Leaving the order the same, connect the amino acids to one another by forming peptide bonds. Note that the order given in the name is the same way the amino acids are connected in the molecule. The first one listed is always the $\ce{N}$-terminus of the polypeptide.
Example $2$
List all of the possible polypeptides that can be formed from cysteine (Cys), leucine (Leu), and arginine (Arg).
Solution
Although there are only three amino acids, the order in which they are bonded changes the identity, properties, and function of the resulting polypeptide. There are six possible polypeptides formed from these three amino acids.
Cys-Leu-Arg
Cys-Arg-Leu
Leu-Cys-Arg
Leu-Arg-Cys
Arg-Cys-Leu
Arg-Leu-Cys
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/13%3A_Amino_Acids_and_Proteins/13.02%3A_Peptides.txt |
Learning Outcomes
• Describe the four levels of protein structure.
• Identify the two types of secondary structure in proteins.
• Describe the interactive forces in each level of protein structure.
• Distinguish between globular and fibrous proteins.
• Define denaturation of proteins.
• Identify ways in which proteins are denatured.
Hemoglobin is a complex protein which has a quaternary structure and contains iron. There are four subunits in the hemoglobin molecule - two alpha subunits and two beta subunits. Each subunit contains one iron ion, whose oxidation state changes from $+2$ to $+3$ and back again, depending upon the environment around the iron. When oxygen binds to the iron, the three-dimensional shape of the molecule changes. Upon release of the oxygen to the cells, the shape changes again.
With hemoglobin of normal structure, this shift in conformation does not present any problems. However, individuals with hemoglobin S do experience serious complications. This hemoglobin has one amino acid in the two beta chains that is different from the amino acid at that point in the primary structure of normal hemoglobin. The result of this one structural change is aggregation of the individual protein molecules when oxygen is released. Adjacent hemoglobin molecules come in contact with one another and clump up, causing the red cells to deform and break.
This abnormality, known as sickle cell, is genetic in nature. A person may inherit the gene from one parent and have sickle cell trait (only some of the hemoglobin is hemoglobin S), which is usually not life-threatening. Inheriting the gene from both parents will result in sickle cell disease, a very serious condition.
Proteins
A polypeptide is a sequence of amino acids between ten and one hundred in length. A protein is a peptide that is greater than one hundred amino acids in length. Proteins are very prevalent in living organisms. Hair, skin, nails, muscles, and the hemoglobin in red blood cells are some of the important parts of your body that are made of different proteins. The wide array of chemical and physiological properties of proteins is a function of their amino acid sequences. Since proteins generally consist of one hundred or more amino acids, the number of amino acid sequences that are possible is virtually limitless.
The three-dimensional structure of a protein is very critical to its function. This structure can be broken down into four levels. The primary structure is the amino acid sequence of the protein. The amino acid sequence of a given protein is unique and defines the function of that protein. Peptide bonds form the connections between the amino acids.
The secondary structure is a highly irregular sub-structure of the protein. The two most common types of protein secondary structure are the alpha helix (see figure below) and the beta sheet (see figure below). An alpha helix consists of amino acids that adopt a spiral shape. A beta pleated sheet (like a fan-folded paper) is alternating rows of amino acids that line up in a side-by-side fashion. In both cases, the secondary structures are stabilized by extensive hydrogen bonding between the side chains. The interaction of the various side chains in the amino acid, specifically the hydrogen bonding, leads to the adoption of a particular secondary structure. The hydrogen bonding occurs between amino acids that are close to each other in the primary structure.
The tertiary structure is the overall three-dimensional structure of the protein. A typical protein consists of several sections of a specific secondary structure (alpha helix or beta sheet) along with other areas in which a more random structure occurs. These areas combine to produce the tertiary structure. The tertiary structure is stabilized by forces similar to the intermolecular forces previously seen between molecules. These attractive forces include London dispersion forces (hydrophobic), hydrogen bonding, dipole-dipole forces, ion-dipole interactions, ionic salt bridges, and disulfide bonds (see figure below).
Some protein molecules consist of multiple protein subunits. The quaternary structure of a protein refers to the specific interaction and orientation of the subunits of that protein. The quaternary structure is a result of the same types of interactions as seen in tertiary structure but between different subunits. Quaternary structures can have different numbers of subunits. For example, hemoglobin contains four subunits while insulin contains two subunits.
Hemoglobin is a very large protein found in red blood cells and whose function is to bind and carry oxygen throughout the bloodstream. Hemoglobin consists of four subunits - two $\alpha$ subunits (yellow) and two $\beta$ subunits (gray) - which then come together in a specific and defined way through interactions of the side chains (see figure below). Hemoglobin also contains four iron atoms, located in the middle of each of the four subunits. The iron atoms are part of a structure called a porphyrin, shown in red in the figure.
Some proteins consist of only one subunit and thus do not have a quaternary structure. The figure below diagrams the interaction of the four levels of protein structure.
Globular and Fibrous Proteins
Once proteins form and have developed all levels of their structure, they can be classified as either fibrous or globular. These classifications give the basic shape of the entire protein molecule. While many proteins are globular proteins (see figure below), keratin proteins are fibrous (see figure below) and make up the hair, nails, and the outer layer of skin.
Denaturation of Proteins
The highly organized structures of proteins are truly masterworks of chemical architecture. But highly organized structures tend to have a certain delicacy, and this is true of proteins. Denaturation is the term used for any change in the three-dimensional structure of a protein that renders it incapable of performing its assigned function. A denatured protein cannot do its job because there is a change in the secondary, tertiary, or quaternary structure (see figure below). A wide variety of reagents and conditions, such as heat, organic compounds, pH changes, and heavy metal ions can cause protein denaturation. Anyone who has fried an egg has observed denaturation. The clear egg white turns opaque as the albumin denatures and coagulates.
The primary structures of proteins are quite sturdy. In general, fairly vigorous conditions are needed to hydrolyze peptide bonds. At the secondary through quaternary levels, however, proteins are quite vulnerable to attack, though they vary in their vulnerability to denaturation. The delicately folded globular proteins are much easier to denature than are the tough, fibrous proteins of hair and skin.
There are a variety of ways to denature proteins including those below.
• Heat above $50^\text{o} \text{C}$
• Strong acids
• Strong bases
• Ionic compounds (i.e. $\ce{NaCl}$)
• Reducing agents
• Detergents
• Heavy metal ions
• Agitation
If you have ever had a hair permanent or chemically straightened your hair, the process involved the denaturation of proteins. The reducing agent (usually an ammonium compound) breaks the disulfide bonds in the hair. The hair is then curled or straightened which aligns the amino acids in a different pattern. An oxidizing agent is applied and the disulfide bonds reform between different amino acids. The change is permanent for the hair that you have at the time but new hair growing in will have the structure of the original proteins and your hair is back to its normal state.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/13%3A_Amino_Acids_and_Proteins/13.03%3A_Protein_Structure.txt |
These are homework exercises to accompany Chapter 13 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health.
Questions
(click here for solutions)
Q13.1.1
Read the material at http://hyperphysics.phy-astr.gsu.edu/hbase/organic/essam.html and answer the following questions:
a. What are essential amino acids?
b. What are nonessential amino acids?
c. What happens if you are deficient in an amino acid?
Q13.1.2
Draw the functional groups present in all amino acids.
Q13.1.3
Complete the following for threonine, lysine, and tyrosine.
a. Draw the amino acid.
b. Circle the side chain.
c. Identify whether it is polar, nonpolar, acidic, or basic.
d. At what pH will it exist as a zwitterion?
e. What is the range of pH values when it will be positively charged?
f. What is the range of pH values when it will be negatively charged?
(click here for solutions)
Q13.2.1
Draw the two dipeptides formed from each pair of amino acids.
a. tyrosine and lysine
b. threonine and gluatmine
c. alanine and histidine
Q13.2.2
Draw and give the full names of the amino acids in the following dipeptides.
Q13.2.3
List of all of the possible polypeptides that can be formed from threonine, alanine, and phenylalanine (use three character abbreviations for each amino acid).
Q13.2.4
Draw the following polypeptides.
a. Ser-Tyr-Gln
b. Lys-Met-Gly
Q13.2.5
Identify each of the amino acids in the polypeptide and then name it using the three character abbreviations.
(click here for solutions)
Q13.3.1
Describe the four levels of protein structure.
Q13.3.2
What levels of structure involve hydrogen bonding?
Q13.3.3
What types of structure is the result of interactions between amino acids that are far apart in the primary structure?
Q13.3.4
What types of interactions hold the secondary structure together?
Q13.3.5
What types of interactions hold the tertiary structure together?
Q13.3.6
What levels of structure are affected by denaturation?
Q13.3.7
A protein has one subunit. Would it have a quaternary structure?
Answers
13.1: Amino Acids
Q13.1.1
a. Essential amino acids are those you get from your diet.
b. Nonessential amino acids are produced in the body.
c. Illness and/or degradation of body's proteins.
Q13.1.2
amine and carboxylic acid
Q13.1.3
Complete the following for threonine, lysine, and tyrosine.
threonine
1. polar
2. 5.60
3. < 5.60
4. > 5.60
lysine
1. basic
2. 9.47
3. < 9.47
4. > 9.47
tyrosine
1. polar
2. 5.63
3. < 5.63
4. > 5.63
13.2: Peptides
Q13.2.1
Draw the two dipeptides formed from each pair of amino acids.
a.
b.
c.
Q13.2.2
a. alanine glycine
b. proline phenylalanine
c. tryptophan lysine
Q13.2.3
Thr-Ala-Phe
Thr-Phe-Ala
Ala-Thr-Phe
Ala-Phe-Thr
Phe-Ala-Thr
Phe-Thr-Ala
Q13.2.4
a.
b.
Q13.2.5
Arg-His-Thr-Glu-Ser
13.3: Protein Structure
Q13.3.1
Primary - sequence of amino acids
Secondary - alpha helix and Beta-pleated sheets held together by hydrogen bonds
Tertiary - third level of structure of protein often forming globular or fibrous structure, held together by variety of attractive forces
Quaternary - complex of multiple proteins held together to function as one, held together by variety of attractive forces (same as tertiary)
Q13.3.2
secondary, tertiary, and quaternary structures
Q13.3.3
tertiary structures
Q13.3.4
hydrogen bonds
Q13.3.5
London dispersion forces, hydrogen bonds, dipole-dipole forces, ion-dipole interactions, salt bridges, and disulfide bonds
Q13.3.6
secondary, tertiary, and quaternary
Q13.3.7
No, a quaternary structure must have multiple subunits. | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/13%3A_Amino_Acids_and_Proteins/13.04%3A_Amino_Acids_and_Proteins_%28Exercises%29.txt |
Biomolecules include large macromolecules (or polyanions) such as proteins, carbohydrates, lipids, and nucleic acids, as well as small molecules such as primary metabolites, secondary metabolites, and natural products.
• 14.1: Enzymes
Most chemical reactions within organisms would be impossible under the conditions in cells. e.g., the body temperature of most organisms is too low for reactions to occur quickly enough to carry out life processes. Reactants may also be present in such low concentrations that it is unlikely they will meet and collide. Therefore, the rate of most biochemical reactions must be increased by a catalyst, which speeds up chemical reactions. In organisms, catalysts are called enzymes.
• 14.2: Lipids and Triglycerides
A lipid is an organic compound such as fat or oil. Organisms use lipids to store energy, but lipids have other important roles as well. Lipids consist of repeating units called fatty acids. There are two types: saturated fatty acids and unsaturated fatty acids. Lipids may consist of fatty acids alone, or they may contain other molecules as well. Some lipids contain alcohol or phosphate groups. Examples include triglycerides, phospholipids, and steroids.
• 14.3: Phospholipids in Cell Membranes
A phospholipid is a lipid that contains a phosphate group and is a major component of cell membranes. A phospholipid consists of a hydrophilic (water-loving) head and hydrophobic (water-fearing) tail (see figure below). The phospholipid is essentially a triglyceride in which a fatty acid has been replaced by a phosphate group of some sort.
• 14.4: Biological Molecules (Exercises)
These are homework exercises to accompany Chapter 14 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health.
14: Biological Molecules
Learning Outcomes
• Explain the role of an enzyme in the body.
• Define active site, substrate, and allosteric site.
• Distinguish between competitive and noncompetitive inhibitors.
• Describe the lock and key vs. induced-fit model of enzymes.
• Provide the characteristics of a cofactor and a coenzyme.
• Describe how substrate concentration, pH, and temperature affect enzyme activity.
• Interpret graphs of reaction rate vs. reaction conditions.
The first enzyme to be isolated was discovered in 1926 by American chemist James Sumner, who crystallized the protein. The enzyme was urease, which catalyzes the hydrolytic decomposition of urea, a component of urine, into ammonia and carbon dioxide.
$\ce{H_2NCON_2} \left( aq \right) + \ce{H_2O} \left( l \right) \overset{\text{urease}}{\rightarrow} 2 \ce{NH_3} \left( g \right) + \ce{CO_2} \left( g \right)$
His discovery was ridiculed at first because nobody believed that enzymes would behave the same way that other chemicals did. Sumner was eventually proven right and won the Nobel Prize in Chemistry in 1946.
Enzymes and Biochemical Reactions
Most chemical reactions within organisms would be impossible under the conditions in cells. For example, the body temperature of most organisms is too low for reactions to occur quickly enough to carry out life processes. Reactants may also be present in such low concentrations that it is unlikely they will meet and collide. Therefore, the rate of most biochemical reactions must be increased by a catalyst. A catalyst is a chemical that speeds up chemical reactions. In organisms, catalysts are called enzymes. Essentially, enzymes are biological catalysts.
Like other catalysts, enzymes are not reactants in the reactions they control. They help the reactants interact but are not used up in the reactions. Instead, they may be used over and over again. Unlike other catalysts, enzymes are usually highly specific for particular chemical reactions. They generally catalyze only one or a few types of reactions.
Enzymes are extremely efficient in speeding up reactions. They can catalyze up to several million reactions per second. As a result, the difference in rates of biochemical reactions with and without enzymes may be enormous. A typical biochemical reaction might take hours or even days to occur under normal cellular conditions without an enzyme, but less than a second with an enzyme.
Figure $1$ diagrams a typical enzymatic reaction. A substrate is the molecule or molecules on which the enzyme acts. In the urease catalyzed reaction, urea is the substrate.
The first step in the reaction is that the substrate binds to a specific part of the enzyme molecule, known as the active site. The binding of the substrate is dictated by the shape of each molecule. Side chains on the enzyme interact with the substrate in a specific way, resulting in the making and breaking of bonds. The active site is the place on an enzyme where the substrate binds. An enzyme folds in such a way that it typically has one active site, usually a pocket or crevice formed by the folding pattern of the protein. Because the active site of an enzyme has such a unique shape, only one particular substrate is capable of binding to that enzyme. In other words, each enzyme catalyzes only one chemical reaction with only one substrate. Once the enzyme/substrate complex is formed, the reaction occurs and the substrate is transformed into products. Finally, the product molecule or molecules are released from the active site. Note that the enzyme is left unaffected by the reaction and is now capable of catalyzing the reaction of another substrate molecule.
For many enzymes, the active site follows a lock and key (A in the figure below) model where the substrate fits exactly into the active site. The enzyme and substrate must be a perfect match so the enzyme only functions as a catalyst for one reaction. Other enzymes have an induced fit (B in the figure below) model. In an induced fit model, the active site can make minor adjustments to accommodate the substrate. This results in an enzyme that is capable of interacting with a small group of similar substrates. Look at the shape of the active site compared to the shape of the substrate in B of the figure below. The active site adjusts to accommodate the substrate.
Inhibitors
An inhibitor is a molecule which interferes with the function of an enzyme, either by slowing or stopping the chemical reaction. Inhibitors can work in a variety of ways, but one of the most common is illustrated in the figure below.
A competitive inhibitor binds competitively at the active site and blocks the substrate from binding. Since no reaction occurs with the inhibitor, the enzyme is prevented from catalyzing the reaction.
A non-competitive inhibitor does not bind at the active site. It attaches at an allosteric site, which is some other site on the enzyme, and changes the shape of the protein. The allosteric site is any site on the enzyme that is not the active site. The attachment of the non-competitive inhibitor to the allosteric site results in a shift in three-dimensional structure that alters the shape of the active site so that the substrate will no longer fit in the active site properly (see figure below).
Cofactors and Coenzymes
Some enzymes require the presence of another substrate as a "helper" molecule in order to function properly. Cofactors and coenzymes serve in this role. Cofactors are inorganic species and coenzymes are small organic molecules. Many vitamins, such as B vitamins, are coenzymes. Some metal ions which function as cofactors for various enzymes include zinc, magnesium, potassium, and iron.
Catalytic Activity of Enzymes
Enzymes generally lower activation energy by reducing the energy needed for reactants to come together and react. One way that enzymes act is to bring reactants (substrates) together so they don't have to expend energy moving about until they collide at random. Enzymes bind both reactant molecules (substrates), tightly and specifically, at a site on the enzyme's active site. Enzymes can also bring molecules to the active site to break them apart. For example, sucrase is the enzyme for the breakdown of sucrose which enters the active site of the enzyme and helps weaken the interactions between the fructose and glucose that make up sucrose. Sucrase is specific to the breakdown of sucrose as are most enzymes. The active site is specific for the reactants of the biochemical reaction the enzyme catalyzes. Similar to puzzle pieces fitting together, the active site can only bind certain substrates. The activities of enzymes also depend on the temperature, concentration, and the pH of the surroudings.
Concentration
As with most reactions, the concentration of the reactant(s) affects the reaction rate. This is also true in enzyme concentration. When either substrate or enzyme concentration is low, the rate of the reaction will be slower than where there are higher concentrations. The two species must interact for a reaction to occur and higher concentrations of one or both will result in more effective interactions between the two.
However, continuing to increase the substrate's concentration will not always increase the reaction rate. This is because at some point, all of the enzymes will be occupied and unavailable to bind with another substrate molecule until the substrate forms a product molecule and is released from the enzyme.
pH
Some enzymes work best at acidic pHs, while others work best in neutral environments. For example, digestive enzymes secreted in the acidic environment (low pH) of the stomach help break down proteins into smaller molecules. The main digestive enzyme in the stomach is pepsin, which works best at a pH of about 1.5. These enzymes would not work optimally at other pHs. Trypsin is another enzyme in the digestive system, which breaks protein chains in food into smaller particles. Trypsin works in the small intestine, which is not an acidic environment. Trypsin's optimum pH is about 8.
Different reactions and different enzymes will achieve their maximum rate at certain pH values. As shown in the figure below, the enzyme achieves a maximum reaction rate at a pH of 4. Notice that the reaction will continue at lower and higher pH values because the enzyme will still function at other pH values but will not be as effective. At very high or very low pH values, denaturation will occur because an enzyme is just a protein with a specific function.
Temperature
As with pH, reactions also have an ideal temperature where the enzyme functions most effectively. It will still function at higher and lower temperatures, but the rate will be less. For many biological reactions, the ideal temperature is at physiological conditions which is around $37^\text{o} \text{C}$ which is normal body temperature. Many enzymes lose function at lower and higher temperatures. At higher temperatures, an enzyme's shape deteriorates. Only when the temperature comes back to normal does the enzyme regain its shape and normal activity unless the temperature was so high that it caused irreversible damage.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/14%3A_Biological_Molecules/14.01%3A_Enzymes.txt |
Learning Outcomes
• Identify the basic structural features of fatty acids.
• Distinguish between saturated and unsaturated fatty acids.
• Define essential fatty acids.
• Identify the structural features of a triglyceride.
• Label the type of bond formed between fatty acids and glycerol in triglyceride.
There is a lot of interest these days on healthy diets as well as concerns about heart problems. There is also a strong market for the sales of omega-3 fatty acids, which are said to help lower fat levels in blood. But too many people rely on the supplements to help their hearts and don't understand the chemistry behind it all. Yes, taking omega-3 fatty acids will give you some of the fatty acids your body requires. No, this is not a substitute for eating a healthy diet and exercising. You can't sit in front of the TV set, eating your large pizza, and expect these pills to keep you health. You've got to do things the hard way - eat your vegetables and get some exercise.
Fatty Acids
A lipid is an organic compound such as fat or oil. Organisms use lipids to store energy, but lipids have other important roles as well. Lipids consist of repeating units called fatty acids. Fatty acids are organic compounds that have the general formula \(\ce{CH_3(CH_2)_{n}COOH}\), where \(n\) usually ranges from 2 to 28 and is always an even number. There are two types of fatty acids: saturated fatty acids and unsaturated fatty acids.
Saturated Fatty Acids
In saturated fatty acids, carbon atoms are bonded to as many hydrogen atoms as possible. This causes the molecules to form straight chains, as shown in the figure below. The straight chains can be packed together very tightly, allowing them to store energy in a compact form. This explains why saturated fatty acids are solids at room temperature. Animals use saturated fatty acids to store energy.
Unsaturated Fatty Acids
In unsaturated fatty acids, some carbon atoms are not bonded to as many hydrogen atoms as possible due to the presence of one or more double bonds in the carbon chain. Instead, they are bonded to other groups of atoms. Wherever carbon binds with these other groups of atoms, it causes chains to bend (see figure above). The bent chains cannot be packed together very tightly, so unsaturated fatty acids are liquids at room temperature. Plants use unsaturated fatty acids to store energy.
Lipids and Diet
Unsaturated fat is generally considered to be healthier because it contains fewer calories than an equivalent amount of saturated fat. Additionally, high consumption of saturated fats is linked to an increased risk of cardiovascular disease. Some examples of foods with high concentrations of saturated fats include butter, cheese, lard, and some fatty meats. Foods with higher concentrations of unsaturated fats include nuts, avocado, and vegetable oils such as canola oil and olive oil.
Humans need lipids for many vital functions, such as storing energy and forming cell membranes. Lipids can also supply cells with energy. In fact, a gram of lipids supplies more than twice as much energy as a gram of carbohydrates or proteins. Lipids are necessary in the diet for most of these functions. Although the human body can manufacture most of the lipids it needs, there are others, called essential fatty acids, that must be consumed in food. Essential fatty acids include omega-3 and omega-6 fatty acids. Both of these fatty acids are needed for important biological processes, not just for energy.
Although some lipids in the diet are essential, excess dietary lipids can be harmful. Because lipids are very high in energy, eating too many may lead to unhealthy weight gain. A high-fat diet may also increase lipid levels in the blood. This, in turn, can increase the risk for health problems such as cardiovascular disease. The dietary lipids of most concern are saturated fatty acids, trans fats, and cholesterol. For example, cholesterol is the lipid mainly responsible for narrowing arteries and causing the disease atherosclerosis.
Types of Lipids
Lipids may consist of fatty acids alone, or they may contain other molecules as well. For example, some lipids contain alcohol or phosphate groups. They include
1. triglycerides: the main form of stored energy in animals.
2. phospholipids: the major components of cell membranes.
3. steroids: serve as chemical messengers and have other roles.
Triglycerides
One type of lipid is called a triglyceride, an ester derived from glycerol combined with three fatty acid molecules.
Glycerol is a triol, an alcohol which contains three hydroxyl functional groups. A fatty acid is a long carbon chain, generally from 12 to 24 carbons in length, with an attached carboxyl group. Each of the three fatty acid molecules undergoes an esterification with one of the hydroxyl groups of the glycerol molecule. The result is a large triester molecule referred to as a triglyceride.
Triglycerides function as a long-term storage form of energy in the human bods. Because of the long carbon chains, triglycerides are nearly nonpolar molecules and thus do not dissolve readily in polar solvents such as water. Instead, oils and fats are soluble in nonpolar organic solvents such as hexane and ethers.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/14%3A_Biological_Molecules/14.02%3A_Lipids_and_Triglycerides.txt |
Learning Outcomes
• Describe the structure of a phospholipid.
• Identify the polar (hydrophilic) and nonpolar (hydrophobic) regions of a phospholipid.
• Explain how the phospholipid molecules form the bilayer of the cell membrane.
When you go to the dentist to get a tooth pulled, you really do not want to feel any pain. The dentist injects an anesthetic into your gum and it eventually becomes numb. One theory as to why anesthetics work deals with the movement of ions across the cell membrane. The anesthetic gets into the membrane structure and causes shifts in how ions move across the membrane. If ion movement is disrupted, nerve impulses will not be transmitted and you will not sense pain - at least not until the anesthetic wears off.
Phospholipids
A phospholipid is a lipid that contains a phosphate group and is a major component of cell membranes. A phospholipid consists of a hydrophilic (water-loving) head and hydrophobic (water-fearing) tail (see figure below). The phospholipid is essentially a triglyceride in which a fatty acid has been replaced by a phosphate group of some sort.
Following the rule of "like dissolves like", the hydrophilic head of the phospholipid molecule dissolves readily in water. The long fatty acid chains of a phospholipid are nonpolar and thus avoid water because of their insolubility. In water, phospholipids spontaneously form a double layer called a lipid bilayer in which the hydrophobic tails of phospholipid molecules are sandwiched between two layers of hydrophilic heads (see figure below). In this way, only the heads of the molecules are exposed to the water, while the hydrophobic tails interact only with each other.
Phospholipid bilayers are critical components of cell membranes. The lipid bilayer acts as a barrier to the passage of molecules and ions into and out of the cell. However, an important function of the cell membrane is to allow selective passage of certain substances into and out of cells. This is accomplished by the embedding of various protein molecules in and through the lipid bilayer (see figure below). These proteins form channels through which certain specific ions and molecules are able to move. Many membrane proteins also contain attached carbohydrates on the outside of the lipid bilayer, allowing it to form hydrogen bonds with water.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky)
14.04: Biological Molecules (Exercises)
These are homework exercises to accompany Chapter 14 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Answers are below the questions.
Questions
(click here for solutions)
Q14.1.1
What do enzymes do in the body?
Q14.1.2
Describe each of the following.
1. active site
2. substrate
3. allosteric site
4. inhibitor
Q14.1.3
How are competitive and noncompetitive inhibitors similar and different from one another?
Q14.1.4
An inhibitor interacts with the enzyme at an allosteric site. Is it competitive or noncompetitive?
Q14.1.5
Describe each of the following enzyme-substrate pairs as using the lock-and-key model or the induced fit model.
Q14.1.6
Each of the following behaves as a cofactor or a coenzyme. Identify each.
1. Zn2+
2. Vitamin B12
3. biotin
4. Fe3+
Vitamin B12
biotin
Q14.1.7
Identify each statement as true or false. Correct the false statement(s).
1. Enzyme activity increases with temperature.
2. Enzyme activity depends on pH.
3. Enzymes are consumed in a chemical reaction.
4. Increasing the concentration will increase the rate of a reaction until all of the enzyme is "occupied".
5. Enzymes can only function at their ideal temperature and pH.
Q14.1.8
The enzyme activity is graphed with respect to the pH of the mixture. Determine the pH at which each enzyme is most effective.
(click here for solutions)
Q14.2.1
What is the functional group in a fatty acid?
Q14.2.2
What is the difference between a fat and an oil?
Q14.2.3
Butter is a fat that is a solid at room temperature. What type of fatty acid does butter contain? How do you know?
Q14.2.4
Describe the difference between saturated and unsaturated fatty acids.
Q14.2.5
Explain why molecules of saturated and unsaturated fatty acids have different shapes.
Q14.2.6
Draw each structure.
1. Saturated fatty acid with 18 carbon atoms.
2. Monounsaturated fatty acid with 14 carbon atoms.
3. Polyunsaturated fatty acid with 14 carbon atoms.
4. Monounsaturated with 16 carbon atoms.
5. Polyunsaturated fatty acid with 18 carbon atoms and three double bonds.
Q14.2.7
Where does the body get essential fatty acids?
Q14.2.8
What molecules react to form a triglyceride?
Q14.2.9
Draw a triglyceride formed from three identical fatty acids.
Q14.2.10
Draw a triglyceride formed from three different fatty acids.
(click here for solutions)
Q14.3.1
What is a phospholipid?
Q14.3.2
Which part of the phospholipid molecule is water-soluble?
Q14.3.3
Which part is not water-soluble?
Q14.3.4
What is the purpose of a semipermeable membrane like the cell membrane?
Answers
14.1: Enzymes
Q14.1.1
Enzymes act as catalysts for biological reactions.
Q14.1.2
Describe each of the following.
1. The active site is where the reaction occurs in the enzyme.
2. The substrate is the molecule that interacts with the enzyme.
3. An allosteric site is on the enzyme away from the active site. Inhibitors can interactive with the enzyme at the allosteric site.
4. An inhibitor is a molecule that interacts with the molecule to slow or stop a reaction.
Q14.1.3
Both competitive and noncompetitive inhibitors slow or stop a reaction. Competitive inhibitors bind with the active site and non-competitive inhibitors bind with an allosteric site.
Q14.1.4
noncompeititve
Q14.1.5
1. lock & key
2. lock & key
3. induced fit
Q14.1.6
1. cofactor
2. coenzyme
3. coenzyme
4. cofactor
Q14.1.7
Identify each statement as true or false. Correct the false statement(s).
1. Enzyme activity increases with temperature and then decreases with increasing temperature beyond a peak temperature.
2. True
3. Enzymes are not consumed in a chemical reaction.
4. True
5. Enzymes
6. can only function most effectively at their ideal temperature and pH, but can function at a range of temperature and pH values.
Q14.1.8
1. pH = 8
2. pH = 6
14.2: Lipids and Triglycerides
Q14.2.1
carboxylic acid
Q14.2.2
Fats are solids at room temperatures while oils are liquids.
Q14.2.3
Saturated fatty acid.
Q14.2.4
Saturated fatty acids contain only single bonds between carbon atoms while unsaturated fatty acids contain one or more double bonds.
Q14.2.5
Saturated fatty acids have a straight chain of carbon atoms while unsaturated fatty acids have a bend at every double bond.
Q14.2.6
a.
b.
Answers will vary. The double bond may be located between any two carbon atoms.
c.
Answers will vary. There should be multiple double bonds and can be located between any pairs of carbon atoms.
d.
Answers will vary. The double bond may be located between any two carbon atoms.
e.
Answers will vary. There should be three double bonds and can be located between any pairs of carbon atoms.
Q14.2.7
Essential fatty acids come from the food we eat.
Q14.2.8
Glycerol and three fatty acids from a triglyceride.
Q14.2.9
Answers will vary but each "tail" should have the same number of carbon atoms.
Q14.2.10
Answers will vary but each "tail" should have a different number of carbon atoms.
14.3: Phospholipids in Cell Membranes
Q14.3.1
A phospholipid is a triglyceride where one of the fatty acid tails is replaced by a phosphate group.
Q14.3.2
The phosphate end is water-soluble.
Q14.3.3
The hydrocarbon tails are not water-soluble.
Q14.3.4
To control the flow of substances in and out of the cell. | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/14%3A_Biological_Molecules/14.03%3A_Phospholipids_in_Cell_Membranes.txt |
Metabolic pathways linked series of chemical reactions occurring within a cell. The reactants, products, and intermediates of an enzymatic reaction are known as metabolites, which are modified by a sequence of chemical reactions catalyzed by enzymes. In a metabolic pathway, the product of one enzyme acts as the substrate for the next. These enzymes often require dietary minerals, vitamins, and other cofactors to function. There are two types of metabolic pathways that are characterized by their ability to either synthesize molecules with the utilization of energy (anabolic pathway) or break down of complex molecules by releasing energy in the process (catabolic pathway). The two pathways complement each other in that the energy released from one is used up by the other.
• 15.1: Glycolysis
Glucose is sliced right in half from a 6-carbon molecule to two 3-carbon molecules. This is the first step and an extremely important part of cellular respiration. It happens all the time, both with and without oxygen. And in the process, transfers some energy to ATP.
• 15.2: The Citric Acid Cycle
• 15.3: Lactic Acid Fermentation
Short spurts of sprinting are sustained by fermentation in muscle cells. This produces just enough ATP to allow these short bursts of increased activity.
• 15.4: The Electron Transport Chain
At the end of the Krebs Cycle, energy from the chemical bonds of glucose is stored in diverse energy carrier molecules: four ATPs, but also two FADH 2 and ten NADH molecules. The primary task of the last stage of cellular respiration, the electron transport chain, is to transfer energy from the electron carriers to even more ATP molecules, the "batteries" which power work within the cell.
• 15.5: Metabolic Cycles (Exercises)
These are homework exercises to accompany Chapter 15 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Solutions are available below the questions.
• 15.6: Homeostasis
The process in which organ systems work to maintain a stable internal environment is called homeostasis. Keeping a stable internal environment requires constant adjustments.
15: Metabolic Cycles
Learning Outcomes
• List the three stages of cellular respiration.
• Explain the purpose of glycolysis.
• Describe the use and formation of ATP during glycolysis.
• Name the products of glycolysis.
• Distinguish between aerobic and anaerobic.
How do you slice a molecule of glucose in half? With sharp knives? Not really. But you lyse it with enzymes during a process named glycolysis. Glucose is sliced right in half from a 6-carbon molecule to two 3-carbon molecules. This is the first step and an extremely important part of cellular respiration. It happens all the time, both with and without oxygen. And in the process, transfers some energy to ATP.
Glycolysis: A Universal and Ancient Pathway for Making ATP
When was the last time you enjoyed yogurt on your breakfast cereal, or had a tetanus shot? These experiences may appear unconnected, but both relate to bacteria which do not use oxygen to make ATP. In fact, tetanus bacteria cannot survive if oxygen is present. However, Lactobacillus acidophilus (bacteria which make yogurt) and Clostridium tetani (bacteria which cause tetanus or lockjaw) share with nearly all organisms the first stage of cellular respiration, glycolysis (see figure below). Because glycolysis is universal, whereas aerobic (oxygen-requiring) cellular respiration is not, most biologists consider it to be the most fundamental and primitive pathway for making ATP.
$\ce{C_6H_{12}O_6} + 6 \ce{O_2} \rightarrow 6 \ce{CO_2} + 6 \ce{H_2O}$
Like photosynthesis, the process represented by this equation is actually many small, individual chemical reactions. We will divide the reactions of cellular respiration into three stages: glycolysis, the Krebs Cycle (also known as the citric acid cycle), and the electron transport chain (see figure below). In this concept, Stage 1, glycolysis, the oldest and most widespread pathway for making ATP, is discussed. Before diving into the details, we must not that this first stage of cellular respiration is unique among the three stages: it does not require oxygen, and it does not take place in the mitochondrion. The chemical reactions of glycolysis occur without oxygen in the cytosol of the cell (see figure below).
The name for Stage 1 clearly indicates what happens during that stage: glyco- refers to glucose, and -lysis means "splitting". In glycolysis, within the cytosol of the cell, a minimum of eight different enzymes break apart glucose into two 3-carbon molecules. The energy released in breaking those bonds is transferred to carrier molecules, ATP and NADH. NADH temporarily holds small amounts of energy which can be used to later build ATP. The 3-carbon product of glycolysis is pyruvate, or pyruvic acid (see figure below). (The sole difference between them is actually a sole hydrogen atom. Pyruvic acid: $\ce{CH_3COCOOH}$, pyruvate: $\ce{CH_3COCOO^-}$). Overall, glycolysis can be represented as:
$\ce{C_6H_{12}O_6} + 2 \ce{NAD^+} + 2 \ce{P_{i}} + 2 \ce{ADP} \rightarrow 2 \: \text{pyruvate} \: + 2 \ce{NADH} + 2 \ce{ATP}$
However, even this equation is deceiving. Just the splitting of glucose requires many steps, each transferring or capturing small amounts of energy. Individual steps appear in the figure below. Studying the pathway in detail reveals that cells must "spend" or "invest" two ATP in order to begin the process of breaking glucose apart. Note that the phosphates produced by breaking apart ATP join with glucose, making it unstable and more likely to break apart. Later steps harness the energy released when glucose splits, and use it to build "hot hydrogens" (NAD$^+$ is reduced to NADH) and ATP (ADP + P$_i \rightarrow$ ATP). If you count the ATP produced, you will find a net yield of two ATP per glucose (4 produced $-$ 2 spent). Remember to double the second set of reactions to account for the two 3-carbon molecules which follow that pathway! The "hot hydrogens" can power other metabolic pathways, or in many organisms, provide energy for further ATP synthesis.
To summarize: In the cytosol of the cell, glycolysis transfers some of the chemical energy stored in one molecule of glucose to two molecules of ATP and two NADH. This makes (some of) the energy in glucose, a universal fuel molecule for cells, available to use in cellular work - moving organelles, transporting molecules across membranes, or building large organic molecules.
Although glycolysis is universal, pathways leading away from glycolysis vary among species depending on the availability of oxygen. If oxygen is unavailable, pyruvate may be converted to lactic acid or ethanol and carbon dioxide in order to regenerate NAD$^+$, called anaerobic respiration. Anaerobic respiration is also called fermentation, which will be discussed in another concept.
If oxygen is present, pyruvate enters the mitochondria for further breakdown, releasing far more energy and producing many additional molecules of ATP in the latter two stages of aerobic respiration - the Krebs Cycle and electron transport chain.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/15%3A_Metabolic_Cycles/15.01%3A_Glycolysis.txt |
Learning Outcomes
• Describe the citric acid cycle (Krebs Cycle).
• Name the products of the citric acid cycle.
• Identify the energy carrier molecules produced in the citric acid cycle.
• Describe what happens to pyruvate before it enters the citric acid cycle.
Aerobic Respiration
Enticing clues - volcanic gases, vast iron ore sediments, and bubbles of ancient air trapped in amber - suggest dramatic changes during the history of earth's atmosphere. Correlating these clues with the fossil record leads to two major conclusions: that early life evolved in the absence of oxygen, and that oxygen first appeared between 2 and 3 billion years ago (see figure below) because of photosynthesis by the blue green bacteria, cyanobacteria. The chemistry of cellular respiration reflects this history. Its first stage, glycolysis, is universal and does not use oxygen.
Absolutely dependent on oxygen gas, we find it difficult to imagine that its appearance must have been disastrous for the anaerobic organisms that evolved in its absence. But oxygen is highly reactive, and at first, its effect on evolution was so negative that some have named this period the "oxygen catastrophe". However, as oxygen gradually formed a protective ozone layer, life rebounded. After the first organisms evolved to use oxygen to their advantage, the diversity of aerobic organisms exploded. According to the Theory of Endosymbiosis, engulfing of some of these aerobic bacteria led to eukaryotic cells with mitochondria, and multicellularity, the evolution of multicellular eukaryotic organisms, followed. Today, we live in an atmosphere which is $21\%$ oxygen, and most of life follows glycolysis with the last two, aerobic stages of cellular respiration.
Recall the purpose of cellular respiration: to release energy from glucose to make ATP, the universal molecule of energy for cellular work. The following equation describes the overall process, although it summarizes many individual chemical reactions.
$\ce{6O2} + \underbrace{\ce{C6H12O6}}_{\text{stored chemical} \ \text{energy, deliverable}} + \ce{398P_i} \ce{->[\text{mitochondia}]} \ce{38 ATP} + \ce{6O2} + \ce{6H2O}$
Once again, the first stage of this process, glycolysis, is ancient, universal, and anaerobic. In the cytoplasm of most cells, glycolysis breaks each 6-carbon molecule of glucose into two 3-carbon molecules of pyruvate. Chemical energy, which had been stored in the now broken bonds, is transferred to 2 ATP and 2 NADH molecules.
The fate of pyruvate depends on the species and the presence or absence of oxygen. If oxygen is present to drive subsequent reaction, pyruvate enters the mitochondria, where the citric acid cycle (also known as the Krebs Cycle) (Stage 2) and electron transport chain (Stage 3) break it down and oxidize it completely to $\ce{CO_2}$ and $\ce{H_2O}$. The energy released builds many more ATP molecules, though of course some is lost as heat. Let's explore the details of how mitochondria use oxygen to make more ATP from glucose by aerobic respiration.
The Citric Acid Cycle: Capturing Energy from Pyruvate
Aerobic respiration begins with the entry of the product of glycolysis, pyruvate, into the mitochondria. For each initial glucose molecule, two pyruvate molecules will enter the mitochondria. Pyruvate, however, is not the molecule that enters the citric acid cycle. Prior to entry into this cycle, pyruvate must be converted into a 2-carbon acetyl-CoenzymeA (acetyl-CoA) unit.
The conversion of pyruvate into acetyl-CoA is referred to as the pyruvate dehydrogenase reaction. It is catalyzed by the pyruvate dehydrogenase complex. This process produces one NADH electron carrier while releasing a $\ce{CO_2}$ molecule. This step is also known as the link reaction or transition step, as it links glycolysis and the citric acid cycle. Of course, as two pyruvates result from glycolysis, two acetyl-CoAs are produced as are 2 NADH molecules.
1. Within the mitochondria, each pyruvate is broken apart and combined with a coenzyme known as CoA to form a 2-carbon molecule, acetyl-CoA, which can enter the Krebs Cycle. A single atom of carbon (per pyruvate) is "lost" as carbon dioxide. The energy released in this breakdown is captured in two NADH molecules. See the figure above. Fatty acids can also break down in to acetyl-CoA. By this means, lipids, like fats, can be "burned" to make ATP using the citric acid cycle.
2. The Krebs Cycle (see figure above) begins by combining each acetyl-CoA with a four-carbon carrier molecule to make a 6-carbon molecule of citric acid (or citrate, its ionized form).
3. The cycle carries citric acid through a series of chemical reactions which gradually release energy and capture it in several carrier molecules. For each acetyl-CoA which enters the cycle, 3 NAD$^+$ are reduced to NADH, one molecule of FAD (another temporary energy carrier) is reduced to $FADH_2$, and one molecule of ATP (actually a precursor, GTP, guanine triphosphate) is produced. Study the figure above to locate each of these energy-capturing events.
4. Note what happens to carbon atoms (black dots in the figure above). For each 2-carbon acetyl-CoA which enters the cycle, two molecules of carbon dioxide are released, completing the breakdown of the original 6-carbon glucose molecule. The final step regenerates the original 4-carbon molecule which began the cycle, so that another acetyl-CoA can enter the cycle.
In summary, the citric acid cycle completes the breakdown of glucose which began with glycolysis. Its chemical reactions oxidize all six of the original carbon atoms to $\ce{CO_2}$, and capture the energy released in 2 ATP, 6 NADH, and 2 FADH$_2$. These energy carriers join the 2 ATP and 2 NADH produced in glycolysis and the 2 NADH produced in the conversion of 2 pyruvates to 2 acetyl-CoA molecules.
At the conclusion of the citric acid cycle, glucose is completely broken down, yet only four ATP have been produced. Moreover, although oxygen is required to drive the citric acid cycle, the cycle's chemical reactions do not themselves consume $\ce{O_2}$. The conclusion of cellular respiration, stage 3, produces the majority of the ATP.
Supplemental Resources
• Citric Acid Cycle (aka Krebs Cycle): virtuallabs.stanford.edu/other/biochem/TCA.swf
• Krebs Cycle (aka Citric Acid Cycle): http://johnkyrk.com/krebs.html
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/15%3A_Metabolic_Cycles/15.02%3A_The_Citric_Acid_Cycle.txt |
Learning Outcomes
• Describe lactic acid fermentation.
• Describe how bacteria, including those we employ to make yogurt, make ATP in the absence of oxygen.
• Discuss how your muscles continue to work for you even when your respiratory and cardiovascular system can no longer keep up a continuous supply of oxygen.
Short spurts of sprinting are sustained by fermentation in muscle cells. This produces just enough ATP to allow these short bursts of increased activity.
Lactic Acid Fermentation: Muscle Cells and Yogurt
For chicken or turkey dinners, do you prefer light meat or dark? Do you consider yourself a sprinter or a long distance runner? What is the biological difference between light meat and dark meat? Or between the two types of runners? Would you believe it has something to do with muscle color?
Are Drumsticks and Athletic Prowess Related?
Muscle color reflects its specialization for aerobic or anaerobic metabolism. Although humans are obligate aerobes (an organism which requires oxygen for cellular respiration), our muscle cells have not given up on ancient pathways which allow them to keep producing ATP quickly when oxygen runs low. The difference is more pronounced in chickens and grouse (see figure below), which stand around all day on their legs. For long periods of time, they carry out aerobic respiration in their "specialized-for-endurance" red muscles. If you are familiar with grouse, you know that these birds "flush" with great speed over short distances. Such "sprinting" flight depends on anaerobic respiration in the white cells of breast and wing muscle, allowing rapid production of ATP in low oxygen situations.
No human muscle is all red or all white, but chances are, if you excel at sprinting short distances or at a sport such as weight lifting, you have more white glycolytic fibers in your leg muscles, allowing anaerobic respiration. If you run marathons, you probably have more red oxidative fibers, performing aerobic respiration.
Lactic Acid Fermentation
You may have not been aware that your muscle cells can ferment. Fermentation is the process of producing ATP in the absence of oxygen, through glycolysis alone. Recall that glycolysis breaks a glucose molecule into two pyruvate molecules, producing a net gain of two ATP and two NADH molecules. Lactic acid fermentation is the type of anaerobic respiration carried out by yogurt bacteria (Lactobacillus and others) and by your own muscle cells when you work them hard and fast.
Lactic acid fermentation converts the 3-carbon pyruvate to the 3-carbon lactic acid $\left( \ce{C_3H_6O_3} \right)$ (see figure below) and regenerates NAD$^+$ in the process, allowing glycolysis to continue to make ATP in low-oxygen conditions. Since there is a limited supply of NAD$^+$ available in any given cell, this electron acceptor must be regenerated to allow ATP production to continue. To achieve this, NADH donates its extra electrons to the pyruvate molecules, regenerating NAD$^+$. Lactic acid is formed by the reduction of pyruvate.
$\ce{C_3H_3O_3} \: \text{(pyruvate)} \: + \ce{NADH} \rightarrow \ce{C_3H_6O_3} \: \text{(lactic acid)} \: + \ce{NAD^+}$
actic acid fermentation converts pyruvate to lactic acid, and regenerates NAD$^+$ from $NADH$.
For Lactobacillus bacteria, the acid resulting from fermentation kills bacterial competitors in buttermilk, yogurt, and some cottage cheese. The benefits extend to humans who enjoy these foods, as well (Figure $5$).
You may have notice this type of fermentation in your own muscles, because muscle fatigue and pain are associated with lactic acid. Lactic acid accumulates in your muscle cells as fermentation proceeds during times of strenuous exercise. During these times, your respiratory and cardiovascular systems cannot transport oxygen to your muscle cells, especially those in your legs, fast enough to maintain aerobic respiration. To allow the continuous production of some ATP, your muscle cells use lactic acid fermentation.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/15%3A_Metabolic_Cycles/15.03%3A_Lactic_Acid_Fermentation.txt |
Learning Outcomes
• Summarize the electron transport chain.
• Recognize that electron transport chain is the third and final stage of aerobic cellular respiration.
• Identify the products of the citric acid cycle.
What do trains, trucks, boats, and planes all have in common? They are ways to transport. And they all use a lot of energy. To make ATP, energy must be "transported" - first from glucose to NADH, and then somehow passed to ATP. How is this done? With an electron transport chain, the third stage of aerobic respiration. This third stage uses energy to make energy.
The Electron Transport Chain: ATP for Life in the Fast Lane
At the end of the Krebs Cycle, energy from the chemical bonds of glucose is stored in diverse energy carrier molecules: four ATPs, but also two FADH$_2$ and ten NADH molecules. The primary task of the last stage of cellular respiration, the electron transport chain, is to transfer energy from the electron carriers to even more ATP molecules, the "batteries" which power work within the cell.
Pathways for making ATP in stage 3 of aerobic respiration closely resemble the electron transport chains used in photosynthesis. In both electron transport chains, energy carrier molecules are arranged in sequence within a membrane so that energy-carrying electrons cascade from one to another, losing a little energy in each step. In both photosynthesis and aerobic respiration, the energy lost is harnessed to pump hydrogen ions into a compartment, creating an electrochemical gradient or chemiosmotic gradient across the enclosing membrane. And in both processes, the energy stored in the chemiosmotic gradient is used with ATP synthase to build ATP.
For aerobic respiration, the electron transport chain or "respiratory chain" is embedded in the inner membrane of the mitochondria (see figure below). The FADH$_2$ and NADH molecules produced in glycolysis and the Krebs Cycle, donate high-energy electrons to energy carrier molecules within the membrane. As they pass from one carrier to another, the energy they lose is used to pump hydrogen ions into the mitochondrial intermembrane space, creating an electrochemical gradient. Hydrogen ions flow "down" the gradient - from outer to inner compartment - through the ion channel/enzyme ATP synthase, which transfers their energy to ATP. Note the paradox that it requires energy to create and maintain a concentration gradient of hydrogen ions that are then used by ATP synthase to create stored energy (ATP). In broad terms, it takes energy to make energy. Coupling the electron transport chain to ATP synthesis with a hydrogen ion gradient is chemiosmosis, first described by Nobel laureate Peter D. Mitchell. This process, the use of energy to phosphorylate ADP and produce ATP is also known as oxidative phosphorylation.
After passing through the electron transport chain, low-energy electrons and low-energy hydrogen ions combine with oxygen to form water. Thus, oxygen's role is to drive the entire set of ATP-producing reactions within the mitochondrion by accepting "spent" hydrogens. Oxygen is the final electron acceptor, no part of the process - from the Krebs Cycle through the electron transport chain- can happen without oxygen.
The electron transport chain can convert the energy from one glucose molecule's worth of $FADH_2$ and $NADH$ + $\ce{H^+}$ into as many as 34 ATP. When the four ATP produced in glycolysis and the Krebs Cycle are added, the total of 38 ATP fits the overall equation for aerobic cellular respiration:
$\ce{6O2} + \underbrace{\ce{C6H12O6}}_{\text{stored chemical energy}} + \ce{38 ADP} + \text{39 P}_\text{i} \rightarrow \underbrace{\ce{38 ATP}}_{\text{stored chemical energy}} + \ce{6CO2} + \ce{6 H2O}$
Aerobic respiration is complete. If oxygen is available, cellular respiration transfers the energy from one molecule of glucose to 38 molecules of ATP, releasing carbon dioxide and water as waste. "Deliverable" food energy has become energy which can be used for work within the cell - transport within the cell, pumping ions and molecules across membranes, and building large organic molecules. Can you see how this could lead to "life in the fast lane" compared to anaerobic respiration (glycolysis alone)?
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky)
15.05: Metabolic Cycles (Exercises)
These are homework exercises to accompany Chapter 15 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health.
15.1: Glycolysis
Q15.1.1
What are the three stages of cellular respiration?
Q15.1.2
What is the purpose of glycolysis?
Q15.1.3
What is the output of glycolysis from a single glucose molecule?
Q15.1.4
How many molecules of ATP are "invested" in glycolysis? How many are produced?
Q15.1.5
Define aerobic and anaerobic.
15.2: The Citric Acid Cycle
Q15.2.1
Where does the Krebs cycle occur in the cell?
Q15.2.2
What happens to the pyruvate produced during glycolysis?
Q15.2.3
How many reactions does it take to complete the cycle?
Q15.2.4
How many "turns" of the citric acid cycle must occur for each molecule of glucose entering glycolysis?
Q15.2.5
What is he output of the citric acid cycle?
Q15.2.6
Trace the six carbon atoms originally from acetyl-CoA through the Krebs Cycle. Trace the flow of energy from the pyruvates produced in glycolysis through the Krebs Cycle.
Q15.2.7
How many energy carriers are produced during the Krebs cycle per acetyl-CoA?
15.3: Lactic Acid Fermentation
Q15.3.1
What is fermentation?
Q15.3.2
Define lactic acid fermentation.
Q15.3.3
Identify yourself as a “sprinter” or an “endurance runner” and predict the type of muscle fiber (red or white) which predominates in your body. Explain your reasoning.
Q15.3.4
What is the chemical equation of lactic acid fermentation?
15.4: The Electron Transport Chain
Q15.4.1
What molecules "feed" the electron transport chain?
Q15.4.2
What is the primary product of the electron transport chain?
Q15.4.3
Where do the reactions of the electron transport chain occur? | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/15%3A_Metabolic_Cycles/15.04%3A_The_Electron_Transport_Chain.txt |
Learning Outcomes
• Describe how homeostasis and equilibrium are different.
Remove one stone and the whole arch collapses. The same is true for the human body. All the systems work together to maintain stability or homeostasis. Disrupt one system, and the whole body may be affected.
Homeostasis
All of the organs and organ systems of the human body work together like a well-oiled machine. This is because they are closely regulated by the nervous and endocrine systems. The nervous system controls virtually all body activities, and the endocrine system secretes hormones that regulate these activities. Functioning together, the organ systems supply body cells with all the substances they need and eliminate their wastes. They also keep temperature, pH, and other conditions at just the right levels to support life processes.
Maintaining Homeostasis
The process in which organ systems work to maintain a stable internal environment is called homeostasis. Keeping a stable internal environment requires constant adjustments. Here are just three of the many ways that human organ systems help the body maintain homeostasis:
• Respiratory system: A high concentration of carbon dioxide in the blood triggers faster breathing. The lungs exhale more frequently, which removes carbon dioxide from the body more quickly.
• Excretory system: A low level of water in the blood triggers retention of water by the kidneys. The kidneys produce more concentration urine, so less water is lost from the body.
• Endocrine system: A high concentration of sugar in the blood triggers secretion of insulin by an endocrine gland called the pancreas. Insulin is a hormone that helps cells absorb sugar from the blood.
So how does your body maintain homeostasis? The regulation of your internal environment is done primarily through negative feedback. Negative feedback is a response to a stimulus that keeps a variable close to a set value (see figure below). Essentially, it "shuts off" or "turns off" a system when it varies from a set value.
For example, your body has an internal thermostat. During a winter day, in your house a thermostat senses the temperature in a room and responds by turning on or off the heater. Your body acts in much the same way. When body temperature rises, receptors in the skin and the brain sense the temperature change. The temperature change triggers a command from the brain. This command can cause several responses. If you are too hot, the skin makes sweat and blood vessels near the skin surface dilate. This response helps decrease body temperature.
Another example of negative feedback has to do with blood glucose levels. When glucose (sugar) levels in the blood are too high, the pancreas secretes insulin to stimulate the absorption of glucose and the conversion of glucose into glycogen, which is stored in the liver. As blood glucose levels decrease, less insulin is produced. When glucose levels are too low, another hormone called glucagon is produced, which causes the liver to convert glycogen back to glucose.
Positive Feedback
Some processes in the body are regulated by positive feedback. Positive feedback is when a response to an event increases the likelihood of the event to continue. An example of positive feedback is milk production in nursing mothers. As the baby drinks her mother's milk, the hormone prolactin, a chemical signal, is released. The more the baby suckles, the more prolactin is released, which causes more milk to be produced. Other examples of positive feedback include contractions during childbirth. When constrictions in the uterus push a baby into the birth canal, additional contractions occur.
Failure of Homeostasis
Many homeostatic mechanisms such as these work continuously to maintain stable conditions in the human body. Sometimes, however, the mechanisms fail. When they do, cells may not get everything they need, or toxic wastes may accumulate in the body. If homeostasis is not restored, the imbalance may lead to disease or even death.
Homeostasis vs. Equilibrium
Homeostasis requires an input of energy to maintain a specific condition necessary for life. Disturbances to homeostasis must be responded to in order to avoid death or disease. For example, a body needs to maintain a certain internal temperature. Go outside in cold weather - body shivers to maintain its body temperature.
Dynamic equilibrium is maintaining a specific condition that minimizes the system's energy, depending on the circumstances. A disturbance to an equilibrium is responded to in order to shift the process to reestablish an equilibrium. For example, if a warm object (say a metal bowl) is placed outside in cold weather - the transfer of heat occurs and the temperature of the bowl equilibrates to the outside temperature. If this happened to a person, it would not be good.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/15%3A_Metabolic_Cycles/15.06%3A_Homeostasis.txt |
Learning Objectives
• Learn the definitions of science, chemistry, and technology.
• Know the principles of green chemistry.
Science is a practice of learning about the natural world. Natural sciences include physics, chemistry, biology, geology and astronomy. Science uses mathematics and logic, which are sometimes called "formal sciences". Natural science makes observations and experiments. Science produces accurate facts, scientific laws and theories. 'Science' also refers to the large amount of knowledge that has been found using this process.
People who study and research science are scientists. Scientists study things by looking at them very carefully, by measuring, and by doing experiments and tests. Scientists try to explain why things act the way that they do, and predict what will happen.
Chemistry is the scientific discipline that involves the properties and behavior of matter, and the changes matter undergoes during a reaction with other substances.
In the scope of its subject, chemistry occupies an intermediate position between physics and biology. It is sometimes called the central science, because it provides a foundation for understanding both basic and applied scientific disciplines at a fundamental level (Figure 1.1.1). For example, chemistry explains aspects of plant chemistry (botany), the formation of igneous rocks (geology), how atmospheric ozone is formed and how environmental pollutants are degraded (ecology), the properties of the soil on the moon (astrophysics), how medications work (pharmacology), and how to collect DNA evidence at a crime scene (forensics).
Science is human understanding of how the universe and the things in it work. On the other hand, technology is the use of resources to solve a problem (such as knowledge, skills, processes, techniques, tools, and raw materials). Paleolithic technology is the oldest known, but people likely used technology long before then. The Paleolithic age began when hominids (early humans) started to use stones as tools for bashing, cutting, and scraping. The age ended when humans began to make small, fine tools (Mesolithic); it came to a close when they started to plant crops and develop other methods of agriculture (Neolithic).
Technology was once quite simple. It was passed on through word of mouth, until written word was invented. Writing allowed technology to develop much quicker. In the present day, people continue to understand more about the world and the universe. The use of the telescope by Galileo, Einstein's theory of relativity, lasers, and computing are all scientific discoveries. Technology is of great importance to science, medicine, and everyday life.
Green chemistry (also called sustainable chemistry) is a type of chemical research and engineering. It supports the design of products and processes that use the smallest amount of dangerous substances possible. In 1990, the Pollution Prevention Act was passed in the United States. This law sought new and original ways to handle pollution. The Pollution Prevention Act aims to avoid problems before they happen.
Green chemistry applies organic chemistry, inorganic chemistry, biochemistry, analytical chemistry, and even physical chemistry. While green chemistry seems to focus on just industrial applications, it does apply to other scientific disciplines. Green chemists aim to reduce the hazards and increase the efficiency of any chemical choice. Green chemistry is distinct from environmental chemistry, which focuses on chemical phenomena in the environment.
The 12 Principles of Green Chemistry
1. Prevent waste.
• Create products without waste, or with minimal wastes, so that the wastes do not need to be taken care of afterwards.
2. Design safer chemicals and products.
• Design chemicals to have little or no toxicity, without altering effectiveness.
3. Design less hazardous chemical syntheses.
• Design a way of synthesizing products, without them being toxic to humans or the environment.
4. Use renewable raw materials.
• Use renewable raw materials, like plant materials; rather than depleting materials, such as fossil fuels.
5. Use catalysts, not stoichiometric reagents.
• Use catalysts, because of their ability to be able to be reused. Catalysts are less harmful than reagents.
6. Avoid chemical derivatives.
• Chemical derivatives generate wastes that can be avoided.
7. Maximize atom economy.
• Maximize the percentage of reactant atoms that convert to usable product atoms in a chemical reaction, so as to minimize or eliminate atomic waste.
8. Use safer solvents and reaction conditions.
• Avoid using harsh solvents; if that cannot be avoided, then use benign chemicals.
9. Increase energy efficiency.
• Use the normal ambient temperature and pressure wherever possible.
10. Design for degradation.
• Design materials to break down into benign substances, by bacterial or other environmentally sound ways.
11. Analyze in real time to prevent pollution.
• Monitor and control the formation of by-products during a reaction.
12. Minimize the potential for accidents.
• Design chemicals to minimize the potential for accidents.
Summary
• Science is the human method to make discoveries about the natural world.
• Chemistry is the branch of science involved with the properties and behavior of matter, and the changes it undergoes during a reaction with other substances.
• Green chemistry applies organic chemistry, inorganic chemistry, biochemistry, analytical chemistry, and physical chemistry to reduce and prevent pollution.
• There are 12 guiding principles in the practice of green chemistry.
• Wikipedia | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/01%3A_Chemistry/1.01%3A_Science_and_Technology_-_The_Roots_of_Knowledge.txt |
Learning Objectives
• Describe the differences between hypothesis and theory as scientific terms.
• Describe the difference between a theory and scientific law.
• Identify the components of the scientific method.
Although many have taken science classes throughout their course of studies, incorrect or misleading ideas about some of the most important and basic principles in science are still commonplace. Most students have heard of hypotheses, theories, and laws, but what do these terms really mean? Before you read this section, consider what you have learned about these terms previously, and what they mean to you. When reading, notice if any of the text contradicts what you previously thought. What do you read that supports what you thought?
What is a Fact?
A fact is a basic statement established by experiment or observation. All facts are true under the specific conditions of the observation.
What is a Hypothesis?
One of the most common terms used in science classes is a "hypothesis". The word can have many different definitions, dependent on the context in which it is being used:
• An educated guess: a scientific hypothesis provides a suggested solution based on evidence.
• Prediction: if you have ever carried out a science experiment, you probably made this type of hypothesis, in which you predicted the outcome of your experiment.
• Tentative or proposed explanation: hypotheses can be suggestions about why something is observed. In order for a hypothesis to be scientific, a scientist must be able to test the explanation to see if it works, and if it is able to correctly predict what will happen in a situation. For example, "if my hypothesis is correct, I should see _____ result when I perform _____ test."
A hypothesis is tentative; it can be easily changed.
What is a Theory?
The United States National Academy of Sciences describes a theory as:
"Some scientific explanations are so well established that no new evidence is likely to alter them. The explanation becomes a scientific theory. In everyday language a theory means a hunch or speculation. Not so in science. In science, the word theory refers to a comprehensive explanation of an important feature of nature supported by facts gathered over time. Theories also allow scientists to make predictions about as yet unobserved phenomena."
"A scientific theory is a well-substantiated explanation of some aspect of the natural world, based on a body of facts that have been repeatedly confirmed through observation and experimentation. Such fact-supported theories are not "guesses," but reliable accounts of the real world. The theory of biological evolution is more than "just a theory." It is as factual an explanation of the universe as the atomic theory of matter (stating that everything is made of atoms) or the germ theory of disease (which states that many diseases are caused by germs). Our understanding of gravity is still a work in progress. But the phenomenon of gravity, like evolution, is an accepted fact."
Note some key features of theories that are important to understand from this description:
• Theories are explanations of natural phenomenon. They aren't predictions (although we may use theories to make predictions). They are explanations of why something is observed.
• Theories aren't likely to change. They have a lot of support and are able to explain many observations satisfactorily. Theories can, indeed, be facts. Theories can change in some instances, but it is a long and difficult process. In order for a theory to change, there must be many observations or evidence that the theory cannot explain.
• Theories are not guesses. The phrase "just a theory" has no room in science. To be a scientific theory carries a lot of weight—it is not just one person's idea about something
Theories aren't likely to change.
What is a Law?
Scientific laws are similar to scientific theories in that they are principles that can be used to predict the behavior of the natural world. Both scientific laws and scientific theories are typically well-supported by observations and/or experimental evidence. Usually, scientific laws refer to rules for how nature will behave under certain conditions, frequently written as an equation. Scientific theories are overarching explanations of how nature works, and why it exhibits certain characteristics. As a comparison, theories explain why we observe what we do, and laws describe what happens.
For example, around the year 1800, Jacques Charles and other scientists were working with gases to, among other reasons, improve the design of the hot air balloon. These scientists found, after numerous tests, that certain patterns existed in their observations of gas behavior. If the temperature of the gas increased, the volume of the gas increased. This is known as a natural law. A law is a relationship that exists between variables in a group of data. Laws describe the patterns we see in large amounts of data, but do not describe why the patterns exist.
Laws vs Theories
A common misconception is that scientific theories are rudimentary ideas that will eventually graduate into scientific laws when enough data and evidence has been accumulated. A theory does not change into a scientific law with the accumulation of new or better evidence. Remember, theories are explanations; laws are patterns seen in large amounts of data, frequently written as an equation. A theory will always remain a theory, a law will always remain a law.
Video \(1\) What is the difference between scientific law and theory?
The Scientific Method
Scientists search for answers to questions and solutions to problems by using a procedure called the scientific method. This procedure consists of making observations, formulating hypotheses, and designing experiments, which in turn lead to additional observations, hypotheses, and experiments in repeated cycles (Figure \(1\)).
Step 1: Make observations.
Observations can be qualitative or quantitative. Qualitative observations describe properties or occurrences in ways that do not rely on numbers. Examples of qualitative observations include the following: "the outside air temperature is cooler during the winter season," "table salt is a crystalline solid," "sulfur crystals are yellow," and "dissolving a penny in dilute nitric acid forms a blue solution and a brown gas." Quantitative observations are measurements, which by definition consist of both a number and a unit. Examples of quantitative observations include the following: "the melting point of crystalline sulfur is 115.21° Celsius," and "35.9 grams of table salt—the chemical name of which is sodium chloride—dissolve in 100 grams of water at 20° Celsius." For the question of the dinosaurs’ extinction, the initial observation was quantitative: iridium concentrations in sediments dating to 66 million years ago were 20–160 times higher than normal.
Step 2: Formulate a hypothesis.
After deciding to learn more about an observation or a set of observations, scientists generally begin an investigation by forming a hypothesis, a tentative explanation for the observation(s). The hypothesis may not be correct, but it puts the scientist’s understanding of the system being studied into a form that can be tested. For example, the observation that we experience alternating periods of light and darkness which correspond to observed movements of the sun, moon, clouds, and shadows, is consistent with either of two hypotheses:
1. Earth rotates on its axis every 24 hours, alternately exposing one side to the sun.
2. The sun revolves around Earth every 24 hours.
Suitable experiments can be designed to choose between these two alternatives. In the case of disappearance of the dinosaurs, the hypothesis was that the impact of a large extraterrestrial object caused their extinction. Unfortunately (or perhaps fortunately), this hypothesis does not lend itself to direct testing by any obvious experiment, but scientists can collect additional data that either supports or refutes it.
Step 3: Design and perform experiments.
After a hypothesis has been formed, scientists conduct experiments to test its validity. Experiments are systematic observations or measurements, preferably made under controlled conditions—that is, under conditions in which a single variable changes.
Step 4: Accept or modify the hypothesis.
A properly designed and executed experiment enables a scientist to determine whether the original hypothesis is valid. In the case of validity, the scientist can proceed to step 5. In other cases, experiments may demonstrate that the hypothesis is incorrect or that it must be modified, thus requiring further experimentation.
Step 5: Development of a law and/or theory.
More experimental data are then collected and analyzed, at which point a scientist may begin to think that the results are sufficiently reproducible (i.e., dependable) to merit being summarized in a law—a verbal or mathematical description of a phenomenon that allows for general predictions. A law simply states what happens; it does not address the question of why.
One example of a law, the law of definite proportions (discovered by the French scientist Joseph Proust [1754–1826]), states that a chemical substance always contains the same proportions of elements by mass. Thus, sodium chloride (table salt) always contains the same proportion by mass of sodium to chlorine—in this case, 39.34% sodium and 60.66% chlorine by mass. Sucrose (table sugar) is always 42.11% carbon, 6.48% hydrogen, and 51.41% oxygen by mass.
Whereas a law states only what happens, a theory attempts to explain why nature behaves as it does. Laws are unlikely to change greatly over time, unless a major experimental error is discovered. A theory, in contrast, is incomplete and imperfect; it evolves with time to explain new facts as they are discovered.
Because scientists can enter the cycle shown in Figure \(1\) at any point, the actual application of the scientific method to different topics can take many different forms. For example, a scientist may start with a hypothesis formed by reading about work done by others in the field, rather than by making direct observations.
Example \(1\)
Classify each statement as a law, theory, experiment, hypothesis, or observation.
1. Ice always floats on liquid water.
2. Birds evolved from dinosaurs.
3. Hot air is less dense than cold air, probably because the components of hot air are moving more rapidly.
4. When 10 g of ice was added to 100 mL of water at 25°C, the temperature of the water decreased to 15.5°C after the ice melted.
5. The ingredients of Ivory soap were analyzed to see whether it really is 99.44% pure, as advertised.
Solution
1. This is a general statement of a relationship between the properties of liquid and solid water, so it is a law.
2. This is a possible explanation for the origin of birds, so it is a hypothesis.
3. This is a statement that tries to explain the relationship between the temperature and the density of air based on fundamental principles, so it is a theory.
4. The temperature is measured before and after a change is made in a system, so these are observations.
5. This is an analysis designed to test a hypothesis (in this case, the manufacturer’s claim of purity), so it is an experiment.
Exercise \(1\)
Classify each statement as a law, theory, experiment, hypothesis, qualitative observation, or quantitative observation.
1. Measured amounts of acid were added to a Rolaids tablet to see whether it really “consumes 47 times its weight in excess stomach acid.”
2. Heat always flows from hot objects to cooler ones, not in the opposite direction.
3. The universe was formed by a massive explosion that propelled matter into a vacuum.
4. Michael Jordan is the greatest pure shooter ever to play professional basketball.
5. Limestone is relatively insoluble in water, but dissolves readily in dilute acid with the evolution of a gas.
Answer 1:
experiment
Answer 2:
law
Answer 3:
theory
Answer 4:
hypothesis
Answer 5:
observation
Summary
• A hypothesis is a tentative explanation that can be tested by further investigation.
• A theory is a well-supported explanation of observations.
• A scientific law is a statement that summarizes the relationship between variables.
• An experiment is a controlled method of testing a hypothesis.
• The scientific method is a method of investigation involving experimentation and observation to acquire new knowledge, solve problems, and answer questions. The key steps in the scientific method are:
• Step 1: Make observations.
• Step 2: Formulate a hypothesis.
• Step 3: Test the hypothesis through experimentation.
• Step 4: Accept or modify the hypothesis.
• Step 5: Development of a law and/or theory.
Contributors and Attributions
• Wikipedia
• Henry Agnew (UC Davis) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/01%3A_Chemistry/1.02%3A_Science-_Reproducible_Testable_Tentative_Predictive_and_Explanatory.txt |
Learning Objective
• Know the meaning and examples of pure and applied research.
How did chemistry develop? What is happening in the field of chemistry today? What career can a chemistry degree apply to? These are all good questions that should be asked by students interested in chemistry. Research in chemistry (or any other field, for that matter) is interesting and challenging. But there are different directions a person can take as they explore research opportunities.
Types of Research
In science, two types of research are generally discussed: pure and applied. Pure research focuses on answering basic questions such as, "how do gases behave?" Applied research is involved in the process of developing specific preparations for a gas, in order for it to be produced and delivered efficiently and economically. This division sounds like it would be easy to make, but sometimes we cannot draw a clear line between what is "pure" and what is "applied".
Examples of "Pure" Research
A lot of "pure" research is of the "what is this?" or "how does it work?" variety. The early history of chemistry contains many examples. The ancient Greek philosophers debated the composition of matter (Earth? Air? Fire? Water? All of the above?). They weren't going to do anything with their knowledge—they just wanted to know.
Studies on the elements (especially after Mendeleev's periodic table was published) were primarily "pure" research types of experiments. Does this element exist? What are its properties? The scientists did not have any practical application in mind, but were curious about the world around them.
Examples of "Applied" Research
There is a great deal of "applied" research taking place today. In general, no new science principles are discovered, but existing knowledge is used to develop new products. Research on laundry detergents will probably not offer any new concepts of soap, but will help us develop materials that get our clothes cleaner, use less water, and create lower amounts of pollution.
A lot of research is done by petroleum companies. They want to find better ways to power vehicles, better lubricants to cut down on engine wear, and better ways to lower air pollution. These companies will use information that is readily available to come up with new products.
Some "In-Between" Examples
Sometimes it is hard to differentiate between pure and applied research. What may start out as simply asking a question may result in some very useful information. If scientists are studying the biochemistry of a microorganism that causes a disease, they may soon find information that would suggest a way to make a chemical that will inactivate the microorganism. The compound could be used to learn more about the biochemistry, but could also be used to cure the disease.
Hemoglobin is a protein in red blood cells that transports oxygen in the bloodstream. Scientists studied hemoglobin simply to learn how it worked. Out of this research came an understanding of how the protein changes shape when oxygen attaches to it. This information was then applied to help patients with sickle cell anemia, a disorder caused by an abnormal hemoglobin structure that makes hemoglobin molecules clump up when oxygen leaves the protein. Basic knowledge of protein structure led to an improved understanding of a wide-spread disease and opened the door for the development of treatments.
Summary
• Pure research focuses on understanding basic properties and processes.
• Applied research focuses on the use of information to create useful materials.
• Sometimes there is no clear line between pure and applied research. | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/01%3A_Chemistry/1.03%3A_Solving_Societys_Problems_-_Scientific_Research.txt |
Learning Objectives
• Separate physical from chemical properties.
• Label a change as chemical or physical.
Chemistry is the study of matter—what it consists of, what its properties are, and how it changes. Matter is anything that has mass and takes up space—that is, anything that is physically real. Some things are easily identified as matter—the screen on which you are reading this book, for example. Others are not so obvious. Because we move so easily through air, we sometimes forget that it, too, is matter. Because of this, chemistry is a science that has its fingers in just about everything. Even a description of the ingredients in a cake, and how those ingredients change when the cake is baked, is chemistry!
Physical and Chemical Properties
All matter has physical and chemical properties. Physical properties are characteristics that scientists can measure without changing the composition of the sample under study, such as mass, color, and volume (the amount of space occupied by a sample). Chemical properties describe the characteristic ability of a substance to react to form new substances; they include its flammability and susceptibility to corrosion. All samples of a pure substance have the same chemical and physical properties. For example, pure copper is always a reddish-brown solid (a physical property) and always dissolves in dilute nitric acid to produce a blue solution and a brown gas (a chemical property).
Physical Property
A physical property is a characteristic of a substance that can be observed or measured without changing the identity of the substance. Silver is a shiny metal that conducts electricity very well. It can be molded into thin sheets—a property called malleability. Salt is dull and brittle and conducts electricity when it has been dissolved into water, which it does quite easily. Physical properties of matter include color, hardness, malleability, solubility, electrical conductivity, density, melting point, and boiling point. In Table $1$, notice that corn oil has a lower mass to volume ratio than water. This means that when added to water, corn oil will “float.”
Table $1$ Densities of Common Substances
Substance Density at 25°C (g/cm3)
blood 1.035
body fat 0.918
whole milk 1.030
corn oil 0.922
mayonnaise 0.910
honey 1.420
Hardness helps determine how an element (especially a metal) might be used. Many elements are fairly soft (silver and gold, for example) while others (such as titanium, tungsten, and chromium) are much harder. Carbon is an interesting example of hardness. In graphite, (the "lead" found in pencils) the carbon is very soft, while the carbon in a diamond is roughly seven times as hard.
Melting and boiling points are somewhat unique identifiers, especially of compounds. In addition to giving some idea of the compound's identity, important information can be obtained about the purity of the material.
Chemical Property
A chemical property of matter describes its "potential" to undergo some chemical change or reaction by virtue of its composition; as well as what elements, electrons, and bonding are present to give the potential for chemical change. It is quite difficult to define a chemical property without using the word "change". Eventually, students of chemistry should be able to look at the formula of a compound and state a chemical property. For example, hydrogen has the potential to ignite and explode, given the right conditions—this is a chemical property. Metals in general have the chemical property of reacting with acids. Zinc reacts with hydrochloric acid to produce hydrogen gas—this is a chemical property.
A chemical property of iron is that it is capable of combining with oxygen to form iron oxide, the chemical name of rust (Figure $2$). The more general term for rusting and other similar processes is corrosion. Other terms that are commonly used in descriptions of chemical changes are burn, rot, explode, decompose, and ferment. Chemical properties are very useful in identifying substances. However, unlike physical properties, chemical properties can only be observed as the substance is in the process of being changed into a different substance.
Table $2$: Contrasting Physical and Chemical Properties
Physical Properties Chemical Properties
Gallium metal melts at 30 oC Iron metal rusts.
Mercury is a very dense liquid. A green banana turns yellow when it ripens.
Gold is shiny. A dry piece of paper burns.
Example $1$
Which of the following is a chemical property of iron?
1. Iron corrodes in moist air.
2. Density = 7.874 g/cm3
3. Iron is soft when pure.
4. Iron melts at 1808 K.
Solution
"Iron corrodes in air" is the only chemical property of iron from the list.
Exercise $\PageIndex{1A}$
Which of the following is a physical property of matter?
