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As scientists explore energy supplies, geothermal sources look very appealing. The natural geysers that exist in some parts of the world could possibly be harnessed to provide power for many purposes. The change in energy content and the release of energy caused by steam condensing to liquid can help fill some of our growing energy needs.
Standard Entropy
All molecular motion ceases at absolute zero $\left( 0 \: \text{K} \right)$. Therefore, the entropy of a pure crystalline substance at absolute zero is defined to be equal to zero. As the temperature of the substance increases, its entropy increases because of an increase in molecular motion. The absolute or standard entropy of substances can be measured. The symbol for entropy is $S$ and the standard entropy of a substance is given by the symbol $S^\text{o}$, indicating that the standard entropy is determined under standard conditions. The units for entropy are $\text{J/K} \cdot \text{mol}$. Standard entropies for a few substances are shown in the table below.
Standard Entropy Values at $25^\text{o} \text{C}$
Table $1$: Standard Entropy Values at $25^\text{o} \text{C}$
Substance $S^\text{o} \left( \text{J/K} \cdot \text{mol} \right)$
$\ce{H_2} \left( g \right)$ 131.0
$\ce{O_2} \left( g \right)$ 205.0
$\ce{H_2O} \left( l \right)$ 69.9
$\ce{H_2O} \left( g \right)$ 188.7
$\ce{C} \: \left( \text{graphite} \right)$ 5.69
$\ce{C} \: \left( \text{diamond} \right)$ 2.4
The knowledge of the absolute entropies of substances allows us to calculate the entropy change $\left( \Delta S^\text{o} \right)$ for a reaction. For example, the entropy change for the vaporization of water can be found as follows:
\begin{align*} \Delta S^\text{o} &= S^\text{o} \left( \ce{H_2O} \left( g \right) \right) - S^\text{o} \left( \ce{H_2O} \left( l \right) \right) \ &= 188.7 \: \text{J/K} \cdot \text{mol} - 69.9 \: \text{J/K} \cdot \text{mol} = 118.8 \: \text{J/K} \cdot \text{mol} \end{align*}\nonumber
The entropy change for the vaporization of water is positive because the gas state has higher entropy than the liquid state.
In general, the entropy change for a reaction can be determined if the standard entropies of each substance are known. The equation below can be applied.
$\Delta S^\text{o} = \Sigma n S^\text{o} \: \left( \text{products} \right) - \Sigma n S^\text{o} \: \left( \text{reactants} \right)\nonumber$
The standard entropy change is equal to the sum of the standard entropies of the products minus the sum of the standard entropies of the reactants. The symbol "$n$" signifies that each entropy must first be multiplied by its coefficient in the balanced equation. The entropy change for the formation of liquid water from gaseous hydrogen and oxygen can be calculated using this equation:
\begin{align*} &2 \ce{H_2} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{H_2O} \left( l \right) \ &\Delta S^\text{o} = 2 \left( 69.9 \right) - \left[ 2 \left( 131.0 \right) + 1 \left( 205.0 \right) \right] = -327 \: \text{J/K} \cdot \text{mol} \end{align*}\nonumber
The entropy change for this reaction is highly negative because three gaseous molecules are being converted into two liquid molecules. According to the drive towards higher entropy, the formation of water from hydrogen and oxygen is an unfavorable reaction. In this case, the reaction is highly exothermic, and the drive towards a decrease in energy allows the reaction to occur. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/20%3A_Entropy_and_Free_Energy/20.02%3A_Standard_Entropy.txt |
Nitroglycerin is a tricky substance. An active ingredient in dynamite (where it is stabilized), "raw" nitroglycerin is very unstable. Physical shock will cause the material to explode. The reaction is shown below.
$4 \ce{C_3H_5(ONO_2)_3} \rightarrow 12 \ce{CO_2} + 10 \ce{H_2O} + 6 \ce{N_2} + \ce{O_2}\nonumber$
The explosion of nitroglycerin releases large volumes of gases and is very exothermic.
Spontaneous Reactions
Reactions are favorable when they result in a decrease in enthalpy and an increase in entropy of the system. When both of these conditions are met, the reaction occurs naturally. A spontaneous reaction is a reaction that favors the formation of products at the conditions under which the reaction is occurring. A roaring bonfire is an example of a spontaneous reaction, since it is exothermic (there is a decrease in the energy of the system as energy is released to the surroundings as heat). The products of a fire are composed partly of gases such as carbon dioxide and water vapor. The entropy of the system increases during a combustion reaction. The combination of energy decrease and entropy increase dictates that combustion reactions are spontaneous reactions.
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, it must be endothermic, accompanied by a decrease in entropy, or both. Our 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:
$\ce{N_2} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{NO} \left( g \right)\nonumber$
Fortunately, this reaction is nonspontaneous at normal temperatures and pressures. It is a highly endothermic reaction with a slightly positive entropy change $\left( \Delta S \right)$. Nitrogen monoxide is capable of being produced at very high temperatures and has been observed to form 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. A piece of paper will not suddenly burst 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.
$\ce{H_2CO_3} \left( aq \right) \rightleftharpoons \ce{CO_2} \left( g \right) + \ce{H_2O} \left( l \right)\nonumber$
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, less than $1\%$ is converted to carbonic acid when the reaction reaches equilibrium. The reverse reaction, as written above, is not spontaneous.
Summary
• Spontaneous and nonspontaneous reactions are defined.
• Examples of both types of reactions are given. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/20%3A_Entropy_and_Free_Energy/20.03%3A_Spontaneous_and_Nonspontaneous_Reactions.txt |
The steam engine pictured below is slowly going out of style, but is still a picturesque part of the modern railroad. The water in a boiler is heated by a fire (usually fueled by coal) and turned to steam. This steam then pushes the pistons that drive the wheels of the train. It is the pressure created by the steam that allows work to be done in moving the train.
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.
$\Delta G^\text{o} = \Delta H^\text{o} - T \Delta S^\text{o}\nonumber$
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. The usual units for $\Delta H$ are $\text{kJ/mol}$, while $\Delta S$ is often reported in $\text{J/K} \cdot \text{mol}$. It is necessary to change the units for $\Delta S$ to $\text{kJ/K} \cdot \text{mol}$, so that the calculation of $\Delta G$ is in $\text{kJ/mol}$.
A spontaneous reaction is one that releases free energy, and so the sign of $\Delta G$ must be negative. Since $\Delta H$ and $\Delta S$ can be either positive or negative, depending on the characteristics of the particular reaction, there are four different general outcomes for $\Delta G$, and these are outlined in the table below.
Enthalpy, Entropy, and Free Energy Changes
Table $1$: Enthalpy, Entropy, and Free Energy Changes
$\Delta H$ $\Delta S$ $\Delta G$
- value (exothermic) + value (disordering) always negative
+ value (endothermic) + value (disordering) negative at higher temperatures
- value (exothermic) - value (ordering) negative at lower temperatures
+ value (endothermic) - value (ordering) never negative
Keep in mind that the temperature in the Gibbs free energy equation is the Kelvin temperature and so can only be positive. 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 that falls into this category is the melting of ice. 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 that 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.
Summary
• Free energy is defined.
• Relationships between enthalpy, entropy, and free energy are described.
20.05: Calculating Free Energy Change (left( Delta Gtexto right))
When you are baking something, you heat the oven to the temperature indicated in the recipe. Then you mix all the ingredients, put them in the proper baking dish, and place them in the oven for a specified amount of time. If you had mixed the ingredients and left them out at room temperature, not much would change. The materials need to be heated to a given temperature, for a set time, in order for the ingredients to react with one another and produce a delicious final product.
Calculating Free Energy $\left( \Delta G^\text{o} \right)$
The free energy change of a reaction can be calculated using the following expression:
$\Delta G^\text{o} = \Delta H^\text{o} - T \Delta S^\text{o}\nonumber$
where $\Delta G =$ free energy change $\left( \text{kJ/mol} \right)$
$\Delta H =$ change in enthalpy $\left( \text{kJ/mol} \right)$
$\Delta S =$ change in entropy $\left( \text{J/K} \cdot \text{mol} \right)$
$T =$ temperature (Kelvin)
Note that all values are for substances in their standard state. In performing calculations, it is necessary to change the units for $\Delta S$ to $\text{kJ/K} \cdot \text{mol}$, so that the calculation of $\Delta G$ is in $\text{kJ/mol}$.
Example $1$
Methane gas reacts with water vapor to produce a mixture of carbon monoxide and hydrogen, according to the balanced equation below.
$\ce{CH_4} \left( g \right) + \ce{H_2O} \left( g \right) \rightarrow \ce{CO} \left( g \right) + 3 \ce{H_2} \left( g \right)\nonumber$
The $\Delta H^\text{o}$ for the reaction is $+206.1 \: \text{kJ/mol}$, while the $\Delta S^\text{o}$ is $+215 \: \text{J/K} \cdot \text{mol}$. Calculate the $\Delta G^\text{o}$ at $25^\text{o} \text{C}$ and determine if the reaction is spontaneous at that temperature.
Known
• $\Delta H^\text{o} = 206.1 \: \text{kJ/mol}$
• $\Delta S^\text{o} = 215 \: \text{J/K} \cdot \text{mol} = 0.215 \: \text{kJ/K} \cdot \text{mol}$
• $T = 25^\text{o} \text{C} = 298 \: \text{K}$
Unknown
Prior to substitution into the Gibbs free energy equation, the entropy change is converted to $\text{kJ/K} \cdot \text{mol}$ and the temperature to Kelvins.
Step 2: Solve.
$\Delta G^\text{o} = \Delta H^\text{o} - T \Delta S^\text{o} = 206.1 \: \text{kJ/mol} - 298 \: \text{K} \left( 0.215 \: \text{kJ/K} \cdot \text{mol} \right) = +142.0 \: \text{kJ/mol}\nonumber$
The resulting positive value of $\Delta G$ indicates that the reaction is not spontaneous at $25^\text{o} \text{C}$.
Step 3: Think about your result.
The unfavorable driving force of increasing enthalpy outweighed the favorable increase in entropy. The reaction will be spontaneous only at some elevated temperature.
Available values for enthalpy and entropy changes are generally measured at the standard conditions of $25^\text{o} \text{C}$ and $1 \: \text{atm}$ pressure. The values are slightly temperature dependent and so we must use caution when calculating specific $\Delta G$ values at temperatures other than $25^\text{o} \text{C}$. However, since the values for $\Delta H$ and $\Delta S$ do not change a great deal, the tabulated values can safely be used when making general predictions about the spontaneity of a reaction at various temperatures. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/20%3A_Entropy_and_Free_Energy/20.04%3A_Free_Energy.txt |
Iron ore $\left( \ce{Fe_2O_3} \right)$ and coke (an impure form of carbon) are heated together to make iron and carbon dioxide. The reaction is non-spontaneous at room temperature, but becomes spontaneous at temperatures above $842 \: \text{K}$. The iron can then be treated with small amounts of other materials to make a variety of steel products.
Temperature and Free Energy
Consider the reversible reaction in which calcium carbonate decomposes into calcium oxide and carbon dioxide gas. The production of $\ce{CaO}$ (called quicklime) has been an important reaction for centuries.
$\ce{CaCO_3} \left( s \right) \rightleftharpoons \ce{CaO} \left( s \right) + \ce{CO_2} \left( g \right)\nonumber$
The $\Delta H^\text{o}$ for the reaction is $177.8 \: \text{kJ/mol}$, while the $\Delta S^\text{o}$ is $160.5 \: \text{J/K} \cdot \text{mol}$. The reaction is endothermic with an increase in entropy due to the production of a gas. We can first calculate the $\Delta G^\text{o}$ at $25^\text{o} \text{C}$ in order to determine if the reaction is spontaneous at room temperature.
$\Delta G^\text{o} = \Delta H^\text{o} - T \Delta S^\text{o} = 177.8 \: \text{kJ/mol} - 298 \: \text{K} \left( 0.1605 \: \text{kJ/K} \cdot \text{mol} \right) = 130.0 \: \text{kJ/mol}\nonumber$
Since the $\Delta G^\text{o}$ is a large positive quantity, the reaction strongly favors the reactants and very little products would be formed. In order to determine a temperature at which $\Delta G^\text{o}$ will become negative, we can first solve the equation for the temperature when $\Delta G^\text{o}$ is equal to zero.
\begin{align*} 0 &= \Delta H^\text{o} - T \Delta S^\text{o} \ T &= \frac{\Delta H^\text{o}}{\Delta S^\text{o}} = \frac{177.8 \: \text{kJ/mol}}{0.1605 \: \text{kJ/K} \cdot \text{mol}} = 1108 \: \text{K} = 835^\text{o} \text{C} \end{align*}\nonumber
So at any temperature higher than $835^\text{o} \text{C}$, the value of $\Delta G^\text{o}$ will be negative and the decomposition reaction will be spontaneous.
Recall that the assumption that $\Delta H^\text{o}$ and $\Delta S^\text{o}$ are independent of temperature means that the temperature at which the sign of $\Delta G^\text{o}$ switches from being positive to negative $\left( 835^\text{o} \text{C} \right)$ is an approximation. It is also important to point out that one should not assume that absolutely no products are formed below $835^\text{o} \text{C}$ and that at that temperature, decomposition suddenly begins. Rather, at lower temperatures, the amount of products formed is simply not great enough to say that the products can be detected by monitoring the pressure of the $\ce{CO_2}$ gas that is produced. Above about $700^\text{o} \text{C}$, measurable amounts of $\ce{CO_2}$ are produced. The pressure of $\ce{CO_2}$ at equilibrium gradually increases with increasing temperature. Above $835^\text{o}$, the pressure of $\ce{CO_2}$ at equilibrium begins to exceed $1 \: \text{atm}$, the standard-state pressure. This is an indication that the products of the reaction are now favored above that temperature. When quicklime is manufactured, the $\ce{CO_2}$ is constantly removed from the reaction mixture as it is produced. This causes the reaction to be driven towards the products, according to Le Chatelier's principle.