1. corrosiveness
2. pH (acidity)
3. density
4. flammability
Answer
3.
Exercise $\PageIndex{1B}$
Which of the following is a chemical property?
1. flammability
2. melting point
3. boiling point
4. density
Answer
1.
Physical and Chemical Changes
Change is happening all around us all of the time. Just as chemists have classified elements and compounds, they have also classified types of changes. Changes are either classified as physical or chemical changes. Chemists learn a lot about the nature of matter by studying the changes that matter can undergo. Chemists make a distinction between two different types of changes that they study—physical changes and chemical changes.
Physical Change
Physical changes are changes in which no bonds are broken or formed. This means that the same types of compounds or elements that were there at the beginning of the change are there at the end of the change. Because the ending materials are the same as the beginning materials, the properties (such as color, boiling point, etc) will also be the same. Physical changes involve moving molecules around, but not changing them. Some types of physical changes include:
• Changes of state (changes from a solid to a liquid or a gas and vice versa).
• Separation of a mixture.
• Physical deformation (cutting, denting, stretching).
• Making solutions (special kinds of mixtures).
As an ice cube melts, its shape changes as it acquires the ability to flow. However, its composition does not change. Melting is an example of a physical change. A physical change is a change to a sample of matter in which some properties of the material change, but the identity of the matter does not. When liquid water is heated, it changes to water vapor. However, even though the physical properties have changed, the molecules are exactly the same as before. Each water molecule still contains two hydrogen atoms covalently bonded to one oxygen atom. When you have a jar containing a mixture of pennies and nickels, and you sort the mixture so that you have one pile of pennies and another pile of nickels, you have not altered the identity of either the pennies or the nickels—you've merely separated them into two groups. This would be an example of a physical change. Similarly, if you have a piece of paper, you don't change it into something other than a piece of paper by ripping it up. What was paper before you started tearing is still paper when you're done. Again, this is an example of a physical change.
Physical changes can further be classified as reversible or irreversible. The melted ice cube may be refrozen, so melting is a reversible physical change. Physical changes that involve a change of state are all reversible. Other changes of state include vaporization (liquid to gas), freezing (liquid to solid), and condensation (gas to liquid). Dissolution is also a reversible physical change. When salt is dissolved into water, the salt is said to have entered the aqueous state. The salt may be regained by boiling off the water, leaving the salt behind.
Chemical Change
Chemical changes occur when bonds are broken and/or formed between molecules or atoms. This means that one substance with a certain set of properties (such as melting point, color, taste, etc.) is turned into a different substance with different properties. Chemical changes are frequently harder to reverse than physical changes.
One good example of a chemical change is burning a candle. The act of burning paper actually results in the formation of new chemicals (carbon dioxide and water, to be exact) from the burning of the wax. Another example of a chemical change is what occurs when natural gas is burned in a furnace. This time, on the left there is a molecule of methane, $\ce{CH_4}$, and two molecules of oxygen, $\ce{O_2}$; on the right are two molecules of water, $\ce{H_2O}$, and one molecule of carbon dioxide, $\ce{CO_2}$. In this case, not only has the appearance changed, but the structure of the molecules has also changed. The new substances do not have the same chemical properties as the original ones. Therefore, this is a chemical change.
We can't actually see molecules breaking and forming bonds, although that's what defines chemical changes. We have to make other observations to indicate that a chemical change has happened. Some of the evidence for chemical change will involve the energy changes that occur in chemical changes, but some evidence involves the fact that new substances with different properties are formed in a chemical change.
Observations that help to indicate chemical change include:
• Temperature changes (temperature increase or decrease).
• Light given off.
• Unexpected color changes (a substance with a different color is made, rather than just mixing the original colors together).
• Bubbles are formed (but the substance is not boiling—you made a substance that is a gas at the temperature of the beginning materials, instead of a liquid).
• Different smell or taste (do not taste your chemistry experiments, though!).
• A solid forms if two clear liquids are mixed (look for floaties—technically called a precipitate).
Example $2$
Label each of the following changes as a physical or chemical change. Give evidence to support your answer.
1. Boiling water.
2. A nail rusting.
3. A green solution and colorless solution are mixed. The resulting mixture is a solution with a pale green color.
4. Two colorless solutions are mixed. The resulting mixture has a yellow precipitate.
Solution
1. Physical: boiling and melting are physical changes. When water boils, no bonds are broken or formed. The change could be written as: $\ce{H_2O} \left( l \right) \rightarrow \ce{H_2O} \left( g \right)$
2. Chemical: The dark grey nail changes color to form an orange flaky substance (the rust); this must be a chemical change. Color changes indicate chemical change. The following reaction occurs: $\ce{Fe} + \ce{O_2} \rightarrow \ce{Fe_2O_3}$
3. Physical: because none of the properties changed, this is a physical change. The green mixture is still green and the colorless solution is still colorless. They have just been combined. No color change, or other evidence of chemical change, occurred.
4. Chemical: the formation of a precipitate and the color change from colorless to yellow indicates a chemical change.
Exercise $2$
Label each of the following changes as a physical or chemical change.
1. A mirror is broken.
2. An iron nail corroded in moist air.
3. Copper metal is melted.
4. A catalytic converter changes nitrogen dioxide to nitrogen gas and oxygen gas.
Answer 1:
physical change
Answer 2:
chemical change
Answer 3:
physical change
Answer 4:
chemical change
Summary
• A physical property is a characteristic of a substance that can be observed or measured without changing the identity of the substance. Physical properties include color, density, hardness, and melting and boiling points.
• A chemical property describes the ability of a substance to undergo a specific chemical change. To identify a chemical property, we look for a chemical change.
• Physical changes are changes that do not alter the identity of a substance.
• Chemical changes are changes that occur when one substance is turned into another substance.
• Chemical changes are frequently harder to reverse than physical changes. Observations that indicate a chemical change occurred include color change, temperature change, light given off, formation of bubbles, and formation of a precipitate.
Contributors and Attributions
• Henry Agnew (UC Davis) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/01%3A_Chemistry/1.04%3A_Chemistry_-_A_Study_of_Matter_and_Its_Changes.txt |
Learning Objectives
• Describe the solid, liquid and gas phases.
• Explain the difference between a pure substance and a mixture.
• Explain the difference between an element and a compound.
• Explain the difference between a homogeneous mixture and a heterogeneous mixture.
The States of Matter
Matter typically exists in one of three states: solid, liquid, or gas. The state that a given substance exhibits is also a physical property. Some substances exist as gases at room temperature (oxygen and carbon dioxide); others, like water and mercury metal, exist as liquids. Most metals exist as solids at room temperature. All substances can exist in any of these three states.
Note
Technically speaking, a fourth state of matter called plasma exists—but it does not naturally occur on earth, so we will omit it from our study here.
Solid
Solids are defined by the following characteristics:
• Definite shape (rigid).
• Definite volume.
• Particles vibrate around fixed axes.
If we were to cool liquid mercury to its freezing point of $-39^\text{o} \text{C}$, we would notice all of the liquid particles go into the solid state, under the right pressure conditions. Mercury can be solidified when its temperature is brought to its freezing point. However, when returned to room temperature conditions, mercury does not exist in a solid state for long, and returns back to its more common liquid form.
Liquid
Liquids have the following characteristics:
• No definite shape (takes the shape of its container).
• Has definite volume.
• Particles are free to move over each other, but are still attracted to each other.
Mercury
Liquid mercury metal is an anomaly. It is the only known metal that is liquid at room temperature. Mercury also has an ability to stick to itself (surface tension)—a property all liquids exhibit. Mercury has a relatively high surface tension, which makes it very unique. In the video below, you see mercury in its common liquid form.
Video $1$: Mercury boiling to become a gas.
If we heat liquid mercury to its boiling point of $357^\text{o} \text{C}$, under the right pressure conditions, all of the particles in the liquid state would go into the gas state.
Gas
Gases have the following characteristics:
• No definite shape (takes the shape of its container).
• No definite volume.
• Particles move in random motion with little or no attraction to each other.
• Highly compressible.
The characteristics of the three states of matter are listed in Table $1$ and an animation of the movement and position of the individual particles is shown in Figure $2$
Table $1$ Characteristics of the Three States of Matter
Characteristics Solids Liquids Gases
shape definite indefinite indefinite
volume definite definite indefinite
relative intermolecular interaction strength strong moderate weak
relative particle positions in contact and fixed in place in contact, but not fixed not in contact, random positions
Table $2$ images of the states of matter.
Images
The gaseous state The liquid state The solid state
Figure $2$ A microscopic model showing particles (atoms or molecules) in the gaseous, liquid, and solid states.
Example $1$
What state(s) of matter does each statement describe?
1. This state has a definite volume, but no definite shape.
2. This state has no definite volume.
3. This state allows the individual particles to move about while remaining in contact.
Solution
1. This statement describes the liquid state.
2. This statement describes the gas state.
3. This statement describes the liquid state.
Exercise $1$
What state or states of matter does each statement describe?
1. This state has individual particles in a fixed position with regard to one another.
2. This state has individual particles far apart from each other in space.
3. This state has a definite shape.
Answer a:
solid
Answer b:
gas
Answer c:
solid
Substances and Mixtures
One useful way of organizing our understanding of matter is to think of a hierarchy that extends down from the most general and complex, to the simplest and most fundamental (Figure $1$). Matter can be classified into two broad categories: pure substances and mixtures. A pure substance is a form of matter that has a constant composition (meaning it's the same everywhere) and properties that are constant throughout the sample (meaning there is only one set of properties such as melting point, color, boiling point, etc. throughout the matter). A material composed of two or more substances is a mixture.
Ordinary table salt is called sodium chloride. It is considered a substance because it has a uniform and definite composition. All samples of sodium chloride are chemically identical. Water is also a pure substance. Salt easily dissolves in water, but salt water cannot be classified as a substance because its composition can vary. You may dissolve a small or large amount of salt into a given amount of water. A mixture is a physical blend of two or more components, each of which retains its own identity and properties in the mixture. Only the form of the salt is changed when it is dissolved into water. It retains its composition and properties.
A homogeneous mixture is a mixture in which the composition is uniform throughout the mixture. The salt water described above is homogeneous because the dissolved salt is evenly distributed throughout the entire salt water sample. Often it is easy to confuse a homogeneous mixture with a pure substance, because they are both uniform. The difference is that the composition of the substance is always the same. The amount of salt in the salt water can vary from one sample to another. All solutions would be considered homogeneous because the dissolved material is present in the same amount throughout the solution.
A heterogeneous mixture is a mixture in which the composition is not uniform throughout the mixture. Vegetable soup is a heterogeneous mixture. Any given spoonful of soup will contain varying amounts of the different vegetables and other components of the soup.
Elements and Compounds
Elements and compounds are both examples of pure substances. A substance that cannot be broken down into chemically simpler components is an element. Aluminum, which is used in soda cans, is an element. A substance that can be broken down into chemically simpler components (because it has more than one element) is a compound. For example, water is a compound composed of the elements hydrogen and oxygen. Today, there are about 118 elements in the known universe. In contrast, scientists have identified tens of millions of different compounds to date.
Atoms, Elements, and Compounds
Video $2$ Atoms, elements, and compounds (first five minutes of video)
Phase
A phase is any part of a sample that has a uniform composition and properties. By definition, a pure substance or a homogeneous mixture consists of a single phase. A heterogeneous mixture consists of two or more phases. When oil and water are combined, they do not mix evenly, but instead form two separate layers. Each of the layers is called a phase.
Example $2$
Identify each substance as a compound, an element, a heterogeneous mixture, or a homogeneous mixture (solution).
1. filtered tea
2. freshly squeezed orange juice
3. a compact disc
4. aluminum oxide (a white powder that contains a 2:3 ratio of aluminum and oxygen atoms)
5. selenium
Given: a chemical substance.
Asked for: its classification.
Strategy:
1. Decide whether a substance is chemically pure. If it is pure, the substance is either an element or a compound. If a substance can be separated into its elements, it is a compound.
2. If a substance is not chemically pure, it is either a heterogeneous mixture or a homogeneous mixture. If its composition is uniform throughout, it is a homogeneous mixture.
Solution:
1. A) Tea is a solution of compounds in water, so it is not chemically pure. It is usually separated from tea leaves by filtration.
B) Because the composition of the solution is uniform throughout, it is a homogeneous mixture.
2. A) Orange juice contains particles of solid (pulp) as well as liquid; it is not chemically pure.
B) Because its composition is not uniform throughout, orange juice is a heterogeneous mixture.
3. A) A compact disc is a solid material that contains more than one element, with regions of different compositions visible along its edge. Hence, a compact disc is not chemically pure.
B) The regions of different composition indicate that a compact disc is a heterogeneous mixture.
4. A) Aluminum oxide is a single, chemically pure compound.
5. A) Selenium is one of the known elements.
Exercise $2$
Identify each substance as a compound, an element, a heterogeneous mixture, or a homogeneous mixture (solution).
1. white wine
2. mercury
3. ranch-style salad dressing
4. table sugar (sucrose)
Answer a:
homogeneous mixture (solution)
Answer b:
element
Answer c:
heterogeneous mixture
Answer d:
compound
Example $3$
How would a chemist categorize each example of matter?
1. saltwater
2. soil
3. water
4. oxygen
Solution
1. Saltwater acts as if it were a single substance even though it contains two substances—salt and water. Saltwater is a homogeneous mixture, or a solution.
2. Soil is composed of small pieces of a variety of materials, so it is a heterogeneous mixture.
3. Water is a substance; more specifically, because water is composed of hydrogen and oxygen, it is a compound.
4. Oxygen, a substance, is an element.
Exercise $3$
How would a chemist categorize each example of matter?
1. filtered coffee
2. hydrogen
3. an egg
Answer a:
a homogeneous mixture (solution)
Answer b:
element
Answer c:
heterogeneous mixture.
Summary
• Three states of matter exist: solid, liquid, and gas.
• Solids have a definite shape and volume.
• Liquids have a definite volume, but take the shape of the container.
• Gases have no definite shape or volume
• Matter can be classified into two broad categories: pure substances and mixtures.
• A pure substance is a form of matter that has a constant composition, and properties that are constant throughout the sample.
• Mixtures are physical combinations of two or more elements and/or compounds.
• Mixtures can be classified as homogeneous or heterogeneous.
• Elements and compounds are both examples of pure substances. Compounds are substances that are made up of more than one type of atom. Elements are the simplest substances made up of only one type of atom. | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/01%3A_Chemistry/1.05%3A_Classification_of_Matter.txt |
Learning Objectives
• Express quantities properly, using a number and a unit.
• State the different measurement systems used in chemistry.
• Express a large number or a small number in scientific notation.
• Learn how to use SI prefixes.
• Perform unit conversions using conversion factors.
A coffee maker’s instructions tell you to fill the coffeepot with 4 cups of water and use 3 scoops of coffee. When you follow these instructions, you are measuring. When you visit a doctor’s office, a nurse checks your temperature, height, weight, and perhaps blood pressure (Figure $1$); the nurse is also measuring.
Chemists measure the properties of matter and express these measurements as quantities. A quantity is an amount of something, and consists of a number and a unit. The number tells us how many (or how much), and the unit tells us what the scale of measurement is. For example, when a distance is reported as “5 kilometers,” we know that the quantity has been expressed in units of kilometers and that the number of kilometers is 5. If you ask a friend how far he or she walks from home to school, and the friend answers “12” without specifying a unit, you do not know whether your friend walks 12 miles, 12 kilometers, 12 furlongs, or 12 yards...etc. Both a number and a unit must be included to express a quantity properly.
To understand chemistry, we need a clear understanding of the units chemists work with and the rules they follow for expressing numbers. The next two sections examine the rules for expressing numbers.
All measurements depend on the use of units that are well known and understood. The English system of measurement units (inches, feet, ounces, etc.) are not used in science because of the difficulty in converting from one unit to another. The metric system is used because all metric units are based on multiples of 10, making conversions very simple. The metric system was originally established in France in 1795. The International System of Units is a system of measurement based on the metric system. The acronym SI is commonly used to refer to this system and stands for the French term, Le Système International d'Unités. The SI was adopted by international agreement in 1960 and is composed of the seven base units shown in Table $1$.
Quantity SI Base Unit Symbol
Table $1$ SI Base Units of Measurement
Length meter $\text{m}$
Mass kilogram $\text{kg}$
Temperature kelvin $\text{K}$
Time second $\text{s}$
Amount of a Substance mole $\text{mol}$
Electric Current ampere $\text{A}$
Luminous Intensity candela $\text{cd}$
The first units are frequently encountered in chemistry. All other measurement quantities, such as volume, force, and energy, can be derived from these seven base units.
Exponential Numbers: Powers of Ten
Chemists often work with numbers that are exceedingly large or small. For example, entering the mass in grams of a hydrogen atom into a calculator would require a display with at least 24 decimal places. A system called scientific notation avoids much of the tedium and awkwardness of manipulating numbers with large or small magnitudes. Furthermore, use of prefixes is another way to express measurements involving large and small numbers.
Scientific Notation
In scientific notation, numbers are expressed in the form
$N \times 10^n \nonumber$
where N is greater than or equal to 1 and less than 10 (1 ≤ N < 10), and n is a positive or negative integer (100 = 1). The number 10 is called the base because it is this number that is raised to the power $n$. Although a base number may have values other than 10, the base number in scientific notation is always 10.
A simple way to convert numbers to scientific notation is to move the decimal point as many places to the left or right as needed to give a number from 1 to 10 (N). The magnitude of n is then determined as follows:
• If the decimal point is moved to the left n places, n is positive.
• If the decimal point is moved to the right n places, n is negative.
Another way to remember this is to recognize that as the number N decreases in magnitude, the exponent increases and vice versa. The application of this rule is illustrated in Example $1$.
Example $1$: Expressing Numbers in Scientific Notation
Convert each number to scientific notation.
1. 637.8
2. 0.0479
3. 12,378
4. 0.00032
Solution
Example $1$ Solution and Explanation
Explanation Answer
a
To convert 637.8 to a number from 1 to 10, we move the decimal point two places to the left: 637.8
Because the decimal point was moved two places to the left, n = 2.
$6.378 \times 10^2$
b
To convert 0.0479 to a number from 1 to 10, we move the decimal point two places to the right: 0.0479
Because the decimal point was moved two places to the right, n = −2.
$4.79 \times 10^{−2}$
c Because the decimal point was moved four places to the left, n = 4. $1.2378 \times 10^4$
d Because the decimal point was moved four places to the right, n = −4. $3.2 \times 10^{−4}$
Exercise $1$
Convert each ordinary number to scientific notation, or vice versa.
1. 67,000,000,000
2. 1,689
3. 12.6
Answer a
6.7 × 1010
Answer b
1.689 × 103
Answer c
1.26 × 101
Exercise $2$
Convert each ordinary number to scientific notation, or vice versa.
1. 0.000006567
2. 6.22 × 10−2
3. 9.9 × 10−9
Answer a
6.567 × 10−6
Answer b
0.0622
Answer c
0.0000000099
Metric Prefixes
Conversions between metric system units are straightforward because the system is based on powers of ten. For example, meters, centimeters, and millimeters are all metric units of length. There are 10 millimeters in 1 centimeter and 100 centimeters in 1 meter. Metric prefixes are used to distinguish between units of different size. These prefixes all derive from either Latin or Greek terms. For example, mega comes from the Greek word $\mu \varepsilon \gamma \alpha \varsigma$, meaning "great". Table $2$ lists the most common metric prefixes and their relationship to the central unit that has no prefix. Length is used as an example to demonstrate the relative size of each prefixed unit.
Prefix Unit Abbreviation Meaning Example
Table $2$ SI Prefixes
giga $\text{G}$ 1,000,000,000 1 gigameter $\left( \text{Gm} \right)=10^9 \: \text{m}$
mega $\text{M}$ 1,000,000 1 megameter $\left( \text{Mm} \right)=10^6 \: \text{m}$
kilo $\text{k}$ 1,000 1 kilometer $\left( \text{km} \right)=1,000 \: \text{m}$
hecto $\text{h}$ 100 1 hectometer $\left( \text{hm} \right)=100 \: \text{m}$
deka $\text{da}$ 10 1 dekameter $\left( \text{dam} \right)=10 \: \text{m}$
1 1 meter $\left( \text{m} \right)$
deci $\text{d}$ 1/10 1 decimeter $\left( \text{dm} \right)=0.1 \: \text{m}$
centi $\text{c}$ 1/100 1 centimeter $\left( \text{cm} \right)=0.01 \: \text{m}$
milli $\text{m}$ 1/1,000 1 millimeter $\left( \text{mm} \right)=0.001 \: \text{m}$
micro $\mu$ 1/1,000,000 1 micrometer $\left( \mu \text{m} \right)=10^{-6} \: \text{m}$
nano $\text{n}$ 1/1,000,000,000 1 nanometer $\left( \text{nm} \right)=10^{-9} \: \text{m}$
pico $\text{p}$ 1/1,000,000,000,000 1 picometer $\left( \text{pm} \right)=10^{-12} \: \text{m}$
There are a couple of odd little practices with the use of metric abbreviations. Most abbreviations are lowercase. We use "$\text{m}$" for meter and not "$\text{M}$". However, when it comes to volume, the base unit "liter" is abbreviated as "$\text{L}$" and not "$\text{l}$". So, 3.5 milliliters is written as $3.5 \: \text{mL}$.
As a practical matter, whenever possible, you should express the units in a small and manageable number. If you are measuring the weight of a material that weighs $6.5 \: \text{kg}$, this is easier than saying it weighs $6500 \: \text{g}$ or $0.65 \: \text{dag}$. All three are correct, but the $\text{kg}$ units in this case make for a small and easily managed number. However, if a specific problem needs grams instead of kilograms, go with the grams for consistency.
Example $2$: Unit Abbreviations
Give the abbreviation for each unit and define the abbreviation in terms of the base unit.
1. kiloliter
2. microsecond
3. decimeter
4. nanogram
Solutions
Example $2$ Solution and Explanation
Explanation Answer
a The prefix kilo means “1,000 ×,” so 1 kL equals 1,000 L kL
b The prefix micro implies 1/1,000,000th of a unit, so 1 µs equals 0.000001 s. µs
c The prefix deci means 1/10th, so 1 dm equals 0.1 m. dm
d The prefix nano means 1/1000000000, so a nanogram is equal to 0.000000001 g ng
Exercise $3$
Give the abbreviation for each unit and define the abbreviation in terms of the base unit.
1. kilometer
2. milligram
3. nanosecond
4. centiliter
Answer a:
km
Answer b:
mg
Answer c:
ns
Answer d:
cL
Mass and Weight
Mass is a measure of the amount of matter that an object contains. The mass of an object is made in comparison to the standard mass of 1 kilogram. The kilogram was originally defined as the mass of $1 \: \text{L}$ of liquid water at $4^\text{o} \text{C}$ (volume of a liquid changes slightly with temperature). In the laboratory, mass is measured with a balance (see below), which must be calibrated with a standard mass so that its measurements are accurate.
Other common units of mass are the gram and the milligram. A gram is 1/1000th of a kilogram, meaning that there are $1000 \: \text{g}$ in $1 \: \text{kg}$. A milligram is 1/1000th of a gram, so there are $1000 \: \text{mg}$ in $1 \: \text{g}$.
Mass is often confused with the term weight. Weight is a measure of force that is equal to the gravitational pull on an object. The weight of an object is dependent on its location. On the moon, the force due to gravity is about one sixth that of the gravitational force on Earth. Therefore, a given object will weigh six times more on Earth than it does on the moon. Since mass is dependent only on the amount of matter present in an object, mass does not change with location. Weight measurements are often made with a spring scale, by reading the distance that a certain object pulls down and stretches a spring.
Length and Volume
Length is the measurement of the extent of something along its greatest dimension. The SI basic unit of length, or linear measure, is the meter $\left( \text{m} \right)$. All measurements of length may be made in meters, though the prefixes listed in various tables will often be more convenient. The width of a room may be expressed as about 5 meters $\left( \text{m} \right)$, whereas a large distance, such as the distance between New York City and Chicago, is better expressed as 1150 kilometers $\left( \text{km} \right)$. Very small distances can be expressed in units such as the millimeter or the micrometer. The width of a typical human hair is about 10 micrometers $\left( \mu \text{m} \right)$.
Volume is the amount of space occupied by a sample of matter. The volume of a regular object can be calculated by multiplying its length by its width and its height. Since each of those is a linear measurement, we say that units of volume are derived from units of length. The SI unit of volume is the cubic meter $\left( \text{m}^3 \right)$, which is the volume occupied by a cube that measures $1 \: \text{m}$ on each side. This very large volume is not very convenient for typical use in a chemistry laboratory. A liter $\left( \text{L} \right)$ is the volume of a cube that measures $10 \: \text{cm}$ $\left( 1 \: \text{dm} \right)$ on each side. A liter is thus equal to both $1000 \: \text{cm}^3$ $\left( 10 \: \text{cm} \times 10 \: \text{cm} \times 10 \: \text{cm} \right)$ and to $1 \: \text{dm}^3$.
1L = $1000 \: \text{cm}^3$ $\left( 10 \: \text{cm} \times 10 \: \text{cm} \times 10 \: \text{cm} \right)$ = $1 \: \text{dm}^3$
A smaller unit of volume that is commonly used is the milliliter ($\text{mL}$). A milliliter is the volume of a cube that measures $1 \: \text{cm}$ on each side. Therefore, a milliliter is equal to a cubic centimeter $\left( \text{cm}^3 \right)$.
1 mL $\left( \text{cm}^3 \right)$
There are $1000 \: \text{mL}$ in $1 \: \text{L}$, which is the same as saying that there are $1000 \: \text{cm}^3$ in $1 \: \text{dm}^3$.
During your studies of chemistry (and physics also), you will note that mathematical equations are used in many different applications. Many of these equations have a number of different variables with which you will need to work. You should also note that these equations will often require you to use measurements with their units. Algebra skills become very important here!
Converting Between Units with Conversion Factors
A conversion factor is a factor used to convert one unit of measurement into another. A simple conversion factor can be used to convert meters into centimeters, or a more complex one can be used to convert miles per hour into meters per second. Since most calculations require measurements to be in certain units, you will find many uses for conversion factors. Always remember that a conversion factor has to represent a fact; this fact can either be simple or much more complex. For instance, you already know that 12 eggs equal one dozen. A more complex fact is that the speed of light is $1.86 \times 10^5$ miles/sec. Either one of these can be used as a conversion factor, depending on what type of calculation you might be working with (Table $1$).
Table $3$ Conversion Factors from SI units to English Units
English Units Metric Units Quantity
1 ounce (oz) 28.35 grams (g) *mass
1 fluid once (oz) 2.96 mL volume
2.205 pounds (lb) 1 kilogram (kg) *mass
1 inch (in) 2.54 centimeters (cm) length
0.6214 miles (mi) 1 kilometer (km) length
1 quarter (qt) 0.95 liters (L) volume
*Pounds and ounces are technically units of force, not mass, but this fact is often ignored by the non-scientific community.
Of course, there are other ratios which are not listed in Table $1$. They may include:
• Ratios embedded in the text of the problem (using words such as per or in each, or using symbols such as / or %).
• Conversions in the metric system, as covered earlier in this chapter.
• Common knowledge ratios (such as 60 seconds $=$ 1 minute).
If you learned the SI units and prefixes described, then you know that 1 cm is 1/100th of a meter.
$1\; \rm{cm} = \dfrac{1}{100} \; \rm{m} = 10^{-2}\rm{m} \nonumber$
or
$100\; \rm{cm} = 1\; \rm{m} \nonumber$
Suppose we divide both sides of the equation by $1 \text{m}$ (both the number and the unit):
$\mathrm{\dfrac{100\:cm}{1\:m}=\dfrac{1\:m}{1\:m}} \nonumber$
As long as we perform the same operation on both sides of the equals sign, the expression remains an equality. Look at the right side of the equation; it now has the same quantity in the numerator (the top) as it has in the denominator (the bottom). Any fraction that has the same quantity in the numerator and the denominator has a value of 1:
$\dfrac{ \text{100 cm}}{\text{1 m}} = \dfrac{ \text{1000 mm}}{\text{1 m}}= \dfrac{ 1\times 10^6 \mu \text{m}}{\text{1 m}}= 1 \nonumber$
We know that 100 cm is 1 m, so we have the same quantity on the top and the bottom of our fraction, although it is expressed in different units.
Performing Dimensional Analysis
Dimensional analysis is amongst the most valuable tools used by physical scientists. Simply put, it is the conversion between an amount in one unit to the corresponding amount in a desired unit, using various conversion factors. This is valuable because certain measurements are more accurate or easier to find than others. The use of units in a calculation to ensure that we obtain the final proper units is called dimensional analysis.
Here is a simple example: How many centimeters are there in 3.55 m? Perhaps you can determine the answer in your head. If there are 100 cm in every meter, then 3.55 m equals 355 cm. To solve the problem more formally with a conversion factor, we first write the quantity we are given, 3.55 m. Then we multiply this quantity by a conversion factor, which is the same as multiplying it by 1. We can write 1 as $\mathrm{\frac{100\:cm}{1\:m}}$ and multiply:
$3.55 \; \rm{m} \times \dfrac{100 \; \rm{cm}}{1\; \rm{m}} \nonumber$
The 3.55 m can be thought of as a fraction with a 1 in the denominator. Because m, the abbreviation for meters, occurs in both the numerator and the denominator of our expression, they cancel out:
$\dfrac{3.55 \; \cancel{\rm{m}}}{ 1} \times \dfrac{100 \; \rm{cm}}{1 \; \cancel{\rm{m}}} \nonumber$
The final step is to perform the calculation that remains once the units have been canceled:
$\dfrac{3.55}{1} \times \dfrac{100 \; \rm{cm}}{1} = 355 \; \rm{cm} \nonumber$
In the final answer, we omit the 1 in the denominator. Thus, by a more formal procedure, we find that 3.55 m equals 355 cm. A generalized description of this process is as follows:
quantity (in old units) × conversion factor = quantity (in new units)
You may be wondering why we use a seemingly complicated procedure for a straightforward conversion. In later studies, the conversion problems you will encounter will not always be so simple. If you can master the technique of applying conversion factors, you will be able to solve a large variety of problems.
In the previous example, we used the fraction $\frac{100 \; \rm{cm}}{1 \; \rm{m}}$ as a conversion factor. Does the conversion factor $\frac{1 \; \rm m}{100 \; \rm{cm}}$ also equal 1? Yes, it does; it has the same quantity in the numerator as in the denominator (except that they are expressed in different units). Why did we not use that conversion factor? If we had used the second conversion factor, the original unit would not have canceled, and the result would have been meaningless. Here is what we would have gotten:
$3.55 \; \rm{m} \times \dfrac{1\; \rm{m}}{100 \; \rm{cm}} = 0.0355 \dfrac{\rm{m}^2}{\rm{cm}} \nonumber$
For the answer to be meaningful, we have to construct the conversion factor in a form that causes the original unit to cancel out. Figure $1$ shows a concept map for constructing a proper conversion.
The General Steps in Performing Dimensional Analysis
1. Identify the "given" information in the problem. Look for a number with units to start this problem with.
2. What is the problem asking you to "find"? In other words, what unit will your answer have?
3. Use ratios and conversion factors to cancel out the units that aren't part of your answer, and leave you with units that are part of your answer.
4. When your units cancel out correctly, you are ready to do the math. You are multiplying fractions, so you multiply the top numbers and divide by the bottom numbers in the fractions.
Steps for Performing Dimensional Analysis and Examples
Example $2$ Example $3$
Steps for Problem Solving The average volume of blood in an adult male is 4.7 L. What is this volume in gallons? A hummingbird can flap its wings once in 18 ms. How many seconds are in 18 ms?
Identify the "given" information and what the problem is asking you to "find." Given: 4.7 L
Find: gal
Given: 18 ms
Find: s
List other known quantities. $1\, L = 3.785 gal$ $1 \,ms = 10^{-3} s$
Prepare a concept map and use the proper conversion factor.
Cancel units and calculate. $4.7 \cancel{\rm{L}} \times \dfrac{1 \; \rm{gal}}{3.785\; \cancel{\rm{L}}} = 1.2\; \rm{gal}$
$18 \; \cancel{\rm{ms}} \times \dfrac{10^{-3}\; \rm{s}}{1 \; \cancel{\rm{ms}}} = 0.018\; \rm{s}$
or
$18 \; \cancel{\rm{ms}} \times \dfrac{1\; \rm{s}}{1,000 \; \cancel{\rm{ms}}} = 0.018\; \rm{s}$
Think about your result. The amount in gal should be slightly less than 4 times smaller than the given amount in L. The amount in s should be 1/1000 the given amount in ms.
Exercise $4$
Perform each conversion.
1. 101,000 ns to seconds
2. 32.08 kg to grams
3. 1.53 grams to cg
Answer a:
$1.01000 x 10^{-4} s$
Answer b:
$3.208 x 10^{4} g$
Answer c:
$1.53 x 10^{2} g$
Multiple Conversions
Sometimes you will have to perform more than one conversion to obtain the desired unit. For example, suppose you want to convert 54.7 km into millimeters. We will set up a series of conversion factors so that each conversion factor produces the next unit in the sequence. We first convert the given amount in km to the base unit, which is meters. We know that 1,000 m =1 km. Then we convert meters to mm, remembering that $1\; \rm{mm}$ = $10^{-3}\; \rm{m}$.
Calculation
\begin{align} 54.7 \; \cancel{\rm{km}} \times \dfrac{1,000 \; \cancel{\rm{m}}}{1\; \cancel{\rm{km}}} \times \dfrac{1\; \cancel{\rm{mm}}}{\cancel{10^{-3} \rm{m}}} & = 54,700,000 \; \rm{mm} \ &= 5.47 \times 10^7\; \rm{mm} \end{align} \nonumber
In each step, the previous unit is canceled and the next unit in the sequence is produced; each successive unit cancels out, until only the unit needed in the answer is left.
Example $4$: Unit Conversion
Convert 58.2 ms to megaseconds in one multi-step calculation.
Solution
Example $4$ Steps for Problem Solving Unit Conversion
Unit Conversion
Identify the "given" information and what the problem is asking you to "find."
Given: 58.2 ms
Find: Ms
List other known quantities.
$1 ms = 10^{-3} s$
$1 Ms = 10^6s$
Prepare a concept map.
Calculate.
\begin{align} 58.2 \; \cancel{\rm{ms}} \times \dfrac{10^{-3} \cancel{\rm{s}}}{1\; \cancel{\rm{ms}}} \times \dfrac{1\; \rm{Ms}}{1,000,000\; \cancel{ \rm{s}}} & =0.0000000582\; \rm{Ms} \nonumber\ &= 5.82 \times 10^{-8}\; \rm{Ms}\nonumber \end{align}\nonumber
Neither conversion factor affects the number of significant figures in the final answer.
Example $5$: Unit Conversion
How many seconds are in a day?
Solution
Example $5$: Steps for Problem Solving Unit Conversions
Unit Conversion
Identify the "given" information and what the problem is asking you to "find."
Given: 1 day
Find: s
List other known quantities.
1 day = 24 hours
1 hour = 60 minutes
1 minute = 60 seconds
Prepare a concept map.
Calculate. $1 \: \text{d} \times \frac{24 \: \text{hr}}{1 \: \text{d}}\times \frac{60 \: \text{min}}{1 \: \text{hr}} \times \frac{60 \: \text{s}}{1 \: \text{min}} = 86,400 \: \text{s} \nonumber$
Exercise $5$
Perform each conversion in one multistep calculation.
1. 43.007 ng to kg
2. 1005 in to ft
3. 12 mi to km
Answer a:
$4.3007 x 10^{-14} kg$
Answer b:
83.75 ft
Answer c:
19 km
Summary
• Metric prefixes derive from Latin or Greek terms. The prefixes are used to make the units manageable.
• The SI system is based on multiples of ten. There are seven basic units in the SI system. Five of these units are commonly used in chemistry.
• Mass is a measure of the amount of matter that an object contains.
• Weight is a measure of force that is equal to the gravitational pull on an object.
• Mass is independent of location, while weight depends on location.
• Length is the measurement of the extent of something along its greatest dimension.
• Volume is the amount of space occupied by a sample of matter.
• Volume can be determined by knowing the length of each side of the item.
• Conversion factors are used to convert one unit of measurement into another.
• Dimensional analysis (unit conversions) involves the use of conversion factors that will cancel unwanted units and produce desired units. | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/01%3A_Chemistry/1.06%3A_The_Measurement_of_Matter.txt |
Learning Objectives
• Define density.
• Determine the density of various substances.
Density ($\rho$) is a physical property found by dividing the mass of an object by its volume. Regardless of the sample size, density is always constant. For example, the density of a pure sample of tungsten is always 19.25 grams per cubic centimeter. This means that whether you have one gram or one kilogram of the sample, the density will never vary. The equation, as we already know, is as follows:
$\text{Density}=\dfrac{\text{Mass}}{\text{Volume}} \nonumber$
or just
$\rho =\dfrac{m}{V} \nonumber$
Table $1$ Densities of Common Substances
Substance Density at 25°C (g/cm3)
blood 1.035
body fat 0.918
whole milk 1.030
corn oil 0.922
mayonnaise 0.910
honey 1.420
Notice in Table $1$ that corn oil has a lower mass to volume ratio than water. This means that when added to water, corn oil will “float.”
Density can be measured for all substances, solids, liquids and gases. For solids and liquids, density is often reported using the units of g/cm3. Densities of gases, which are significantly lower than the densities of solids and liquids, are often given using units of g/L.
Example $1$: Ethyl Alcohol
Calculate the density of a 30.2 mL sample of ethyl alcohol with a mass of 23.71002 g
Solution
$\rho = \dfrac{23.71002\,g}{30.2\, mL} = 0.785\, g/mL \nonumber$
Exercise $1$
1. Find the density (in kg/L) of a sample that has a volume of 36.5 L and a mass of 10.0 kg.
2. If you have a 2.130 mL sample of acetic acid with mass 2.234 g, what is the density in g/mL?
Answer a
$0.274 \,kg/L$
Answer b
$1.049 \,g/mL$
Summary
• Density is defined as the mass of an object divided by its volume.
Contributors and Attributions
• Henry Agnew (UC Davis)
1.08: Energy - Heat and Temperature
Learning Objectives
• Identify the difference between temperature and heat.
• Recognize the different scales used to measure temperature.
• Perform temperature conversions from one unit to another.
The concept of temperature may seem familiar to you, but many people confuse temperature with heat. Temperature is a measure of how hot or cold an object is relative to another object (its thermal energy content), whereas heat is the flow of thermal energy between objects with different temperatures. Temperature is a measure of the average kinetic energy of the particles in matter. In everyday usage, temperature indicates a measure of how hot or cold an object is. Temperature is an important parameter in chemistry. When a substance changes from solid to liquid, it is because there was in increase in the temperature of the material. Chemical reactions usually proceed faster if the temperature is increased. Many unstable materials (such as enzymes) will be viable longer at lower temperatures.
Three different scales are commonly used to measure temperature: Fahrenheit (expressed as °F), Celsius (°C), and Kelvin (K). Thermometers measure temperature by using materials that expand or contract when heated or cooled. Mercury or alcohol thermometers, for example, have a reservoir of liquid that expands when heated and contracts when cooled, so the liquid column lengthens or shortens as the temperature of the liquid changes.
The Fahrenheit Scale
The first thermometers were glass and contained alcohol, which expanded and contracted as the temperature changed. The German scientist Daniel Gabriel Fahrenheit used mercury in the tube, an idea put forth by Ismael Boulliau. The Fahrenheit scale was first developed in 1724, and tinkered with for some time after that. The main problem with this scale is the arbitrary definitions of temperature. The freezing point of water was defined as $32^\text{o} \text{F}$ and the boiling point as $212^\text{o} \text{F}$. The Fahrenheit scale is typically not used for scientific purposes.
The Celsius Scale
The Celsius scale of the metric system is named after Swedish astronomer Anders Celsius (1701-1744). The Celsius scale sets the freezing point and boiling point of water at $0^\text{o} \text{C}$ and $100^\text{o} \text{C}$, respectively. The distance between those two points is divided into 100 equal intervals, each of which is one degree. Another term sometimes used for the Celsius scale is "centigrade" because there are 100 degrees between the freezing and boiling points of water on this scale. However, the preferred term is "Celsius".
The Kelvin Scale
The Kelvin temperature scale is named after Scottish physicist and mathematician Lord Kelvin (1824-1907). It is based on molecular motion, with the temperature of $0 \: \text{K}$, also known as absolute zero, being the point where all molecular motion ceases. The freezing point of water on the Kelvin scale is $273.15 \: \text{K}$, while the boiling point is $373.15 \: \text{K}$. Notice that there is no "degree" used in the temperature designation. Unlike the Fahrenheit and Celsius scales where temperatures are referred to as "degrees $\text{F}$" or "degrees $\text{C}$", we simply designated temperatures in the Kelvin scale as kelvins.
Converting between Scales
The Kelvin is the same size as the Celsius degree, so measurements are easily converted from one to the other. The freezing point of water is 0°C = 273.15 K; the boiling point of water is 100°C = 373.15 K. The Kelvin and Celsius scales are related as follows:
$T \,\text{(in °C)} + 273.15 = T \, \text{(in K)} \label{1.8.1}$
$T \; \text{ (in K)} − 273.15 = T \; \text{(in °C)} \label{1.8.2}$
Degrees on the Fahrenheit scale, however, are based on an English tradition of using 12 divisions, just as 1 ft = 12 in. The relationship between degrees Fahrenheit and degrees Celsius is as follows:
$°C = \dfrac{5}{9} \times (°F-32) \label{1.8.3}$
$°F = \dfrac{9}{5} \times (°C)+32 \label{1.8.4}$
There is only one temperature for which the numerical value is the same on both the Fahrenheit and Celsius scales: −40°C = −40°F.
Example $1$: Temperature Conversions
A student is ill with a temperature of 103.5°F. What is her temperature in °C and K?
Solution
Converting from Fahrenheit to Celsius requires the use of Equation $\ref{1.8.3}$:
\begin{align*} °C &= \dfrac{5}{9} \times (103.5°F - 32) \ &= 39.7 \,°C\end{align*}
Converting from Celsius to Kelvin requires the use of Equation $\ref{1.8.1}$:
\begin{align*} K &= 39.7 \,°C + 273.15 \ &= 312.9\,K \end{align*}
Exercise $1$
Convert each temperature to °C and °F.
1. the temperature of the surface of the sun (5800 K)
2. the boiling point of gold (3080 K)
3. the boiling point of liquid nitrogen (77.36 K)
Answer (a)
5527 °C, 9980 °F
Answer (b)
2807 °C, 5084 °F
Answer (c)
-195.8 °C, -320.4 °F
Summary
• Three different scales are commonly used to measure temperature: Fahrenheit (expressed as °F), Celsius (°C), and Kelvin (K).
• The freezing point of water is 0°C = 273.15 K; the boiling point of water is 100°C = 373.15 K.
• Both a degree Celsius and a kelvin are 9/5 the size of a degree Fahrenheit.
Contributors and Attributions
• Henry Agnew (UC Davis) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/01%3A_Chemistry/1.07%3A_Density.txt |
Thumbnail: Spinning Buckminsterfullerene (\(\ce{C60}\)). (CC BY-SA 3.0; unported; Sponk).
02: Atoms
Learning Objectives
• Give a short history of the concept of the atom.
• Describe the contributions of Democritus to atomic theory.
You learned earlier that all matter in the universe is made out of tiny building blocks called atoms; all modern scientists accept the concept of the atom. However, when the concept of the atom was first proposed about 2,500 years ago, ancient philosophers laughed at the idea. It has always been difficult to convince people of the existence of things that are too small to see. We will spend some time considering the evidence (observations) that convince scientists of the existence of atoms.
Democritus and the Greek Philosophers
About 2,500 years ago, early Greek philosophers believed the entire universe was a single, huge entity. In other words, "everything was one." They believed that all objects, all matter, and all substances were connected as a single, unchangeable "thing." One of the first people to propose "atoms" was a man known as Democritus. As an alternative to the beliefs of the Greek philosophers, he suggested that atomos, or atomon—tiny, indivisible, solid objects—make up all matter in the universe.
Democritus then reasoned that changes occur when the many atomos in an object were reconnected or recombined in different ways. Democritus even extended this theory, suggesting that there were different varieties of atomos with different shapes, sizes, and masses. He thought, however, that shape, size, and mass were the only properties differentiating the different types of atomos. According to Democritus, other characteristics, like color and taste, did not reflect properties of the atomos themselves, but rather, resulted from the different ways in which the atomos were combined and connected to one another.
The early Greek philosophers tried to understand the nature of the world through reason and logic, but not through experiment and observation. As a result, they had some very interesting ideas, but they felt no need to justify their ideas based on life experiences. In a lot of ways, you can think of the Greek philosophers as being "all thought and no action." It's truly amazing how much they achieved using their minds, but because they never performed any experiments, they missed or rejected a lot of discoveries that they could have made otherwise. Greek philosophers dismissed Democritus' theory entirely. Sadly, it took over two millennia before the theory of atomos (or "atoms," as they're known today) was fully appreciated.
Summary
• 2,500 years ago, Democritus suggested that all matter in the universe was made up of tiny, indivisible, solid objects he called "atomos." However, other Greek philosophers disliked Democritus' "atomos" theory because they felt it was illogical.
Contributors and Attributions
• Henry Agnew (UC Davis)
2.02: Scientific Laws - Conservation of Mass and Definite Proportions
Learning Objectives
• Explain the law of conservation of mass.
• Explain the law of definite proportions.
The Law of Conservation of Mass
It may seem as though burning destroys matter, but the same amount, or mass, of matter still exists after a campfire as before. Figure $2$ (below) shows that when wood burns, it combines with oxygen and changes not only to ashes, but also to carbon dioxide and water vapor. The gases float off into the air, leaving behind just the ashes. Suppose you had measured the mass of the wood before it burned and the mass of the ashes after it burned. Also assume that you had been able to measure the oxygen used by the fire and the gases produced by the fire. What would you find? The total mass of matter after the fire would be the same as the total mass of matter before the fire.
The law of conservation of mass was created in 1789 by French chemist Antoine Lavoisier. The law of conservation of mass states that matter cannot be created or destroyed in a chemical reaction. For example—in Figure $3$—when wood burns, the mass of the soot, ashes, and gases equals the original mass of the wood and the oxygen when it first reacted.
So, the mass of the product(s) equals the mass of the reactants. Reactants are two or more elements chemically interacting to make a new substance; product is the substance(s) formed as the result of a chemical reaction (Video $1$). Matter (and mass) may not be able to be created or destroyed, but can change forms to other substances like liquids, gases, solids, etc.
Video $1$ This is a nice little demonstration showing the conservation of mass in action.
It's important to know the law well. If a 300 kg tree is burning down—when the process is complete—there are only ashes left, and all of them together weigh 10 kg. You may wonder where the other 290 kg went. The missing 290 kg was released into the atmosphere as smoke, so the only visible matter left is the 10 kg of ash. If you understand the law of conservation of mass, then you know that the other 290 kg had to go somewhere, because it had to equal the mass of the tree before it burnt down.
Example $1$
If heating 10.0 grams of calcium carbonate (CaCO3) produces 4.4 g of carbon dioxide (CO2) and 5.6 g of calcium oxide (CaO), show that these observations are in agreement with the law of conservation of mass.
Solution
\begin{align*} \text{Mass of the reactants} &= \text{Mass of the products} \[4pt] 10.0\, \text{g of } \ce{CaCO3} &= 4.4 \,\text{g of }\ce{CO2} + 5.6\, \text{g of } \ce{ CaO} \[4pt] 10.0\,\text{g of reactant} &= 10.0\, \text{g of products} \end{align*} \nonumber
Because the mass of the reactant is equal to the mass of the products, the observations are in agreement with the law of conservation of mass.
Exercise $1$
Potassium hydroxide ($\ce{KOH}$) readily reacts with carbon dioxide ($\ce{CO2}$) to produce Potassium carbonate ($\ce{K2CO3}$) and water ($\ce{H2O}$). How many grams of potassium carbonate are produced if 224.4 g of $\ce{KOH}$ react with 88.0 g of $\ce{CO2}$? The reaction also produces 36.0 g of water.
Answer
276.4 g of potassium carbonate
The Law of Definite Proportions
Joseph Proust (1754-1826) formulated the law of definite proportions (also called the Law of Constant Composition or Proust's Law). This law states that if a compound is broken down into its constituent elements, the masses of the constituents will always have the same proportions, regardless of the quantity or source of the original substance. Joseph Proust based this law primarily on his experiments with basic copper carbonate. The illustration below depicts this law in action.
Law of Definite Proportions states that in a given type of chemical substance, the elements are always combined in the same proportions by mass.
The Law of Definite Proportions applies when elements are reacted together to form the same product. Therefore, while the Law of Definite Proportions can be used to compare two experiments in which hydrogen and oxygen react to form water, the Law of Definite Proportions can not be used to compare one experiment in which hydrogen and oxygen react to form water, and another experiment in which hydrogen and oxygen react to form hydrogen peroxide (peroxide is another material that can be made from hydrogen and oxygen).
Example $2$: water
Oxygen makes up 88.8% of the mass of any sample of pure water, while hydrogen makes up the remaining 11.2% of the mass. You can get water by melting ice or snow; or by condensing steam from a river, sea, pond, etc. It can be from different places: USA, UK, Australia, or anywhere. Water can be made by chemical reactions, such as burning hydrogen in oxygen.
However, if the water is pure, it will always consist of 88.8 % oxygen by mass and 11.2 % hydrogen by mass, irrespective of its source or method of preparation.
Video $2$ Law of definite proportions.
Summary
• Burning and other changes in matter do not destroy matter.
• The mass of matter is always the same before and after the changes occur.
• The law of conservation of mass states that matter cannot be created or destroyed.
• The law of definite proportions states that a given chemical compound always contains the same elements in the exact same proportions by mass. | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/02%3A_Atoms/2.01%3A_Atoms_-_Ideas_from_the_Ancient_Greeks.txt |
Learning Objective
• Summarize Dalton's atomic theory.
• Explain the Law of multiple proportions.
While it must be assumed that many more scientists, philosophers, and others studied composition of matter after Democritus, a major leap forward in our understanding of the composition of matter took place in the 1800s with the work of the British scientist John Dalton. He started teaching school at age twelve, and was primarily known as a teacher. In his twenties, he moved to the growing city of Manchester, where he was able to pursue some scientific studies. His work in several areas of science brought him a number of honors. When he died, over 40,000 people in Manchester marched at his funeral.
Dalton's modern atomic theory, proposed around 1803, is a fundamental concept that states that all elements are composed of atoms. Previously, we defined an atom as the smallest part of an element that maintains the identity of that element. Individual atoms are extremely small; even the largest atom has an approximate diameter of only 5.4 × 10−10 m. At that size, it takes over 18 million of these atoms, lined up side by side, to equal the width of a human pinky finger (about 1 cm).
Dalton studied the weights of various elements and compounds. He noticed that matter always combined in fixed ratios based on weight, or volume in the case of gases. Chemical compounds always contain the same proportion of elements by mass, regardless of amount, which provided further support for Proust's law of definite proportions. Dalton also observed that there could be more than one combination of two elements.
From his experiments and observations, as well as the work from peers of his time, Dalton proposed his new theory of the atom. This later became known as Dalton's atomic theory. The general tenets of this theory were as follows:
• All matter is composed of extremely small particles called atoms.
• Atoms of a given element are identical in size, mass, and other properties (Figure $2$ ). Atoms of different elements differ in size, mass, and other properties.
• Atoms cannot be subdivided, created, or destroyed.
• Atoms of different elements can combine in simple whole number ratios to form chemical compounds (Figure $3$).
• In chemical reactions, atoms are combined, separated, or rearranged.
Note
Dalton's atomic theory has been largely accepted by the scientific community, with the exception of three changes. Present-day science acknowledges that:
(1) An atom can be further subdivided.
(2) All atoms of an element are not identical in mass.
(3) Using nuclear fission and fusion techniques, we can create or destroy atoms by changing them into other atoms.
Example $1$
In the following drawing, the green spheres represent atoms of a certain element. The purple spheres represent atoms of another element. If the spheres touch, they are part of a single unit of a compound. Does the following chemical change represented by these symbols violate any of the ideas of Dalton’s atomic theory? If so, which one?
Solution
The starting materials consist of two green spheres and two purple spheres. The products consist of only one green sphere and one purple sphere. This violates Dalton’s postulate that atoms are neither created nor destroyed during a chemical change, but are merely redistributed. (In this case, atoms appear to have been destroyed.)
Exercise $1$
In the following drawing, the green spheres represent atoms of a certain element. The purple spheres represent atoms of another element. If the spheres touch, they are part of a single unit of a compound. Does the following chemical change represented by these symbols violate any of the ideas of Dalton’s atomic theory? If so, which one?
Answer
The starting materials consist of four green spheres and two purple spheres. The products consist of four green spheres and two purple spheres. This does not violate any of Dalton’s postulates—atoms are neither created nor destroyed, but are redistributed in small, whole-number ratios.
Law of Multiple Proportions
Many combinations of elements can react to form more than one compound. In such cases, the masses of a particular element that combine with a fixed mass of a second element are demonstrable in ratios of small whole numbers. It's easy to say this, but please make sure that you understand how it works. Nitrogen forms a very large number of oxides, five of which are shown here.
• Line 1 shows the ratio of the relative weights of the two elements in each compound. These ratios were calculated by simply taking the molar mass of each element, and multiplying by the number of atoms of that element per mole of the compound. Thus for NO2, we have (1 × 14) : (2 × 16) = 14:32. (These numbers were not known in the early days of Chemistry because atomic weights [i.e., molar masses] of most elements were not reliably known.)
• The numbers in Line 2 are just the mass ratios of O:N, found by dividing the corresponding ratios in line 1. But someone who depends solely on experiment would work these out by finding the mass of O that combines with unit mass (1 g) of nitrogen.
• Line 3 is obtained by dividing the figures of the Line 2 by the smallest O:N ratio in Line 2 (which is the ratio for N2O). Note that just as the law of multiple proportions says, the weight of oxygen that combines with unit weight of nitrogen work out to small integers; there is a typo in Line 3—where there is a 3, there should be a 4.
• Of course, we just as easily could have illustrated the law by considering the mass of nitrogen that combines with one gram of oxygen; it works both ways!
The law of multiple proportions states that if two elements form more than one compound between them, the masses of one element combined with a fixed mass of the second element form in ratios of small integers.
Video $1$ The law of multiple proportions
Example $1$: Law Multiple Proportions
A sample of compound A (a clear, colorless gas) is analyzed and found to contain 4.27 g carbon and 5.69 g oxygen. A sample of compound B (also a clear, colorless gas) is analyzed and found to contain 5.19 g carbon and 13.84 g oxygen. Are these data an example of the law of definite proportions, the law of multiple proportions, or neither? What do these data tell you about substances A and B?
Solution
In compound A, the mass ratio of carbon to oxygen is:
$\mathrm{\dfrac{1.33\: g\: O}{1\: g\: C}} \nonumber$
In compound B, the mass ratio of carbon to oxygen is:
$\mathrm{\dfrac{2.67\: g\: O}{1\: g\: C}} \nonumber$
The ratio of these ratios is:
$\mathrm{\dfrac{\dfrac{1.33\: g\: O}{1\: g\: C}}{\dfrac{2.67\: g\: O}{1\: g\: C}}=\dfrac{1}{2}} \nonumber$
This supports the law of multiple proportions. This means that A and B are different compounds, with A having one-half as much carbon per amount of oxygen (or twice as much oxygen per amount of carbon) as B. A possible pair of compounds that would fit this relationship would be A = CO2 and B = CO.
Exercise $1$
A sample of compound X (a clear, colorless, combustible liquid with a noticeable odor) is analyzed and found to contain 14.13 g carbon and 2.96 g hydrogen. A sample of compound Y (a clear, colorless, combustible liquid with a noticeable odor that is slightly different from X’s odor) is analyzed and found to contain 19.91 g carbon and 3.34 g hydrogen. Are these data an example of the law of definite proportions, the law of multiple proportions, or neither? What do these data tell you about substances X and Y?
Answer
In compound X, the mass ratio of carbon to hydrogen is $\mathrm{\dfrac{14.13\: g\: C}{2.96\: g\: H}}$.
In compound Y, the mass ratio of carbon to hydrogen is $\mathrm{\dfrac{19.91\: g\: C}{3.34\: g\: H}}$.
The ratio of these ratios is
$\mathrm{\dfrac{\dfrac{14.13\: g\: C}{2.96\: g\: H}}{\dfrac{19.91\: g\: C}{3.34\: g\: H}}=\dfrac{4.77\: g\: C/g\: H}{5.96\: g\: C/g\: H}=0.800=\dfrac{4}{5}}. \nonumber$
This small, whole-number ratio supports the law of multiple proportions. This means that X and Y are different compounds.
The evidence for atoms is so great that few doubt their existence. In fact, individual atoms are now routinely observed with state-of-the art technologies. Moreover, they can even be used for making pretty images or, as IBM research demonstrates in Video $1$, control of individual atoms can be used to create animations.
Video$1$: A Boy And His Atom: The World's Smallest Movie
A Boy and His Atom is a 2012 stop-motion animated short film released by IBM Research. The movie tells the story of a boy and a wayward atom who meet and become friends. It depicts a boy playing with an atom that takes various forms. It was made by moving carbon monoxide molecules viewed with a scanning tunneling microscope, a device that magnifies them 100 million times. These molecules were moved to create images, which were then saved as individual frames to make the film.
Summary
• Dalton's Atomic Theory is the first scientific theory to relate chemical changes to the structure, properties, and behavior of the atom. The general tenets of this theory were as follows:
• All matter is composed of extremely small particles called atoms.
• Atoms of a given element are identical in size, mass, and other properties. Atoms of different elements differ in size, mass, and other properties.
• Atoms cannot be subdivided, created, or destroyed.
• Atoms of different elements can combine in simple whole number ratios to form chemical compounds.
• The law of multiple proportions states that whenever the same two elements form more than one compound, the different masses of one element that combine with the same mass of the other element are in the ratio of small whole numbers. | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/02%3A_Atoms/2.03%3A_John_Dalton_and_the_Atomic_Theory_of_Matter.txt |
Learning Objectives
• Define mole and molar mass.
• Perform calculations to convert between moles and mass of a substance.
• Perform calculations to convert between mass of a substance and number of particles.
The Mole and Avogadro's Number
It certainly is easy to count bananas or to count elephants (as long as you stay out of their way). However, you would be counting grains of sugar from your sugar canister for a long, long time. Atoms and molecules are extremely small—far, far smaller than a grain of sugar. Counting atoms or molecules is not only unwise, it is absolutely impossible. One drop of water contains about $10^{22}$ molecules of water. If you counted 10 molecules every second for 50 years without stopping, you would have counted only $1.6 \times 10^{10}$ molecules. Put another way, at that counting rate, it would take you over 30 trillion years to count the water molecules in one tiny drop.
Chemists of the past needed a name that stood for a very large number of items. Amadeo Avogadro (1776-1856), an Italian scientist, provided such a number. He is responsible for the counting unit of measure called the mole. A mole $\left( \text{mol} \right)$ is the amount of a substance that contains $6.02 \times 10^{23}$ representative particles of that substance. The mole is the SI unit for amount of a substance. Just like the dozen and the gross, it is a name that stands for a number. There are therefore $6.02 \times 10^{23}$ water molecules in a mole of water molecules. There also would be $6.02 \times 10^{23}$ bananas in a mole of bananas—if such a huge number of bananas ever existed.
The number $6.02 \times 10^{23}$ is called Avogadro's number, the number of representative particles in a mole. It is an experimentally determined number. A representative particle is the smallest unit in which a substance naturally exists. For the majority of elements, the representative particle is the atom. Iron, carbon, and helium consist of iron atoms, carbon atoms, and helium atoms, respectively. Seven elements exist in nature as diatomic molecules and they are $\ce{H_2}$, $\ce{N_2}$, $\ce{O_2}$, $\ce{F_2}$, $\ce{Cl_2}$, $\ce{Br_2}$, and $\ce{I_2}$. The representative particle for these elements is the molecule. Likewise, all molecular compounds such as $\ce{H_2O}$ and $\ce{CO_2}$ exist as molecules, and so the molecule is their representative particle. For ionic compounds such as $\ce{NaCl}$ and $\ce{Ca(NO_3)_2}$, the representative particle is the formula unit. A mole of any substance contains Avogadro's number $\left( 6.02 \times 10^{23} \right)$ of representative particles.
Converting Between Number of Atoms to Moles and Vice Versa
We can use Avogadro's number as a conversion factor, or ratio, in dimensional analysis problems. If we are given the number of atoms of an element X, we can convert it into moles by using the relationship
$\text{1 mol X} = 6.022 \times 10^{23} \text{ X atoms} \nonumber$
Example $1$: Moles of Carbon
The element carbon exists in two primary forms: graphite and diamond. How many moles of carbon atoms is $4.72 \times 10^{24}$ atoms of carbon?
Solution
Example $1$: Steps for Problem Solving; Molar Conversion of Carbon
Steps for Problem Solving The element carbon exists in two primary forms: graphite and diamond. How many moles of carbon atoms is $4.72 \times 10^{24}$ atoms of carbon?
Identify the "given" information and what the problem is asking you to "find." Given: $4.72 \times 10^{24}$ C atoms
Find: mol C
List other known quantities. $1\, mol = 6.022 \times 10^{23}$ C atoms
Prepare a concept map and use the proper conversion factor.
Cancel units and calculate. $4.72 \times 10^{24} \: \cancel{\text{C} \: \ce{atoms}} \times \frac{1 \: \text{mol} \: \ce{C}}{6.02 \times 10^{23} \: \cancel{\text{C} \: \ce{atoms}}} = 7.84 \: \text{mol} \: \ce{C} \nonumber$
Think about your result. The given number of carbon atoms was greater than Avogadro's number, so the number of moles of $\ce{C}$ atoms is greater than 1 mole. Since Avogadro's number is a measured quantity with three significant figures, the result of the calculation is rounded to three significant figures.
Molar Mass
The molar mass of a substance is defined as the mass in grams of 1 mole of that substance. One mole of isotopically pure carbon-12 has a mass of 12 g. For an element, the molar mass is the mass of 1 mol of atoms of that element; for a covalent molecular compound, it is the mass of 1 mol of molecules of that compound; for an ionic compound, it is the mass of 1 mol of formula units. That is, the molar mass of a substance is the mass (in grams per mole) of 6.022 × 1023 atoms, molecules, or formula units of that substance. In each case, the number of grams in 1 mol is the same as the number of atomic mass units that describe the atomic mass, the molecular mass, or the formula mass, respectively.
The molar mass of any substance is its atomic mass, molecular mass, or formula mass in grams per mole.
The periodic table lists the atomic mass of carbon as 12.011 amu; the average molar mass of carbon—the mass of 6.022 × 1023 carbon atoms—is therefore 12.011 g/mol:
Table $1$ Molar Mass of Select Substances
Substance (formula) Basic Unit Atomic, Molecular, or Formula Mass (amu) Molar Mass (g/mol)
carbon (C) atom 12.011 (atomic mass) 12.011
ethanol (C2H5OH) molecule 46.069 (molecular mass) 46.069
calcium phosphate [Ca3(PO4)2] formula unit 310.177 (formula mass) 310.177
Converting Between Grams and Moles
The molar mass of any substance is the mass in grams of one mole of representative particles of that substance. The representative particles can be atoms, molecules, or formula units of ionic compounds. This relationship is frequently used in the laboratory. Suppose that for a certain experiment you need 3.00 moles of calcium chloride $\left( \ce{CaCl_2} \right)$. Since calcium chloride is a solid, it would be convenient to use a balance to measure the mass that is needed. Dimensional analysis will allow you to calculate the mass of $\ce{CaCl_2}$ that you should measure, as shown in Example $2$.
Example $2$: Calcium Chloride
Calculate the mass of 3.00 moles of calcium chloride (CaCl2).
Solution
Example $2$:Steps for Problem Solving; Calculate the mass of 3.00 moles of calcium chloride (CaCl2).
Steps for Problem Solving Calculate the mass of 3.00 moles of calcium chloride (CaCl2).
Identify the "given" information and what the problem is asking you to "find." Given: 3.00 moles of CaCl2
Find: g CaCl2
List other known quantities. 1 ml CaCl2 = 110.98 g CaCl2
Prepare a concept map and use the proper conversion factor.
Cancel units and calculate. $3.00 \: \cancel{\text{mol} \: \ce{CaCl_2}} \times \dfrac{110.98 \: \text{g} \: \ce{CaCl_2}}{1 \: \cancel{\text{mol} \: \ce{CaCl_2}}} = 333 \: \text{g} \: \ce{CaCl_2}$
Think about your result.
Example $3$: Water
How many moles are present in 108 grams of water?
Solution
Example $3$: Steps for Problem Solving; how many moles are present in 108 grams of water
Steps for Problem Solving How many moles are present in 108 grams of water?
Identify the "given" information and what the problem is asking you to "find." Given: 108 g H2O
Find: mol H2O
List other known quantities. $1 \: \text{mol} \: \ce{H_2O} = 18.02 \: \text{g}$ H2O
Prepare a concept map and use the proper conversion factor.
Cancel units and calculate. $108 \: \cancel{\text{g} \: \ce{H_2O}} \times \dfrac{1 \: \text{mol} \: \ce{H_2O}}{18.02 \: \cancel{\text{g} \: \ce{H_2O}}} = 5.99 \: \text{mol} \: \ce{H_2O}$
Think about your result.
Exercise $1$: Nitrogen Gas
What is the mass of $7.50 \: \text{mol}$ of Nitrogen gas $\ce{N2}$?
Answer:
210 g
Conversions Between Mass and Number of Particles
In "Conversions Between Moles and Mass", you learned how to convert back and forth between moles and the number of representative particles. Now you have seen how to convert back and forth between moles and mass of a substance in grams. We can combine the two types of problems into one. Mass and number of particles are both related to moles. To convert from mass to number of particles or vice-versa, it will first require a conversion to moles, as shown in Figure $1$ and Example $4$.
Example $4$: Chlorine
How many molecules is $20.0 \: \text{g}$ of chlorine gas, $\ce{Cl_2}$?
Solution
Example $4$:Steps for Problem Solving how many molecules is $20.0 \: \text{g}$ of chlorine gas, $\ce{Cl_2}$?
Steps for Problem Solving How many molecules is $20.0 \: \text{g}$ of chlorine gas, $\ce{Cl_2}$?
Identify the "given" information and what the problem is asking you to "find." Given: 20.0 g Cl2
Find: # Cl2 molecules
List other known quantities. 1 mol Cl2 = 70.90 g Cl2, 1mol Cl2 = 6.022 x 1023 Cl2 molecules
Prepare a concept map and use the proper conversion factor.
Cancel units and calculate. $20.0 \: \cancel{\text{g} \: \ce{Cl_2}} \times \dfrac{1 \: \cancel{\text{mol} \: \ce{Cl_2}}}{70.90 \: \cancel{\text{g} \: \ce{Cl_2}}} \times \dfrac{6.02 \times 10^{23} \: \text{molecules} \: \ce{Cl_2}}{1 \: \cancel{\text{mol} \: \ce{Cl_2}}} = 1.70 \times 10^{23} \: \text{molecules} \: \ce{Cl_2}$
Think about your result. Since the given mass is less than half of the molar mass of chlorine, the resulting number of molecules is less than half of Avogadro's number.
Exercise $2$: Calcium Chloride
How many formula units are in 25.0 g of CaCl2 ?
Answer:
1.36 x 1023 CaCl2 formula units
Summary
• A mole of any substance contains Avogadro's number $\left( 6.02 \times 10^{23} \right)$ of representative particles.
• The molar mass of a substance is defined as the mass in grams of 1 mole of that substance.
• Calculations involving conversions between moles of a material and the mass of that material are described.
• Calculations are illustrated for conversions between mass and number of particles.
Contributors and Attributions
• Henry Agnew (UC Davis) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/02%3A_Atoms/2.04%3A_The_Mole_and_Molar_Mass.txt |
Learning Objectives
• Explain how elements are organized into the periodic table.
• Describe how some characteristics of elements relate to their positions on the periodic table.
The Periodic Table is the symbol of Chemistry for many. It is a single image that contains all of the known elements in the universe combined into an easily legible table. There are many patterns present in the table as well. All of the elements seem to fit together, connecting to form the "image of chemistry". The idea of elements first came about in 3000 B.C. The great Greek philosopher Aristotle conceived an idea that everything on earth was made up of these elements. In ancient times, elements like gold and silver were readily accessible, however, the elements that Aristotle chose were Earth, Water, Fire, and Air.
Emerging Patterns and Periodicity within the Known Elements
The modern periodic table has evolved through a long history of attempts by chemists to arrange the elements according to their properties, as an aid in predicting chemical behavior. One of the first to suggest such an arrangement was the German chemist Johannes Dobereiner (1780–1849), who noticed that many of the known elements could be grouped into triads. A triad is a set of three elements that have similar properties—for example, chlorine, bromine, and iodine; copper, silver, and gold. Dobereiner proposed that all elements could be grouped in such triads, but subsequent attempts to expand his concept were unsuccessful.
By the mid-19th century, the atomic masses of many of the elements had been determined. Scientists like John Newlands and Alexandre-Emile Béguyer de Chancourtois formed their own versions of periodic tables. The English chemist John Newlands (1838–1898), who hypothesized that the chemistry of the elements might be related to their masses, arranged the known elements in order of increasing atomic mass and discovered that every seventh element had similar properties. Newlands therefore suggested that the elements could be classified into octaves. He described octaves as a group of seven elements, which correspond to the horizontal rows in the main groups of today's periodic table. There were seven elements, because the noble gases were not known at the time. Unfortunately, Newlands’s “law of octaves” did not seem to work for elements heavier than calcium, and his idea was publicly ridiculed.
The Periodic Table: Mendeleev and Meyer
The periodic table achieved its modern form through the work of the German chemist Julius Lothar Meyer (1830–1895) and the Russian chemist Dimitri Mendeleev (1834–1907), both of whom focused on the relationships between atomic mass and various physical and chemical properties. In 1869, they independently proposed essentially identical arrangements of the elements. Meyer aligned the elements in his table according to periodic variations in simple atomic properties, such as “atomic volume”, which he obtained by dividing the atomic mass (molar mass) in grams per mole by the density of the element in grams per cubic centimeter. This property is equivalent to what is today defined as molar volume—the molar mass of an element divided by its density (measured in cubic centimeters per mole):
$\frac{molar\; mass\left ( \cancel{g}/mol \right )}{density\left ( \cancel{g}/cm^{3} \right )}=molar\; volume\left ( cm^{3}/mol \right ) \tag{3.1.1}$
Mendeleev, who first published his periodic table in 1869 (Figure $1$ ), is usually credited with the origin of the modern periodic table. The key difference between his arrangement of the elements, and that of Meyer and others, is that Mendeleev did not assume that all the elements had been discovered (in fact, only about two-thirds of the naturally occurring elements were known at the time). Instead, he deliberately left blanks in his table at atomic masses 44, 68, 72, and 100—in the expectation that elements with those atomic masses would be discovered. Those blanks correspond to the elements we now know as scandium, gallium, germanium, and technetium.
The groups in Mendeleev's table are determined by how many oxygen or hydrogen atoms are needed to form compounds with each element. For example, in Group I, two atoms of hydrogen (H), lithium (Li), sodium (Na), and potassium (K) form compounds with one atom of oxygen. In Group VII, one atom of fluorine (F), chlorine (Cl), and bromine (Br), react with one atom of hydrogen. Notice how this approach has trouble with the transition metals. Until roughly 1960, a rectangular table based on reactivity—and developed from Mendeleev's table—was standard at the front of chemistry lecture halls.
The most convincing evidence in support of Mendeleev’s arrangement of the elements was the discovery of two previously unknown elements whose properties closely corresponded with his predictions (Table $1$ ). Two of the blanks Mendeleev had left in his original table were below aluminum and silicon, awaiting the discovery of two as-yet-unknown elements, eka-aluminum and eka-silicon (from the Sanskrit eka, meaning “one,” as in “one beyond aluminum”). The observed properties of gallium and germanium matched those of eka-aluminum and eka-silicon so well that once they were discovered, Mendeleev’s periodic table rapidly gained acceptance.
Video $1$: The genius of Mendeleev's periodic table.
Table $1$: Comparison of the Properties Predicted by Mendeleev in 1869 for eka-Aluminum and eka-Silicon with the Properties of Gallium (discovered in 1875) and Germanium (discovered in 1886).
Property eka-Aluminum (predicted) Gallium (observed) eka-Silicon (predicted) Germanium (observed)
atomic mass 68 69.723 72 72.64
element metal metal dirty-gray metal gray-white metal
low mp* mp = 29.8°C high mp mp = 938°C
d = 5.9 g/cm3 d = 5.91 g/cm3 d = 5.5 g/cm3 d = 5.323 g/cm3
oxide E2O3 Ga2O3 EO2 GeO2
d = 5.5 g/cm3 d = 6.0 g/cm3 d = 4.7 g/cm3 d = 4.25 g/cm3
chloride ECl3 GaCl3 ECl4 GeCl4
volatile
mp = 78°C
bp* = 201°C
bp < 100°C bp = 87°C
*mp = melting point; bp = boiling point.
When the chemical properties of an element suggested that it might have been assigned the wrong place in earlier tables, Mendeleev carefully reexamined its atomic mass. He discovered, for example, that the atomic masses previously reported for beryllium, indium, and uranium were incorrect. The atomic mass of indium had originally been reported as 75.6, based on an assumed stoichiometry of InO for its oxide. If this atomic mass were correct, then indium would have to be placed in the middle of the nonmetals, between arsenic (atomic mass 75) and selenium (atomic mass 78). Because elemental indium is a silvery-white metal, however, Mendeleev postulated that the stoichiometry of its oxide was really In2O3 rather than InO. This would mean that indium’s atomic mass was actually 113, placing the element between two other metals, cadmium and tin.
One group of elements that was absent from Mendeleev’s table is the noble gases, all of which were discovered more than 20 years later—between 1894 and 1898—by Sir William Ramsay (1852–1916; Nobel Prize in Chemistry 1904). Initially, Ramsay did not know where to place these elements in the periodic table. Argon, the first to be discovered, had an atomic mass of 40. This was greater than chlorine’s and comparable to that of potassium; so Ramsay, using the same kind of reasoning as Mendeleev, decided to place the noble gases between the halogens and the alkali metals. In 1913, however, young British physicist H. G. J. Moseley (1887–1915) analyzed the frequencies of x-rays emitted by the elements, and discovered that the underlying foundation of the order of the elements was by the atomic number—not the atomic mass. Moseley hypothesized that the placement of each element in his series corresponded to its atomic number Z, which is the number of positive charges (protons) in its nucleus. Moseley left his research work at the University of Oxford to join the British army as a telecommunications officer during World War I. He was killed during the Battle of Gallipoli in Turkey. Finally, in 1945, the Manhattan Project yielded the discovery of many new radioactive elements. Glenn T. Seaborg suggested an addition of the actinide and lanthanide series at the bottom of the table. This idea came with the discovery of Americium and Curium, and their unique properties. The change was not accepted at first, but is now included in all periodic tables. Figure $2$ shows the time period for the discovery of the different elements.
• Before 1800 (36 elements): Discoveries during and before the Age of Enlightenment.
• 1800-1849 (+22 elements): Impulse from Scientific Revolution and Atomic theory and Industrial Revolution.
• 1850-1899 (+23 elements): The age of Classifying Elements received an impulse from the Spectrum analysis.
• 1900-1949 (+13 elements): Impulse from the old quantum theory, the Refinements to the periodic table, and quantum mechanics.
• 1950-1999 (+15 elements): Manhattan Project and Particle physics issues, for atomic numbers 97 and above.
Each element on the modern periodic table is represented by its atomic number and atomic mass (Figure $3$). The atomic mass of each element is found under the element symbol in the periodic table. Examples are shown below. The atomic mass of tin (Sn) is 118.71 u while the atomic mass of carbon (C) is 12.011 u. On the other hand, the atomic number (Z) of each element is found above the atomic symbol.
The periodic table is found vis this link:
https://pubchem.ncbi.nlm.nih.gov/periodic-table/png/Periodic_Table_of_Elements_w_Atomic_Mass_PubChem.png
Summary
• The modern periodic table was based on empirical correlations of properties such as atomic mass; early models using limited data noted the existence of triads and octaves of elements with similar properties.
• The periodic table achieved its current form through the work of Dimitri Mendeleev and Julius Lothar Meyer, both of whom focused on the relationship between atomic mass and chemical properties.
• The correlation with the electronic structure of atoms was discovered when H. G. J. Moseley showed that the periodic arrangement of the elements was determined by atomic number, not atomic mass. | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/02%3A_Atoms/2.05%3A_Mendeleev_and_Periodic_Table.txt |
Learning Objective
• Know the difference between an atom and a molecule.
An atom is the smallest unit of ordinary matter that has the properties of a chemical element. Every solid, liquid, gas, and plasma is composed of either neutral (un-ionized), or ionized atoms. Atoms are extremely small; typical sizes are around 100 picometers (1×10−10 m, a ten-millionth of a millimeter, or 1/254,000,000 of an inch). They are small enough that attempting to predict their behavior using classical physics—as if they were billiard balls, for example—gives noticeably incorrect predictions due to quantum effects. Current atomic models have incorporated quantum mechanics to better explain and predict this behavior.
Every atom is composed of a nucleus and one or more electrons bound to the nucleus. The nucleus is made of one or more protons and a number of neutrons; only the most common variety (isotope) of hydrogen has no neutrons.
A molecule is an electrically neutral group of two or more atoms held together by chemical bonds. Molecules are distinguished from ions by their lack of electrical charge. However, in quantum physics, organic chemistry, and biochemistry, the term molecule is often used less strictly—as it is also applied to polyatomic ions.
In the kinetic theory of gases, the term molecule is often used for any gas, regardless of its composition. According to this definition, noble gas atoms are considered molecules because they are monatomic molecules.
A molecule may be homonuclear—that is, it consists of atoms of one chemical element, as with oxygen (O2). A heteronuclear molecule is a chemical compound composed of more than one element, as with water (H2O). Atoms and complexes connected by non-covalent interactions, such as hydrogen bonds or ionic bonds, are typically not considered single molecules.
Molecules as components of matter are common in organic substances (and therefore biochemistry). They also make up most of the oceans and atmosphere. However, the majority of familiar solid substances on Earth—including most of the minerals that make up the crust, mantle, and core of the Earth—contain many chemical bonds, but are not made of identifiable molecules. Also, no typical molecule can be defined for ionic crystals (salts) and covalent crystals (network solids), although these are often composed of repeating unit cells that extend either in a plane (such as in graphene) or three-dimensionally (such as in diamond, quartz, or sodium chloride). The theme of repeated unit-cellular-structure also holds for most condensed phases with metallic bonding, which means that solid metals are also not made of molecules. In glasses (solids that exist in a vitreous disordered state), atoms may also be held together by chemical bonds with no presence of any definable molecule, nor any of the regularity of repeating units that characterizes crystals.
Three common molecules shown below.
Video \(1\): The difference between an atom and a molecule.
Summary
• An atom is the smallest unit of matter that has the properties of a chemical element
• A molecule is an electrically neutral group of two or more atoms held together by chemical bonds.
Contributors and Attributions
• Wikipedia
• Henry Agnew (UC Davis) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/02%3A_Atoms/2.06%3A_Atoms_and_Molecules-_Real_and_Relevant.txt |
Thumbnail: Electron shell diagram for Sodium, the 19th element in the periodic table of elements. (CC BY-SA; 2.5; Pumbaa(opens in new window)).
03: Atomic Structure
Learning Objectives
• Understand the significance of the experiments performed by Thomson and Millikan.
Alessandro Giuseppe Antonio Anastasio Volta (1745–1827) was an Italian physicist, chemist, and pioneer of electricity and power; who is credited as the inventor of the electric battery and the discoverer of methane. He invented the Voltaic pile in 1799; with this invention, Volta proved that electricity could be generated chemically, and debunked the prevalent theory that electricity was generated solely by living beings. Volta's invention sparked a great amount of scientific excitement and led others to conduct similar experiments, which eventually led to the development of the field of electrochemistry.
Electrolysis
Sir Humphry Davy, 1st Baronet (1778–1829) was a Cornish chemist and inventor. He was a pioneer in the field of electrolysis, which uses the voltaic pile to split common compounds and thus prepare many new elements. He went on to electrolyse molten salts and discovered several new metals, including sodium and potassium in 1807. The following year, he discovered calcium, strontium, barium, magnesium and boron, as well as the elemental nature of chlorine and iodine. He also studied the forces involved in these separations, inventing the new field of electrochemistry.
Davy's laboratory assistant, Michael Faraday, went on to enhance Davy's work and would become the more famous and influential scientist. Davy is supposed to have even claimed Faraday as his greatest discovery. Faraday (1791–1867) was an English scientist who contributed to the study of electromagnetism and electrochemistry. His main discoveries include the principles underlying electromagnetic induction, diamagnetism and electrolysis.
As a chemist, Faraday discovered benzene, investigated the clathrate hydrate of chlorine, invented an early form of the Bunsen burner and the system of oxidation numbers, and popularized terminology such as "anode", "cathode", "electrode" and "ion". Faraday ultimately became the first and foremost Fullerian Professor of Chemistry at the Royal Institution, a lifetime position. His work on electrolysis paved the way for subsequent experiments performed using cathode ray tubes.
Cathode Ray Tubes
The TV set seen below is becoming harder and harder to find these days. The main reason is because they are older and based on outdated technology. The new TV sets are flat screen technology that take up less space and give better picture quality, especially with the advent of high-definition broadcasting. The technology used in the older TV sets used cathode ray tubes. A beam of electrons was sprayed to a picture tube which was treated to react with the electrons to produce an image. Similar CRT (cathode ray tube) devices were used in computer monitors, now also replaced by flat screen monitors.
The first cathode ray tube prototype was developed by Heinrich Geissler, a German glassblower and physicist. He used a mercury pump to create a vacuum in a tube. Geissler explored a number of techniques to remove air from the tube and to prevent leaks, as well as ways to get good connections of the wires in the tubes. In 1878, Sir William Crookes, a British scientist, displayed the first cathode rays using a modification of the Geissler apparatus. His major contribution to construction of the tube was to develop ways to evacuate almost all the air from the tube. Crookes also carried out many experiments using more reliable equipment to confirm earlier finding about the properties of cathode rays. He discovered two things which supported the hypothesis that the cathode ray consisted of a stream of particles.
• When an object was placed between the cathode and the opposite end of the tube, it cast a shadow on the glass. The shadow caused by the object indicates that particles were being blocked on their way from the cathode to the anode.
• A cathode ray tube was constructed with a small metal rail between the two electrodes. Attached to the rail was a paddle wheel capable of rotating along the rail. Upon starting up the cathode ray tube, the wheel rotated from the cathode towards the anode. Notice that the cathode and anode are positioned so that the rays will strike the top of the paddle wheel. Crookes concluded that the cathode ray was made of particles which must have mass.
The cathode ray tube was first invented by Sir William Crookes.
Further Research with the Crookes Tube
Crookes' work opened the door to a number of important discoveries. Other scientists were able to demonstrate that the "cathode ray" was actually a stream of electrons. In 1897, Karl Ferdinand Braun developed the first oscilloscope, using a cathode ray tube to see an electrical pulse as it passed through the instrument. The invention of television would not have been possible without the cathode ray tube. Work with a modified system led to the discovery of x-rays in 1895 by the German physicist Wilhelm Roentgen. This simple device has led to major advances in science and technology.
Thomson's Experiment: Mass to Charge Ratio
In 1897, the British physicist J. J. Thomson (1856–1940) proved that atoms were not the most basic form of matter. He demonstrated that cathode rays could be deflected, or bent, by magnetic or electric fields, which indicated that cathode rays consist of charged particles (Figure 1.17c). More important, by measuring the extent of the deflection of the cathode rays in magnetic or electric fields of various strengths, Thomson was able to calculate the mass-to-charge ratio of the particles. These particles were emitted by the negatively charged cathode and repelled by the negative terminal of an electric field. Because like charges repel each other and opposite charges attract, Thomson concluded that the particles had a net negative charge; these particles are now called electrons. Most relevant to the field of chemistry, Thomson found that the mass-to-charge ratio of cathode rays is independent of the nature of the metal electrodes or the gas, which suggested that electrons were fundamental components of all atoms.
Cathode ray tube
Video $1$ A cathode ray tube.
Thomson's experiment
Video $2$Video from Davidson College demonstrating Thompson's e/m experiment. You can also use an applet from the Google Group Physics Flash to simulate the experiment and here is another applet, from Kings Center for Visual Exploration in Alberta (KCVE) with a detailed explanation
Millikan's Oil-Drop Experiment: Electron Charge
In 1909, more information about the electron was uncovered by American physicist Robert A. Millikan via his “oil drop” experiments. Millikan created microscopic oil droplets, which could be electrically charged by friction as they formed or by using X-rays. These droplets initially fell due to gravity, but their downward progress could be slowed or even reversed by an electric field lower in the apparatus. By adjusting the electric field strength and making careful measurements and appropriate calculations, Millikan was able to determine the charge on individual drops (Figure $2$) .
He then concluded that the charge of a single electron is 1.6 $\times$ 10−19 C. With this information and Thomson’s mass-to-charge ratio (1.759 $\times$ 1011 C/kg)., Millikan determined the mass of an electron:
$Mass\: of\: electron = {charge}\times \dfrac {mass}{charge} \nonumber$
$\mathrm{=1.602\times 10^{-19}\:\cancel{C}\times \dfrac{1\: kg}{1.759\times 10^{11}\:\cancel{C}}=9.107\times 10^{-31}\:kg} \tag{3.1.2}$
Scientists had now established that the atom was not indivisible as Dalton had believed, and due to the work of Thomson, Millikan, and others, the charge and mass of the negative, subatomic particles—the electrons—were known. However, the positively charged part of an atom was not yet well understood.
Summary
• The cathode ray tube was first invented by Sir William Crookes.
• Experiments showed that the rays had mass.
• Thomson’s cathode ray tube showed that atoms contain small, negatively charged particles called electrons.
• Millikan discovered that there is a fundamental electric charge—the charge of an electron. | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/03%3A_Atomic_Structure/3.01%3A_Electricity_and_the_Atom.txt |
Learning Objective
• Understand the significance of the discovery of X-rays and radioactivity to the modern view of the structure of the atom.
The discovery of radioactivity took place over several years beginning with the discovery of x-rays in 1895 by Wilhelm Conrad Roentgen and continuing with such people as Henri Becquerel and the Curie family. The application of x-rays and radioactive materials is far reaching in medicine and industry. Radioactive material is used in everything from nuclear reactors to isotope infused saline solutions. These technologies allow us to utilize great amounts of energy and observe biological systems in ways which were unthinkable less than a century ago. As will be discussed in section3.4, the British physicist Ernest Rutherford (1871-1937) conducted experiments (based on the Curies' work) that led to the modern view of the structure of the atom.
Roentgen: The Discovery of X-Rays
From Section 3.1, we saw how J. J. Thomson discovered the electron by using cathode ray tubes (CRT). In November of 1895, a German physicist named Wilhelm Roentgen (Röntgen) used this same technology to discover the x-ray. Roentgen's ray originated from the CRT and reacted with a barium platinocyanide screen to fluoresce. After seeing this, Roentgen placed objects between this ray and the screen. He noted that the ray would still penetrate substances and leave an image on photographic film.
When naming this type of radiation, Roentgen reflected back to his experiments. He could not confidently explain his observations. For this reason, he decided to call this new type of radiation the x-ray. Throughout his lifetime, he continue to explore this technology and even used his wife as a test subject (refer to chapter 1 for her x-ray image). By never patenting this invention, Roentgen freely shared the x-ray with other researchers and medical professionals. In 1901, he was awarded the Nobel Prize in physics for his work with the x-ray.
As with any new technology, people looked for ways to apply the x-ray to day to day life. Thomas Edison, the inventor of the light bulb, thought the average person should have an x-ray machine in their home. He designed x-ray machines to be smaller and portable. Unfortunately, many of his technicians died from radiation poisoning.
Hand mit Ringen (Hand with Rings): print of Wilhelm Röntgen's first "medical" X-ray, of his wife's hand, taken on 22 December 1895.
Figure $2$: (left) Photo of Wilhelm Conrad Roengten, the discoverer of x-rays. (right) Surgical operation during World War I using a fluoroscope that used x-rays to find embedded bullets
During Edison's time, people would host x-ray parties in their homes. The host of these gatherings would allow guests to x-ray different parts of their body. People would leave these events with their framed souvenirs. X-rays were even used to measure foot size. Devices like the fluoroscope were placed in shoe stores. Technicians, consumers, and observers could look in oculars while the foot was being x-rayed. Unnecessary exposure to radiation was not considered as a hazard during these times. In this era, most believed this was the most accurate way to measure the foot.
The harmful effects and different uses of X-rays
In November 1896, an article entitled "the harmful effects of X-rays" was published in Nature. The witness was an X-ray demonstrator during the summer in London. He therefore paid for himself throughout the summer at the rate of several hours per day of exposure. He testified: "In the first two or three weeks I felt no inconvenience, but after a while appeared on the fingers of my right hand many dark spots which pierced under the skin, and gradually they became very painful; the rest of the skin was red and strongly inflamed My hand was so bad that I was constantly forced to bathe it in very cold water An ointment momentarily calm the pain but the epidermis had dried up had become hard and yellow like parchment and completely insensible, so I was not surprised when my hand began to peel. "
"Soon the skin and nails fall off and the fingers swelled, the pain remaining constant. I lost the skin of my right and left hands, and four of my nails have disappeared from the right hand and two left and three others were ready to fall off. During more than six weeks I was unable to hold anything in my right hand and I can not hold a pen since the loss of my nails …"
Figure $3$: Radiation poisoning due to over exposure of x-rays.
Different applications use different parts of the X-ray spectrum.
Figure $4$: X-rays are part of the electromagnetic spectrum, with wavelengths shorter than visible light.
The Discovery of Radioactivity
When Becquerel heard about Roentgen's discovery, he wondered if his fluorescent minerals would give the same x-rays. Becquerel placed some of his rock crystals on top of a well-covered photographic plate and sat them in the sunlight. The sunlight made the crystals glow with a bright fluorescent light, but when Becquerel developed the film he was very disappointed. He found that only one of his minerals, a uranium salt, had fogged the photographic plate. He decided to try again, and this time, to leave them out in the sun for a longer period of time. Fortunately, the weather did not cooperate and Becquerel had to leave the crystals and film stored in a drawer for several cloudy days. Before continuing his experiments, Becquerel decided to check one of the photographic plates to make sure the chemicals were still good. To his amazement, he found that the plate had been exposed in spots where it had been near the uranium containing rocks and some of these rocks had not been exposed to sunlight at all. In later experiments, Becquerel confirmed that the radiation from the uranium had no connection with light or fluorescence, but the amount of radiation was directly proportional to the concentration of uranium in the rock. Becquerel had discovered radioactivity.
Figure $5$: Image of Becquerel's photographic plate which has been fogged by exposure to radiation from a uranium salt. The shadow of a metal Maltese Cross placed between the plate and the uranium salt is clearly visible (Public Domain).
The Curies and Radium
One of Becquerel's assistants, a young Polish scientist named Maria Sklowdowska (to become Marie Curie after she married Pierre Curie), became interested in the phenomenon of radioactivity. With her husband, she decided to find out if chemicals other than uranium were radioactive. The Austrian government was happy to send the Curies a ton of pitchblende from the mining region of Joachimstahl because it was waste material that had to be disposed of anyway. The Curies wanted the pitchblende because it was the residue of uranium mining. From the ton of pitchblende, the Curies separated $0.10 \: \text{g}$ of a previously unknown element, radium, in the form of the compound radium chloride. This radium was many times more radioactive than uranium.
By 1902, the world was aware of a new phenomenon called radioactivity and of new elements which exhibited natural radioactivity. For this work, Becquerel and the Curies shared the 1903 Nobel Prize in Physics. These three researchers continued to work with hazardous radioactive materials. They experienced various ailments included burns and weakness by touching these substances. In 1906, Pierre Curie accidently fell under a horse drawn cart. He immediately died from his wounds and left Marie with two small children (Irene, 9 years old and Eve, 2 years old). After his death, Marie continued her research by identifying a new element, Polonium. In 1911, she was awarded the Nobel Prize in chemistry for the discoveries of radium and polonium. As of today, Marie Curie is the only person ever to have received two Nobel Prizes in the sciences.
Video$2$: Everyone knows that winning the Nobel Prize is a big deal, but why do we even have a Nobel Prize? And why does it matter?
Summary
• Henri Becquerel, Marie Curie, and Pierre Curie shared the discovery of radioactivity.
• Wilhelm Roengtgen used cathode ray tubes to discover the x-ray. | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/03%3A_Atomic_Structure/3.02%3A_Serendipity_in_Science-_X-Rays_and_Radioactivity.txt |
Learning Objective
Differentiate between alpha particles, beta particles and gamma rays.
Ernest Rutherford, (30 August 1871 – 19 October 1937), was a New Zealand who came to be known as the father of . While working in Thomson’s laboratory shortly after Thomson discovered the electron, Rutherford showed that compounds of uranium and other elements emitted at least two distinct types of radiation. One was readily absorbed by matter and seemed to consist of particles that had a positive charge and were massive compared to electrons. Because it was the first kind of radiation to be discovered, Rutherford called these substances alpha (α) particles. Rutherford also showed that the particles in the second type of radiation, beta (β) particles, had the same charge and mass-to-charge ratio as Thomson’s electrons; they are now known to be high-speed electrons. In 1908 was awarded the "for his investigations into the disintegration of the elements, and the chemistry of radioactive substances", for which he was the first Canadian and Oceanian Nobel laureate. A third type of radiation, gamma (γ) rays, was discovered somewhat later and found to be similar to the lower-energy form of radiation called x-rays, now used to produce images of bones and teeth.
Table \(1\): Names, Symbols, Representations, and Descriptions of the most common types of radioactivity.
Summary
An alpha particle is a high energy helium nucleus.
A beta particle is a high energy electron.
A gamma ray is a very high energy electromagnetic radiation
Contributors and Attributions
• TextMap: Chemistry-The Central Science (Brown et al.)
3.04: Rutherford's Experiment- The Nuclear Model of the Atom
Learning Objectives
• Describe Thomson's "plum pudding" model of the atom and the evidence for it.
• Describe Rutherford's gold foil experiment and explain how this experiment altered the "plum pudding" model.
The electron was discovered by J.J. Thomson in 1897. The existence of protons was also known, as was the fact that atoms were neutral in charge. Since the intact atom had no net charge and the electron and proton had opposite charges, the next step after the discovery of subatomic particles was to figure out how these particles were arranged in the atom. This is a difficult task because of the incredibly small size of the atom. Therefore, scientists set out to design a model of what they believed the atom could look like. The goal of each atomic model was to accurately represent all of the experimental evidence about atoms in the simplest way possible.
Following the discovery of the electron, J.J. Thomson developed what became known as the "plum pudding" model (Figure \(1\)) in 1904. Plum pudding is an English dessert similar to a blueberry muffin. In Thomson's plum pudding model of the atom, the electrons were embedded in a uniform sphere of positive charge like blueberries stuck into a muffin. The positive matter was thought to be jelly-like or a thick soup. The electrons were somewhat mobile. As they got closer to the outer portion of the atom, the positive charge in the region was greater than the neighboring negative charges and the electron would be pulled back more toward the center region of the atom.
However, this model of the atom soon gave way to a new model developed by New Zealander Ernest Rutherford (1871 - 1937) about five years later. Thomson did still receive many honors during his lifetime, including being awarded the Nobel Prize in Physics in 1906 and a knighthood in 1908.
Atoms and Gold
In 1911, Rutherford and coworkers Hans Geiger and Ernest Marsden initiated a series of groundbreaking experiments that would completely change the accepted model of the atom. They bombarded very thin sheets of gold foil with fast moving alpha particles.
According to the accepted atomic model, in which an atom's mass and charge are uniformly distributed throughout the atom, the scientists expected that all of the alpha particles would pass through the gold foil with only a slight deflection or none at all. Surprisingly, as shown in Figure \(2\) (while most of the alpha particles were indeed undeflected, a very small percentage (about 1 in 8000 particles) bounced off the gold foil at very large angles. Some were even redirected back toward the source. No prior knowledge had prepared them for this discovery. In a famous quote, Rutherford exclaimed that it was "as if you had fired a 15-inch [artillery] shell at a piece of tissue and it came back and hit you."
Rutherford needed to come up with an entirely new model of the atom in order to explain his results. Because the vast majority of the alpha particles had passed through the gold, he reasoned that most of the atom was empty space. In contrast, the particles that were highly deflected must have experienced a tremendously powerful force within the atom. He concluded that all of the positive charge and the majority of the mass of the atom must be concentrated in a very small space in the atom's interior, which he called the nucleus. The nucleus is the tiny, dense, central core of the atom and is composed of protons and neutrons.
Rutherford's atomic model became known as the nuclear model. In the nuclear atom, the protons and neutrons, which comprise nearly all of the mass of the atom, are located in the nucleus at the center of the atom. The electrons are distributed around the nucleus and occupy most of the volume of the atom. It is worth emphasizing just how small the nucleus is compared to the rest of the atom. If we could blow up an atom to be the size of a large professional football stadium, the nucleus would be about the size of a marble.
Rutherford's model proved to be an important step towards a full understanding of the atom. However, it did not completely address the nature of the electrons and the way in which they occupied the vast space around the nucleus. For this and other insights, Rutherford was awarded the Nobel Prize in Chemistry in 1908. Unfortunately, Rutherford would have preferred to receive the Nobel Prize in Physics because he considered physics superior to chemistry. In his opinion, “All science is either physics or stamp collecting.”
Summary
• The plum pudding model is an early attempt to show what an atom looks like.
• Bombardment of gold foil with alpha particles showed that some particles were deflected.
• The nuclear model of the atom consists of a small and dense positively charged interior surrounded by a cloud of electrons.
Contributors and Attributions
• Henry Agnew (UC Davis)
• TextMap: Chemistry - The Central Science | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/03%3A_Atomic_Structure/3.03%3A_Three_Types_of_Radioactivity.txt |
Learning Objectives
• List the properties of the three main subatomic particles.
• Define atomic mass unit (amu).
• Define atomic number and mass number.
• Define isotopes.
• Determine the number of protons, neutrons, and electrons in an atom with a given mass number.
Figure $1$ A summary of the historical development of models of the components and structure of the atom. The dates in parentheses are the years in which the key experiments were performed. (CC BY-SA-NC).
Figure $2$The Evolution of Atomic Theory, as Illustrated by Models of the Oxygen Atom. Bohr’s model and the current model are described in Chapter 6, "The Structure of Atoms." Image used with Permission (CC BY-SA-NC).
The nucleus (plural, nuclei) is a positively charged region at the center of the atom. It consists of two types of subatomic particles packed tightly together. The particles are protons, which have a positive electric charge, and neutrons, which are neutral in electric charge. Outside of the nucleus, an atom is mostly empty space, with orbiting negative particles called electrons whizzing through it. Figure $3$ below shows these parts of the atom.
The nucleus of the atom is extremely small. Its radius is only about 1/100,000 of the total radius of the atom. If an atom were the size of a football stadium, the nucleus would be about the size of a pea! Electrons have virtually no mass, but protons and neutrons have a lot of mass for their size. As a result, the nucleus has virtually all the mass of an atom. Given its great mass and tiny size, the nucleus is very dense. If an object the size of a penny had the same density as the nucleus of an atom, its mass would be greater than 30 million tons!
Holding It All Together
Particles with opposite electric charges attract each other. This explains why negative electrons orbit the positive nucleus. Particles with the same electric charge repel each other. This means that the positive protons in the nucleus push apart from one another. So why doesn't the nucleus fly apart? An even stronger force - called the strong nuclear force - holds protons and neutrons together in the nucleus.
Table $1$ gives the properties and locations of electrons, protons, and neutrons. The third column shows the masses of the three subatomic particles in "atomic mass units." An atomic mass unit ($\text{amu}$) is defined as one-twelfth the mass of a carbon-12 atom. Atomic mass units ($\text{amu}$) are useful, because, as you can see, the mass of a proton and the mass of a neutron are almost exactly $1$ in this unit system. Table $1$: Properties of Subatomic Particles
Particle Symbol Mass (amu) Relative Mass (proton = 1) Relative Charge Location
proton p+ 1 1 +1 inside the nucleus
electron e 5.45× 10−4 0.00055 −1 outside nucleus
neutron n0 1 1 0 inside the nucleus
amu in gram and kilogram
1 amu = 1.6605 × 10−24 g = 1.6605 × 10−27 kg
Atomic Number
Negative and positive charges of equal magnitude cancel each other out. This means that the negative charge on an electron perfectly balances the positive charge on the proton. In other words, a neutral atom must have exactly one electron for every proton. If a neutral atom has 1 proton, it must have 1 electron. If a neutral atom has 2 protons, it must have 2 electrons. If a neutral atom has 10 protons, it must have 10 electrons. You get the idea. In order to be neutral, an atom must have the same number of electrons and protons.
Scientists distinguish between different elements by counting the number of protons in the nucleus (Table $2$). If an atom has only one proton, we know it's a hydrogen atom. An atom with two protons is always a helium atom. If scientists count four protons in an atom, they know it's a beryllium atom. An atom with three protons is a lithium atom, an atom with five protons is a boron atom, an atom with six protons is a carbon atom . . . the list goes on.
Since an atom of one element can be distinguished from an atom of another element by the number of protons in its nucleus, scientists are always interested in this number, and how this number differs between different elements. The number of protons in an atom is called its atomic number ($Z$). This number is very important because it is unique for atoms of a given element. All atoms of an element have the same number of protons, and every element has a different number of protons in its atoms. For example, all helium atoms have two protons, and no other elements have atoms with two protons.
Name Protons Neutrons Electrons Atomic Number (Z) Mass Number (A)
Table $2$: Atoms of the First Six Elements
Hydrogen 1 0 1 1 1
Helium 2 2 2 2 4
Lithium 3 4 3 3 7
Beryllium 4 5 4 4 9
Boron 5 6 5 5 11
Carbon 6 6 6 6 12
Of course, since neutral atoms have to have one electron for every proton, an element's atomic number also tells you how many electrons are in a neutral atom of that element. For example, hydrogen has an atomic number of 1. This means that an atom of hydrogen has one proton, and, if it's neutral, one electron as well. Gold, on the other hand, has an atomic number of 79, which means that an atom of gold has 79 protons, and, if it's neutral, 79 electrons as well.
Neutral Atoms
Atoms are neutral in electrical charge because they have the same number of negative electrons as positive protons (Table $2$). Therefore, the atomic number of an atom also tells you how many electrons the atom has. This, in turn, determines many of the atom's chemical properties.
Mass Number
The mass number ($A$) of an atom is the total number of protons and neutrons in its nucleus. The mass of the atom is a unit called the atomic mass unit $\left( \text{amu} \right)$. One atomic mass unit is the mass of a proton, or about $1.67 \times 10^{-27}$ kilograms, which is an extremely small mass. A neutron has just a tiny bit more mass than a proton, but its mass is often assumed to be one atomic mass unit as well. Because electrons have virtually no mass, just about all the mass of an atom is in its protons and neutrons. Therefore, the total number of protons and neutrons in an atom determines its mass in atomic mass units (Table $2$).
Consider helium again. Most helium atoms have two neutrons in addition to two protons. Therefore the mass of most helium atoms is 4 atomic mass units ($2 \: \text{amu}$ for the protons + $2 \: \text{amu}$ for the neutrons). However, some helium atoms have more or less than two neutrons. Atoms with the same number of protons but different numbers of neutrons are called isotopes. Because the number of neutrons can vary for a given element, the mass numbers of different atoms of an element may also vary. For example, some helium atoms have three neutrons instead of two (these are called isotopes and are discussed in detail later on)
Why do you think that the "mass number" includes protons and neutrons, but not electrons? You know that most of the mass of an atom is concentrated in its nucleus. The mass of an atom depends on the number of protons and neutrons. You have already learned that the mass of an electron is very, very small compared to the mass of either a proton or a neutron (like the mass of a penny compared to the mass of a bowling ball). Counting the number of protons and neutrons tells scientists about the total mass of an atom. An atom's mass number is very easy to calculate provided you know the number of protons and neutrons in an atom. The mass number of a carbon atom with 6 protons and 7 neutrons is calculated and shown as follows:
$\text{mass number} \: A = \left( \text{number of protons} \right) + \left( \text{number of neutrons} \right) \nonumber$
$\text{mass number} = \text{6} + \text{6} = \text{12} \nonumber$
Example $1$
What is the mass number of an atom of helium that contains 2 neutrons?
Solution
$\left( \text{number of protons} \right) = 2$ (Remember that an atom of helium always has 2 protons.)
$\left( \text{number of neutrons} \right) = 2$
$\text{mass number} = \left( \text{number of protons} \right) + \left( \text{number of neutrons} \right)$
Isotopes
All atoms of the same element have the same number of protons, but some may have different numbers of neutrons. For example, all carbon atoms have six protons, and most have six neutrons as well. But some carbon atoms have seven or eight neutrons instead of the usual six. Atoms of the same element that differ in their numbers of neutrons are called isotopes. Many isotopes occur naturally. Usually one or two isotopes of an element are the most stable and common. Different isotopes of an element generally have the same physical and chemical properties. That's because they have the same numbers of protons and electrons.
An Example: Hydrogen Isotopes
Hydrogen is an example of an element that has isotopes. Three isotopes of hydrogen are modeled in Figure $4$. Most hydrogen atoms have just one proton and one electron and lack a neutron. These atoms are just called hydrogen. Some hydrogen atoms have one neutron as well. These atoms are the isotope named deuterium. Other hydrogen atoms have two neutrons. These atoms are the isotope named tritium.
For most elements other than hydrogen, isotopes are named for their mass number. For example, carbon atoms with the usual 6 neutrons have a mass number of 12 (6 protons + 6 neutrons = 12), so they are called carbon-12. Carbon atoms with 7 neutrons have atomic mass of 13 (6 protons + 7 neutrons = 13). These atoms are the isotope called carbon-13.
Example $2$: Lithium Isotopes
1. What is the atomic number and the mass number of an isotope of lithium containing 3 neutrons.
2. What is the atomic number and the mass number of an isotope of lithium containing 4 neutrons?
Solution
A lithium atom contains 3 protons in its nucleus irrespective of the number of neutrons or electrons.
a.
\begin{align}\text{atomic number} = \left( \text{number of protons} \right) &= 3 \nonumber \ \left( \text{number of neutrons} \right) &= 3 \nonumber\end{align} \nonumber
\begin{align} \text{mass number} & = \left( \text{number of protons} \right) + \left( \text{number of neutrons} \right) \nonumber\ \text{mass number} & = 3 + 3 \nonumber\ &= 6 \nonumber \end{align}\nonumber
b.
\begin{align}\text{atomic number} = \left( \text{number of protons} \right) &= 3 \nonumber\ \left( \text{number of neutrons} \right) & = 4\nonumber\end{align}\nonumber
\begin{align}\text{mass number} & = \left( \text{number of protons} \right) + \left( \text{number of neutrons} \right)\nonumber \ \text{mass number} & = 3 + 4\nonumber \ &= 7 \nonumber \end{align}\nonumber
Notice that because the lithium atom always has 3 protons, the atomic number for lithium is always 3. The mass number, however, is 6 in the isotope with 3 neutrons, and 7 in the isotope with 4 neutrons. In nature, only certain isotopes exist. For instance, lithium exists as an isotope with 3 neutrons, and as an isotope with 4 neutrons, but it doesn't exist as an isotope with 2 neutrons or as an isotope with 5 neutrons.
\begin{alignat*}{3} &[x \mapsto s]x &&= s && \ &[x \mapsto s]y &&= y \qquad &&\text{als } y\neq x \ &[x \mapsto s](\lambda(y)t_1) &&= \lambda(y)[x \mapsto s]t_1 \qquad &&\text{als } y \neq x \text{ en } y \not \in FV(s) \ &[x \mapsto s](t_1 \; t_2)&&= ([x \mapsto s]t_1)\;([x \mapsto s]t_2) && \end{alignat*} \nonumber
Exercise $1$
The number of protons in the nucleus of a tin atom is 50, while the number of neutrons in the nucleus is 68. What are the atomic number and the mass number of this isotope?
Answer
Atomic number = 50, mass number = 118
Symbols for Isotopes
There are two main ways in which scientists frequently show the mass number of an atom they are interested in. It is important to note that the mass number is not given on the periodic table. These two ways include writing a nuclear symbol or by giving the name of the element with the mass number written.
To write a nuclear symbol, the mass number is placed at the upper left (superscript) of the chemical symbol and the atomic number is placed at the lower left (subscript) of the symbol. The complete nuclear symbol for helium-4 is drawn below:
The following nuclear symbols are for a nickel nucleus with 31 neutrons and a uranium nucleus with 146 neutrons.
$\ce{^{59}_{28}Ni} \nonumber$
$\ce{ ^{238}_{92}U} \nonumber$
In the nickel nucleus represented above, the atomic number 28 indicates the nucleus contains 28 protons, and therefore, it must contain 31 neutrons in order to have a mass number of 59. The uranium nucleus has 92 protons as do all uranium nuclei and this particular uranium nucleus has 146 neutrons.
Another way of representing isotopes is by adding a hyphen and the mass number to the chemical name or symbol. Thus the two nuclei would be Nickel-59 or Ni-59 and Uranium-238 or U-238, where 59 and 238 are the mass numbers of the two atoms, respectively. Note that the mass numbers (not the number of neutrons) are given to the side of the name.
Example $3$: POTASSIUM-40
How many protons, electrons, and neutrons are in an atom of $^{40}_{19}\ce{K}$?
Solution
$\text{atomic number} = \left( \text{number of protons} \right) = 19 \nonumber$
For all atoms with no charge, the number of electrons is equal to the number of protons.
$\text{number of electrons} = 19 \nonumber$
The mass number, 40 is the sum of the protons and the neutrons.
To find the number of neutrons, subtract the number of protons from the mass number.
$\text{number of neutrons} = 40 - 19 = 21. \nonumber$
Example $4$: Zinc-65
How many protons, electrons, and neutrons are in an atom of zinc-65?
Solution
$\text{number of protons} = 30 \nonumber$
For all atoms with no charge, the number of electrons is equal to the number of protons.
$\text{number of electrons} = 30 \nonumber$
The mass number, 65 is the sum of the protons and the neutrons.
To find the number of neutrons, subtract the number of protons from the mass number.
$\text{number of neutrons} = 65 - 30 = 35 \nonumber$
Exercise $2$
How many protons, electrons, and neutrons are in each atom?
1. $^{60}_{27}\ce{Co}$
2. Na-24
3. $^{45}_{20}\ce{Ca}$
4. Sr-90
Answer a:
27 protons, 27 electrons, 33 neutrons
Answer b:
11 protons, 11 electrons, 13 neutrons
Answer c:
20 protons, 20 electrons, 25 neutrons
Answer d:
38 protons, 38 electrons, 52 neutrons
Table $3$: Properties of Isotopes of the First Six Elements.
Element Symbol Atomic Number Number of Protons Number of Neutrons Mass (amu) % Natural Abundance
hydrogen $\ce{^1_1H}$
(protium)
1 1 0 1.0078 99.989
$\ce{^2_1H}$
(deuterium)
1 1 1 2.0141 0.0115
$\ce{^3_1H}$
(tritium)
1 1 2 3.01605 — (trace)
helium $\ce{^3_2He}$ 2 2 1 3.01603 0.00013
$\ce{^4_2He}$ 2 2 2 4.0026 100
lithium $\ce{^6_3Li}$ 3 3 3 6.0151 7.59
$\ce{^7_3Li}$ 3 3 4 7.0160 92.41
beryllium $\ce{^9_4Be}$ 4 4 5 9.0122 100
boron $\ce{^{10}_5B}$ 5 5 5 10.0129 19.9
$\ce{^{11}_5B}$ 5 5 6 11.0093 80.1
carbon $\ce{^{12}_6C}$ 6 6 6 12.0000 98.89
$\ce{^{13}_6C}$ 6 6 7 13.0034 1.11
$\ce{^{14}_6C}$ 6 6 8 14.0032 — (trace)
Summary
• The atom consists of discrete particles that govern its chemical and physical behavior.
• Each atom of an element contains the same number of protons, which is the atomic number (Z).
• Neutral atoms have the same number of electrons and protons.
• Atoms of an element that contain different numbers of neutrons are called isotopes.
• Each isotope of a given element has the same atomic number but a different mass number (A), which is the sum of the numbers of protons and neutrons.
• The relative masses of atoms are reported using the atomic mass unit (amu) which is defined as one-twelfth of the mass of one atom of carbon-12, with 6 protons, 6 neutrons, and 6 electrons. The nuclear model of the atom consists of a small and dense positively charged interior surrounded by a cloud of electrons.
Contributors and Attributions
• Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected]).
• TextMap: Chemistry-The Central Science (Brown et al.)
• TextMap: Fundamentals of General, Organicn, and Biological Chemistry (McMurry et al.)
• Henry Agnew (UC Davis) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/03%3A_Atomic_Structure/3.05%3A_The_Atomic_Nucleus.txt |
Learning Objectives
• Know the properties of different types of electromagnetic radiation.
• Define an energy level in terms of the Bohr model.
• Discuss how the Bohr model can be used to explain atomic spectra.
• Describe the arrangement of electrons using the shell model.
Electromagnetic waves have an extremely wide range of wavelengths, frequencies, and energies. The highest energy form of electromagnetic waves are gamma (γ) rays and the lowest energy form are radio waves.
The figure below shows the electromagnetic spectrum, which is all forms of electromagnetic radiation. On the far left of Figure $1$ are the highest energy electromagnetic waves. These are called gamma rays and can be quite dangerous, in large numbers, to living systems. The next lower energy form of electromagnetic waves are called x-rays. Most of you are familiar with the penetration abilities of these waves. They can also be dangerous to living systems. Humans are advised to limit as much as possible the number of medical x-rays they have per year. Next lower, in energy, are ultraviolet rays. These rays are part of sunlight and the upper end of the ultraviolet range can cause sunburn and perhaps skin cancer. The tiny section next in the spectrum is the visible range of light … this section has been greatly expanded in the bottom half of the figure so it can be discussed in more detail. The visible range of electromagnetic radiation are the frequencies to which the human eye responds. Lower in the spectrum are infrared rays and radio waves.
The light energies that are in the visible range are electromagnetic waves that cause the human eye to respond when those frequencies enter the eye. The eye sends a signal to the brain and the individual "sees" various colors. The highest energy waves in the visible region cause the brain to see violet and as the energy decreases, the colors change to blue, green, yellow, orange, and red. When the energy of the wave is above or below the visible range, the eye does not respond to them. When the eye receives several different frequencies at the same time, the colors are blended by the brain. If all frequencies of light strike the eye together, the brain sees white and if there are no visible frequencies striking the eye, the brain sees black. The objects that you see around you are light absorbers - that is, the chemicals on the surface of the object will absorb certain frequencies and not others. Your eyes detect the frequencies that strike your eye. Therefore, if your friend is wearing a red shirt, it means the dye in that shirt absorbs every frequency except red and the red frequencies are reflected. If your only light source was one exact frequency of blue light and you shined it on a shirt that was red in sunlight, the shirt would appear black because no light would be reflected. The light from fluorescent types of lights do not contain all the frequencies of sunlight and so clothes inside a store may appear to be a slightly different color than when you get them home.
Continuous and Line Spectra
Electric light bulbs contain a very thin wire in them that emits light when heated. The wire is called a filament. The particular wire used in light bulbs is made of tungsten. A wire made of any metal would emit light under these circumstances but tungsten was chosen because the light it emits contains virtually every frequency and therefore, the light emitted by tungsten appears white. A wire made of some other element would emit light of some color that was not convenient for our uses. Every element emits light when energized by heating or passing electric current through it. Elements in solid form begin to glow when they are heated sufficiently and elements in gaseous form emit light when electricity passes through them. This is the source of light emitted by neon signs and is also the source of light in a fire.
Each Element Has a Unique Spectrum
The light frequencies emitted by atoms are mixed together by our eyes so that we see a blended color. Several physicists, including Angstrom in 1868 and Balmer in 1875, passed the light from energized atoms through glass prisms in such a way that the light was spread out so they could see the individual frequencies that made up the light. The emission spectrum (or atomic spectrum) of a chemical element is the unique pattern of light obtained when the element is subjected to heat or electricity.
When hydrogen gas is placed into a tube and electric current passed through it, the color of emitted light is pink. But when the color is spread out, we see that the hydrogen spectrum is composed of four individual frequencies. The pink color of the tube is the result of our eyes blending the four colors. Every atom has its own characteristic spectrum; no two atomic spectra are alike. The image below shows the emission spectrum of iron. Because each element has a unique emission spectrum, elements can be defined using them.
You may have heard or read about scientists discussing what elements are present in the sun or some more distant star, and after hearing that, wondered how scientists could know what elements were present in a place no one has ever been. Scientists determine what elements are present in distant stars by analyzing the light that comes from stars and finding the atomic spectrum of elements in that light. If the exact four lines that compose hydrogen's atomic spectrum are present in the light emitted from the star, that element contains hydrogen.
Bohr's Explanation of Line Spectra
In 1913, the Danish physicist Niels Bohr proposed a model of the electron cloud of an atom in which electrons orbit the nucleus and were able to produce atomic spectra. Understanding Bohr's model requires some knowledge of electromagnetic radiation (or light). Bohr's key idea in his model of the atom is that electrons occupy definite orbitals that require the electron to have a specific amount of energy. In order for an electron to be in the electron cloud of an atom, it must be in
one of the allowable orbitals and it must have the precise energy required for that orbit. Orbits closer to the nucleus would require smaller amounts of energy for an electron and orbits farther from the nucleus would require the electrons to have a greater amount of energy. The possible orbits are known as energy levels (n). One of the weaknesses of Bohr's model was that he could not offer a reason why only certain energy levels or orbits were allowed.
Figure $5$: Niels Bohr with Albert Einstein at Paul Ehrenfest's home in Leiden (December 1925).
Bohr hypothesized that the only way electrons could gain or lose energy would be to move from one energy level to another, thus gaining or losing precise amounts of energy. The energy levels are quantized, meaning that only specific amounts are possible. It would be like a ladder that had rungs only at certain heights. The only way you can be on that ladder is to be on one of the rungs and the only way you could move up or down would be to move to one of the other rungs. Suppose we had such a ladder with 10 rungs. Other rules for the ladder are that only one person can be on a rung in normal state and the ladder occupants must be on the lowest rung available. If the ladder had five people on it, they would be on the lowest five rungs. In this situation, no person could move down because all the lower rungs are full. Bohr worked out rules for the maximum number of electrons that could be in each energy level in his model and required that an atom in its normal state (ground state) had all electrons in the lowest energy levels available. Under these circumstances, no electron could lose energy because no electron could move down to a lower energy level. In this way, Bohr's model explained why electrons circling the nucleus did not emit energy and spiral into the nucleus.
Figure $6$ The energy levels (n= 1,2,3...) of the electrons can be viewed as rungs on a ladder.
The evidence used to support Bohr's model came from the atomic spectra. He suggested that an atomic spectrum is made by the electrons in an atom moving energy levels.
Ground States and Excited States
The electrons typically have the lowest energy possible, called the ground state. If the electrons are given energy (through heat, electricity, light, etc.) the electrons in an atom could absorb energy by jumping to a higher energy level, or excited state. The electrons then give off the energy in the form of a piece of light, called a photon, they had absorbed to fall back to a lower energy level. The energy emitted by electrons dropping back to lower energy levels would always be precise amounts of energy because the differences in energy levels were precise. This explains why you see specific lines of light when looking at an atomic spectrum - each line of light matches a specific "step down" that an electron can take in that atom. This also explains why each element produces a different atomic spectrum. Because each element has different acceptable energy levels for their electrons, the possible steps each element's electrons can take differ from all other elements.
Based on the wavelengths of the spectral lines, Bohr was able to calculate the energies that the hydrogen electron would have in each of its allowed energy levels. He then mathematically showed which energy level transitions correspond to the spectral lines in the atomic emission spectrum (see below).
He found that the four visible spectral lines corresponded to transitions from higher energy levels down to the second energy level $\left( n=2 \right)$. This is called the Balmer series (Figure $8$ ). Transitions ending in the ground state $\left( n=1 \right)$ are called the Lyman series, but the energies released are so large that the spectral lines are all in the ultraviolet region of the spectrum. The transitions called the Paschen series and the Brackett series both result in spectral lines in the infrared region because the energies are too small.
Bohr's model was a tremendous success in explaining the spectrum of the hydrogen atom. Unfortunately, when the mathematics of the model was applied to atoms with more than one electron, it was not able to correctly predict the frequencies of the spectral lines. While Bohr's model represented a great advancement in the atomic model and the concept of electron transitions between energy levels is valid, improvements were needed in order to fully understand all atoms and their chemical behavior.
Different metal electrons emit different wavelengths of light to return to their respective ground states, so the flame colors are varied. These flames can be used to produce atomic emmision spectra of the elements combusted. Using known values of emission spectra, one can perform a flame test on un unknown substance, gather an emmision spectrum from it, and determine which elements are in the unknown substance.
For example, in the case of copper ion, there are multiple different "paths" that the excited electrons can follow to emit photon of certain discrete energy. This produces multiple spectra lines because each discrete energy level difference will yield a specific wavelength of light, which determines the color.
Building Atoms: Main Shells
An electron shell is the outside part of an atom around the atomic nucleus. It is a group of atomic orbitals with the same value of the principal quantum number $n$. Electron shells have one or more electron subshells, or sublevels. The name for electron shells comes from the Bohr model, in which groups of electrons were believed to go around the nucleus at certain distances, so that their orbits formed "shells".
An electron shell may be thought of as an orbit followed by electrons around an atom nucleus. Because each shell can contain only a fixed number of electrons, each shell is associated with a particular range of electron energy, and thus each shell must fill completely before electrons can be added to an outer shell. The electrons in the outermost shell determine the chemical properties of the atom (see Valence shell). For an explanation of why electrons exist in these shells see electron configuration.
Figure $10$ A shell diagram of lithium (left) and Sodium (right)
The electron shells are labeled K, L, M, N, O, P, and Q; or 1, 2, 3, 4, 5, 6, and 7; going from innermost shell outwards. Electrons in outer shells have higher average energy and travel farther from the nucleus than those in inner shells. This makes them more important in determining how the atom reacts chemically and behaves as a conductor, because the pull of the atom's nucleus upon them is weaker and more easily broken. In this way, a given element's reactivity is highly dependent upon its electronic configuration.
Drawing Shell Models
Video $1$ How to draw the shell model for sulfur.
Note: The number of electrons that can occupy each energy level are 2 (first level), 8 (2nd level), 18 (3rd level), and 32 (4th level) based on the formula: # of electrons = 2(n)2, wherein n = principle energy level.
Table $1$ shows the number of electrons that fill each shell for neutral atoms of several elements. As mentioned earlier, the innermost shell (corresponding to lowest energy) is filled first and only a fixed number of electrons is allowed in each shell. The only electron in hydrogen (Z=1) goes to the first shell. In lithium atom (Z=3), the two electrons fill the first shell, and the third electron goes to the second shell. An argon atom (Z=18) has 18 electrons. The 10 electrons fill the first and second shells, and the remaining 8 electrons go to the third shell. The electron configuration for elements pass argon are covered in more detail in section 3.7.
Table $1$ Electron Configuraion (Arrangement) of Several Elements.
Element
Symbol
Atomic Number*
(Z)
First Shell
n=1
(2 electrons allowed)
Second Shell
n=2
(8 electrons allowed)
Third Shell
n=3
(18 electrons allowed)
H 1 1
He 2 2
C 6 2 4
N 7 2 5
Na 11 2 8 1
Mg 12 2 8 2
Cl 17 2 8 7
Ar 18 2 8 8
Note:* In a neutral atom he number of protons is equal to the number of electrons.
Summary
• Electromagnetic radiation has a wide spectrum, including gamma rays, X-rays, UV rays, visible light, IR radiation, microwaves, and radio waves.
• The different colors of light differ in their frequencies (or wavelengths).
• Bohr's model suggests each atom has a set of unchangeable energy levels and electrons in the electron cloud of that atom must be in one of those energy levels.
• Bohr's model suggests that the atomic spectra of atoms is produced by electrons gaining energy from some source, jumping up to a higher energy level, then immediately dropping back to a lower energy level and emitting the energy different between the two energy levels.
• The existence of the atomic spectra is support for Bohr's model of the atom.
• Bohr's model was only successful in calculating energy levels for the hydrogen atom.
• The shell model is a good representation of electron arrangement only for elements 1-18.
Contributors and Attributions
• Connections (John Hutchinson)
• Wikipedia
• Henry Agnew (UC Davis) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/03%3A_Atomic_Structure/3.06%3A_Electron_Arrangement-_The_Bohr_Model_%28Orbits%29.txt |
Learning Objectives
• Represent the organization of electrons by an electron configuration based on the quantum mechanical model of an atom.
Quantum mechanics is the study of the motion of objects that are atomic or subatomic in size and thus demonstrate wave-particle duality. One of the fundamental (and hardest to understand) principles of quantum mechanics is that the electron is both a particles and a wave. In the everyday macroscopic world of things we can see, something cannot be both. But this duality can exist in the quantum world of the submicroscopic at the atomic scale.
At the heart of quantum mechanics is the idea that we cannot specify accurately the location of an electron. All we can say is that there is a probability that it exists within this certain volume of space. The scientist Erwin Schrödinger developed an equation that deals with these calculations, which we will not pursue at this time. Recall that in the Bohr model, the exact path of the electron was restricted to very well-defined circular orbits around the nucleus. An orbital is the quantum mechanical refinement of Bohr’s orbit. In contrast to his concept of a simple circular orbit with a fixed radius, orbitals are mathematically derived regions of space with different probabilities of having an electron.
Figure $1$ The Erwin Schrödinger.
Building Atoms by Orbital Filling
In the quantum-mechanical model of an atom, electrons in the same atom that have the same principal quantum number (n) or principal energy level are said to occupy an electron shell of the atom. Orbitals define regions in space where you are likely to find electrons. As shown in Figure $2$ s orbitals are spherical shaped, and p orbitals are dumb-bell shaped. Orbitals within a shell are divided into subshells (sublevels). There are two subshells in the second shell namely the 2s subshell (with one orbital) and the 2p subshell (with three orbitals), see Table $1$. It should be noted that the diagrams in Figure $2$ are estimates of the electron distribution in space, not surfaces electrons are fixed on.
Table $1$ summarizes the number and type orbitals that can be found within each principal energy level as well as the number of electrons allowed. Table $1$ Electron Arrangement Within Energy Levels
Principal Quantum Number $\left( n \right)$
(Principal Shell)
Sublevels
(Subshell)
Number of Orbitals per Sublevel
Number of Electrons per Sublevel
(Maximum of
2 electrons/orbital)
1 1s
1
(1s)
2
2 2s
1
(2s)
2
2p
3
(2px 2py 2pz)
6
3 3s
1
(3s)
2
3p
3
(3px 3py 3pz)
6
3d
5
(refer to Figure $\PageIndex{2c}$)
10
In contrast to the shell model discussed in section 3.6, chemists more often use a more detailed electron configuration to represent the organization of electrons in shells and subshells in an atom. An atom's electron configuration includes the shell and subshell labels, with a right superscript giving the number of electrons in that subshell. The shells and subshells are listed in the order of filling. Electrons are typically organized around an atom by starting at the lowest possible quantum numbers first, which are the shells-subshells with lower energies.
For example, an H atom has a single electron in the 1s subshell. Its electron configuration is H: 1s1
He has two electrons in the 1s subshell. Its electron configuration is He: 1s2
The three electrons for Li are arranged in the 1s subshell (two electrons) and the 2s subshell (one electron). The electron configuration of Li is
Li: 1s22s1
Figure $3$ The diagram of an electron configuration of hydrogen right
specifies the subshell and superscript number of electrons.
Element Name Symbol Atomic Number Electron Configuration
Table $2$: Electron Configurations of First and Second-Period Elements
Hydrogen $\ce{H}$ 1 $1s^1$
Helium $\ce{He}$ 2 $1s^2$
Lithium $\ce{Li}$ 3 $1s^2 2s^1$
Beryllium $\ce{Be}$ 4 $1s^2 2s^2$
Boron $\ce{B}$ 5 $1s^2 2s^2 2p^1$
Carbon $\ce{C}$ 6 $1s^2 2s^2 2p^2$
Nitrogen $\ce{N}$ 7 $1s^2 2s^2 2p^3$
Oxygen $\ce{O}$ 8 $1s^2 2s^2 2p^4$
Fluorine $\ce{F}$ 9 $1s^2 2s^2 2p^5$
Neon $\ce{Ne}$ 10 $1s^2 2s^2 2p^6$
In order to create ground state electron configurations for any element, it is necessary to know the way in which the atomic sublevels are organized in order of increasing energy. Figure $4$ shows the order of increasing energy of the sublevels.
Figure $5$ illustrates the traditional way to remember the filling order for atomic orbitals.
Example $1$: Nitrogen Atoms
Nitrogen has 7 electrons. Write the electron configuration for nitrogen.
Solution:
Take a close look at Figure 9.6.5, and use it to figure out how many electrons go into each sublevel, and also the order in which the different sublevels get filled.
1. Begin by filling up the 1s sublevel. This gives 1s2. Now all of the orbitals in the red n = 1 block are filled.
Since we used 2 electrons, there are 7 − 2 = 5 electrons left
2. Next, fill the 2s sublevel. This gives 1s22s2. Now all of the orbitals in the s sublevel of the orange n = 2 block are filled.
Since we used another 2 electrons, there are 5 − 2 = 3 electrons left
3. Notice that we haven't filled the entire n = 2 block yet… there are still the p orbitals!
The final 3 electrons go into the 2p sublevel. This gives 1s22s22p3
The overall electron configuration is: 1s22s22p3.
Example $2$: Potassium Atoms
Potassium has 19 electrons. Write the electron configuration code for potassium.
Solution
This time, take a close look at Figure 9.6.5.
1. Begin by filling up the 1s sublevel. This gives 1s2. Now the n = 1 level is filled.
Since we used 2 electrons, there are 19 − 2 = 17 electrons left
2. Next, fill the 2s sublevel. This gives 1s22s2
Since we used another 2 electrons, there are 17 − 2 = 15 electrons left
3. Next, fill the 2p sublevel. This gives 1s22s22p6. Now the n = 2 level is filled.
Since we used another 6 electrons, there are 15 − 6 = 9 electrons left
4. Next, fill the 3s sublevel. This gives 1s22s22p63s2
Since we used another 2 electrons, there are 9 − 2 = 7 electrons left
5. Next, fill the 3p sublevel. This gives 1s22s22p63s23p6
Since we used another 6 electrons, there are 7 − 6 = 1 electron left
Here's where we have to be careful – right after 3p6!!
Remember, 4s comes before 3d!
6. The final electron goes into the 4s sublevel. This gives 1s22s22p63s23p64s1
The overall electron configuration is: 1s22s22p63s23p64s1.
Exercise $1$: Magnesium and Sodium Atoms
What is the electron configuration for Mg and Na?
Answer Mg
Mg: 1s22s22p63s2
Answer Na
Na: 1s22s22p63s1
Summary
• Orbitals define regions in space where you are likely to find electrons. s orbitals are spherical shaped. p orbitals are dumb-bell shaped. The three possible p orbitals are always perpendicular to each other.
• Electron configuration notation simplifies the indication of where electrons are located in a specific atom. Superscripts are used to indicate the number of electrons in a given sublevel.
• Electrons are added to atomic orbitals in order from low energy to high energy. | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/03%3A_Atomic_Structure/3.07%3A_Electron_Arrangement-_The_Quantum_Model.txt |
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Learning Objectives
• Know the classification of the different elements.
• Define valence electrons.
• Describe how the number of valence electrons are related the arrangement of the main group elements and its chemical behavior.
Elements that have similar chemical properties are grouped in columns called groups (or families). As well as being numbered, some of these groups have names—for example, alkali metals (the first column of elements), alkaline earth metals (the second column of elements), halogens (the next-to-last column of elements), and noble gases (the last column of elements). In and atomic , the main group is the group of elements whose lightest members are represented by , , , , , , , and as arranged in the of the elements. The main group includes the elements (except , which is sometimes not included) in groups 1 and 2, and groups 13 to 18. Main-group elements (with some of the lighter ) are the most abundant elements on , in the , and in the . They are sometimes also called the representative elements. In older nomenclature the main-group elements are groups IA and IIA, and groups IIIB to 0 (CAS groups IIIA to VIIIA). Group 12 is labelled as group IIB in both systems. Group 3 is labelled as group IIIA in the older nomenclature (CAS group IIIB).
Each row of elements on the periodic table is called a period. Periods have different lengths; the first period has only 2 elements (hydrogen and helium), while the second and third periods have 8 elements each. The fourth and fifth periods have 18 elements each, and later periods are so long that a segment from each is removed and placed beneath the main body of the table.
One way to categorize the elements of the periodic table is shown in Figure \(1\). The first two columns on the left and the last six columns on the right as mentioned earlier are the main group elements. The ten-column block between these columns contains the transition metals. The two rows beneath the main body of the periodic table contain the inner transition metals. The elements in these two rows are also referred to as, respectively, the lanthanide metals and the actinide metals.
Family Features: Outer Electron Configurations
Valence Electrons
The valence shell is the outermost shell of an atom in its uncombined state, which contains the electrons most likely to account for the nature of any reactions involving the atom and of the bonding interactions it has with other atoms. Valence electrons are the outer-shell electrons of an atom. These are electrons that can participate in the formation of a chemical bond. The presence of valence electrons can determine the element's chemical properties and whether it may bond with other elements.
An atom with a closed shell of valence electrons tends to be chemically inert. An atom with one or two valence electrons more than a closed shell is highly reactive, because the extra valence electrons are easily removed to form a positive ion. An atom with one or two valence electrons fewer than a closed shell is also highly reactive, because of a tendency either to gain the missing valence electrons (thereby forming a negative ion), or to share valence electrons (thereby forming a covalent bond).
The number of valence electrons of an element can be determined by the periodic table group (vertical column) in which the element is categorized (Table \(1\)). With the exception of groups 3–12 (the transition metals), the units digit of the group number identifies how many valence electrons are associated with a neutral atom of an element listed under that particular column.
Table \(1\): Number of Valence Electrons for Main Periodic Table Groups.
Periodic table group Valence electrons
Group 1 (I) (alkali metals) 1
Group 2 (II) (alkaline earth metals) 2
Groups 3-12 (transition metals) 2*
Group 13 (III) (boron group) 3
Group 14 (IV) (carbon group) 4
Group 15 (V) (pnictogens) 5
Group 16 (VI) (chalcogens) 6
Group 17 (VII) (halogens) 7
Group 18 (VIII or 0) (noble gases) 8**
** Except for helium, which has only two valence electrons.
Valence Electrons and the Periodic Table
Video \(1\) Finding the valence electron for an element.
Family Groups
As previously noted, the periodic table is arranged so that elements with similar chemical behaviors are in the same group. Chemists often make general statements about the properties of the elements in a group using descriptive names with historical origins.
Group 1: The Alkali Metals
The alkali metals are lithium, sodium, potassium, rubidium, cesium, and francium. Hydrogen is unique in that it is generally placed in Group 1, but it is not a metal.
The compounds of the alkali metals are common in nature and daily life. One example is table salt (sodium chloride); lithium compounds are used in greases, in batteries, and as drugs to treat patients who exhibit manic-depressive, or bipolar, behavior. Although lithium, rubidium, and cesium are relatively rare in nature, and francium is so unstable and highly radioactive that it exists in only trace amounts, sodium and potassium are the seventh and eighth most abundant elements in Earth’s crust, respectively.
Group 2: The Alkaline Earth Metals
The alkaline earth metals are beryllium, magnesium, calcium, strontium, barium, and radium. Beryllium, strontium, and barium are rare, and radium is unstable and highly radioactive. In contrast, calcium and magnesium are the fifth and sixth most abundant elements on Earth, respectively; they are found in huge deposits of limestone and other minerals.
Group 17: The Halogens
The halogens are fluorine, chlorine, bromine, iodine, and astatine. The name halogen is derived from the Greek words for “salt forming,” which reflects that all the halogens react readily with metals to form compounds, such as sodium chloride and calcium chloride (used in some areas as road salt).
Compounds that contain the fluoride ion are added to toothpaste and the water supply to prevent dental cavities. Fluorine is also found in Teflon coatings on kitchen utensils. Although chlorofluorocarbon propellants and refrigerants are believed to lead to the depletion of Earth’s ozone layer and contain both fluorine and chlorine, the latter is responsible for the adverse effect on the ozone layer. Bromine and iodine are less abundant than chlorine, and astatine is so radioactive that it exists in only negligible amounts in nature.
Group 18: The Noble Gases
The noble gases are helium, neon, argon, krypton, xenon, and radon. Because the noble gases are composed of only single atoms, they are called monatomic. At room temperature and pressure, they are unreactive gases. Because of their lack of reactivity, for many years they were called inert gases or rare gases. However, the first chemical compounds containing the noble gases were prepared in 1962. Although the noble gases are relatively minor constituents of the atmosphere, natural gas contains substantial amounts of helium. Because of its low reactivity, argon is often used as an unreactive (inert) atmosphere for welding and in light bulbs. The red light emitted by neon in a gas discharge tube is used in neon lights.
Example \(1\): Groups
Provide the family or group name of each element.
1. Li
2. Ar
3. Cl
Solution
1. Lithium is an alkali metal (Group 1)
2. Argon is a noble gas (Group 18)
3. Chlorine is a halogen (Group 17)
Exercise \(1\): Groups
Provide the family or group name of each element.
1. F
2. Ca
3. Kr
Answer a:
Fluorine is a halogen (Group 17)
Answer b:
Calcium is a alkaline earth metal (Group 2)
Answer c:
Krypton is a noble gas (Group 18)
Metals and Nonmetals
Certain elemental properties become apparent in a survey of the periodic table as a whole. Every element can be classified as either a metal, a nonmetal, or a metalloid (or semi metal), as shown in Figure \(2\). A metal is a substance that is shiny, typically (but not always) silvery in color, and an excellent conductor of electricity and heat. Metals are also malleable (they can be beaten into thin sheets) and ductile (they can be drawn into thin wires). A nonmetal is typically dull and a poor conductor of electricity and heat. Solid nonmetals are also very brittle. As shown in Figure \(2\), metals occupy the left three-fourths of the periodic table, while nonmetals (except for hydrogen) are clustered in the upper right-hand corner of the periodic table. The elements with properties intermediate between those of metals and nonmetals are called metalloids (or semi-metals). Elements adjacent to the bold line in the right-hand portion of the periodic table have semimetal properties.
Example \(2\)
Based on its position in the periodic table, classify each element below as metal, a nonmetal, or a metalloid.
1. Se
2. Mg
3. Ge
Solution
1. In Figure \(1\), selenium lies above and to the right of the diagonal line marking the boundary between metals and nonmetals, so it should be a nonmetal.
2. Magnesium lies to the left of the diagonal line marking the boundary between metals and nonmetals, so it should be a metal.
3. Germanium lies within the diagonal line marking the boundary between metals and nonmetals, so it should be a metalloid.
Exercise \(2\)
Based on its location in the periodic table, do you expect indium to be a nonmetal, a metal, or a metalloid?
Answer
Indium is a metal.
Example \(3\): Classification of Elements
Classify each element as metal, non metal, transition metal or inner transition metal.
1. Li
2. Ar
3. Am
4. Fe
Solution
1. Lithium is a metal.
2. Argon is a non metal
3. Americium is an inner transition metal
4. Iron is a transition metal.
Exercise \(3\): Classification of Elements
Classify each element as metal, non metal, transition metal or inner transition metal.
1. F
2. U
3. Cu
Answer a:
Fluorine is a nonmetal.
Answer b:
Uranium is a metal (and a inner transition metal too)
Answer c:
Copper is a metal (and a transition metal too)
Alkali Metals
Video \(2\) Alkali metals in water.
Summary
• Valence electrons are the outer-shell electrons of an atom. These are electrons that can participate in the formation of a chemical bond.
• Elements that exhibit similar chemistry appear in vertical columns called groups (numbered 1–18 from left to right).
• The seven horizontal rows are called periods. Some of the groups have widely-used common names, including the alkali metals (Group 1) and the alkaline earth metals (Group 2) on the far left, and the halogens (Group 17) and the noble gases (Group 18) on the far right.
• An important grouping of elements in the periodic table are the main group elements, the transition metals, and the inner transition metals (the lanthanides, and the actinides).
• The elements can be broadly divided into metals, nonmetals, and semi metals. Semi metals exhibit properties intermediate between those of metals and nonmetals. Metals are located on the left of the periodic table, and nonmetals are located on the upper right. They are separated by a diagonal band of semi metals.
• Metals are lustrous, good conductors of electricity, and readily shaped (they are ductile and malleable). Solid nonmetals are generally brittle and poor electrical conductors.
Contributors and Attributions
• Henry Agnew (UC Davis)
• Wikipedia | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/03%3A_Atomic_Structure/3.08%3A_Electron_Configurations_and_the_Periodic_Table.txt |
Thumbnail: Non-polar covalent bonds in methane (\(\ce{CH4}\)). The Lewis structure shows electrons shared between C and H atoms. (CC BY-sa 2.5; via ).
04: Chemical Bonds
Learning Objective
• Define the octet rule.
The octet rule is a chemical rule of thumb that reflects observation that elements tend to bond in such a way that each atom has eight electrons in its valence shell, giving it the same electronic configuration as a noble gas.
Referring to the octet rule, atoms attempt to get a noble gas electron configuration, which is eight valence electrons. Metals have few electrons in their outer-most orbitals. By losing those electrons, these metals can achieve noble gas configuration and satisfy the octet rule. Sodium has one valence electron, so giving it up would result in the same electron configuration as neon. Due to the loss of electron, a positively charged cation (Na+) called sodium ion is formed.
Similarly, nonmetals that have close to 8 electrons in their valence shells tend to readily accept electrons to achieve noble gas configuration. Chlorine has seven valence electrons, so if it takes one it will have eight (an octet). Chlorine has the electron configuration of argon when it gains an electron. The resulting ion (Cl-) is called chloride ion.
The octet rule could have been satisfied if chlorine gave up all seven of its valence electrons and sodium took them. In that case, both would have the electron configurations of noble gasses, with a full valence shell. However, their charges would be much higher. It would be Na7- and Cl7+, which is much less stable than Na+ and Cl-. Atoms are more stable when they have no charge, or a small charge.
Summary
The octet rule refers to the tendency of atoms to prefer to have eight electrons in the valence shell.
Contributor
Organic Chemistry Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Ionic_and_Covalent_Bonds
4.02: Lewis (Electron-Dot) Symbols
Learning Objective
• Draw a Lewis electron dot diagram for an atom.
• Know the importance of Lewis dot in bonding.
At the beginning of the 20th century, an American physical chemist G. N. Lewis (1875–1946) devised a system of symbols—now called Lewis electron dot symbols (often shortened to Lewis dot symbols) that can be used for predicting the number of bonds formed by most elements in their compounds.
Lewis Dot symbols
• convenient representation of valence electrons
• allows scientists to keep track of valence electrons during bond formation
• consists of the atomic symbol for the element plus a dot for each valence electron
To write an element’s Lewis dot symbol, we place dots representing its valence electrons, one at a time, around the element’s chemical symbol. Up to four dots are placed above, below, to the left, and to the right of the symbol (in any order, as long as elements with four or fewer valence electrons have no more than one dot in each position). The next dots, for elements with more than four valence electrons, are again distributed one at a time, each paired with one of the first four. For example, the element sulfur has six valence electrons (note roman numeral above group on the periodic table) and its Lewis symbol would be:
Fluorine, for example, has seven valence electrons, so its Lewis dot symbol is constructed as follows:
Lewis eventually published his theory of chemical bonding in 1916. Formation of chemical bonds to complete the requirement of eight electrons for the atom becomes a natural tendency. Lewis dot symbols of the first two periods are given here to illustrate this point. In fact, the entire group (column) of elements have the same Lewis dot symbols, because they have the same number of valence electrons.
By going through the periodic table, we see that the Lewis electron dot symbols of atoms will never have more than eight dots around the atomic symbo see Table $1$.
Table $1$ Lewis Dot Symbols of the Main Group Elements
$\ce{H\cdot}$ $\textrm{He:}$
$\underset{\:}{\ce{Li\cdot}}$
$\underset{\:}{\ce{\cdot Be \cdot}}$
$\ce{ \cdot \underset{\:}{\overset{\Large{\cdot}}{B}} \cdot}$
$\ce{ \cdot \underset{\Large{\cdot}}{\overset{\Large{\cdot}}{C}} \cdot}$
$\underset{\Large{\cdot\,}} {\overset{\Large{\cdot}} {\textrm{:N}\cdot}}$
$\underset{\Large{\cdot\cdot\,}} {\overset{\Large{\cdot}} {\textrm{:O}\cdot}}$
$\underset{\Large{\cdot\cdot}} {\overset{\Large{\cdot\cdot}} {\textrm{:F}\cdot}}$
$\underset{\Large{\cdot\cdot}} {\overset{\Large{\cdot\cdot}}{\textrm{:Ne:}}}$
$\ce{Na}$
$\ce{K}$
$\ce{Rb}$
$\ce{Cs}$
$\ce{Mg}$
$\ce{Ca}$
$\ce{Sr}$
$\ce{Ba}$
$\ce{Al}$
$\ce{Ga}$
$\ce{In}$
$\ce{Tl}$
$\ce{Si}$
$\ce{Ge}$
$\ce{Sn}$
$\ce{Pb}$
$\ce{P}$
$\ce{As}$
$\ce{Sb}$
$\ce{Bi}$
$\ce{S}$
$\ce{Se}$
$\ce{Te}$
$\ce{Po}$
$\ce{Cl}$
$\ce{Br}$
$\ce{I}$
$\ce{At}$
$\ce{Ar}$
$\ce{Kr}$
$\ce{At}$
$\ce{Rn}$
Example $1$: Lewis Dot Diagrams
What is the Lewis electron dot diagram for each element?
1. aluminum
2. selenium
Solution
1. The valence electron configuration for aluminum is 3s23p1. So it would have three dots around the symbol for aluminum, two of them paired to represent the 3s electrons:
$\dot{Al:} \nonumber$
1. The valence electron configuration for selenium is 4s24p4. In the highest-numbered shell, the n = 4 shell, there are six electrons. Its electron dot diagram is as follows:
$\mathbf{\cdot }\mathbf{\dot{\underset{.\: .}Se}}\mathbf{:} \nonumber$
Exercise $1$
What is the Lewis electron dot diagram for each element?
1. phosphorus
2. argon
Answer a
$\mathbf{\cdot }\mathbf{\dot{\underset{.}P}}\mathbf{:} \nonumber$
Answer b
$\mathbf{:}\mathbf{\ddot{\underset{.\, .}Ar}}\mathbf{:} \nonumber$
Summary
• Lewis electron dot diagrams use dots to represent valence electrons around an atomic symbol.
• Lewis dot symbols can be used to predict the number of bonds formed by elements in a compound.
4.03: The Reaction of Sodium with Chlorine
Learning Objectives
• Explain the bonding nature of ionic compounds.
• Relate microscopic bonding properties to macroscopic solid properties.
Neutral atoms and their associated ions have very different physical and chemical properties. Sodium atoms form sodium metal, a soft, silvery-white metal that burns vigorously in air and reacts explosively with water. Chlorine atoms form chlorine gas, Cl2, a yellow-green gas that is extremely corrosive to most metals and very poisonous to animals and plants. The vigorous reaction between the elements sodium and chlorine forms the white, crystalline compound sodium chloride, common table salt, which contains sodium cations and chloride anions (Figure \(1\)). The compound composed of these ions exhibits properties entirely different from the properties of the elements sodium and chlorine. Chlorine is poisonous, but sodium chloride is essential to life; sodium atoms react vigorously with water, but sodium chloride simply dissolves in water.
Ionic Bonds
When two atoms approach each other, they have the potential to bond (or connect). If a metal and a nonmetal interact, then an ionic bond will result. These types of bonds involve the metal donating it(s) valence electron(s) to a nonmetal. As the electronic transfer occurs, both atoms will achieve more stable confirmations. The end result will be a less reactive compound. These type of species are composed of both cations and anions. In addition, they are crystalline and solid in nature (Figure \(2\)). A few examples of real-world ionic compounds would be NaCl (table salt) and NaF (active ingredient in toothpaste).
The formula for table salt is NaCl. It is the result of Na+ ions and Cl- ions bonding together (Figure \(3\)). If sodium metal and chlorine gas mix under the right conditions, they will form salt. The sodium loses an electron, and the chlorine gains that electron. This reaction is highly favorable because of the electrostatic attraction between the particles. In the process, a great amount of light and heat is released. The resulting salt is mostly unreactive — it is stable. It will not undergo any explosive reactions, unlike the sodium and chlorine that it is made of.
Figure \(2\) NaCl crystals.
(Public Domain; NASA).
The reaction is represented with Lewis dot symbols below.
Chloride Salts
The sodium atom is donating its 1 valence electron to the chlorine atom. This creates a sodium cation and a chlorine anion. Notice that the net charge of the resulting compound is 0.
Summary
• An ionic bond is formed when a metal donates it(s) valence electron(s) to a nonmetal.
• The resulting ionic compound is more stable and less reactive. | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/04%3A_Chemical_Bonds/4.01%3A_The_Art_of_Deduction-_Stable_Electron_Configurations.txt |
Learning Objectives
• State the octet rule.
• Define cations and anions.
• Define ionic bond.
• Draw Lewis structures for ionic compounds.
The Octet Rule: The Drive for Eight
The octet rule refers to the tendency of atoms to prefer to have eight electrons in the valence shell. When atoms have fewer than eight electrons, they tend to react and form more stable compounds. When discussing the octet rule, we do not consider d or f electrons. Only the s and p electrons are involved in the octet rule, making it useful for the main group elements (elements not in the transition metal or inner-transition metal blocks); an octet in these atoms corresponds to an electron configurations ending with $s^2p^6$.
Note
The noble gases rarely form compounds. They have the most stable configuration (full octet, no charge), so they have no reason to react and change their configuration. All other elements attempt to gain, lose, or share electrons to achieve a noble gas configuration.
Most atoms do not have eight electrons in their valence electron shell. Some atoms have only a few electrons in their outer shell, while some atoms lack only one or two electrons to have an octet. In cases where an atom has three or fewer valence electrons, the atom may lose those valence electrons quite easily until what remains is a lower shell that contains an octet. Atoms that lose electrons acquire a positive charge as a result because they are left with fewer negatively charged electrons to balance the positive charges of the protons in the nucleus. Positively charged ions are called cations. Most metals become cations when they make ionic compounds.
Cations
A neutral sodium atom is likely to achieve an octet in its outermost shell by losing its one valence electron.
$\ce{Na \rightarrow Na^{+} + e^{-}} \nonumber$
The cation produced in this way, Na+, is called the sodium ion to distinguish it from the element. The outermost shell of the sodium ion is the second electron shell, which has eight electrons in it. The octet rule has been satisfied. F
Anions
Some atoms have nearly eight electrons in their valence shell and can gain additional valence electrons until they have an octet. When these atoms gain electrons, they acquire a negative charge because they now possess more electrons than protons. Negatively charged ions are called anions. Most nonmetals become anions when they make ionic compounds.
A neutral chlorine atom has seven electrons in its outermost shell. Only one more electron is needed to achieve an octet in chlorine’s valence shell. (In table salt, this electron comes from the sodium atom.)
$\ce{e^{-} +Cl -> Cl^{-}} \nonumber$
In this case, the ion has the same outermost shell as the original atom, but now that shell has eight electrons in it. Once again, the octet rule has been satisfied. The resulting anion, Cl, is called the chloride ion; note the slight change in the suffix (-ide instead of -ine) to create the name of this anion.
The names for positive and negative ions are pronounced CAT-eye-ons and ANN-eye-ons, respectively.