20.07: Changes of State and Free Energy
Energy in a body of water can be gained or lost, dependent on conditions. When water is heated above a certain temperature, steam is generated. The increase in heat energy creates a higher level of disorder in the water molecules as they boil off and leave the liquid.
Changes of State and Free Energy
At the temperature at which a change of state occurs, the two states are in equilibrium with one another. For an ice-water system, equilibrium takes place at $0^\text{o} \text{C}$, so $\Delta G^\text{o}$ is equal to 0 at that temperature. The heat of fusion of water is known to be equal to $6.01 \: \text{kJ/mol}$, and so the Gibbs free energy equation can be solved for the entropy change that occurs during the melting of ice. The symbol $\Delta S_\text{fus}$ represents the entropy change during the melting process, while $T_\text{f}$ is the freezing point of water.
\begin{align*} \Delta G &= 0 = \Delta H - T \Delta S \ \Delta S_\text{fus} &= \frac{\Delta H_\text{fus}}{T_\text{f}} = \frac{6.01 \: \text{kJ/mol}}{273 \: \text{K}} = 0.0220 \: \text{kJ/K} \cdot \text{mol} = 22.0 \: \text{J/K} \cdot \text{mol} \end{align*}\nonumber
The entropy change is positive as the solid state changes into the liquid state. If the transition went from the liquid to the solid state, the numerical value for $\Delta S$ would be the same, but the sign would be reversed, since the phase changes indicates going from a less ordered to a more ordered situation.
A similar calculation can be performed for the vaporization of liquid to gas. In this case, we use the molar heat of vaporization. This value is $40.79 \: \text{kJ/mol}$. The $\Delta S_\text{vap}$ is as follows:
$\Delta S = \frac{40.79 \: \text{kJ/mol}}{373 \: \text{K}} = 0.1094 \: \text{kJ/K} \cdot \text{mol} = 109.4 \: \text{J/K} \cdot \text{mol}\nonumber$
The value is positive, reflecting the increase in disorder going from liquid to vapor. Condensation from vapor to liquid would give a negative value for $\Delta S$. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/20%3A_Entropy_and_Free_Energy/20.06%3A_Temperature_and_Free_Energy.txt |
Formation of stalactites (pointing down) and stalagmites (pointing up) is a complex process. Solutions of minerals drip down and absorb carbon dioxide as water flows through the cave. Calcium carbonate dissolves in this liquid and re-deposits on the rock as the carbon dioxide is dissipated into the environment.
Equilibrium Constant and $\Delta G$
At equilibrium, the $\Delta G$ for a reversible reaction is equal to zero. $K_\text{eq}$ relates the concentrations of all substances in the reaction at equilibrium. Through a more advanced treatment of thermodynamics, we can write the following equation:
$\Delta G^\text{o} = -RT \: \text{ln} \: K_\text{eq}\nonumber$
The variable $R$ is the ideal gas constant $\left( 8.314 \: \text{J/K} \cdot \text{mol} \right)$, $T$ is the Kelvin temperature, and $\text{ln} \: K_\text{eq}$ is the natural logarithm of the equilibrium constant.
When $K_\text{eq}$ is large, the products of the reaction are favored and the negative sign in the equation means that the $\Delta G^\text{o}$ is negative. When $K_\text{eq}$ is small, the reactants of the reaction are favored. The natural logarithm of a number less than one is negative, and so the sign of $\Delta G^\text{o}$ is positive. The table below summarizes the relationship of $\Delta G^\text{o}$ to $K_\text{eq}$:
Relationship of $\Delta G^\text{o}$ and $K_\text{eq}$
Table $1$: Relationship of $\Delta G^\text{o}$ and $K_\text{eq}$
$K_\text{eq}$ $\text{ln} \: K_\text{eq}$ $\Delta G^\text{o}$ Description
>1 positive negative Products are favored at equilibrium.
1 0 0 Reactants and products are equally favored.
<1 negative positive Reactants are favored at equilibrium.
Knowledge of either the standard free energy change or the equilibrium constant for a reaction allows for the calculation of the other. The following two sample problems illustrate each case.
Example $1$
The formation of nitrogen monoxide from nitrogen and oxygen gases is a reaction that strongly favors the reactants at $25^\text{o} \text{C}$.
$\ce{N_2} \left( g \right) + \ce{O_2} \left( g \right) \rightleftharpoons 2 \ce{NO} \left( g \right)\nonumber$
The actual concentrations of each gas would be difficult to measure, and so the $K_\text{eq}$ for the reaction can more easily be calculated from the $\Delta G^\text{o}$, which is equal to $173.4 \: \text{kJ/mol}$. Find the $K_\text{eq}$.
Known
• $\Delta G^\text{o} = +173.4 \: \text{kJ/mol}$
• $R = 8.314 \: \text{J/K} \cdot \text{mol}$
• $T = 25^\text{o} \text{C} = 298 \: \text{K}$
Unknown
In order to make the units agree, the value of $\Delta G^\text{o}$ will need to be converted to $\text{J/mol}$ $\left( 173,400 \: \text{J/mol} \right)$. To solve for $K_\text{eq}$, the inverse of the natural logarithm, $e^x$, will be used.
Step 2: Solve.
\begin{align*} \Delta G^\text{o} &= -RT \: \text{ln} \: K_\text{eq} \ \text{ln} \: K_\text{eq} &= \frac{-\Delta G^\text{o}}{RT} \ K_\text{eq} &= e^{\frac{-\Delta G^\text{o}}{RT}} = e^{\frac{-173,400 \: \text{J/mol}}{8.314 \: \text{J/K} \cdot \text{mol} \left( 298 \: \text{K} \right)}} = 4.0 \times 10^{-31} \end{align*}\nonumber
Step 3: Think about your result.
The large positive free energy change leads to a $K_\text{eq}$ that is extremely small. Both lead to the conclusion that the reactants are highly favored and very few product molecules are present at equilibrium.
Example $2$
The solubility product constant $\left( K_\text{sp} \right)$ of lead (II) iodide is $1.4 \times 10^{-8}$ at $25^\text{o} \text{C}$. Calculate $\Delta G^\text{o}$ for the dissociation of lead (II) iodide in water.
$\ce{PbI_2} \left( s \right) \rightleftharpoons \ce{Pb^{2+}} \left( aq \right) + 2 \ce{I^-} \left( aq \right)\nonumber$
Known
• $K_\text{eq} = K_\text{sp} = 1.4 \times 10^{-8}$
• $R = 8.314 \: \text{J/K} \cdot \text{mol}$
• $T = 25^\text{o} \text{C} = 298 \: \text{K}$
Unknown
The equation relating $\Delta G^\text{o}$ to $K_\text{eq}$ can be solved directly.
Step 2: Solve.
\begin{align*} \Delta G^\text{o} &= -RT \: \text{ln} \: K_\text{eq} \ &= -8.314 \: \text{J/K} \cdot \text{mol} \left( 298 \: \text{K} \right) \: \text{ln} \left( 1.4 \times 10^{-8} \right) \ &= 45,000 \: \text{J/mol} \ &= 45 \: \text{kJ/mol} \end{align*}\nonumber
Step 3: Think about your result.
The large, positive $\Delta G^\text{o}$ indicates that the solid lead (II) iodide is nearly insoluble and so very little of the solid is dissociated at equilibrium.
Summary
• The relationship between $\Delta G$ and $K_\text{eq}$ is described.
• Calculations involving these two parameters are shown. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/20%3A_Entropy_and_Free_Energy/20.08%3A_Calculations_of_Free_Energy_and_%28K_texteq%29.txt |
A cup of coffee first thing in the morning can help start the day. However, 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!
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. These properties are:
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 indicates. 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:
$\ce{Zn} \left( s \right) + \ce{H_2SO_4} \left( aq \right) \rightarrow \ce{ZnSO_4} \left( aq \right) + \ce{H_2} \left( g \right)\nonumber$
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.
21.02: Properties of Bases
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.
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.
Figure \(1\): Phenolphthalein indicator in presence of base.
Please note that tasting chemicals and touching them are NOT good lab practices and should be avoided—in other words, don't do this at home.
Bases are less common as foods, but they are nonetheless present in many household products. Many cleaners contain ammonia, a base. Sodium hydroxide is found in drain cleaner. Antacids, which combat excess stomach acid, are comprised of bases such as magnesium hydroxide or sodium hydrogen carbonate. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/21%3A_Acids_and_Bases/21.01%3A_Properties_of_Acids.txt |
Venus is the planet nearest to Earth, but has a very different and hostile environment. It has a surface temperature that averages around $450^\text{o} \text{C}$. The atmosphere is composed of carbon dioxide, but clouds of sulfuric acid move through the upper atmosphere, helping to create the extremely unfriendly conditions. Because of these conditions, Venus is not a place you want to visit on vacation.
Arrhenius Acids
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 that ionizes to yield hydrogen ions $\left( \ce{H^+} \right)$ in aqueous solution.
Acids are molecular compounds with ionizable hydrogen atoms. Only hydrogen atoms that are part of a highly polar covalent bond are ionizable. Hydrogen chloride $\left( \ce{HCl} \right)$ is a gas at room temperature and under normal pressure. The $\ce{H-Cl}$ bond in hydrogen chloride is a polar bond. The hydrogen atom is electron deficient because of the higher electronegativity of the chlorine atom. Consequently, 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 extra attached hydrogen ion.
Equations showing the ionization of an acid in water are frequently simplified by omitting the water molecule:
$\ce{HCl} \left( g \right) \rightarrow \ce{H^+} \left( aq \right) + \ce{Cl^-} \left( aq \right)\nonumber$
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.
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 $\left( \ce{CH_3COOH} \right)$ belongs to a class of acids called organic 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:
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 $\ce{H_3PO_4}$).
Summary
• An Arrhenius acid is a compound that ionizes to yield hydrogen ions $\left( \ce{H^+} \right)$ in aqueous solution.
• Examples of Arrhenius acids are given. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/21%3A_Acids_and_Bases/21.03%3A_Arrhenius_Acids.txt |
Sodium hydroxide is a versatile chemical. It can be used for such mundane purposes as cleaning clogged drains. Several commercial preparations contain sodium hydroxide for this purpose. It also has a number of applications in the food processing field. Ice cream is thickened using $\ce{NaOH}$. If olives are soaked in a solution containing sodium hydroxide and other chemicals, the olives will turn black. Soft pretzels are made chewy by the application of sodium hydroxide to the food. This compound has been widely used in the synthesis of plastics, for etching aluminum, for paint removal, and is employed in the dehorning of cattle.
Arrhenius Bases
An Arrhenius base is a compound that ionizes to yield hydroxide ions $\left( \ce{OH^-} \right)$ in aqueous solution. The table below lists several of the more common bases.
Base Name Formula
Table $1$: Common Bases
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.
$\ce{NaOH} \left( s \right) \rightarrow \ce{Na^+} \left( aq \right) + \ce{OH^-} \left( aq \right)\nonumber$
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).
Sodium belongs to the group of elements called the alkali metals. An alkaline solution is another name for a solution that is basic. All alkali metals react readily with water to produce the metal hydroxide and hydrogen gas. The resulting solutions are basic.
$2 \ce{K} \left( s \right) + 2 \ce{H_2O} \left( l \right) \rightarrow 2 \ce{KOH} \left( aq \right) + \ce{H_2} \left( g \right)\nonumber$
Bases that consist of an alkali metal cation and the hydroxide anion are all very soluble in water. Compounds of the Group 2 metals (the alkaline earth metals) are also basic. However, these compounds are generally not as soluble in water. Therefore, the dissociation reactions for these compounds are shown as equilibrium reactions:
$\ce{Mg(OH)_2} \left( s \right) \overset{\ce{H_2O}}{\rightleftharpoons} \ce{Mg^{2+}} \left( aq \right) + 2 \ce{OH^-} \left( aq \right)\nonumber$
These relatively insoluble hydroxides were some of the compounds discussed in the context of the solubility product constant $\left( K_\text{sp} \right)$. The solubility of magnesium hydroxide is $0.0084 \: \text{g}$ per liter of water at $25^\text{o} \text{C}$. Because of its low solubility, magnesium hydroxide is not as dangerous as sodium hydroxide. In fact, magnesium hydroxide is the active ingredient in a product called milk of magnesia, which is used as an antacid or a mild laxative.
Summary
• Arrhenius base is a compound that ionizes to yield hydroxide ions $\left( \ce{OH^-} \right)$ in aqueous solution.
• Examples of Arrhenius bases are given.
21.05: Brnsted-Lowry Acids and Bases
The Arrhenius concept of acids and bases was a significant contribution to the scientific understanding of acids and bases. It replaced and expanded Lavoisier's original idea that all acids contained oxygen. However, the Arrhenius theory also had its shortcomings. It did not take into account the role of the solvent. In addition, this concept did not deal with acid-base behavior in solvents such as benzene, where there could be no ionization. So, formation of a new theory was imperative, which built on the findings of Arrhenius, but also went beyond them.
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 $\left( \ce{NH_3} \right)$. Its aqueous solution turns litmus blue, it reacts with acids, and displays all the other properties of a base. However, it does not contain the hydroxide ion. In 1923, a broader definition of acids and bases was independently proposed by Danish chemist Johannes Brønsted (1879-1947) and 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. A hydrogen ion is commonly referred to as a proton, and so acids and bases are 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 defined as such under the Brønsted-Lowry definition. The new definition, however, includes some substances that are left out according to the Arrhenius definition.
What kind of molecule would qualify as a Brønsted-Lowry base? These molecules need to be able to accept a hydrogen ion (or proton). Two possibilities come to mind: an anion that can form a neutral compound with a proton, or a molecule in which one or more atoms has lone-pair electrons. The most obvious anion is the Arrhenius base $\ce{OH^-}$. This ion can form a water molecule with a proton by accepting the proton. The acetate anion $\ce{CH_3COO^-}$ is another anion that can combine with a proton to form acetic acid. Lone-pair electron groups would include the nitrogen atom (see figure below). The two electrons at the top of the nitrogen atom are not connected in any type of bond, but they do interact readily with a bare proton.
Oxygen is another atom with lone pair electrons that can function as a Brønsted-Lowry base.
The two single electrons (to the left and bottom of the atom) can form single covalent bonds with other atoms, while the two pairs of double electrons (top and right) are available to interact with a hydrogen ion.
Summary
• 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. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/21%3A_Acids_and_Bases/21.04%3A_Arrhenius_Bases.txt |
The Roman god Janus was considered the god of gates, doors, beginnings, and endings. He is portrayed with two faces, looking in two directions at once. Janus would have been in a good position to look at the acid-base reactions we see in this concept, since they are equilibrium reactions involving two different forms of both acids and bases.
Brønsted-Lowry Acid-Base Reactions
An acid-base reaction, according to 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.
$\begin{array}{ccccccc} \ce{NH_3} \left( aq \right) & + & \ce{H_2O} \left( l \right) & \rightleftharpoons & \ce{NH_4^+} \left( aq \right) & + & \ce{OH^-} \left( aq \right) \ \text{base} & & \text{acid} & & \text{acid} & & \text{base} \end{array}\nonumber$
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 ammonia 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. One acid and one base are reactants, and one acid and one base are 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.
$\begin{array}{ccccccc} \ce{HSO_4^-} \left( aq \right) & + & \ce{H_2O} \left( l \right) & \rightleftharpoons & \ce{H_3O^+} \left( aq \right) & + & \ce{SO_4^{2-}} \left( aq \right) \ \text{acid} & & \text{base} & & \text{acid} & & \text{base} \end{array}\nonumber$
So, 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.
In the reaction illustrated below, water serves both as acid and base simultaneously. One water molecule serves as an acid and donates a proton. The other water molecule functions as a base by accepting the proton.
A typical Brønsted-Lowry acid-base reaction contains two conjugate acid-base pairs, as shown below.
$\ce{HNO_2} \left( aq \right) + \ce{PO_4^{3-}} \left( aq \right) \rightleftharpoons \ce{NO_2^-} \left( aq \right) + \ce{HPO_4^{2-}} \left( aq \right)\nonumber$
One conjugate acid-base pair is $\ce{NHO_2}$/$\ce{NO_2^-}$, while the other pair is $\ce{HPO_4^{2-}}$/$\ce{PO_4^{3-}}$.
Summary
• An acid-base reaction, according to the Brønsted-Lowry definition, is a transfer of a proton from one molecule or ion to another.
• 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.
• A conjugate base is the particle produced when an acid donates a proton.
• Examples of conjugate acid-base pairs are given.
21.07: Lewis Acids and Bases
Ideas in science do not stay static. One discovery builds upon another. The concept of acids and bases has grown from the fundamental ideas of Arrhenius to Brønsted-Lowry to Lewis. Each step adds to our understanding of the surrounding world, and makes the "big picture" even bigger.
Lewis Acids and Bases
Gilbert Lewis (1875-1946) proposed a third theory of acids and bases that is even more general than either the Arrhenius or Brønsted-Lowry theories. A Lewis acid is a substance that accepts a pair of electrons to form a covalent bond. A Lewis base is a substance that donates a pair of electrons to form a covalent bond. So, a Lewis acid-base reaction is represented by the transfer of a pair of electrons from a base to an acid. A hydrogen ion, which lacks any electrons, accepts a pair of electrons. It is an acid under both the Brønsted-Lowry and Lewis definitions. Ammonia consists of a nitrogen atom as the central atom with a lone pair of electrons. The reaction between ammonia and the hydrogen ion can be depicted as shown in the figure below.
The lone pair on the nitrogen atom is transferred to the hydrogen ion, making the \(\ce{NH_3}\) a Lewis base while the \(\ce{H^+}\) is a Lewis acid.
Some reactions that do not qualify as acid-base reactions under the other definitions do so under only the Lewis definition. An example is the reaction of ammonia with boron trifluoride.
Boron trifluoride is the Lewis acid, while ammonia is again the Lewis base. As there is no hydrogen ion involved in this reaction, it qualifies as an acid-base reaction only under the Lewis definition. The table below summarizes the three acid-base theories.
Acid-Base Definitions
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
Lewis electron-pair acceptor electron-pair donor
Summary
• A Lewis acid is a substance that accepts a pair of electrons to form a covalent bond.
• A Lewis base is a substance that donates a pair of electrons to form a covalent bond.
• Examples of Lewis acids and bases are given. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/21%3A_Acids_and_Bases/21.06%3A_Brnsted-Lowry_Acid-Base_Reactions.txt |
At one time, you could take the little caps off the top of a car battery and check the condition of the sulfuric acid inside. If it got low, you could add more acid. But, sulfuric acid is hazardous stuff, so the batteries are now sealed to protect people. Because of the acid's dangerous nature, it is not a good idea to cut into a battery to see what it looks like—you could get acid burns.
The Ion-Product of Water
The self-ionization of water (the process in which water ionizes to hydronium ions and hydroxide ions) occurs to a very limited extent. 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.
$\ce{H_2O} \left( l \right) + \ce{H_2O} \left( l \right) \rightleftharpoons \ce{H_3O^+} \left( aq \right) + \ce{OH^-} \left( aq \right)\nonumber$
We often use the simplified form of the reaction:
$\ce{H_2O} \left( l \right) \rightleftharpoons \ce{H^+} \left( aq \right) + \ce{OH^-} \left( aq \right)\nonumber$
The equilibrium constant for the self-ionization of water is referred to as the ion-product for water and is given the symbol $K_\text{w}$.
$K_\text{w} = \left[ \ce{H^+} \right] \left[ \ce{OH^-} \right]\nonumber$
The ion-product of water $\left( K_\text{w} \right)$ is the mathematical product of the concentration of hydrogen ions and hydroxide ions. Note that $\ce{H_2O}$ is not included in the ion-product expression because it is a pure liquid. The value of $K_\text{w}$ is very small, in accordance with a reaction that favors the reactants. At $25^\text{o} \text{C}$, the experimentally determined value of $K_\text{w}$ in pure water is $1.0 \times 10^{-14}$.
$K_\text{w} = \left[ \ce{H^+} \right] \left[ \ce{OH^-} \right] = 1.0 \times 10^{-14}\nonumber$
In pure water, the concentrations of hydrogen and hydroxide ions are equal to one another. Pure water or any other aqueous solution in which this ratio holds is said to be neutral. To find the molarity of each ion, the square root of $K_\text{w}$ is taken.
$\left[ \ce{H^_+} \right] = \left[ \ce{OH^-} \right] = 1.0 \times 10^{-7}\nonumber$
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.
$\ce{HCl} \left( g \right) \rightarrow \ce{H^+} \left( aq \right) + \ce{Cl^-} \left( aq \right)\nonumber$
This increases the concentration of $\ce{H^+}$ ions in the solution. According to Le Chatelier's principle, the equilibrium represented by $\ce{H_2O} \left( l \right) \rightleftharpoons \ce{H^+} \left( aq \right) + \ce{OH^-} \left( aq \right)$ is forced to the left, towards the reactant. As a result, the concentration of the hydroxide ion decreases.
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.
$\ce{KOH} \left( s \right) \rightarrow \ce{K^+} \left( aq \right) + \ce{OH^-} \left( aq \right)\nonumber$
The increase in concentration of the $\ce{OH^-}$ ions causes a decrease in the concentration of the $\ce{H^+}$ ions and the ion-product of $\left[ \ce{H^+} \right] \left[ \ce{OH^-} \right]$ remains constant.
Example $1$
Hydrochloric acid $\left( \ce{HCl} \right)$ is a strong acid, meaning it is $100\%$ ionized in solution. What is the $\left[ \ce{H^+} \right]$ and the $\left[ \ce{OH^-} \right]$ in a solution of $2.0 \times 10^{-3} \: \text{M} \: \ce{HCl}$?
Known
• $\left[ \ce{HCl} \right] = 2.0 \times 10^{-3} \: \text{M}$
• $K_\text{w} = 1.0 \times 10^{-14}$
Unknown
• $\left[ \ce{H^+} \right] = ? \: \text{M}$
• $\left[ \ce{OH^-} \right] = ? \: \text{M}$
Because $\ce{HCl}$ is $100\%$ ionized, the concentration of $\ce{H^+}$ ions in solution will be equal to the original concentration of $\ce{HCl}$. Each $\ce{HCl}$ molecule that was originally present ionizes into one $\ce{H^+}$ ion and one $\ce{Cl^-}$ ion. The concentration of $\ce{OH^-}$ can then be determined from the $\left[ \ce{H^+} \right]$ and $K_\text{w}$.
Step 2: Solve.
\begin{align*} \left[ \ce{H^+} \right] &= 2.0 \times 10^{-3} \: \text{M} \ K_\text{w} &= \left[ \ce{H^+} \right] \left[ \ce{OH^-} \right] = 1.0 \times 10^{-14} \ \left[ \ce{OH^-} \right] &= \frac{K_\text{w}}{\left[ \ce{H^+} \right]} = \frac{1.0 \times 10^{-14}}{2.0 \times 10^{-3}} = 5.0 \times 10^{-12} \: \text{M} \end{align*}\nonumber
Step 3: Think about your result.
The $\left[ \ce{H^+} \right]$ is much higher than the $\left[ \ce{OH^-} \right]$ because the solution is acidic. As with other equilibrium constants, the unit for $K_\text{w}$ is customarily omitted.
Summary
• The self-ionization of water is described and an ionization constant for the process is stated.
• Acidic and basic solutions are defined.
• Calculations using $K_\text{w}$ are illustrated.
21.09: The pH Scale
Grapefruit juice has a pH of somewhere between 2.9 and 3.3, depending on the specific product. Excessive exposure to this juice can cause erosion of tooth enamel and can lead to tooth damage. The acids in grapefruit juice are carbon-based, with citric acid being one of the major constituents. This compound has three ionizable hydrogens on each molecule which contribute to the relatively low pH of the juice. Another component of grapefruit juice is malic acid, containing two ionizable hydrogens per molecule.
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:
$\text{pH} = -\text{log} \: \left[ \ce{H^+} \right]\nonumber$
In pure water or a neutral solution, the $\left[ \ce{H^+} \right] = 1.0 \times 10^{-7} \: \text{M}$. Substituting into the pH expression:
$\text{pH} = -\text{log} \left[ 1.0 \times 10^{-7} \right] = -\left( -7.00 \right) = 7.00\nonumber$
The pH of pure water or any neutral solution is thus 7.00. For recording purposes, the numbers to 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. Consider a solution with $\left[ \ce{H^+} \right] = 1.0 \times 10^{-4} \: \text{M}$. That is a hydrogen-ion concentration that is 1000 times higher than the concentration in pure water. The pH of such a solution is 4.00, a difference of just 3 pH units. Notice that when the $\left[ \ce{H^+} \right]$ 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 with $\left[ \ce{H^+} \right] = 1 \times 10^{-2} \: \text{M}$ is 2 and the pH of a solution with $\left[ \ce{H^+} \right] = 1 \times 10^{-10} \: \text{M}$ is 10.
As we saw earlier, a solution with $\left[ \ce{H^+} \right]$ higher than $1.0 \times 10^{-7}$ is acidic, while a solution with $\left[ \ce{H^+} \right]$ lower than $1.0 \times 10^{-7}$ is basic. Consequently, solutions with a pH of less than 7 are acidic, while those with a pH higher than 7 are basic. Figure $1$ illustrates this relationship, along with some examples of various solutions.
Summary
• The pH of a solution is the negative logarithm of the hydrogen-ion concentration.
• Solutions with a pH of less than 7 are acidic, while those with a pH higher than 7 are basic.
• pH values for several common materials are listed. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/21%3A_Acids_and_Bases/21.08%3A_Ion-Product_of_Water.txt |
It's not uncommon to see a tropical fish tank in homes or businesses. These brightly-colored creatures are relaxing to watch, but do require a certain amount of maintenance in order for them to survive. Tap water is usually too alkaline when it comes out of the faucet, so some adjustments need to be made. The pH of the water will change over time while it is in the tank, which means it needs to be tested every so often. Any fish tank caretaker has a chance to be a chemist for their fish!
Calculating pH of Acids and Bases
Calculation of pH is simple when there is a $1 \times 10^\text{power}$ problem. However, in real life, this is rarely the situation. If the coefficient is not equal to 1, a calculator must be used to find the pH. For example, the pH of a solution with $\left[ \ce{H^+} \right] = 2.3 \times 10^{-5} \: \text{M}$ can be found as shown below.
$\text{pH} = -\text{log} \left[ 2.3 \times 10^{-5} \right] = 4.64\nonumber$
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.
$\left[ \ce{H^+} \right] = 10^{-\text{pH}}\nonumber$
For example, suppose that you have a solution with a pH of 9.14. To find the $\left[ \ce{H^+} \right]$ use the $10^x$ key.