Note the usefulness of the periodic table in predicting likely ion formation and charge (Figure $1$). Moving from the far left to the right on the periodic table, main-group elements tend to form cations with a charge equal to the group number. That is, group 1 elements form 1+ ions; group 2 elements form 2+ ions, and so on. Moving from the far right to the left on the periodic table, elements often form anions with a negative charge equal to the number of groups moved left from the noble gases. For example, group 17 elements (one group left of the noble gases) form 1− ions; group 16 elements (two groups left) form 2− ions, and so on. This trend can be used as a guide in many cases, but its predictive value decreases when moving toward the center of the periodic table. In fact, transition metals and some other metals often exhibit variable charges that are not predictable by their location in the table. For example, copper can form ions with a 1+ or 2+ charge, and iron can form ions with a 2+ or 3+ charge.
Lewis Symbols for Ionic Compounds
In the section above, we saw how ions are formed by losing electrons to make cations or by gaining electrons to form anions. The astute reader may have noticed something: Many of the ions that form have eight electrons in their valence shell. Either atoms gain enough electrons to have eight electrons in the valence shell and become the appropriately charged anion, or they lose the electrons in their original valence shell; the lower shell, now the valence shell, has eight electrons in it, so the atom becomes positively charged. For whatever reason, having eight electrons in a valence shell is a particularly energetically stable arrangement of electrons. When atoms form compounds, the octet rule is not always satisfied for all atoms at all times, but it is a very good rule of thumb for understanding the kinds of bonding arrangements that atoms can make.
It is not impossible to violate the octet rule. Consider lithium: in its elemental form, it has one valence electron and is stable. It is rather reactive, however, and does not require a lot of energy to remove that electron to make the Li+ ion. We could remove another electron by adding even more energy to the ion, to make the Li2+ ion. However, that requires much more energy than is normally available in chemical reactions, so sodium stops at a 1+ charge after losing a single electron. It turns out that the Li+ ion has a complete octet in its new valence shell, the n = 2 shell, which satisfies the octet rule. The octet rule is a result of trends in energies and is useful in explaining why atoms form the ions that they do.
Now consider an Li atom in the presence of a Cl atom. The two atoms have these Lewis electron dot diagrams and electron configurations:
$\mathbf{Li\, \cdot }\; \; \; \; \; \; \; \; \; \; \mathbf{\cdot }\mathbf{\ddot{\underset{.\: .}Cl}}\mathbf{\: :} \nonumber$
$\left [ Ne \right ]3s^{1}\; \; \; \; \left [ Ne \right ]3s^{2}3p^{5} \nonumber$
For the Li atom to obtain an octet, it must lose an electron; for the Cl atom to gain an octet, it must gain an electron. An electron transfers from the Na atom to the Cl atom:
$\mathbf{Li\, \cdot }\curvearrowright \mathbf{\cdot }\mathbf{\ddot{\underset{.\: .}Cl}}\mathbf{\: :} \nonumber$
resulting in two ions—the Na+ ion and the Cl ion:
$\mathbf{Li\, \cdot }^{+}\; \; \; \; \; \; \; \; \mathbf{:}\mathbf{\ddot{\underset{.\: .}Cl}}\mathbf{\: :}^{-} \nonumber$
$\left [ Ne \right ]\; \; \; \; \; \left [ Ne \right ]3s^{2}3p^{6} \nonumber$
Both species now have complete octets, and the electron shells are energetically stable. From basic physics, we know that opposite charges attract. This is what happens to the Na+ and Cl ions:
$\mathbf{Li\, \cdot }^{+}\; + \; \mathbf{:}\mathbf{\ddot{\underset{.\: .}Cl}}\mathbf{\: :}^{-}\rightarrow Na^{+}Cl^{-}\; \; or\; \; NaCl \nonumber$
where we have written the final formula (the formula for sodium chloride) as per the convention for ionic compounds, without listing the charges explicitly. The attraction between oppositely charged ions is called an ionic bond, and it is one of the main types of chemical bonds in chemistry. Ionic bonds are caused by electrons transferring from one atom to another.
In electron transfer, the number of electrons lost must equal the number of electrons gained. We saw this in the formation of NaCl. A similar process occurs between Mg atoms and O atoms, except in this case two electrons are transferred:
The two ions each have octets as their valence shell, and the two oppositely charged particles attract, making an ionic bond:
$\mathbf{Mg\,}^{2+}\; + \; \left[\mathbf{:}\mathbf{\ddot{\underset{.\: .}O}}\mathbf{\: :}\right]^{2-}\; \; \; \; \; Mg^{2+}O^{2-}\; or\; MgO \nonumber$
Remember, in the final formula for the ionic compound, we do not write the charges on the ions.
What about when an Na atom interacts with an O atom? The O atom needs two electrons to complete its valence octet, but the Na atom supplies only one electron:
$\mathbf{Na\, \cdot }\curvearrowright \mathbf{\cdot }\mathbf{\ddot{\underset{.}O}}\mathbf{\: :} \nonumber$
The O atom still does not have an octet of electrons. What we need is a second Na atom to donate a second electron to the O atom:
These three ions attract each other to give an overall neutral-charged ionic compound, which we write as Na2O. The need for the number of electrons lost being equal to the number of electrons gained explains why ionic compounds have the ratio of cations to anions that they do. This is required by the law of conservation of matter as well.
Example $1$: Synthesis of calcium chloride from Elements
With arrows, illustrate the transfer of electrons to form calcium chloride from $Ca$ atoms and $Cl$ atoms.
Solution
A $Ca$ atom has two valence electrons, while a $Cl$ atom has seven electrons. A $Cl$ atom needs only one more to complete its octet, while $Ca$ atoms have two electrons to lose. Thus we need two Cl atoms to accept the two electrons from one $Ca$ atom. The transfer process looks as follows:
The oppositely charged ions attract each other to make CaCl2.
Example $2$: Ionic Formula
Find the formula of the ionic compound formed from O and Al.
Solution: We first write down Lewis diagrams for each atom involved:
We now see that each O atom needs 2 electrons to make up an octet, while each Al atom has 3 electrons to donate. In order that the same number of electrons would be donated as accepted, we need 2 Al atoms (2 × 3e donated) and 3 O atoms (3 × 2e accepted). The whole process is then
The resultant oxide consists of aluminum ions, Al3+, and oxide ions, O2–, in the ratio of 2:3. The formula is Al2O3.
Exercise $1$
With arrows, illustrate the transfer of electrons to form potassium sulfide from $K$ atoms and $S$ atoms.
Answer:
Summary
• The tendency to form species that have eight electrons in the valence shell is called the octet rule.
• The attraction of oppositely charged ions caused by electron transfer is called an ionic bond.
• The strength of ionic bonding depends on the magnitude of the charges and the sizes of the ions.
Contributors and Attributions
• Henry Agnew (UC Davis)
• OpenSTAX | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/04%3A_Chemical_Bonds/4.04%3A_Using_Lewis_Symbols_for_Ionic_Compounds.txt |
Learning Objectives
• Use the rules for naming ionic compounds.
• Write the correct formula for an ionic compound.
Names of Binary ionic Compounds
After learning a few more details about the names of individual ions, you will be a step away from knowing how to name ionic compounds. This section begins the formal study of nomenclature, the systematic naming of chemical compounds.
Naming Ions
The name of a monatomic cation is simply the name of the element followed by the word ion. Thus, Na+ is the sodium ion, Al3+ is the aluminum ion, Ca2+ is the calcium ion, and so forth.
We have seen that some elements lose different numbers of electrons, producing ions of different charges (Figure 4.4.1) . Iron, for example, can form two cations, each of which, when combined with the same anion, makes a different compound with unique physical and chemical properties. Thus, we need a different name for each iron ion to distinguish Fe2+ from Fe3+. The same issue arises for other ions with more than one possible charge.
There are two ways to make this distinction. In the simpler, more modern approach, called the Stock system (Table \(1\)), an ion’s positive charge is indicated by a roman numeral in parentheses after the element name, followed by the word ion. Thus, Fe2+ is called the iron(II) ion, while Fe3+ is called the iron(III) ion. This system is used only for elements that form more than one common positive ion. We do not call the Na+ ion the sodium(I) ion because (I) is unnecessary. Sodium forms only a 1+ ion, so there is no ambiguity about the name sodium ion.
Table \(1\): The Modern and Common System of Cation Names
Element Stem Charge
Modern Name
(Stock System)
Common Name
iron ferr- 2+ iron(II) ion ferrous ion
3+ iron(III) ion ferric ion
copper cupr- 1+ copper(I) ion cuprous ion
2+ copper(II) ion cupric ion
tin stann- 2+ tin(II) ion stannous ion
4+ tin(IV) ion stannic ion
lead plumb- 2+ lead(II) ion plumbous ion
4+ lead(IV) ion plumbic ion
chromium chrom- 2+ chromium(II) ion chromous ion
3+ chromium(III) ion chromic ion
gold aur- 1+ gold(I) ion aurous ion
3+ gold(III) ion auric ion
The second system, called the common system, is not conventional but is still prevalent and used in the health sciences. This system recognizes that many metals have two common cations. The common system uses two suffixes (-ic and -ous) that are appended to the stem of the element name. The -ic suffix represents the greater of the two cation charges, and the -ous suffix represents the lower one. In many cases, the stem of the element name comes from the Latin name of the element. Table \(1\) lists the elements that use the common system, along with their respective cation names.
Table \(2\): Some Monatomic Anions
Ion Name
F fluoride ion
Cl chloride ion
Br bromide ion
I iodide ion
O2− oxide ion
S2− sulfide ion
P3− phosphide ion
N3− nitride ion
The name of a monatomic anion consists of the stem of the element name, the suffix -ide, and then the word ion. Thus, as we have already seen, Cl is “chlor-” + “-ide ion,” or the chloride ion. Similarly, O2− is the oxide ion, Se2 is the selenide ion, and so forth. Table \(2\) lists the names of some common monatomic ions.
Example \(1\)
Name each ion.
1. Ca2+
2. S2−
3. Cu+
Solution
1. the calcium ion
2. the sulfide ion
3. the copper(I) ion or the cuprous ion
Name each ion.
1. Fe2+
2. Fe3+
3. Ba2+
Answer a:
iron(II) ion
Answer b:
iron(III) ion
Answer c:
barium ion
Example \(2\)
Write the formula for each ion.
1. the bromide ion
2. the cupric ion
3. the magnesium ion
1. Br
2. Cu2+
3. Mg2+
Exercise \(2\)
Write the formula for each ion.
1. the fluoride ion
2. the stannous ion
3. the potassium ion
Answer a:
F-
Answer b:
Sn 2+
Answer c:
K+
Naming Binary Ionic Compounds with a Metal that Forms Only One Type of Cation
Now that we know how to name ions, we are ready to name ionic compounds. A binary ionic compound is a compound composed of a monatomic metal cation and a monatomic nonmetal anion. The metal cation is named first, followed by the nonmetal anion as illustrated in Figure \(1\) for the compound BaCl2. The word ion is dropped from both parts.
Subscripts in the formula do not affect the name.
Example \(3\): Naming Ionic Compounds
Name each ionic compound.
1. CaCl2
2. AlF3
3. KCl
Solution
1. Using the names of the ions, this ionic compound is named calcium chloride.
2. The name of this ionic compound is aluminum fluoride.
3. The name of this ionic compound is potassium chloride
Exercise \(3\)
Name each ionic compound.
1. AgI
2. MgO
3. Ca3P2
Answer a:
silver iodide
Answer b:
magnesium oxide
Answer c:
calcium phosphide
Naming Binary Ionic Compounds with a Metal That Forms More Than One Type of Cation
If you are given a formula for an ionic compound whose cation can have more than one possible charge, you must first determine the charge on the cation before identifying its correct name. For example, consider FeCl2 and FeCl3 . In the first compound, the iron ion has a 2+ charge because there are two Cl ions in the formula (1− charge on each chloride ion). In the second compound, the iron ion has a 3+ charge, as indicated by the three Cl ions in the formula. These are two different compounds that need two different names. By the Stock system, the names are iron(II) chloride and iron(III) chloride (Figure \(2\)).
Figure \(2\) Naming the \(FeCl_2\) and \(FeCl_3\) compounds in the modern/stock system.
Name of cation (metal) + (Roman Numeral in parenthesis) + Base name of anion (nonmetal) and -ide
If we were to use the stems and suffixes of the common system, the names would be ferrous chloride and ferric chloride, respectively (Figure \(3\)) .
Figure \(3\) Naming the \(FeCl_2\) and \(FeCl_3\) compounds in the old/common System.
"Old" base name of cation (metal) and -ic or -ous + Base name of anion (nonmetal) and -ide
-ous (for ions with lower charge)
-ic (for ions with higher charge)
Example \(4\):
Name each ionic compound.
1. Co2O3
2. FeCl2
Solution
Example \(4\): Explanation of the names of ionic compounds
Ionic Compound Explanation Answer
Co2O3
We know that cobalt can have more than one possible charge; we just need to determine what it is.
• Oxide always has a 2− charge, so with three oxide ions, we have a total negative charge of 6−.
• This means that the two cobalt ions have to contribute 6+, which for two cobalt ions means that each one is 3+.
• Therefore, the proper name for this ionic compound is cobalt(III) oxide.
cobalt(III) oxide
FeCl2
Iron can also have more than one possible charge.
• Chloride always has a 1− charge, so with two chloride ions, we have a total negative charge of 2−.
• This means that the one iron ion must have a 2+charge.
• Therefore, the proper name for this ionic compound is iron(II) chloride.
iron(II) chloride
Exercise \(4\)
Name each ionic compound.
1. AuCl3
2. PbO2
3. CuO
Answer a:
gold(III) chloride
Answer b:
lead(IV) oxide
Answer c:
copper(II) oxide
Figure \(4\) is a synopsis of how to name simple ionic compounds.
Exercise \(5\)
Name each ionic compound.
1. ZnBr2
2. Al2O3
3. AuF3
4. AgF
Answer a:
zinc bromide
Answer b:
aluminum oxide
Answer c:
gold(III) fluoride or auric fluoride
Answer d:
silver fluoride
Writing Formulas of Ionic Compounds
Ionic compounds do not exist as molecules. In the solid state, ionic compounds are in crystal lattice containing many ions each of the cation and anion. An ionic formula, like \(\ce{NaCl}\), is an empirical formula. This formula merely indicates that sodium chloride is made of an equal number of sodium and chloride ions. Sodium sulfide, another ionic compound, has the formula \(\ce{Na_2S}\). This formula indicates that this compound is made up of twice as many sodium ions as sulfide ions. This section will teach you how to find the correct ratio of ions, so that you can write a correct formula.
If you know the name of a binary ionic compound, you can write its chemical formula. Start by writing the metal ion with its charge, followed by the nonmetal ion with its charge. Because the overall compound must be electrically neutral, decide how many of each ion is needed in order for the positive and negative charge to cancel each other out.
Example \(5\): Aluminum Nitride and Lithium Oxide
Write the formula for aluminum nitride and lithium oxide.
Solution
Example \(5\): Steps for Problem Solving Writing Formulas
Steps for Problem Solving Write the formula for aluminum nitride Write the formula for lithium oxide
1. Write the symbol and charge of the cation (metal) first and the anion (nonmetal) second. \(\ce{Al^{3+}} \: \: \: \: \: \ce{N^{3-}}\) \(\ce{Li^+} \: \: \: \: \: \ce{O^{2-}}\)
2. Use a multiplier to make the total charge of the cations and anions equal to each other.
total charge of cations = total charge of anions
1(3+) = 1(3-)
+3 = -3
total charge of cations = total charge of anions
2(1+) = 1(2-)
+2 = -2
3. Use the multipliers as subscript for each ion. \(\ce{Al_1N_1}\) \(\ce{Li_2O_1}\)
4. Write the final formula. Leave out all charges and all subscripts that are 1. \(\ce{AlN}\) \(\ce{Li_2O}\)
An alternative way to writing a correct formula for an ionic compound is to use the crisscross method. In this method, the numerical value of each of the ion charges is crossed over to become the subscript of the other ion. Signs of the charges are dropped.
Example \(6\): The Crisscross Method for lead (IV) oxide
Write the formula for lead (IV) oxide.
Solution
Example \(6\): Steps For Problem Solving the Crisscross Method for writing formulas.
Crisscross Method Example: Write the formula for lead (IV) oxide
1. Write the symbol and charge of the cation (metal) first and the anion (nonmetal) second. \(\ce{Pb^{4+}} \: \: \: \: \: \ce{O^{2-}}\)
2. Transpose only the number of the positive charge to become the subscript of the anion and the number only of the negative charge to become the subscript of the cation.
3. Reduce to the lowest ratio. \(\ce{Pb_2O_4}\)
4. Write the final formula. Leave out all subscripts that are 1. \(\ce{PbO_2}\)
Exercise \(6\)
Write the chemical formula for an ionic compound composed of each pair of ions.
1. the calcium ion and the oxygen ion
2. the 2+ copper ion and the sulfur ion
3. the 1+ copper ion and the sulfur ion
Answer a:
CaO
Answer b:
CuS
Answer c:
Cu2S
Be aware that ionic compounds are empirical formulas and so must be written as the lowest ratio of the ions.
Example \(7\): Sulfur Compound
Write the formula for sodium combined with sulfur.
Solution
Example \(7\): Steps For Problem Solving the Crisscross Method for writing formulas.
Crisscross Method Write the formula for sodium combined with sulfur
1. Write the symbol and charge of the cation (metal) first and the anion (nonmetal) second. \(\ce{Na^{+}} \: \: \: \: \: \ce{S^{2-}}\)
2. Transpose only the number of the positive charge to become the subscript of the anion and the number only of the negative charge to become the subscript of the cation.
3. Reduce to the lowest ratio. This step is not necessary.
4. Write the final formula. Leave out all all subscripts that are 1. \(\ce{Na_2S}\)
Exercise \(7\)
Write the formula for each ionic compound.
1. sodium bromide
2. lithium chloride
3. magnesium oxide
Answer a:
NaBr
Answer b:
LiCl
Answer c:
MgO
Recognizing Ionic Compounds
There are two ways to recognize ionic compounds. First, compounds between metal and nonmetal elements are usually ionic. For example, CaBr2 contains a metallic element (calcium, a group 2 (or 2A) metal) and a nonmetallic element (bromine, a group 17 (or 7A) nonmetal). Therefore, it is most likely an ionic compound. (In fact, it is ionic.) In contrast, the compound NO2 contains two elements that are both nonmetals (nitrogen, from group 15 (or 5A), and oxygen, from group 16 (or 6A). It is not an ionic compound; it belongs to the category of covalent compounds discussed elsewhere. Also note that this combination of nitrogen and oxygen has no electric charge specified, so it is not the nitrite ion.
Second, if you recognize the formula of a polyatomic ion in a compound, the compound is ionic. For example, if you see the formula Ba(NO3)2, you may recognize the “NO3” part as the nitrate ion, NO3. (Remember that the convention for writing formulas for ionic compounds is not to include the ionic charge.) This is a clue that the other part of the formula, Ba, is actually the Ba2+ ion, with the 2+ charge balancing the overall 2− charge from the two nitrate ions. Thus, this compound is also ionic.
Example \(8\)
Identify each compound as ionic or not ionic.
1. Na2O
2. PCl3
3. OF2
Solution
Example \(8\): Steps For Problem Solving Identifying Ionic Compounds
Explanation Answer
a. Sodium is a metal, and oxygen is a nonmetal. Therefore, Na2O is expected to be ionic. \(Na_2O\), ionic
b. Both phosphorus and chlorine are nonmetals. Therefore, PCl3 is not ionic. \(PCl_3\), not ionic
c. Both oxygen and fluorine are nonmetals. Therefore, OF2 is not ionic. \(OF_2\), ionic
Exercise \(8\)
Identify each compound as ionic or not ionic.
1. N2O
2. FeCl3
Answer a:
not ionic
Answer b:
ionic
Summary
• Ionic compounds are named by stating the cation first, followed by the anion.
• Positive and negative charges must balance.
• Some anions have multiple forms and are named accordingly with the use of roman numerals in parenthesis.
• Formulas for ionic compounds contain the symbols and number of each atom present in a compound in the lowest whole number ratio.
Contributors and Attributions
• Henry Agnew (UC Davis) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/04%3A_Chemical_Bonds/4.05%3A_Formulas_and_Names_of_Binary_Ionic_Compounds.txt |
Learning Objectives
• Define covalent bond.
• Illustrate covalent bond formation with Lewis electron dot diagrams.
Ionic bonding typically occurs when it is easy for one atom to lose one or more electrons and another atom to gain one or more electrons. However, some atoms won’t give up or gain electrons easily. Yet they still participate in compound formation. How? There is another mechanism for obtaining a complete valence shell: sharing electrons. When electrons are shared between two atoms, they make a bond called a covalent bond.
Chemists frequently use Lewis diagrams to represent covalent bonding in molecular substances. For example, the Lewis diagrams of two separate hydrogen atoms are as follows:
The Lewis diagram of two hydrogen atoms sharing electrons looks like this:
We can use circles to show that each H atom has two electrons around the nucleus, completely filling each atom’s valence shell:
Because each H atom has a filled valence shell, this bond is stable, and we have made a diatomic hydrogen molecule. (This explains why hydrogen is one of the diatomic elements.) For simplicity’s sake, it is not unusual to represent the covalent bond with a dash, instead of with two dots:
Because two atoms are sharing one pair of electrons, this covalent bond is called a single bond. As another example, consider fluorine. F atoms have seven electrons in their valence shell:
These two atoms can do the same thing that the H atoms did; they share their unpaired electrons to make a covalent bond.
Note that each F atom has a complete octet around it now:
We can also write this using a dash to represent the shared electron pair:
There are two different types of electrons in the fluorine diatomic molecule. The bonding electron pair makes the covalent bond. Each F atom has three other pairs of electrons that do not participate in the bonding; they are called lone pair electrons. Each F atom has one bonding pair and three lone pairs of electrons.
Covalent bonds can be made between different elements as well. One example is HF. Each atom starts out with an odd number of electrons in its valence shell:
The two atoms can share their unpaired electrons to make a covalent bond:
In this molecule, the hydrogen atom does not have nonbonding electrons, while the fluorine atom has six nonbonding electrons (three lone electron pairs). The circles show how the valence electron shells are filled for both atoms.
Example \(1\):
Use Lewis electron dot diagrams to illustrate the covalent bond formation in HBr.
Solution
HBr is very similar to HF, except that it has Br instead of F. The atoms are as follows:
The two atoms can share their unpaired electron:
Exercise \(1\)
Use Lewis electron dot diagrams to illustrate the covalent bond formation in Cl2.
Answer:
When working with covalent structures, it sometimes looks like you have leftover electrons. You apply the rules you learned so far and there are still some electrons hanging out there unattached. You can't just leave them there. So where do you put them?
Multiple Covalent Bonds
The sharing of a pair of electrons represents a single covalent bond, usually just referred to as a single bond. However, in many molecules atoms attain complete octets by sharing more than one pair of electrons between them:
• Two electron pairs shared a double bond
• Three electron pairs shared a triple bond
Because each nitrogen contains 5 valence electrons, they need to share 3 pairs to each achieve a valence octet. N2 is fairly inert, due to the strong triple bond between the two nitrogen atoms.
Summary
• Covalent bonds are formed when atoms share electrons.
• Lewis electron dot diagrams can be drawn to illustrate covalent bond formation.
• Double bonds or triple bonds between atoms may be necessary to properly illustrate the bonding in some molecules.
Contributors and Attributions
• Anonymous
• Henry Agnew (UC Davis) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/04%3A_Chemical_Bonds/4.06%3A_Covalent_Bonds-_Shared_Electron_Pairs.txt |
Learning Objectives
• Define electronegativity
• Determine the polarity of a covalent bond.
Electronegativity
The ability of an atom in a molecule to attract shared electrons is called electronegativity. When two atoms combine, the difference between their electronegativities is an indication of the type of bond that will form. If the difference between the electronegativities of the two atoms is small, neither atom can take the shared electrons completely away from the other atom and the bond will be covalent. If the difference between the electronegativities is large, the more electronegative atom will take the bonding electrons completely away from the other atom (electron transfer will occur) and the bond will be ionic. This is why metals (low electronegativities) bonded with nonmetals (high electronegativities) typically produce ionic compounds.
A bond may be so polar that an electron actually transfers from one atom to another, forming a true ionic bond. How do we judge the degree of polarity? Scientists have devised a scale called electronegativity, a scale for judging how much atoms of any element attract electrons. Electronegativity is a unitless number; the higher the number, the more an atom attracts electrons. A common scale for electronegativity is shown in Figure $1$.
Bond Polarity
The polarity of a covalent bond can be judged by determining the difference of the electronegativities of the two atoms involved in the covalent bond, as summarized in the following table:
Table $1$:Electronegativity differences and bond type
Electronegativity Difference Bond Type
0 nonpolar covalent
0–0.4 slightly polar covalent
0.5–2.0 definitely polar covalent
>2.0 likely ionic
Nonpolar Covalent Bonds
A bond in which the electronegativity difference is less than 1.9 is considered to be mostly covalent in character. However, at this point we need to distinguish between two general types of covalent bonds. A nonpolar covalent bond is a covalent bond in which the bonding electrons are shared equally between the two atoms. In a nonpolar covalent bond, the distribution of electrical charge is balanced between the two atoms.
The two chlorine atoms share the pair of electrons in the single covalent bond equally, and the electron density surrounding the $\ce{Cl_2}$ molecule is symmetrical. Also note that molecules in which the electronegativity difference is very small (<0.5) are also considered nonpolar covalent. An example would be a bond between chlorine and bromine ($\Delta$EN $=3.0 - 2.8 = 0.2$).
Polar Covalent Bonds
A bond in which the electronegativity difference between the atoms is between 0.5 and 2.0 is called a polar covalent bond. A polar covalent bond is a covalent bond in which the atoms have an unequal attraction for electrons and so the sharing is unequal. In a polar covalent bond, sometimes simply called a polar bond, the distribution of electrons around the molecule is no longer symmetrical.
An easy way to illustrate the uneven electron distribution in a polar covalent bond is to use the Greek letter delta $\left( \delta \right)$.
The atom with the greater electronegativity acquires a partial negative charge, while the atom with the lesser electronegativity acquires a partial positive charge. The delta symbol is used to indicate that the quantity of charge is less than one. A crossed arrow can also be used to indicate the direction of greater electron density.
Electronegativity differences in bonding using Pauling scale. Using differences in electronegativity to classify bonds as covalent, polar covalent, or ionic.
Example $1$: Bond Polarity
What is the polarity of each bond?
1. C–H
2. O–H
Solution
Using Figure $1$, we can calculate the difference of the electronegativities of the atoms involved in the bond.
1. For the C–H bond, the difference in the electronegativities is 2.5 − 2.1 = 0.4. Thus we predict that this bond will be non polar covalent.
2. For the O–H bond, the difference in electronegativities is 3.5 − 2.1 = 1.4, so we predict that this bond will be definitely polar covalent.
Exercise $1$
What is the polarity of each bond?
1. Rb–F
2. P–Cl
Answer a
likely ionic
Answer b
polar covalent
Summary
• The type of bond (polar covalent,non polar covalent or ionic) between two atoms is determined by the differences in electronegativity.
• For atoms sharing a polar covalent bond, the atom with the greater electronegativity acquires a partial negative charge, while the atom with the lesser electronegativity acquires a partial positive charge.
Contributors and Attributions
• StackExchange (thomij).
• Henry Agnew (UC Davis)
4.08: Polyatomic Molecules- Water Ammonia and Methane
Learning Objectives
• Determine the number of bonds formed by common nonmetal elements.
• Illustrate covalent bond formation with Lewis electron dot diagrams.
More than two atoms can participate in covalent bonding, although any given covalent bond will be between two atoms only. Water, ammonia, and methane are common examples that will be discussed in detail below. Carbon is unique in the extent to which it forms single, double, and triple bonds to itself and other elements. The number of bonds formed by an atom in its covalent compounds is not arbitrary. Hydrogen, oxygen, nitrogen, and carbon have very strong tendencies to form substances in which they have one, two, three, and four bonds to other atoms, respectively (Table \(1\)).
Table \(1\) The Number of Bonds That Selected Atoms Commonly Form to Other Atoms
Atom Number of Bonds
H (group 1) 1
O (group 16) 2
N (group 15) 3
C (group 14) 4
Water
Consider H and O atoms:
The H and O atoms can share an electron to form a covalent bond:
The H atom has a complete valence shell. However, the O atom has only seven electrons around it, which is not a complete octet. We fix this by including a second H atom, whose single electron will make a second covalent bond with the O atom:
(It does not matter on what side the second H atom is positioned.) Now the O atom has a complete octet around it, and each H atom has two electrons, filling its valence shell. This is how a water molecule, H2O, is made.
Ammonia
The N atom has the following Lewis electron dot diagram:
It has three unpaired electrons, each of which can make a covalent bond by sharing electrons with an H atom. The electron dot diagram of NH3 is as follows:
Methane
The C atom has the following Lewis electron dot diagram:
It has four unpaired electrons, each of which can make a covalent bond by sharing electrons with an H atom. The electron dot diagram of CH4 is as follows:
Summary
In polyatomic molecules, there is a pattern of covalent bonds that different atoms can form.
Contributors and Attributions
• TextMap: Beginning Chemistry (Ball et al.) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/04%3A_Chemical_Bonds/4.07%3A_Unequal_Sharing-_Polar_Covalent_Bonds.txt |
Learning Objectives
• Recognize polyatomic ions in chemical formulas.
• Write the correct formula for ionic compounds with polyatomic ions.
• Use the rules for naming ionic compounds with polyatomic ions.
Polyatomic Ions
Some ions consist of groups of atoms bonded together and have an overall electric charge. Because these ions contain more than one atom, they are called polyatomic ions. Polyatomic ions have characteristic formulas, names, and charges that should be memorized. For example, NO3 is the nitrate ion; it has one nitrogen atom and three oxygen atoms and an overall 1− charge. Table \(1\) lists the most common polyatomic ions.
Table \(1\) Some Polyatomic Ions
Name Formula
ammonium ion NH4+
acetate ion C2H3O2 (also written CH3CO2)
carbonate ion CO32
chromate ion CrO42
dichromate ion Cr2O72
hydrogen carbonate ion (bicarbonate ion) HCO3
cyanide ion CN
hydroxide ion OH
nitrate ion NO3
nitrite ion NO2
permanganate ion MnO4
phosphate ion PO43
hydrogen phosphate ion HPO42
dihydrogen phosphate ion H2PO4
sulfate ion SO42
hydrogen sulfate ion (bisulfate ion) HSO4
sulfite ion SO32
The rule for constructing formulas for ionic compounds containing polyatomic ions is the same as for formulas containing monatomic (single-atom) ions: the positive and negative charges must balance. If more than one of a particular polyatomic ion is needed to balance the charge, the entire formula for the polyatomic ion must be enclosed in parentheses, and the numerical subscript is placed outside the parentheses. This is to show that the subscript applies to the entire polyatomic ion. An example is Ba(NO3)2.
Writing Formulas for Ionic Compounds Containing Polyatomic Ions
Writing a formula for ionic compounds containing polyatomic ions also involves the same steps as for a binary ionic compound. Write the symbol and charge of the cation followed by the symbol and charge of the anion.
Example \(1\): Calcium Nitrate
Write the formula for calcium nitrate.
Solution
Example \(1\): Steps For Problem Solving, Write the formula for calcium nitrate
Criss Cross Method Write the formula for calcium nitrate
1. Write the symbol and charge of the cation (metal) first and the anion (nonmetal) second. \(\ce{Ca^{2+}} \: \: \: \: \: \ce{NO_3^-}\)
2. Transpose only the number of the positive charge to become the subscript of the anion and the number only of the negative charge to become the subscript of the cation.
3. Reduce to the lowest ratio. \(\ce{Ca_1(NO_3)_2}\)
4. Write the final formula. Leave out all subscripts that are 1. If there is only 1 of the polyatomic ion, leave off parentheses. \(\ce{Ca(NO_3)_2}\)
Example \(2\)
Write the chemical formula for an ionic compound composed of the potassium ion and the sulfate ion
Solution
Example \(2\): Explanation for Writing the Chemical Formula for an Ionic Compound Composed of the Potassium Ion and the Sulfate Ion
Explanation Answer
Potassium ions have a charge of 1+, while sulfate ions have a charge of 2−. We will need two potassium ions to balance the charge on the sulfate ion, so the proper chemical formula is K2SO4. \(K_2SO_4\)
Exercise \(1\)
Write the chemical formula for an ionic compound composed of each pair of ions.
1. the magnesium ion and the carbonate ion
2. the aluminum ion and the acetate ion
Answer a:
MgCO3
Answer b:
Al(CH3COO)3
Naming Ionic Compounds with Polyatomic Ions
The process of naming ionic compounds with polyatomic ions is the same as naming binary ionic compounds. The cation is named first, followed by the anion. One example is the ammonium Sulfate compound in Figure \(6\).
Example \(3\): Naming Ionic Compounds
Write the proper name for each ionic compound.
1. (NH4)2S
2. AlPO4,
3. Fe3(PO4)2
Solution
Example \(3\): Explanation for Naming Ionic Compounds
Ionic Compound Explanation Answer
1. (NH4)2S
a. The ammonium ion has a 1+ charge and the sulfide ion has a 2− charge.
Two ammonium ions need to balance the charge on a single sulfide ion.
The compound’s name is ammonium sulfide.
ammonium sulfide
b. AlPO4,
b. The ions have the same magnitude of charge, one of each (ion) is needed to balance the charges.
The name of the compound is aluminum phosphate.
aluminum phosphate
c. Fe3(PO4)2
c. Neither charge is an exact multiple of the other, so we have to go to the least common multiple of 6.
To get 6+, three iron(II) ions are needed, and to get 6−, two phosphate ions are needed .
The compound’s name is iron(II) phosphate.
iron(II) phosphate
Exercise \(2\)
Write the proper name for each ionic compound.
1. (NH4)3PO4
2. Co(NO2)3
Answer a:
ammonium phosphate
Answer b:
cobalt(III) nitrite
Summary
• Formulas for ionic compounds contain the symbols and number of each atom and/or polyatomic ion present in a compound in the lowest whole number ratio.
• The process of naming ionic compounds with polyatomic ions are the same as binary ionic compounds.
Contributors and Attributions
• Henry Agnew (UC Davis) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/04%3A_Chemical_Bonds/4.09%3A_Polyatomic_Ions.txt |
Learning Objective
• Write Lewis structures for molecules and ions.
• Know the exceptions to the octet rule.
The following procedure can be used to construct Lewis electron structures for more complex molecules and ions:
How-to: Constructing Lewis electron structures
1. Determine the total number of valence electrons in the molecule or ion.
• Add together the valence electrons from each atom. (Recall that the number of valence electrons is indicated by the position of the element in the periodic table.)
• If the species is a polyatomic ion, remember to add or subtract the number of electrons necessary to give the total charge on the ion.
For CO32, for example, we add two electrons to the total because of the −2 charge.
2. Arrange the atoms to show specific connections.
• When there is a central atom, it is usually the least electronegative element in the compound. Chemists usually list this central atom first in the chemical formula (as in CCl4 and CO32, which both have C as the central atom), which is another clue to the compound’s structure.
• Hydrogen and the halogens are almost always connected to only one other atom, so they are usually terminal rather than central.
3. Place a bonding pair of electrons between each pair of adjacent atoms to give a single bond.
• In H2O, for example, there is a bonding pair of electrons between oxygen and each hydrogen.
4. Beginning with the terminal atoms, add enough electrons to each atom to give each atom an octet (two for hydrogen).
• These electrons will usually be lone pairs.
5. If any electrons are left over, place them on the central atom.
• We will explain later that some atoms are able to accommodate more than eight electrons.
6. If the central atom has fewer electrons than an octet, use lone pairs from terminal atoms to form multiple (double or triple) bonds to the central atom to achieve an octet.
• This will not change the number of electrons on the terminal atoms.
7. Final check
• Always make sure all valence electrons are accounted for and each atom has an octet of electrons except for hydrogen (with two electrons).
The central atom is usually the least electronegative element in the molecule or ion; hydrogen and the halogens are usually terminal.
Now let’s apply this procedure to some particular compounds, beginning with one we have already discussed.
Example $1$: Water
Write the Lewis Structure for H2O.
Solution
Example $1$: steps for writing the Lewis structure of water
Steps for Writing Lewis Structures Example $1$:
1. Determine the total number of valence electrons in the molecule or ion. Each H atom (group 1) has 1 valence electron, and the O atom (group 16) has 6 valence electrons, for a total of 8 valence electrons.
2. Arrange the atoms to show specific connections.
H O H
Because H atoms are almost always terminal, the arrangement within the molecule must be HOH.
3. Place a bonding pair of electrons between each pair of adjacent atoms to give a single bond.
4. Beginning with the terminal atoms, add enough electrons to each atom to give each atom an octet (two for hydrogen).
Placing one bonding pair of electrons between the O atom and each H atom gives
H -O- H
with 4 electrons left over.
Each H atom has a full valence shell of 2 electrons.
5. If any electrons are left over, place them on the central atom.
Adding the remaining 4 electrons to the oxygen (as two lone pairs) gives the following structure:
6. If the central atom has fewer electrons than an octet, use lone pairs from terminal atoms to form multiple (double or triple) bonds to the central atom to achieve an octet. Not necessary
7. Final check The Lewis structure gives oxygen an octet and each hydrogen two electrons,
Example $2$
Write the Lewis structure for the $CH_2O$ molecule
Solution
Steps are writing the Lewis structure of the $CH_2O$ molecule
Steps for Writing Lewis Structures Example $2$
1. Determine the total number of valence electrons in the molecule or ion. Each hydrogen atom (group 1) has one valence electron, carbon (group 14) has 4 valence electrons, and oxygen (group 16) has 6 valence electrons, for a total of [(2)(1) + 4 + 6] = 12 valence electrons.
2. Arrange the atoms to show specific connections.
Because carbon is less electronegative than oxygen and hydrogen is normally terminal, C must be the central atom.
3. Place a bonding pair of electrons between each pair of adjacent atoms to give a single bond.
Placing a bonding pair of electrons between each pair of bonded atoms gives the following:
Six electrons are used, and 6 are left over.
4. Beginning with the terminal atoms, add enough electrons to each atom to give each atom an octet (two for hydrogen).
Adding all 6 remaining electrons to oxygen (as three lone pairs) gives the following:
Although oxygen now has an octet and each hydrogen has 2 electrons, carbon has only 6 electrons.
5. If any electrons are left over, place them on the central atom.
Not necessary
There are no electrons left to place on the central atom.
6. If the central atom has fewer electrons than an octet, use lone pairs from terminal atoms to form multiple (double or triple) bonds to the central atom to achieve an octet.
To give carbon an octet of electrons, we use one of the lone pairs of electrons on oxygen to form a carbon–oxygen double bond:
7. Final check Both the oxygen and the carbon now have an octet of electrons, so this is an acceptable Lewis electron structure. The O has two bonding pairs and two lone pairs, and C has four bonding pairs. This is the structure of formaldehyde, which is used in embalming fluid.
Exercise $1$
Write Lewis electron structures for CO2 and SCl2, a vile-smelling, unstable red liquid that is used in the manufacture of rubber.
Answer CO2
.
Answer SCl2
.
The United States Supreme Court has the unenviable task of deciding what the law is. This responsibility can be a major challenge when there is no clear principle involved or where there is a new situation not encountered before. Chemistry faces the same challenge in extending basic concepts to fit a new situation. Drawing of Lewis structures for polyatomic ions uses the same approach, but tweaks the process a little to fit a somewhat different set of circumstances.
Lewis Structures for Polyatomic Ions
Recall that a polyatomic ion is a group of atoms that are covalently bonded together and which carry an overall electrical charge. The ammonium ion, $\ce{NH_4^+}$, is formed when a hydrogen ion $\left( \ce{H^+} \right)$ attaches to the lone pair of an ammonia $\left( \ce{NH_3} \right)$ molecule in a coordinate covalent bond.
When drawing the Lewis structure of a polyatomic ion, the charge of the ion is reflected in the number of total valence electrons in the structure. In the case of the ammonium ion:
$1 \: \ce{N}$ atom $= 5$ valence electrons
$4 \: \ce{H}$ atoms $= 4 \times 1 = 4$ valence electrons
subtract 1 electron for the $1+$charge of the ion
total of 8 valence electrons in the ion
It is customary to put the Lewis structure of a polyatomic ion into a large set of brackets, with the charge of the ion as a superscript outside the brackets.
Exercise $2$
Draw the Lewis electron dot structure for the sulfate ion.
Answer
Exceptions to the Octet Rule
As important and useful as the octet rule is in chemical bonding, there are some well-known violations. This does not mean that the octet rule is useless—quite the contrary. As with many rules, there are exceptions, or violations.
There are three violations to the octet rule. Odd-electron molecules represent the first violation to the octet rule. Although they are few, some stable compounds have an odd number of electrons in their valence shells. With an odd number of electrons, at least one atom in the molecule will have to violate the octet rule. Examples of stable odd-electron molecules are NO, NO2, and ClO2. The Lewis electron dot diagram for NO is as follows:
Although the O atom has an octet of electrons, the N atom has only seven electrons in its valence shell. Although NO is a stable compound, it is very chemically reactive, as are most other odd-electron compounds.
Electron-deficient molecules represent the second violation to the octet rule. These stable compounds have less than eight electrons around an atom in the molecule. The most common examples are the covalent compounds of beryllium and boron. For example, beryllium can form two covalent bonds, resulting in only four electrons in its valence shell:
Boron commonly makes only three covalent bonds, resulting in only six valence electrons around the B atom. A well-known example is BF3:
The third violation to the octet rule is found in those compounds with more than eight electrons assigned to their valence shell. These are called expanded valence shell molecules. Such compounds are formed only by central atoms in the third row of the periodic table or beyond that have empty d orbitals in their valence shells that can participate in covalent bonding. One such compound is PF5. The only reasonable Lewis electron dot diagram for this compound has the P atom making five covalent bonds:
Formally, the P atom has 10 electrons in its valence shell.
Example $3$: Octet Violations
Identify each violation to the octet rule by drawing a Lewis electron dot diagram.
1. ClO
2. SF6
Solution
a. With one Cl atom and one O atom, this molecule has 6 + 7 = 13 valence electrons, so it is an odd-electron molecule. A Lewis electron dot diagram for this molecule is as follows:
b. In SF6, the central S atom makes six covalent bonds to the six surrounding F atoms, so it is an expanded valence shell molecule. Its Lewis electron dot diagram is as follows:
Exercise $3$: Xenon Difluoride
Identify the violation to the octet rule in XeF2 by drawing a Lewis electron dot diagram.
Answer:
The Xe atom has an expanded valence shell with more than eight electrons around it.
Summary
• Lewis dot symbols (Lewis structures) provide a simple rationalization of why elements form compounds with the observed stoichiometries.
• In Lewis electron structures, we encounter bonding pairs, which are shared by two atoms, and lone pairs, which are not shared between atoms.
• Lewis structures for polyatomic ions follow the same rules as those for other covalent compounds.
• There are three violations to the octet rule: odd-electron molecules, electron-deficient molecules, and expanded valence shell molecules | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/04%3A_Chemical_Bonds/4.10%3A_Rules_for_Writing_Lewis_Formulas.txt |
Learning Objective
• Determine the shape of simple molecules and polyatomic ions.
Molecules have shapes. There is an abundance of experimental evidence to that effect—from their physical properties to their chemical reactivity. Small molecules—molecules with a single central atom—have shapes that can be easily predicted. The basic idea in molecular shapes is called valence shell electron pair repulsion (VSEPR). It basically says that electron pairs, being composed of negatively charged particles, repel each other to get as far away from each other as possible. VSEPR makes a distinction between electron group geometry, which expresses how electron groups (bonds and nonbonding electron pairs) are arranged, and molecular geometry, which expresses how the atoms in a molecule are arranged. However, the two geometries are related.
There are two types of electron groups: any type of bond—single, double, or triple—and lone electron pairs. When applying VSEPR to simple molecules, the first thing to do is to count the number of electron groups around the central atom. Remember that a multiple bond counts as only one electron group.
Any molecule with only two atoms is linear. A molecule whose central atom contains only two electron groups orients those two groups as far apart from each other as possible—180° apart. When the two electron groups are 180° apart, the atoms attached to those electron groups are also 180° apart, so the overall molecular shape is linear. Examples include BeH2 and CO2:
The two molecules, shown in the figure below in a "ball and stick" model.
A molecule with three electron groups orients the three groups as far apart as possible. They adopt the positions of an equilateral triangle—120° apart and in a plane. The shape of such molecules is trigonal planar. An example is BF3:
Some substances have a trigonal planar electron group distribution but have atoms bonded to only two of the three electron groups. An example is GeF2:
From an electron group geometry perspective, GeF2 has a trigonal planar shape, but its real shape is dictated by the positions of the atoms. This shape is called bent or angular.
A molecule with four electron groups about the central atom orients the four groups in the direction of a tetrahedron, as shown in Figure \(1\) Tetrahedral Geometry. If there are four atoms attached to these electron groups, then the molecular shape is also tetrahedral. Methane (CH4) is an example.
This diagram of CH4 illustrates the standard convention of displaying a three-dimensional molecule on a two-dimensional surface. The straight lines are in the plane of the page, the solid wedged line is coming out of the plane toward the reader, and the dashed wedged line is going out of the plane away from the reader.
NH3 is an example of a molecule whose central atom has four electron groups but only three of them are bonded to surrounding atoms.
Although the electron groups are oriented in the shape of a tetrahedron, from a molecular geometry perspective, the shape of NH3 is trigonal pyramidal.
H2O is an example of a molecule whose central atom has four electron groups but only two of them are bonded to surrounding atoms.
Although the electron groups are oriented in the shape of a tetrahedron, the shape of the molecule is bent or angular. A molecule with four electron groups about the central atom but only one electron group bonded to another atom is linear because there are only two atoms in the molecule.
Double or triple bonds count as a single electron group. The Lewis electron dot diagram of formaldehyde (CH2O) is shown in Figure \(9\).
The central C atom has three electron groups around it because the double bond counts as one electron group. The three electron groups repel each other to adopt a trigonal planar shape.
(The lone electron pairs on the O atom are omitted for clarity.) The molecule will not be a perfect equilateral triangle because the C–O double bond is different from the two C–H bonds, but both planar and triangular describe the appropriate approximate shape of this molecule. Figure \(11\) illustrates several representations of the water, ammonia, and methane molecules.
Table \(1\) summarizes the shapes of molecules based on their number of electron groups and surrounding atoms and the common shapes are shown in figure \(1\). Table \(1\): Summary of Molecular Shapes
Number of Electron Groups on Central Atom Number of Bonding Groups Number of Lone Pairs Electron Geometry Molecular Shape
2 2 0 linear linear
3 3 0 trigonal planar trigonal planar
3 2 1 trigonal planar bent
4 4 0 tetrahedral tetrahedral
4 3 1 tetrahedral trigonal pyramidal
4 2 2 tetrahedral bent
Figure \(12\): Common structures for molecules and Ppolyatomic ions that consist of a central atom bonded to two or three other atoms.
Example \(1\):
What is the approximate shape of each molecule?
1. PCl3
2. NOF
Solution
The first step is to draw the Lewis structure of the molecule.
1. For PCl3, the electron dot diagram is as follows:
The lone electron pairs on the Cl atoms are omitted for clarity. The P atom has four electron groups with three of them bonded to surrounding atoms, so the molecular shape is trigonal pyramidal.
• The electron dot diagram for NOF is as follows:
The N atom has three electron groups on it, two of which are bonded to other atoms. The molecular shape is bent.
• Exercise \(1\)
What is the approximate molecular shape of CH2Cl2?
Answer
Tetrahedral
Exercise \(2\)
Ethylene (C2H4) has two central atoms. Determine the geometry around each central atom and the shape of the overall molecule. Hint, hydrogen is a terminal atom.
Answer
Trigonal planar about both central C atoms
Summary
The approximate shape of a molecule can be predicted from the number of electron groups and the number of surrounding atoms.
Contributors and Attributions
• TextMap: Beginning Chemistry (Ball et al.)
• Henry Agnew (UC Davis) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/04%3A_Chemical_Bonds/4.11%3A_Molecular_Shapes-_The_VSEPR_Theory.txt |
Learning Objective
• Determine if a molecule is polar or nonpolar.
Molecular Polarity
To determine if a molecule is polar or nonpolar, it is frequently useful to look at Lewis structures. Nonpolar compounds will be symmetric, meaning all of the sides around the central atom are identical - bonded to the same element with no unshared pairs of electrons. Notice that a tetrahedral molecule such as $\ce{CCl_4}$ is nonpolar Figure ($1$. Another non polar molecule shown below is boron trifluoride, BF3. BF3 is a trigonal planar molecule and all three peripheral atoms are the same.
Polar molecules are asymmetric, either containing lone pairs of electrons on a central atom or having atoms with different electronegativities bonded. This works pretty well - as long as you can visualize the molecular geometry. That's the hard part. To know how the bonds are oriented in space, you have to have a strong grasp of Lewis structures and VSEPR theory. Assuming you do, you can look at the structure of each one and decide if it is polar or not - whether or not you know the individual atom electronegativity. This is because you know that all bonds between dissimilar elements are polar, and in these particular examples, it doesn't matter which direction the dipole moment vectors are pointing (out or in).
A polar molecule is a molecule in which one end of the molecule is slightly positive, while the other end is slightly negative. A diatomic molecule that consists of a polar covalent bond, such as $\ce{HF}$, is a polar molecule.
As mentioned in section 4.7, because the electrons in the bond are nearer to the F atom, this side of the molecule takes on a partial negative charge, which is represented by δ− (δ is the lowercase Greek letter delta). The other side of the molecule, the H atom, adopts a partial positive charge, which is represented by δ+. The two electrically charged regions on either end of the molecule are called poles, similar to a magnet having a north and a south pole. A molecule with two poles is called a dipole (see figure below). Hydrogen fluoride is a dipole.
For molecules with more than two atoms, the molecular geometry must also be taken into account when determining if the molecule is polar or nonpolar. The figure below shows a comparison between carbon dioxide and water. Carbon dioxide $\left( \ce{CO_2} \right)$ is a linear molecule. The oxygen atoms are more electronegative than the carbon atom, so there are two individual dipoles pointing outward from the $\ce{C}$ atom to each $\ce{O}$ atom. However, since the dipoles are of equal strength and are oriented this way, they cancel out and the overall molecular polarity of $\ce{CO_2}$ is zero.
Water is a bent molecule because of the two lone pairs on the central oxygen atom. The individual dipoles point from the $\ce{H}$ atoms toward the $\ce{O}$ atom. Because of the shape, the dipoles do not cancel each other out and the water molecule is polar. In the figure below, the net dipole is shown in blue and points upward.
Three other polar molecules are shown below with the arrows pointing to the more electron dense atoms. Just like the water molecule, none of the bond moments cancel out.
To summarize, to be polar, a molecule must:
1. Contain at least one polar covalent bond.
2. Have a molecular structure such that the sum of the vectors of each bond dipole moment does not cancel.
Steps to Identify Polar Molecules
1. Draw the Lewis structure
2. Figure out the geometry (using VSEPR theory)
3. Visualize or draw the geometry
4. Find the net dipole moment (you don't have to actually do calculations if you can visualize it)
5. If the net dipole moment is zero, it is non-polar. Otherwise, it is polar.
Example $1$:
Label each of the following as polar or nonpolar.
1. Water, H2O:
2. Methanol, CH3OH:
3. Hydrogen Cyanide, HCN:
4. Oxygen, O2:
5. Propane, C3H8:
Solution
1. Water is polar. Any molecule with lone pairs of electrons around the central atom is polar.
2. Methanol is polar. This is not a symmetric molecule. The $\ce{-OH}$ side is different from the other 3 $\ce{-H}$ sides.
3. Hydrogen cyanide is polar. The molecule is not symmetric. The nitrogen and hydrogen have different electronegativities, creating an uneven pull on the electrons.
4. Oxygen is nonpolar. The molecule is symmetric. The two oxygen atoms pull on the electrons by exactly the same amount.
5. Propane is nonpolar, because it is symmetric, with $\ce{H}$ atoms bonded to every side around the central atoms and no unshared pairs of electrons.
Exercise $1$
Label each of the following as polar or nonpolar.
a. SO3
b. NH3
Answer a
non polar
Answer b
polar
Summary
• Non polar molecules are symmetric with no unshared electrons.
• Polar molecules are asymmetric, either containing lone pairs of electrons on a central atom or having atoms with different electronegativities bonded.
Contributors and Attributions
• StackExchange (thomij).
• Henry Agnew (UC Davis) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/04%3A_Chemical_Bonds/4.12%3A_Shapes_and_Properties-_Polar_and_Nonpolar_Molecules.txt |
Learning Objectives
• Identify the reactants and products in any chemical reaction.
• Convert word equations into chemical equations.
• Use the common symbols, $\left( s \right)$, $\left( l \right)$, $\left( g \right)$, $\left( aq \right)$, and $\rightarrow$ appropriately when writing a chemical reaction.
• Explain the roles of subscripts and coefficients in chemical equations.
• Balance a chemical equation when given the unbalanced equation.
• Explain the role of the Law of Conservation of Mass in a chemical reaction.
In a chemical change, new substances are formed. In order for this to occur, the chemical bonds of the substances break, and the atoms that compose them separate and rearrange themselves into new substances with new chemical bonds. When this process occurs, we call it a chemical reaction. A chemical reaction is the process in which one or more substances are changed into one or more new substances.
Reactants and Products
To describe a chemical reaction, we need to indicate what substances are present at the beginning and what substances are present at the end. The substances that are present at the beginning are called reactants and the substances present at the end are called products.
Sometimes when reactants are put into a reaction vessel, a reaction will take place to produce products. Reactants are the starting materials, that is, whatever we have as our initial ingredients. The products are just that, wheat is produced or the result of what happens to the reactants when we put them together in the reaction vessel. If we think about baking chocolate chip cookies, our reactants would be flour, butter, sugar, vanilla, some baking soda, salt, egg, and chocolate chips. What would be the products? Cookies! The reaction vessel would be our mixing bowl.
$\underbrace{\text{Flour} + \text{Butter} + \text{Sugar} + \text{Vanilla} + \text{Baking Soda} + \text{Eggs} + \text{Chocolate Chips}}_{\text{Ingredients = Reactants}} \rightarrow \underbrace{\text{Cookies}}_{\text{Product}} \nonumber$
Writing Chemical Equations
When sulfur dioxide is added to oxygen, sulfur trioxide is produced. Sulfur dioxide and oxygen, $\ce{SO_2} + \ce{O_2}$, are reactants and sulfur trioxide, $\ce{SO_3}$, is the product.
$\underbrace{\ce{2 SO2(g) + O2(g) }}_{\text{Reactants}} \rightarrow \underbrace{\ce{2SO3(g)}}_{\text{Products}} \nonumber$
In chemical reactions, the reactants are found before the symbol "$\rightarrow$" and the products are found after the symbol "$\rightarrow$". The general equation for a reaction is:
$\text{Reactants } \rightarrow \text{Products} \nonumber$
There are a few special symbols that we need to know in order to "talk" in chemical shorthand. In the table below is the summary of the major symbols used in chemical equations. You will find there are others but these are the main ones that we need to know. Table $1$ shows a listing of symbols used in chemical equations.
Table $1$ Symbols Used in Chemical Equations
Symbol Description
$+$ used to separate multiple reactants or products
$\rightarrow$ yield sign; separates reactants from products
$\rightleftharpoons$ replaces the yield sign for reversible reactions that reach equilibrium
$\overset{\ce{Pt}}{\rightarrow}$ formula written above the arrow is used as a catalyst in the reaction
$\overset{\Delta}{\rightarrow}$ triangle indicates that the reaction is being heated
$\left( s \right)$ reactant or product in the solid state
$\left( l \right)$ reactant or product in the liquid state
$\left( g \right)$ reactant or product in the gas state
$\left( aq \right)$ reactant or product in an aqueous solution (dissolved in water)
Chemists have a choice of methods for describing a chemical reaction.
1. They could draw a picture of the chemical reaction.
2. They could write a word equation for the chemical reaction:
"Two molecules of hydrogen gas react with one molecule of oxygen gas to produce two molecules of water vapor."
3. They could write the equation in chemical shorthand.
$2 \ce{H_2} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{H_2O} \left( g \right) \nonumber$
In the symbolic equation, chemical formulas are used instead of chemical names for reactants and products and symbols are used to indicate the phase of each substance. It should be apparent that the chemical shorthand method is the quickest and clearest method for writing chemical equations.
We could write that an aqueous solution of calcium nitrate is added to an aqueous solution of sodium hydroxide to produce solid calcium hydroxide and an aqueous solution of sodium nitrate. Or in shorthand we could write:
$\ce{Ca(NO_3)_2} \left( aq \right) + 2 \ce{NaOH} \left( aq \right) \rightarrow \ce{Ca(OH)_2} \left( s \right) + 2 \ce{NaNO_3} \left( aq \right) \nonumber$
How much easier is that to read? Let's try it in reverse? Look at the following reaction in shorthand and write the word equation for the reaction:
$\ce{Cu} \left( s \right) + \ce{AgNO_3} \left( aq \right) \rightarrow \ce{Cu(NO_3)_2} \left( aq \right) + \ce{Ag} \left( s \right) \nonumber$
The word equation for this reaction might read something like "solid copper reacts with an aqueous solution of silver nitrate to produce a solution of copper (II) nitrate with solid silver."
To turn word equations into symbolic equations, we need to follow the given steps:
1. Identify the reactants and products. This will help you know what symbols go on each side of the arrow and where the $+$ signs go.
2. Write the correct formulas for all compounds. You will need to use the rules you learned in Chapter 5 (including making all ionic compounds charge balanced).
3. Write the correct formulas for all elements. Usually this is given straight off of the periodic table. However, there are seven elements that are considered diatomic, meaning they are always found in pairs in nature. They include those elements listed in the table.
Table $2$ Diatomic Elements
Element Name Hydrogen Nitrogen Oxygen Fluorine Chlorine Bromine Iodine
Formula $H_2$ $N_2$ $O_2$ $F_2$ $Cl_2$ $Br_2$ $I_2$
Example $1$
Transfer the following symbolic equations into word equations or word equations into symbolic equations.
1. $\ce{HCl} \left( aq \right) + \ce{NaOH} \left( aq \right) \rightarrow \ce{NaCl} \left( aq \right) + \ce{H_2O} \left( l \right)$
2. Gaseous propane, $\ce{C_3H_8}$, burns in oxygen gas to produce gaseous carbon dioxide and liquid water.
3. Hydrogen fluoride gas reacts with an aqueous solution of potassium carbonate to produce an aqueous solution of potassium fluoride, liquid water, and gaseous carbon dioxide.
Solution
a. An aqueous solution of hydrochloric acid reacts with an aqueous solution of sodium hydroxide to produce an aqueous solution of sodium chloride and liquid water.
b. Reactants: propane ($\ce{C_3H_8}$) and oxygen ($\ce{O_2}$)
Product: carbon dioxide ($\ce{CO_2}$) and water ($\ce{H_2O}$)
$\ce{C_3H_8} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + \ce{H_2O} \left( l \right) \nonumber$
c. Reactants: hydrogen fluoride and potassium carbonate
Products: potassium fluoride, water, and carbon dioxide
$\ce{HF} \left( g \right) + \ce{K_2CO_3} \left( aq \right) \rightarrow \ce{KF} \left( aq \right) + \ce{H_2O} \left( l \right) + \ce{CO_2} \left( g \right) \nonumber$
Exercise $1$
Transfer the following symbolic equations into word equations or word equations into symbolic equations.
1. hydrogen gas reacts with nitrogen gas to produce gaseous ammonia
2. $\ce{HCl} \left( aq \right) + \ce{LiOH} \left( aq \right) \rightarrow \ce{NaCl} \left( aq \right) + \ce{H_2O} \left( l \right)$
3. copper metal is heated with oxygen gas to produce solid copper(II) oxide.
Answer a
$H_2 (g) + N_2 (g) \rightarrow NH_3 (g)$
Answer b
An aqueous solution of hydrochloric acid reacts with an aqueous solution of lithium hydroxide to produce an aqueous solution of lithium chloride and liquid water.
Answer c
$Cu (s) + O_2 (g) \rightarrow CuO (s)$
Even though chemical compounds are broken up and new compounds are formed during a chemical reaction, atoms in the reactants do not disappear nor do new atoms appear to form the products. In chemical reactions, atoms are never created or destroyed. The same atoms that were present in the reactants are present in the products - they are merely reorganized into different arrangements. In a complete chemical equation, the two sides of the equation must be present on the reactant and the product sides of the equation.
Coefficients and Subscripts
There are two types of numbers that appear in chemical equations. There are subscripts, which are part of the chemical formulas of the reactants and products and there are coefficients that are placed in front of the formulas to indicate how many molecules of that substance is used or produced.
The subscripts are part of the formulas and once the formulas for the reactants and products are determined, the subscripts may not be changed. The coefficients indicate the number of each substance involved in the reaction and may be changed in order to balance the equation. The equation above indicates that one mole of solid copper is reacting with two moles of aqueous silver nitrate to produce one mole of aqueous copper (II) nitrate and two atoms of solid silver.
Balancing a Chemical Equation
Because the identities of the reactants and products are fixed, the equation cannot be balanced by changing the subscripts of the reactants or the products. To do so would change the chemical identity of the species being described, as illustrated in Figure $2$.
The simplest and most generally useful method for balancing chemical equations is “inspection,” better known as trial and error. The following is an efficient approach to balancing a chemical equation using this method.
Steps in Balancing a Chemical Equation
1. Identify the most complex substance.
2. Beginning with that substance, choose an element(s) that appears in only one reactant and one product, if possible. Adjust the coefficients to obtain the same number of atoms of this element(s) on both sides.
3. Balance polyatomic ions (if present on both sides of the chemical equation) as a unit.
4. Balance the remaining atoms, usually ending with the least complex substance and using fractional coefficients if necessary. If a fractional coefficient has been used, multiply both sides of the equation by the denominator to obtain whole numbers for the coefficients.
5. Count the numbers of atoms of each kind on both sides of the equation to be sure that the chemical equation is balanced.
Example $2$: Combustion of Heptane
Balancing the chemical Equation for the combustion of Heptane ($\ce{C_7H_{16}}$)
$\ce{C_7H_{16} (l) + O_2 (g) → CO_2 (g) + H_2O (g) } \nonumber$
Solution
Example $2$: Steps for Problem Solving the Combustion Of Heptane
Steps Explanation
1. Identify the most complex substance. The most complex substance is the one with the largest number of different atoms, which is $C_7H_{16}$. We will assume initially that the final balanced chemical equation contains 1 molecule or formula unit of this substance.
2. Adjust the coefficients.
a. Because one molecule of n-heptane contains 7 carbon atoms, we need 7 CO2 molecules, each of which contains 1 carbon atom, on the right side:
$\ce{C7H16 (l) + O2 (g) → } \underline{7} \ce{CO2 (g) + H2O (g) } \nonumber$
Reactants and Products of Elements
Reactants Element/Polyatomic Ion Products
7 C 7
b. Because one molecule of n-heptane contains 16 hydrogen atoms, we need 8 H2O molecules, each of which contains 2 hydrogen atoms, on the right side:
$\ce{C7H16 (l) + O2 (g) → 7 CO2 (g) + } \underline{8} \ce{H2O (g) } \nonumber$
Reactants and Products of Elements
Reactants Element/Polyatomic Ion Products
7 C 7
16 H 16
3. Balance polyatomic ions as a unit. There are no polyatomic ions to be considered in this reaction.
4. Balance the remaining atoms.
The carbon and hydrogen atoms are now balanced, but we have 22 oxygen atoms on the right side and only 2 oxygen atoms on the left. We can balance the oxygen atoms by adjusting the coefficient in front of the least complex substance, O2, on the reactant side:
$\ce{C7H16 (l) + }\underline{11} \ce{ O2 (g) → 7 CO2 (g) + 8H2O (g) } \nonumber$
Reactants and Products of Elements
Reactants Element/Polyatomic Ion Products
7 C 7
16 H 16
22 O 22
5. Check your work. The equation is now balanced, and there are no fractional coefficients: there are 7 carbon atoms, 16 hydrogen atoms, and 22 oxygen atoms on each side. Always check to be sure that a chemical equation is balanced.
Example $3$: Combustion of Isooctane
Combustion of Isooctane ($\ce{C_8H_{18}}$)
$\ce{C8H18 (l) + O2 (g) -> CO_2 (g) + H_2O(g)} \nonumber$
Solution
The assumption that the final balanced chemical equation contains only one molecule or formula unit of the most complex substance is not always valid, but it is a good place to start. The combustion of any hydrocarbon with oxygen produces carbon dioxide and water.
Example $3$: Solution for the combustion of isooctane
steps examples
1. Identify the most complex substance. The most complex substance is the one with the largest number of different atoms, which is $\ce{C8H18}$. We will assume initially that the final balanced chemical equation contains 1 molecule or formula unit of this substance.
2. Adjust the coefficients.
a. The first element that appears only once in the reactants is carbon: 8 carbon atoms in isooctane means that there must be 8 CO2 molecules in the products:
$\ce{C8H18 (l) + O2 (g) -> }\underline{8} \ce{ CO2 (g) + H2O(g)}\nonumber$
Reactants and Products of Elements
Reactants Element/Polyatomic Ion Products
8 C 8
b. Eighteen hydrogen atoms in isooctane means that there must be 9 H2O molecules in the products:
$\ce{C8H18 (l) + O2 (g) -> 8CO2 (g) + }\underline{9} \ce{ H2O(g)} \nonumber$
Reactants and Products of Elements
Reactants Element/Polyatomic Ion Products
8 C 8
18 H 18
3. Balance polyatomic ions as a unit. There are no polyatomic ions to be considered in this reaction.
4. Balance the remaining atoms.
The carbon and hydrogen atoms are now balanced, but we have 25 oxygen atoms on the right side and only 2 oxygen atoms on the left. We can balance the least complex substance, O2, but because there are 2 oxygen atoms per O2 molecule, we must use a fractional coefficient ($\dfrac{25}{2}$) to balance the oxygen atoms:
$\ce{C8H18 (l) + } \underline{ \dfrac{25}{2} } \ce{O2 (g)→ 8CO2 (g) + 9H2O(g) }\nonumber$
Reactants and Products of Elements
Reactants Element/Polyatomic Ion Products
8 C 8
18 H 18
25 O 25
The equation is now balanced, but we usually write equations with whole-number coefficients. We can eliminate the fractional coefficient by multiplying all coefficients on both sides of the chemical equation by 2:
$\underline{2} \ce{C8H18(l) + } \underline{25} \ce{O2(g) ->} \underline{16} \ce{CO2(g) + } \underline{18} \ce{H2O(g)} \nonumber$
5. Check your work.
The balanced chemical equation has 16 carbon atoms, 36 hydrogen atoms, and 50 oxygen atoms on each side.
Reactants and Products of Elements
Reactants Element/Polyatomic Ion Products
16 C 16
36 H 36
50 O 50
Balancing equations requires some practice on your part as well as some common sense. If you find yourself using very large coefficients or if you have spent several minutes without success, go back and make sure that you have written the formulas of the reactants and products correctly.
Example $4$: Precipitation of Lead (II) Chloride
Aqueous solutions of lead (II) nitrate and sodium chloride are mixed. The products of the reaction are an aqueous solution of sodium nitrate and a solid precipitate of lead (II) chloride. Write the balanced chemical equation for this reaction.
Solution
Example $4$: Solution for the combustion of isooctane
Steps for Problem Solving Example $4$
1. Identify the most complex substance.
The most complex substance is lead (II) chloride.
$\ce{Pb(NO3)2(aq) + NaCl(aq) → NaNO3(aq) + PbCl2(s)} \nonumber$
2. Adjust the coefficients.
There are twice as many chloride ions in the reactants than in the products. Place a 2 in front of the NaCl in order to balance the chloride ions.
$\ce{Pb(NO3)2(aq) + }\underline{ 2} \ce{NaCl(aq) → NaNO3(aq) + PbCl2(s)} \nonumber$
Reactants and Products of Elements
Reactants Element/Polyatomic Ion Products
1 Pb 1
2 Na 1
2 Cl 2
3. Balance polyatomic ions as a unit.
The nitrate ions are still unbalanced. Place a 2 in front of the NaNO3. The result is:
$\ce{Pb(NO3)2(aq) + 2NaCl(aq) → } \underline {2} \ce{NaNO3(aq) + PbCl2(s)} \nonumber$
Reactants and Products of Elements
Reactants Element/Polyatomic Ion Products
1 Pb 1
2 Na 2
2 Cl 2
2 NO3- 2
4. Balance the remaining atoms. There is no need to balance the remaining atoms because they are already balanced.
5. Check your work.
$\ce{Pb(NO3)2(aq) + 2NaCl(aq) → 2NaNO3(aq) + PbCl2(s)} \nonumber$
Reactants and Products of Elements
Reactants Element/Polyatomic Ion Products
1 Pb 1
2 Na 2
2 Cl 2
2 NO3- 2
Exercise $2$
Is each chemical equation balanced?
1. $\ce{2Hg(ℓ)+ O_2(g) \rightarrow Hg_2O_2(s)}$
2. $\ce{C_2H_4(g) + 2O_2(g)→ 2CO_2(g) + 2H_2O(g)}$
3. $\ce{Mg(NO_3)_2(s) + 2Li (s) \rightarrow Mg(s)+ 2LiNO_3(s)}$
Answer a
yes
Answer b
no
Answer c
yes
Exercise $3$
Balance the following chemical equations.
1. $\ce{N2 (g) + O2 (g) → NO2 (g) }$
2. $\ce{Pb(NO3)2(aq) + FeCl3(aq) → Fe(NO3)3(aq) + PbCl2(s)}$
3. $\ce{C6H14(l) + O2(g)→ CO2(g) + H2O(g)}$
Answer a
N2 (g) + 2O2 (g) → 2NO2 (g)
Answer b
3Pb(NO3)2(aq) + 2FeCl3(aq) → 2Fe(NO3)3(aq) + 3PbCl2(s)
Answer c
2C6H14(l) + 19O2(g)→ 12CO2(g) + 14H2O(g)
Summary
• A chemical reaction is the process in which one or more substances are changed into one or more new substances.
• Chemical reactions are represented by chemical equations.
• Chemical equations have reactants on the left, an arrow that is read as "yields", and the products on the right.
• To be useful, chemical equations must always be balanced. Balanced chemical equations have the same number and type of each atom on both sides of the equation.
• The coefficients in a balanced equation must be the simplest whole number ratio. Mass is always conserved in chemical reactions.
Contributors and Attributions
• Henry Agnew (UC Davis) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/05%3A_Chemical_Accounting/5.01%3A_Chemical_Sentences-_Equations.txt |
Learning Objectives
• Identify the relative volume of gas consumed or produced based on a balanced chemical equation for a reaction.
The behavior of gases can be represented by gas laws as will be discussed in more detail in Chapter 6. Avogadro's law states that all gases (that show ideal behavior) contain the same number of molecules in the same volume (at the same temperature and pressure). Thus, the ratios of volumes of gases involved in a chemical reaction are given by the coefficients in the equation for the reaction, provided that the gas volumes are measured at the same temperature and pressure.
We can extend Avogadro’s law (that the volume of a gas is directly proportional to the number of moles of the gas) to chemical reactions with gases: Gases combine, or react, in definite and simple proportions by volume, provided that all gas volumes are measured at the same temperature and pressure. For example, since nitrogen and hydrogen gases react to produce ammonia gas according to the balanced equation below
$\ce{N2}(g)+\ce{3H2}(g)⟶\ce{2NH3}(g) \nonumber$
a given volume of nitrogen gas reacts with three times that volume of hydrogen gas to produce two times that volume of ammonia gas, if pressure and temperature remain constant.
The explanation for this is illustrated in Figure $1$. According to Avogadro’s law, equal volumes of gaseous N2, H2, and NH3, at the same temperature and pressure, contain the same number of molecules. Because one molecule of N2 reacts with three molecules of H2 to produce two molecules of NH3, the volume of H2 required is three times the volume of N2, and the volume of NH3 produced is two times the volume of N2.
Example $5$: Reaction of Gases
Propane, C3H8(g), is used in gas grills to provide the heat for cooking. What volume of O2(g) measured at 25 °C and 760 torr is required to react with 2.7 L of propane measured under the same conditions of temperature and pressure? Assume that the propane undergoes complete combustion.
Solution
The ratio of the volumes of C3H8 and O2 will be equal to the ratio of their coefficients in the balanced equation for the reaction:
\begin{align} &\ce{C3H8}(g)+\ce{5O2}(g) ⟶ &&\ce{3CO2}(g)+\ce{4H2O}(l)\ \ce{&1\: volume + 5\: volumes &&3\: volumes + 4\: volumes} \end{align} \nonumber
From the equation, we see that one volume of C3H8 will react with five volumes of O2:
$\mathrm{2.7\cancel{L\:C_3H_8}×\dfrac{5\: L\:\ce{O2}}{1\cancel{L\:C_3H_8}}=13.5\: L\:\ce{O2}} \nonumber$
A volume of 13.5 L of O2 will be required to react with 2.7 L of C3H8.
Exercise $5$
An acetylene tank for an oxyacetylene welding torch provides 9340 L of acetylene gas, C2H2, at 0 °C and 1 atm. How many tanks of oxygen, each providing 7.00 × 103 L of O2 at 0 °C and 1 atm, will be required to burn the acetylene?
$\ce{2C2H2 + 5O2⟶4CO2 + 2H2O} \nonumber$
Answer
3.34 tanks (2.34 × 104 L)
Example $6$: Volumes of Reacting Gases
Ammonia is an important fertilizer and industrial chemical. Suppose that a volume of 683 billion cubic feet of gaseous ammonia, measured at 25 °C and 1 atm, was manufactured. What volume of H2(g), measured under the same conditions, was required to prepare this amount of ammonia by reaction with N2?
$\ce{N2}(g)+\ce{3H2}(g)⟶\ce{2NH3}(g) \nonumber$
Solution
Because equal volumes of H2 and NH3 contain equal numbers of molecules and each three molecules of H2 that react produce two molecules of NH3, the ratio of the volumes of H2 and NH3 will be equal to 3:2. Two volumes of NH3, in this case in units of billion ft3, will be formed from three volumes of H2:
$\mathrm{683\cancel{billion\:ft^3\:NH_3}×\dfrac{3\: billion\:ft^3\:H_2}{2\cancel{billion\:ft^3\:NH_3}}=1.02×10^3\:billion\:ft^3\:H_2} \nonumber$
The manufacture of 683 billion ft3 of NH3 required 1020 billion ft3 of H2. (At 25 °C and 1 atm, this is the volume of a cube with an edge length of approximately 1.9 miles.)
Exercise $6$
What volume of O2(g) measured at 25 °C and 760 torr is required to react with 17.0 L of ethylene, C2H4(g), measured under the same conditions of temperature and pressure? The products are CO2 and water vapor.
Answer
51.0 L
Summary
Avogadro's law may be used to determine the volume of gaseous reactant(s) or product(s) based on a balanced chemical equation. | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/05%3A_Chemical_Accounting/5.02%3A_Volume_Relationships_in_Chemical_Equations.txt |
Learning Objectives
• Use Avogadro's number to convert to moles and vice versa given the number of particles of an element.
• Know the definition of the mole.
• Determine the formula mass of an ionic or molecular compound.
• Determine the percent composition of each element in a compound from the chemical formula.
When objects are very small, it is often inconvenient or inefficient, or even impossible to deal with the objects one at a time. For these reasons, we often deal with very small objects in groups, and have even invented names for various numbers of objects. The most common of these is "dozen" which refers to 12 objects. We frequently buy objects in groups of 12, like doughnuts or pencils. Even smaller objects such as straight pins or staples are usually sold in boxes of 144, or a dozen dozen. A group of 144 is called a "gross".
This problem of dealing with things that are too small to operate with as single items also occurs in chemistry. Atoms and molecules are too small to see, let alone to count or measure. Chemists needed to select a group of atoms or molecules that would be convenient to operate with.