$\left[ \ce{H^+} \right] = 10^{-\text{pH}} = 10^{-9.14} = 7.24 \times 10^{-10} \: \text{M}\nonumber$
Hydroxide Ion Concentration and pH
As we saw earlier, the hydroxide ion concentration of any aqueous solution is related to the hydrogen ion concentration through the value of $K_\text{w}$. We can use that relationship to calculate the pH of a solution of a base.
Example $1$
Sodium hydroxide is a strong base. Find the pH of a solution prepared by dissolving $1.0 \: \text{g}$ of $\ce{NaOH}$ into enough water to make $1.0 \: \text{L}$ of solution.
Known
• Mass $\ce{NaOH} = 1.0 \: \text{g}$
• Molar mass $\ce{NaOH} = 40.00 \: \text{g/mol}$
• Volume solution $= 1.0 \: \text{L}$
• $K_\text{w} = 1.0 \times 10^{-14}$
Unknown
First, convert the mass of $\ce{NaOH}$ to moles. Second, calculate the molarity of the $\ce{NaOH}$ solution. Because $\ce{NaOH}$ is a strong base and is soluble, the $\left[ \ce{OH^-} \right]$ will be equal to the concentration of the $\ce{NaOH}$. Third, use $K_\text{w}$ to calculate the $\left[ \ce{H^+} \right]$ in the solution. Lastly, calculate the pH.
Step 2: Solve.
\begin{align*} &1.00 \: \cancel{\text{g} \: \ce{NaOH}} \times \frac{1 \: \text{mol} \: \ce{NaOH}}{40.00 \: \cancel{\text{g} \: \ce{NaOH}}} = 0.025 \: \text{mol} \: \ce{NaOH} \ &\text{Molarity} = \frac{0.025 \: \text{mol} \: \ce{NaOH}}{1.00 \: \text{L}} = 0.025 \: \text{M} \: \ce{NaOH} = 0.025 \: \text{M} \: \ce{OH^-} \ &\left[ \ce{H^+} \right] = \frac{K_\text{w}}{\left[ \ce{OH^-} \right]} = \frac{1.0 \times 10^{-14}}{0.025 \: \text{M}} = 4.0 \times 10^{-13} \: \text{M} \ &\text{pH} = -\text{log} \left[ \ce{H^+} \right] = -\text{log} \left( 4.0 \times 10^{-13} \right) = 12.40 \end{align*}
Step 3: Think about your result.
The solution is basic and so its pH is greater than 7. The reported pH is rounded to two decimal places because the original mass and volume has two significant figures.
21.11: The pOH Concept
Soap Lake has a long history as a healing place. Indian tribes would put aside their rivalries when they came to the lake to enjoy the high mineral content of the water. In the days before good antibiotics, thousands of visitors would come and enjoy the soothing waters of the lake. Soap Lake is alkaline, with water quality though to be similar to that of the moons of the planet Jupiter.
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:
$\text{pOH} = -\text{log} \left[ \ce{OH^-} \right]\nonumber$
The pH of a solution can be related to the pOH. Consider a solution with a pH $= 4.0$. The $\left[ \ce{H^+} \right]$ of the solution would be $1.0 \times 10^{-4} \: \text{M}$. Dividing $K_\text{w}$ by this yields a $\left[ \ce{OH^-} \right]$ of $1.0 \times 10^{-10} \: \text{M}$. Finally the pOH of the solution equals $-\text{log} \left( 1.0 \times 10^{-10} \right) = 10$. This example illustrates the following relationship.
$\text{pH} + \text{pOH} = 14\nonumber$
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 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.
Known
• pH $= 4.42$
• pH $+$ pOH $= 14$
Unknown
First, the pOH is calculated, followed by the $\left[ \ce{OH^-} \right]$.
Step 2: Solve.
\begin{align*} \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} \end{align*}\nonumber
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.
Summary
• The pOH of a solution is the negative logarithm of the hydroxide-ion concentration.
• Calculations involving pOH are described. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/21%3A_Acids_and_Bases/21.10%3A_Calculating_pH_of_Acids_and_Bases.txt |
The etching of glass is a slow process that can produce beautiful artwork. Traditionally, the glass has been treated with dilute hydrofluoric acid which gradually dissolves the glass under it. Parts of the piece that should not be etched are covered with wax or some other non-reactive material. In more recent times, compounds such as ammonium bifluoride have been used. Whichever chemical is employed, the artist must be very careful not to get any on their skin.
Strong and Weak Acids and Acid Ionization Constant
Acids are classified as either strong or weak, based on their ionization in water. A strong acid is an acid which is completely ionized in an aqueous solution. Hydrogen chloride $\left( \ce{HCl} \right)$ ionizes completely into hydrogen ions and chloride ions in water.
$\ce{HCl} \left( g \right) \rightarrow \ce{H^+} \left( aq \right) + \ce{Cl^-} \left( aq \right)\nonumber$
A weak acid is an acid that ionizes only slightly in an aqueous solution. Acetic acid (found in vinegar) is a very common weak acid. Its ionization is shown below.
$\ce{CH_3COOH} \left( aq \right) \rightleftharpoons \ce{H^+} \left( aq \right) + \ce{CH_3COO^-} \left( aq \right)\nonumber$
The ionization of acetic acid is incomplete, and so the equation is shown with a double arrow. The extent of ionization of weak acids varies, but is generally less than $10\%$. A $0.10 \: \text{M}$ solution of acetic acid is only about $1.3\%$ ionized, meaning that the equilibrium strongly favors the reactants.
Weak acids, like strong acids, ionize to yield the $\ce{H^+}$ ion and a conjugate base. Because $\ce{HCl}$ is a strong acid, its conjugate base $\left( \ce{Cl^-} \right)$ is extremely weak. The chloride ion is incapable of accepting the $\ce{H^+}$ ion and becoming $\ce{HCl}$ again. In general, the stronger the acid, the weaker its conjugate base. Likewise, the weaker the acid, the stronger its conjugate base.
Acid Conjugate Base
Table $1$: Relative Strengths of Acids and their Conjugate Bases
Strong Acids
$\ce{HCl}$ (hydrochloric acid) (strongest) $\ce{Cl^-}$ (chloride ion) (weakest)
$\ce{H_2SO_4}$ (sulfuric acid) $\ce{HSO_4^-}$ (hydrogen sulfate ion)
$\ce{HNO_3}$ (nitric acid) $\ce{NO_3^-}$ (nitrate ion)
Weak Acids
$\ce{H_3PO_4}$ (phosphoric acid) $\ce{H_2PO_4^-}$ (dihydrogen phosphate ion)
$\ce{CH_3COOH}$ (acetic acid) $\ce{CH_3COO^-}$ (acetate ion)
$\ce{H_2CO_3}$ (carbonic acid) $\ce{HCO_3^-}$ (hydrogen carbonate ion)
$\ce{HCN}$ (hydrocyanic acid) (weakest) $\ce{CN^-}$ (cyanide ion) (strongest)
Strong acids are $100\%$ ionized in solution. Weak acids are only slightly ionized. Phosphoric acid is stronger than acetic acid, and so is ionized to a greater extent. Acetic acid is stronger than carbonic acid, and so on.
The Acid Ionization Constant, $K_\text{a}$
The ionization for a general weak acid, $\ce{HA}$, can be written as follows:
$\ce{HA} \left( aq \right) \rightarrow \ce{H^+} \left( aq \right) + \ce{A^-} \left( aq \right)\nonumber$
Because the acid is weak, an equilibrium expression can be written. An acid ionization constant $\left( K_\text{a} \right)$ is the equilibrium constant for the ionization of an acid.
$K_\text{a} = \frac{\left[ \ce{H^+} \right] \left[ \ce{A^-} \right]}{\left[ \ce{HA} \right]}\nonumber$
The acid ionization represents the fraction of the original acid that has been ionized in solution. Therefore, the numerical value of $K_\text{a}$ is a reflection of the strength of the acid. Weak acids with relatively higher $K_\text{a}$ values are stronger than acids with relatively lower $K_\text{a}$ values. Because strong acids are essentially $100\%$ ionized, the concentration of the acid in the denominator is nearly zero and the $K_\text{a}$ value approaches infinity. For this reason, $K_\text{a}$ values are generally reported for weak acids only.
The table below is a listing of acid ionization constants for several acids. Note that polyprotic acids have a distinct ionization constant for each ionization step, with each successive ionization constant being smaller than the previous one.
Name of Acid Ionization Equation $K_\text{a}$
Table $2$: Acid Ionization Constants at $25^\text{o} \text{C}$
Sulfuric acid
$\ce{H_2SO_4} \rightleftharpoons \ce{H^+} + \ce{HSO_4^-}$
$\ce{HSO_4} \rightleftharpoons \ce{H^+} + \ce{SO_4^{2-}}$
very large
$1.3 \times 10^{-2}$
Oxalic acid
$\ce{H_2C_2O_4} \rightleftharpoons \ce{H^+} + \ce{HC_2O_4^-}$
$\ce{HC_2O_4} \rightleftharpoons \ce{H^+} + \ce{C_2O_4^{2-}}$
$6.5 \times 10^{-2}$
$6.1 \times 10^{-5}$
Phosphoric acid
$\ce{H_3PO_4} \rightleftharpoons \ce{H^+} + \ce{H_2PO_4^-}$
$\ce{H_2PO_4^-} \rightleftharpoons \ce{H^+} + \ce{HPO_4^{2--}}$
$\ce{HPO_4^{2-}} \rightleftharpoons \ce{H^+} + \ce{PO_4^{3-}}$
$7.5 \times 10^{-3}$
$6.2 \times 10^{-8}$
$4.8 \times 10^{-13}$
Hydrofluoric acid $\ce{HF} \rightleftharpoons \ce{H^+} + \ce{F^-}$ $7.1 \times 10^{-4}$
Nitrous acid $\ce{HNO_2} \rightleftharpoons \ce{H^+} + \ce{NO_2^-}$ $4.5 \times 10^{-4}$
Benzoic acid $\ce{C_6H_5COOH} \rightleftharpoons \ce{H^+} + \ce{C_6H_5COO^-}$ $6.5 \times 10^{-5}$
Acetic acid $\ce{CH_3COOH} \rightleftharpoons \ce{H^+} + \ce{CH_3COO^-}$ $1.8 \times 10^{-5}$
Carbonic acid
$\ce{H_2CO_3} \rightleftharpoons \ce{H^+} + \ce{HCO_3^-}$
$\ce{HCO_3^-} \rightleftharpoons \ce{H^+} + \ce{CO_3^{2-}}$
$4.2 \times 10^{-7}$
$4.8 \times 10^{-11}$
Hydrocyanic acid $\ce{HCN} \rightleftharpoons \ce{H^+} + \ce{CN^-}$ $4.9 \times 10^{-10}$
Summary
• A strong acid is an acid which is completely ionized in an aqueous solution.
• A weak acid is an acid that ionizes only slightly in an aqueous solution.
• The acid ionization constant $\left( K_\text{a} \right)$ is defined. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/21%3A_Acids_and_Bases/21.12%3A_Strong_and_Weak_Acids_and_Acid_Ionization_Constant_%28K_texta%29.txt |
All the complex electronics and apparatuses in a space shuttle generate heat, as do the astronauts. The shuttles have a complex arrangement of systems to dissipate that heat into outer space. One of the components of this system is a series of coils filled with ammonia that are located on the outside of the shuttle. Ammonia absorbs the heat and then releases it into space as the gas circulates through the coils. This approach is both inexpensive and effective.
Strong and Weak Bases and Base Ionization Constant, $K_\text{b}$
As with acids, bases can either be strong or weak, depending on the extent of their ionization. A strong base is a base that ionizes completely in an aqueous solution. The most common strong bases are soluble metal hydroxide compounds such as potassium hydroxide. Some metal hydroxides are not as strong, simply because they are not as soluble. Calcium hydroxide is only slightly soluble in water, but the portion that does dissolve also dissociates into ions.
A weak base is a base that ionizes only slightly in an aqueous solution. Recall that a base can be defined as a substance that accepts a hydrogen ion from another substance. When a weak base such as ammonia is dissolved in water, it accepts an $\ce{H^+}$ ion from water, forming the hydroxide ion and the conjugate acid of the base, the ammonium ion.
$\ce{NH_3} \left( aq \right) + \ce{H_2O} \left( l \right) \rightleftharpoons \ce{NH_4^+} \left( aq \right) + \ce{OH^-} \left( aq \right)\nonumber$
The equilibrium greatly favors the reactants and the extent of ionization of the ammonia molecule is very small.
An equilibrium expression can be written for the reactions of weak bases with water. Because the concentration of water is extremely large and virtually constant, the water is not included in the expression. A base ionization constant $\left( K_\text{b} \right)$ is the equilibrium constant for the ionization of a base. For ammonia, the expression is:
$K_\text{b} = \frac{\left[ \ce{NH_4^+} \right] \left[ \ce{OH^-} \right]}{\left[ \ce{NH_3} \right]}\nonumber$
The numerical value of $K_\text{b}$ is a reflection of the strength of the base. Weak bases with relatively high $K_\text{b}$ values are stronger than bases with relatively low $K_\text{b}$ values. The table below is a listing of base ionization constants for several weak bases.
Table $1$: Base Ionization Constants at $25^\text{o} \text{C}$
Name of Base Ionization Equation $K_\text{b}$
Methylamine $\ce{CH_3NH_2} + \ce{H_2O} \rightleftharpoons \ce{CH_3NH_3^+} + \ce{OH^-}$ $5.6 \times 10^{-4}$
Ammonia $\ce{NH_3} + \ce{H_2O} \rightleftharpoons \ce{NH_4^+} + \ce{OH^-}$ $1.8 \times 10^{-5}$
Pyridine $\ce{C_5H_5N} + \ce{H_2O} \rightleftharpoons \ce{C_5H_5NH^+} + \ce{OH^-}$ $1.7 \times 10^{-9}$
Acetate ion $\ce{CH_3COO^-} + \ce{H_2O} \rightleftharpoons \ce{CH_3COOH} + \ce{OH^-}$ $5.6 \times 10^{-10}$
Fluoride ion $\ce{F^-} + \ce{H_2O} \rightleftharpoons \ce{HF} + \ce{OH^-}$ $1.4 \times 10^{-11}$
Urea $\ce{H_2NCONH_2} + \ce{H_2O} \rightleftharpoons \ce{H_2NCONH_3^+} + \ce{OH^-}$ $1.5 \times 10^{-14}$
Notice that the conjugate base of a weak acid is also a strong base. For example, the acetate ion has a small tendency to accept a hydrogen ion from water to form acetic acid and the hydroxide ion.
Summary
• A strong base is a base that ionizes completely in an aqueous solution.
• A weak base is a base that ionizes only slightly in an aqueous solution.
• A base ionization constant $\left( K_\text{b} \right)$ is the equilibrium constant for the ionization of a base.
• The conjugate base of a weak acid is also a strong base. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/21%3A_Acids_and_Bases/21.13%3A_Strong_and_Weak_Bases_and_Base_Ionization_Constant_%28left%28_K_textb_right%29%29.txt |
The pH meter was invented because Florida orange growers needed a way to test the acidity of their fruit. The first meter was invented by Arnold Beckman, who went on to form Beckman Instruments. Beckman's business was very successful, and he used much of his fortune to fund science education and research. The Beckman family donated \$40 million to build the Beckman Institute at the University of Illinois.
Calculating $K_\text{a}$ and $K_\text{b}$
The numerical value of $K_\text{a}$ and $K_\text{b}$ can be determined from an experiment. A solution of known concentration is prepared and its pH is measured with an instrument called a pH meter.
Example $1$
A $0.500 \: \text{M}$ solution of formic acid is prepared and its pH is measured to be 2.04. Determine the $K_\text{a}$ for formic acid.
Known
• Initial $\left[ \ce{HCOOH} \right] = 0.500 \: \text{M}$
• pH $= 2.04$
Unknown
First, the pH is used to calculate the $\left[ \ce{H^+} \right]$ at equilibrium. An ICE table is set up in order to determine the concentrations of $\ce{HCOOH}$ and $\ce{HCOO^-}$ at equilibrium. All concentrations are then substituted into the $K_\text{a}$ expression and the $K_\text{a}$ value is calculated.
Step 2: Solve.
$\left[ \ce{H^+} \right] = 10^{-\text{pH}} = 10^{-2.04} = 9.12 \times 10^{-3} \: \text{M}\nonumber$
Since each formic acid molecule that ionizes yields one $\ce{H^+}$ ion and one formate ion $\left( \ce{HCOO^-} \right)$, the concentrations of $\ce{H^+}$ and $\ce{HCOO^-}$ are equal at equilibrium. We assume that the initial concentrations of each ion are zero, resulting in the following ICE table.
$\begin{array}{l|ccc} & \ce{HCOOH} & \ce{H^+} & \ce{HCOO^-} \ \hline \text{Initial} & 0.500 & 0 & 0 \ \text{Change} & -9.12 \times 10^{-3} & +9.12 \times 10^{-3} & +9.12 \times 10^{-3} \ \text{Equilibrium} & 0.491 & 9.12 \times 10^{-3} & 9.12 \times 10^{-3} \end{array}\nonumber$
Now, substituting into the $K_\text{a}$ expression gives:
$K_\text{a} = \frac{\left[ \ce{H^+} \right] \left[ \ce{HCOO^-} \right]}{\left[ \ce{HCOOH} \right]} = \frac{\left( 9.12 \times 10^{-3} \right) \left( 9.12 \times 10^{-3} \right)}{0.491} = 1.7 \times 10^{-4}\nonumber$
Step 3: Think about your result.
The value of $K_\text{a}$ is consistent with that of a weak acid. Two significant figures are appropriate for the answer, since there are two digits after the decimal point in the reported pH.
Similar steps can be taken to determine the $K_\text{b}$ of a base. For example, a $0.750 \: \text{M}$ solution of the weak base ethylamine $\left( \ce{C_2H_5NH_2} \right)$ has a pH of 12.31.
$\ce{C_2H_5NH_2} + \ce{H_2O} \rightleftharpoons \ce{C_2H_5NH_3^+} + \ce{OH^-}\nonumber$
Since one of the products of the ionization reaction is the hydroxide ion, we need to first find the $\left[ \ce{OH^-} \right]$ at equilibrium. The pOH is $14 - 12.31 = 1.69$. The $\left[ \ce{OH^-} \right]$ is then found from $10^{-1.69} = 2.04 \times 10^{-2} \: \text{M}$. The ICE table is then set up as shown below.
$\begin{array}{l|ccc} & \ce{C_2H_5NH_2} & \ce{C_2H_5NH_3^+} & \ce{OH^-} \ \hline \text{Initial} & 0.750 & 0 & 0 \ \text{Change} & -2.04 \times 10^{-2} & +2.04 \times 10^{-2} & +2.04 \times 10^{-2} \ \text{Equilibrium} & 0.730 & 2.04 \times 10^{-2} & 2.04 \times 10^{-2} \end{array}\nonumber$
Substituting into the $K_\text{b}$ expression yields the $K_\text{b}$ for ethylamine.
$K_\text{b} = \frac{\left[ \ce{C_2H_5NH_3^+} \right] \left[ \ce{OH^-} \right]}{\left[ \ce{C_2H_5NH_2} \right]} = \frac{\left( 2.04 \times 10^{-2} \right) \left( 2.04 \times 10^{-2} \right)}{0.730} = 5.7 \times 10^{-4}\nonumber$ | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/21%3A_Acids_and_Bases/21.14%3A_Calculating_%28K_texta%29_and_%28K_textb%29.txt |
Bees are beautiful creatures that help plants flourish. They carry pollen from one plant to another to facilitate plant growth and development. However, they can be troublesome when they sting! For those who are allergic to bee venom, this can be a serious, life-threatening problem. For all other humans, it can be a painful experience. When stung by a bee, one first-aid treatment is to apply a paste of baking soda (sodium bicarbonate) to the stung area. This weak base helps with the itching and swelling that accompanies the bee sting.
Calculating pH of Weak Acid and Base Solutions
The $K_\text{a}$ and $K_\text{b}$ values have been determined for a great many acids and bases, as shown in Tables 21.12.2 and 21.13.1. These can be used to calculate the pH of any solution of a weak acid or base whose ionization constant is known.
Example $1$
Calculate the pH of a $2.00 \: \text{M}$ solution of nitrous acid $\left( \ce{HNO_2} \right)$. The $K_\text{a}$ for nitrous acid is $4.5 \times 10^{-4}$.
Known
• Initial $\left[ \ce{HNO_2} \right] = 2.00 \: \text{M}$
• $K_\text{a} = 4.5 \times 10^{-4}$
Unknown
First, an ICE table is set up with the variable $x$ used to signify the change in concentration of the substance due to ionization of the acid. Then the $K_\text{a}$ expression is used to solve for $x$ and calculate the pH.
Step 2: Solve.
$\begin{array}{l|ccc} & \ce{HNO_2} & \ce{H^+} & \ce{NO_2^-} \ \hline \text{Initial} & 2.00 & 0 & 0 \ \text{Change} & -x & +x & +x \ \text{Equilibrium} & 2.00 - x & x & x \end{array}\nonumber$
The $K_\text{a}$ expression and value are used to set up an equation to solve for $x$.
$K_\text{a} = 4.5 \times 10^{-4} = \frac{\left( x \right) \left( x \right)}{2.00 - x} = \frac{x^2}{2.00 - x}\nonumber$
The quadratic equation is required to solve this equation for $x$. However, a simplification can be made of the fact that the extent of ionization of weak acids is small. The value of $x$ will be significantly less than 2.00, so the "$-x$" in the denominator can be dropped.
\begin{align*} 4.5 \times 10^{-4} &= \frac{x^2}{2.00 - x} \approx \frac{x^2}{2.00} \ x &= \sqrt{ 4.5 \times 10^{-4} \left( 2.00 \right)} = 2.9 \times 10^{-2} \: \text{M} = \left[ \ce{H^+} \right] \end{align*}\nonumber
Since the variable $x$ represents the hydrogen-ion concentration, the pH of the solution can now be calculated.
$\text{pH} = -\text{log} \left[ \ce{H^+} \right] = -\text{log} \left[ 2.9 \times 10^{-2} \right] = 1.54\nonumber$
Step 3: Think about your result.
The pH of a $2.00 \: \text{M}$ solution of a strong acid would be equal to $-\text{log} \left( 2.00 \right) = -0.30$. The higher pH of the $2.00 \: \text{M}$ nitrous acid is consistent with it being a weak acid and therefore not as acidic as a strong acid would be.
The procedure for calculating the pH of a solution of a weak base is similar to that of the weak acid in the example. However, the variable $x$ will represent the concentration of the hydroxide ion. The pH is found by taking the negative logarithm to get the pOH, followed by subtracting from 14 to get the pH.
Summary
• The procedure for calculating the pH of a weak acid or base is illustrated. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/21%3A_Acids_and_Bases/21.15%3A_Calculating_pH_of_Weak_Acid_and_Base_Solutions.txt |
Pouring concrete and working it are messy jobs. In the process, a lot of wastewater with an alkaline pH is generated. Often, regulations require that this wastewater be cleaned up at the site. One practical way to neutralize the basic pH is to bubble $\ce{CO_2}$ into the water. The carbon dioxide forms a weak acid (carbonic acid, $\ce{H_2CO_3}$) in solution which serves to bring the alkaline pH down to something closer to neutral.
Neutralization Reactions and Net Ionic Equations for Neutralization Reactions
A neutralization reaction is a reaction in which an acid and a base react in an aqueous solution to produce a salt and water. The aqueous sodium chloride that is produced in the reaction is called a salt. A salt is an ionic compound composed of a cation from a base and an anion from an acid. A salt is essentially any ionic compound that is neither an acid nor a base.
Strong Acid-Strong Base Reactions
When equal amounts of a strong acid such as hydrochloric acid are mixed with a strong base such as sodium hydroxide, the result is a neutral solution. The products of the reaction do not have the characteristics of either an acid or a base. The balanced molecular equation is:
$\ce{HCl} \left( aq \right) + \ce{NaOH} \left( aq \right) \rightarrow \ce{NaCl} \left( aq \right) + \ce{H_2O} \left( l \right)\nonumber$
Chemical reactions occurring in aqueous solution are more accurately represented with a net ionic equation. The full ionic equation for the neutralization of hydrochloric acid by sodium hydroxide is written as follows:
$\ce{H^+} \left( aq \right) + \ce{Cl^-} \left( aq \right) + \ce{Na^+} \left( aq \right) + \ce{OH^-} \left( aq \right) \rightarrow \ce{Na^+} \left( aq \right) + \ce{Cl^-} \left( aq \right) + \ce{H_2O} \left( l \right)\nonumber$
Since the acid and base are both strong, they are fully ionized and so are written as ions, as is the $\ce{NaCl}$ formed as a product. The sodium and chloride ions are spectator ions in the reaction, leaving the following as the net ionic reaction.
$\ce{H^+} \left( aq \right) + \ce{OH^-} \left( aq \right) \rightarrow \ce{H_2O} \left( l \right)\nonumber$
All neutralization reactions of a strong acid with a strong base simplify to the net ionic reaction of hydrogen ion combining with hydroxide ion to produce water.
What if the acid is a diprotic acid such as sulfuric acid? The balanced molecular equation now involves a 1:2 ratio between acid and base.
$\ce{H_2SO_4} \left( aq \right) + 2 \ce{NaOH} \left( aq \right) \rightarrow \ce{Na_2SO_4} \left( aq \right) + \ce{H_2O} \left( l \right)\nonumber$
In order for the reaction to be a full neutralization, twice as many moles of $\ce{NaOH}$ must react with the $\ce{H_2SO_4}$. The sodium sulfate salt is soluble, and so the net ionic reaction is again the same. Different mole ratios occur for other polyprotic acids or bases with multiple hydroxides such as $\ce{Ca(OH)_2}$.
Reactions Involving a Weak Acid or Weak Base
Reactions where at least one of the components is weak do not generally result in a neutral solution. The reaction between weak nitrous acid and strong potassium hydroxide is shown below.
$\ce{HNO_2} \left( aq \right) + \ce{KOH} \left( aq \right) \rightarrow \ce{KNO_2} \left( aq \right) + \ce{H_2O} \left( l \right)\nonumber$
In order to write the net ionic equation, the weak acid must be written as a molecule since it does not ionize to a great extent in water. The base and the salt are fully dissociated.
$\ce{HNO_2} \left( aq \right) + \ce{K^+} \left( aq \right) + \ce{OH^-} \left( aq \right) \rightarrow \ce{K^+} \left( aq \right) + \ce{NO_2^-} \left( aq \right) + \ce{H_2O} \left( l \right)\nonumber$
The only spectator ion is the potassium ion, resulting in the net ionic equation:
$\ce{HNO_2} \left( aq \right) + \ce{OH^-} \left( aq \right) \rightarrow \ce{NO_2^-} \left( aq \right) + \ce{H_2O} \left( l \right)\nonumber$
The strong hydroxide ion essentially "forces" the weak nitrous acid to become ionized. The hydrogen ion from the acid combines with the hydroxide ion to form water, leaving the nitrite ion as the other product. The resulting solution is not neutral (pH $= 7$), but instead is slightly basic.