Avogadro's Number:Counting Atoms
Owing to their tiny size, atoms and molecules cannot be counted by direct observation. But much as we do when "counting" beans in a jar, we can estimate the number of particles in a sample of an element or compound if we have some idea of the volume occupied by each particle and the volume of the container. Once this has been done, we know the number of formula units (to use the most general term for any combination of atoms we wish to define) in any arbitrary weight of the substance. The number will of course depend both on the formula of the substance and on the weight of the sample. However, if we consider a weight of substance that is the same as its formula (molecular) weight expressed in grams, we have only one number to know: Avogadro's number.
Avogadro's number
Avogadro's number is known to ten significant digits:
$N_A = 6.022141527 \times 10^{23}. \nonumber$
However, you only need to know it to three significant figures:
$N_A \approx 6.02 \times 10^{23}. \label{3.2.1}$
So $6.02 \times 10^{23}$ of what? Well, of anything you like: apples, stars in the sky, burritos. However, the only practical use for $N_A$ is to have a more convenient way of expressing the huge numbers of the tiny particles such as atoms or molecules that we deal with in chemistry. Avogadro's number is a collective number, just like a dozen. Students can think of $6.02 \times 10^{23}$ as the "chemist's dozen".
Before getting into the use of Avogadro's number in problems, take a moment to convince yourself of the reasoning embodied in the following examples.
Things to understand about Avogadro's number
• It is a number, just as is "dozen", and thus is dimensionless.
• It is a huge number, far greater in magnitude than we can visualize
• Its practical use is limited to counting tiny things like atoms, molecules, "formula units", electrons, or photons.
• The value of NA can be known only to the precision that the number of atoms in a measurable weight of a substance can be estimated. Because large numbers of atoms cannot be counted directly, a variety of ingenious indirect measurements have been made involving such things as Brownian motion and X-ray scattering.
The Mole: "A Dozen Eggs and a Mole of Sugar, Please"
The mole (symbol: mol) is the base unit of amount of substance ("number of substance") in the International System of Units or System International (SI), defined as exactly 6.02214076×1023 particles, e.g., atoms, molecules, ions or electrons. The current definition was adopted in November 2018, revising its old definition based on the number of atoms in 12 grams of carbon-12 (12C) (the isotope of carbon with relative atomic mass 12 Daltons by definition).
It is not obvious why eggs come in dozens rather than 10s or 14s, or why a ream of paper contains 500 sheets rather than 400 or 600. The definition of a mole—that is, the decision to base it on 12 g of carbon-12—is also arbitrary. The important point is that 1 mole of carbon—or of anything else, whether atoms, compact discs, or houses—always has the same number of objects: 6.02 × 1023.
The Mole
Video $1$ How big is a mole?
Converting Between Number of Atoms to Moles and Vice Versa
We can use Avogadro's number as a conversion factor, or ratio, in dimensional analysis problems. If we are given the number of atoms of an element X, we can convert it into moles of by using the relationship
$\text{1 mol X} = 6.022 \times 10^{23} \text{ X atoms}. \nonumber$
An example on the use of Avogadro's number as a conversion factor is given below for carbon.
Example $1$: Moles of Carbon
The element carbon exists in two primary forms: graphite and diamond. How many moles of carbon atoms is $4.72 \times 10^{24}$ atoms of carbon?
Solution
Example $1$:Steps for Problem Solving how many moles of carbon atoms is $4.72 \times 10^{24}$ atoms of carbon?
Steps for Problem Solving Example
Identify the "given"information and what the problem is asking you to "find." Given: $4.72 \times 10^{24}$ C atoms
Find: mol C
List other known quantities $1\, mol = 6.022 \times 10^{23}$ C atoms
Prepare a concept map and use the proper conversion factor.
Cancel units and calculate. $4.72 \times 10^{24} \: \cancel{\text{C} \: \ce{atoms}} \times \frac{1 \: \text{mol} \: \ce{C}}{6.02 \times 10^{23} \: \cancel{\text{C} \: \ce{atoms}}} = 7.84 \: \text{mol} \: \ce{C} \nonumber$
Think about your result. The given number of carbon atoms was greater than Avogadro's number,so the number of moles of $\ce{C}$ atoms is greater than 1 mole. Since Avogadro's number is a measured quantity with three significant figures, the result of the calculation is rounded to three significant figures
Formula Mass
One skill needed in future chapters is the ability to determine the mass of the formula of various chemical substances. This quantity is called the formula mass. The formula mass is obtained by adding the masses of each individual atom in the formula of the substance. Because a proper formula is electrically neutral (with no net electrons gained or lost), the ions can be considered atoms for the purpose of calculating the formula mass.
Let us start by calculating the formula mass of sodium chloride (NaCl). This formula mass is the sum of the atomic masses of one sodium atom and one chlorine atom, which we find from the periodic table; here, we use the masses to two decimal places:
To two decimal places, the formula mass of NaCl is 58.44 amu.
For covalent substances, the formula represents the numbers and types of atoms composing a single molecule of the substance; therefore, the formula mass may be correctly referred to as a molecular mass. Consider chloroform (CHCl3), a covalent compound once used as a surgical anesthetic and now primarily used in the production of tetrafluoroethylene, the building block for the “anti-stick” polymer, Teflon. The molecular formula of chloroform indicates that a single molecule contains one carbon atom, one hydrogen atom, and three chlorine atoms. The average molecular mass of a chloroform molecule is therefore equal to the sum of the average atomic masses of these atoms.
To two decimal places, the formula mass of CHCl3 is 119.37 amu.
For ionic compounds with polyatomic ions, the sum must include the number and mass of each atom in the formula for the polyatomic ion. as shown in the example below for aluminum sulfate, Al2(SO4)3.
Example $2$ Formula Mass for an Ionic Compound
Aluminum sulfate, Al2(SO4)3, is an ionic compound that is used in the manufacture of paper and in various water purification processes. What is the formula mass (amu) of this compound?
Solution
The formula for this compound indicates it contains Al3+ and SO42 ions combined in a 2:3 ratio. For purposes of computing a formula mass, it is helpful to rewrite the formula in the simpler format, Al2S3O12. Following the approach outlined above, the formula mass for this compound is calculated as follows:
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The formula mass for Al2(SO4)3, is 342.14 amu.
Exercise $1$
Use the atomic masses (rounded to two decimal places) to determine the formula mass for each ionic compound.
1. TiO2
2. AgBr
3. Au(NO3)3
4. Fe3(PO4)2
a. 79.87 amu
b. 187.77 amu
c. 383.0 amu
Percent Composition of a Compound from a Chemical Formula
The percent composition of a compound can also be determined from the formula of the compound. The subscripts in the formula are first used to calculate the mass of each element in one mole of the compound. That is divided by the molar mass of the compound and multiplied by $100\%$.
$\% \: \text{by mass} = \frac{\text{mass of element in} \: 1 \: \text{mol}}{\text{molar mass of compound}} \times 100\% \nonumber$
The percent composition of a given compound is always the same as long as the compound is pure.
Example $3$
Dichlorine heptoxide $\left( \ce{Cl_2O_7} \right)$ is a highly reactive compound used in some organic synthesis reactions. Calculate the percent composition of dichlorine heptoxide.
Solution
Example $3$:Steps for Problem Solving calculate the percent composition of dichlorine heptoxide $\left( \ce{Cl_2O_7} \right)$.
Steps for Problem Solving Example
Identify the "given"information and what the problem is asking you to "find."
Given : Cl2O7
Find: % Composition (% Cl and %O)
List other known quantities
Mass of Cl in 1 mol Cl2O7 , 2 Cl : 2 x 35.45 g = 70.90 g
Mass of O in 1 mol Cl2O7 , 7 O: 7 x 16.00 g = 112.00 g
Molar mass of Cl2O7 = 182.90 g/mol
Cancel units and calculate.
$\% \ce{Cl} = \frac{70.90 \: \text{g} \: \ce{Cl}}{182.90 \: \text{g}} \times 100\% = 38.76\% \: \ce{Cl} \nonumber$
$\% \: \ce{O} = \frac{112.00 \: \text{g} \: \ce{O}}{182.90 \: \text{g}} \times 100\% = 61.24\% \: \ce{O} \nonumber$
Calculate the percent by mass of each element by dividing the mass of that element in 1 mole of the compound by the molar mass of the compound and multiplying by $100\%$.
Think about your result. The percentages add up to $100\%$.
Percent composition can also be used to determine the mass of a certain element that is contained in any mass of a compound. In the previous sample problem, it was found that the percent composition of dichlorine heptoxide is $38.76\% \: \ce{Cl}$ and $61.24\% \: \ce{O}$. Suppose that you needed to know the masses of chlorine and oxygen present in a $12.50 \: \text{g}$ sample of dichlorine heptoxide. You can set up a conversion factor based on the percent by mass of each element.
$12.50 \: \text{g} \: \ce{Cl_2O_7} \times \frac{38.76 \: \text{g} \: \ce{Cl}}{100 \: \text{g} \: \ce{Cl_2O_7}} = 4.845 \: \text{g} \: \ce{Cl} \nonumber$
$12.50 \: \text{g} \: \ce{Cl_2O_7} \times \frac{61.24 \: \text{g} \: \ce{O}}{100 \: \text{g} \: \ce{Cl_2O_7}} = 7.655 \: \text{g} \: \ce{O} \nonumber$
The sum of the two masses is $12.50 \: \text{g}$, the mass of the sample size.
Exercise $2$
Barium fluoride is a transparent crystal that can be found in nature as the mineral frankdicksonite. Determine the percent composition of barium fluoride.
Answer a:
78.32% Ba and 21.67% F
Summary
• The mole (symbol: mol) is the base unit of amount of substance ("number of substance") in the International System of Units or System International (SI), defined as exactly 6.02214076×1023 particles, e.g., atoms, molecules, ions or electrons.
• Avogadro's number is related to moles of any substance X as follows:
$\text{1 mol X} = 6.022 \times 10^{23} \text{ X atoms}. \nonumber$
• Formula masses of ionic and molecular compounds can be determined from the masses of the atoms in their formulas.
• Processes are described for calculating the percent composition of a compound based on the chemical formula. | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/05%3A_Chemical_Accounting/5.03%3A_Avogadro%27s_Number_and_the_Mole.txt |
Learning Objectives
• Perform conversions between mass and moles of a substance.
• Convert from mass or moles of one substance to mass or moles of another substance in a chemical reaction.
• Use a balanced chemical equation to determine molar relationships between substances.
Molar Mass
The molar mass of any substance is the mass in grams of one mole of representative particles of that substance. The representative particles can be atoms, molecules, or formula units of ionic compounds. This relationship is frequently used in the laboratory. The simplest type of manipulation using molar mass as a conversion factor is a mole-mass conversion (or its reverse, a mass-mole conversion). In such a conversion, we use the molar mass of a substance as a conversion factor to convert mole units into mass units (or, conversely, mass units into mole units).
We also established that 1 mol of Al has a mass of 26.98 g (Example). Stated mathematically,
1 mol Al = 26.98 g Al
We can divide both sides of this expression by either side to get one of two possible conversion factors:
$\mathrm{\dfrac{1\: mol\: Al}{26.98\: g\: Al}\, and\, \dfrac{26.98\: g\: Al}{1\: mol\: Al}} \label{Eq1}$
The first conversion factor can be used to convert from mass to moles, and the second converts from moles to mass. Both can be used to solve problems that would be hard to do “by eye.”
Example $1$
What is the mass of 3.987 mol of Al?
Solution
Example $1$: Steps for Problem Solving the mass of 3.987 moles of aluminum (Al).
Steps for Problem Solving Example
Identify the "given"information and what the problem is asking you to "find." Given: 3.987 mol of Al
Find: g Al
List other known quantities 1 mol Al = 26.98 g Al
Prepare a concept map and use the proper conversion factor.
Cancel units and calculate. $3.987 \: \cancel{\text{mol} \: \ce{Al}} \times \dfrac{26.98 \: \text{g} \: \ce{Al}}{1 \: \cancel{\text{mol} \: \ce{Al}}} = 107.6 \: \text{g} \: \ce{Al}$
Think about your result. The calculated value makes sense because it is almost four times times the mass for 1 mole of aluminum. Our final answer is expressed to four significant figures.
Exercise $1$
How many moles are present in 100. g of Al? (Hint: you will have to use the other conversion factor we obtained for aluminum.)
Answer
3.71 g Al
Conversions like this are possible for any substance, as long as the proper atomic mass, formula mass, or molar mass is known (or can be determined) and expressed in grams per mole. Figure $1$ is a chart for determining what conversion factor is needed, and Figure $2$ is a flow diagram for the steps needed to perform a conversion.
Suppose that for a certain experiment you need 3.00 moles of calcium chloride $\left( \ce{CaCl_2} \right)$. Since calcium chloride is a solid, it would be convenient to use a balance to measure the mass that is needed. Dimensional analysis will allow you to calculate the mass of $\ce{CaCl_2}$ that you should measure as show in Example $3$.
Example $2$: Calcium Chloride
Calculate the mass of 3.00 moles of calcium chloride (CaCl2).
Solution
Example $2$: Steps for Problem Solving the mass of 3.00 moles of calcium chloride (CaCl2).
Steps for Problem Solving Example
Identify the "given"information and what the problem is asking you to "find." Given: 3.00 moles of CaCl2
Find: g CaCl2
List other known quantities 1 ml CaCl2 = 110.98 g CaCl2
Prepare a concept map and use the proper conversion factor.
Cancel units and calculate. $3.00 \: \cancel{\text{mol} \: \ce{CaCl_2}} \times \dfrac{110.98 \: \text{g} \: \ce{CaCl_2}}{1 \: \cancel{\text{mol} \: \ce{CaCl_2}}} = 333 \: \text{g} \: \ce{CaCl_2}$
Think about your result.
Exercise $2$: Calcium Oxide
What is the mass of $7.50 \: \text{mol}$ of (calcium oxide) $\ce{CaO}$?
Answer:
420.60 g
Example $3$: Water
How many moles are present in 108 grams of water?
Solution
Example $3$: Steps for Problem Solving moles in 108 grams of water?
Steps for Problem Solving Example
Identify the "given"information and what the problem is asking you to "find." Given: 108 g H2O
Find: mol H2O
List other known quantities $1 \: \text{mol} \: \ce{H_2O} = 18.02 \: \text{g}$ H2O
Prepare a concept map and use the proper conversion factor.
Cancel units and calculate. $108 \: \cancel{\text{g} \: \ce{H_2O}} \times \dfrac{1 \: \text{mol} \: \ce{H_2O}}{18.02 \: \cancel{\text{g} \: \ce{H_2O}}} = 5.99 \: \text{mol} \: \ce{H_2O}$
Think about your result.
Exercise $3$: Nitrogen Gas
What is the mass of $7.50 \: \text{mol}$ of Nitrogen gas $\ce{N2}$?
Answer:
210 g
Mole and Mass Relationships in Chemical Equations
We have established that a balanced chemical equation is balanced in terms of moles as well as atoms or molecules. We have used balanced equations to set up ratios, now in terms of moles of materials, that we can use as conversion factors to answer stoichiometric questions, such as how many moles of substance A react with so many moles of reactant B. We can extend this technique even further. Recall that we can relate a molar amount to a mass amount using molar mass. We can use that ability to answer stoichiometry questions in terms of the masses of a particular substance, in addition to moles. We do this using the following sequence:
Collectively, these conversions are called mole-mass calculations.
As an example, consider the balanced chemical equation
$Fe_2O_3 + 3SO_3 \rightarrow Fe_2(SO_4)_3 \nonumber$
If we have 3.59 mol of Fe2O3, how many grams of SO3 can react with it? Using the mole-mass calculation sequence, we can determine the required mass of SO3 in two steps. First, we construct the appropriate molar ratio, determined from the balanced chemical equation, to calculate the number of moles of SO3 needed. Then using the molar mass of SO3 as a conversion factor, we determine the mass that this number of moles of SO3 has.
As usual, we start with the quantity we were given:
$\mathrm{3.59\: \cancel{ mol\: Fe_2O_3 } \times \left( \dfrac{3\: mol\: SO_3}{1\: \cancel{ mol\: Fe_2O_3}} \right) =10.77\: mol\: SO_3} \label{Eq2}$
The mol Fe2O3 units cancel, leaving mol SO3 unit. Now, we take this answer and convert it to grams of SO3, using the molar mass of SO3 as the conversion factor:
$\mathrm{10.77\: \bcancel{mol\: SO_3} \times \left( \dfrac{80.06\: g\: SO_3}{1\: \bcancel{ mol\: SO_3}} \right) =862\: g\: SO_3} \label{Eq3}$
Our final answer is expressed to three significant figures. Thus, in a two-step process, we find that 862 g of SO3 will react with 3.59 mol of Fe2O3. Many problems of this type can be answered in this manner.
The same two-step problem can also be worked out in a single line, rather than as two separate steps, as follows:
$3.59 \cancel{\, mol \, Fe_2O_3} \times \underbrace{\left( \dfrac{ 3 \bcancel{ \, mol\, SO_3}}{ 1 \cancel{\, mol\, Fe_2O_3}} \right)}_{\text{converts to moles of SO}_3} \times \underbrace{ \left( \dfrac{ 80.06 {\, g \, SO_3}}{ 1 \, \bcancel{ mol\, SO_3}} \right)}_{\text{converts to grams of SO}_3} = 862\, g\, SO_3 \nonumber$
We get exactly the same answer when combining all the math steps together as we do when we calculate one step at a time.
Example $4$: Generation of Aluminum Oxide
How many moles of HCl will be produced when 249 g of AlCl3 are reacted according to this chemical equation?
$2AlCl_3 + 3H_2O(ℓ) → Al_2O_3 + 6HCl(g) \nonumber$
Solution
Example $4$: Steps for Problem Solving how many moles are in 249 g of AlCl3 ?
Steps for Problem Solving Example
Identify the "given"information and what the problem is asking you to "find." Given: 249 g AlCl3
Find: moles HCl
List other known quantities 1 mol AlCl3 = 133.33 g/mol
6 mol of HCl to 2 mol AlCl3
Prepare a concept map and use the proper conversion factor.
Cancel units and calculate. $249\, \cancel{g\, AlCl_{3}}\times \dfrac{1\, \cancel{mol\, AlCl_{3}}}{133.33\, \cancel{g\, AlCl_{3}}}\times \dfrac{6\, mol\, HCl}{2\, \cancel{mol\, AlCl_{3}}}=5.60\, mol\, HCl$
Think about your result. Since 249 g of AlCl3 is less than 266.66 g, the mass for 2 moles of AlCl3 and the relationship is 6 mol of HCl to 2 mol AlCl3 , the answer should be less than 6 moles of HCl.
Exercise $4$: Generation of Aluminum Oxide
How many moles of Al2O3 will be produced when 23.9 g of H2O are reacted according to this chemical equation?
$2AlCl_3 + 3H_2O(ℓ) → Al_2O_3 + 6HCl(g) \nonumber$
Answer
0.442 mol Al2O3
Molar Relationships in Chemical Equations
Previously, you learned to balance chemical equations by comparing the numbers of each type of atom in the reactants and products. The coefficients in front of the chemical formulas represent the numbers of molecules or formula units (depending on the type of substance). Here, we will extend the meaning of the coefficients in a chemical equation.
Consider the simple chemical equation
$2H_2 + O_2 → 2H_2O \nonumber$
The convention for writing balanced chemical equations is to use the lowest whole-number ratio for the coefficients. However, the equation is balanced as long as the coefficients are in a 2:1:2 ratio. For example, this equation is also balanced if we write it as
$4H_2 + 2O_2 → 4H_2O \nonumber$
The ratio of the coefficients is 4:2:4, which reduces to 2:1:2. The equation is also balanced if we were to write it as
$22H_2 + 11O_2 → 22H_2O \nonumber$
because 22:11:22 also reduces to 2:1:2.
Suppose we want to use larger numbers. Consider the following coefficients:
$12.044 \times 10^{23}\; H_2 + 6.022 \times 10^{23}\; O_2 → 12.044 \times 10^{23}\; H_2O \nonumber$
These coefficients also have the ratio 2:1:2 (check it and see), so this equation is balanced. But 6.022 × 1023 is 1 mol, while 12.044 × 1023 is 2 mol (and the number is written that way to make this more obvious), so we can simplify this version of the equation by writing it as
$2 \;mol\; H_2 + 1\; mol\; O_2 → 2 \;mol\; H_2O \nonumber$
We can leave out the word mol and not write the 1 coefficient (as is our habit), so the final form of the equation, still balanced, is
$2H_2 + O_2 → 2H_2O \nonumber$
Now we interpret the coefficients as referring to molar amounts, not individual molecules. The lesson? Balanced chemical equations are balanced not only at the molecular level but also in terms of molar amounts of reactants and products. Thus, we can read this reaction as “two moles of hydrogen react with one mole of oxygen to produce two moles of water.”
By the same token, the ratios we constructed to describe molecules reaction can also be constructed in terms of moles rather than molecules. For the reaction in which hydrogen and oxygen combine to make water, for example, we can construct the following ratios:
$\mathrm{\dfrac{2\: mol\: H_2}{1\: mol\: O_2}\: or\: \dfrac{1\: mol\: O_2}{2\: mol\: H_2}} \nonumber$
$\mathrm{\dfrac{2\: mol\: H_2O}{1\: mol\: O_2}\: or\: \dfrac{1\: mol\: O_2}{2\: mol\: H_2O}} \nonumber$
$\mathrm{\dfrac{2\: mol\: H_2}{2\: mol\: H_2O}\: or\: \dfrac{2\: mol\: H_2O}{2\: mol\: H_2}} \nonumber$
We can use these ratios to determine what amount of a substance, in moles, will react with or produce a given number of moles of a different substance. The study of the numerical relationships between the reactants and the products in balanced chemical reactions is called stoichiometry.
Example $5$
How many moles of oxygen react with hydrogen to produce 27.6 mol of H2O?
Solution
Example $5$: Steps for Problem Solving how many moles of oxygen react with hydrogen to produce 27.6 mol of H2O?
Steps for Problem Solving Example
Find a balanced equation that describes the reaction
Unbalanced: H2 + O2 → H2O
Balanced: 2H2 + O22H2O
Identify the "given"information and what the problem is asking you to "find." Given: moles H2O Find: moles oxygen
List other known quantities 1 mol O2 = 2 mol H2O
Prepare a concept map and use the proper conversion factor.
Cancel units and calculate.
$\mathrm{\cancel{27.6\: mol\: H_2O}\times\dfrac{1\: mol\: O_2}{\cancel{2\: mol\: H_2O}}=13.8\: mol\: O_2}$
To produce 27.6 mol of H2O, 13.8 mol of O2 react.
Think about your result. Since each mole of oxygen produces twice as many moles of water, it makes sense that the produced amount is greater than the reactant amount
Example $6$
How many moles of ammonia are produced if 4.20 moles of hydrogen are reacted with an excess of nitrogen?
Solution
Example $6$: Steps for Problem Solving how many moles of ammonia are produced if 4.20 moles of hydrogen are reacted with an excess of nitrogen
Steps for Problem Solving Example
Find a balanced equation that describes the reaction
Unbalanced: N2 + H2 → NH3
Balanced: N2 + 3H2 2NH3
Identify the "given"information and what the problem is asking you to "find."
Given: $\ce{H_2} = 4.20 \: \text{mol}$
Find: $\text{mol}$ of $\ce{NH_3}$
List other known quantities 3 mol H2 = 2 mol NH3
Prepare a concept map and use the proper conversion factor.
Cancel units and calculate.
$\cancel{4.20 \: \text{mol} \: H_2} \times \dfrac{2 \: \text{mol} \: NH_3}{\cancel{3 \: \text{mol} \: H_2}} = 2.80 \: \text{mol} \: NH_3$
The reaction of $4.20 \: \text{mol}$ of hydrogen with excess nitrogen produces $2.80 \: \text{mol}$ of ammonia
Think about your result. The result corresponds to the 3:2 ratio of hydrogen to ammonia from the balanced equation.
Exercise $5$
1. Given the following balanced chemical equation,$\ce{C5H12 + 8O2 → 5CO2 + 6H2O} \nonumber$how many moles of H2O can be formed if 0.0652 mol of C5H12 were to react?
2. Balance the following unbalanced equation and determine how many moles of H2O are produced when 1.65 mol of NH3 react.$\ce{NH3 + O2 → N2 + H2O} \nonumber$
Answer a:
3.14 mol H2O
Answer b:
4NH3 + 3O2 → 2N2 + 6H2O; 2.48 mol H2O
Mass Relationships in Chemical Equations
It is a small step from mole-mass calculations to mass-mass calculations. If we start with a known mass of one substance in a chemical reaction (instead of a known number of moles), we can calculate the corresponding masses of other substances in the reaction. The first step in this case is to convert the known mass into moles, using the substance’s molar mass as the conversion factor. Then—and only then—we use the balanced chemical equation to construct a conversion factor to convert that quantity to moles of another substance, which in turn can be converted to a corresponding mass. Sequentially, the process is as follows:
This three-part process can be carried out in three discrete steps or combined into a single calculation that contains three conversion factors. The following example illustrates both techniques.
Example $7$: Decomposition of Ammonium Nitrate
Ammonium nitrate decomposes to dinitrogen monoxide and water according to the following equation.
$\ce{NH_4NO_3} \left( s \right) \rightarrow \ce{N_2O} \left( g \right) + 2 \ce{H_2O} \left( l \right) \nonumber$
In a certain experiment, $45.7 \: \text{g}$ of ammonium nitrate is decomposed. Find the mass of each of the products formed.
Example $7$: Steps for Problem Solving the Decomposition of Ammonium Nitrate
Steps for Problem Solving Example $2$
Identify the "given"information and what the problem is asking you to "find."
Given: $45.7 \: \text{g} \: \ce{NH_4NO_3}$
Find:
Mass $\ce{N_2O} = ? \: \text{g}$
Mass $\ce{H_2O} = ? \: \text{g}$
List other known quantities
1 mol $\ce{NH_4NO_3} = 80.06 \: \text{g/mol}$
1 mol $\ce{N_2O} = 44.02 \: \text{g/mol}$
1 mol $\ce{H_2O} = 18.02 \: \text{g/mol}$
1 mol NH4NO3 to 1 mol N2O to 2 mol H2O
Prepare two concept maps and use the proper conversion factor.
Cancel units and calculate.
$45.7 \: \text{g} \: \ce{NH_4NO_3} \times \dfrac{1 \: \text{mol} \: \ce{NH_4NO_3}}{80.06 \: \text{g} \: \ce{NH_4NO_3}} \times \dfrac{1 \: \text{mol} \: \ce{N_2O}}{1 \: \text{mol} \: \ce{NH_4NO_3}} \times \dfrac{44.02 \: \text{g} \: \ce{N_2O}}{1 \: \text{mol} \: \ce{N_2O}} = 25.1 \: \text{g} \: \ce{N_2O}$
$45.7 \: \text{g} \: \ce{NH_4NO_3} \times \dfrac{1 \: \text{mol} \: \ce{NH_4NO_3}}{80.06 \: \text{g} \: \ce{NH_4NO_3}} \times \dfrac{2 \: \ce{H_2O}}{1 \: \text{mol} \: \ce{NH_4NO_3}} \times \dfrac{18.02 \: \text{g} \: \ce{H_2O}}{1 \: \text{mol} \: \ce{H_2O}} = 20.6 \: \text{g} \: \ce{H_2O}$
Think about your result. The total mass of the two products is equal to the mass of ammonium nitrate which decomposed, demonstrating the law of conservation of mass. Each answer has three significant figures.
Exercise $6$: Carbon Tetrachloride
Methane can react with elemental chlorine to make carbon tetrachloride ($\ce{CCl_4}$). The balanced chemical equation is as follows.
$\ce{CH4 (g) + 4 Cl2 (g) → CCl2 (l) + 4 HCl (l) } \nonumber$
How many grams of HCl are produced by the reaction of 100.0g of $\ce{CH4}$?
Answer
908.7g HCl
Summary
• Calculations involving conversions between moles of a substance and the mass of that substance are described.
• The balanced chemical reaction can be used to determine molar and mass relationships between substances.
Contributors and Attributions
• TextMap: The Basics of GOB Chemistry (Ball et al.)
• Henry Agnew (UC Davis) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/05%3A_Chemical_Accounting/5.04%3A_Molar_Mass-_Mole-to-Mass_and_Mass-to-Mole_Conversions.txt |
Learning Objectives
• Learn some terminology involving solutions.
• Express the amount of solute in a solution in various concentration units.
When one substance dissolves into another, a solution is formed. A solution is a homogenous mixture consisting of a solute dissolved into a solvent. The solute is the substance that is being dissolved, while the solvent is the dissolving medium. Solutions can be formed with many different types and forms of solutes and solvents. In this chapter, we will focus on solution where the solvent is water. An aqueous solution is water that contains one or more dissolved substance. The dissolved substances in an aqueous solution may be solids, gases, or other liquids.
In order to be a true solution, a mixture must be stable. When sugar is fully dissolved into water, it can stand for an indefinite amount of time, and the sugar will not settle out of the solution. Further, if the sugar-water solution is passed through a filter, it will remain with the water. This is because the dissolved particles in a solution are very small, usually less than $1 \: \text{nm}$ in diameter. Solute particles can be atoms, ions, or molecules, depending on the type of substance that has been dissolved. Water typically dissolves most ionic compounds and polar molecules. Nonpolar molecules, such as those found in grease or oil, do not dissolve in water.
Solution Concentrations
A concentrated solution is one in which there is a large amount of solute in a given amount of solvent. A dilute solution is one in which there is a small amount of solute in a given amount of solvent. A dilute solution is a concentrated solution that has been, in essence, watered down. Think of the frozen juice containers you buy in the grocery store. What you have to do is take the frozen juice from inside these containers and usually empty it into 3 or 4 times the container size full of water to mix with the juice concentrate and make your container of juice. Therefore, you are diluting the concentrated juice. When we talk about solute and solvent, the concentrated solution
has a lot of solute versus the dilute solution that would have a smaller amount of solute.
The terms "concentrated" and "dilute" provide qualitative methods of describing concentration. Although qualitative observations are necessary and have their place in every part of science, including chemistry, we have seen throughout our study of science that there is a definite need for quantitative measurements in science. This is particularly true in solution chemistry. In this section, we will explore some quantitative methods of expressing solution concentration.
To define a solution precisely, we need to state its concentration: how much solute is dissolved in a certain amount of solvent. Words such as dilute or concentrated are used to describe solutions that have a little or a lot of dissolved solute, respectively, but these are relative terms whose meanings depend on various factors. When we say that vinegar is $5\%$ acetic acid in water, we are giving the concentration. If we said the mixture was $10\%$ acetic acid, this would be more concentrated than the vinegar solution and we know exactly by how much.
Molarity
One way of expressing concentration is to give the number of moles of solute per unit volume of solution. Of all the quantitative measures of concentration, molarity is the one used most frequently by chemists. Molarity is defined as the number of moles of solute per liter of solution.
$\mathrm{molarity=\dfrac{number\: of\: moles\: of\: solute}{number\: of\: liters\: of\: solution}} \label{defMolarity}$
The symbol for molarity is $\text{M}$ or moles/liter. Chemists also use square brackets to indicate a reference to the molarity of a substance. For example, the expression $\left[ \ce{Ag^+} \right]$ refers to the molarity of the silver ion in solution. Solution concentrations expressed in molarity are the easiest to calculate with but the most difficult to make in the lab. Such concentration units are useful for discussing chemical reactions in which a solute is a product or a reactant. Molar mass can then be used as a conversion factor to convert amounts in moles to amounts in grams.
It is important to remember that “mol” in this expression refers to moles of solute and that “L” refers to liters of solution. For example, if you have 1.5 mol of NaCl dissolved in 0.500 L of solution, its molarity is therefore
$\mathrm{\dfrac{1.5\: mol\: NaCl}{0.500\: L\: solution}=3.0\: M\: NaCl} \nonumber$
Sometimes (aq) is added when the solvent is water, as in “3.0 M NaCl (aq).” You would read as "a 3.00 molar sodium chloride solution," meaning that there are 3.00 moles of NaOH dissolved per one liter of solution.
Be sure to note that molarity is calculated as the total volume of the entire solution, not just volume of solvent! The solute contributes to total volume.
If the quantity of the solute is given in mass units, you must convert mass units to mole units before using the definition of molarity to calculate concentration. For example, what is the molar concentration of a solution of 22.4 g of HCl dissolved in 1.56 L?
First, convert the mass of solute to moles using the molar mass of HCl (36.5 g/mol):
$22.4\cancel{gHCl}\times \dfrac{1\: mol\: HCl}{36.5\cancel{gHCl}}=0.614\, mol\; HCl \nonumber$
Now we can use the definition of molarity to determine a concentration:
$M \: =\: \dfrac{0.614\: mol\: HCl}{1.56L\: solution}=0.394\, M HCl \nonumber$
Before a molarity concentration can be calculated, the amount of the solute must be expressed in moles, and the volume of the solution must be expressed in liters, as demonstrated in the following example.
Example $1$
A solution is prepared by dissolving $42.23 \: \text{g}$ of $\ce{NH_4Cl}$ into enough water to make $500.0 \: \text{mL}$ of solution. Calculate its molarity.
Solution
Example $1$: Steps for Problem Solving Molarity
Steps for Problem Solving
Example
Identify the "given"information and what the problem is asking you to "find."
Given:
Mass $= 42.23 \: \text{g} \: \ce{NH_4Cl}$
Volume solution $= 500.0 \: \text{mL} = 0.5000 \: \text{L}$
Find: Molarity = ? M
List other known quantities. Molar mass $\ce{NH_4Cl} = 53.50 \: \text{g/mol}$
Plan the problem.
1. The mass of the ammonium chloride is first converted to moles.
2. Then the molarity is calculated by dividing by liters. Note the given volume has been converted to liters.
$\mathrm{M=\frac{mol\: NH_4Cl}{L\: solution}}$
Cancel units and calculate.
Now substitute the known quantities into the equation and solve.
\begin{align} 42.23 \ \cancel{\text{g} \: \ce{NH_4Cl}} \times \dfrac{1 \: \text{mol} \: \cancel{\ce{NH_4Cl}}}{53.50 \: \text{g} \: \ce{NH_4Cl}} &= 0.7893 \: \text{mol} \: \ce{NH_4Cl} \ \dfrac{0.7893 \: \text{mol} \: \ce{NH_4Cl}}{0.5000 \: \text{L solution}} &= 1.579 \: \text{M} \end{align}
Think about your result. The molarity is $1.579 \: \text{M}$, meaning that a liter of the solution would contain $1.579 \: \text{mol} \: \ce{NH_4Cl}$. Four significant figures are appropriate.
Exercise $1$
What is the molarity of a solution made when 66.2 g of C6H12O6 are dissolved to make 235 mL of solution?
Answer
1.57 M C6H12O6
Exercise $2$
What is the concentration, in $\text{mol/L}$, where $137 \: \text{g}$ of $\ce{NaCl}$ has been dissolved in enough water to make $500 \: \text{mL}$ of solution?
Answer
4.69 M NaCl
Figure $1$ illustrates how a solution is prepared with the measured amount of a solid solute and the desired final volume of the solution.
Percent Concentrations
There are several ways of expressing the concentration of a solution by using a percentage. The percent can further be determined in one of two ways: (1) the ratio of the mass of the solute divided by the mass of the solution or (2) the ratio of the volume of the solute divided by the volume of the solution.
Percent by Volume
The percentage of solute in a solution can more easily be determined by volume when the solute and solvent are both liquids. The volume of the solute divided by the volume of the solution expressed as a percent yields the percent by volume $\left( \frac{\text{volume}}{\text{volume}} \right)$ of the solution. If a solution is made by adding $40 \: \text{mL}$ of ethanol to $20 \: \text{mL}$ of water, the percent by volume is:
\begin{align} \text{Percent by volume} &= \frac{\text{volume of solute}}{\text{volume of solution}} \times 100\% \ &= \frac{40 \: \text{mL ethanol}}{240 \: \text{mL solution}} \times 100\% \ &= 16.7\% \: \text{ethanol} \end{align} \nonumber
Frequently, ingredient labels on food products and medicines have amounts listed as percentages (see figure below).
Percent by Mass
The mass/mass percent (% m/m) is defined as the mass of a solute divided by the mass of a solution times 100:
$\mathrm{\% \:m/m = \dfrac{mass\: of\: solute}{mass\: of\: solution}\times100\%} \nonumber$
mass of solution = mass of solute + mass solvent
If you can measure the masses of the solute and the solution, determining the mass/mass percent is easy. Each mass must be expressed in the same units to determine the proper concentration.
Suppose that a solution was prepared by dissolving $25.0 \: \text{g}$ of sugar into $100.0 \: \text{g}$ of water.
The mass of the solution is
mass of solution = 25.0g sugar + 100.0g water = 125.0 g
The percent by mass would be calculated by:
$\text{Percent by mass} = \frac{25.0 \: \text{g sugar}}{125.0 \: \text{g solution}} \times 100\% = 20.0\% \: \text{sugar} \nonumber$
Example $2$
A saline solution with a mass of 355 g has 36.5 g of NaCl dissolved in it. What is the mass/mass percent concentration of the solution?
Solution
We can substitute the quantities given in the equation for mass/mass percent:
$\mathrm{\%\: m/m=\dfrac{36.5\: g}{355\: g}\times100\%=10.3\%}$
Exercise $3$
A dextrose (also called D-glucose, C6H12O6) solution with a mass of 2.00 × 102 g has 15.8 g of dextrose dissolved in it. What is the mass/mass percent concentration of the solution?
Answer
7.90 %
Using Mass Percent in Calculations
Sometimes you may want to make up a particular mass of solution of a given percent by mass and need to calculate what mass of the solute to use. Using mass percent as a conversion can be useful in this type of problem. The mass percent can be expressed as a conversion factor in the form $\frac{g \; \rm{solute}}{100 \; \rm{g solution}}$ or $\frac{100 \; \rm g solution}{g\; \rm{solute}}$
For example, if you need to make $3000.0 \: \text{g}$ of a $5.00\%$ solution of sodium chloride, the mass of solute needs to be determined.
Solution
Given: 3000.0 g NaCl solution
5.00% NaCl solution
Find: mass of solute = ? g NaCl
Other known quantities: 5.00 g NaCl is to 100 g solution
To solve for the mass of NaCl, the given mass of solution is multiplied by the conversion factor.
$g NaCl = 3,000.0 \cancel{g \: NaCl \:solution} \times \frac{5.00 \:g \: NaCl}{100\cancel{g \: NaCl \: solution}} = 150.0g \: NaCl \nonumber$
You would need to weigh out $150 \: \text{g}$ of $\ce{NaCl}$ and add it to $2850 \: \text{g}$ of water. Notice that it was necessary to subtract the mass of the $\ce{NaCl}$ $\left( 150 \: \text{g} \right)$ from the mass of solution $\left( 3000 \: \text{g} \right)$ to calculate the mass of the water that would need to be added.
Exercise $4$
What is the amount (in g) of hydrogen peroxide (H2O2) needed to make a 6.00 kg , 3.00 % (by mass) H2O2 solution?
180. g H2O2
Summary
• Solutions are composed of a solvent (major component) and a solute (minor component).
• Concentration of solutions can be more precisely expressed in terms of molarity, percent by volume, and percent by mass.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky)
• Henry Agnew (UC Davis) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/05%3A_Chemical_Accounting/5.05%3A_Solutions.txt |
Thumbnail: A water drop. (CC BY 2.0; José Manuel Suárez via Wikipedia).
06: Gases Liquids Solids ... and Intermolecular Forces
Learning Objectives
• Describe the different states of matter.
• Describe the different changes that occur between solid, liquid, and gas.
Matter can exist in one of several different states, including a gas, liquid, or solid state. The amount of energy in molecules of matter determines the state of matter.
• A gas is a state of matter in which atoms or molecules have enough energy to move freely. The molecules come into contact with one another only when they randomly collide.
• A liquid is a state of matter in which atoms or molecules are constantly in contact but have enough energy to keep changing positions relative to one another.
• A solid is a state of matter in which atoms or molecules do not have enough energy to move. They are constantly in contact and in fixed positions relative to one another.
In Figure \(2\), all three phases are shown at the submicroscopic level in animations. Notice how the movement and freedom of molecules steadily increases as attractive forces decrease from solid to liquid to gas phase.
SOLID LIQUID GAS
Figure \(2\) Animation of all three phases at the submicroscopic level.
The change from solid to liquid usually does not significantly change the volume of a substance. However, the change from a liquid to a gas significantly increases the volume of a substance, by a factor of 1,000 or more. Figures \(3\) and \(4\) show the differences among solids, liquids, and gases at the molecular level, while Table \(1\) lists the different characteristics of these states.
Table \(1\) Characteristics of the Three States of Matter
Characteristic Solid Liquid Gas
shape definite indefinite indefinite
volume definite definite indefinite
relative intermolecular interaction strength strong moderate weak
relative particle positions in contact and fixed in place in contact but not fixed not in contact, random positions
Example \(1\)
What state or states of matter does each statement, describe?
1. This state has a definite volume.
2. This state has no definite shape.
3. This state allows the individual particles to move about while remaining in contact.
Solution
1. This statement describes either the liquid state or the solid state.
2. This statement describes either the liquid state or the gas state.
3. This statement describes the liquid state.
Exercise \(1\)
What state or states of matter does each statement describe?
1. This state has individual particles in a fixed position with regard to each other.
2. This state has individual particles far apart from each other in space.
3. This state has indefinite shape.
Answer a
solid
Answer b
gas
Answer c
liquid or gas
Phase Changes
A phase change is a physical process in which a substance goes from one phase to another. Usually the change occurs when adding or removing heat at a particular temperature, known as the melting point or the boiling point of the substance. The melting point is the temperature at which the substance goes from a solid to a liquid (or from a liquid to a solid). The boiling point is the temperature at which a substance goes from a liquid to a gas (or from a gas to a liquid). The nature of the phase change depends on the direction of the heat transfer. Heat going into a substance changes it from a solid to a liquid or a liquid to a gas. Removing heat from a substance changes a gas to a liquid or a liquid to a solid.
Two key points are worth emphasizing. First, at a substance’s melting point or boiling point, two phases can exist simultaneously. Take water (H2O) as an example. On the Celsius scale, H2O has a melting point of 0°C and a boiling point of 100°C. At 0°C, both the solid and liquid phases of H2O can coexist. However, if heat is added, some of the solid H2O will melt and turn into liquid H2O. If heat is removed, the opposite happens: some of the liquid H2O turns into solid H2O. A similar process can occur at 100°C: adding heat increases the amount of gaseous H2O, while removing heat increases the amount of liquid H2O (Figure \(3\)).
Water is a good substance to use as an example because many people are already familiar with it. Other substances have melting points and boiling points as well.
Second, as shown in Figure \(3\), the temperature of a substance does not change as the substance goes from one phase to another. In other words, phase changes are isothermal (isothermal means “constant temperature”). Again, consider H2O as an example. Solid water (ice) can exist at 0°C. If heat is added to ice at 0°C, some of the solid changes phase to make liquid, which is also at 0°C. Remember, the solid and liquid phases of H2O can coexist at 0°C. Only after all of the solid has melted into liquid does the addition of heat change the temperature of the substance.
Summary of Phase Changes
All of the changes of state that occur between solid, liquid, and gas are summarized in the diagram in the figure below. Freezing is the opposite of melting, and both represent the equilibrium between the solid and liquid states. Evaporation occurs when a liquid turns to a gas. Condensation is the opposite of vaporization, and both represent the equilibrium between the liquid and gas states. Deposition is the opposite of sublimation, and both represent the equilibrium between the solid and gas states.
Summary
• Solids, liquids, and gases are different phases that have their own unique properties.
• A change of state (or phase chnage) can be brought about by putting heat into a system or removing it from the system.
Contributors and Attributions
• Libretexts: The Basics of GOB (Ball et.al.)
• Henry Agnew (UC Davis) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/06%3A_Gases_Liquids_Solids_..._and_Intermolecular_Forces/6.01%3A_Solids_Liquids_and_Gases.txt |
Learning Objectives
• Know the physical properties of ionic and molecular substances.
The physical state and properties of a particular compound depend in large part on the type of chemical bonding it displays. Molecular compounds, sometimes called
covalent compounds, display a wide range of physical properties due to the different types of intermolecular attractions such as different kinds of polar interactions. The melting and boiling points of molecular compounds are generally quite low compared to those of ionic compounds. This is because the energy required to disrupt the intermolecular forces between molecules is far less than the energy required to break the ionic bonds in a crystalline ionic compound (Figure $1$) . Ionic solids typically melt at high temperatures and boil at even higher temperatures. For example, sodium chloride melts at 801 °C and boils at 1413 °C. (As a comparison, the molecular compound water melts at 0 °C and boils at 100 °C.). The water solubility of molecular compounds is variable and depends primarily on the type of intermolecular forces involved.
Figure $1$ Interactions in Ionic and Covalent Solids.
Since molecular compounds are composed of neutral molecules, their electrical conductivity is generally quite poor, whether in the solid or liquid state. In solid form, an ionic compound is not electrically conductive because its ions are unable to flow (“electricity” is the flow of charged particles). When molten, however, it can conduct electricity because its ions are able to move freely through the liquid (Figure $2$; Video $1$).
Conductivity of Molten Salt
Video $1$ Watch this video to see a mixture of salts melt and conduct electricity.
The table below summarizes some of the differences between ionic and molecular compounds.
Table $1$ Comparison of Ionic and Molecular Compounds
Property Ionic Compounds Molecular Compounds
Type of elements Metal and nonmetal Nonmetals only
Bonding Ionic - transfer of electron(s) between atoms Covalent - sharing of pair(s) of electrons between atoms
Representative unit Formula unit Molecule
Physical state at room temperature Solid Gas, liquid, or solid
Water solubility Usually high Variable
Melting and boiling temperatures Generally high Generally low
Electrical conductivity Good when molten or in solution Poor
One type of molecular compound behaves quite differently than that described so far. A covalent network solid is a compound in which all of the atoms are connected to one another by covalent bonds. Diamond is composed entirely of carbon atoms, each bonded to four other carbon atoms in a tetrahedral geometry. Melting a covalent network solid is not accomplished by overcoming the relatively weak intermolecular forces. Rather, all of the covalent bonds must be broken, a process that requires extremely high temperatures. Diamond, in fact, does not melt at all. Instead, it vaporizes to a gas at temperatures above $3500^\text{o} \text{C}$.
Summary
• The physical properties of a material are affected by the intermolecular forces holding the molecules together.
• Ionic compounds usually form hard crystalline solids with high melting points. Covalent molecular compounds, in contrast, consist of discrete molecules held together by weak intermolecular forces and can be gases, liquids, or solids at room temperature and pressure.
• Ionic compounds in molten form or in solution can conduct electricity while molecular compounds do not.. | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/06%3A_Gases_Liquids_Solids_..._and_Intermolecular_Forces/6.02%3A_Comparing_Ionic_and_Molecular_Substances.txt |
Learning Objectives
• Describe the types of intermolecular forces possible between atoms or molecules in condensed phases (dispersion forces, dipole-dipole attractions, and hydrogen bonding)
• Identify the types of intermolecular forces experienced by specific molecules based on their structures
Under appropriate conditions, the attractions between all gas molecules will cause them to form liquids or solids. This is due to intermolecular forces, not intramolecular forces. Intramolecular forces are those within the molecule that keep the molecule together, for example, the bonds between the atoms. Intermolecular forces are the attractions between molecules, which determine many of the physical properties of a substance. Figure \(1\) illustrates these different molecular forces. The strengths of these attractive forces vary widely, though usually the IMFs between small molecules are weak compared to the intramolecular forces that bond atoms together within a molecule. For example, to overcome the IMFs in one mole of liquid HCl and convert it into gaseous HCl requires only about 17 kilojoules. However, to break the covalent bonds between the hydrogen and chlorine atoms in one mole of HCl requires about 25 times more energy—430 kilojoules.
Dipole–Dipole Forces
Dipole-dipole forces are the attractive forces that occur between polar molecules (see figure below). A molecule of hydrogen chloride has a partially positive hydrogen atom and a partially negative chlorine atom. A collection of many hydrogen chloride molecules will align themselves so that the oppositely charged regions of neighboring molecules are near each other.
The strength of the intermolecular interactions increases as the dipole moment of the molecules increases, as shown in Table \(1\).
Table \(1\): Relationships between the Dipole Moment and the Boiling Point for Organic Compounds of Similar Molar Mass
Compound Molar Mass (g/mol) Dipole Moment (D) Boiling Point (K)
C3H6 (cyclopropane) 42 0 240
CH3OCH3 (dimethyl ether) 46 1.30 248
CH3CN (acetonitrile) 41 3.9 355
The attractive energy between two ions is proportional to 1/r, whereas the attractive energy between two dipoles is proportional to 1/r6.
Example \(1\) Dipole-Dipole Forces and Their Effects
Predict which will have the higher boiling point: N2 or CO. Explain your reasoning.
Solution
CO and N2 are both diatomic molecules with masses of about 28 amu, so they experience similar London dispersion forces. Because CO is a polar molecule, it experiences dipole-dipole attractions. Because N2 is nonpolar, its molecules cannot exhibit dipole-dipole attractions. The dipole-dipole attractions between CO molecules are comparably stronger than the dispersion forces between nonpolar N2 molecules, so CO is expected to have the higher boiling point.
A common method for preparing oxygen is the decomposition
Exercise \(1\)
Predict which will have the higher boiling point: \(\ce{ICl}\) or \(\ce{Br2}\). Explain your reasoning.
Answer
ICl. ICl and Br2 have similar masses (~160 amu) and therefore experience similar London dispersion forces. ICl is polar and thus also exhibits dipole-dipole attractions; Br2 is nonpolar and does not. The relatively stronger dipole-dipole attractions require more energy to overcome, so ICl will have the higher boiling point
Dispersion Forces
Dispersion forces are the weakest of all intermolecular forces. They are often called London forces after Fritz London (1900 - 1954), who first proposed their existence in 1930. London dispersion forces are intermolecular forces that occur between all atoms and molecules due to the random motion of electrons.
For example, the electron cloud of a helium atom contains two electrons, and, when averaged over time, these electrons will distribute themselves evenly around the nucleus. However, at any given moment, the electron distribution may be uneven, resulting in an instantaneous dipole. This weak and temporary dipole can subsequently influence neighboring helium atoms through electrostatic attraction and repulsion. The formation of an induced dipole is illustrated below.
Dispersion forces that develop between atoms in different molecules can attract the two molecules to each other. The forces are relatively weak, however, and become significant only when the molecules are very close. Larger and heavier atoms and molecules exhibit stronger dispersion forces than do smaller and lighter atoms and molecules. F2 and Cl2 are gases at room temperature (reflecting weaker attractive forces); Br2 is a liquid, and I2 is a solid (reflecting stronger attractive forces). Trends in observed melting and boiling points for the halogens clearly demonstrate this effect, as seen in Table \(1\).
Table \(2\): Melting and Boiling Points of the Halogens
Halogen Molar Mass Atomic Radius Melting Point Boiling Point
fluorine, F2 38 g/mol 72 pm 53 K 85 K
chlorine, Cl2 71 g/mol 99 pm 172 K 238 K
bromine, Br2 160 g/mol 114 pm 266 K 332 K
iodine, I2 254 g/mol 133 pm 387 K 457 K
astatine, At2 420 g/mol 150 pm 575 K 610 K
The increase in melting and boiling points with increasing atomic/molecular size may be rationalized by considering how the strength of dispersion forces is affected by the electronic structure of the atoms or molecules in the substance. In a larger atom, the valence electrons are, on average, farther from the nuclei than in a smaller atom. Thus, they are less tightly held and can more easily form the temporary dipoles that produce the attraction. The measure of how easy or difficult it is for another electrostatic charge (for example, a nearby ion or polar molecule) to distort a molecule’s charge distribution (its electron cloud) is known as polarizability. A molecule that has a charge cloud that is easily distorted is said to be very polarizable and will have large dispersion forces; one with a charge cloud that is difficult to distort is not very polarizable and will have small dispersion forces.
For similar substances, London dispersion forces get stronger with increasing atomic or molecular size.
The strengths of London dispersion forces also depend significantly on molecular shape because shape determines how much of one molecule can interact with its neighboring molecules at any given time. For example, part (b) in Figure \(4\) shows 2,2-dimethylpropane (neopentane) and n-pentane, both of which have the empirical formula C5H12. Neopentane is almost spherical, with a small surface area for intermolecular interactions, whereas n-pentane has an extended conformation that enables it to come into close contact with other n-pentane molecules. As a result, the boiling point of neopentane (9.5°C) is more than 25°C lower than the boiling point of n-pentane (36.1°C).
London dispersion forces
All molecules, whether polar or nonpolar, are attracted to one another by London dispersion forces in addition to any other attractive forces that may be present.
Example \(2\) London Forces and Their Effects
Order the following compounds of a group 14 element and hydrogen from lowest to highest boiling point: CH4, SiH4, GeH4, and SnH4. Explain your reasoning.
Given: compounds
Asked for: order of increasing boiling points
Strategyy:
Determine the intermolecular forces in the compound. Also, compare the molar masses and surface area of compounds with similar types of intermolecular force. The substance with the weakest forces and the lowest molar masses and least surface area have the lowest boiling point.
Solution
Applying the skills acquired in the chapter on chemical bonding and molecular geometry, all of these compounds are predicted to be nonpolar, so they may experience only dispersion forces: the smaller the molecule, the less polarizable and the weaker the dispersion forces; the larger the molecule, the larger the dispersion forces. The molar masses of CH4, SiH4, GeH4, and SnH4 are approximately 16 g/mol, 32 g/mol, 77 g/mol, and 123 g/mol, respectively. Therefore, CH4 is expected to have the lowest boiling point and SnH4 the highest boiling point. The ordering from lowest to highest boiling point is expected to be
CH4 < SiH4 < GeH4 < SnH4
A graph of the actual boiling points of these compounds versus the period of the group 14 elements shows this prediction to be correct:
Exercise \(2\)
Order the following hydrocarbons from lowest to highest boiling point: C2H6, C3H8, and C4H10.
Answer
All of these compounds are nonpolar and only have London dispersion forces: the larger the molecule, the larger the dispersion forces and the higher the boiling point. The ordering from lowest to highest boiling point is therefore
C2H6 < C3H8 < C4H10.
Hydrogen Bonds
The attractive force between water molecules is an unusually strong type of dipole-dipole interaction. Water contains hydrogen atoms that are bound to a highly electronegative oxygen atom, making for very polar bonds. The partially positive hydrogen atom of one molecule is then attracted to the oxygen atom of a nearby water molecule (see figure below).
A hydrogen bond is an intermolecular attractive force in which a hydrogen atom, that is covalently bonded to a small, highly electronegative atom, is attracted to a lone pair of electrons on an atom in a neighboring molecule. Hydrogen bonds are very strong compared to other dipole-dipole interactions, but still much weaker than a covalent bond. A typical hydrogen bond is about \(5\%\) as strong as a covalent bond.
Hydrogen bonding occurs only in molecules where hydrogen is covalently bonded to one of three elements: fluorine, oxygen, or nitrogen. These three elements are so electronegative that they withdraw the majority of the electron density from the covalent bond with hydrogen, leaving the \(\ce{H}\) atom very electron-deficient. Because the hydrogen atom does not have any electrons other than the ones in the covalent bond, its positively charged nucleus is almost completely exposed, allowing strong attractions to other nearby lone pairs of electrons.
The hydrogen bonding that occurs in water leads to some unusual, but very important properties. Most molecular compounds that have a mass similar to water are gases at room temperature. However, because of the strong hydrogen bonds, water molecules are able to stay condensed in the liquid state. The figure below shows how its bent shape and the presence of two hydrogen atoms per molecule allows each water molecule to hydrogen bond with several other molecules.
In the liquid state, the hydrogen bonds of water can break and reform as the molecules flow from one place to another. When water is cooled, the molecules begin to slow down. Eventually, when water is frozen to ice, the hydrogen bonds become more rigid and form a well-defined network (Figure \(7\)). The bent shape of the molecules leads to gaps in the hydrogen bonding network of ice. Ice has the very unusual property that its solid state is less dense than its liquid state. As a result, ice floats in liquid water. Virtually all other substances are denser in the solid state than in the liquid state. Hydrogen bonds also play a very important biological role in the physical structures of proteins and nucleic acids.
Hydrogen bond formation requires both a hydrogen bond donor and a hydrogen bond acceptor.
Because ice is less dense than liquid water, rivers, lakes, and oceans freeze from the top down. In fact, the ice forms a protective surface layer that insulates the rest of the water, allowing fish and other organisms to survive in the lower levels of a frozen lake or sea. If ice were denser than the liquid, the ice formed at the surface in cold weather would sink as fast as it formed. Bodies of water would freeze from the bottom up, which would be lethal for most aquatic creatures. The expansion of water when freezing also explains why automobile or boat engines must be protected by “antifreeze” and why unprotected pipes in houses break if they are allowed to freeze.
Strength of Intermolecular Forces
Molecules with hydrogen atoms bonded to electronegative atoms such as O, N, and F tend to exhibit unusually strong intermolecular interactions due to hydrogen bonds as illustrated for the covalent hydrides of elements of groups 14–17 in Figure \(8\). These result in much higher boiling points than are observed for substances in which dipole-dipole forces or London dispersion forces dominate. All the polar molecules (with predominantly dipole-dipole forces) in group 16 (H2S to H2Te) have lower boiling points than H2O (with hydrogen bonds). Methane and its heavier congeners in group 14 form a series whose boiling points increase smoothly with increasing molar mass. This is the expected trend in nonpolar molecules, for which London dispersion forces are the exclusive intermolecular forces. In contrast, the hydrides of the lightest members of groups 15–17 have boiling points that are more than 100°C greater than predicted on the basis of their molar masses. The effect is most dramatic for water: if we extend the straight line connecting the points for H2Te and H2Se to the line for period 2, we obtain an estimated boiling point of −130°C for water! Imagine the implications for life on Earth if water boiled at −130°C rather than 100°C.
Example \(3\)
Considering CH3OH, C2H6, Xe, and (CH3)3N, which can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures.
Given: compounds
Asked for: formation of hydrogen bonds and structure
Strategy:
1. Identify the compounds with a hydrogen atom attached to O, N, or F. These are likely to be able to act as hydrogen bond donors.
2. Of the compounds that can act as hydrogen bond donors, identify those that also contain lone pairs of electrons, which allow them to be hydrogen bond acceptors. If a substance is both a hydrogen donor and a hydrogen bond acceptor, draw a structure showing the hydrogen bonding.
Solution:
A Of the species listed, xenon (Xe), ethane (C2H6), and trimethylamine [(CH3)3N] do not contain a hydrogen atom attached to O, N, or F; hence they cannot act as hydrogen bond donors.
B The one compound that can act as a hydrogen bond donor, methanol (CH3OH), contains both a hydrogen atom attached to O (making it a hydrogen bond donor) and two lone pairs of electrons on O (making it a hydrogen bond acceptor); methanol can thus form hydrogen bonds by acting as either a hydrogen bond donor or a hydrogen bond acceptor. The hydrogen-bonded structure of methanol is as follows:
Exercise \(3\)
Considering CH3CO2H, (CH3)3N, NH3, and CH3F, which can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures.
Answer
CH3CO2H and NH3;
Although hydrogen bonds are significantly weaker than covalent bonds, with typical dissociation energies of only 15–25 kJ/mol, they have a significant influence on the physical properties of a compound. Compounds such as HF can form only two hydrogen bonds at a time as can, on average, pure liquid NH3. Consequently, even though their molecular masses are similar to that of water, their boiling points are significantly lower than the boiling point of water, which forms four hydrogen bonds at a time.
Example \(4\): Buckyballs
Arrange C60 (buckminsterfullerene, which has a cage structure), NaCl, He, Ar, and N2O in order of increasing boiling points.
Given: compounds
Asked for: order of increasing boiling points
Strategy:
Identify the intermolecular forces in each compound and then arrange the compounds according to the strength of those forces. The substance with the weakest forces will have the lowest boiling point.
Solution:
Electrostatic interactions are strongest for an ionic compound, so we expect NaCl to have the highest boiling point. To predict the relative boiling points of the other compounds, we must consider their polarity (for dipole–dipole interactions), their ability to form hydrogen bonds, and their molar mass (for London dispersion forces). Helium is nonpolar and by far the lightest, so it should have the lowest boiling point. Argon and N2O have very similar molar masses (40 and 44 g/mol, respectively), but N2O is polar while Ar is not. Consequently, N2O should have a higher boiling point. A C60 molecule is nonpolar, but its molar mass is 720 g/mol, much greater than that of Ar or N2O. Because the boiling points of nonpolar substances increase rapidly with molecular mass, C60 should boil at a higher temperature than the other nonionic substances. The predicted order is thus as follows, with actual boiling points in parentheses:
He (−269°C) < Ar (−185.7°C) < N2O (−88.5°C) < C60 (>280°C) < NaCl (1465°C).
Exercise \(4\)
Arrange 2,4-dimethylheptane, Ne, CS2, Cl2, and KBr in order of decreasing boiling points.
Answer
KBr (1435°C) > 2,4-dimethylheptane (132.9°C) > CS2 (46.6°C) > Cl2 (−34.6°C) > Ne (−246°C)
Example \(5\):
Identify the most significant intermolecular force in each substance.
1. C3H8
2. CH3OH
3. H2S
Solution
a. Although C–H bonds are polar, they are only minimally polar. The most significant intermolecular force for this substance would be dispersion forces.
b. This molecule has an H atom bonded to an O atom, so it will experience hydrogen bonding.
c. Although this molecule does not experience hydrogen bonding, the Lewis electron dot diagram and VSEPR indicate that it is bent, so it has a permanent dipole. The most significant force in this substance is dipole-dipole interaction.
Exercise \(5\)
Identify the most significant intermolecular force in each substance.
1. HF
2. HCl
Answer a
hydrogen bonding
Answer b
dipole-dipole interactions
Summary
• Molecules in liquids are held to other molecules by intermolecular interactions, which are weaker than the intramolecular interactions that hold the atoms together within molecules and polyatomic ions.
• Dipole–dipole interactions arise from the electrostatic interactions of the positive and negative ends of molecules with permanent dipole moments.
• London dispersion forces are due to the formation of instantaneous dipole moments in polar and nonpolar molecules as a result of short-lived fluctuations of electron charge distribution, which in turn cause the temporary formation of an induced dipole in adjacent molecules.
• Hydrogen bonds are especially strong dipole–dipole interactions between molecules that have hydrogen bonded to a highly electronegative atom, such as O, N, or F. The resulting partially positively charged H atom on one molecule (the hydrogen bond donor) can interact strongly with a lone pair of electrons of a partially negatively charged O, N, or F atom on adjacent molecules (the hydrogen bond acceptor).
Contributors and Attributions
• TextMap: Chemistry the Central Science (Brown et al.
• Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected]).
• Libretext: Chemistry for Allied Health (Soult) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/06%3A_Gases_Liquids_Solids_..._and_Intermolecular_Forces/6.03%3A_Forces_between_Molecules.txt |
Learning Objectives
• Learn some terminology involving solutions.
• Explain the significance of the statement "like dissolves like".
• Explain why certain substances dissolve in other substances.
A solution is another name for a homogeneous mixture. A mixture as a material composed of two or more substances. In a solution, the combination is so intimate that the
different substances cannot be differentiated by sight, even with a microscope. Compare, for example, a mixture of salt and pepper and another mixture consisting of salt and water. In the first mixture, we can readily see individual grains of salt and the flecks of pepper. A mixture of salt and pepper is not a solution. However, in the second mixture, no matter how carefully we look, we cannot see two different substances. Salt dissolved in water is a solution.
The major component of a solution, called the solvent, is typically the same phase as the solution itself. Each minor component of a solution (and there may be more than one) is called the solute. In most of the solutions we will describe in this textbook, there will be no ambiguity about whether a component is the solvent or the solute.) For example, in a solution of salt in water, the solute is salt, and solvent is water Figure $1$.
Solutions come in all phases, and the solvent and the solute do not have to be in the same phase to form a solution (such as salt and water). For example, air is a gaseous solution of about 80% nitrogen and about 20% oxygen, with some other gases present in much smaller amounts. An alloy is a solid solution consisting of a metal (like iron) with some other metals or nonmetals dissolved in it. Steel, an alloy of iron and carbon and small amounts of other metals, is an example of a solid solution. Table $1$ lists some common types of solutions, with examples of each.
Table $1$ Types of Solutions
Solvent Phase Solute Phase Example
gas gas air
liquid gas carbonated beverages
liquid liquid ethanol (C2H5OH) in H2O (alcoholic beverages)
liquid solid salt water
solid gas H2 gas absorbed by Pd metal
solid liquid Hg(ℓ) in dental fillings
solid solid steel alloys
Example $1$: Sugar and Water
A solution is made by dissolving 1.00 g of sucrose ($\ce{C12H22O11}$) in 100.0 g of liquid water. Identify the solvent and solute in the resulting solution.
Solution
Either by mass or by moles, the obvious minor component is sucrose, so it is the solute. Water—the majority component—is the solvent. The fact that the resulting solution is the same phase as water also suggests that water is the solvent.
Exercise $1$
A solution is made by dissolving 3.33 g of $\ce{HCl(g)}$ in 40.0 g of liquid methyl alcohol ($\ce{CH3OH}$). Identify the solvent and solute in the resulting solution.
Answer
solute: HCl(g); solvent: CH3OH
Like Dissolves Like
A simple way to predict which compounds will dissolve in other compounds is the phrase "like dissolves like". What this means is that polar compounds dissolve polar compounds, nonpolar compounds dissolve nonpolar compounds, but polar and nonpolar do not dissolve in each other.
Even some nonpolar substances dissolve in water but only to a limited degree. Have you ever wondered why fish are able to breathe? Oxygen gas, an nonpolar molecules, does dissolve in water and it is this oxygen that the fish take in through their gills. Or, one more example of a nonpolar compound that dissolves in water is the reason we can enjoy carbonated sodas. Pepsi-cola and all the other sodas have carbon dioxide gas, $\ce{CO_2}$, a nonpolar compound, dissolved in a sugar-water solution. In this case, to keep as much gas in solution as possible, the sodas are kept under pressure.
This general trend of "like dissolves like" is summarized in the following table:
Table $2$ Summary of Solubilities
Solute Solvent Is Solution Formed?
Polar Covalent Polar yes
Non-polar Covalent Non-polar yes
Polar Covalent Non-polar no
Non-polar Covalent Polar no
Ionic Polar yes
Ionic Non-polar no
Note that every time charged particles (ionic compounds or polar substances) are mixed, a solution is formed. When particles with no charges (nonpolar compounds) are mixed, they will form a solution. However, if substances with charges are mixed with other substances without charges a solution does not form.
When an ionic compound is considered "insoluble", it doesn't necessarily mean the compound is completely untouched by water. All ionic compounds dissolve to some extent. An insoluble compound just doesn't dissolve in any noticeable or appreciable amount.
What is it that makes a solute soluble in some solvents but not others?
The answer is intermolecular interactions. The intermolecular interactions include London dispersion forces, dipole-dipole interactions, and hydrogen bonding (as described in
the previous section). From experimental studies, it has been determined that if molecules of a solute experience the same intermolecular forces that the solvent does, the solute will likely dissolve in that solvent. So, NaCl—a very polar substance because it is composed of ions—dissolves in water, which is very polar, but not in oil, which is generally nonpolar. Nonpolar wax dissolves in nonpolar hexane, but not in polar water. Liquids that dissolve in one another in all proportions are said to be miscible. Liquids that do not dissolve in one another are called immiscible.
Figure $2$ Water (clear liquid) and oil (yellow) do not form liquid solutions. (CC BY-SA 1.0 Generic; Victor Blacus)
Example $2$: Polar and Nonpolar Solvents
Would I2 be more soluble in CCl4 or H2O? Explain your answer.
Solution
I2 is nonpolar. Of the two solvents, CCl4 is nonpolar and H2O is polar, so I2 would be expected to be more soluble in CCl4.
Exercise $2$
Would C3H7OH be more soluble in CCl4 or H2O? Explain your answer.
Answer
H2O because both experience hydrogen bonding
Example $3$
Water is considered a polar solvent. Which substances should dissolve in water?
1. methanol (CH3OH)
2. sodium sulfate (Na2SO4)
3. octane (C8H18)
Solution
Because water is polar, substances that are polar or ionic will dissolve in it.
1. Because of the OH group in methanol, we expect its molecules to be polar. Thus, we expect it to be soluble in water. As both water and methanol are liquids, the word miscible can be used in place of soluble.
2. Sodium sulfate is an ionic compound, so we expect it to be soluble in water.
3. Like other hydrocarbons, octane is nonpolar, so we expect that it would not be soluble in water.
Exercise $3$
Toluene (C6H5CH3) is widely used in industry as a nonpolar solvent. Which substances should dissolve in toluene?
1. water (H2O)
2. sodium sulfate (Na2SO4)
3. octane (C8H18)
Answer
octane (C8H18) will dissolve. It is also non-polar.
The Dissolving Process
Water typically dissolves most ionic compounds and polar molecules. Nonpolar molecules, such as those found in grease or oil, do not dissolve in water. We will first examine the process that occurs when an ionic compound, such as table salt (sodium chloride), dissolves in water.
Water molecules move about continuously due to their kinetic energy. When a crystal of sodium chloride is placed into water, the water's molecules collide with the crystal lattice. Recall that the crystal lattice is composed of alternating positive and negative ions. Water is attracted to the sodium chloride crystal because water is polar; it has both a positive and a negative end. The positively charged sodium ions in the crystal attract the oxygen end of the water molecules because they are partially negative. The negatively charged chloride ions in the crystal attract the hydrogen end of the water molecules because they are partially positive. The action of the polar water molecules takes the crystal lattice apart (see figure below).
After coming apart from the crystal, the individual ions are then surrounded by solvent particles in a process called solvation. Note in the figure above that the individual $\ce{Na^+}$ ions are surrounded by water molecules with the oxygen atom oriented near the positive ion. Likewise, the chloride ions are surrounded by water molecules with the opposite orientation. Thus, numerous ion-dipole interactions are formed. Hydration is the process of solute particles being surrounded by water molecules arranged in a specific manner. Hydration helps to stabilize aqueous solutions by preventing the positive and negative ions from coming back together and forming a precipitate.
Table sugar is made of the molecular compound sucrose $\left( \ce{C_{12}H_{22}O_{11}} \right)$. Solid sugar consists of individual sugar molecules held together by intermolecular attractive forces. When water dissolves sugar, it separates the individual sugar molecules by disrupting the attractive forces, but it does not break the covalent bonds between the carbon, hydrogen, and oxygen atoms. Dissolved sugar molecules are also hydrated. The hydration shell around a molecule of sucrose is arranged so that its partially negative oxygen atoms are near the partially positive hydrogen atoms in the solvent, and vice versa.
Summary
• Solutions are composed of a solvent (major component) and a solute (minor component).
• “Like dissolves like” is a useful rule for deciding if a solute will be soluble in a solvent.
• Liquids that dissolve in one another in all proportions are said to be miscible.
• Liquids that do not dissolve in one another are called immiscible.
• Ion-dipole interactions are formed when ionic compounds dissolve in water.
Contributors and Attributions
• Henry Agnew (UC Davis)
• Libretext: Chemistry for Allied Health (Soult)
• Libretext: The Basics of GOB Chemistry (Ball et al.)
6.05: Gases- The Kinetic-Molecular Theory
Learning Objectives
• State the major concepts behind the kinetic theory of gases.
• Relate the general properties of gases to the kinetic theory.
Gases were among the first substances studied in terms of the modern scientific method, which was developed in the 1600s. It did not take long to recognize that gases all shared certain physical behaviors, suggesting that all gases could be described by one all-encompassing theory. Today, that theory is the kinetic theory of gases. It is based on the following statements:
1. Gases consist of tiny particles of matter that are in constant motion.
2. Gas particles are constantly colliding with each other and the walls of a container. These collisions are elastic; that is, there is no net loss of energy from the collisions.
3. Gas particles are separated by large distances, with the size of a gas particle tiny compared to the distances that separate them.
4. There are no interactive forces (i.e., attraction or repulsion) between the particles of a gas.
5. The average speed of gas particles is dependent on the temperature of the gas.
Figure \(1\) shows a representation of how we mentally picture the gas phase.
This model of gases explains some of the physical properties of gases. Because most of a gas is empty space, a gas has a low density and can expand or contract under the appropriate influence. The fact that gas particles are in constant motion means that two or more gases will always mix, as the particles from the individual gases move and collide with each other.
An ideal gas is a gas that exactly follows the statements of the kinetic theory. Unfortunately, real gases are not ideal. Many gases deviate slightly from agreeing perfectly with the kinetic theory of gases. However, most gases adhere to the statements so well that the kinetic theory of gases is well accepted by the scientific community.
• The physical behavior of gases is explained by the kinetic theory of gases.
• An ideal gas adheres exactly to the kinetic theory of gases.
Contributors and Attributions
• Henry Agnew (UC Davis) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/06%3A_Gases_Liquids_Solids_..._and_Intermolecular_Forces/6.04%3A_Forces_in_Solutions.txt |
Learning Objectives
• Learn what is meant by the term gas laws.
• Know the different units of pressure.
• Learn and apply Boyle’s law.
• Learn and apply Charles's law.
• Learn and apply Gay-Lussac's law.
The behavior of gases can be modeled with gas laws. Boyle’s law relates a gas’s pressure and volume at constant temperature and amount. Charles’s law relates a gas’s volume and temperature at constant pressure and amount. In gas laws, temperatures must always be expressed in Kelvins.
Pressure Units
A barometer measures gas pressure by the height of the column of mercury. One unit of gas pressure is the millimeter of mercury $\left( \text{mm} \: \ce{Hg} \right)$. An equivalent unit to the $\text{mm} \: \ce{Hg}$ is called the $\text{torr}$, in honor of the inventor of the barometer, Evangelista Torricelli. The pascal $\left( \text{Pa} \right)$ is the standard unit of pressure. A pascal is a very small amount of pressure, so the most useful unit for everyday gas pressures is the kilopascal $\left( \text{kPa} \right)$. A kilopascal is equal to 1000 pascals. Another commonly used unit of pressure is the atmosphere $\left( \text{atm} \right)$. Standard atmospheric pressure is called $1 \: \text{atm}$ of pressure and is equal to $760 \: \text{mm} \: \ce{Hg}$ and $101.3 \: \text{kPa}$. Atmospheric pressure is also often stated as pounds per square inch $\left( \text{psi} \right)$. The atmospheric pressure at sea level is $14.7 \: \text{psi}$.
$1 \: \text{atm} = 760 \: \text{mm} \: \ce{Hg} = 760 \: \text{torr} = 101.3 \: \text{kPa} = 14.7 \: \text{psi} \nonumber$
Boyle's Law: Pressure and Volume
When seventeenth-century scientists began studying the physical properties of gases, they noticed some simple relationships between some of the measurable properties of the gas. Take pressure (P) and volume (V), for example. Scientists noted that for a given amount of a gas (usually expressed in units of moles [n]), if the temperature (T) of the gas was kept constant, pressure and volume were related: As one increases, the other decreases. As one decreases, the other increases. We say that pressure and volume are inversely related.
There is more to it, however: pressure and volume of a given amount of gas at constant temperature are numerically related. If you take the pressure value and multiply it by the volume value, the product is a constant for a given amount of gas at a constant temperature:
$P × V = \text{ constant at constant n and T} \nonumber$
If either volume or pressure changes while amount and temperature stay the same, then the other property must change so that the product of the two properties still equals that same constant. That is, if the original conditions are labeled $P_1$ and $V_1$ and the new conditions are labeled $P_2$ and $V_2$, we have
$P_1V_1 = \text{constant} = P_2V_2 \nonumber$
where the properties are assumed to be multiplied together. Leaving out the middle part, we have simply
$P_1V_1 = P_2V_2 \text{ at constant n and T} \nonumber$
This equation is an example of a gas law. A gas law is a simple mathematical formula that allows you to model, or predict, the behavior of a gas. This particular gas law ia called Boyle's law, after the English scientist Robert Boyle, who first announced it in 1662. Figure $1$ shows two representations of how Boyle’s law works.