Reactions can also involve a weak base and strong acid, resulting in a solution that is slightly acidic. The molecular and net ionic equations for the reaction of hydrochloric acid and ammonia are shown below.
\begin{align*} &\ce{HCl} \left( aq \right) + \ce{NH_3} \left( aq \right) \rightarrow \ce{NH_4Cl} \left( aq \right) \ &\ce{H^+} \left( aq \right) + \ce{NH_3} \left( aq \right) \rightarrow \ce{NH_4^+} \left( aq \right) \: \: \: \: \: \: \: \: \: \: \left( \ce{Cl^-} \: \text{is a spectator ion} \right) \end{align*}\nonumber
Reactions between acids and bases that are both weak may result in solutions that are neutral, acidic, or basic.
Summary
• A neutralization reaction is a reaction in which an acid and a base react in an aqueous solution to produce a salt and water.
• Equations for acid-base neutralizations are given.
• Net ionic equations for neutralization reactions are given. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/21%3A_Acids_and_Bases/21.16%3A_Neutralization_Reaction_and_Net_Ionic_Equations_for_Neutralization_Reactions.txt |
A large amount of current research involves the development of biodiesel fuels. Often this material can be made from used vegetable oils. The vegetable oil is treated with lye to create the biofuel. In the oils is a variable amount of acid that needs to be determined, so that the workers will know how much lye to add to make the final fuel. Before the lye is added, the native vegetable oil is titrated to find out how much free acid is present. Then, the amount of lye added can be adjusted to take into account the amount needed to neutralize these free acids.
Titration Experiment
In the neutralization of hydrochloric acid by sodium hydroxide, the mole ratio of acid to base is 1:1.
$\ce{HCl} \left( aq \right) + \ce{NaOH} \left( aq \right) \rightarrow \ce{NaCl} \left( aq \right) + \ce{H_2O} \left( l \right)\nonumber$
One mole of $\ce{HCl}$ would be fully neutralized by one mole of $\ce{NaOH}$. If instead the hydrochloric acid was reacted with barium hydroxide, the mole ratio would be 2:1.
$2 \ce{HCl} \left( aq \right) + \ce{Ba(OH)_2} \left( aq \right) \rightarrow \ce{BaCl_2} \left( aq \right) + 2 \ce{H_2O} \left( l \right)\nonumber$
Now two moles of $\ce{HCl}$ would be required to neutralize one mole of $\ce{Ba(OH)_2}$. The mole ratio ensures that the number of moles of $\ce{H^+}$ ions supplied by the acid is equal to the number of $\ce{OH^-}$ ions supplied by the base. This must be the case for neutralization to occur. The equivalence point is the point in a neutralization reaction where the number of moles of hydrogen ions is equal to the number of moles of hydroxide ions.
In the laboratory, it is useful to have an experiment where the unknown concentration of an acid or a base can be determined. This can be accomplished by performing a controlled neutralization reaction. A titration is an experiment where a volume of a solution of known concentration is added to a volume of another solution in order to determine its concentration. Many titrations are acid-base neutralization reactions, though other types of titrations can also be performed.
In order to perform an acid-base titration, the chemist must have a way to visually detect that the neutralization reaction has occurred. An indicator is a substance that has a distinctly different color when in an acidic or basic solution. A commonly used indicator for strong acid-strong base titrations is phenolphthalein. Solutions in which a few drops of phenolphthalein have been added turn from colorless to brilliant pink as the solution turns from acidic to basic. The steps in a titration reaction are outlined below.
1. A measured volume of an acid of unknown concentration is added to an Erlenmeyer flask.
2. Several drops of an indicator are added to the acid and mixed by swirling the flask.
3. A buret is filled with a base solution of known molarity.
4. The stopcock of the buret is opened and base is slowly added to the acid, while the flask is constantly swirled to ensure mixing. The stopcock is closed at the exact point at which the indicator just changes color.
The standard solution is the solution in a titration whose concentration is known. In the titration described above, the base solution is the standard solution. It is very important in a titration to add the solution from the buret slowly, so that the point at which the indicator changes color can be found accurately. The end point of a titration is the point at which the indicator changes color. When phenolphthalein is the indicator, the end point will be signified by a faint pink color.
Summary
• The equivalence point is the point in a neutralization reaction where the number of moles of hydrogen ions is equal to the number of moles of hydroxide ions.
• A titration is an experiment where a volume of a solution of known concentration is added to a volume of another solution in order to determine its concentration. (Many titrations are acid-base neutralization reactions.)
• An indicator is a substance that has a distinctly different color when in an acidic or basic solution.
• The process for carrying out a titration is described. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/21%3A_Acids_and_Bases/21.17%3A_Titration_Experiment.txt |
The manufacture of soap requires a number of chemistry techniques. One necessary piece of information is the saponification number. This is the amount of base needed to hydrolyze a certain amount of fat to produce the free fatty acids that are an essential part of the final product. The fat is heated with a known amount of base (usually $\ce{NaOH}$ or $\ce{KOH}$). After hydrolysis is complete, the leftover base is titrated to determine how much was needed to hydrolyze the fat sample.
Titration Calculations
At the equivalence point in a neutralization, the moles of acid are equal to the moles of base.
$\text{moles acid} = \text{moles base}\nonumber$
Recall that the molarity $\left( \text{M} \right)$ of a solution is defined as the moles of the solute divided by the liters of solution $\left( \text{L} \right)$. So the moles of solute are therefore equal to the molarity of a solution multiplied by the volume in liters.
$\text{moles solute} = \text{M} \times \text{L}\nonumber$
We can then set the moles of acid equal to the moles of base.
$\text{M}_A \times \text{V}_A = \text{M}_B \times \text{V}_B\nonumber$
$\text{M}_A$ is the molarity of the acid, while $\text{M}_B$ is the molarity of the base. $\text{V}_A$ and $\text{V}_B$ are the volumes of the acid and base, respectively.
Suppose that a titration is performed and $20.70 \: \text{mL}$ of $0.500 \: \text{M} \: \ce{NaOH}$ is required to reach the end point when titrated against $15.00 \: \text{mL}$ of $\ce{HCl}$ of unknown concentration. The above equation can be used to solve for the molarity of the acid.
$\text{M}_A = \frac{\text{M}_B \times \text{V}_B}{\text{V}_A} = \frac{0.500 \: \text{M} \times 20.70 \: \text{mL}}{15.00 \: \text{mL}} = 0.690 \: \text{M}\nonumber$
The higher molarity of the acid compared to the base in this case means that a smaller volume of the acid is required to reach the equivalence point.
The above equation works only for neutralizations in which there is a 1:1 ratio between the acid and the base. The example below demonstrates the technique to solve a titration problem for a titration of sulfuric acid with sodium hydroxide.
Example $1$
In a titration of sulfuric acid against sodium hydroxide, $32.20 \: \text{mL}$ of $0.250 \: \text{M} \: \ce{NaOH}$ is required to neutralize $26.60 \: \text{mL}$ of $\ce{H_2SO_4}$. Calculate the molarity of the sulfuric acid.
$\ce{H_2SO_4} \left( aq \right) + 2 \ce{NaOH} \left( aq \right) \rightarrow \ce{Na_2SO_4} \left( aq \right) + 2 \ce{H_2O} \left( l \right)\nonumber$
Step 1: List the known values and plan the problem.
• Molarity $\ce{NaOH} = 0.250 \: \text{M}$
• Volume $\ce{NaOH} = 32.20 \: \text{mL}$
• Volume $\ce{H_2SO_4} = 26.60 \: \text{mL}$
Unknown
First determine the moles of $\ce{NaOH}$ in the reaction. From the mole ratio, calculate the moles of $\ce{H_2SO_4}$ that reacted. Finally, divide the moles of $\ce{H_2SO_4}$ by its volume to get the molarity.
Step 2: Solve.
\begin{align*} &\text{mol} \: \ce{NaOH} = \text{M} \times \text{L} = 0.250 \: \text{M} \times 0.03220 \: \text{L} = 8.05 \times 10^{-3} \: \text{mol} \: \ce{NaOH} \ &8.05 \times 10^{-3} \: \text{mol} \: \ce{NaOH} \times \frac{1 \: \text{mol} \: \ce{H_2SO_4}}{2 \: \text{mol} \: \ce{NaOH}} = 4.03 \times 10^{-3} \: \text{mol} \: \ce{H_2SO_4} \ &\frac{4.03 \times 10^{-3} \: \text{mol} \: \ce{H_2SO_4}}{0.02660 \: \text{L}} = 0.151 \: \text{M} \: \ce{H_2SO_4} \end{align*}\nonumber
Step 3: Think about your result.
The volume of $\ce{H_2SO_4}$ required is smaller than the volume of $\ce{NaOH}$ because of the two hydrogen ions contributed by each molecule.
Summary
• The process of calculating concentration from titration data is described and illustrated. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/21%3A_Acids_and_Bases/21.18%3A_Titration_Calculations.txt |
The \(x\)-\(y\) plot that we know of as a graph was the brainchild of the French mathematician-philosopher Rene Descartes (1596-1650). His studies in mathematics led him to develop what was known as "Cartesian geometry", including the concept of today's graphs. The coordinates are often referred to as Cartesian coordinates.
Titration Curves
As base is added to acid at the beginning of a titration, the pH rises very slowly. Nearer to the equivalence point, the pH begins to rapidly increase. If the titration is a strong acid with a strong base, the pH at the equivalence point is equal to 7. A bit past the equivalence point, the rate of change of the pH again slows down. A titration curve is a graphical representation of the pH of a solution during a titration. The figure below shows two different examples of a strong acid-strong base titration curve. On the left is a titration in which the base is added to the acid, and so the pH progresses from low to high. On the right is a titration in which the acid is added to the base. In this case, the pH starts out high and decreases during the titration. In both cases, the equivalence point is reached when the moles of acid and base are equal and the pH is 7. This also corresponds to the color change of the indicator.
Titration curves can also be generated in the case of a weak acid-strong base titration or a strong acid-weak base titration. The general shape of the titration curve is the same, but the pH at the equivalence point is different. In a weak acid-strong base titration, the pH is greater than 7 at the equivalence point. In a strong acid-weak base titration, the pH is less than 7 at the equivalence point.
Summary
• A titration curve is a graphical representation of the pH of a solution during a titration.
• In a strong acid-strong base titration, the equivalence point is reached when the moles of acid and base are equal and the pH is 7.
• In a weak acid-strong base titration, the pH is greater than 7 at the equivalence point.
• In a strong acid-weak base titration, the pH is less than 7 at the equivalence point.
21.20: Indicators
"Boil Them Cabbage Down" is an old bluegrass song. Many people enjoy the music, but chemistry students also enjoy the product of boiled cabbage. Extracting the anthocyanin dye from cabbage leaves with boiling water gives a solution that is red when acidic, purple when neutral, and green to yellow when basic.
Indicators
An acid-base indicator is a substance that displays different colors when in the presence of an acid or a base. How does that work? An indicator is a weak acid that ionizes within a known pH range, usually about 2 pH units. We can represent the protonated form of the indicator molecule as $\ce{HIn}$ and the deprotonated form as $\ce{In^-}$. The following equilibrium exists for the indicator:
$\ce{HIn} \left( aq \right) \rightleftharpoons \ce{H^+} \left( aq \right) + \ce{In^-} \left( aq \right)\nonumber$
According to Le Chatelier's principle, the addition of $\ce{H^+}$ ions (as in a low pH solution) drives the equilibrium to the left and the protonated $\ce{HIn}$ predominates. The addition of $\ce{OH^-}$ (as in a high pH solution) decreases the $\ce{H^+}$ concentration and drives the equilibrium to the right and the deprotonated $\ce{In^-}$ predominates. To be useful as an indicator, the two forms must be different colors. In the case of phenolphthalein, the protonated form is colorless, while the deprotonated form is pink. The figure below shows a variety of acid-base indicators that can be used in titration experiments.
Depending on the pH at the equivalence point, the appropriate indicator must be chosen. For example, bromphenol blue has a yellow color below a pH of about 3 and a blue-violet color above a pH of about 4. Bromphenol blue would not be a good choice as the indicator for a strong acid-strong base titration, because the pH is 7 at the equivalence point. Instead, it could be used for a strong acid-weak base titration, where the pH at the equivalence point is lower.
Most indicators have two colored forms. Universal indicator displays the entire rainbow of colors from low pH to high pH (see figure below). Universal indicator is used to make pH paper, which can be used to quickly test solutions for their approximate pH.
Summary
• An acid-base indicator is a substance that displays different colors when in the presence of an acid or a base.
• Examples of acid-base indicators are given. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/21%3A_Acids_and_Bases/21.19%3A_Titration_Curves.txt |
Baking seems easy with all the pre-mixed items available. ("just add water and stir"). However, there is a good amount of chemistry involved in baking with ingredients that are measured out. One important ingredient is baking powder. The fluffiness in the final product of a non-yeast recipe is usually due to the carbon dioxide formed by baking powder. One popular brand uses a mix of sodium bicarbonate and sodium aluminum sulfate to produce the $\ce{CO_2}$. The reaction is:
$3 \ce{NaHCO_3} + \ce{NaAl(SO_4)_2} \rightarrow \ce{Al(OH)_3} + 2 \ce{Na_2SO_4} + 3 \ce{CO_2}\nonumber$
If all goes well, the biscuits rise, the pancakes are fluffy, and everybody is happy.
Hydrolysis of Salts: Equations
A salt is an ionic compound that is formed when an acid and a base neutralize each other. While it may seem that salt solutions are always neutral, they can frequently be either acidic or basic.
Consider the salt formed when the weak acid hydrofluoric acid is neutralized by the strong base sodium hydroxide. The molecular and net ionic equations are shown below.
\begin{align*} &\ce{HF} \left( aq \right) + \ce{NaOH} \left( aq \right) \rightarrow \ce{NaF} \left( aq \right) + \ce{H_2O} \left( l \right) \ &\ce{HF} \left( aq \right) + \ce{OH^-} \left( aq \right) \rightarrow \ce{F^-} \left( aq \right) + \ce{H_2O} \left( l \right) \end{align*}\nonumber
Since sodium fluoride is soluble, the sodium ion is a spectator ion in the neutralization reaction. The fluoride ion is capable of reacting, to a small extent, with water, accepting a proton.
$\ce{F^-} \left( aq \right) + \ce{H_2O} \left( l \right) \rightleftharpoons \ce{HF} \left( aq \right) + \ce{OH^-} \left( aq \right)\nonumber$
The fluoride ion is acting as a weak Brønsted-Lowry base. The hydroxide ion that is produced as a result of the above reaction makes the solution slightly basic. Salt hydrolysis is a reaction in which one of the ions from a salt reacts with water, forming either an acidic or basic solution.
Salts That Form Basic Solutions
When solid sodium fluoride is dissolved into water, it completely dissociates into sodium ions and fluoride ions. The sodium ions do not have any capability of hydrolyzing, but the fluoride ions hydrolyze to produce a small amount of hydrofluoric acid and hydroxide ion.
$\ce{F^-} \left( aq \right) + \ce{H_2O} \left( l \right) \rightleftharpoons \ce{HF} \left( aq \right) + \ce{OH^-} \left( aq \right)\nonumber$
Salts that are derived from the neutralization of a weak acid $\left( \ce{HF} \right)$ by a strong base $\left( \ce{NaOH} \right)$ will always produce salt solutions that are basic.
Salts That Form Acidic Solutions
Ammonium chloride $\left( \ce{NH_4Cl} \right)$ is a salt that is formed when the strong acid $\ce{HCl}$ is neutralized by the weak base $\ce{NH_3}$. Ammonium chloride is soluble in water. The chloride ion produced is incapable of hydrolyzing because it is the conjugate base of the strong acid $\ce{HCl}$. In other words, the $\ce{Cl^-}$ ion cannot accept a proton from water to form $\ce{HCl}$ and $\ce{OH^-}$, as the fluoride ion did in the previous section. However, the ammonium ion is capable of reacting slightly with water, donating a proton and so acting as an acid.
$\ce{NH_4^+} \left( aq \right) + \ce{H_2O} \left( l \right) \rightleftharpoons \ce{H_3O^+} \left( aq \right) + \ce{NH_3} \left( aq \right)\nonumber$
Salts That Form Neutral Solutions
A salt that is derived from the reaction of a strong acid with a strong base forms a solution that has a pH of 7. An example is sodium chloride, formed from the neutralization of $\ce{HCl}$ by $\ce{NaOH}$. A solution of $\ce{NaCl}$ in water has no acidic or basic properties, since neither ion is capable of hydrolyzing. Other salts that form neutral solutions include potassium nitrate $\left( \ce{KNO_3} \right)$ and lithium bromide $\left( \ce{LiBr} \right)$. The table below summarizes how to determine the acidity or basicity of a salt solution.
Table $1$
Salt formed from: Salt Solution
Strong acid $+$ Strong base Neutral
Strong acid $+$ Weak base Acidic
Weak acid $+$ Strong base Basic
Salts formed from the reaction of a weak acid and a weak base are more difficult to analyze due to competing hydrolysis reactions between the cation and the anion. These salts are not considered in this chapter's concept.
Summary
• Salt hydrolysis is a reaction in which one of the ions from a salt reacts with water, forming either an acidic or basic solution.
• Salts that are derived from the neutralization of a weak acid by a strong base will always produce salt solutions that are basic.
• Salts that are derived from the neutralization of a strong acid by a weak base will always produce salt solutions that are acidic.
• A salt that is derived from the reaction of a strong acid with a strong base forms a solution that has a pH of 7. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/21%3A_Acids_and_Bases/21.21%3A_Hydrolysis_of_Salts_-_Equations.txt |
Many enjoy a cool dip in a swimming pool on a hot day, but may not realize the work needed to keep that water safe and healthy. The ideal pH for a swimming pool is around 7.2. The pH will change as a result of many factors. Adjustment can be accomplished with different chemicals, depending on the tested pH. High pH can be lowered with liquid $\ce{HCl}$ (unsafe material) or sodium bisulfate. The bisulfate anion is a weak acid and can dissociate partially in solution. To increase pH, use sodium carbonate. The carbonate anion forms an equilibrium with protons that results in some formation of carbon dioxide.
Calculating pH of Salt Solutions
It is often helpful to be able to predict the effect a salt solution will have on the pH of a certain solution. Knowledge of the relevant acidity or basicity constants allows us to carry out the necessary calculations.
Example $1$
If we dissolve $\ce{NaF}$ in water, we get the following equilibrium:
$\ce{F^-} \left( aq \right) + \ce{H_2O} \left( l \right) \rightleftharpoons \ce{HF} \left( aq \right) + \ce{OH^-} \left( aq \right)\nonumber$
The pH of the resulting solution can be determined if the $K_\text{b}$ of the fluoride ion is known. $20.0 \: \text{g}$ of sodium fluoride is dissolved in enough water to make $500.0 \: \text{mL}$ of solution. Calculate the pH of the solution. The $K_\text{b}$ of the fluoride ion is $1.4 \times 10^{-11}$.
Known
• Mass $\ce{NaF} = 20.0 \: \text{g}$
• Molar mass $\ce{NaF} = 41.99 \: \text{g/mol}$
• Volume solution $= 0.5000 \: \text{L}$
• $K_\text{b}$ of $\ce{F^-} = 1.4 \times 10^{-11}$
Unknown
The molarity of the $\ce{F^-}$ solution can be calculated from the mass, molar mass, and solution volume. Since $\ce{NaF}$ completely dissociates, the molarity of the $\ce{NaF}$ is equal to the molarity of the $\ce{F^-}$ ion. An ICE table (below) can be used to calculate the concentration of $\ce{OH^-}$ produced and then the pH of the solution.
Step 2: Solve.
\begin{align*} 20.0 \: \cancel{\text{g} \: \ce{NaF}} \times \frac{1 \: \cancel{\text{mol} \: \ce{NaF}}}{41.99 \: \cancel{\text{g} \: \ce{NaF}}} \times \frac{1 \: \text{mol} \: \ce{F^-}}{1 \: \cancel{\text{mol} \: \ce{NaF}}} &= 0.476 \: \text{mol} \: \ce{F^-} \ \frac{0.476 \: \text{mol} \: \ce{F^-}}{0.5000 \: \text{L}} &= 0.953 \: \text{M} \: \ce{F^-} \end{align*}\nonumber
$\text{Hydrolysis equation:} \: \: \: \ce{F^-} \left( aq \right) + \ce{H_2O} \left( l \right) \rightleftharpoons \ce{HF} \left( aq \right) + \ce{OH^-} \left( aq \right)\nonumber$
$\begin{array}{l|ccc} & \ce{F^-} & \ce{HF} & \ce{OH^-} \ \hline \text{Initial} & 0.953 & 0 & 0 \ \text{Change} & -x & +x & +x \ \text{Equilibrium} & 0.953 - x & x & x \end{array}\nonumber$
\begin{align*} K_\text{b} &= 1.4 \times 10^{-11} = \frac{\left( x \right) \left( x \right)}{0.953 - x} = \frac{x^2}{0.953 - x} \approx \frac{x^2}{0.953} \ x &= \left[ \ce{OH^-} \right] = \sqrt{1.4 \times 10^{-11} \left( 0.953 \right)} = 3.65 \times 10^{-6} \: \text{M} \ \text{pOH} &= -\text{log} \left( 3.65 \times 10^{-6} \right) = 5.44 \ \text{pH} &= 14 - 5.44 = 8.56 \end{align*}\nonumber
Step 3: Think about your result.
The solution is slightly basic due to the hydrolysis of the fluoride ion.
Salts That Form Acidic Solutions
When the ammonium ion dissolves in water, the following equilibrium exists:
$\ce{NH_4^+} \left( aq \right) + \ce{H_2O} \left( l \right) \rightleftharpoons \ce{H_3O^+} \left( aq \right) + \ce{NH_3} \left( aq \right)\nonumber$
The production of hydronium ions causes the resulting solution to be acidic. The pH of a solution of ammonium chloride can be found in a very similar way to the sodium fluoride solution in the previous example. However, since the ammonium chloride is acting as an acid, it is necessary to know the $K_\text{a}$ of $\ce{NH_4^+}$, which is $5.6 \times 10^{-10}$. We will find the pH of a $2.00 \: \text{M}$ solution of $\ce{NH_4Cl}$. Because the $\ce{NH_4Cl}$ completely ionizes, the concentration of the ammonium ion is $2.00 \: \text{M}$.
$\ce{NH_4Cl} \left( s \right) \rightarrow \ce{NH_4^+} \left( aq \right) + \ce{Cl^-} \left( aq \right)\nonumber$
Again, an ICE table (below) is set up in order to solve for the concentration of the hydronium (or $\ce{H^+}$) ion produced.
$\begin{array}{l|ccc} & \ce{NH_4^+} & \ce{H^+} & \ce{NH_3} \ \hline \text{Initial} & 2.00 & 0 & 0 \ \text{Change} & -x & +x & +x \ \text{Equilibrium} & 2.00 - x & x & x \end{array}\nonumber$
Now substituting into the $K_\text{a}$ expression gives:
\begin{align*} K_\text{a} &= 5.6 \times 10^{-10} = \frac{x^2}{2.00 - x} \approx \frac{x^2}{2.00} \ x &= \left[ \ce{H^+} \right] = \sqrt{ 5.6 \times 10^{-10} \left( 2.00 \right)} = 3.3 \times 10^{-5} \: \text{M} \ \text{pH} &= -\text{log} \left( 3.3 \times 10^{-5} \right) = 4.48 \end{align*}\nonumber
A salt produced from a strong acid and a weak base yields a solution that is acidic. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/21%3A_Acids_and_Bases/21.22%3A_Calculating_pH_of_Salt_Solutions.txt |
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 it. 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. Commercial buffer solutions, which have a wide variety of pH values, can be obtained.
Some common buffer systems are listed in the table below.
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 (the salt). 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.
$\ce{CH_3COO^-} \left( aq \right) + \ce{H^+} \left( aq \right) \rightarrow \ce{CH_3COOH} \left( aq \right)\nonumber$
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.
$\ce{CH_3COOH} \left( aq \right) + \ce{OH^-} \left( aq \right) \rightarrow \ce{CH_3COO^-} \left( aq \right) + \ce{H_2O} \left( l \right)\nonumber$
Again the ratio of acetate ion to acetic acid changes slightly, this time causing a very small increase in the pH.
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.
Summary
• A buffer is a solution of a weak acid or a base and its salt.
• 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.
• Reactions showing how buffers regulate pH are described. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/21%3A_Acids_and_Bases/21.23%3A_Buffers.txt |
People worry a lot about their smiles. Over the years, teeth do discolor some, so the procedure of teeth bleaching has become more and more popular. Best done in a dentist's office, various chemical preparations containing peroxides are used to whiten teeth. Less effective, but easier to use are "teeth-whitening" toothpastes (also containing peroxides) that promise to give you a brighter smile.
Oxygen in Reactions
Many elements simply combine with oxygen to form the oxide of that element. The heating of magnesium in air allows it to combine with oxygen from the air to form magnesium oxide:
$2 \ce{Mg} \left( s \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{MgO} \left( s \right)\nonumber$
Compounds can also react with oxygen, possibly creating oxides of more than one element. 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 oxidized.
Oxidation is also defined as a loss of hydrogen atoms. In the following equation, ethanol is oxidized to acetaldehyde by the loss of two hydrogen atoms:
$\ce{CH_3CH_2OH} \rightarrow \ce{CH_3CHO} + 2 \ce{H} \: \text{atoms}\nonumber$
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 $\left( \ce{NaClO} \right)$, which releases oxygen that oxidizes stains. Hydrogen peroxide $\left( \ce{H_2O_2} \right)$ releases oxygen as it spontaneously decomposes. It acts as a bleach and an antiseptic that kills bacteria by oxidizing them.
The chemical reaction that is the opposite of oxidation is called reduction. Following from the notion that oxidation was originally thought to mean only the addition of oxygen, reduction was thought to be only the removal of oxygen from a substance. Many naturally occurring metal ores are present as oxides. The pure metals can be extracted by reduction. Iron is obtained from iron (III) oxide by reacting with carbon at high temperatures:
$2 \ce{Fe_2O_3} \left( s \right) + 3 \ce{C} \left( s \right) \rightarrow 4 \ce{Fe} \left( s \right) + 3 \ce{CO_2} \left( g \right)\nonumber$
The removal of oxygen from the $\ce{Fe_2O_3}$ means that it is being reduced to $\ce{Fe}$. Note that an oxidation process is simultaneously occurring. The carbon reactant is being oxidized to $\ce{CO_2}$. This is an important concept. Oxidation and reduction must happen together. Neither can happen alone in a reaction.