Boyle’s law is an example of a second type of mathematical problem we see in chemistry—one based on a mathematical formula. Tactics for working with mathematical formulas are different from tactics for working with conversion factors. First, most of the questions you will have to answer using formulas are word-type questions, so the first step is to identify what quantities are known and assign them to variables. Second, in most formulas, some mathematical rearrangements (i.e., algebra) must be performed to solve for an unknown variable. The rule is that to find the value of the unknown variable, you must mathematically isolate the unknown variable by itself and in the numerator of one side of the equation. Finally, units must be consistent. For example, in Boyle’s law there are two pressure variables; they must have the same unit. There are also two volume variables; they also must have the same unit. In most cases, it won’t matter what the unit is, but the unit must be the same on both sides of the equation.
Example $1$
A sample of gas has an initial pressure of 2.44 atm and an initial volume of 4.01 L. Its pressure changes to 1.93 atm. What is the new volume if temperature and amount are kept constant?
Solution
Example $1$: Steps for Problem Solving volume and pressure changes.
Steps for Problem Solving
Example $1$
Identify the "given"information and what the problem is asking you to "find."
Given: P1 = 2.44 atm and V1 = 4.01 L
P2 = 1.93 atm
Find: V2 = ? L
List other known quantities. none
Plan the problem.
First, rearrange the equation algebraically to solve for $V_2$.
$V_2 = \frac{P_1 \times V_1}{P_2} \nonumber$
Cancel units and calculate.
Now substitute the known quantities into the equation and solve.
$V_2 = \frac{2.44 \: \cancel{\text{atm}} \times 4.01 \: \text{L}}{1.93 \: \cancel{atm}} = 5.07 \: \text{L} \nonumber$
Think about your result. We know that pressure and volume are inversely related; as one decreases, the other increases. Pressure is decreasing (from 2.44 atm to 1.93 atm), so volume should be increasing to compensate, and it is (from 4.01 L to 5.07 L). So the answer makes sense based on Boyle’s law.
Exercise $1$
If P1 = 334 torr, V1 = 37.8 mL, and P2 = 102 torr, what is V2?
Answer
124 mL
As mentioned, you can use any units for pressure or volume, but both pressures must be expressed in the same units, and both volumes must be expressed in the same units.
Example $2$:
A sample of gas has an initial pressure of 722 torr and an initial volume of 88.8 mL. Its volume changes to 0.663 L. What is the new pressure?
Solution
Example $2$:Steps for Problem Solving volume and pressure changes
Steps for Problem Solving
Example $2$
Identify the "given"information and what the problem is asking you to "find."
Given: P1 = 722 torr and V1 = 88.8 mL
V2 = 0.633 L
Find: P2 = ? torr
List other known quantities. 1 L = 1000 mL to have the same units for volume.
Plan the problem.
1. Perform the conversion of the second volume unit from L to mL.
2. Rearrange the equation algebraically to solve for $P_2$.
$P_2 = \frac{P_1 \times V_1}{V_2} \nonumber$
Cancel units and calculate.
1. $0.663\, \cancel{L}\times \frac{1000\, ml}{1\, \cancel{L}}=663\, ml \nonumber$
2. Substitute the known quantities into the equation and solve. $P_2 = \frac{722 \: \text{torr} \times 88.8 \: \cancel{\text{mL}}}{663 \: \cancel{\text{mL}}} = 96.7 \: \text{torr} \nonumber$
Think about your result. When the volume increased, the pressure decreased, which is as expected for Boyle’s law.
Exercise $2$
If V1 = 456 mL, P1 = 308 torr, and P2 = 1.55 atm, what is V2?
Answer
119 mL
Charles's Law: Temperature and Volume
French physicist Jacques Charles (1746 - 1823) studied the effect of temperature on the volume of a gas at constant pressure. Charles's Law states that the volume of a given mass of gas varies directly with the absolute temperature of the gas when pressure is kept constant. The absolute temperature is temperature measured with the Kelvin scale. The Kelvin scale must be used because zero on the Kelvin scale corresponds to a complete stoppage of molecular motion.
Mathematically, the direct relationship of Charles's Law can be represented by the following equation:
$\frac{V}{T} = k \nonumber$
As with Boyle's Law, $k$ is constant only for a given gas sample. The table below shows temperature and volume data for a set amount of gas at a constant pressure. The third column is the constant for this particular data set and is always equal to the volume divided by the Kelvin temperature.
Temperature $\left( \text{K} \right)$ Volume $\left( \text{mL} \right)$ $\frac{V}{T} = k$ $\left( \frac{\text{mL}}{\text{K}} \right)$
Table $1$ Temperature-Volume Data
50 20 0.40
100 40 0.40
150 60 0.40
200 80 0.40
300 120 0.40
500 200 0.40
1000 400 0.40
When this data is graphed, the result is a straight line, indicative of a direct relationship, shown in the figure below.
Notice that the line goes exactly toward the origin, meaning that as the absolute temperature of the gas approaches zero, its volume approaches zero. However, when a gas is brought to extremely cold temperatures, its molecules would eventually condense into the liquid state before reaching absolute zero. The temperature at which this change into the liquid state occurs varies for different gases.
Charles's Law can also be used to compare changing conditions for a gas. Now we use $V_1$ and $T_1$ to stand for the initial volume and temperature of a gas, while $V_2$ and $T_2$ stand for the final volume and temperature. The mathematical relationship of Charles's Law becomes:
$\frac{V_1}{T_1} = \frac{V_2}{T_2} \nonumber$
This equation can be used to calculate any one of the four quantities if the other three are known. The direct relationship will only hold if the temperatures are expressed in Kelvin. Temperatures in Celsius will not work. Recall the relationship that $\text{K} = \: ^\text{o} \text{C} + 273$.
Example $3$:
A balloon is filled to a volume of $2.20 \: \text{L}$ at a temperature of $22^\text{o} \text{C}$. The balloon is then heated to a temperature of $71^\text{o} \text{C}$. Find the new volume of the balloon.
Steps for Problem Solving
Example$3$
Identify the "given"information and what the problem is asking you to "find."
Given:
$V_1 = 2.20 \: \text{L}$ and
$T_1 = 22^\text{o} \text{C} = 295 \: \text{K}$
$T_2 = 71^\text{o} \text{C} = 344 \: \text{K}$
Find: V2 = ? L
List other known quantities. The temperatures have first been converted to Kelvin.
Plan the problem.
First, rearrange the equation algebraically to solve for $V_2$.
$V_2 = \frac{V_1 \times T_2}{T_1} \nonumber$
Cancel units and calculate.
Now substitute the known quantities into the equation and solve.
$V_2 = \frac{2.20 \: \text{L} \times 344 \: \cancel{\text{K}}}{295 \: \cancel{\text{K}}} = 2.57 \: \text{L} \nonumber$
Think about your result. The volume increases as the temperature increases. The result has three significant figures.
Exercise $3$
If V1 = 3.77 L and T1 = 255 K, what is V2 if T2 = 123 K?
Answer
1.82 L
Example $4$:
A sample of a gas has an initial volume of 34.8 L and an initial temperature of −67°C. What must be the temperature of the gas for its volume to be 25.0 L?
Solution
Example $4$: Steps for Problem Solving Volume and Tempature Changes
Steps for Problem Solving
Example $4$
Identify the "given"information and what the problem is asking you to "find."
Given:
Given:T1 = -27oC and V1 = 34.8 L
V2 = 25.0 L
Find: T2 = ? K
List other known quantities. K = -27oC + 273
Plan the problem.
1. Convert the initial temperature to Kelvin
2. Rearrange the equation algebraically to solve for $T_2$.
$T_2 = \frac{V_2 \times T_1}{V_1} \nonumber$
Cancel units and calculate.
1. −67°C + 273 = 206 K
2. Substitute the known quantities into the equation and solve.
$T_2 = \frac{25.0 \: \cancel{\text{L}} \times 206 \: \text{K}}{34.8 \: \cancel{\text{L}}} = 148 \: \text{K} \nonumber$
Think about your result. This is also equal to −125°C. As temperature decreases, volume decreases, which it does in this example.
Exercise $4$
If V1 = 623 mL, T1 = 255°C, and V2 = 277 mL, what is T2?
Answer
235 K, or −38°
Gay-Lussac's Law: Temperature and Pressure
When the temperature of a sample of gas in a rigid container is increased, the pressure of the gas increases as well. The increase in kinetic energy results in the molecules of gas striking the walls of the container with more force, resulting in a greater pressure. The French chemist Joseph Gay-Lussac (1778 - 1850) discovered the relationship between the pressure of a gas and its absolute temperature. Gay-Lussac's Law states that the pressure of a given mass of gas varies directly with the absolute temperature of the gas, when the volume is kept constant. Gay-Lussac's Law is very similar to Charles's Law, with the only difference being the type of container. Whereas the container in a Charles's Law experiment is flexible, it is rigid in a Gay-Lussac's Law experiment.
The mathematical expressions for Gay-Lussac's Law are likewise similar to those of Charles's Law:
$\frac{P}{T} \: \: \: \text{and} \: \: \: \frac{P_1}{T_1} = \frac{P_2}{T_2} \nonumber$
A graph of pressure vs. temperature also illustrates a direct relationship. As a gas is cooled at constant volume its pressure continually decreases until the gas condenses to a liquid.
Example $5$
The gas in an aerosol can is under a pressure of $3.00 \: \text{atm}$ at a temperature of $25^\text{o} \text{C}$. It is dangerous to dispose of an aerosol can by incineration. What would the pressure in the aerosol can be at a temperature of $845^\text{o} \text{C}$?
Solution
Example $5$: Steps for Problem Solving Volume and Tempature Changes
Steps for Problem Solving
Example $5$
Identify the "given"information and what the problem is asking you to "find."
Given:
$P_1 = 3.00 \: \text{atm}$
$T_1 = 25^\text{o} \text{C} = 298 \: \text{K}$
$T_2 = 845^\text{o} \text{C} = 1118 \: \text{K}$
Find: $P_2 = ? \: \text{atm}$
List other known quantities The temperatures have first been converted to Kelvin.
Plan the problem
First, rearrange the equation algebraically to solve for $P_2$.
$P_2 = \frac{P_1 \times T_2}{T_1} \nonumber$
Calculate
Now substitute the known quantities into the equation and solve.
$P_2 = \frac{3.00 \: \text{atm} \times 1118 \: \cancel{\text{K}}}{298 \: \cancel{\text{K}}}= 11.3 \: \text{atm} \nonumber$
Think about your result. The pressure increases dramatically due to large increase in temperature.
Molar Volume
It should be obvious by now that some physical properties of gases depend strongly on the conditions. What we need is a set of standard conditions so that properties of gases can be properly compared to each other. Standard Temperature and Pressure (STP) is defined as exactly 100 kPa of pressure (0.986 atm) and 273 K (0°C). For simplicity, we will use 1 atm as standard pressure. Defining STP allows us to compare more directly the properties of gases that differ from each other.
One property shared among gases is a molar volume. The molar volume is the volume of 1 mol of a gas. At STP, the molar volume of a gas can be easily determined by using the ideal gas law:
$(1\, atm)V=(1\, mol)(0.08205\frac{L.atm}{mol.K})(273\, K) \nonumber$
All the units cancel except for L, the unit of volume. So V = 22.4 L
Note that we have not specified the identity of the gas; we have specified only that the pressure is 1 atm and the temperature is 273 K. This makes for a very useful approximation: any gas at STP has a volume of 22.4 L per mole of gas; that is, the molar volume at STP is 22.4 L/mol (Fig. 6.6.1 Molar Volume). This molar volume makes a useful conversion factor in stoichiometry problems if the conditions are at STP. If the conditions are not at STP, a molar volume of 22.4 L/mol is not applicable. However, if the conditions are not at STP, the combined gas law can be used to calculate what the volume of the gas would be if at STP; then the 22.4 L/mol molar volume can be used.
Example $6$:
How many moles of Ar are present in 38.7 L at STP?
Solution
We can use the molar volume, 22.4 L/mol, as a conversion factor, but we need to reverse the fraction so that the L units cancel and mol units are introduced. It is a one-step conversion:
$38.7\, \not{L}\times \frac{1\, mol}{22.4\not{L}}=1.73\, mol \nonumber$
Exercise $5$
What volume does 4.87 mol of Kr have at STP?
109 L
Note
Chemistry Is Everywhere: Breathing
Breathing (more properly called respiration) is the process by which we draw air into our lungs so that our bodies can take up oxygen from the air. Let us apply the gas laws to breathing.
Start by considering pressure. We draw air into our lungs because the diaphragm, a muscle underneath the lungs, moves down to reduce pressure in the lungs, causing external air to rush in to fill the lower-pressure volume. We expel air by the diaphragm pushing against the lungs, increasing pressure inside the lungs and forcing the high-pressure air out. What are the pressure changes involved? A quarter of an atmosphere? A tenth of an atmosphere? Actually, under normal conditions, it’s only 1 or 2 torr of pressure difference that makes us breathe in and out.
A normal breath is about 0.50 L. If room temperature is about 22°C, then the air has a temperature of about 295 K. With normal pressure being 1.0 atm, how many moles of air do we take in for every breath? The ideal gas law gives us an answer:
$(1.0\, atm)(0.50\, L)=n(0.08205\frac{L.atm}{mol.K})(295\, K) \nonumber$
Solving for the number of moles, we get
n = 0.021 mol air
This ends up being about 0.6 g of air per breath—not much but enough to keep us alive.
Summary
• The behavior of gases can be modeled with gas laws.
• Boyle’s law relates a gas’s pressure and volume at constant temperature and amount.
• Charles’s law relates a gas’s volume and temperature at constant pressure and amount.
• Gay-Lussac's Law states that the pressure of a given mass of gas varies directly with the absolute temperature of the gas, when the volume is kept constant.
• In gas laws, temperatures must always be expressed in Kelvins.
Contributors and Attributions
• Henry Agnew (UC Davis) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/06%3A_Gases_Liquids_Solids_..._and_Intermolecular_Forces/6.06%3A_The_Simple_Gas_Laws.txt |
Learning Objectives
• Learn and apply the ideal gas law.
• Learn and apply the combined gas law.
There are a number of chemical reactions that require ammonia. In order to carry out the reaction efficiently, we need to know how much ammonia we have for stoichiometric purposes. Using gas laws, we can determine the number of moles present in the tank if we know the volume, temperature, and pressure of the system.
Ideal Gas Law
As with the other gas laws, we can also say that $\frac{\left( P \times V \right)}{\left( T \times n \right)}$ is equal to a constant. The constant can be evaluated provided that the gas being described is considered to be ideal.
The ideal gas law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas. If we substitute in the variable $R$ for the constant, the equation becomes:
$\dfrac{P \times V}{T \times n} = R \nonumber$
The ideal gas law is conveniently rearranged to look this way, with the multiplication signs omitted:
$PV = nRT \nonumber$
The variable $R$ in the equation is called the ideal gas constant.
Evaluating the Ideal Gas Constant
The value of $R$, the ideal gas constant, depends on the units chosen for pressure, temperature, and volume in the ideal gas equation. It is necessary to use Kelvin for the temperature and it is conventional to use the SI unit of liters for the volume. However, pressure is commonly measured in one of three units: $\text{kPa}$, $\text{atm}$, or $\text{mm} \: \ce{Hg}$. Therefore, $R$ can have three different values.
We will demonstrate how $R$ is calculated when the pressure is measured in $\text{kPa}$. The volume of $1.00 \: \text{mol}$ of any gas at STP (Standard temperature, 273.15 K and pressure, 1 atm)is measured to be $22.414 \: \text{L}$. We can substitute $101.325 \: \text{kPa}$ for pressure, $22.414 \: \text{L}$ for volume, and $273.15 \: \text{K}$ for temperature into the ideal gas equation and solve for $R$.
\begin{align*} R &= \frac{PV}{nT} \[4pt] &= \frac{101.325 \: \text{kPa} \times 22.414 \: \text{L}}{1.000 \: \text{mol} \times 273.15 \: \text{K}} \[4pt] &= 8.314 \: \text{kPa} \cdot \text{L/K} \cdot \text{mol} \end{align*} \nonumber
This is the value of $R$ that is to be used in the ideal gas equation when the pressure is given in $\text{kPa}$. The table below shows a summary of this and the other possible values of $R$. It is important to choose the correct value of $R$ to use for a given problem.
Unit of $P$ Unit of $V$ Unit of $n$ Unit of $T$ Value and Unit of $R$
Table $1$ Values of the Ideal Gas Constant
$\text{kPa}$ $\text{L}$ $\text{mol}$ $\text{K}$ $8.314 \: \text{J/K} \cdot \text{mol}$
$\text{atm}$ $\text{L}$ $\text{mol}$ $\text{K}$ $0.08206 \: \text{L} \cdot \text{atm/K} \cdot \text{mol}$
$\text{mm} \: \ce{Hg}$ $\text{L}$ $\text{mol}$ $\text{K}$ $62.36 \: \text{L} \cdot \text{mm} \: \ce{Hg}/\text{K} \cdot \text{mol}$
Notice that the unit for $R$ when the pressure is in $\text{kPa}$ has been changed to $\text{J/K} \cdot \text{mol}$. A kilopascal multiplied by a liter is equal to the SI unit for energy, a joule $\left( \text{J} \right)$.
Example $1$ Oxygen Gas
What volume is occupied by $3.76 \: \text{g}$ of oxygen gas at a pressure of $88.4 \: \text{kPa}$ and a temperature of $19^\text{o} \text{C}$? Assume the oxygen is ideal.
Solution
Example $1$ Steps for Problem Solving volume of oxygen gas
Steps for Problem Solving
Example $1$
Identify the "given"information and what the problem is asking you to "find."
Given:
• $P = 88.4 \: \text{kPa}$
• $T = 19^\text{o} \text{C} = 292 \: \text{K}$
Mass $\ce{O_2} = 3.76 \: \text{g}$
Find: V = ? L
List other known quantities
$\ce{O_2} = 32.00 \: \text{g/mol}$
$R = 8.314 \: \text{J/K} \cdot \text{mol}$
Plan the problem
1. First, determine the number of moles of O2 from the given mass and the molar mass.
2. Then, rearrange the equation algebraically to solve for V
$V = \frac{nRT}{P} \nonumber$
Calculate
1.
$3.76 \: \cancel{\text{g}} \times \frac{1 \: \text{mol} \: \ce{O_2}}{32.00 \: \cancel{\text{g}} \: \ce{O_2}} = 0.1175 \: \text{mol} \: \ce{O_2} \nonumber$
2. Now substitute the known quantities into the equation and solve.
$V = \frac{nRT}{P} = \frac{0.1175 \: \cancel{\text{mol}} \times 8.314 \: \cancel{\text{J/K}} \cdot \cancel{\text{mol}} \times 292 \: \cancel{\text{K}}}{88.4 \: \cancel{\text{kPa}}} = 3.23 \: \text{L} \: \ce{O_2} \nonumber$
Think about your result. The number of moles of oxygen is far less than one mole, so the volume should be fairly small compared to molar volume $\left( 22.4 \: \text{L/mol} \right)$ since the pressure and temperature are reasonably close to standard. The result has three significant figures because of the values for $T$ and $P$. Since a joule $\left( \text{J} \right) = \text{kPa} \cdot \text{L}$, the units cancel out correctly, leaving a volume in liters.
Example $2$: Argon Gas
A 4.22 mol sample of Ar has a pressure of 1.21 atm and a temperature of 34°C. What is its volume?
Solution
Example $2$: Steps for Problem Solving volume of argon gas
Steps for Problem Solving
Example $2$
Identify the "given"information and what the problem is asking you to "find."
Given:
n = 4.22 mol
P = 1.21 atm
T = 34°C
Find: V = ? L
List other known quantities none
Plan the problem
1. The first step is to convert temperature to kelvin.
2. Then, rearrange the equation algebraically to solve for V
$V = \frac{nRT}{P} \nonumber$
Calculate
1. 34 + 273 = 307 K
2. Now substitute the known quantities into the equation and solve.
\begin{align*} V=\frac{(4.22\, \cancel{mol})(0.08205\frac{L.\cancel{atm}}{\cancel{mol.K}})(307\, \cancel{K)}}{1.21\cancel{atm}} \[4pt] &= 87.9 \,L \end{align*} \nonumber
Think about your result. The number of moles of Ar is large so the expected volume should also be large.
Exercise $1$
A 0.0997 mol sample of O2 has a pressure of 0.692 atm and a temperature of 333 K. What is its volume?
3.94 L
Exercise $2$
For a 0.00554 mol sample of H2, P = 23.44 torr and T = 557 K. What is its volume?
Answer
8.21 L
One thing we notice about all the gas laws is that, collectively, volume and pressure are always in the numerator, and temperature is always in the denominator. This suggests that we can propose a gas law that combines pressure, volume, and temperature. This gas law is known as the combined gas law, and its mathematical form is
$\frac{P_{1}V_{1}}{T_{1}}=\dfrac{P_{2}V_{2}}{T_{2}}\; at\; constant\; n \nonumber$
This allows us to follow changes in all three major properties of a gas. Again, the usual warnings apply about how to solve for an unknown algebraically (isolate it on one side of the equation in the numerator), units (they must be the same for the two similar variables of each type), and units of temperature must be in Kelvin.
Example $3$:
A sample of gas at an initial volume of 8.33 L, an initial pressure of 1.82 atm, and an initial temperature of 286 K simultaneously changes its temperature to 355 K and its volume to 5.72 L. What is the final pressure of the gas?
Solution
Example $3$: Steps for Problem Solving pressure of gas
Steps for Problem Solving
Example $3$
Identify the "given"information and what the problem is asking you to "find."
Given:
V1 = 8.33 L, P1 = 1.82 atm, and T1 = 286 K
V2 = 5.72 L and T2 = 355 K
Find: P2 = ? atm
List other known quantities none
Plan the problem
First, rearrange the equation algebraically to solve for $V_2$.
$P_2 = \frac{P_1 V_1 T_2 }{T_1V_2}$
Calculate
Now substitute the known quantities into the equation and solve.
$P_2 = \frac{(1.82\, atm)(8.33\, \cancel{L})(355\, \cancel{K})}{(286\, \cancel{K})(5.72\, \cancel{L})}=3.22 atm \nonumber$
Think about your result. Ultimately, the pressure increased, which would have been difficult to predict because two properties of the gas were changing.
Exercise $3$
If P1 = 662 torr, V1 = 46.7 mL, T1 = 266 K, P2 = 409 torr, and T2 = 371 K, what is V2?
Answer
105 mL
As with other gas laws, if you need to determine the value of a variable in the denominator of the combined gas law, you can either cross-multiply all the terms or just take the reciprocal of the combined gas law. Remember, the variable you are solving for must be in the numerator and all by itself on one side of the equation.
Summary
• The ideal gas constant is calculated to be $8.314 \: \text{J/K} \cdot \text{mol}$ when the pressure is in kPa.
• The ideal gas law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas.
• The combined gas law relates pressure, volume, and temperature of a gas.
Contributors and Attributions
• Henry Agnew (UC Davis) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/06%3A_Gases_Liquids_Solids_..._and_Intermolecular_Forces/6.07%3A_The_Ideal_Gas_Law.txt |
Learning Objectives
• Examine properties of acids and bases.
Many people enjoy drinking coffee. A cup first thing in the morning helps start the day. But keeping the coffee maker clean can be a problem. Lime deposits build up after a while and slow down the brewing process. The best cure for this is to put vinegar (dilute acetic acid) in the pot and run it through the brewing cycle. The vinegar dissolves the deposits and cleans the maker, which will speed up the brewing process back to its original rate. Just be sure to run water through the brewing process after the vinegar, or you will get some really horrible coffee.
Perhaps you have eaten too much pizza and felt very uncomfortable hours later. This feeling is due to excess stomach acid being produced. The discomfort can be dealt with by taking an antacid. The base in the antacid will react with the $\ce{HCl}$ in the stomach and neutralize it, taking care of that unpleasant feeling. Figure $1$ provides some examples of acids and bases.
Acids
Acids are very common in some of the foods that we eat. Citrus fruits such as oranges and lemons contain citric acid and ascorbic acid, which is better known as vitamin C. Carbonated sodas contain phosphoric acid. Vinegar contains acetic acid. Your own stomach utilizes hydrochloric acid to digest food. Acids are a distinct class of compounds because of the properties of their aqueous solutions as outlined below:
1. Aqueous solutions of acids are electrolytes, meaning that they conduct electrical current. Some acids are strong electrolytes because they ionize completely in water, yielding a great many ions. Other acids are weak electrolytes that exist primarily in a non-ionized form when dissolved in water.
2. Acids have a sour taste. Lemons, vinegar, and sour candies all contain acids.
3. Acids change the color of certain acid-base indicators. Two common indicators are litmus and phenolphthalein. Blue litmus turns red in the presence of an acid, while phenolphthalein turns colorless.
4. Acids react with active metals to yield hydrogen gas. Recall that an activity series is a list of metals in descending order of reactivity. Metals that are above hydrogen in the activity series will replace the hydrogen from an acid in a single-replacement reaction, as shown below:
$\ce{Zn} \left( s \right) + \ce{H_2SO_4} \left( aq \right) \rightarrow \ce{ZnSO_4} \left( aq \right) + \ce{H_2} \left( g \right) \label{eq1}$
5. Acids react with bases to produce a salt compound and water. When equal moles of an acid and a base are combined, the acid is neutralized by the base. The products of this reaction are an ionic compound, which is labeled as a salt, and water.
Common examples: Lemons, oranges, vinegar, urine, sulfuric acid, hydrochloric acid
Bases
Bases have properties that mostly contrast with those of acids.
1. Aqueous solutions of bases are also electrolytes. Bases can be either strong or weak, just as acids can.
2. Bases often have a bitter taste and are found in foods less frequently than acids. Many bases, like soaps, are slippery to the touch.
3. Bases also change the color of indicators. Litmus turns blue in the presence of a base while phenolphthalein turns pink.
4. Bases do not react with metals in the way that acids do.
5. Bases react with acids to produce a salt and water.
Common Examples: Soap, toothpaste, bleach, cleaning agents, limewater, ammonia water, sodium hydroxide.
Warning
Tasting chemicals and touching them are NOT good lab practices and should be avoided - in other words, don't do this at home.
Summary
• A brief summary of properties of acids and bases are given.
• Properties of acids and bases mostly contrast each other.
Contributors and Attributions
• Peggy Lawson (Oxbow Prairie Heights School). Funded by Saskatchewan Educational Technology Consortium.
• Wikipedia
• Henry Agnew (UC Davis) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/07%3A_Acids_and_Bases/7.01%3A_Acids_and_Bases_-_Experimental_Definitions.txt |
Learning Objectives
• Identify an Arrhenius acid and an Arrhenius base.
• Identify a Brønsted-Lowry acid and a Brønsted-Lowry base.
• Write chemical reactions between an Arrhenius acid and an Arrhenius base and between Brønsted-Lowry acid and a Brønsted-Lowry base.
There are three major classifications of substances known as acids or bases. The Arrhenius definition states that an acid produces H+ in solution and a base produces OH-. This theory was developed by Svante Arrhenius in 1883. Later, two more sophisticated and general theories were proposed. These are the Brønsted-Lowry and the Lewis definitions of acids and bases. The Lewis theory is discussed elsewhere.
The Arrhenius Theory of Acids and Bases
In 1884, the Swedish chemist Svante Arrhenius proposed two specific classifications of compounds, termed acids and bases. When dissolved in an aqueous solution, certain ions were released into the solution. An Arrhenius acid is a compound that increases the concentration of $\ce{H^{+}}$ ions that are present when added to water. These H+ ions form the hydronium ion (H3O+) when they combine with water molecules. This process is represented in a chemical equation by adding H2O to the reactants side.
$\ce{HCl(aq) \rightarrow H^{+}(aq) + Cl^{-}(aq) } \nonumber$
In this reaction, hydrochloric acid ($HCl$) dissociates completely into hydrogen (H+) and chlorine (Cl-) ions when dissolved in water, thereby releasing H+ ions into solution. Formation of the hydronium ion equation:
$\ce{ HCl(aq) + H_2O(l) \rightarrow H_3O^{+}(aq) + Cl^{-}(aq)} \nonumber$
An Arrhenius base is a compound that increases the concentration of $\ce{OH^{-}}$ ions that are present when added to water. The dissociation is represented by the following equation:
$\ce{ NaOH \; (aq) \rightarrow Na^{+} \; (aq) + OH^{-} \; (aq) } \nonumber$
In this reaction, sodium hydroxide (NaOH) disassociates into sodium ($Na^+$) and hydroxide ($OH^-$) ions when dissolved in water, thereby releasing OH- ions into solution.
Arrhenius acids are substances which produce hydrogen ions in solution and Arrhenius bases are substances which produce hydroxide ions in solution.
Limitations to the Arrhenius Theory
The Arrhenius theory has many more limitations than the other two theories. The theory does not explain the weak base ammonia (NH3), which in the presence of water, releases hydroxide ions into solution, but does not contain OH- itself. Also, the Arrhenius definition of acid and base is limited to aqueous (i.e., water) solutions.
The Brønsted-Lowry Theory of Acids and Bases
In 1923, Danish chemist Johannes Brønsted and English chemist Thomas Lowry independently proposed new definitions for acids and bases, ones that focus on proton transfer. A Brønsted-Lowry acid is any species that can donate a proton (H+) to another molecule. A Brønsted-Lowry base is any species that can accept a proton from another molecule. In short, a Brønsted-Lowry acid is a proton donor (PD), while a Brønsted-Lowry base is a proton acceptor (PA).
A Brønsted-Lowry acid is a proton donor, while a Brønsted-Lowry base is a proton acceptor.
Let us use the reaction of ammonia in water to demonstrate the Brønsted-Lowry definitions of an acid and a base. Ammonia and water molecules are reactants, while the ammonium ion and the hydroxide ion are products:
$\ce{NH3(aq) + H2O (ℓ) <=> NH^{+}4(aq) + OH^{−}(aq) }\label{Eq1}$
What has happened in this reaction is that the original water molecule has donated a hydrogen ion to the original ammonia molecule, which in turn has accepted the hydrogen ion. We can illustrate this as follows (Figure $2$):
Because the water molecule donates a hydrogen ion to the ammonia, it is the Brønsted-Lowry acid, while the ammonia molecule—which accepts the hydrogen ion—is the Brønsted-Lowry base. Thus, ammonia acts as a base in both the Arrhenius sense and the Brønsted-Lowry sense.
Is an Arrhenius acid like hydrochloric acid still an acid in the Brønsted-Lowry sense? Yes, but it requires us to understand what really happens when HCl is dissolved in water. Recall that the hydrogen atom is a single proton surrounded by a single electron. To make the hydrogen ion, we remove the electron, leaving a bare proton. Do we really have bare protons floating around in aqueous solution? No, we do not. What really happens is that the H+ ion attaches itself to H2O to make H3O+, which is called the hydronium ion. For most purposes, H+ and H3O+ represent the same species, but writing H3O+ instead of H+ shows that we understand that there are no bare protons floating around in solution. Rather, these protons are actually attached to solvent molecules.
The Hydronium Ion
A proton in aqueous solution may be surrounded by more than one water molecule, leading to formulas like $\ce{H5O2^{+}}$ or $\ce{H9O4^{+}}$ rather than $\ce{H3O^{+}}$. It is simpler, however, to use $\ce{H3O^{+}}$ to represent the hydronium ion.
With this in mind, how do we define HCl as an acid in the Brønsted-Lowry sense? Consider what happens when HCl is dissolved in H2O:
$\ce{HCl(g) + H_2O (ℓ) \rightarrow H_3O^{+}(aq) + Cl^{−}(aq) }\label{Eq2}$
We can depict this process using Lewis electron dot diagrams (Figure $4$):
Now we see that a hydrogen ion is transferred from the HCl molecule to the H2O molecule to make chloride ions and hydronium ions. As the hydrogen ion donor, HCl acts as a Brønsted-Lowry acid; as a hydrogen ion acceptor, H2O is a Brønsted-Lowry base. So HCl is an acid not just in the Arrhenius sense but also in the Brønsted-Lowry sense. Moreover, by the Brønsted-Lowry definitions, H2O is a base in the formation of aqueous HCl. So the Brønsted-Lowry definitions of an acid and a base classify the dissolving of HCl in water as a reaction between an acid and a base—although the Arrhenius definition would not have labeled H2O a base in this circumstance.
A summary of the acid-base definitions based on the Arrhenius and Bronstes-Lowry theories is given in Table $1$.
Table $1$ Acid-Base Definitions
Type Acid Base
Arrhenius $\ce{H^+}$ ions in solution $\ce{OH^-}$ ions in solution
Brønsted-Lowry $\ce{H^+}$ donor $\ce{H^+}$ acceptor
All Arrhenius acids and bases are Brønsted-Lowry acids and bases as well. However, not all Brønsted-Lowry acids and bases are Arrhenius acids and bases.
Example $1$
Aniline (C6H5NH2) is slightly soluble in water. It has a nitrogen atom that can accept a hydrogen ion from a water molecule just like the nitrogen atom in ammonia does. Write the chemical equation for this reaction and identify the Brønsted-Lowry acid and base.
Solution
C6H5NH2 and H2O are the reactants. When C6H5NH2 accepts a proton from H2O, it gains an extra H and a positive charge and leaves an OH ion behind. The reaction is as follows:
$\ce{C6H5NH2(aq) + H2O(ℓ) <=> C6H5NH3^{+}(aq) + OH^{−}(aq)} \nonumber$
Because C6H5NH2 accepts a proton, it is the Brønsted-Lowry base. The H2O molecule, because it donates a proton, is the Brønsted-Lowry acid.
Exercise $1$
Identify the Brønsted-Lowry acid and the Brønsted-Lowry base in this chemical equation.
$\ce{H2PO4^{-} + H_2O <=> HPO4^{2-} + H3O^{+}} \nonumber$
Answer
Brønsted-Lowry acid: H2PO4-; Brønsted-Lowry base: H2O
Exercise $2$
Which of the following compounds is a Bronsted-Lowry base?
1. HCl
2. HPO42-
3. H3PO4
4. NH4+
5. CH3NH3+
Answer:
A Brønsted-Lowry Base is a proton acceptor, which means it will take in an H+. This eliminates $\ce{HCl}$, $\ce{H3PO4}$, $\ce{NH4^{+}}$ and $\ce{CH_3NH_3^{+}}$ because they are Bronsted-Lowry acids. They all give away protons. In the case of $\ce{HPO4^{2-}}$, consider the following equation:
$\ce{HPO4^{2-} (aq) + H2O (l) \rightarrow PO4^{3-} (aq) + H3O^{+}(aq) } \nonumber$
Here, it is clear that HPO42- is the acid since it donates a proton to water to make H3O+ and PO43-. Now consider the following equation:
$\ce{ HPO4^{2-}(aq) + H2O(l) \rightarrow H2PO4^{-} + OH^{-}(aq)} \nonumber$
In this case, HPO42- is the base since it accepts a proton from water to form H2PO4- and OH-. Thus, HPO42- is an acid and base together, making it amphoteric.
Since HPO42- is the only compound from the options that can act as a base, the answer is (b) HPO42-.
Summary
• An Arrhenius acid is a compound that increases the H+ ion concentration and an Arrhenius base is a compound that increases the OH ion concentration in aqueous solution.
• A Brønsted-Lowry acid is a proton donor; a Brønsted-Lowry base is a proton acceptor.
• All Arrhenius acids and bases are Brønsted-Lowry acids and bases as well. However, not all Brønsted-Lowry acids and bases are Arrhenius acids and bases.
Contributors and Attributions
• Libretext: The Basics of GOB Chemistry (Ball et al.)
• Libretext: Chemistry for Allied Health (Soult) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/07%3A_Acids_and_Bases/7.02%3A_Acids_Bases_and_Salts.txt |
Learning Objectives
• Identify acid anyhdrides and basic anhydrides.
• Know the reactions of acid and basic anhydrides with water.
Metal oxides are substances used in the manufacture of commonly used bases like calcium hydroxide (traditionally known as slaked lime), sodium hydroxide and potassium hydroxide. Acid anhydrides (non-metal oxides) are environmentally relevant. Sulfur and nitrogen oxides are considered air pollutants as they react with atmospheric water vapor to produce acid rain (Chapter 13).
Nonmetal Oxides: Acid Anhydrides
Acid anhydrides are compounds that turn into an acid when you add water. They are non-metal oxides. The oxides of period three elements demonstrate periodicity with respect to acidity. As you move across the period, the oxides become more acidic. Sodium and magnesium oxides are alkaline. Aluminium oxides are amphoteric (reacting both as a base or acid). Silicon, phosphorus, sulfur, and chlorine oxides are acidic. Some non-metal oxides, such as nitrous oxide (N2O) and carbon monoxide (CO), do not display any acid/base characteristics. These are a little more complicated than basic anhydrides, so don't worry too much about them right now.
Several examples of reactions involving nonmetal oxides are as follows:
1. Carbon dioxide is the anhydride of carbonic acid:
CO2(g) + H2O(l) → H2CO3 (aq)
2. Sulfur dioxide reacts with water to form the weak acid, sulfurous acid:
SO2(g) + H2O (l)→ H2SO3 (aq)
3. Sulfur trioxide forms the strong sulfuric acid with water (so, sulfur trioxide is the anhydride of sulfuric acid):
SO3(g) + H2O(l) → H2SO4 (aq)
This reaction is important in the manufacture of the acid.
Metal Oxides: Basic Anhydrides
Oxides of more electropositive elements like the alkali or alkaline earth metals tend to be basic. They are called basic anhydrides. Exposed to water, they may form bases (basic hydroxides). Sodium oxide is basic—when hydrated, it forms sodium hydroxide. Here's another example:
CaO(s) + H2O(l) → Ca(OH)2 (aq)
If the metal is an alkali or alkaline earth, the reaction probably happens quickly and produces a lot of heat. If the metal is a transition metal, the reaction might not happen so easily or at all.
Summary
• Acid anhydrides are nonmetal oxides that react with water to form acids.
• Basic anhydrides are metal oxides that react with water to form bases.
• Wikipedia
7.04: Strong and Weak Acids and Bases
Learning Objectives
• Define a strong and a weak acid and base.
• Recognize an acid or a base as strong or weak.
Strong and Weak Acids
Except for their names and formulas, so far we have treated all acids as equals, especially in a chemical reaction. However, acids can be very different in a very important way. Consider HCl(aq). When HCl is dissolved in H2O, it completely dissociates into H+(aq) and Cl(aq) ions; all the HCl molecules become ions:
$HCl\overset{100\%}{\rightarrow}H^{+}(aq)+Cl^{-}(aq) \nonumber$
Any acid that dissociates 100% into ions is called a strong acid. If it does not dissociate 100%, it is a weak acid. HC2H3O2 is an example of a weak acid:
$HC_{2}H_{3}O_{2}\overset{\sim 5\%}{\longrightarrow}H^{+}(aq)+C_{2}H_{3}O_{2}^{-}(aq) \nonumber$
Because this reaction does not go 100% to completion, it is more appropriate to write it as a reversible reaction:
$HC_{2}H_{3}O_{2}\rightleftharpoons H^{+}(aq)+C_{2}H_{3}O_{2}^{-}(aq) \nonumber$
As it turns out, there are very few strong acids, which are given in Table $1$. If an acid is not listed here, it is a weak acid. It may be 1% ionized or 99% ionized, but it is still classified as a weak acid.
Any acid that dissociates 100% into ions is called a strong acid. If it does not dissociate 100%, it is a weak acid.
Table $1$ Strong Acids and Bases
Acids Bases
HCl LiOH
HBr NaOH
HI KOH
HNO3 RbOH
H2SO4 CsOH
HClO3 Mg(OH)2
HClO4 Ca(OH)2
Sr(OH)2
Ba(OH)2
Strong and Weak Bases
The issue is similar with bases: a strong base is a base that is 100% ionized in solution. If it is less than 100% ionized in solution, it is a weak base. There are very few strong bases (Table $1$); any base not listed is a weak base. All strong bases are OH compounds. So a base based on some other mechanism, such as NH3 (which does not contain OH ions as part of its formula), will be a weak base.
Example $1$: Identifying Strong and Weak Acids and Bases
Identify each acid or base as strong or weak.
1. HCl
2. Mg(OH)2
3. C5H5N
Solution
1. Because HCl is listed in Table $1$, it is a strong acid.
2. Because Mg(OH)2 is listed in Table $1$, it is a strong base.
3. The nitrogen in C5H5N would act as a proton acceptor and therefore can be considered a base, but because it does not contain an OH compound, it cannot be considered a strong base; it is a weak base.
Exercise $1$
Identify each acid or base as strong or weak.
1. $\ce{RbOH}$
2. $\ce{HNO_2}$
Answer a
strong base
Answer b
weak acid
Example $2$: Characterizing Base Ionization
Write the balanced chemical equation for the dissociation of Ca(OH)2 and indicate whether it proceeds 100% to products or not.
Solution
This is an ionic compound of Ca2+ ions and OH ions. When an ionic compound dissolves, it separates into its constituent ions:
$\ce{Ca(OH)2 → Ca^{2+}(aq) + 2OH^{−}(aq)} \nonumber$
Because Ca(OH)2 is listed in Table $1$, this reaction proceeds 100% to products.
Exercise $2$
Write the balanced chemical equation for the dissociation of hydrazoic acid (HN3) and indicate whether it proceeds 100% to products or not.
Answer a
The reaction is as follows:
$\ce{HN3 → H^{+}(aq) + N3^{−}(aq)} \nonumber$
It does not proceed 100% to products because hydrazoic acid is not a strong acid.
Summary
• Strong acids and bases are 100% ionized in aqueous solution.
• Weak acids and bases are less than 100% ionized in aqueous solution.
Contributors and Attributions
• Libretext: Beginning Chemistry (Ball et al.)
• Henry Agnew (UC Davis) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/07%3A_Acids_and_Bases/7.03%3A_Acidic_and_Basic_Anhydrides.txt |
Learning Objective
• Write acid-base neutralization reactions.
What happens when an acid such as HCl is mixed with a base such as NaOH:
$\ce{HCl (aq) + NaOH (aq) → NaCl (aq) + H_2O (l)} \nonumber$
When an acid and a base are combined, water and a salt are the products. Double displacement reactions of this type are called neutralization reactions. We can write an expanded version of this equation, with aqueous substances written in their longer form:
$\ce{H^{+} (aq) + Cl^{-} (aq) + Na^{+} (aq) + OH^{-} (aq) → Na^{+} (aq) + Cl^{-} (aq) + H_2O (l)} \nonumber$
Removing the spectator ions we get the net ionic equation:
$\ce{H^{+} (aq) + OH^{-} (aq) → H_2O (l)} \nonumber$
When a strong acid and a strong base are combined in the proper amounts - when $[\ce{H^{+}}]$ equals $[\ce{OH^{-}}$]\) - a neutral solution results in which pH = 7. The acid and base have neutralized each other, and the acidic and basic properties are no longer present.
Salt solutions do not always have a pH of 7, however. Through a process known as hydrolysis, the ions produced when an acid and base combine may react with the water molecules to produce a solution that is slightly acidic or basic. We will not go into details here, but generally if a strong acid is mixed with a weak base there the resulting solution will be slightly acidic; if a strong base is mixed with a weak acid the solution will be slightly basic.
Neutralization
Video $1$ Acid-base neutralization reaction.
Example $1$: propionic acid + calcium hydroxide
Calcium propionate is used to inhibit the growth of molds in foods, tobacco, and some medicines. Write a balanced chemical equation for the reaction of aqueous propionic acid (CH3CH2CO2H) with aqueous calcium hydroxide [Ca(OH)2].
Solution
Example $1$: steps for problem solving the reaction of propionic acid + calcium hydroxide
Steps Reaction
Write the unbalanced equation.
This is a double displacement reaction, so the cations and anions swap to create the water and the salt.
CH3CH2CO2H(aq) + Ca(OH)2(aq)→(CH3CH2CO2)2Ca(aq) + H2O(l)
Balance the equation.
Because there are two OH ions in the formula for Ca(OH)2, we need two moles of propionic acid, CH3CH2CO2H to provide H+ ions.
2CH3CH2CO2H(aq) + Ca(OH)2(aq)→(CH3CH2CO2)2Ca(aq) +2H2O(l)
Exercise $1$
Write a balanced chemical equation for the reaction of solid barium hydroxide with dilute acetic acid.
Answer
$\ce{Ba(OH)2(s) + 2CH3CO2H (aq)→Ba(CH3CO2)2 (aq) + 2H2O(l)} \nonumber$
Applications
In analytical laboratories, chemical methods are used for determining the amounts of acids or bases in foods and household and industrial chemicals.
In , chemical neutralization methods are often applied to reduce the damage that an effluent may cause upon release to the environment.
There are many uses of neutralization reactions that are acid-alkali reactions. A very common use is antacid tablets (section 7.8).
In chemical synthesis of nanomaterials, the heat of neutralization reaction can be used to facilitate the chemical reduction of metal precursors.
Also in the digestive tract, neutralization reactions are used when food is moved from the stomach to the intestines. In order for the nutrients to be absorbed through the intestinal wall, an alkaline environment is needed, so the pancreas produce an antacid bicarbonate to cause this transformation to occur.
Another common use, though perhaps not as widely known, is in fertilizers and control of soil pH. Slaked lime (calcium hydroxide) or limestone (calcium carbonate) may be worked into soil that is too acidic for plant growth. Fertilizers that improve plant growth are made by neutralizing sulfuric acid (H2SO4) or nitric acid (HNO3) with ammonia gas (NH3), making or . These are salts utilized in the fertilizer.
Industrially, a by-product of the burning of coal, gas, may combine with water vapor in the air to eventually produce sulfuric acid, which falls as acid rain. To prevent the sulfur dioxide from being released, a device known as a scrubber gleans the gas from smoke stacks. This device first blows calcium carbonate into the combustion chamber where it decomposes into calcium oxide (lime) and carbon dioxide. This lime then reacts with the sulfur dioxide produced forming calcium sulfite. A suspension of lime is then injected into the mixture to produce a slurry, which removes the and any remaining unreacted sulfur dioxide.
Fluffy Cakes
Q: How do acid-base reactions make cake fluffy?
A: Usually cakes include an acidic ingredient (this varies) and sodium bicarbonate, a base. When they react, the proton from the acid is transferred to the bicarbonate, making the weak acid carbonic acid. Carbonic acid is the product of an acid anhydride reaction between carbon dioxide and water. This reaction can be reversed, or carbonic acid can decompose into water and carbon dioxide. Especially at the high temperatures inside a baking cake, this decomposition will happen, and produce carbon dioxide gas. The pressure of the hot gas will form bubbles inside the cake, making it fluffy.
Summary
• Acids and bases combine and produce salt and water.
Contributors and Attributions
• Peggy Lawson (Oxbow Prairie Heights School). Funded by Saskatchewan Educational Technology Consortium.
• Henry Agnew (UC Davis)
• Wikipedia | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/07%3A_Acids_and_Bases/7.05%3A_Neutralization.txt |
Learning Objectives
• Define $pH$.
• Determine the pH of acidic and basic solutions.
• Determine the hydrogen (hydronium) ion concentration from pH and vice versa.
Molar concentration values of hydrogen $[H^+]$ can be markedly different from one aqueous solution to another. So chemists defined a new scale that succinctly indicates the concentrations of either of these two ions. This is known as the $pH$ scale. The range of values from 0 to 14 that describes the acidity or basicity of a solution. You can use $pH$ to make a quick determination whether a given aqueous solution is acidic, basic, or neutral.
pH is a logarithmic scale. A solution that has a pH of 1.0 has 10 times the [H+] as a solution with a pH of 2.0, which in turn has 10 times the [H+] as a solution with a pH of 3.0 and so forth.
Warning: The pH scale has no limits
pH is usually (but not always) between 0 and 14. Knowing the dependence of pH on $[H+]$, we can summarize as follows:
• If pH < 7, then the solution is acidic.
• If pH = 7, then the solution is neutral.
• If pH > 7, then the solution is basic.
Figure $2$ illustrates the relationship between pH and the hydrogen ion concentration, along with some examples of various solutions. Because hydrogen ion concentrations are generally less than one (for example $1.3 \times 10^{-3}\,M$), the log of the number will be a negative number. To make pH even easier to work with, pH is defined as the negative log of $[H^+]$, which will give a positive value for pH.
The general formula for determining [H+] from pH is as follows:
[H+] = 10−pH
Example $1$
Label each solution as acidic, basic, or neutral based only on the stated $pH$.
1. milk of magnesia, pH = 10.5
2. pure water, pH = 7
3. wine, pH = 3.0
Answer
1. With a pH greater than 7, milk of magnesia is basic. (Milk of magnesia is largely Mg(OH)2.)
2. Pure water, with a pH of 7, is neutral.
3. With a pH of less than 7, wine is acidic.
Exercise $1$
Identify each substance as acidic, basic, or neutral based only on the stated $pH$.
1. human blood with $pH$ = 7.4
2. household ammonia with $pH$ = 11.0
3. cherries with $pH$ = 3.6
Answer a
basic
Answer b
basic
Answer c
acidic
Table $1$: gives the typical pH values of some common substances. Note that several food items are on the list, and most of them are acidic. Table $1$ Typical pH Values of Various Substances*
Substance pH
stomach acid 1.7
lemon juice 2.2
vinegar 2.9
soda 3.0
wine 3.5
coffee, black 5.0
milk 6.9
pure water 7.0
blood 7.4
seawater 8.5
milk of magnesia 10.5
ammonia solution 12.5
1.0 M NaOH 14.0
*Actual values may vary depending on conditions
Example $2$:
What is the [H+] for an aqueous solution whose pH is 6?
Solution
The pH value of 6 denotes that the exponent of 10 is -6. Therefore the answer is
[H+] = 1.0 x 10−6M
Exercise $2$
What is the [H+] for an aqueous solution whose pH is 11?
Answer
[H+] = 1.0 × 10−11 M
Example $3$:
What is the pH of an aqueous solution whose hydrogen ion concentration is 1.0 x 10−7M? Is the solution acidic, basic, or neutral.
Solution
The hydrogen ion concentration is 1.0 x 10−7M. The exponent of 10 is -7, which denoted that pH= -(-7) = 7. The solution is neutral.
Exercise $3$
What is the pH of an aqueous solution whose hydrogen ion concentration is 1.0 x 10−3M? Is the solution acidic, basic, or neutral.
Solution
The hydrogen ion concentration is 1.0 x 10−3M. The exponent of 10 is -3, which denoted that pH= -(-3) = 3. The solution is acidic.
Summary
To make pH even easier to work with, pH is defined as the negative log of $[H^+]$, which will give a positive value for pH.
pH is usually (but not always) between 0 and 14.
• If pH < 7, then the solution is acidic.
• If pH = 7, then the solution is neutral.
• If pH > 7, then the solution is basic.
Contributors and Attributions
• Peggy Lawson (Oxbow Prairie Heights School). Funded by Saskatchewan Educational Technology Consortium.
• Template:OpenStax
• Henry Agnew (UC Davis) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/07%3A_Acids_and_Bases/7.06%3A_The_pH_Scale.txt |
Learning Objectives
• Identify conjugate acid base pair.
• Define buffers and know the composition of different buffer systems.
• Describe how buffers work.
Conjugate Acid-Base Pair
In reality, all acid-base reactions involve the transfer of protons between acids and bases. For example, consider the acid-base reaction that takes place when ammonia is dissolved in water. A water molecule (functioning as an acid) transfers a proton to an ammonia molecule (functioning as a base), yielding the conjugate base of water, $\ce{OH^-}$, and the conjugate acid of ammonia, $\ce{NH4+}$:
In the reaction of ammonia with water to give ammonium ions and hydroxide ions, ammonia acts as a base by accepting a proton from a water molecule, which in this case means that water is acting as an acid. In the reverse reaction, an ammonium ion acts as an acid by donating a proton to a hydroxide ion, and the hydroxide ion acts as a base. The conjugate acid–base pairs for this reaction are $NH_4^+/NH_3$ and $H_2O/OH^−$.
The strongest acids are at the bottom left, and the strongest bases are at the top right. The conjugate base of a strong acid is a very weak base, and, conversely, the conjugate acid of a strong base is a very weak acid.
Example $1$
Identify the conjugate acid-base pairs in this equilibrium.
$\ce{CH3CO2H + H2O <=> H3O^{+} + CH3CO2^{-}} \nonumber$
Solution
Similarly, in the reaction of acetic acid with water, acetic acid donates a proton to water, which acts as the base. In the reverse reaction, $H_3O^+$ is the acid that donates a proton to the acetate ion, which acts as the base.
Once again, we have two conjugate acid–base pairs:
• the parent acid and its conjugate base ($CH_3CO_2H/CH_3CO_2^−$) and
• the parent base and its conjugate acid ($H_3O^+/H_2O$).
Example $2$
Identify the conjugate acid-base pairs in this equilibrium.
$(CH_{3})_{3}N + H_{2}O\rightleftharpoons (CH_{3})_{3}NH^{+} + OH^{-} \nonumber$
Solution
One pair is H2O and OH, where H2O has one more H+ and is the conjugate acid, while OH has one less H+ and is the conjugate base.
The other pair consists of (CH3)3N and (CH3)3NH+, where (CH3)3NH+ is the conjugate acid (it has an additional proton) and (CH3)3N is the conjugate base.
Exercise $1$
Identify the conjugate acid-base pairs in this equilibrium.
$\ce{NH2^{-} + H2O\rightleftharpoons NH3 + OH^{-}} \nonumber$
Answer:
H2O (acid) and OH (base); NH2 (base) and NH3 (acid)
Buffer Solutions
Weak acids are relatively common, even in the foods we eat. But we occasionally encounter a strong acid or base, such as stomach acid, which has a strongly acidic pH of 1.7. By definition, strong acids and bases can produce a relatively large amount of H+ or OH ions and consequently have marked chemical activities. In addition, very small amounts of strong acids and bases can change the pH of a solution very quickly. If 1 mL of stomach acid [approximated as 0.1 M HCl(aq)] were added to the bloodstream and no correcting mechanism were present, the pH of the blood would decrease from about 7.4 to about 4.7—a pH that is not conducive to continued living. Fortunately, the body has a mechanism for minimizing such dramatic pH changes.
This mechanism involves a buffer, a solution that resists dramatic changes in pH. Buffers do so by being composed of certain pairs of solutes: either a weak acid plus a salt derived from that weak acid, or a weak base plus a salt of that weak base. For example, a buffer can be composed of dissolved HC2H3O2 (a weak acid) and NaC2H3O2 (the salt derived from that weak acid). Another example of a buffer is a solution containing NH3 (a weak base) and NH4Cl (a salt derived from that weak base).
Let us use an HC2H3O2/NaC2H3O2 buffer to demonstrate how buffers work. If a strong base—a source of OH(aq) ions—is added to the buffer solution, those OH ions will react with the HC2H3O2 in an acid-base reaction:
$\ce{HC2H3O2(aq) + OH^{-}(aq) \rightarrow H2O(ℓ) + C2H3O^{-}2(aq)} \label{Eq1}$
Rather than changing the pH dramatically by making the solution basic, the added OH ions react to make H2O, so the pH does not change much.
If a strong acid—a source of H+ ions—is added to the buffer solution, the H+ ions will react with the anion from the salt. Because HC2H3O2 is a weak acid, it is not ionized much. This means that if lots of H+ ions and C2H3O2 ions are present in the same solution, they will come together to make HC2H3O2:
$\ce{H^{+}(aq) + C2H3O^{−}2(aq) \rightarrow HC2H3O2(aq)} \label{Eq2}$
Rather than changing the pH dramatically and making the solution acidic, the added H+ ions react to make molecules of a weak acid. Figure $2$ illustrates both actions of a buffer.
Buffers made from weak bases and salts of weak bases act similarly. For example, in a buffer containing NH3 and NH4Cl, NH3 molecules can react with any excess H+ ions introduced by strong acids:
NH3(aq) + H+(aq) → NH4+(aq)
while the NH4+(aq) ion can react with any OH ions introduced by strong bases:
NH4+(aq) + OH(aq) → NH3(aq) + H2O(ℓ)
Some common buffer systems are listed in the table below.
Table $1$ Some Common Buffers
Table $1$ Some Common Buffers
Buffer System Buffer Components pH of buffer (equal molarities of both components)
Acetic acid/acetate ion $\ce{CH_3COOH}$/$\ce{CH_3COO^-}$ 4.74
Carbonic acid/hydrogen carbonate ion $\ce{H_2CO_3}$/$\ce{HCO_3^-}$ 6.38
Dihydrogen phosphate ion/hydrogen phosphate ion $\ce{H_2PO_4^-}$/$\ce{HPO_4^{2-}}$ 7.21
Ammonia/ammonium ion $\ce{NH_3}$/$\ce{NH_4^+}$ 9.25
Buffers work well only for limited amounts of added strong acid or base. Once either solute is completely reacted, the solution is no longer a buffer, and rapid changes in pH may occur. We say that a buffer has a certain capacity. Buffers that have more solute dissolved in them to start with have larger capacities, as might be expected.
Human blood has a buffering system to minimize extreme changes in pH. One buffer in blood is based on the presence of HCO3 and H2CO3 [the second compound is another way to write CO2(aq)]. With this buffer present, even if some stomach acid were to find its way directly into the bloodstream, the change in the pH of blood would be minimal. Inside many of the body's cells, there is a buffering system based on phosphate ions.
Example $3$:
Which combinations of compounds can make a buffer solution?
1. HCHO2 and NaCHO2
2. HCl and NaCl
3. CH3NH2 and CH3NH3Cl
4. NH3 and NaOH
Solution
1. HCHO2 is formic acid, a weak acid, while NaCHO2 is the salt made from the anion of the weak acid (the formate ion [CHO2]). The combination of these two solutes would make a buffer solution.
2. HCl is a strong acid, not a weak acid, so the combination of these two solutes would not make a buffer solution.
3. CH3NH2 is methylamine, which is like NH3 with one of its H atoms substituted with a CH3 group. Because it is not listed in Table $1$, we can assume that it is a weak base. The compound CH3NH3Cl is a salt made from that weak base, so the combination of these two solutes would make a buffer solution.
4. NH3 is a weak base, but NaOH is a strong base. The combination of these two solutes would not make a buffer solution.
Exercise $2$
Which combinations of compounds can make a buffer solution?
1. NaHCO3 and NaCl
2. H3PO4 and NaH2PO4
3. NH3 and (NH4)3PO4
4. NaOH and NaCl
Answer a
Yes.
Answer b
No. Need a weak acid or base and a salt of its conjugate base or acid.
Answer c
Yes.
Answer d
No. Need a weak base or acid.
Food and Drink Application: the Acid that Eases Pain
Although medicines are not exactly "food and drink," we do ingest them, so let's take a look at an acid that is probably the most common medicine: acetylsalicylic acid, also known as aspirin. Aspirin is well known as a pain reliever and antipyretic (fever reducer).
The structure of aspirin is shown in the accompanying figure. The acid part is circled; it is the H atom in that part that can be donated as aspirin acts as a Brønsted-Lowry acid. Because it is not given in Table $1$, acetylsalicylic acid is a weak acid. However, it is still an acid, and given that some people consume relatively large amounts of aspirin daily, its acidic nature can cause problems in the stomach lining, despite the stomach's defenses against its own stomach acid.
Because the acid properties of aspirin may be problematic, many aspirin brands offer a "buffered aspirin" form of the medicine. In these cases, the aspirin also contains a buffering agent—usually MgO—that regulates the acidity of the aspirin to minimize its acidic side effects.
As useful and common as aspirin is, it was formally marketed as a drug starting in 1899. The US Food and Drug Administration (FDA), the governmental agency charged with overseeing and approving drugs in the United States, wasn't formed until 1906. Some have argued that if the FDA had been formed before aspirin was introduced, aspirin may never have gotten approval due to its potential for side effects—gastrointestinal bleeding, ringing in the ears, Reye's syndrome (a liver problem), and some allergic reactions. However, recently aspirin has been touted for its effects in lessening heart attacks and strokes, so it is likely that aspirin will remain on the market.
Summary
• A buffer is a solution that resists sudden changes in pH.
• Reactions showing how buffers regulate pH are described.
Contributors and Attributions
• TextMap: Beginning Chemistry (Ball et al.) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/07%3A_Acids_and_Bases/7.07%3A_Buffers_and_Conjugate_Acid-Base_Pairs.txt |
Learning Objective
• Know the many uses of common acids and bases.
Antacids: A Basic Remedy
Antacids contain alkaline ions that chemically neutralize stomach gastric acid, reducing damage and relieving pain. For example, Tums is a very commonly used antacid that can be bought over the counter. Other antacids, such as Alka Seltzer, work in similar ways. Below, show the mechanism and products for reaction in which calcium carbonate (the active ingredient in Tums) neutralizes HCl in a simple proton transfer reaction.
CaCO3 (s) + HCl (aq) → H2CO3 (aq) + CaCl2 (aq)
Following this acid-base reaction, carbonic acid quickly degrades into CO2 and H2O.
H2CO3 (aq) CO2(g) + H2O(l)
Figure $1$ A comparison of a healthy condition to GERD. I
mage used with permission (Cc BY-SA 4.0; BruceBlaus .
Under the generic name algeldrate, aluminium hydroxide is used as an antacid. Aluminium hydroxide is preferred over other alternatives such as sodium bicarbonate because Al(OH)3, being insoluble, does not increase the pH of stomach above 7 and hence, does not trigger secretion of excess stomach acid. Brand names include Alu-Cap, Aludrox, Gaviscon, and Pepsamar. In 2016 Gaviscon was one of the biggest selling branded over-the-counter medications sold in Great Britain, with sales of £62 million. It reacts with excess acid in the stomach, reducing the acidity of the stomach content, which may relieve the symptoms of ulcers, heartburn or dyspepsia. Such products can cause constipation, because the aluminum ions inhibit the contractions of smooth muscle cells in the gastrointestinal tract, slowing peristalsis and lengthening the time needed for stool to pass through the colon. Some such products are formulated to minimize such effects through the inclusion of equal concentrations of magnesium hydroxide or magnesium carbonate, which have counterbalancing laxative effects.
Antacids are distinct from acid-reducing drugs like H2-receptor antagonists or proton pump inhibitors and they do not kill the bacteria Helicobacter pylori, which causes most ulcers. Ranitidine (Zantac), famotidine (Pepcid AC), and cimetidine (Tagamet) are in a class of medications called H2 blockers. It decreases the amount of acid made in the stomach and are used to treat ulcers; gastroesophageal reflux disease (GERD), a condition in which backward flow of acid from the stomach causes heartburn and injury of the food pipe (esophagus); and conditions where the stomach produces too much acid, such as Zollinger-Ellison syndrome. These over-the-counter medications are used to prevent and treat symptoms of heartburn associated with acid indigestion and sour stomach. Omeprazole (Prilosec) also works by decreasing the amount of acid made in the stomach and is in a class of medications called proton-pump inhibitors.
Acids and Bases in Industry and at Home
Common Acids
It should not be hard for you to name several common acids, but you might find that listing bases is just a little more difficult. Here's a partial list of some common acids, along with some chemical formulas:
Table $1$ Common Acids and Their Uses
Chemist Name
Common Name Uses
hydrochloric acid, HCl muriatic acid (used in pools) and stomach acid Used in cleaning (refining) metals, in maintenance of swimming pools, and for household cleaning.
sulfuric acid, H2SO4 Used in car batteries, and in the manufacture of fertilizers.
nitric acid, HNO3 Used in the manufacture of fertilizers, explosives and in extraction of gold.
acetic acid, HC2H3O2 vinegar Main ingredient in vinegar.
carbonic acid, H2CO3 responsible for the "fiz" in carbonated drinks As an ingredient in carbonated drinks.
citric acid, C6H8O7 Used in food and dietary supplements. Also added as an acidulant in creams, gels, liquids, and lotions.
acetylsalicylic acid, C6H4(OCOCH3)CO2H aspirin The active ingredient in aspirin.
Hydrochloric Acid
Hydrochloric acid is a corrosive, strong mineral acid with many industrial uses. One of the most important applications of hydrochloric acid is in the pickling of steel, to remove rust or iron oxide scale from iron or steel before subsequent processing. Another major use of hydrochloric acid is in the production of organic compounds, such as vinyl chloride and dichloroethane for PVC. Other organic compounds produced with hydrochloric acid include bisphenol A for polycarbonate, activated carbon, and ascorbic acid, as well as numerous pharmaceutical products. Other inorganic compounds produced with hydrochloric acid include road application salt calcium chloride, nickel(II) chloride for electroplating, and zinc chloride for the galvanizing industry and battery production.
Sulfuric Acid
Sulfuric acid is a very important commodity chemical, and indeed, a nation's sulfuric acid production is a good indicator of its industrial strength. World production in the year 2004 was about 180 million tonnes, with the following geographic distribution: Asia 35%, North America (including Mexico) 24%, Africa 11%, Western Europe 10%, Eastern Europe and Russia 10%, Australia and Oceania 7%, South America 7%. Most of this amount (≈60%) is consumed for fertilizers, particularly superphosphates, ammonium phosphate and ammonium sulfates. About 20% is used in chemical industry for production of detergents, synthetic resins, dyestuffs, pharmaceuticals, petroleum catalysts, insecticides and antifreeze, as well as in various processes such as oil well acidicizing, aluminium reduction, paper sizing, water treatment. About 6% of uses are related to pigments and include paints, enamels, printing inks, coated fabrics and paper, and the rest is dispersed into a multitude of applications such as production of explosives, cellophane, acetate and viscose textiles, lubricants, non-ferrous metals, and batteries.
Because the hydration of sulfuric acid is thermodynamically favorable (and is highly exothermic) and the affinity of it for water is sufficiently strong, sulfuric acid is an excellent dehydrating agent. Concentrated sulfuric acid has a very powerful dehydrating property, removing water ($\ce{H2O}$) from other compounds including sugar and other carbohydrates and producing carbon, heat, steam. Sulfuric acid behaves as a typical acid in its reaction with most metals by generating hydrogen gas (Equation \ref{Eq1}).
Nitric Acid
Nitric acid ($\ce{HNO3}$) is a highly corrosive mineral acid and is also commonly used as a strong oxidizing agent. Nitric acid is normally considered to be a strong acid at ambient temperatures. The main industrial use of nitric acid is for the production of fertilizers. Nitric acid is neutralized with ammonia to give ammonium nitrate. This application consumes 75–80% of the 26 million tonnes produced annually (1987). The other main applications are for the production of explosives, nylon precursors, and specialty organic compounds.
Table $2$ Common Bases and Its Uses
Some Common Bases
Uses
sodium hydroxide, NaOH
(lye or caustic soda)
Used in the manufacture of soaps and detergents and as the main ingredient in oven and drain cleaners.
potassium hydroxide, KOH
(lye or caustic potash)
Used in the production of liquid soaps and soft soaps. Used in alkaline batteries.
magnesium hydroxide, Mg(OH)2
(milk of magnesia)
Used as an ingredient in laxatives, antacids, and deodorants. Also used in the neutralization of acidic wastewater.
calcium hydroxide, Ca(OH)2
(slaked lime)
Used in the manufacture of cement and lime water. Also, added to neutralize acidic soil.
aluminum hydroxide Used in water purification and as an ingredient in antacids.
ammonia, NH3 Used as a building block for the synthesis of many pharmaceutical products and in many commercial cleaning products. Used in the manufacture of fertilizers.
Sodium Hydroxide
Sodium hydroxide, also known as lye and caustic soda, is an inorganic compound with formula $\ce{NaOH}$.
It is a popular strong base used in industry. Around 56% of sodium hydroxide produced is used by industry, 25% of which is used in the paper industry. Sodium hydroxide is also used in the manufacture of sodium salts and detergents, pH regulation, and organic synthesis. It is used in the Bayer process of aluminium production. In bulk, it is most often handled as an aqueous solution, since solutions are cheaper and easier to handle.
Sodium hydroxide is used in many scenarios where it is desirable to increase the alkalinity of a mixture, or to neutralize acids.
Calcium Hydroxide
Calcium hydroxide (traditionally called slaked lime) is an inorganic compound with the chemical formula $\ce{Ca(OH)2}$. It is a colorless crystal or white powder. It has many names including hydrated lime, caustic lime, builders' lime, slaked lime, cal, or pickling lime.
One significant application of calcium hydroxide is as a flocculant, in water and sewage treatment. It is also used in fresh water treatment for raising the pH of the water so that pipes will not corrode where the base water is acidic, because it is self-regulating and does not raise the pH too much.
It is also used in the preparation of ammonia gas (NH3), using the following reaction:
Ca(OH)2 + 2NH4Cl → 2NH3 + CaCl2 + 2H2O
Another large application is in the paper industry, where it is an intermediate in the reaction in the production of sodium hydroxide. Because of its low toxicity and the mildness of its basic properties, slaked lime is widely used in the food industry:
Ammonia
Ammonia is a compound of nitrogen and hydrogen with the formula $\ce{NH3}$ and is a colorless gas with a characteristic pungent smell. It is the active product of “smelling salts,” and can quickly revive the faint of heart and light of head. Although common in nature and in wide use, ammonia is both caustic and hazardous in its concentrated form.
Globally, approximately 88% (as of 2014) of ammonia is used as fertilizers either as its salts, solutions or anhydrously. When applied to soil, it helps provide increased yields of crops such as maize and wheat. 30% of agricultural nitrogen applied in the US is in the form of anhydrous ammonia and worldwide 110 million tonnes are applied each year. Ammonia is also a building block for the synthesis of many pharmaceutical products and is used in many commercial cleaning products.
Summary
Common acids and bases have many and various uses in industry and in daily life.
Contributors and Attributions
• Peggy Lawson (Oxbow Prairie Heights School). Funded by Saskatchewan Educational Technology Consortium.
• Wikipedia
• TextMap: Chemistry-A Molecular Approach (Tro)
• Henry Agnew (UC Davis)
• MedlinePlus | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/07%3A_Acids_and_Bases/7.08%3A_Acids_and_Bases_in_Industry_and_in_Daily_Life.txt |
Thumbnail: Batteries are empowered by REDOX chemistry (Pixabay License; Didgeman via Pixabay).
08: Oxidation and Reduction
Learning Objectives
• Define oxidation and reduction.
• Assign oxidation numbers to atoms in simple compounds.
• Recognize a reaction as an oxidation-reduction reaction.
Oxygen is an element that has been known for centuries. In its pure elemental form, oxygen is highly reactive, and it readily makes compounds with most other elements. It is also the most abundant element by mass in the Earth's crust. The class of reactions called oxidation and reduction were originally defined with respect to the element oxygen.
Many elements simply combine with oxygen to form the oxide of that element. Heating magnesium in air allows it to combine with oxygen to form magnesium oxide.
$2 \ce{Mg} \left( s \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{MgO} \left( s \right) \nonumber$
Many compounds react with oxygen as well, often in very exothermic processes that are generally referred to as combustion reactions. For example, when methane burns, carbon dioxide and water are produced.
$\ce{CH_4} \left( g \right) + 2 \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + 2 \ce{H_2O} \left( g \right) \nonumber$
Carbon dioxide is an oxide of carbon, while water is an oxide of hydrogen. Early scientists viewed oxidation as a process in which a substance was reacted with oxygen to produce one or more oxides. In the previous examples, magnesium and methane are being oxidized.
The definitions of oxidation and reduction were eventually broadened to include similar types of reactions that do not necessarily involve oxygen. Oxygen is more electronegative than any element except for fluorine. Therefore, when oxygen is bonded to any element other than fluorine, electrons from the other atom are shifted away from that atom and toward the oxygen atom. An oxidation-reduction reaction (sometimes abbreviated as a redox reaction) is a reaction that involves the full or partial transfer of electrons from one reactant to another. Oxidation involves a full or partial loss of electrons, while reduction involves a full or partial gain of electrons.
Redox reactions require that we keep track of the electrons assigned to each atom in a chemical reaction. How do we do that? We use oxidation numbers to keep track of electrons in atoms. Oxidation numbers are assigned to atoms based on four rules. Oxidation numbers are not necessarily equal to the charge on the atom (although sometimes they can be); we must keep the concepts of charge and oxidation numbers separate.
Assigning Oxidation Numbers
The oxidation number is a positive or negative number that is assigned to an atom to indicate its degree of oxidation or reduction. In oxidation-reduction processes, the driving force for chemical change is in the exchange of electrons between chemical species. A series of rules have been developed to help us.
Assigning Oxidation Numbers
The rules for assigning oxidation numbers to atoms are as follows:
1. Atoms in their elemental state are assigned an oxidation number of 0.
Examples: $\ce{H_2}$, $\ce{Br_2}$, $\ce{Na}$, $\ce{Be}$, $\ce{K}$, $\ce{O_2}$, $\ce{P_4}$, all have oxidation number of 0.
2. Atoms in monatomic (i.e., one-atom) ions are assigned an oxidation number equal to their charge.
Example: In MgCl2, magnesium has an oxidation number of +2, while each chlorine atom has an oxidation number of −1
3. In compounds, fluorine is assigned a −1 oxidation number; oxygen is usually assigned a −2 oxidation number [except in peroxide compounds (where it is −1) and in binary compounds with fluorine (where it is positive)]; and hydrogen is usually assigned a +1 oxidation number [except when it exists as the hydride ion (H), in which case rule 2 prevails].
Example: In H2O, the H atoms each have an oxidation number of +1, while the O atom has an oxidation number of −2, even though hydrogen and oxygen do not exist as ions in this compound.
4. In compounds, all other atoms are assigned an oxidation number so that the sum of the oxidation numbers on all the atoms in the species equals the charge on the species (which is zero if the species is neutral).
Example: In SO2. Each O atom has an oxidation number of −2; for the sum of the oxidation numbers to equal the charge on the species (which is zero), the S atom is assigned an oxidation number of +4. Does this mean that the sulfur atom has a 4+ charge on it? No, it means only that the S atom is assigned a +4 oxidation number by our rules of apportioning electrons among the atoms in a compound.
Note
Oxidation numbers are usually written with the sign first, then the magnitude, to differentiate them from charges.
Example $1$
Assign oxidation numbers to the atoms in each substance.
1. Cl2
2. GeO2
3. CaCl2
Solution
1. Cl2 is the elemental form of chlorine. Rule 1 states each atom has an oxidation number of 0.
2. By rule 3, oxygen is normally assigned an oxidation number of −2. For the sum of the oxidation numbers to equal the charge on the species (zero), the Ge atom is assigned an oxidation number of +4.
3. The Ca2+ ion has an oxidation number of +2 by rule 2. Also according to rule 2, each chlorine atom is assigned an oxidation number of −1.
Exercise $1$: Phosphoric Acid
Assign oxidation numbers to the atoms in H3PO4.
Answer
H: +1; O: −2; P: +5
Oxidation and Reduction Reactions
The processes of oxidation and reduction are commonly viewed in terms of the gain or loss of electrons or the increase or decrease in oxidation numbers. However, many reactions in organic chemistry that involve the gain or loss of oxygen or hydrogen are also considered as redox reactions.
The table below summarizes the processes of oxidation and reduction.
Table $1$ The Processes of Oxidation and Reduction.
Oxidation Reduction
Complete loss of electrons (ionic reaction) Complete gain of electrons (ionic reaction)
Increase in oxidation number Decrease in oxidation number
Gain of oxygen Loss of oxygen
Loss of hydrogen in a molecular compound Gain of hydrogen in a molecular compound
1. Oxidation is complete loss of electrons. Reduction is complete gain of electrons
To understand electron-transfer reactions like the one between zinc metal and hydrogen ions, chemists separate them into two parts: one part focuses on the loss of electrons, and one part focuses on the gain of electrons. The loss of electrons is called oxidation. The gain of electrons is called reduction.
Electrons that are lost are written as products; electrons that are gained are written as reactants. For example, the reaction below represents when acid is added to zinc metal.
$\ce{Zn (s) + 2H^{+} (aq) → Zn^{2+}(aq) + H2(g)} \nonumber$
zinc atoms are oxidized to Zn2+. The half reaction for the oxidation reaction, omitting phase labels, is as follows:
$\ce{Zn → Zn^{2+} + 2e^{−}} \nonumber$
This half reaction is balanced in terms of the number of zinc atoms, and it also shows the two electrons that are needed as products to account for the zinc atom losing two negative charges to become a 2+ ion. With half reactions, there is one more item to balance: the overall charge on each side of the reaction. If you check each side of this reaction, you will note that both sides have a zero net charge.
Hydrogen is reduced in the reaction. The balanced reduction half reaction is as follows:
$\ce{2H^{+} + 2e^{−} → H2} \nonumber$
There are two hydrogen atoms on each side, and the two electrons written as reactants serve to neutralize the 2+ charge on the reactant hydrogen ions. Again, the overall charge on both sides is zero.
Note: "LEO says GER"
LEO stands for Losing Electrons is Oxidation, while GER stands for Gaining Electrons is Reduction.
This guy knows his redox reactions.
2. Oxidation is an increase in oxidation number. Reduction is a decrease in oxidation number.
Consider the reaction below between elemental iron and copper sulfate:
$\ce{Fe} + \ce{CuSO_4} \rightarrow \ce{FeSO_4} + \ce{Cu} \nonumber$
In the course of the reaction, the oxidation number of $\ce{Fe}$ increases from zero to $+2$. The oxidation number of copper decreases from $+2$ to $0$. This result is in accordance with the activity series. Iron is above copper in the series, so will be more likely to form $\ce{Fe^{2+}}$ while converting the $\ce{Cu^{2+}}$ to metallic copper $\left( \ce{Cu^0} \right)$.
3. Oxidation is gain of oxygen. Reduction is loss of oxygen.
Oxidation and reduction can also be defined in terms of changes in composition. The original meaning of oxidation was “adding oxygen,” so when oxygen is added to a molecule, the molecule is being oxidized. The reverse is true for reduction: if a molecule loses oxygen atoms, the molecule is being reduced. For example, the acetaldehyde ($CH3CHO$) molecule takes on an oxygen atom to become acetic acid ($\ce{CH3COOH}$).
$\ce{2CH3CHO + O2 → 2CH_3COOH} \nonumber$
Thus, acetaldehyde is being oxidized.
4. Oxidation is loss of hydrogen in a molecular compound. Reduction is gain of hydrogen in a molecular compound.
Oxidation and reduction can be defined in terms of the gain or loss of hydrogen atoms. If a molecule adds hydrogen atoms, it is being reduced. If a molecule loses hydrogen atoms, the molecule is being oxidized. For example, in the conversion of acetaldehyde into ethanol CH3CHO, hydrogen atoms are added to acetaldehyde, so the acetaldehyde is being reduced:
$\ce{CH3CHO + H2 → CH3CH2OH} \nonumber$
Example $2$
In each conversion, indicate whether oxidation or reduction is occurring.
1. N2 → NH3
2. CH3CH2OHCH3 → CH3COCH3
3. HCHO → HCOOH
Solution
1. Hydrogen is being added to the original reactant molecule, so reduction is occurring.
2. Hydrogen is being removed from the original reactant molecule, so oxidation is occurring.
3. Oxygen is being added to the original reactant molecule, so oxidation is occurring.
Exercise $2$
In each conversion, indicate whether oxidation or reduction is occurring.
1. CH4 → CO2 + H2O
2. NO2 → N2
3. CH2=CH2 → CH3CH3
1. oxidation
2. reduction
3. reduction
Example $3$:
Identify what is being oxidized and reduced in this redox reaction.
$\ce{2Na + Br2 → 2NaBr} \nonumber$
Solution
Both reactants are the elemental forms of their atoms, so the Na and Br atoms have oxidation numbers of 0. In the ionic product, the Na+ ions have an oxidation number of +1, while the Br ions have an oxidation number of −1.
$2\underset{0}{Na}+\underset{0}{Br_{2}}\rightarrow 2\underset{+1 -1}{NaBr} \nonumber$
Sodium is increasing its oxidation number from 0 to +1, so it is being oxidized; bromine is decreasing its oxidation number from 0 to −1, so it is being reduced:
Because oxidation numbers are changing, this is a redox reaction. The total number of electrons being lost by sodium (two, one lost from each Na atom) is gained by bromine (two, one gained for each Br atom).
Exercise $3$
Identify what is being oxidized and reduced in this redox reaction.
$\ce{C + O2 → CO2}\nonumber$
Answer
C is being oxidized from 0 to +4; O is being reduced from 0 to −2.
Example $4$: Gain or Loss of Electrons
For each half reaction, identify if there was a loss or gain of electrons.
1. 2N3- → N2
2. Cr6+ → Cr3+
Solutions
a. When we compare the overall charges on each side of the equation, we find a charge of 2(-3)=-6 on the left but a charge of 0 on the right. This equation is not properly balanced. To balance it, six electrons must be added on the right side as follows:
$\ce{2N^{3-} → N2 + 6e^{-}}\nonumber$
So there was a loss of six electrons. Now the equation is balanced, not only in terms of elements but also in terms of charge.
b. When we compare the overall charges on each side of the equation, we find a charge of +6 on the left and a charge of +3 on the right. This equation is not properly balanced. To end up with an oxidation number of +3 on the right side 3 electrons must be added on the left side as follows:
$\ce{Cr^{6+} + 3e^{-} → Cr^{3+}\nonumber$
So there was a gain of three electrons. Now the equation is balanced in terms of charge.
Exercise $4$ Gain or Loss of Electrons
For each half reaction, identify if there was a loss or gain of electrons.
1. O2 → 2O2-
2. Cu+ → Cu2+
Answers
a. gain of four electrons
b. loss of one electron
Example $5$: Reducing Silver Ions
Write and balance the redox reaction that has silver ions and aluminum metal as reactants and silver metal and aluminum ions as products.
Solution
We start by using symbols of the elements and ions to represent the reaction:
$\ce{Ag^{+} + Al → Ag + Al^{3+}} \nonumber$
The equation looks balanced as it is written. However, when we compare the overall charges on each side of the equation, we find a charge of +1 on the left but a charge of +3 on the right. This equation is not properly balanced. To balance it, let us write the two half reactions. Silver ions are reduced, and it takes one electron to change Ag+ to Ag:
$\ce{Ag^{+} + e^{−} → Ag} \nonumber$
Aluminum is oxidized, losing three electrons to change from Al to Al3+:
$\ce{Al → Al^{3+} + 3e^{−}} \nonumber$
To combine these two half reactions and cancel out all the electrons, we need to multiply the silver reduction reaction by 3:
Now the equation is balanced, not only in terms of elements but also in terms of charge.
Exercise $5$
Write and balance the redox reaction that has calcium ions and potassium metal as reactants and calcium metal and potassium ions as products.
Summary
• Rules for determining oxidation numbers are listed.
• Examples of oxidation number determinations are provided.
• Chemical reactions in which electrons are transferred are called oxidation-reduction, or redox, reactions.
• Oxidation is defined in terms of the loss of electrons, gain of oxygen, loss of hydrogen, or an increase in oxidation number.
• Reduction is defined in terms of gain of electrons, loss of oxygen, gain of hydrogen, or a decrease in oxidation number.
• Oxidation and reduction always occur together, even though they can be written as separate chemical equations.
Contributors and Attributions
• Libretext: Beginning Chemistry (Ball et al.)
• Libretext: The Basics of GOB Chemistry (Ball et al.)
• Libretext: Chemistry for Allied Health (Soult)
• Ed Vitz (Kutztown University), John W. Moore (UW-Madison), Justin Shorb (Hope College), Xavier Prat-Resina (University of Minnesota Rochester), Tim Wendorff, and Adam Hahn. | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/08%3A_Oxidation_and_Reduction/8.01%3A_Oxidation_and_Reduction-_Four_Views.txt |
Learning Objective
• Identify oxidizing and reducing agents.
Life on planet Earth is a complicated and well-organized set of processes. Animals are designed to breathe oxygen and plants are designed to produce oxygen. Photosynthesis is the means by which we get the oxygen we need for life. Light striking a plant pigment known as chlorophyll initiates a complex series of reactions, many of which involve redox processes complete with movement of electrons. In this series of reactions, water is converted to oxygen gas, and we have something to sustain our lives.
The reaction below is a redox reaction that produces zinc sulfide:
$\ce{Zn} + \ce{S} \rightarrow \ce{ZnS}\nonumber$
The half-reactions can be written:
\begin{align*} &\text{Oxidation:} \: \ce{Zn} \rightarrow \ce{Zn^{2+}} + 2 \ce{e^-} \ &\text{Reduction:} \: \ce{S} + 2 \ce{e^-} \rightarrow \ce{S^{2-}} \end{align*} \nonumber
In the reaction above, zinc is being oxidized by losing electrons. However, there must be another substance present that gains those electrons and in this case that is the sulfur. In other words, the sulfur is causing the zinc to be oxidized. Sulfur is called the oxidizing agent. The zinc causes the sulfur to gain electrons and become reduced and so the zinc is called the reducing agent. The oxidizing agent is a substance that causes oxidation by accepting electrons; therefore, its oxidation state decreases. The reducing agent is a substance that causes reduction by losing electrons; therefore its oxidation state increases. The simplest way to think of this is that the oxidizing agent is the substance that is reduced, while the reducing agent is the substance that is oxidized as shown in Figure $1$ and summarized in Table $1$.
Note
Both the oxidizing and reducing agents are the reactants and therefore appear on the left-hand side of an equation.
Table $1$: A Comparison of Oxidizing and Reducing Agents.
Oxidizing Agents Reducing Agents
Oxidation State Decreases Increases
# of Electrons Gained Lost
Substance is... Reduced Oxidized
The examples below show how to analyze a redox reaction and identify oxidizing and reducing agents.
Example $1$ Half-equations
When chlorine gas is bubbled into a solution of sodium bromide, a reaction occurs which produces aqueous sodium chloride and bromine. Determine what is being oxidized and what is being reduced. Identify the oxidizing and reducing agents.
$\ce{Cl_2} \left( g \right) + 2 \ce{NaBr} \left( aq \right) \rightarrow 2 \ce{NaCl} \left( aq \right) + \ce{Br_2} \left( l \right) \nonumber$
Step 1: Plan the problem.
Break the reaction down into a net ionic equation and then into half-reactions. The substance that loses electrons is being oxidized and is the reducing agent. The substance that gains electrons is being reduced and is the oxidizing agent.
Step 2: Solve.
\begin{align*} \ce{Cl_2} \left( g \right) + \cancel{2 \ce{Na^+} \left( aq \right)} + 2 \ce{Br^-} \left( aq \right) &\rightarrow \cancel{2 \ce{Na^+} \left( aq \right)} + 2 \ce{Cl^-} \left( aq \right) + \ce{Br_2} \left( l \right) \ \ce{Cl_2} \left( g \right) + 2 \ce{Br^-} \left( aq \right) &\rightarrow 2 \ce{Cl^-} \left( aq \right) + \ce{Br_2} \left(l \right) \: \: \: \: \: \left( \text{net ionic equation} \right) \end{align*} \nonumber
\begin{align*} &\text{Reduction:} \: \ce{Cl_2} \left( g \right) + 2 \ce{e^-} \rightarrow 2 \ce{Cl^-} \left( aq \right) \ &\text{Oxidation:} \: 2 \ce{Br^-} \left( aq \right) \rightarrow \ce{Br_2} \left( l \right) + 2 \ce{e^-} \end{align*} \nonumber
The $\ce{Cl_2}$ is being reduced and is the oxidizing agent. The $\ce{Br^-}$ is being oxidized and is the reducing agent.
Exercise $1$ : Half-equations
Write the following reaction in the form of half-equations. Identify each half-equation as an oxidation or a reduction. Also identify the oxidizing agent and the reducing agent in the overall reaction
$\ce{Zn + 2Fe^{3+} -> Zn^{2+} +2Fe^{2+}} \nonumber \nonumber$
Answer
The half-equations are
$\ce{Zn -> Zn^{2+} + 2e^{-}}$ oxidation—loss of electrons
$\ce{2e^{-} + 2Fe^{3+} -> 2Fe^{2+}}$ reduction—gain of electrons
Zinc has been oxidized, the oxidizing agent must have been the other reactant, namely, iron(III).
Iron(III) ion has been reduced, the zinc must be the reducing agent.
Example $2$: Identify reducing and oxidizing agents
Identify the reducing and oxidizing agents in the balanced redox reaction:
$Cl_2 (aq) + 2Br^- (aq) \rightarrow 2Cl^- (aq) + Br_2 (aq)\nonumber$
Oxidation half reaction
$2 Br^- (aq) \rightarrow Br_2 (aq)\nonumber$
Oxidation States: -1 to 0
Reduction Half Reaction
$Cl_2 (aq) \rightarrow 2 Cl^- (aq)\nonumber$
Oxidation States: 0 to -1
Overview
• Br- loses an electron; it is oxidized from Br- to Br2; thus, Br- is the reducing agent.
• Cl2 gains one electron; it is reduced from Cl2 to 2 Cl-; thus, Cl2 is the oxidizing agent.
Exercise $2$: Identify reducing and oxidizing agents
Identify the oxidizing agent and the reducing agent in the following redox reaction:
$MnO_4^- + SO_3^{2-} \rightarrow Mn^{2+} + SO_4^{2-}\nonumber$
Solution
$S$ is the reducing agent and $Mn$ is the oxidizing agent.
Summary
• An oxidizing agent is a substance that causes oxidation by accepting electrons; therefore, it gets reduced.
• A reducing agent is a substance that causes reduction by losing electrons; therefore it gets oxidized.
• Examples of how to identify oxidizing and reducing agents are shown. | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/08%3A_Oxidation_and_Reduction/8.02%3A_Oxidizing_and_Reducing_Agents.txt |
Learning Objectives
• Define electrochemistry.
• Describe the basic components of electrochemical cells.
• List some of the characteristics, applications and limitations of cells and batteries.
• Know the difference between galvanic and electrolytic cells.
• Define electrolysis and list several of its applications.
Metal exposed to the outside elements will usually corrode if not protected. The corrosion process is a series of redox reactions involving the metal of the sculpture. In some situations, the metals are deliberately left unprotected so that the surface will undergo changes that may enhance the esthetic value of the work.
Electrochemical Reactions
Chemical reactions either absorb or release energy, which can be in the form of electricity. Electrochemistry is a branch of chemistry that deals with the interconversion of chemical energy and electrical energy. Electrochemistry has many common applications in everyday life. All sorts of batteries, from those used to power a flashlight to a calculator to an automobile, rely on chemical reactions to generate electricity. Electricity is used to plate objects with decorative metals like gold or chromium. Electrochemistry is important in the transmission of nerve impulses in biological systems. Redox chemistry, the transfer of electrons, is behind all electrochemical processes.
An electrochemical cell is any device that converts chemical energy into electrical energy or electrical energy into chemical energy. There are three components that make up an electrochemical reaction. There must be a solution where redox reactions can occur. These reactions generally take place in water to facilitate electron and ion movement. A conductor must exist for electrons to be transferred. This conductor is usually some kind of wire so that electrons can move from one site to another. Ions also must be able to move through some form of salt bridge that facilitates ion migration.
Luigi Galvani
Luigi Galvani (1737 - 1798) was an Italian physician and scientist who did research on nerve conduction in animals. His accidental observation of the twitching of frog legs when they were in contact with an iron scalpel while the legs hung on copper hooks led to studies on electrical conductivity in muscles and nerves. He believed that animal tissues contained an "animal electricity" similar to the natural electricity that caused lightning to form.
Alessandro Volta developed the first "voltaic cell" in 1800. This battery consisted of alternating disks of zinc and silver with pieces of cardboard soaked in brine separating the disks. Since there were no voltmeters at the time (and no idea that the electric current was due to electron flow), Volta had to rely on another measure of battery strength: the amount of shock produced (it's never a good idea to test things on yourself). He found that the intensity of the shock increased with the number of metal plates in the system. Devices with twenty plates produced a shock that was quite painful. It's a good thing we have voltmeters today to measure electric current instead of the "stick your finger on this and tell me what you feel" method.
Galvanic (Voltaic) Cells
Galvanic cells, also known as voltaic cells, are electrochemical cells in which spontaneous oxidation-reduction reactions produce electrical energy. In writing the equations, it is often convenient to separate the oxidation-reduction reactions into half-reactions to facilitate balancing the overall equation and to emphasize the actual chemical transformations.
Consider what happens when a clean piece of copper metal is placed in a solution of silver nitrate (Figure $1$). As soon as the copper metal is added, silver metal begins to form and copper ions pass into the solution. The blue color of the solution on the far right indicates the presence of copper ions. The reaction may be split into its two half-reactions. Half-reactions separate the oxidation from the reduction, so each can be considered individually.
\begin{align*} &\textrm{oxidation: }\ce{Cu}(s)⟶\ce{Cu^2+}(aq)+\ce{2e-}\ &\underline{\textrm{reduction: }2×(\ce{Ag+}(aq)+\ce{e-}⟶\ce{Ag}(s))\hspace{40px}\ce{or}\hspace{40px}\ce{2Ag+}(aq)+\ce{2e-}⟶\ce{2Ag}(s)}\ &\textrm{overall: }\ce{2Ag+}(aq)+\ce{Cu}(s)⟶\ce{2Ag}(s)+\ce{Cu^2+}(aq) \end{align*} \nonumber
The equation for the reduction half-reaction had to be doubled so the number electrons “gained” in the reduction half-reaction equaled the number of electrons “lost” in the oxidation half-reaction.
Galvanic or voltaic cells involve spontaneous electrochemical reactions in which the half-reactions are separated (Figure $2$) so that current can flow through an external wire. The beaker on the left side of the figure is called a half-cell, and contains a 1 M solution of copper(II) nitrate [Cu(NO3)2] with a piece of copper metal partially submerged in the solution. The copper metal is an electrode. The copper is undergoing oxidation; therefore, the copper electrode is the anode. The anode is connected to a voltmeter with a wire and the other terminal of the voltmeter is connected to a silver electrode by a wire. The silver is undergoing reduction; therefore, the silver electrode is the cathode. The half-cell on the right side of the figure consists of the silver electrode in a 1 M solution of silver nitrate (AgNO3). At this point, no current flows—that is, no significant movement of electrons through the wire occurs because the circuit is open. The circuit is closed using a salt bridge, which transmits the current with moving ions.
The salt bridge consists of a concentrated, nonreactive, electrolyte solution such as the sodium nitrate (NaNO3) solution used in this example. As electrons flow from left to right through the electrode and wire, nitrate ions (anions) pass through the porous plug on the left into the copper(II) nitrate solution. This keeps the beaker on the left electrically neutral by neutralizing the charge on the copper(II) ions that are produced in the solution as the copper metal is oxidized. At the same time, the nitrate ions are moving to the left, sodium ions (cations) move to the right, through the porous plug, and into the silver nitrate solution on the right. These added cations “replace” the silver ions that are removed from the solution as they were reduced to silver metal, keeping the beaker on the right electrically neutral.
When the electrochemical cell is constructed in this fashion, a positive cell potential (0.46 V) indicates a spontaneous reaction and that the electrons are flowing from the left to the right.
Key Features of a Standard Galvanic Cell (as shown in Figure $2$
• Electrons flow from the anode to the cathode: left to right in the standard galvanic cell.
• The electrode in the left half-cell is the anode because oxidation occurs here. The name refers to the flow of anions in the salt bridge toward it.
• The electrode in the right half-cell is the cathode because reduction occurs here. The name refers to the flow of cations in the salt bridge toward it.
• Oxidation occurs at the anode (the left half-cell in the figure).
• Reduction occurs at the cathode (the right half-cell in the figure).
• The salt bridge must be present to close (complete) the circuit and both an oxidation and reduction must occur for current to flow.
Example $1$: Chromium-Copper Galvanic Cell
Consider a galvanic cell consisting of
$\ce{2Cr}(s)+\ce{3Cu^2+}(aq)⟶\ce{2Cr^3+}(aq)+\ce{3Cu}(s)$
Write the oxidation and reduction half-reactions. Which reaction occurs at the anode? The cathode?
Solution
By inspection, Cr is oxidized when three electrons are lost to form Cr3+, and Cu2+ is reduced as it gains two electrons to form Cu. Balancing the charge gives
\begin{align*} &\textrm{oxidation: }\ce{2Cr}(s)⟶\ce{2Cr^3+}(aq)+\ce{6e-}\ &\textrm{reduction: }\ce{3Cu^2+}(aq)+\ce{6e-}⟶\ce{3Cu}(s)\ &\overline{\textrm{overall: }\ce{2Cr}(s)+\ce{3Cu^2+}(aq)⟶\ce{2Cr^3+}(aq)+\ce{3Cu}(s)} \end{align*}
Cell notation uses the simplest form of each of the equations, and starts with the reaction at the anode. No concentrations were specified so:
$\ce{Cr}(s)│\ce{Cr^3+}(aq)║\ce{Cu^2+}(aq)│\ce{Cu}(s).$
Oxidation occurs at the anode and reduction at the cathode.
Exercise $1$ Copper-Zinc Galvanic Cell
Consider a galvanic cell consisting of
$\ce{Zn}(s)+\ce{Cu^2+}(aq)⟶\ce{Zn^2+}(aq)+\ce{Cu}(s)$
Write the oxidation and reduction half-reactions. Which reaction occurs at the anode? The cathode?
Answer
From the information given in the problem:
\begin{align*} &\textrm{anode (oxidation): }\ce{Zn}(s)⟶\ce{Zn^2+}(aq)+\ce{2e-}\ &\textrm{cathode (reduction): }\ce{Cu^2+}(aq)+\ce{2e-}⟶\ce{Cu}(s)\ &\overline{\textrm{overall: }\ce{Zn}(s)+\ce{Cu^2+}(aq)⟶\ce{Zn^2+}(aq)+\ce{Cu}(s)} \end{align*}
Batteries
A battery is an electrochemical cell or series of cells that produces an electric current. In principle, any galvanic cell could be used as a battery. An ideal battery would never run down, produce an unchanging voltage, and be capable of withstanding environmental extremes of heat and humidity. Real batteries strike a balance between ideal characteristics and practical limitations. For example, the mass of a car battery is about 18 kg or about 1% of the mass of an average car or light-duty truck. This type of battery would supply nearly unlimited energy if used in a smartphone, but would be rejected for this application because of its mass. Thus, no single battery is “best” and batteries are selected for a particular application, keeping things like the mass of the battery, its cost, reliability, and current capacity in mind. There are two basic types of batteries: primary and secondary. A few batteries of each type are described next.
Dry Cells (Primary Batteries)
Primary batteries are single-use batteries because they cannot be recharged. A common primary battery is the dry cell (Figure $1$). The dry cell is a zinc-carbon battery. The zinc can serves as both a container and the negative electrode. The positive electrode is a rod made of carbon that is surrounded by a paste of manganese(IV) oxide, zinc chloride, ammonium chloride, carbon powder, and a small amount of water. The reaction at the anode can be represented as the ordinary oxidation of zinc:
$\ce{Zn}(s)⟶\ce{Zn^2+}(aq)+\ce{2e-} \nonumber$
The reaction at the cathode is more complicated, in part because more than one reaction occurs. The series of reactions that occurs at the cathode is approximately
$\ce{2MnO2}(s)+\ce{2NH4Cl}(aq)+\ce{2e-}⟶\ce{Mn2O3}(s)+\ce{2NH3}(aq)+\ce{H2O}(l)+\ce{2Cl-} \nonumber$
The overall reaction for the zinc–carbon battery can be represented as
$\ce{2MnO2}(s) + \ce{2NH4Cl}(aq) + \ce{Zn}(s) ⟶ \ce{Zn^2+}(aq) + \ce{Mn2O3}(s) + \ce{2NH3}(aq) + \ce{H2O}(l) + \ce{2Cl-} \nonumber$
with an overall cell potential which is initially about 1.5 V, but decreases as the battery is used. It is important to remember that the voltage delivered by a battery is the same regardless of the size of a battery. For this reason, D, C, A, AA, and AAA batteries all have the same voltage rating. However, larger batteries can deliver more moles of electrons.
Take Note
The dry cell is not very efficient in producing electrical energy because only the relatively small fraction of the $MnO_2$ that is near the cathode is actually reduced and only a small fraction of the zinc cathode is actually consumed as the cell discharges. In addition, dry cells have a limited shelf life because the $Zn$ anode reacts spontaneously with $NH_4Cl$ in the electrolyte, causing the case to corrode and allowing the contents to leak out (Figure $2$).
Alkaline batteries (Figure $4$) were developed in the 1950s partly to address some of the performance issues with zinc–carbon dry cells. They are manufactured to be exact replacements for zinc-carbon dry cells. As their name suggests, these types of batteries use alkaline electrolytes, often potassium hydroxide.
An alkaline battery can deliver about three to five times the energy of a zinc-carbon dry cell of similar size. Alkaline batteries are prone to leaking potassium hydroxide, so these should also be removed from devices for long-term storage. While some alkaline batteries are rechargeable, most are not.
Warning
Attempts to recharge an alkaline battery that is not rechargeable often leads to rupture of the battery and leakage of the potassium hydroxide electrolyte.
Lead Storage Batteries (Secondary Batteries)
The lead acid battery (Figure $5$) is the type of secondary battery used in your automobile. Secondary batteries are rechargeable. The lead acid battery is inexpensive and capable of producing the high current required by automobile starter motors. The reactions for a lead acid battery are
\begin{align*} &\textrm{anode: }\ce{Pb}(s)+\ce{HSO4-}(aq)⟶\ce{PbSO4}(s)+\ce{H+}(aq)+\ce{2e-}\ &\underline{\textrm{cathode: } \ce{PbO2}(s)+\ce{HSO4-}(aq)+\ce{3H+}(aq)+\ce{2e-}⟶\ce{PbSO4}(s)+\ce{2H2O}(l)}\ &\textrm{overall: }\ce{Pb}(s)+\ce{PbO2}(s)+\ce{2H2SO4}(aq)⟶\ce{2PbSO4}(s)+\ce{2H2O}(l) \end{align*} \nonumber
Each cell produces 2 V, so six cells are connected in series to produce a 12-V car battery. Lead acid batteries are heavy and contain a caustic liquid electrolyte, but are often still the battery of choice because of their high current density. The lead acid battery in your automobile consists of six cells connected in series to give 12 V. Their low cost and high current output makes these excellent candidates for providing power for automobile starter motors.
As mentioned earlier, unlike a dry cell, the lead storage battery is rechargeable. Note that the forward redox reaction generates solid lead (II) sulfate which slowly builds up on the plates. Additionally, the concentration of sulfuric acid decreases. When the car is running normally, its generator recharges the battery by forcing the above reactions to run in the opposite, or non spontaneous direction.
$2 \ce{PbSO_4} \left( s \right) + 2 \ce{H_2O} \left( l \right) \rightarrow \ce{Pb} \left( s \right) + \ce{PbO_2} \left( s \right) + 4 \ce{H+} \left( aq \right) + 2 \ce{SO_4^{2-}} \left( aq \right) \nonumber$
This reaction regenerates the lead, lead (IV) oxide, and sulfuric acid needed for the battery to function properly. Theoretically, a lead storage battery should last forever. In practice, the recharging is not $100\%$ efficient because some of the lead (II) sulfate falls from the electrodes and collects on the bottom of the cells.
Warning
Lead storage batteries contain a significant amount of lead so they must always be disposed of properly.
Other Batteries
The nickel–cadmium, or NiCad, battery (Figure $6$) is used in small electrical appliances and devices like drills,
portable vacuum cleaners, and AM/FM digital tuners. It consists of a nickel-plated cathode, cadmium-plated anode, and a potassium hydroxide electrode. The positive and negative plates, which are prevented from shorting by the separator, are rolled together and put into the case. This is a “jelly-roll” design and allows the NiCd cell to deliver much more current than a similar-sized alkaline battery.
The voltage is about 1.2 V to 1.25 V as the battery discharges. When properly treated, a NiCd battery can be recharged about 1000 times.
Warning
Cadmium is a toxic heavy metal so NiCd batteries should never be opened or put into the regular trash.
Lithium ion batteries are among the most popular rechargeable batteries and are used in many portable electronic devices. The battery voltage is about 3.7 V. Lithium batteries are popular because they can provide a large amount current, are lighter than comparable batteries of other types, produce a nearly constant voltage as they discharge, and only slowly lose their charge when stored. Specialized lithium-iodide (polymer) batteries find application in many long-life, critical devices, such as pacemakers and other implantable electronic medical devices. These devices are designed to last 15 or more years.
Disposable primary lithium batteries must be distinguished from secondary lithium-ion or a lithium-polymer. The term "lithium battery" refers to a family of different lithium-metal chemistries, comprising many types of cathodes and electrolytes but all with metallic lithium as the anode. Lithium batteries are widely used in portable consumer electronic devices, and in electric vehicles ranging from full sized vehicles to radio controlled toys.
(0.197 to 0.984 in) in diameter and 1 to 6 mm (0.039 to 0.236 in) high — like a button on a garment, hence the name. A metal can forms the bottom body and positive terminal of the cell. An insulated top cap is the negative terminal. Button cells are single cells, usually disposable primary cells. Common anode materials are zinc or lithium. Common cathode materials are manganese dioxide, silver oxide, carbon monofluoride, cupric oxide or oxygen from the air. Mercuric oxide button cells were formerly common, but are no longer available due to the toxicity and environmental effects of mercury.
Warning
Button cells are very dangerous for small children. Button cells that are swallowed can cause severe internal burns and significant injury or death.
Fuel Cells
A fuel cell is a device that converts chemical energy into electrical energy. Fuel cells are similar to
batteries but require a continuous source of fuel, often hydrogen. They will continue to produce electricity as long as fuel is available. Hydrogen fuel cells have been used to supply power for satellites, space capsules, automobiles, boats, and submarines (Figure $8$).
The voltage is about 0.9 V. The efficiency of fuel cells is typically about 40% to 60%, which is higher than the typical internal combustion engine (25% to 35%) and, in the case of the hydrogen fuel cell, produces only water as exhaust. Although fuel cells are an essentially pollution-free means of obtaining electrical energy, their expense and technological complexity have thus far limited their applications.
Electrolysis
In this chapter, we have described various galvanic cells in which a spontaneous chemical reaction is used to generate electrical energy. In an electrolytic cell, however, the opposite process, called electrolysis, occurs: an external voltage is applied to drive a nonspontaneous reaction. Electrolysis has many commercial and industrial applications. Electrometallurgy is the process of reduction of metals from metallic compounds to obtain the pure form of metal using electrolysis. Aluminium, lithium, sodium, potassium, magnesium, calcium, and in some cases copper, are produced in this way. Chlorine and sodium hydroxide are the major products of the electrolysis of brine, a water/sodium chloride mixture. Water electrolysis is used to generate oxygen for spacecraft and nuclear submarines.
Electrolysis is also used in the cleaning and preservation of old artifacts. Because the process separates the non-metallic particles from the metallic ones, it is very useful for cleaning a wide variety of metallic objects, from old coins to even larger objects including rusted cast iron cylinder blocks and heads when rebuilding automobile engines. Rust removal from small iron or steel objects by electrolysis can be done in a home workshop using simple materials such as a plastic bucket, tap water, lengths of rebar, washing soda, baling wire, and a battery charger.
An important use for electrolytic cells is in electroplating. Electroplating results in a thin coating of one metal on top of a conducting surface. Reasons for electroplating include making the object more corrosion resistant, strengthening the surface, producing a more attractive finish, or for purifying metal. The metals commonly used in electroplating include cadmium, chromium, copper, gold, nickel, silver, and tin. Common consumer products include silver-plated or gold-plated tableware, chrome-plated automobile parts, and jewelry. We can get an idea of how this works by investigating how silver-plated tableware is produced (Figure $9$).
Summary
• Electrochemistry is a branch of chemistry that deals with the interconversion of chemical energy and electrical energy.
• Batteries are galvanic cells, or a series of cells, that produce an electric current.
• There are two basic types of batteries: primary and secondary.
• Primary batteries are “single use” and cannot be recharged. Dry cells and (most) alkaline batteries are examples of primary batteries.
• The second type is rechargeable and is called a secondary battery. Examples of secondary batteries include nickel-cadmium (NiCd), lead acid, and lithium ion batteries.
• Fuel cells are similar to batteries in that they generate an electrical current, but require continuous addition of fuel and oxidizer. The hydrogen fuel cell uses hydrogen and oxygen from the air to produce water, and is generally more efficient than internal combustion engines.
• Electrolysis occurs when an external voltage is applied to drive a nonspontaneous reaction. | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/08%3A_Oxidation_and_Reduction/8.03%3A_Electrochemistry-_Cells_and_Batteries.txt |
Learning Objectives
• Understand the process of corrosion.
• Understand the process of explosion.
Corrosion is a galvanic process by which metals deteriorate through oxidation—usually but not always to their oxides. For example, when exposed to air, iron rusts, silver tarnishes, and copper and brass acquire a bluish-green surface called a patina. Of the various metals subject to corrosion, iron is by far the most important commercially. An estimated \$100 billion per year is spent in the United States alone to replace iron-containing objects destroyed by corrosion.
Explosions can occur in nature due to a large influx of energy. Most natural explosions arise from volcanic processes of various sorts. Chemical explosions, usually involve a rapid and violent oxidation reaction that produces large amounts of hot gas. Gunpowder was the first explosive to be discovered and put to use. Other notable early developments in chemical explosive technology were Frederick Augustus Abel's development of nitrocellulose in 1865 and Alfred Nobel's invention of dynamite in 1866. Chemical explosions (both intentional and accidental) are often initiated by an electric spark or flame in the presence of oxygen. Accidental explosions may occur in fuel tanks, rocket engines, etc.
The Rusting of Iron
Perhaps the most familiar example of corrosion is the formation of rust on iron. Iron will rust when it is exposed to oxygen and water. The main steps in the rusting of iron appear to involve the following. Once exposed to the atmosphere, iron rapidly oxidizes.
$\ce{Fe(s) -> Fe^{2+}(aq) + 2e^{-}} \tag{cathode}$
The electrons reduce oxygen in the air in acidic solutions.
$\ce{O2(g) + 4H^{+}(aq) + 4e^{-} -> 2H_2O(l)} \tag{anode}$
$\ce{2Fe(s) + O2(g) + 4H^{+}(aq) -> 2Fe^{2+}(aq) + 2H2O(l)} \tag{over all}$
What we call rust is hydrated iron(III) oxide, which forms when iron(II) ions react further with oxygen.
$\ce{4Fe^{2+}(aq) + O2(g) + (4+2x)H2O(l) -> 2Fe2O3 \cdot x H2O(s) + 8H^{+}(aq)} \nonumber$
The number of water molecules is variable, so it is represented by x. Unlike the patina on copper, the formation of rust does not create a protective layer and so corrosion of the iron continues as the rust flakes off and exposes fresh iron to the atmosphere.
One way to keep iron from corroding is to keep it painted. Other strategies include alloying the iron with other metals. For example, stainless steel is mostly iron with a bit of chromium. Zinc-plated or galvanized iron uses a different strategy. Zinc is more easily oxidized than iron because zinc has a lower reduction potential. Since zinc has a lower reduction potential, it is a more active metal. Thus, even if the zinc coating is scratched, the zinc will still oxidize before the iron. This suggests that this approach should work with other active metals.
Changing Colors
The Statue of Liberty is a landmark every American recognizes. The Statue of Liberty is easily identified by its height, stance, and unique blue-green color. When this statue was first delivered from France, its appearance was not green. It was brown, the color of its copper “skin.” So how did the Statue of Liberty change colors? The change in appearance was a direct result of corrosion. The copper that is the primary component of the statue slowly underwent oxidation from the air. The oxidation-reduction reactions of copper metal in the environment occur in several steps. Copper metal is oxidized to copper(I) oxide ($\ce{Cu_2O}$), which is red, and then to copper(II) oxide, which is black.
$\ce{2Cu(s)} +\ce{1/2 O2(g)} \rightarrow \underset{\text{red}}{\ce{Cu2O(s)} } \nonumber$
$\ce{Cu2O(s)} +\ce{1/2 O2(g)}\rightarrow \underset{\text{black}}{\ce{2CuO(s)} } \nonumber$
Coal, which was often high in sulfur, was burned extensively in the early part of the last century. As a result, sulfur trioxide, carbon dioxide, and water all reacted with the $\ce{CuO}$.
$\ce{2CuO(s)} + \ce{CO2(g)} + \ce{H2O(l)} \rightarrow \underset{\text{green}}{\ce{Cu2CO3(OH)2(s)} } \nonumber$
$\ce{3CuO(s)} + \ce{2CO2(g)}+\ce{H2O(l)} \rightarrow \underset{\text{blue}}{\ce{Cu_2(CO_3)2(OH)2(s)} } \nonumber$
$\ce{4CuO(s)} + \ce{SO3(g)}+\ce{3H2O(l)} \rightarrow \underset{\text{green}}{\ce{Cu_4SO4(OH)6(s)} } \nonumber$
These three compounds are responsible for the characteristic blue-green patina seen today. Fortunately, formation of the patina created a protective layer on the surface, preventing further corrosion of the copper skin. The formation of the protective layer is a form of passivation, which is discussed elsewhere.
Silver Tarnishes
Even though coinage metals are known to be resistant to oxygen in air, silver will tarnish due to $\ce{H2S(g)}$
$\ce{2Ag(s) + H_2S(g) -> Ag2S(s) + H2(g)} \nonumber$
Explosive Reactions
TNT is one of the most commonly used explosives for military, industrial, and mining applications. TNT has been used in conjunction with hydraulic fracturing, a process used to recover oil and gas from shale formations. The technique involves displacing and detonating nitroglycerin in hydraulically induced fractures followed by wellbore shots using pelletized TNT.
TNT is valued partly because of its insensitivity to shock and friction, with reduced risk of accidental detonation compared to more sensitive explosives such as nitroglycerin. TNT melts at 80 °C (176 °F), far below the temperature at which it will spontaneously detonate, allowing it to be poured or safely combined with other explosives. TNT neither absorbs nor dissolves in water, which allows it to be used effectively in wet environments. To detonate, TNT must be triggered by a pressure wave from a starter explosive, called an explosive booster.
Although blocks of TNT are available in various sizes (e.g. 250 g, 500 g, 1,000 g), it is more commonly encountered in synergistic explosive blends comprising a variable percentage of TNT plus other ingredients
Upon detonation, TNT decomposes as follows:
$\ce{2 C7H5N3O6 → 3 N2 + 5 H2O + 7 CO + 7 C} \nonumber$
$\ce{2 C7H5N3O6 → 3 N2 + 5 H2 + 12 CO + 2 C} \nonumber$
The reaction is exothermic but has a high activation energy in the gas phase (~62 kcal/mol). The condensed phases (solid or liquid) show markedly lower activation energies of roughly 35 kcal/mol due to unique bimolecular decomposition routes at elevated densities. Because of the production of carbon, TNT explosions have a sooty appearance. Because TNT has an excess of carbon, explosive mixtures with oxygen-rich compounds can yield more energy per kilogram than TNT alone. During the 20th century, amatol, a mixture of TNT with ammonium nitrate was a widely used military explosive.
TNT can be detonated with a high velocity initiator or by efficient concussion. For many years, TNT used to be the reference point for the Figure of Insensitivity. TNT had a rating of exactly 100 on the "F of I" scale. The reference has since been changed to a more sensitive explosive called RDX, which has an F of I rating of 80.
ANFO (or AN/FO, for ammonium nitrate/fuel oil) is a widely used bulk industrial explosive. Its name is commonly pronounced as "an-fo". It consists of 94% porous prilled ammonium nitrate ($\ce{NH4NO3}$) (AN), which acts as the oxidizing agent and absorbent for the fuel, and 6% number 2 fuel oil (FO). The fuel component of ANFO is typically diesel, but kerosene, coal dust, racing fuel, or even molasses have been used instead. Finely powdered aluminium in the mixture will sensitise it to detonate more readily.
The chemistry of ANFO detonation is the reaction of ammonium nitrate with a long-chain alkane ($\ce{C_nH_{2n+2}}$) to form nitrogen, carbon dioxide, and water.
$\ce{NH4NO3 + C_nH2_{n+2} → xN2 + yH2O + zCO2} \nonumber$
In an ideal stoichiometrically balanced reaction, ANFO is composed of about 94.3% AN and 5.7% FO by \(\ce{weight. In practice, a slight excess of fuel oil is added, as underdosing results in reduced performance while overdosing merely results in more post-blast fumes. When detonation conditions are optimal, the aforementioned gases are the only products. In practical use, such conditions are impossible to attain, and blasts produce moderate amounts of toxic gases such as carbon monoxide and nitrogen oxides (NOx).
$\ce{NH4NO3 + C_nH_{2n+2} → vCO + wNOx + yH2O + zCO2} \nonumber$
ANFO has found wide use in coal mining, quarrying, metal mining, and civil construction in applications where its low cost and ease of use may outweigh the benefits of other explosives, such as water resistance, oxygen balance, higher detonation velocity, or performance in small-diameter columns. ANFO is also widely used in avalanche hazard mitigation.
Summary
• Corrosion and explosion are two major redox processes.
• Examples of corrosion (metal deterioration) and explosion (using TNT and ANFO) are given. | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/08%3A_Oxidation_and_Reduction/8.04%3A_Corrosion_and_Explosion.txt |
Learning Objectives
• Know the occurrence and properties of oxygen.
• Know reactions involving oxygen as an oxidizing agent.
Elemental oxygen is a strong oxidizing agent. It reacts with most other elements and many compounds. Oxygen was isolated by Michael Sendivogius before 1604, but it is commonly believed that the element was discovered independently by Carl Wilhelm Scheele, in Uppsala, in 1773 or earlier, and Joseph Priestley in Wiltshire, in 1774. The name oxygen was coined in 1777 by Antoine Lavoisier, who first recognized oxygen as a chemical element and correctly characterized the role it plays in combustion.
Oxygen: Occurrence and Properties
By mass, oxygen is the third-most abundant element in the universe, after hydrogen and helium. Oxygen is the most abundant element on the earth’s crust. About 50% of the mass of the earth’s crust consists of oxygen (combined with other elements, principally silicon). Oxygen occurs as O2 molecules and, to a limited extent, as O3 (ozone) molecules in air. It forms about 20% of the mass of the air. About 89% of water by mass consists of combined oxygen. In combination with carbon, hydrogen, and nitrogen, oxygen is a large part of plants and animals.
Oxygen is a colorless, odorless, and tasteless gas at ordinary temperatures. It is slightly denser than air. Although it is only slightly soluble in water (49 mL of gas dissolves in 1 L at STP), oxygen’s solubility is very important to aquatic life.
Oxygen is essential in combustion processes such as the burning of fuels. All combustion reactions are also examples of redox reactions. A combustion reaction occurs when a substance reacts with oxygen to create heat. One example is the combustion of octane, the principle component of gasoline:
$\ce{2 C_8H_{18} (l) + 25 O_2 (g) -> 16 CO_2 (g) + 18 H_2O (g) } \nonumber$
Combustion reactions are a major source of energy for modern industry. Plants and animals use the oxygen from the air in respiration. The administration of oxygen-enriched air is an important medical practice when a patient is receiving an inadequate supply of oxygen because of shock, pneumonia, or some other illness.
The chemical industry employs oxygen for oxidizing many substances. A significant amount of oxygen produced commercially is important in the removal of carbon from iron during steel production. Large quantities of pure oxygen are also necessary in metal fabrication and in the cutting and welding of metals with oxyhydrogen and oxyacetylene torches. Liquid oxygen is important to the space industry. It is an oxidizing agent in rocket engines. It is also the source of gaseous oxygen for life support in space.
As we know, oxygen is very important to life. The energy required for the maintenance of normal body functions in human beings and in other organisms comes from the slow oxidation of chemical compounds. Oxygen is the final oxidizing agent in these reactions. In humans, oxygen passes from the lungs into the blood, where it combines with hemoglobin, producing oxyhemoglobin. In this form, blood transports the oxygen to tissues, where it is transferred to the tissues. The ultimate products are carbon dioxide and water. The blood carries the carbon dioxide through the veins to the lungs, where the blood releases the carbon dioxide and collects another supply of oxygen. Digestion and assimilation of food regenerate the materials consumed by oxidation in the body; the energy liberated is the same as if the food burned outside the body.
Oxygen: Reactions with Other Elements
Oxygen reacts directly at room temperature or at elevated temperatures with all other elements except the noble gases, the halogens, and few second- and third-row transition metals of low reactivity (those with higher reduction potentials than copper). Rust is an example of the reaction of oxygen with iron. The more active metals form peroxides or superoxides. Less active metals and the nonmetals give oxides. Two examples of these reactions are:
$\ce{2Mg (s) + O2(g) -> 2MgO(s)} \nonumber$
$\ce{P4(s)+ 5O2 (g) -> P4O10(s)} \nonumber$
The oxides of halogens, at least one of the noble gases, and metals with higher reduction potentials than copper do not form by the direct action of the elements with oxygen.
Oxygen: Reaction with Compounds
Elemental oxygen also reacts with some compounds. If it is possible to oxidize any of the elements in a given compound, further oxidation by oxygen can occur. For example, hydrogen sulfide, H2S, contains sulfur with an oxidation state of 2−. Because the sulfur does not exhibit its maximum oxidation state, we would expect H2S to react with oxygen. It does, yielding water and sulfur dioxide. The reaction is:
$\ce{2H2S}(g)+\ce{3O2}(g)⟶\ce{2H2O}(l)+\ce{2SO2}(g) \nonumber$
It is also possible to oxidize oxides such as CO and P4O6 that contain an element with a lower oxidation state. The ease with which elemental oxygen picks up electrons is mirrored by the difficulty of removing electrons from oxygen in most oxides. Of the elements, only the very reactive fluorine can oxidize oxides to form oxygen gas.
Most nonmetals react with oxygen to form nonmetal oxides. Depending on the available oxidation states for the element, a variety of oxides might form. Fluorine will combine with oxygen to form fluorides such as OF2, where the oxygen has a 2+-oxidation state.
Sulfur Oxygen Compounds
The two common oxides of sulfur are sulfur dioxide, SO2, and sulfur trioxide, SO3. The odor of burning sulfur comes from sulfur dioxide. Sulfur dioxide, occurs in volcanic gases Figure $1$ and in the atmosphere near industrial plants that burn fuel containing sulfur compounds.
Ozone: Another Form of Oxygen
When dry oxygen is passed between two electrically charged plates, ozone (O3, illustrated in Figure $2$), an allotrope of oxygen possessing a distinctive odor, forms. The formation of ozone from oxygen is an endothermic reaction, in which the energy comes from an electrical discharge, heat, or ultraviolet light:
$\ce{3O2}(g)\xrightarrow{\ce{electric\: discharge}}\ce{2O3}(g) \nonumber$
The sharp odor associated with sparking electrical equipment is due, in part, to ozone.
Ozone forms naturally in the upper atmosphere by the action of ultraviolet light from the sun on the oxygen there. Most atmospheric ozone occurs in the stratosphere, a layer of the atmosphere extending from about 10 to 50 kilometers above the earth’s surface. This ozone acts as a barrier to harmful ultraviolet light from the sun by absorbing it via a chemical decomposition reaction:
$\ce{O3}(g)\xrightarrow{\ce{ultraviolet\: light}}\ce{O}(g)+\ce{O2}(g) \nonumber$
The reactive oxygen atoms recombine with molecular oxygen to complete the ozone cycle. The presence of stratospheric ozone decreases the frequency of skin cancer and other damaging effects of ultraviolet radiation. It has been clearly demonstrated that chlorofluorocarbons, CFCs (known commercially as Freons), which were present as aerosol propellants in spray cans and as refrigerants, caused depletion of ozone in the stratosphere. This occurred because ultraviolet light also causes CFCs to decompose, producing atomic chlorine. The chlorine atoms react with ozone molecules, resulting in a net removal of O3 molecules from stratosphere. This process is explored in detail in our coverage of chemical kinetics. There is a worldwide effort to reduce the amount of CFCs used commercially, and the ozone hole is already beginning to decrease in size as atmospheric concentrations of atomic chlorine decrease. While ozone in the stratosphere helps protect us, ozone in the troposphere is a problem. This ozone is a toxic component of photochemical smog.
The uses of ozone depend on its reactivity with other substances. It can be used as a bleaching agent for oils, waxes, fabrics, and starch: It oxidizes the colored compounds in these substances to colorless compounds. It is an alternative to chlorine as a disinfectant for water.
Low level ozone (or tropospheric ozone) is an atmospheric pollutant.[42] It is not emitted directly by car engines or by industrial operations, but formed by the reaction of sunlight on air containing hydrocarbons and nitrogen oxides that react to form ozone directly at the source of the pollution or many kilometers down wind.
Ozone reacts directly with some hydrocarbons such as aldehydes and thus begins their removal from the air, but the products are themselves key components of smog.
Ozone can be used to remove iron and manganese from water, forming a precipitate which can be filtered:
$2 Fe^{2+} + O_3 + 5 H_2O → 2 Fe(OH)_3(s) + O_2 + 4 H^+ \nonumber$
$2 Mn^{2+} + 2 O_3 + 4 H_2O → 2 MnO(OH)_2(s) + 2 O_2 + 4 H^+ \nonumber$
Ozone will also oxidize dissolved hydrogen sulfide in water to sulfurous acid:
$3 O_3 + H_2S → H_2SO_3 + 3 O_2 \nonumber$
These three reactions are central in the use of ozone based well water treatment.
Ozone will also detoxify cyanides by converting them to cyanates.
$CN^− + O_3 → CNO^− + O_2 \nonumber$
Ozone will also completely decompose urea:
$(NH_2)_2CO + O_3 → N_2 + CO_2 + 2 H_2O \nonumber$
Other Common Oxidizing Agents
Chlorine is usually used (in the form of hypochlorous acid) to kill bacteria and other microbes in drinking water supplies and public swimming pools. In most private swimming pools, chlorine itself is not used, but rather sodium hypochlorite, formed from chlorine and sodium hydroxide, or solid tablets of chlorinated isocyanurates. The drawback of using chlorine in swimming pools is that the chlorine reacts with the proteins in human hair and skin. The distinctive 'chlorine aroma' associated with swimming pools is not the result of chlorine itself, but of chloramine, a chemical compound produced by the reaction of free dissolved chlorine with amines in organic substances. As a disinfectant in water, chlorine is more than three times as effective against Escherichia coli as bromine, and more than six times as effective as iodine. Increasingly, chloramine itself is being directly added to drinking water for purposes of disinfection, a process known as chloramination.
It is often impractical to store and use poisonous chlorine gas for water treatment, so alternative methods of adding chlorine are used. These include hypochlorite solutions, which gradually release chlorine into the water, and compounds like sodium dichloro-s-triazinetrione (dihydrate or anhydrous), sometimes referred to as "dichlor", and trichloro-s-triazinetrione, sometimes referred to as "trichlor". These compounds are stable while solid and may be used in powdered, granular, or tablet form. When added in small amounts to pool water or industrial water systems, the chlorine atoms hydrolyze from the rest of the molecule forming hypochlorous acid (HOCl), which acts as a general biocide, killing germs, micro-organisms, algae, and so on.
Breathalizers
The breathalyzer is a redox reaction. When the potassium dichromate reacts with ethanol it loses an oxygen atom (gets reduced), going from the orange dichromate to the green chromium sulfate. At the same time dichromate is being reduced, ethanol gains an oxygen atom (gets oxidized), forming acetic acid. The sulfuric acid helps to remove the ethanol from the exhaled air into the test solution and also provides the necessary acidic conditions.
In this reaction, the chromium atom is reduced from $Cr^{6+}$ to $Cr^{3+}$, and the ethanol is oxidized to acetic acid.
Hydrogen peroxide can be used for the sterilization of various surfaces, including surgical tools and may be deployed as a vapor (VHP) for room sterilization. $\ce{H2O2}$ demonstrates broad-spectrum efficacy against viruses, bacteria, yeasts, and bacterial spores. In general, greater activity is seen against Gram-positive than Gram-negative bacteria; however, the presence of catalase or other peroxidases in these organisms can increase tolerance in the presence of lower concentrations. Higher concentrations of $\ce{H2O2}$ (10 to 30%) and longer contact times are required for sporicidal activity.
Hydrogen peroxide is seen as an environmentally safe alternative to chlorine-based bleaches, as it degrades to form oxygen and water and it is generally recognized as safe as an antimicrobial agent by the U.S. Food and Drug Administration (FDA).
Historically hydrogen peroxide was used for disinfecting wounds, partly because of its low cost and prompt availability compared to other antiseptics. It is now thought to inhibit healing and to induce scarring because it destroys newly formed skin cells. Only a very low concentration of
$\ce{H2O2}$ can induce healing, and only if not repeatedly applied. Surgical use can lead to gas embolism formation. Despite this, it is still used for wound treatment in many countries but is also prevalent as a major first aid antiseptic in the United States
Diluted $\ce{H2O2}$ (between 1.9% and 12%) mixed with aqueous ammonia is used to bleach human hair. The chemical's bleaching property lends its name to the phrase "peroxide blonde". Hydrogen peroxide is also used for tooth whitening. It can be found in most whitening toothpastes.
Almost all applications of potassium permanganate ($\ce{KMnO4}$) exploit its oxidizing properties. As a strong oxidant that does not generate toxic byproducts, $\ce{KMnO4}$ has many niche uses. Potassium permanganate is used for a number of skin conditions. This includes fungal infections of the foot, impetigo, pemphigus, superficial wounds, dermatitis, and tropical ulcers. It is on the WHO Model List of Essential Medicines, the most important medications needed in a basic health system. Potassium permanganate is used extensively in the water treatment industry.
Benzoyl peroxide (BPO) is a medication and industrial chemical. As a medication, it is used to treat mild to moderate acne. For more severe cases, it may be used with other treatments. Some versions are sold mixed with antibiotics such as clindamycin. Other uses include bleaching flour, hair bleaching, teeth whitening, and textile bleaching.[5] It is also used in the plastic industry
Sodium hypochlorite is most often encountered as a pale greenish-yellow dilute solution commonly known as liquid bleach or simply bleach, a household chemical widely used (since the 18th century) as a disinfectant or a bleaching agent. The compound in solution is unstable and easily decomposes, liberating chlorine, which is the active principle of such products. Indeed, sodium hypochlorite is the oldest and still most important chlorine-based bleach.
Warning
While sodium hypochlorite is non-toxic, its corrosive properties, common availability, and reaction products make it a significant safety risk. In particular, mixing liquid bleach with other cleaning products, such as acids or ammonia, may produce toxic fumes.
Summary
• Oxygen is one of the most reactive elements. This reactivity, coupled with its abundance, makes the chemistry of oxygen very rich and well understood.
• Many compounds of the representative metals and non metals react with oxygen to form oxides.
• Other than oxygen, ozone, chlorine, potassium dichromate, hydrogen peroxide, potassium peroxide, benzoyl peroxide, sodium hypochlorite are common oxidizing agents mentioned in this section, that have many beneficial uses. | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/08%3A_Oxidation_and_Reduction/8.05%3A_Oxygen_-_An_Abundant_and_Essential_Oxidizing_Agent.txt |
Learning Objectives
• Know the occurrence and properties of hydrogen.
• Know reactions involving common reducing agents such as carbon, hydrogen, and antioxidants.
Metal Extraction
Smelting is a process of applying heat to ore in order to extract a base metal. It uses heat and a chemical reducing agent to decompose the ore, leaving the metal base behind. It is used to extract many metals from their ores, including silver, iron, copper, and other base metals. The reducing agent is commonly a source of carbon, such as coke—or, in earlier times, charcoal. Coke is a grey, hard, and porous fuel with a high carbon content and few impurities, made by heating coal or oil in the absence of air.
Coke is used in smelting iron ore for example in a blast furnace. The reaction can be represented as follows:
$2Fe_2O_3 + 3C → 4 Fe + 3 CO_2 \nonumber$
Antioxidants
In food chemistry, the substances known as antioxidants are reducing agents. Ascorbic acid (vitamin C; $\ce{C6H8O6}$) is thought to retard potentially damaging oxidation of living cells. In the process, it is oxidized to dehydroascorbic acid ($\ce{C6H6O6}$). In the stomach, ascorbic acid reduces the nitrite ion ($\ce{NO_2^{−}}$) to nitric oxide ($\ce{NO}$):
$\ce{C_6H_8O_6 + 2H^{+} + 2NO_2^{−} \rightarrow C_6H_6O_6 + 2H_2O + 2NO} \label{Eq7}$
If reaction in Equation $\ref{Eq7}$ did not occur, nitrite ions from foods would oxidize the iron in hemoglobin, destroying its ability to carry oxygen.
Tocopherol (vitamin E) is also an antioxidant. In the body, vitamin E is thought to act by scavenging harmful by-products of metabolism, such as the highly reactive molecular fragments called free radicals. In foods, vitamin E acts to prevent fats from being oxidized and thus becoming rancid. Vitamin C is also a good antioxidant (Figure $2$).
Hydrogen as a Reducing Agent
Large quantities of H2 are needed in the petroleum and chemical industries. The largest application of H2 is for the processing ("upgrading") of fossil fuels, and in the production of ammonia. Mass production of Ammonia mostly uses the Haber–Bosch process, reacting hydrogen (H2) and nitrogen (N2) at a moderately-elevated temperature (450 °C) and high pressure (100 standard atmospheres (10,000 kPa)):
$3 H_2(g)+ N_2(g) → 2 NH_3(g) \nonumber$
H2 has several other important uses. H2 is used as a hydrogenating agent (Chapter 17), particularly in increasing the level of saturation of unsaturated fats and oils (found in items such as margarine), and in the production of methanol. It is similarly the source of hydrogen in the manufacture of hydrochloric acid. H2 is also used as a reducing agent of metallic ores.
Hydrogen gas is highly flammable producing a large amount of heat when it reacts with oxygen gas as shown below.
$2 H_2(g) + O_2(g) → 2 H_2O(l) + 572 kJ \nonumber$
Hydrogen gas forms explosive mixtures with air in concentrations from 4–74% and with chlorine at 5–95%. The explosive reactions may be triggered by spark, heat, or sunlight.
A Closer Look at Hydrogen
Hydrogen is among the ten most abundant elements on the planet, but very little is found in elemental form due to its low density and reactivity. Much of the terrestrial hydrogen is locked up in water molecules and organic compounds like hydrocarbons.
Hydrogen is the fuel for reactions of the Sun and other stars (fusion reactions). Hydrogen is the lightest and most abundant element in the universe. About 70%- 75% of the universe is composed of hydrogen by mass. All stars are essentially large masses of hydrogen gas that produce enormous amounts of energy through the fusion of hydrogen atoms at their dense cores. In smaller stars, hydrogen atoms collided and fused to form helium and other light elements like nitrogen and carbon(essential for life). In the larger stars, fusion produces the lighter and heavier elements like calcium, oxygen, and silicon.
On Earth, hydrogen is mostly found in association with oxygen; its most abundant form being water (H2O). Hydrogen is only .9% by mass and 15% by volume abundant on the earth, despite water covering about 70% of the planet. Because hydrogen is so light, there is only 0.5 ppm (parts per million) in the atmosphere, which is a good thing considering it is EXTREMELY flammable.
Hydrogen gas can be prepared by reacting a dilute strong acid like hydrochloric acids with an active metal. The metal becomes oxides, while the H+ (from the acid) is reduced to hydrogen gas. This method is only practical for producing small amounts of hydrogen in the lab, but is much too costly for industrial production:
$Zn_{(s)} + 2H^+_{(aq)} \rightarrow Zn^{2+}_{(aq)} + H_{2(g)} \nonumber$
The purest form of H2(g) can come from electrolysis of H2O(l), the most common hydrogen compound on this plant. This method is also not commercially viable because it requires a significant amount of energy ($\Delta H = 572 \;kJ$):
$2H_2O_{(l)} \rightarrow 2H_{2(g)} + O_{2(g)} \nonumber$
$H_2O$ is the most abundant form of hydrogen on the planet, so it seems logical to try to extract hydrogen from water without electrolysis of water. To do so, we must reduce the hydrogen with +1 oxidation state to hydrogen with 0 oxidation state (in hydrogen gas).
Three commonly used reducing agents are carbon (in coke or coal), carbon monoxide, and methane. These react with water vapor form H2(g):
$C_{(s)} + 2H_2O_{(g)} \rightarrow CO(g) + H_{2(g)} \nonumber$
$CO_{(g)} + 2H_2O_{(g)} \rightarrow CO2 + H_{2(g)} \nonumber$
Reforming of Methane:
$CH_{4(g)} + H_2O_{(g)} \rightarrow CO(g) + 3H_{2(g)} \nonumber$
These three methods are most industrially feasible (cost effective) methods of producing H2(g).
Summary
• Common reducing agents include carbon (in the form of coke or coal), hydrogen gas, as well as those substances referred to in the food chemistry as antioxidants (e.g. ascorbic acid and vitamin E).
• Various reactions involving reducing agents were given.
Contributors and Attributions
• Libretext:The Basics of GOB Chemistry (Ball et al.)
• Wikipedia
• Ridhi Sachdev (UC Davis)
• Stephen R. Marsden
8.07: Oxidations Reduction and Living Things
Learning Objective
• Know the overall reactions involved in respiration and photosynthesis.
Oxidation and reduction reactions are especially crucial in biological processes such as cellular respiration and photosynthesis.
Cellular respiration is a set of metabolic reactions and processes that take place in the cells of organisms to convert biochemical energy from nutrients into adenosine triphosphate (ATP), and then release waste products. Cellular respiration is considered an exothermic redox reaction which releases heat. The overall reaction occurs in a series of biochemical steps, most of which are redox reactions themselves. Although cellular respiration is technically a combustion reaction, it clearly does not resemble one when it occurs in a living cell because of the slow release of energy from the series of reactions.
Nutrients that are commonly used by animal and plant cells in respiration include sugar, amino acids and fatty acids, and the most common oxidizing agent (electron acceptor) is molecular oxygen (O2).
The reaction for aerobic respiration is essentially the reverse of photosynthesis and is simplified as:
$C_6H_{12}O_6 + 6 O_2 → 6 CO_2 + 6 H_2O + 2880 kJ/mol \nonumber$
Green plants carry out the redox reaction that makes possible almost all life on Earth. They do this through a process called photosynthesis, in which carbon dioxide and water are converted to glucose ($\ce{C6H12O6}$). The synthesis of glucose requires a variety of proteins called enzymes and a green pigment called chlorophyll that converts sunlight into chemical energy
A simplified overall formula for photosynthesis is:
$6 CO_2 + 6 H_2O + photons → C_6H_{12}O_6 + 6 O_2 \nonumber$
or simply
carbon dioxide + water + sunlight → glucose + oxygen gas
In this reaction, carbon dioxide is reduced to glucose, and water is oxidized to oxygen gas. Other reactions convert the glucose to more complex carbohydrates, plant proteins, and oils.
Summary
• Redox reactions are common in organic and biological chemistry, including aerobic respiration, and photosynthesis.
Contributors and Attributions
• Libretext: Basics of GOB Chemistry (Ball et al.)
• Wikipedia | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/08%3A_Oxidation_and_Reduction/8.06%3A_Some_Common_Reducing_Agents.txt |
Scientists of the 18th and early 19th centuries studied compounds obtained from plants and animals and labeled them organic because they were isolated from “organized” (living) systems. Compounds isolated from nonliving systems, such as rocks and ores, the atmosphere, and the oceans, were labeled inorganic. For many years, scientists thought organic compounds could be made by only living organisms because they possessed a vital force found only in living systems. The vital force theory began to decline in 1828, when the German chemist Friedrich Wöhler synthesized urea from inorganic starting materials. He reacted silver cyanate (AgOCN) and ammonium chloride (NH4Cl), expecting to get ammonium cyanate (NH4OCN). What he expected is described by the following equation.
$AgOCN + NH_4Cl \rightarrow AgCl + NH_4OCN \label{Eq1}$
Instead, he found the product to be urea (NH2CONH2), a well-known organic material readily isolated from urine. This result led to a series of experiments in which a wide variety of organic compounds were made from inorganic starting materials. The vital force theory gradually went away as chemists learned that they could make many organic compounds in the laboratory.
Today organic chemistry is the study of the chemistry of the carbon compounds, and inorganic chemistry is the study of the chemistry of all other elements. It may seem strange that we divide chemistry into two branches—one that considers compounds of only one element and one that covers the 100-plus remaining elements. However, this division seems more reasonable when we consider that of tens of millions of compounds that have been characterized, the overwhelming majority are carbon compounds.
The word organic has different meanings. Organic fertilizer, such as cow manure, is organic in the original sense; it is derived from living organisms. Organic foods generally are foods grown without synthetic pesticides or fertilizers. Organic chemistry is the chemistry of compounds of carbon.
Carbon is unique among the other elements in that its atoms can form stable covalent bonds with each other and with atoms of other elements in a multitude of variations. The resulting molecules can contain from one to millions of carbon atoms. We previously surveyed organic chemistry by dividing its compounds into families based on functional groups. We begin with the simplest members of a family and then move on to molecules that are organic in the original sense—that is, they are made by and found in living organisms. These complex molecules (all containing carbon) determine the forms and functions of living systems and are the subject of biochemistry.
Organic compounds, like inorganic compounds, obey all the natural laws. Often there is no clear distinction in the chemical or physical properties among organic and inorganic molecules. Nevertheless, it is useful to compare typical members of each class, as in Table $1$.
Table $1$: General Contrasting Properties and Examples of Organic and Inorganic Compounds
Organic Hexane Inorganic NaCl
low melting points −95°C high melting points 801°C
low boiling points 69°C high boiling points 1,413°C
low solubility in water; high solubility in nonpolar solvents insoluble in water; soluble in gasoline greater solubility in water; low solubility in nonpolar solvents soluble in water; insoluble in gasoline
flammable highly flammable nonflammable nonflammable
aqueous solutions do not conduct electricity nonconductive aqueous solutions conduct electricity conductive in aqueous solution
exhibit covalent bonding covalent bonds exhibit ionic bonding ionic bonds
Keep in mind, however, that there are exceptions to every category in this table. To further illustrate typical differences among organic and inorganic compounds, Table $1$ also lists properties of the inorganic compound sodium chloride (common table salt, NaCl) and the organic compound hexane (C6H14), a solvent that is used to extract soybean oil from soybeans (among other uses). Many compounds can be classified as organic or inorganic by the presence or absence of certain typical properties, as illustrated in Table $1$.
Today, organic compounds are key components of plastics, soaps, perfumes, sweeteners, fabrics, pharmaceuticals, and many other substances that we use every day. The value to us of organic compounds ensures that organic chemistry is an important discipline within the general field of chemistry. In this chapter, we discuss why the element carbon gives rise to a vast number and variety of compounds, how those compounds are classified, and the role of organic compounds in representative biological and industrial settings.
Contributors and Attributions
• Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected]).
• TextMap: The Basics of GOB Chemistry (Ball et al.)
09: Organic Chemistry
Scientists of the 18th and early 19th centuries studied compounds obtained from plants and animals and labeled them organic because they were isolated from “organized” (living) systems. Compounds isolated from nonliving systems, such as rocks and ores, the atmosphere, and the oceans, were labeled inorganic. For many years, scientists thought organic compounds could be made by only living organisms because they possessed a vital force found only in living systems. The vital force theory began to decline in 1828, when the German chemist Friedrich Wöhler synthesized urea from inorganic starting materials. He reacted silver cyanate (AgOCN) and ammonium chloride (NH4Cl), expecting to get ammonium cyanate (NH4OCN). What he expected is described by the following equation.
$AgOCN + NH_4Cl \rightarrow AgCl + NH_4OCN \label{Eq1}$
Instead, he found the product to be urea (NH2CONH2), a well-known organic material readily isolated from urine. This result led to a series of experiments in which a wide variety of organic compounds were made from inorganic starting materials. The vital force theory gradually went away as chemists learned that they could make many organic compounds in the laboratory.
Today organic chemistry is the study of the chemistry of the carbon compounds, and inorganic chemistry is the study of the chemistry of all other elements. It may seem strange that we divide chemistry into two branches—one that considers compounds of only one element and one that covers the 100-plus remaining elements. However, this division seems more reasonable when we consider that of tens of millions of compounds that have been characterized, the overwhelming majority are carbon compounds.
The word organic has different meanings. Organic fertilizer, such as cow manure, is organic in the original sense; it is derived from living organisms. Organic foods generally are foods grown without synthetic pesticides or fertilizers. Organic chemistry is the chemistry of compounds of carbon.
Carbon is unique among the other elements in that its atoms can form stable covalent bonds with each other and with atoms of other elements in a multitude of variations. The resulting molecules can contain from one to millions of carbon atoms. We previously surveyed organic chemistry by dividing its compounds into families based on functional groups. We begin with the simplest members of a family and then move on to molecules that are organic in the original sense—that is, they are made by and found in living organisms. These complex molecules (all containing carbon) determine the forms and functions of living systems and are the subject of biochemistry.
Organic compounds, like inorganic compounds, obey all the natural laws. Often there is no clear distinction in the chemical or physical properties among organic and inorganic molecules. Nevertheless, it is useful to compare typical members of each class, as in Table $1$.
Table $1$: General Contrasting Properties and Examples of Organic and Inorganic Compounds
Organic Hexane Inorganic NaCl
low melting points −95°C high melting points 801°C
low boiling points 69°C high boiling points 1,413°C
low solubility in water; high solubility in nonpolar solvents insoluble in water; soluble in gasoline greater solubility in water; low solubility in nonpolar solvents soluble in water; insoluble in gasoline
flammable highly flammable nonflammable nonflammable
aqueous solutions do not conduct electricity nonconductive aqueous solutions conduct electricity conductive in aqueous solution
exhibit covalent bonding covalent bonds exhibit ionic bonding ionic bonds
Keep in mind, however, that there are exceptions to every category in this table. To further illustrate typical differences among organic and inorganic compounds, Table $1$ also lists properties of the inorganic compound sodium chloride (common table salt, NaCl) and the organic compound hexane (C6H14), a solvent that is used to extract soybean oil from soybeans (among other uses). Many compounds can be classified as organic or inorganic by the presence or absence of certain typical properties, as illustrated in Table $1$.
Today, organic compounds are key components of plastics, soaps, perfumes, sweeteners, fabrics, pharmaceuticals, and many other substances that we use every day. The value to us of organic compounds ensures that organic chemistry is an important discipline within the general field of chemistry. In this chapter, we discuss why the element carbon gives rise to a vast number and variety of compounds, how those compounds are classified, and the role of organic compounds in representative biological and industrial settings. | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/09%3A_Organic_Chemistry/9.01%3A_Prelude_to_Organic_Chemistry.txt |
Learning Objectives
• Define and identify alkanes, alkenes, alkynes, and cyclic hydrocarbons.
• List some properties of hydrocarbons.
• Identify and name simple (straight-chain) alkanes given formulas and write formulas for straight-chain alkanes given their names.
• Define structural and geometric isomers.
• Write condensed structural formulas and draw line-angle formulas given complete structural formulas for alkanes,alkenes, alkynes, and cyclic hydrocarbons.
The simplest organic compounds are those composed of only two elements: carbon and hydrogen. These compounds are called hydrocarbons. Hydrocarbons themselves are separated into two types: aliphatic hydrocarbons and aromatic hydrocarbons. Aliphatic hydrocarbons are hydrocarbons based on chains of C atoms. There are three types of aliphatic hydrocarbons. Alkanes are aliphatic hydrocarbons with only single covalent bonds. Alkenes are hydrocarbons that contain at least one C–C double bond, and alkynes are hydrocarbons that contain a C–C triple bond. Occasionally, we find an aliphatic hydrocarbon with a ring of C atoms; these hydrocarbons are called cycloalkanes (or cycloalkenes or cycloalkynes). Because alkanes have the maximum number of H atoms possible according to the rules of covalent bonds, alkanes are also referred to as saturated hydrocarbons.
Note
The word saturated has the same meaning for hydrocarbons as it does for the dietary fats and oils: the molecule has no carbon-to-carbon double bonds (C=C).
Alkanes
The simplest alkanes have their C atoms bonded in a straight chain; these are called normal alkanes. They are named according to the number of C atoms in the chain. The smallest alkane is methane:
The diagrams representing alkanes are called structural formulas because they show the structure of the molecule. As molecules get larger, structural formulas become more and more complex. One way around this is to use a condensed structural formula, which lists the formula of each C atom in the backbone of the molecule.
condensed structural formula
The condensed formulas show hydrogen atoms right next to the carbon atoms to which they are attached, as illustrated for butane:
The ultimate condensed formula is a line-angle formula (or line drawing) , in which carbon atoms are implied at the corners and ends of lines, and each carbon atom is understood to be attached to enough hydrogen atoms to give each carbon atom four bonds. For example, we can represent pentane (CH3CH2CH2CH2CH3) and isopentane [(CH3)2CHCH2CH3] as follows:
Table $1$ Structural formulas for methane, ethane , and propane
Structural Formula
Ball and Stick Model
Space Filling model
Name Methane Ethane Propane
Condensed Structural Formula CH4 CH3CH3 CH3CH2CH3
Molecular Formula CH4 C2H6 C3H8
Line angle drawing
Example $1$
Provide the molecular formula, the complete structural formula, the condensed structural formula and the line angle formula for a straight chain alkane with 8 carbon atoms.
Solutions
Example $2$ writing formulas for a straight chain alkane with eight carbon atoms.
structural formula Explanation Answer
Molecular Formula
The general formula for an alkane is CnH2n + 2
For an alkane with 8 carbon atoms, C8H(2 × 8) + 2 = C8H18.
C8H18.
Complete Structural Formula Eight carbons first linked together and all remaining bonds to hydrogen to fulfill the octet rule. C1-C2-C3-C4-C5-C6-C7-C8
Condensed Structural Formula
The terminal carbon atoms (C1 and C8) are bonded to three hydrogen atoms each and the six middle carbon atoms are bonded to two hydrogen atoms each.
The six middle carbon atoms are all bonded to two hydrogen atoms so the structural formula can be simplified further.
CH3CH2CH2CH2CH2CH2CH2CH3
CH3(CH2)6CH3
Line Angle Drawing Eight carbon atoms are implied at the corners and ends of lines, and each carbon atom is understood to be attached to enough hydrogen atoms to give each carbon atom four bonds.
Exercise $1$
Provide the molecular formula, the complete structural formula, the condensed structural formula, and the line angle formula for a straight chain alkane with 7 carbon atoms.
Answer
molecular formula: C7H16
complete structural formula:
condensed structural formula: CH3(CH2)5CH3
line angle drawing:
Homologous Series
Methane (CH4), ethane (C2H6), and propane (C3H8) are the beginning of a series of compounds in which any two members in a sequence differ by one carbon atom and two hydrogen atoms—namely, a CH2 unit. The first 10 members of this series are given in Table $1$.
Table $1$ The First 10 Straight-Chain Alkanes
Molecular Formula Condensed Structural Formula Name
CH4 CH4 methane
C2H6 CH3CH3 ethane
C3H8 CH3CH2CH3 propane
C4H10 CH3CH2CH2CH3 butane
C5H12 CH3CH2CH2CH2CH3 pentane
C6H14 CH3(CH2)4CH3 hexane
C7H16 CH3(CH2)5CH3 heptane
C8H18 CH3(CH2)6CH3 octane
C9H20 CH3(CH2)7CH3 nonane
C10H22 CH3(CH2)8CH3 decane
Note
Parentheses in condensed structural formulas indicate that the enclosed grouping of atoms is attached to the adjacent carbon atom.
Consider the series in Figure $2$. The sequence starts with C3H8, and a CH2 unit is added in each step moving up the series. Any family of compounds in which adjacent members differ from each other by a definite factor (here a CH2 group) is called a homologous series. The members of such a series, called homologs, have properties that vary in a regular and predictable manner. The principle of homology gives organization to organic chemistry in much the same way that the periodic table gives organization to inorganic chemistry. Instead of a bewildering array of individual carbon compounds, we can study a few members of a homologous series and from them deduce some of the properties of other compounds in the series.
The principle of homology allows us to write a general formula for alkanes: CnH2n + 2. Using this formula, we can write a molecular formula for any alkane with a given number of carbon atoms. For example, an alkane with eight carbon atoms has the molecular formula C8H(2 × 8) + 2 = C8H18.
Naming (IUPAC Nomenclature)
A systematic way of naming hydrocarbons and other organic compounds has been devised by the International Union of Pure and Applied Chemistry (IUPAC). These rules, used worldwide, are known as the IUPAC System of Nomenclature. A stem name in Table $1$ indicates the number of carbon atoms in the longest continuous chain (LCC). Atoms or groups attached to this carbon chain, called substituents, are then named, with their positions indicated by numbers.
Table $2$Stems That Indicate the Number of Carbon Atoms in Organic Molecules
Stem Number of C Atoms
meth- 1
eth- 2
prop- 3
but- 4
pent- 5
hex- 6
hept- 7
oct- 8
non- 9
dec- 10
Note
A continuous (unbranched) chain of carbon atoms is often called a straight chain even though the tetrahedral arrangement about each carbon gives it a zigzag shape. Straight-chain alkanes are sometimes called normal alkanes, and their names are given the prefix n-. For example, butane is called n-butane. We will not use that prefix here because it is not a part of the system established by the International Union of Pure and Applied Chemistry.
Example $2$
What is the name of an eight carbon alkane?
Solution
The stem that refers to eight carbon atoms is oct plus the suffix -ane for an alkane, so the answer is Octane.
Exercise $2$
Provide the name for the following compounds:
a. seven carbon alkane
b. six carbon alkane
Answer
a. heptane
b. hexane
Isomerism-Branched Chain Alkanes
One of the interesting aspects of organic chemistry is that it is three-dimensional. A molecule can have a shape in space that may contribute to its properties. Molecules can differ in the way the atoms are arranged - the same combination of atoms can be assembled in more than one way. These compounds are known as isomers. Isomers are molecules with the same molecular formulas, but different arrangements of atoms. We will look at some isomer possibilities for alkanes and alkenes.
We can write the structure of butane (C4H10) by stringing four carbon atoms in a row,
–C–C–C–C–
and then adding enough hydrogen atoms to give each carbon atom four bonds:
The compound butane has this structure, but there is another way to put 4 carbon atoms and 10 hydrogen atoms together. Place 3 of the carbon atoms in a row and then branch the fourth one off the middle carbon atom:
Now we add enough hydrogen atoms to give each carbon four bonds.
There is a hydrocarbon that corresponds to this structure, which means that two different compounds have the same molecular formula: C4H10. The two compounds have different properties—for example, one boils at −0.5°C; the other at −11.7°C. Different compounds having the same molecular formula are called isomers.
Notice that C4H10 is depicted with a bent chain in Figure $4$. The four-carbon chain may be bent in various ways because the groups can rotate freely about the C–C bonds. However, this rotation does not change the identity of the compound. It is important to realize that bending a chain does not change the identity of the compound; all of the following represent the same compound, butane:
The structure of isobutane shows a continuous chain of three carbon atoms only, with the fourth attached as a branch off the middle carbon atom of the continuous chain, which is different from the structures of butane (compare the two structures in Figure $4$.
Unlike C4H10, the compounds methane (CH4), ethane (C2H6), and propane (C3H8) do not exist in isomeric forms because there is only one way to arrange the atoms in each formula so that each carbon atom has four bonds.
Next beyond C4H10 in the homologous series is pentane. Each compound has the same molecular formula: C5H12. The compound at the far left is pentane because it has all five carbon atoms in a continuous chain. The compound in the middle is isopentane; like isobutane, it has a one CH3 branch off the second carbon atom of the continuous chain. The compound at the far right, discovered after the other two, was named neopentane (from the Greek neos, meaning “new”). Although all three have the same molecular formula, they have different properties, including boiling points: pentane, 36.1°C; isopentane, 27.7°C; and neopentane, 9.5°C.
Naming Branched Alkanes
You have already learned the basics of nomenclature-the names of the first 10 normal hydrocarbons. Here, we will add some steps to the procedure so you can name branched hydrocarbons.
First, given the structure of an alkane, identify the longest continuous chain of C atoms. Note that the longest chain may not be drawn in a straight line. The longest chain determines the parent name of the hydrocarbon. For example, in the molecule
the longest chain of carbons has six C atoms. Therefore, it will be named as a hexane. However, in the molecule
the longest chain of C atoms is not six, but seven, as shown. So this molecule will be named as a heptane.
The next step is to identify the branches, or substituents, on the main chain. The names of the substituents, or alkyl groups, are derived from the names of the parent hydrocarbons; however, rather than having the ending -ane, the substituent name has the ending -yl. Table $3$, lists the substituent names for the five smallest substituents.
Table $3$ Substituent Names.
Substituent Formula Number of C Atoms Name of Substituent
CH3 1 methyl-
CH3CH2 2 ethyl-
CH3CH2CH2 3 propyl-
CH3CH2CH2CH2 4 butyl-
CH3CH2CH2CH2CH2 5 pentyl-
and so forth and so forth and so forth
In naming the branched hydrocarbon, the name of the substituent is combined with the parent name of the hydrocarbon without spaces. However, there is likely one more step. The longest chain of the hydrocarbon must be numbered, and the numerical position of the substituent must be included to account for possible isomers. As with double and triple bonds, the main chain is numbered to give the substituent the lowest possible number. For example, in this alkane
the longest chain is five C atoms long, so it is a pentane. There is a one-carbon substituent on the third C atom, so there is a methyl group at position 3. We indicate the position using the number, which is followed by a hyphen, the substituent name, and the parent hydrocarbon name-in this case, 3-methylpentane. That name is specific to that particular hydrocarbon and no other molecule. Organic chemistry nomenclature is very specific!
It is common to write the structural formula of a hydrocarbon without the H atoms, for clarity. So we can also represent 3-methylpentane as
where it is understood that any unwritten covalent bonds are bonds with H atoms. With this understanding, we recognize that the structural formula for 3-methylpentane refers to a molecule with the formula of C6H14.
Example $3$:
Name this molecule.
Solution
The longest continuous carbon chain has seven C atoms, so this molecule will be named as a heptane. There is a two-carbon substituent on the main chain, which is an ethyl group. To give the substituent the lowest numbering, we number the chain from the right side and see that the substituent is on the third C atom. So this hydrocarbon is 3-ethylheptane.
Exercise $3$
Name this molecule.
Answer
2-methylpentane
Branched hydrocarbons may have more than one substituent. If the substituents are different, then give each substituent a number (using the smallest possible numbers) and list the substituents in alphabetical order, with the numbers separated by hyphens and with no spaces in the name. So the molecule
is 3-ethyl-2-methylpentane.
If the substituents are the same, then use the name of the substituent only once, but use more than one number, separated by a comma. Also, put a numerical prefix before the substituent name that indicates the number of substituents of that type. The numerical prefixes are listed in Table $4$. The number of the position values must agree with the numerical prefix before the substituent.
Table $4$ Numerical Prefixes to Use for Multiple Substituents
Number of Same Substituent Numerical Prefix
2 di-
3 tri-
4 tetra-
5 penta-
and so forth and so forth
Consider this molecule:
The longest chain has four C atoms, so it is a butane. There are two substituents, each of which consists of a single C atom; they are methyl groups. The methyl groups are on the second and third C atoms in the chain (no matter which end the numbering starts from), so we would name this molecule 2,3-dimethylbutane. Note the comma between the numbers, the hyphen between the numbers and the substituent name, and the presence of the prefix di- before the methyl. Other molecules-even with larger numbers of substituents-can be named similarly.
Example $4$:
Name this molecule.
Solution
The longest chain has seven C atoms, so we name this molecule as a heptane. We find two one-carbon substituents on the second C atom and a two-carbon substituent on the third C atom. So this molecule is named 3-ethyl-2,2-dimethylheptane.
Exercise $4$
Name this molecule.
Answer
4,4,5-tripropyloctane
Properties of Alkanes
All alkanes are composed of carbon and hydrogen atoms, and have similar bonds, structures, and formulas; noncyclic alkanes all have a formula of CnH2n+2. The number of carbon atoms present in an alkane has no limit. Greater numbers of atoms in the molecules will lead to stronger intermolecular attractions (dispersion forces) and correspondingly different physical properties of the molecules. Properties such as melting point and boiling point (Table $5$) usually change smoothly and predictably as the number of carbon and hydrogen atoms in the molecules change.
Table $5$ Properties of Some Alkanes
Alkane Molecular Formula Melting Point (°C) Boiling Point (°C) Phase at STP4 Number of Structural Isomers
methane CH4 –182.5 –161.5 gas 1
ethane C2H6 –183.3 –88.6 gas 1
propane C3H8 –187.7 –42.1 gas 1
butane C4H10 –138.3 –0.5 gas 2
pentane C5H12 –129.7 36.1 liquid 3
hexane C6H14 –95.3 68.7 liquid 5
heptane C7H16 –90.6 98.4 liquid 9
octane C8H18 –56.8 125.7 liquid 18
nonane C9H20 –53.6 150.8 liquid 35
decane C10H22 –29.7 174.0 liquid 75
tetradecane C14H30 5.9 253.5 solid 1858
octadecane C18H38 28.2 316.1 solid 60,523
Alkanes are relatively stable molecules, but heat or light will activate reactions that involve the breaking of C–H or C–C single bonds. Combustion is one such reaction:
$\ce{CH4}(g)+\ce{2O2}(g)⟶\ce{CO2}(g)+\ce{2H2O}(g) \nonumber$
Alkanes burn in the presence of oxygen, a highly exothermic oxidation-reduction reaction that produces carbon dioxide and water. As a consequence, alkanes are excellent fuels. For example, methane, CH4, is the principal component of natural gas. Butane, C4H10, used in camping stoves and lighters is an alkane. Gasoline is a liquid mixture of continuous- and branched-chain alkanes, each containing from five to nine carbon atoms, plus various additives to improve its performance as a fuel. Kerosene, diesel oil, and fuel oil are primarily mixtures of alkanes with higher molecular masses. The main source of these liquid alkane fuels is crude oil, a complex mixture that is separated by fractional distillation. Fractional distillation takes advantage of differences in the boiling points of the components of the mixture (Figure $10$). You may recall that boiling point is a function of intermolecular interactions, which was discussed in the chapter on solutions and colloids.
In a substitution reaction, another typical reaction of alkanes, one or more of the alkane’s hydrogen atoms is replaced with a different atom or group of atoms. No carbon-carbon bonds are broken in these reactions, and the hybridization of the carbon atoms does not change. For example, the reaction between ethane and molecular chlorine depicted here is a substitution reaction:
The C–Cl portion of the chloroethane molecule is an example of a functional group, the part or moiety of a molecule that imparts a specific chemical reactivity. The types of functional groups present in an organic molecule are major determinants of its chemical properties and are used as a means of classifying organic compounds as detailed in the remaining sections of this chapter.
Cyclic Hydrocarbons
A cyclic hydrocarbon is a hydrocarbon in which the carbon chain joins to itself in a ring. A cycloalkane is a cyclic hydrocarbon in which all of the carbon-carbon bonds are single bonds. Like other alkanes, cycloalkanes are saturated compounds. Cycloalkanes have the general formula $\ce{C_{n}H_{2n}}$. The simplest cycloalkane shown below is cyclopropane, a three-carbon ring.
The structural formulas of cyclic hydrocarbons can be represented in multiple ways, two of which are shown above. Each atom can be shown as in the structure on the left from the figure above. A convenient shorthand is to omit the element symbols and only show the shape, as in the triangle on the right. Carbon atoms are understood to be the vertices of the triangle.
The carbon atoms in cycloalkanes are still $sp^3$ hybridized, with an ideal bond angle of $109.5^\text{o}$. However, an examination of the cyclopropane structure shows that the triangular structure results in a $\ce{C-C-C}$ bond angle of $60^\text{o}$. This deviation from the ideal angle is called ring strain and makes cyclopropane a fairly unstable and reactive molecule. Ring strain is decreased for cyclobutane, with a bond angle of $90^\text{o}$, but is still significant. Cyclopentane has a bond angle of about $108^\text{o}$. This minimal ring strain for cyclopentane makes it a more stable compound.
Cyclohexane is a six-carbon cycloalkane shown below
All three of the depictions of cyclohexane are somewhat misleading because the molecule is not planar. In order to reduce the ring strain and attain a bond angle of approximately $109.5^\text{o}$, the molecule is puckered. The puckering of the ring means that every other carbon atom is above and below the plane. The figure below shows two possibilities for the puckered cyclohexane molecule. Each of the structures is called a conformation. The conformation on the right is called the boat conformation, while the one on the left is called the chair conformation.
While both conformations reduce the ring strain compared to a planar molecule, the chair is preferred. This is because the chair conformation results in fewer repulsive interactions between the hydrogen atoms. However, interconversion readily occurs between the two conformations.
Larger cycloalkanes also exist, but are less common. Cyclic hydrocarbons may also be unsaturated. A cycloalkene is a cyclic hydrocarbon with at least one carbon-carbon double bond. A cycloalkyne is a cyclic hydrocarbon with at least one carbon-carbon triple bond. Shown below are the simplified structural formulas for cyclohexene and cyclooctyne.
Unsaturated Hydrocarbons: Alkenes and Alkynes
Alkenes
Organic compounds that contain one or more double or triple bonds between carbon atoms are described as unsaturated. Unsaturated hydrocarbons have less than the maximum number of H atoms possible. Unsaturated hydrocarbon molecules that contain one or more double bonds are called alkenes. Carbon atoms linked by a double bond are bound together by two bonds, one σ bond and one π bond. Double and triple bonds give rise to a different geometry around the carbon atom that participates in them, leading to important differences in molecular shape and properties. The differing geometries are responsible for the different properties of unsaturated versus saturated fats.
Note
The word unsaturated has the same meaning for hydrocarbons as it does for the dietary fats and oils: the molecule has at least one carbon-to-carbon double (C=C) or triple bond (C=C).
Ethene, C2H4, is the simplest alkene. Each carbon atom in ethene, commonly called ethylene, has a trigonal planar structure. The second member of the series is propene (propylene) (Figure $16$); the butene isomers follow in the series. Four carbon atoms in the chain of butene allows for the formation of isomers based on the position of the double bond, as well as a new form of isomerism.
Structural formulas of Alkenes
Structural Formula
Ball and Stick Model
Space Filling model
Name ethene propene 1-butene
Condensed Structural Formula CH2CH2 CH2CHCH3 CH3CH2CHCH2
Molecular Formula C2H2 C3H6 C4H8
Line angle drawing
The name of an alkene is derived from the name of the alkane with the same number of carbon atoms. The presence of the double bond is signified by replacing the suffix -ane with the suffix -ene. The location of the double bond is identified by naming the smaller of the numbers of the carbon atoms participating in the double bond:
Alkynes
Hydrocarbon molecules with one or more triple bonds are called alkynes; they make up another series of unsaturated hydrocarbons. Two carbon atoms joined by a triple bond are bound together by one σ bond and two π bonds. The sp-hybridized carbons involved in the triple bond have bond angles of 180°, giving these types of bonds a linear, rod-like shape.
The simplest member of the alkyne series is ethyne, C2H2, commonly called acetylene. The Lewis structure for ethyne, a linear molecule, is:
The IUPAC nomenclature for alkynes is similar to that for alkenes except that the suffix -yne is used to indicate a triple bond in the chain. For example, $\mathrm{CH_3CH_2C≡CH}$ is called 1-butyne.
Naming Alkenes and Alkynes
Alkenes and alkynes are named in a similar fashion. The biggest difference is that when identifying the longest carbon chain, it must contain the C–C double or triple bond. Furthermore, when numbering the main chain, the double or triple bond gets the lowest possible number. This means that there may be longer or higher-numbered substituents than may be allowed if the molecule were an alkane. For example, this molecule
is 2,4-dimethyl-3-heptene (note the number and the hyphens that indicate the position of the double bond).
Example $5$
Name this molecule.
Solution
The longest chain that contains the C–C triple bond has six C atoms, so this is a hexyne molecule. The triple bond starts at the third C atom, so this is a 3-hexyne. Finally, there are two methyl groups on the chain; to give them the lowest possible number, we number the chain from the left side, giving the methyl groups the second position. So the name of this molecule is 2,2-dimethyl-3-hexyne.
Exercise $5$
Name this molecule.
Answer
2,3,4-trimethyl-2-pentene
Example $6$
Identify the molecule as an alkane, an alkene, or an alkyne.
Solution
1. The molecule has at least one carbon-to-carbon triple bond (C=C) so it is an alkyne.
2. The molecule has a one carbon-to-carbon double bond (C=C) so it is an alkene.
3. The molecule only has carbon-to-carbon single bonds so it is an alkane.
Exercise $6$
Identify the molecule as an alkane, an alkene, or an alkyne.
Answer
1. alkane, 2. alkene, 3. alkyne
Properties of Alkenes and Alkynes
Ethylene (the common industrial name for ethene) is a basic raw material in the production of polyethylene and other important compounds. Over 135 million tons of ethylene were produced worldwide in 2010 for use in the polymer, petrochemical, and plastic industries. Ethylene is produced industrially in a process called cracking, in which the long hydrocarbon chains in a petroleum mixture are broken into smaller molecules.
Halogens can also react with alkenes and alkynes, but the reaction is different. In these cases, the halogen reacts with the C–C double or triple bond and inserts itself onto each C atom involved in the multiple bonds. This reaction is called an addition reaction. One example is
The reaction conditions are usually mild; in many cases, the halogen reacts spontaneously with an alkene or an alkyne.
Hydrogen can also be added across a multiple bond; this reaction is called a hydrogenation reaction. In this case, however, the reaction conditions may not be mild; high pressures of H2 gas may be necessary. A platinum or palladium catalyst is usually employed to get the reaction to proceed at a reasonable pace:
$CH_{2}=CH_{2}+H_{2}\overset{metal\: catalyst}{\rightarrow} CH_{3}CH_{3} \nonumber$
Geometric Isomers (Cis-Trans) Isomers
With a molecule such as 2-butene, a different type of isomerism called geometric isomerism can be observed. Geometric isomers are isomers in which the order of atom bonding is the same but the arrangement of atoms in space is different. The double bond in an alkene is not free to rotate because of the nature of the pi bond. Therefore, there are two different ways to construct the 2-butene molecule. The image below shows the two geometric isomers, called cis-2-butene and trans-2-butene.
The cis isomer has the two single hydrogen atoms on the same side of the molecule, while the trans isomer has them on opposite sides of the molecule. In both molecules, the bonding order of the atoms is the same. In order for geometric isomers to exist, there must be a rigid structure in the molecule to prevent free rotation around a bond. If the double bond in an alkene was capable of rotating, the two geometric isomers above would not exist. In addition, the two carbon atoms must each have two different groups attached in order for there to be geometric isomers. Propene has no geometric isomers because one of the carbon atoms has two single hydrogens bonded to it.
Physical and chemical properties of geometric isomers are generally different. While cis-2-butene is a polar molecule, trans-2-butene is nonpolar. Heat or irradiation with light can be used to bring about the conversion of one geometric isomer to another. The input of energy must be large enough to break the pi bond between the two carbon atoms, which is weaker than the sigma bond. At that point,the now single bond is free to rotate and the isomers can interconvert.
As with alkenes, alkynes display structural isomerism beginning with 1-butyne and 2-butyne. However, there are no geometric isomers with alkynes because there is only one other group bonded to the carbon atoms that are involved in the triple bond.
Summary
• Simple alkanes exist as a homologous series, in which adjacent members differ by a CH2 unit.
• Condensed chemical formulas show the hydrogen atoms (or other atoms or groups) right next to the carbon atoms to which they are attached.
• Line-angle formulas imply a carbon atom at the corners and ends of lines. Each carbon atom is understood to be attached to enough hydrogen atoms to give each carbon atom four bonds.
• Alkanes have both common names and systematic names, specified by IUPAC.
• Names and structures of typical cyclic hydrocarbons are given.
• Structural and geometric isomers are defined.
• Examples of alkane and alkene isomers are given.
Contributors and Attributions
• OpenSTAX
• TextMap: Beginning Chemistry (Ball et al.)
• Henry Agnew (UC Davis) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/09%3A_Organic_Chemistry/9.02%3A_Aliphatic_Hydrocarbons.txt |
Learning Objective
• Name and identify aromatic compounds.
Benzene is the parent compound of the large family of organic compounds known as aromatic compounds. Unlike cyclohexane, benzene only contains six hydrogen atoms, giving the impression that the ring is unsaturated and each carbon atom participates in one double bond. Two different structures with alternating single and double bonds around the ring can be written for benzene.
In benzene, the true bonding between carbon atoms is neither a single nor a double bond. Rather, all of the bonds are a hybrid of a single and double bond. In benzene, the pi bonding electrons are free to move completely around the ring. Delocalized electrons are electrons that are not confined to the bond between two atoms, but are instead allowed to move between three or more. The delocalization of the electrons in benzene can best be shown by showing benzene with a ring inside the hexagon, with the hydrogen atoms understood.
Delocalization of the electrons makes for a more stable molecule than a similar molecule that does not have delocalized electrons. Benzene is a more stable and less reactive compound than straight-chain hexenes. The $sp^2$ hybridization of the carbon atoms results in a planar molecule as opposed to the puckered structure of cyclohexane.
There are many derivatives of benzene. The hydrogen atoms can be replaced by many different substituents. Aromatic compounds more readily undergo substitution reactions than addition reactions; replacement of one of the hydrogen atoms with another substituent will leave the delocalized double bonds intact. The following are typical examples of substituted benzene derivatives:
Toluene and xylene are important solvents and raw materials in the chemical industry. Styrene is used to produce the polymer polystyrene. The figure below shows the structural formulas for vanilla and naphthalene. Naphthalene is a chemical which is commonly used in mothballs. The vanilla molecule is extracted from vanilla beans.
Nomenclature of Aromatic Compounds
The simplest aromatic compounds are benzene rings with one substituent replacing one of the hydrogen atoms. If this substituent is an alkyl group, it is named first, followed in one word with "benzene". The molecule shown below is therefore called ethylbenzene.
Substituents can be groups other than alkyl groups. If a chlorine atom were substituted for a hydrogen, the name becomes chlorobenzene. An $\ce{-NH_2}$ group is called an amino group, so the corresponding molecule is called aminobenzene, often referred to as an aniline. An $\ce{-NO_2}$ group is called a nitro group and so the third example below is nitrobenzene.
If more than one substituent is present, their location relative to each other can be indicated by numbering the positions on the benzene ring.
The number of the carbon location then precedes the name of the substituent in the overall name, with the numbers separated by a comma. As with branched alkanes, the system requires that the numbers be the lowest possible and that prefixes be used for more than one of the same substituent. If there are different substituents, the first in alphabetical order is given the lower number and listed first. The structures below are called 1,2-dimethylbenzene and 1-ethyl-4-methylbenzene.
An alternate system for naming di-substituted benzene rings uses three different prefixes: ortho, meta, and para. If two groups are in the ortho position, they are on adjacent carbon atoms. The meta positioning refers to being in a 1,3 arrangement. The para positioning refers to being in a 1,4 arrangement. Shown below are the three possibilities for dimethylbenzene, also called xylene.
Lastly, a benzene ring missing one hydrogen atom $\left( \ce{-C_6H_5} \right)$ can itself be considered the substituent on a longer chain of carbon atoms. That group is called a phenyl group and so the molecule below is called 2-phenylbutane.
Summary
• Aromatic hydrocarbons contain ring structures with delocalized π electron systems.
• Benzene is the parent compound of the large family of organic compounds known as aromatic compounds.
• Naming of aromatic compound i.e. benzene rings with different substituents is described. | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/09%3A_Organic_Chemistry/9.03%3A_Aromatic_Compounds-_Benzene_and_Its_Relatives.txt |
Learning Objectives
• Identify haloalkanes (alkyl halides) and know their occurrence, uses, and properties.
• Learn how to name simple haloalkanes.
Haloalkanes also known as alkyl halides or halogenoalkanes, are a group of organic compounds in which one or more halogen atoms are substituted for one or more hydrogen atoms in a hydrocarbon. Haloalkanes are widely used commercially and, consequently, are known under many chemical and commercial names. Although these compounds were once widely used as dry cleaning solvents, coolants in refrigerators and air conditioners, and propellants in hairsprays and deodorants, increasing awareness of their toxicity and impact on the environment has led to a widespread decrease in applications for these materials.
The general formulas for organic molecules with functional groups use the letter $\ce{R}$ to stand for the rest of the molecule outside of the functional group. Because there are four possible halogen atoms (fluorine, chlorine, bromine, or iodine) that can act as the functional group, we use the general formula $\ce{R-X}$ to represent an alkyl halide.
Haloalkanes have higher boiling points than alkanes containing the same number of carbons. They are at best only slightly soluble in water but tend to dissolve in organic solvents.
Chloromethane (or methylchloride, CH3Cl) was a widely used refrigerant, but its use has been discontinued due to its toxicity and flammability. Methyl chloride was also once used for producing lead-based gasoline additives (tetramethyllead).
The most important use of chloromethane today is as a chemical intermediate in the production of silicone polymers. Smaller quantities are used as a solvent in the manufacture of butyl rubber and in petroleum refining.
Large amounts of methyl chloride are produced naturally in the oceans by the action of sunlight on biomass and chlorine in sea foam. However, all methyl chloride that is used in industry is produced synthetically. Most methyl chloride is prepared by reacting methanol with hydrogen chloride, according to the chemical equation
$CH_3OH + HCl → CH_3Cl + H_2O \nonumber$
This can be carried out either by bubbling hydrogen chloride gas through boiling methanol with or without a zinc chloride catalyst, or by passing combined methanol and hydrogen chloride vapors over an alumina catalyst at 350 °C (662 °F).
A smaller amount of chloromethane is produced by heating a mixture of methane and chlorine to over 400 °C (752 °F). However, this method also results in more highly chlorinated compounds such as dichloromethane, chloroform, and carbon tetrachloride and is usually only used when these other products are also desired.
Short chain haloalkanes such as dichloromethane (CH2Cl2), trichloromethane (chloroform, CHCl3) and tetrachloromethane (carbon tetrachloride, CCl4) are commonly used as hydrophobic solvents in chemistry. They were formerly very common in industry; however, their use has been greatly curtailed due to their toxicity and harmful environmental effects.
Some specific compounds are still used. Halothane (2-bromo-2-chloro-1,1,1-trifluoroethane) is still used in some situations as an inhalation anesthetic. The compound DDT is a very effective pesticide, but is only used when nothing else works because of its harmful effects on the environment.
Naming Alkyl Halides
IUPAC System: The rules for naming simple alkyl halides are listed below.
1. Name the parent compound by finding the longest continuous carbon atom chain that also contains the halogen. Add a prefix for the particular halogen atom. The prefixes for each of the four halogens are fluoro-, chloro-, bromo-, and iodo-. If more than one kind of halogen atom is present, put them in alphabetical order. If there is more than one of the same halogen on a given carbon atom, use the prefixes di-, tri-, or tetra- before the prefix for the halogen.
2. As with hydrocarbons, number the carbon chain in a way that makes the sum of halogen numbers as low as possible. If different halogens are in equivalent positions, give the lower number to the one that comes first in alphabetical order.
3. Add the numerical prefix into the name before the halogen prefix.
4. Separate numbers with commas and separate numbers from names or prefixes with a hyphen. There are no spaces in the name.
Listed below are some examples of names and structural formulas of a few alkyl halides.
Note that for the structure based on methane, no number needs to be used since there is only one carbon atom. In the third example, the chloro- is listed first alphabetically and the chain is numbered so that the sum of the numbers is as low as possible.
Common Names: The common names of alkyl halides consist of two parts: the name of the alkyl group plus the stem of the name of the halogen, with the ending -ide.
Alkyl halides with simple alkyl groups (one to four carbon atoms) are often called by common names. Those with a larger number of carbon atoms are usually given IUPAC names.
Example $1$
Give the common and IUPAC names for each compound.
1. CH3CH2CH2Br
2. (CH3)2CHCl
Solution
1. The alkyl group (CH3CH2CH2–) is a propyl group, and the halogen is bromine (Br). The common name is therefore propyl bromide. For the IUPAC name, the prefix for bromine (bromo) is combined with the name for a three-carbon chain (propane), preceded by a number identifying the carbon atom to which the Br atom is attached, so the IUPAC name is 1-bromopropane.
2. The alkyl group [(CH3)2CH–] has three carbon atoms, with a chlorine (Cl) atom attached to the middle carbon atom. The alkyl group is therefore isopropyl, and the common name of the compound is isopropyl chloride. For the IUPAC name, the Cl atom (prefix chloro-) attached to the middle (second) carbon atom of a propane chain results in 2-chloropropane.
Exercise $1$
Give common and IUPAC names for each compound.
1. CH3CH2I
2. CH3CH2CH2CH2F
Answers
a. 1-iodoethane, ethyl iodide
b. 1-fluorobutane, butyl fluoride
Example $2$
Give the IUPAC name for each compound.
• Solution
1. The parent alkane has five carbon atoms in the longest continuous chain; it is pentane. A bromo (Br) group is attached to the second carbon atom of the chain. The IUPAC name is 2-bromopentane.
2. The parent alkane is hexane. Methyl (CH3) and bromo (Br) groups are attached to the second and fourth carbon atoms, respectively. Listing the substituents in alphabetical order gives the name 4-bromo-2-methylhexane.
Exercise $2$
Give the IUPAC name for each compound.
• Answers
a. 2-chloro-3-methylbutane
b. 1-bromo-2-chloro-4-methylpentane
• A wide variety of interesting and often useful compounds have one or more halogen atoms per molecule. For example, methane (CH4) can react with chlorine (Cl2), replacing one, two, three, or all four hydrogen atoms with Cl atoms. Several halogenated products derived from methane and ethane (CH3CH3) are listed in Table $1$, along with some of their uses.
Table $1$: Some Halogenated Hydrocarbons
Formula Common Name IUPAC Name Some Important Uses
Derived from CH4
CH3Cl methyl chloride chloromethane refrigerant; the manufacture of silicones, methyl cellulose, and synthetic rubber
CH2Cl2 methylene chloride dichloromethane laboratory and industrial solvent
CHCl3 chloroform trichloromethane industrial solvent
CCl4 carbon tetrachloride tetrachloromethane dry-cleaning solvent and fire extinguishers (but no longer recommended for use)
CBrF3 halon-1301 bromotrifluoromethane fire extinguisher systems
CCl3F chlorofluorocarbon-11 (CFC-11) trichlorofluoromethane foaming plastics
CCl2F2 chlorofluorocarbon-12 (CFC-12) dichlorodifluoromethane refrigerant
Derived from CH3CH3
CH3CH2Cl ethyl chloride chloroethane local anesthetic
ClCH2CH2Cl ethylene dichloride 1,2-dichloroethane solvent for rubber
CCl3CH3 methylchloroform 1,1,1-trichloroethane solvent for cleaning computer chips and molds for shaping plastics
Chlorofluorocarbons and Fluorocarbons
Chlorofluorocarbons (CFCs) and hydrochlorofluorocarbons (HCFCs) are fully or partly halogenated paraffin hydrocarbons that contain only carbon (C), hydrogen (H), chlorine (Cl), and fluorine (F), produced as volatile derivative of methane, ethane, and propane. They are also commonly known by the DuPont brand name Freon. The most common representative is dichlorodifluoromethane (R-12 or Freon-12). Many CFCs have been widely used as refrigerants, propellants (in aerosol applications), and solvents. Because CFCs contribute to ozone depletion in the upper atmosphere, the manufacture of such compounds has been phased out under the Montreal Protocol, and they are being replaced with other products such as hydrofluorocarbons (HFCs)
Chlorinated or fluorinated alkenes undergo polymerization. Important halogenated polymers include polyvinyl chloride (PVC), and polytetrafluoroethene (PTFE, or Teflon). Billions of kilograms of chlorodifluoromethane are produced annually as precursor to tetrafluoroethylene, the monomer that is converted into Teflon. The production of these materials releases substantial amounts of wastes. PVC and PTFE will be discussed in more detail in Chapter 10.
Summary
• Haloalkanes also known as alkyl halides or halogenoalkanes, are a group of organic compounds in which one or more halogen atoms are substituted for one or more hydrogen atoms in a hydrocarbon.
• These compounds are used as dry cleaning, industrial, and laboratory solvents, coolants in refrigerators and air conditioners, and propellants in hairsprays and deodorants
• The use of haloalkanes has been greatly curtailed due to their toxicity and harmful environmental effects.
• Naming of simple haloalkanes is described.
Contributors and Attributions
• Libretext: The Basics of GOB Chemistry (Ball et al.)
• Wikipedia
9.05: The Functional Group
Learning Objectives
• Define functional group.
• Identify the functional group(s) present in organic compounds.
With over twenty million known organic compounds in existence, it would be very challenging to memorize chemical reactions for each one. Fortunately, molecules with similar functional groups tend to undergo similar reactions. A functional group is defined as an atom or group of atoms within a molecule that has similar chemical properties whenever it appears in various compounds. Even if other parts of the molecule are quite different, certain functional groups tend to react in certain ways.
We've already looked at alkanes, but they are generally unreactive. We primarily use alkanes as a source of energy when they are combusted. While the majority of functional groups involve atoms other than carbon and hydrogen, we will also look at some that include only carbon and hydrogen. Some of the most common functional groups are presented in the following sections.
Organic molecules vary greatly in size and when focusing on functional groups, we want to direct our attention to the atoms involved in the functional group. As a result, the abbreviation R is used in some examples. The letter R is used in molecular structures to represent the “Rest of the molecule”. It consists of a group of carbon and hydrogen atoms of any size. It is used as an abbreviation since a group of carbon and hydrogen atoms does not affect the functionality of the compound. In some molecules, you will see R, R’, or R’’ which indicates that the R groups in the molecule can be different from one another. For example, R might be –CH2CH3 while R’ is –CH2CH2CH2CH3.
The structure of capsaicin, the compound responsible for the heat in peppers, incorporates several functional groups, labeled in the figure below and explained throughout this section.
The table below summarizes the structures that will be discussed in this chapter:
Table \(1\) Selected Organic Functional Groups
Example \(1\)
Identify the common functional groups for ATP.
Solution
The common functional groups for ATP are hydroxyl, ether and amine. The other functional groups are covered in higher organic chemistry courses.
Exercise \(1\)
Identify the functional groups (other than alkanes) in the following organic compounds. State whether alcohols and amines are primary, secondary, or tertiary.
Answer
Summary
• Functional groups consist of a single atom (such as Cl) or a group of atoms (such as CO2H). It can determine the chemical reactivity of a molecule under a given set of conditions
• The major families of organic compounds are characterized by their functional groups.
Contributors and Attributions
Libretext: Chemistry for Allied Health (Soult)
TextMap: Organic Chemistry (Wade). | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/09%3A_Organic_Chemistry/9.04%3A_Halogenated_Hydrocarbons-_Many_Uses_Some_Hazards.txt |
Learning Objectives
• Describe the structure and properties of alcohols, ethers and phenols.
• Know the name and uses of simple alcohols, phenols and ethers.
An alcohol is an organic compound with a hydroxyl (OH) functional group on an aliphatic carbon atom. Because OH is the functional group of all alcohols, we often represent alcohols by the general formula ROH, where R is an alkyl group. Alcohols are common in nature. Most people are familiar with ethyl alcohol (ethanol), the active ingredient in alcoholic beverages, but this compound is only one of a family of organic compounds known as alcohols. The family also includes such familiar substances as cholesterol and the carbohydrates. Methanol (CH3OH) and ethanol (CH3CH2OH) are the first two members of the homologous series of alcohols.
Nomenclature of Alcohols
Alcohols with one to four carbon atoms are frequently called by common names, in which the name of the alkyl group is followed by the word alcohol:
According to the International Union of Pure and Applied Chemistry (IUPAC), the name of an alcohol comes from the hydrocarbon from which it was derived. The final -e in the name of the hydrocarbon is replaced by -ol, and the carbon atom to which the –OH group is bonded is indicated by a number placed before the name.
Example $1$: Naming Alcohols
Consider the following example. How should it be named?
Solution
The carbon chain contains five carbon atoms. If the hydroxyl group was not present, we would have named this molecule pentane. To address the fact that the hydroxyl group is present, we change the ending of the name to -ol. In this case, since the –OH is attached to carbon 2 in the chain, we would name this molecule 2-pentanol.
Exercise $1$
Name the following molecule:
Answer
2-methyl-2-pentanol
Note
The IUPAC adopted new nomenclature guidelines in 2013 that require this number to be placed as an “infix” rather than a prefix. For example, the new name for 2-propanol would be propan-2-ol. Widespread adoption of this new nomenclature will take some time, and students are encouraged to be familiar with both the old and new naming protocols.
Example $2$: Naming Alcohols
Consider the following example. How should it be named according to the guidelines adopted in 2013?
Solution
The carbon chain contains four carbon atoms. If the hydroxyl group was not present, we would have named this molecule butane. To address the fact that the hydroxyl group is present, we change the ending of the name to -ol. In this case, since the –OH is attached to carbon 2 in the chain, we would name this molecule butan-2-ol.
Note
"Always place the OH on the lowest possible number for the chain."
In the previous example, if the carbon atoms were numbered from left to right then the OH would have been placed on the third carbon and the molecule would have been named butan-3-ol, but butan-2-ol is the more preferred name.
Exercise $2$: Naming Alcohols
Name the following molecules according to the guidelines adopted in 2013?
a.
b.
Answer
a. propan-2-ol
b. hexan-2-ol
Table $1$ names and classifies some of the simpler alcohols. Some of the common names reflect a compound’s classification as secondary (sec-) or tertiary (tert-). These designations are not used in the IUPAC nomenclature system for alcohols. Note that there are four butyl alcohols in the table, corresponding to the four butyl groups: the butyl group (CH3CH2CH2CH2) discussed before, and three others:
Table $1$: Classification and Nomenclature of Some Alcohols
Condensed Structural Formula Common Name IUPAC Name
CH3OH wood alcohol (methyl alcohol) methanol
CH3CH2OH grain alcohol (ethyl alcohol) ethanol
CH3CH2CH2OH propyl alcohol 1-propanol
(CH3)2CHOH rubbing alcohol (isopropyl alcohol) 2-propanol
CH3CH2CH2CH2OH butyl alcohol 1-butanol
CH3CH2CHOHCH3 sec-butyl alcohol 2-butanol
(CH3)2CHCH2OH isobutyl alcohol 2-methyl-1-propanol
(CH3)3COH tert-butyl alcohol 2-methyl-2-propanol
cyclohexyl alcohol cyclohexanol
Methanol (Methyl Alcohol)
Methanol, also known as methyl alcohol among others, is a chemical with the formula CH3OH (a methyl group linked to a hydroxyl group, often abbreviated MeOH). Methanol acquired the name wood alcohol because it was once produced chiefly by the destructive distillation of wood. Today, methanol is mainly produced industrially by hydrogenation of carbon monoxide.
CO(g) + 2H2(g) → CH3OH(l)
Methanol is the simplest alcohol, consisting of a methyl group linked to a hydroxyl group. It is a light, volatile, colorless, flammable liquid with a distinctive odor similar to that of ethanol (drinking alcohol). At room temperature, it is a polar liquid. With more than 20 million tons produced annually, it is used as a fuel additive and as a precursor to other commodity chemicals, including formaldehyde, acetic acid, methyl tert-butyl ether, as well as a host of more specialized chemicals.
Ethanol (Ethyl Alcohol)
Ethanol (also called ethyl alcohol, grain alcohol, drinking alcohol, or simply alcohol) is a chemical compound, a simple alcohol with the chemical formula C2H6O. Its formula can be also written as CH3−CH2−OH or C2H5OH (an ethyl group linked to a hydroxyl group), and is often abbreviated as EtOH. Ethanol is a volatile, flammable, colorless liquid with a slight characteristic odor. It is a psychoactive substance and is the principal type of alcohol found in alcoholic drinks.
Ethanol is naturally produced by the fermentation of sugars by yeasts or via petrochemical processes, and is most commonly consumed as a popular recreational drug. Ethanol is the alcohol produced by some species of yeast that is found in wine, beer, and distilled drinks. It has long been prepared by humans harnessing the metabolic efforts of yeasts in fermenting various sugars:
Large quantities of ethanol (for industrial use) are synthesized from the addition reaction of water with ethylene using an acid as a catalyst:
Ethanol also has medical applications as an antiseptic and disinfectant. The compound is widely used as a chemical solvent, either for scientific chemical testing or in synthesis of other organic compounds, and is a vital substance used across many different kinds of manufacturing industries. Ethanol is also used as a clean-burning fuel source.
Toxicity of Alcohols
With respect to acute toxicity, simple alcohols have low acute toxicities. Doses of several milliliters are tolerated. For pentanols, hexanols, octanols and longer alcohols, LD50 (lethal dose, 50%) range from 2–5 g/kg (rats, oral). Methanol and ethanol are less acutely toxic. Methanol is however far more toxic than ethanol. All alcohols are mild skin irritants.
The metabolism of methanol (and ethylene glycol) is affected by the presence of ethanol, which has a higher affinity for liver alcohol dehydrogenase. In this way methanol will be excreted intact in urine.
The immediate effect of alcohol depends on the drinker's blood alcohol concentration (BAC). BAC can be different for each person depending on their age, sex, pre-existing health condition, even if they drink the same amount of alcohol.
Different BACs have different effects. Table $2$ list the common effects of alcohol on the body depending on the BAC. However, tolerance varies considerably between individuals, as does individual response to a given dosage; the effects of alcohol differ widely between people. Hence in this context, BAC percentages are just estimates used for illustrative purposes.
Table $2$ Blood Alcohol Levels and Effects Source: Wikipedia.
mg/dL mM
Blood Alcohol Level
% v/v
Effects
50 11 0.05% Euphoria, talkativeness, relaxation
100 22 0.1% Central nervous system depression, nausea, possible vomiting, impaired motor and sensory function, impaired cognition
>140 30 >0.14% Decreased blood flow to brain
300 65 0.3% Stupefaction, possible unconsciousness
400 87 0.4% Possible death
500 109 >0.55% Death
Rubbing alcohol refers to either isopropyl alcohol (propan-2-ol) or ethanol based liquids. Rubbing alcohol is undrinkable even if it is ethanol based, due to the bitterants added. Product labels for rubbing alcohol include a number of warnings about the chemical, including the flammability hazards and its intended use only as a topical antiseptic and not for internal wounds or consumption. It should be used in a well-ventilated area due to inhalation hazards. Poisoning can occur from ingestion, inhalation, absorption, or consumption of rubbing alcohol.
Multifunctional Alcohols
A polyol is an organic compound containing multiple hydroxyl groups. Examples of polyols discussed in this section include ethylene glycol, propylene glycol, and glycerol.
Alcohols containing two or more hydroxyl groups can be made. Examples include 1,2-ethanediol (ethylene glycol, used in antifreeze) and 1,2,3-propanetriol (glycerine, used as a solvent for cosmetics and medicines):
Ethylene glycol (IUPAC name: ethane-1,2-diol) is an organic compound with the formula (CH2OH)2. It is mainly used for two purposes, as a raw material in the manufacture of polyester fibers and for antifreeze formulations. It is an odorless, colorless, sweet-tasting, viscous liquid. Ethylene glycol is toxic. Household pets are especially susceptible to ethylene glycol poisoning from vehicle antifreeze leaks.
Propylene glycol (IUPAC name: propane-1,2-diol) is a organic compound with the chemical formula CH3CH(OH)CH2OH. It is a viscous, colorless liquid which is nearly odorless but possesses a faintly sweet taste. Containing two alcohol groups, it is classed as a diol. It is misciblewith a broad range of solvents, including water, acetone, and chloroform. In general, glycols are non-irritating, have very low volatility and very low toxicity.
Forty-five percent of propylene glycol produced is used as chemical feedstock for the production of unsaturated polyester resins. Propylene glycol is also used in various edible items such as coffee-based drinks, liquid sweeteners, ice cream, whipped dairy products and soda.Vaporizers used for delivery of pharmaceuticals or personal-care products often include propylene glycol among the ingredients. In alcohol-based hand sanitizers, it is used as a humectant to prevent the skin from drying. Propylene glycol is used as a solvent in many pharmaceuticals, including oral, injectable and topical formulations, such as for diazepam and lorazepam which are insoluble in water. Certain formulations of artificial tears, such as Systane, use proplyene glycol as an ingredient.Propylene glycol is frequently used as a substitute for ethylene glycol in low toxicity, environmentally friendly automotive antifreeze. It is also used to winterize the plumbing systems in vacant structures.
Glycerol (/ˈɡlɪsərɒl/; also called glycerine or glycerin; see spelling differences) is a simple polyol compound. It is a colorless, odorless, viscous liquid that is sweet-tasting and non-toxic. The glycerol backbone is found in many lipids which are known as glycerides. It is widely used in the food industry as a sweetener and humectant in pharmaceutical formulations. Glycerol has three hydroxyl groups that are responsible for its solubility in water and its hygroscopic nature.
In food and beverages, glycerol serves as a humectant, solvent, and sweetener, and may help preserve foods. It is also used as filler in commercially prepared low-fat foods (e.g., cookies), and as a thickening agent in liqueurs. Glycerol and water are used to preserve certain types of plant leaves. As a sugar substitute, it has approximately 27 kilocalories per teaspoon (sugar has 20) and is 60% as sweet as sucrose. It does not feed the bacteria that form plaques and cause dental cavities.[citation needed] As a food additive, glycerol is labeled as E number E422. It is added to icing (frosting) to prevent it from setting too hard.
Glycerol is used in medical, pharmaceutical and personal care preparations, often as a means of improving smoothness, providing lubrication, and as a humectant. Ichthyosis and xerosis have been relieved by the topical use glycerin. It is found in allergen immunotherapies, cough syrups, elixirs and expectorants, toothpaste, mouthwashes, skin care products, shaving cream, hair care products, soaps, and water-based personal lubricants (Figure $10$). In solid dosage forms like tablets, glycerol is used as a tablet holding agent.
Phenols
Compounds in which an OH group is attached directly to an aromatic ring are designated ArOH and called phenols. Phenols differ from alcohols in that they are slightly acidic in water. They react with aqueous sodium hydroxide (NaOH) to form salts.
$\ce{ArOH (aq) + NaOH (aq) \rightarrow ArONa (aq) + H_2O} \nonumber$
The parent compound, C6H5OH, is itself called phenol. (An old name, emphasizing its slight acidity, was carbolic acid.) Phenol is a white crystalline compound that has a distinctive (“hospital smell”) odor.
To Your Health: Phenols and Us
Phenols are widely used as antiseptics (substances that kill microorganisms on living tissue) and as disinfectants (substances intended to kill microorganisms on inanimate objects such as furniture or floors). The first widely used antiseptic was phenol. Joseph Lister used it for antiseptic surgery in 1867. Phenol is toxic to humans, however, and can cause severe burns when applied to the skin. In the bloodstream, it is a systemic poison—that is, one that is carried to and affects all parts of the body. Its severe side effects led to searches for safer antiseptics, a number of which have been found.
One safer phenolic antiseptic is 4-hexylresorcinol (4-hexyl-1,3-dihydroxybenzene; resorcinol is the common name for 1,3-dihydroxybenzene, and 4-hexylresorcinol has a hexyl group on the fourth carbon atom of the resorcinol ring). It is much more powerful than phenol as a germicide and has fewer undesirable side effects. Indeed, it is safe enough to be used as the active ingredient in some mouthwashes and throat lozenges.
In addition to acting as an antiseptic, phenol is also a useful precursor in many chemical syntheses to produce pharmaceuticals, food preservatives, polymers, resins and adhesives. Phenolics are also present in a number of biological systems and natural products such as neurotransmitters, flavouring agents, and vitamins to name a few.
Ethers
Ethers are compounds that contain the functional group –O–. Ethers do not have a designated suffix like the other types of molecules we have named so far. In the IUPAC system, the oxygen atom and the smaller carbon branch are named as an alkoxy substituent and the remainder of the molecule as the base chain, as in alkanes. As shown in the following compound, the red symbols represent the smaller alkyl group and the oxygen atom, which would be named “methoxy.” The larger carbon branch would be ethane, making the molecule methoxyethane. Many ethers as shown below, are referred to with common names instead of the IUPAC system names. For common names, the two branches connected to the oxygen atom are named separately and followed by “ether.”
In the general formula for ethers, R—O—R, the hydrocarbon groups (R) may be the same or different.
Ether molecules have no hydrogen atom on the oxygen atom (that is, no OH group). Therefore there is no intermolecular hydrogen bonding between ether molecules, and ethers therefore have quite low boiling points for a given molar mass. Indeed, ethers have boiling points about the same as those of alkanes of comparable molar mass and much lower than those of the corresponding alcohols (Table $3$).
Table $3$ Comparison of Boiling Points of Alkanes, Alcohols, and Ethers
Condensed Structural Formula Name Molar Mass Boiling Point (°C) Intermolecular Hydrogen Bonding in Pure Liquid?
CH3CH2CH3 propane 44 –42 no
CH3OCH3 dimethyl ether 46 –25 no
CH3CH2OH ethyl alcohol 46 78 yes
CH3CH2CH2CH2CH3 pentane 72 36 no
CH3CH2OCH2CH3 diethyl ether 74 35 no
CH3CH2CH2CH2OH butyl alcohol 74 117 yes
Ether molecules do have an oxygen atom, however, and engage in hydrogen bonding with water molecules. Consequently, an ether has about the same solubility in water as the alcohol that is isomeric with it. For example, dimethyl ether and ethanol (both having the molecular formula C2H6O) are completely soluble in water, whereas diethyl ether and 1-butanol (both C4H10O) are barely soluble in water (8 g/100 mL of water).
Diethyl ether, the most widely used compound of this class, is a colorless, volatile liquid that is highly flammable. It was first used in 1846 as an anesthetic, but better anesthetics have now largely taken its place. Diethyl ether and other ethers are presently used primarily as solvents for gums, fats, waxes, and resins. Tertiary-butyl methyl ether, C4H9OCH3 (abbreviated MTBE—italicized portions of names are not counted when ranking the groups alphabetically—so butyl comes before methyl in the common name), is used as an additive for gasoline. MTBE belongs to a group of chemicals known as oxygenates due to their capacity to increase the oxygen content of gasoline.
To Your Health: Ethers as General Anesthetics
A general anesthetic acts on the brain to produce unconsciousness and a general insensitivity to feeling or pain. Diethyl ether (CH3CH2OCH2CH3) was the first general anesthetic to be used.
Diethyl ether is relatively safe because there is a fairly wide gap between the dose that produces an effective level of anesthesia and the lethal dose. However, because it is highly flammable and has the added disadvantage of causing nausea, it has been replaced by newer inhalant anesthetics, including the fluorine-containing compounds halothane, enflurane, and isoflurane. Unfortunately, the safety of these compounds for operating room personnel has been questioned. For example, female operating room workers exposed to halothane suffer a higher rate of miscarriages than women in the general population.
A list of important ethers and their uses are given in Table $4$
Table $4$ Important Ethers and Their Uses
Ethylene oxide Also the simplest epoxide. Most ethylene oxide is used for synthesis of ethylene glycols (antifreeze), including diethylene glycol and triethylene glycol, that accounts for up to 75% of global consumption.
Dimethyl ether An aerosol spray propellant. A potential renewable alternative fuel for diesel engines with a cetane rating as high as 56–57.
Diethyl ether A common low boiling solvent (b.p. 34.6 °C) and an early anaesthetic. Used as starting fluid for diesel engines. Also used as a refrigerant and in the manufacture of smokeless gunpowder, along with use in perfumery.
Dimethoxyethane(DME) A water miscible solvent often found in lithium batteries (b.p. 85 °C):
Dioxane A cyclic ether and high-boiling solvent (b.p. 101.1 °C).
Example $2$: classifying functional groups
Identify the following compounds as alcohol, phenol or ether.
a.
b.
Solution
a. The compound has the general formula R—O—R, so it is an ether. b. The compound has two hydroxyl (–OH) groups, so it is an alcophol.
Exercise $2$
Identify the following compounds as alcohol, phenol or ether.
a.
b.
Answer
a. alcohol b. ether
Summary
• The –OH group is the functional group of an alcohol. Various alcohols have a wide range of applications in the medical field as well as in the transportation, food, cosmetic industries.
• The name of an alcohol comes from the hydrocarbon from which it was derived.
• The –R–O–R– group is the functional group of an ether. Diethyl ether and other ethers are presently used primarily as solvents for gums, fats, waxes, and resins. Ethylene oxide in particular is used in the manufacture of ethylene glycol (antifreeze).
• Phenols are compounds in which an OH group is attached directly to an aromatic ring. Many phenols are used as antiseptics. | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/09%3A_Organic_Chemistry/9.06%3A_Alcohols_Phenols_and_Ethers.txt |
Learning Objectives
• Describe the structure and properties of aldehydes and ketones.
• Name common aldehydes and ketones.
Aldehydes and ketones are widespread in nature and are often combined with other functional groups. Examples of naturally occurring molecules which contain a aldehyde or ketone functional group are shown in the following two figures. The compounds in Figure $1$ are found chiefly in plants or microorganisms and those in Figure $2$ have animal origins. Aldehydes and ketones are known for their sweet and sometimes pungent odors. The odor from vanilla extract comes from the molecule vanillin. Likewise, benzaldehyde provides a strong scent of almonds and is this author’s favorite chemical smell. Because of their pleasant fragrances aldehyde and ketone containing molecules are often found in perfumes. However, not all of the fragrances are pleasing. In particular, 2-Heptanone provides part of the sharp scent from blue cheese and (R)-Muscone is part of the musky smell from the Himalayan musk deer. Lastly, ketones show up in many important hormones such as progesterone (a female sex hormone) and testosterone (a male sex hormone). Notice how subtle differences in structure can cause drastic changes in biological activity. The ketone functionality also shows up in the anti-inflammatory steroid, Cortisone.
Aldehydes and ketones are two related categories of organic compounds that both contain the carbonyl group, shown below
The difference between aldehydes and ketones is the placement of the carbonyl group within the molecule. An aldehyde is an organic compound in which the carbonyl group is attached to a carbon atom at the end of a carbon chain. A ketone is an organic compound in which the carbonyl group is attached to a carbon atom within the carbon chain. The general formulas for each are shown below.
For aldehydes, the $\ce{R}$ group may be a hydrogen atom or any length carbon chain.
Naming Simple Aldehydes and Keones
Aldehydes are named by finding the longest continuous chain that contains the carbonyl group. Change the -e at the end of the name of the alkane to -al.
For the common name of aldehydes start with the common parent chain name and add the suffix -aldehyde.
The IUPAC system names are given on top while the common name is given on the bottom in parentheses.
For ketones, $\ce{R}$ and $\ce{R'}$ must be carbon chains, of either the same or different lengths. The steps for naming ketones, followed by two examples, are shown below.
1. Name the parent compound by finding the longest continuous chain that contains the carbonyl group. Change the -e at the end of the name of the alkane to -one.
2. Number the carbon atoms in the chain in a way that the carbonyl group has the lowest possible number.
3. Add the numerical prefix into the name before the name of the ketone.
4. Use a hyphen between the number and the name of the ketone.
The common name for ketones are simply the substituent groups listed alphabetically + ketone.
Some common ketones are known by their generic names. Such as the fact that propanone is commonly referred to as acetone.
The IUPAC system names are given on top while the common name is given on the bottom in parentheses.
Properties of Aldehydes and Ketones
Aldehydes and ketones can work weak hydrogen bonds with water through the carbonyl oxygen atom. The lower members of both series (3 carbons or fewer) are soluble in water in all proportions. As the length of the carbon chain increases, water solubility decreases. Similar to ethers, neither aldehydes nor ketones can hydrogen bond with themselves. As a result, their boiling points are generally lower than those of alcohols. Unlike alkanes however, aldehydes and ketones are polar molecules due to the more electronegative oxygen atom. The dipole-dipole interactions are stronger than the dispersion forces present in alkanes. The boiling points of aldehydes and ketones are intermediate between those of alkanes and alcohols. For example, the boiling point of ethane is $-89^\text{o} \text{C}$, ethanal is $20^\text{o} \text{C}$, and ethanol is $78^\text{o} \text{C}$.
Methanal, commonly known as formaldehyde, was once commonly used as a biological preservative for dead animals. In recent years formaldehyde has been shown to be a carcinogen and so has been replaced for the purpose by safer alternatives. Aldehydes are currently used in the production of resins and plastics. The simplest ketone, propanone, is commonly called acetone. Acetone is a common organic solvent that was one used in most nail polish removers, but has largely been replaced by other solvents.
Some Common Aldehydes
Formaldehyde, an aldehyde with the formula HCHO, is a colorless gas with a pungent and irritating odor. It is sold in an aqueous solution called formalin, which contains about 37% formaldehyde by weight. Formaldehyde causes coagulation of proteins, so it kills bacteria (and any other living organism) and stops many of the biological processes that cause tissue to decay. Thus, formaldehyde is used for preserving tissue specimens and embalming bodies. It is also used to sterilize soil or other materials. Formaldehyde is used in the manufacture of Bakelite, a hard plastic having high chemical and electrical resistance.
Formaldehyde is produced industrially by the catalytic oxidation of methanol according to the chemical equation:
2 CH3OH + O2 → 2 CH2O + 2 H2O
Acetaldehyde (systematic name ethanal) is an organic chemical compound with the formula CH3CHO, sometimes abbreviated by chemists as MeCHO (Me = methyl). It is one of the most important aldehydes, occurring widely in nature and being produced on a large scale in industry. Acetaldehyde occurs naturally in coffee, bread, and ripe fruit, and is produced by plants.
In 2003, global production was about 1 million tonnes. Before 1962, ethanol and acetylene were the major sources of acetaldehyde. Since then, ethylene is the dominant feedstock.
The main method of production is the oxidation of ethylene by the Wacker process, which involves oxidation of ethylene using a homogeneous palladium/copper system:
2 CH2=CH2 + O2 → 2 CH3CHO
Propionaldehyde or propanal is the organic compound with the formula CH3CH2CHO. It is a saturated 3-carbon aldehyde and is a structural isomer of acetone. It is a colorless liquid with a slightly irritating, fruity odor. Butyraldehyde, also known as butanal, is an organic compound with the formula CH3(CH2)2CHO. This compound is the aldehyde derivative of butane. It is a colourless flammable liquid with an unpleasant smell. It is miscible with most organic solvents. A major use of butyraldehyde is in the production of bis(2-ethylhexyl) phthalate, a major plasticizer. Benzaldehyde (C6H5CHO) is an organic compound consisting of a benzene ring with a formyl substituent. It is the simplest aromatic aldehyde and one of the most industrially useful.It is a colorless liquid with a characteristic almond-like odor. The primary component of bitter almond oil, benzaldehyde can be extracted from a number of other natural sources. Synthetic benzaldehyde is the flavoring agent in imitation almond extract, which is used to flavor cakes and other baked goods
Traces of many aldehydes are found in essential oils and often contribute to their favorable odors, e.g. cinnamaldehyde, cilantro, and vanillin.
Some Common Ketones
In terms of scale, the most important ketones are acetone, methylethyl ketone, and cyclohexanone. They are also common in biochemistry, but less so than in organic chemistry in general.
Dimethyl ketone, CH3COCH3, commonly called acetone, is the simplest ketone. Acetone is a colorless liquid. Among its many uses are as a solvent for lacquer (including fingernail polish), cellulose acetate, cellulose nitrate, acetylene, plastics, and varnishes; as a paint and varnish remover; and as a solvent in the manufacture of pharmaceuticals and chemicals.
Acetone is produced directly or indirectly from propylene. Approximately 83% of acetone is produced via the cumene process; as a result, acetone production is tied to phenol production. In the cumene process, benzene is alkylated with propylene to produce cumene, which is oxidized by air to produce phenol and acetone:
Other processes involve the direct oxidation of propylene (Wacker-Hoechst process), or the hydration of propylene to give 2-propanol, which is oxidized to acetone.
Butanone, also known as methyl ethyl ketone (MEK), is an organic compound with the formula CH3C(O)CH2CH3. This colorless liquid ketone has a sharp, sweet odor reminiscent of butterscotch and acetone. It is produced industrially on a large scale, and also occurs in trace amounts in nature. It is soluble in water and is commonly used as an industrial solvent. Butanone is an effective and common solvent and is used in processes involving gums, resins, cellulose acetate and nitrocellulose coatings and in vinyl films. For this reason it finds use in the manufacture of plastics, textiles, in the production of paraffin wax, and in household products such as lacquer, varnishes, paint remover, a denaturing agent for denatured alcohol, glues, and as a cleaning agent. It has similar solvent properties to acetone but boils at a higher temperature and has a significantly slower evaporation rate.
Cyclohexanone is the organic compound with the formula (CH2)5CO. The molecule consists of six-carbon cyclic molecule with a ketonefunctional group. This colorless oil has an odor reminiscent of that of acetone. Over time, samples of cyclohexanone assume a yellow color. Cyclohexanone is slightly soluble in water and miscible with common organic solvents. Billions of kilograms are produced annually, mainly as a precursor to nylon.
Example $1$
Classify each compound as an aldehyde or a ketone.
a.
b.
c.
Solution
a. This compound has the carbonyl group on an end carbon atom, so it is an aldehyde.
b. This compound has the carbonyl group on an interior carbon atom, so it is a ketone.
c. This compound has the carbonyl group between two alkyl groups, so it is a ketone.
Exercise $1$
Classify each compound as an aldehyde or a ketone.
a.
b.
c.
Answer
a. ketone b. aldehyde c. ketone
Summary
• An aldehyde is an organic compound in which the carbonyl group is attached to a carbon atom at the end of a carbon chain.
• A ketone is an organic compound in which the carbonyl group is attached to a carbon atom within the carbon chain.The common names of aldehydes are taken from the names of the corresponding carboxylic acids: formaldehyde, acetaldehyde, and so on. The common names of ketones, like those of ethers, consist of the names of the groups attached to the carbonyl group, followed by the word ketone.
• Stem names of aldehydes and ketones are derived from those of the parent alkanes, using an -al ending for an aldehydes and an -one ending for a ketone. | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/09%3A_Organic_Chemistry/9.07%3A_Aldehydes_and_Ketones.txt |
Learning Objectives
• Describe the structure and properties of carboxylic acids and esters.
• Name common carboxylic acids and esters.
The odor of vinegar is caused by the presence of acetic acid, a carboxylic acid, in the vinegar. The odor of ripe bananas and many other fruits is due to the presence of esters, compounds that can be prepared by the reaction of a carboxylic acid with an alcohol. Because esters do not have hydrogen bonds between molecules, they have lower vapor pressures than the alcohols and carboxylic acids from which they are derived.
Both carboxylic acids and esters contain a carbonyl group with a second oxygen atom bonded to the carbon atom in the carbonyl group by a single bond. In a carboxylic acid, the second oxygen atom also bonds to a hydrogen atom. In an ester, the second oxygen atom bonds to another carbon atom. The names for carboxylic acids and esters include prefixes that denote the lengths of the carbon chains in the molecules and are derived following nomenclature rules similar to those for inorganic acids and salts (see these examples):
The functional groups for an acid and for an ester are shown in red in these formulas.
Carboxylic Acids
Carboxylic acids occur widely in nature, often combined with alcohols or other functional groups, as in fats, oils, and waxes. They are components of many foods, medicines, and household products (Figure $1$). Not surprisingly, many of them are best known by common names based on Latin and Greek words that describe their source.
The carboxyl group contains the $\ce{C=O}$ of the carbonyl group, with the carbon atom also being bonded to a hydroxyl $\left( \ce{-OH} \right)$ group. A carboxylic acid is an organic compound that contains the carboxyl functional group. The general formula for a carboxylic acid can be abbreviated as $\ce{R-COOH}$. The carbon atom of the carboxyl group may be attached to a hydrogen atom or to a carbon chain. The naming of a carboxylic acid is as follows: Name the parent compound by finding the longest continuous chain that contains the carboxyl group. Change the -e at the end of the name of the alkane to -oic acid.
Carboxylic acids are weak acids, meaning they are not 100% ionized in water. Generally only about 1% of the molecules of a carboxylic acid dissolved in water are ionized at any given time. The remaining molecules are undissociated in solution.
We prepare carboxylic acids by the oxidation of aldehydes or alcohols whose –OH functional group is located on the carbon atom at the end of the chain of carbon atoms in the alcohol:
The simplest carboxylic acid, formic acid (HCOOH, methanoic acid), was first obtained by the distillation of ants (Latin formica, meaning “ant”). The bites of some ants inject formic acid, and the stings of wasps and bees contain formic acid (as well as other poisonous materials).
The next higher homolog is acetic acid (CH3COOH, ethanoic acid), which is made by fermenting cider and honey in the presence of oxygen. This fermentation produces vinegar, a solution containing 4%–10% acetic acid, plus a number of other compounds that add to its flavor. Acetic acid is probably the most familiar weak acid used in educational and industrial chemistry laboratories.
Pure acetic acid solidifies at 16.6°C, only slightly below normal room temperature. In the poorly heated laboratories of the late 19th and early 20th centuries in northern North America and Europe, acetic acid often “froze” on the storage shelf. For that reason, pure acetic acid (sometimes called concentrated acetic acid) came to be known as glacial acetic acid, a name that survives to this day.
The third homolog, propionic acid (CH3CH2COOH, propionic acid), is seldom encountered in everyday life. The fourth homolog, butyric acid (CH3CH2CH2COOH), is one of the most foul-smelling substances imaginable. It is found in rancid butter and is one of the ingredients of body odor. By recognizing extremely small amounts of this and other chemicals, bloodhounds are able to track fugitives.
Many carboxylic acids occur naturally in plants and animals. Citrus fruits such as oranges and lemons contain citric acid (Figure $4$). Ethanoic and citric acids are frequently added to foods to give them a tart flavor.
Benzoic, propanoic, and sorbic acids are used as food preservatives because of their ability to kill microorganisms that can lead to spoilage. Methanoic and ethanoic acids are widely used in industry as starting points for the manufacture of paints, adhesives, and coatings.
Esters: The Sweet Smell of RCOOR'
An ester is an organic compound that is a derivative of a carboxylic acid in which the hydrogen atom of the hydroxyl group has been replaced with an alkyl group. The structure is the product of a carboxylic acid (the $\ce{R}$-portion) and an alcohol (the $\ce{R'}$-portion). The general formula for an ester is shown below.
The $\ce{R}$ group can either be a hydrogen or a carbon chain. The $\ce{R'}$ group must be a carbon chain since a hydrogen atom would make the molecule a carboxylic acid.
Esters are produced by the reaction of acids with alcohols. For example, the ester ethyl acetate, CH3CO2CH2CH3, is formed when acetic acid reacts with ethanol:
Chemistry Is Everywhere: Esters, Fragrances, and Flavorings
Esters are very interesting compounds, in part because many have very pleasant odors and flavors. (Remember, never taste anything in the chemistry lab!) Many esters occur naturally and contribute to the odor of flowers and the taste of fruits. Other esters are synthesized industrially and are added to food products to improve their smell or taste; it is likely that if you eat a product whose ingredients include artificial flavorings, those flavorings are esters. Here are some esters and their uses, thanks to their odors, flavors, or both:
Table $1$: Esters, Fragrances, and Flavorings
Ester Tastes/Smells Like Ester Tastes/Smells Like
allyl hexanoate pineapple isobutyl formate raspberry
benzyl acetate pear isobutyl acetate pear
butyl butanoate pineapple methyl phenylacetate honey
ethyl butanoate banana nonyl caprylate orange
ethyl hexanoate pineapple pentyl acetate apple
ethyl heptanoate apricot propyl ethanoate pear
ethyl pentanoate apple propyl isobutyrate rum
Among the most important of the natural esters are fats (such as lard, tallow, and butter) and oils (such as linseed, cottonseed, and olive oils), which are esters of the trihydroxyl alcohol glycerine, C3H5(OH)3, with large carboxylic acids, such as palmitic acid, CH3(CH2)14CO2H, stearic acid, CH3(CH2)16CO2H, and oleic acid, $\mathrm{CH_3(CH_2)_7CH=CH(CH_2)_7CO_2H}$. Oleic acid is an unsaturated acid; it contains a $\mathrm{C=C}$ double bond. Palmitic and stearic acids are saturated acids that contain no double or triple bonds.
Note
Fats and vegetable oils are esters of long-chain fatty acids and glycerol. Esters of phosphoric acid are of the utmost importance to life.
Esters are common solvents. Ethyl acetate is used to extract organic solutes from aqueous solutions—for example, to remove caffeine from coffee. It also is used to remove nail polish and paint. Cellulose nitrate is dissolved in ethyl acetate and butyl acetate to form lacquers. The solvent evaporates as the lacquer “dries,” leaving a thin film on the surface. High boiling esters are used as softeners (plasticizers) for brittle plastics.
Summary
• A carboxylic acid is an organic compound that contains the carboxyl functional group.
• The general formula for a carboxylic acid can be abbreviated as $\ce{R-COOH}$.
• Many carboxylic acids are used in the food and beverage industry for flavoring and/or as preservatives.
• An ester has an OR group attached to the carbon atom of a carbonyl group.
• Fats and vegetable oils are esters of long-chain fatty acids and glycerol.
• Esters occur widely in nature and generally have pleasant odors and are often responsible for the characteristic fragrances of fruits and flowers.
Contributors and Attributions
• Libretext : The Basics of GOB Chemistry (Ball et al.)
• TextMap: Beginning Chemistry (Ball et al.)
• OpenSTAX | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/09%3A_Organic_Chemistry/9.08%3A_Carboxylic_Acids_and_Esters.txt |
Learning Objectives
• Describe the structure and properties of amines and amides.
• Name simple amines and amides.
The addition of nitrogen into an organic framework leads to two families of molecules. Compounds containing a nitrogen atom bonded in a hydrocarbon framework are classified as amines. Compounds that have a nitrogen atom bonded to one side of a carbonyl group are classified as amides.
Amines
An amine is an organic compound that can be considered to be a derivative of ammonia $\left( \ce{NH_3} \right)$. Amines are molecules that contain carbon-nitrogen bonds. Amines are named by naming the alkyl groups attached to the nitrogen atom, followed by the suffix -amine as illustrated here for a few simple examples:
The name of larger molecules involve the class-identifying suffix –ine as you will see later in this section(e.g. caffeine and nicotine).
Methylamine and ethylamine are gases at room temperature, while larger amines are liquids. As with other organic compounds that form hydrogen bonds, water solubility is reflected in the length of the carbon chains. Smaller amines are soluble, while larger ones are less soluble.
Amines generally have rather pungent or noxious odors. Ammonia can be considered the simplest amine and has a very distinctive odor. Methylamine has an unpleasant odor associated with dead fish. Amines are often formed biologically during the breakdown of proteins in animal cells, and so many have the smell of death and decay. Putrescine and cadaverine are two such amines and are aptly named for their foul odors. The toxins which many animals use as a form of defense are frequently amines. Amines are used industrially as dyes and in many drugs.
To Your Health: Amines in Death and Life
Amines have “interesting” odors. The simple ones smell very much like ammonia. Higher aliphatic amines smell like decaying fish. Or perhaps we should put it the other way around: Decaying fish give off odorous amines. The stench of rotting fish is due in part to two diamines: putrescine and cadaverine. They arise from the decarboxylation of ornithine and lysine, respectively, amino acids that are found in animal cells.
Aromatic amines generally are quite toxic. They are readily absorbed through the skin, and workers must exercise caution when handling these compounds. Several aromatic amines, including β-naphthylamine, are potent carcinogens.
Amides
The amide functional group has an nitrogen atom attached to a carbonyl carbon atom. If the two remaining bonds on the nitrogen atom are attached to hydrogen atoms, the compound is a simple amide. If one or both of the two remaining bonds on the atom are attached to alkyl or aryl groups, the compound is a substituted amide.
Simple amides are named as derivatives of carboxylic acids. The -ic ending of the common name or the -oic ending of the International Union of Pure and Applied Chemistry (IUPAC) name of the carboxylic acid is replaced with the suffix -amide.
Amides can be produced when carboxylic acids react with amines or ammonia in a process called amidation. A water molecule is eliminated from the reaction, and the amide is formed from the remaining pieces of the carboxylic acid and the amine (note the similarity to formation of an ester from a carboxylic acid and an alcohol discussed in the previous section):
The reaction between amines and carboxylic acids to form amides is biologically important. It is through this reaction that amino acids (molecules containing both amine and carboxylic acid substituents) link together in a polymer to form proteins.
The carbonyl carbon-to-nitrogen bond is called an amide linkage. This bond is quite stable and is found in the repeating units of protein molecules, where it is called a peptide linkage.
Amides are pervasive in nature and technology as structural materials. The amide linkage is easily formed, confers structural rigidity, and resists hydrolysis. Nylons are polyamides, as are the very resilient materials Aramid, Twaron, and Kevlar. Amide linkages constitute a defining molecular feature of proteins, the secondary structure of which is due in part to the hydrogen bonding abilities of amides. Amide linkages in a biochemicalcontext are called peptide bonds when they occur in the main chain of a protein and isopeptide bonds when they occur to a side-chain of the protein. Proteins can have structural roles, such as in hair or spider silk, but also nearly all enzymes are proteins. Low molecular weight amides, such as dimethylformamide (HC(O)N(CH3)2), are common solvents. Many drugs are amides, including paracetamol, penicillin and LSD. Moreover, plant N-alkylamides have a wide range of biological functionalities.
Drugs with the Amide Group
(Figure $3$) Paracetamol (acetaminophen) LSD (Lysergic diethylamide)
Heterocyclic Compounds: Alkaloids and Others
Looking back at the various cyclic hydrocarbons discussed previously, we see that all the atoms in the rings of these compounds are carbon atoms. In other cyclic compounds, called heterocyclic compounds (Greek heteros, meaning “other”), nitrogen, oxygen, sulfur, or some other atom is incorporated in the ring. Many heterocyclic compounds are important in medicine and biochemistry. Some compose part of the structure of the nucleic acids, which in turn compose the genetic material of cells and direct protein synthesis.
Many heterocyclic amines occur naturally in plants. Like other amines, these compounds are basic. Such a compound is an alkaloid, a name that means “like alkalis.” Many alkaloids are physiologically active, including the familiar drugs caffeine, nicotine, and cocaine.
To Your Health: Three Well-Known Alkaloids
Caffeine is a stimulant found in coffee, tea, and some soft drinks. Its mechanism of action is not well understood, but it is thought to block the activity of adenosine, a heterocyclic base that acts as a neurotransmitter, a substance that carries messages across a tiny gap (synapse) from one nerve cell (neuron) to another cell. The effective dose of caffeine is about 200 mg, corresponding to about two cups of strong coffee or tea.
Nicotine acts as a stimulant by a different mechanism; it probably mimics the action of the neurotransmitter acetylcholine. People ingest this drug by smoking or chewing tobacco. Its stimulant effect seems transient, as this initial response is followed by depression. Nicotine is highly toxic to animals. It is especially deadly when injected; the lethal dose for a human is estimated to be about 50 mg. Nicotine has also been used in agriculture as a contact insecticide.
Cocaine acts as a stimulant by preventing nerve cells from taking up dopamine, another neurotransmitter, from the synapse. High levels of dopamine are therefore available to stimulate the pleasure centers of the brain. The enhancement of dopamine action is thought to be responsible for cocaine’s “high” and its addictive properties. After the binge, dopamine is depleted in less than an hour. This leaves the user in a pleasureless state and (often) craving more cocaine.
Cocaine is used as the salt cocaine hydrochloride and in the form of broken lumps of the free (unneutralized) base, which is called crack cocaine.
Because it is soluble in water, cocaine hydrochloride is readily absorbed through the watery mucous membranes of the nose when it is snorted. Crack cocaine is more volatile than cocaine hydrochloride. It vaporizes at the temperature of a burning cigarette. When smoked, cocaine reaches the brain in 15 s.
Morphine and Heroin
Morphine, a strong narcotic used to relieve pain, contains two hydroxyl functional groups, located at the bottom of the
molecule in this structural formula. Changing one of these hydroxyl groups to a methyl ether group forms codeine, a less potent drug used as a local anesthetic. If both hydroxyl groups are converted to esters of acetic acid, the powerfully addictive drug heroin results (Figure $4$).
Example $1$: Amines and Amides
Identify whether each compound is an amide or an amine.
a.
b.
c.
Solutions
a. The compound has the CONH2 functional group so it is an amide.
b. The compound has the CONH2 functional group so it is an amide.
c. The compound has the NH functional group so it is an amine.
Exercise $1$
a.
b. CH3CH2CH2CH2CH2CH2NH2
c.
Answer
a. amine b. amine c. amide
Summary
• An amine is a derivative of ammonia in which one, two, or all three hydrogen atoms are replaced by hydrocarbon groups. The amine functional group is as follows:
• Amines are named by naming the alkyl groups attached to the nitrogen atom, followed by the suffix -amine.
• Amides have a general structure in which a nitrogen atom is bonded to a carbonyl carbon atom.
• Like amines, various nomenclature rules may be used to name amides, but all include use of the class-specific suffix -amide:
• Heterocyclic amines are cyclic compounds with one or more nitrogen atoms in the ring.
Contributors and Attributions
Libretext: The Basics of GOB Chemistry (Ball et al.)
OpenSTAX
Wikipedia | textbooks/chem/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/09%3A_Organic_Chemistry/9.09%3A_Nitrogen-Containing_Compounds-_Amines_and_Amides.txt |
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