Reduction can also be considered as a gain of hydrogen. The reverse of the ethanol $\rightarrow$ acetaldehyde reaction shown above is a reduction reaction:
$\ce{CH_3CHO} + 2 \ce{H} \: \text{atoms} \rightarrow \ce{CH_3CH_2OH}\nonumber$
Summary
• Oxidation is a process in which a substance is reacted with oxygen to produce one or more oxides; it is also defined as a loss of hydrogen atoms.
• Reduction is the removal of oxygen from a substance, or the gain of hydrogen atoms.
• Oxidation and reduction reactions must occur together; neither can happen alone in a reaction.
22.02: Redox Reactions and Ionic Compounds
Nitric acid has many uses in the manufacture of fertilizers and explosives. Most nitric acid is manufactured from ammonia using a three-step process. The ammonia is oxidized to $\ce{HNO_3}$ through the formation of several nitrogen oxides, finally resulting in the acid.
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 mixed with sulfur and heated, the compound zinc sulfide is produced. Two valence electrons from each zinc atom are transferred to each sulfur atom.
Since the zinc is losing electrons in the reaction, it is being oxidized. The sulfur is gaining electrons and is thus being reduced. An oxidation-reduction reaction is a reaction that involves the full or partial transfer of electrons from one reactant to another. Oxidation is the full or partial loss of electrons or the gain of oxygen. Reduction is the full or partial gain of electrons or the loss of oxygen. A redox reaction is another term for an oxidation-reduction reaction.
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*}\nonumber
It is important to remember that the two half-reactions occur simultaneously. The resulting ions that are formed are then attracted to one another in an ionic bond.
Another example of an oxidation-reduction reaction involving electron transfer is the well-known combination of metallic sodium and chlorine gas to form sodium chloride:
$2 \ce{Na} + \ce{Cl_2} \rightarrow 2 \ce{NaCl}\nonumber$
The half reactions are as follows:
\begin{align*} &\text{Oxidation:} \: \ce{Na} \rightarrow \ce{Na^+} + \ce{e^-} \ &\text{Reduction:} \: \ce{Cl} + \ce{e^-} \rightarrow \ce{Cl^-} \end{align*}\nonumber
We will concern ourselves with the balancing of these equations at another time.
Summary
• An oxidation-reduction reaction, or redox reaction, is a reaction that involves the full or partial transfer of electrons from one reactant to another.
• Oxidation is the full or partial loss of electrons or the gain of oxygen.
• Reduction is the full or partial gain of electrons or the loss of oxygen.
• A half-reaction is an equation that shows either the oxidation or the reduction reaction that occurs during a redox reaction.
• Examples of oxidation-reduction reactions are shown. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/22%3A_Oxidation-Reduction_Reactions/22.01%3A_Oxygen_in_Reactions.txt |
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 plants produce the oxygen that animals 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 the movement of electrons. In this series of reactions, water is converted to oxygen gas, and we have something to sustain our lives.
Oxidizing and Reducing Agents
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. The reducing agent is a substance that causes reduction by losing electrons. 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. The example below shows how to analyze a redox reaction.
Example $1$
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{Oxidation:} \: \ce{Cl_2} \left( g \right) + 2 \ce{e^-} \rightarrow 2 \ce{Cl^-} \left( aq \right) \ &\text{Reduction:} \: 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.
Summary
• The oxidizing agent is a substance that causes oxidation by accepting electrons.
• The reducing agent is a substance that causes reduction by losing electrons.
• Examples of oxidizing and reducing agents are shown.
22.04: Molecular Redox Reactions
Acetone is a versatile chemical used both in manufacture of plastics and as a solvent. It is a major constituent of such products as nail polish remover, paints, and cleaning fluids. The manufacture of acetone involves formation of an intermediate peroxide compound by oxidation, followed by formation of the final product.
Molecular Redox Reactions
The electron loss and gain is easy to see in a reaction in which ions are formed. However, in many reactions, no such electron transfer occurs. In a molecular compound, electrons are shared between atoms in a type of bond called a covalent bond. Yet, it is still common for reactions involving molecular compounds to be classified as redox reactions.
When hydrogen gas is reacted with oxygen gas, water is formed as the product.
$2 \ce{H_2} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{H_2O} \left( l \right)\nonumber$
In the individual hydrogen molecules, a pair of bonding electrons is shared equally between the hydrogen atoms (a nonpolar covalent bond). Likewise, the bonding electrons in the oxygen molecule are also shared equally between the two oxygen atoms. However, when the atoms are rearranged to form the water molecule, the electron sharing is no longer equal. In each hydrogen-oxygen bond in the water molecule, the bonding electrons are more attracted to the oxygen atom than they are to the hydrogen atom. We know this because oxygen has a higher electronegativity than hydrogen.
In the course of this reaction, electrons are shifted away from each hydrogen atom and towards the oxygen atom. The hydrogen is oxidized because it undergoes a partial loss of electrons. Even though the loss is not complete enough to form ions, the hydrogen atoms in water have less electron density near them than they did in the $\ce{H_2}$ molecule. The oxygen is reduced because it undergoes a partial gain of electrons. The oxygen atom in water has greater electron density here than it did in the $\ce{O_2}$ molecule.
Another approach to this type of problem is to go back to our earlier definitions of oxidation being the gain of oxygen or loss of hydrogen, and reduction being the gain of hydrogen or loss of oxygen. This makes the decision about redox reactions much easier. The hydrogen is oxidized because it added oxygen to form water. Conversely, the oxygen is reduced because it added hydrogen to form water. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/22%3A_Oxidation-Reduction_Reactions/22.03%3A_Oxidizing_and_Reducing_Agents.txt |
The tractor was once a very useful piece of farm equipment. But tractors got old, likely broke down a lot, and were eventually difficult to find parts for. Farmers bought new, more efficient tractors with all the latest gadgets. There is not much of a market for old tractors, so they often sit parked out of the way, exposed to the weather, and quietly left to rust.
Corrosion
Rust is a combination of several different oxides of iron. The equations below show the steps involved in one of the many processes of rust formation.
\begin{align*} 2 \ce{Fe} \left( s \right) + \ce{O_2} \left( g \right) + 4 \ce{H^+} \left( aq \right) &\rightarrow 2 \ce{Fe^{2+}} \left( aq \right) + 2 \ce{H_2O} \left( l \right) \ 4 \ce{Fe^{2+}} \left( aq \right) + \ce{O_2} \left( g \right) + 6 \ce{H_2O} \left( l \right) &\rightarrow 2 \ce{Fe_2O_3} \cdot \ce{H_2O} \left( s \right) + 8 \ce{H^+} \left( aq \right) \end{align*}\nonumber
Iron is first oxidized to iron (II) ions by oxygen. In the second step, the iron (II) ions are further oxidized and combine with water and oxygen gas to produce a hydrated form of iron (III) oxide known as rust. Rusting is one of many examples of corrosion. Corrosion is the deterioration of metals by redox processes. Corrosion causes enormous amounts of damage to buildings, bridges, ships, cars, and other objects. It has been estimated that corrosion costs the U.S. economy over \$100 billion each year. A great amount of time and effort is spent to try to limit or prevent corrosion.
Corrosion Resistance
Some metals, such as gold and platinum, do not corrode easily because they are very resistant to oxidation by common substances. Some other metals begin to be oxidized, but are further protected from additional corrosion by a coating formed on the surface. Aluminum reacts with oxygen to form aluminum oxide, which remains tightly packed on the surface. The aluminum oxide prevents the interior of the aluminum from corroding. Not all corrosion is a result of reaction with oxygen. Copper corrodes by reaction with carbon dioxide to form copper (II) carbonate. This distinctive green compound is also called patina and prevents the copper underneath from further corrosion (see Statue of Liberty in the figure below).
Conversely, the iron oxides that form during the corrosion of iron form a layer that flakes off easily, rather than remaining tight to the surface. This allows the iron to corrode completely through until the iron object is destroyed.
One way to prevent corrosion is to protect the surface of metal. Covering the surface of a metal object with paint or oil will prevent corrosion by not allowing oxygen to contact it. Unfortunately, scratches in the paint or wearing off of the oil will allow the corrosion to begin. Corrosion-sensitive metals can also be coated with another metal that is resistant to corrosion. A "tin can" is actually made of iron coated with a thin layer of tin, which protects the iron.
Corrosion can also be controlled by connecting the object to be protected to another object made of a metal that is corroded even more easily. When an iron nail is wrapped with a strip of zinc and exposed to water, the zinc (being a more active metal than iron) is oxidized while the iron remains intact. This technique, called cathodic protection, is commonly used to prevent the hulls of steel ships from rusting. Blocks of zinc are attached to the underside of the hull (see figure below). The zinc blocks preferentially corrode, keeping the hull intact. The zinc blocks must be periodically replaced to maintain the protection of the iron.
Summary
• Corrosion is the deterioration of metals by redox processes.
• Examples of corrosion are given. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/22%3A_Oxidation-Reduction_Reactions/22.05%3A_Corrosion.txt |
Moving from studying the element iron to iron compounds, we need to be able to clearly designate the form of the iron ion. An example of this is iron that has been oxidized to form iron oxide during the process of rusting. Although Antoine Lavoisier first began the idea of oxidation as a concept, it was Wendell Latimer (1893-1955) who gave us the modern concept of oxidation numbers. His 1938 book The Oxidation States of the Elements and Their Potentials in Aqueous Solution laid out the concept in detail. Latimer was a well-known chemist who later became a member of the National Academy of Sciences. Not bad for a gentleman who started college planning on being a lawyer.
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 determine oxidation numbers:
1. For free elements (uncombined state), each atom has an oxidation number of zero. $\ce{H_2}$, $\ce{Br_2}$, $\ce{Na}$, $\ce{Be}$, $\ce{K}$, $\ce{O_2}$, $\ce{P_4}$, all have an oxidation number of 0.
2. Monatomic ions have oxidation numbers equal to their charge. $\ce{Li^+} = +1$, $\ce{Ba^{2+}} = +2$, $\ce{Fe^{3+}} = +3$, $\ce{I^-} = -1$, $\ce{O^{2-}} = -2$, etc. Alkali metal oxidation numbers $= +1$. Alkaline earth oxidation numbers $= +2$. Aluminum $= +3$ in all of its compounds. Oxygen's oxidation number $= -2$ except when in hydrogen peroxide $\left( \ce{H_2O_2} \right)$, or a peroxide ion $\left( \ce{O_2^{2-}} \right)$ where it is $-1$.
3. Hydrogen's oxidation number is $+1$, except for when bonded to metals as the hydride ion forming binary compounds. In $\ce{LiH}$, $\ce{NaH}$, and $\ce{CaH_2}$, the oxidation number is $-1$.
4. Fluorine has an oxidation number of $-1$ in all of its compounds.
5. Halogens ($\ce{Cl}$, $\ce{Br}$, $\ce{I}$) have negative oxidation numbers when they form halide compounds. When combined with oxygen, they have positive numbers. In the chlorate ion $\left( \ce{ClO_3^-} \right)$, the oxidation number of $\ce{Cl}$ is $+5$, and the oxidation number of $\ce{O}$ is $-2$.
6. In a neutral atom or molecule, the sum of the oxidation numbers must be 0. In a polyatomic ion, the sum of the oxidation numbers of all the atoms in the ion must be equal to the charge on the ion.
Example $1$
What is the oxidation number for manganese in the compound potassium permanganate $\left( \ce{KMnO_4} \right)$?
Solution
The oxidation number for $\ce{K}$ is $+1$ (rule 2).
The oxidation number for $\ce{O}$ is $-2$ (rule 2).
Since this is a compound (there is no charge indicated on the molecule), the net charge on the molecule is zero (rule 6).
So we have:
\begin{align*} +1 + \ce{Mn} + 4 \left( -2 \right) &= 0 \ \ce{Mn} - 7 &= 0 \ \ce{Mn} &= +7 \end{align*}\nonumber
When dealing with oxidation numbers, we must always include the charge on the atom.
Another way to determine the oxidation number of $\ce{Mn}$ in this compound is to recall that the permanganate anion $\left( \ce{MnO_4^-} \right)$ has a charge of $-1$. In this case:
\begin{align*} \ce{Mn} + 4 \left( -2 \right) &= -1 \ \ce{Mn} - 8 &= -1 \ \ce{Mn} &= +7 \end{align*}\nonumber
Example $2$
What is the oxidation number for iron in $\ce{Fe_2O_3}$?
Solution
\begin{align*} &\ce{O} \: \text{is} \: -2 \: \left( \text{rule 2} \right) \ &2 \ce{Fe} + 3 \left( -2 \right) = 0 \ &2 \ce{Fe} = 6 \ &\ce{Fe} = 3 \end{align*}\nonumber
If we have the compound $\ce{FeO}$, then $\ce{Fe} + \left( -2 \right) = 0$ and $\ce{Fe} = 2$. Iron is one of those materials that can have more than one oxidation number.
The halogens (except for fluorine) can also have more than one number. In the compound $\ce{NaCl}$, we know that $\ce{Na}$ is $+1$, so $\ce{Cl}$ must be $-1$. But what about $\ce{Cl}$ in $\ce{NaClO_3}$?
\begin{align*} \ce{Na} &= 1 \ \ce{O} &= -2 \ 1 + \ce{Cl} + 3 \left( -2 \right) &= 0 \ 1 + \ce{Cl} - 6 &= 0 \ \ce{Cl} - 5 &= 0 \ \ce{Cl} &= +5 \end{align*}\nonumber
Not quite what we expected, but $\ce{Cl}$, $\ce{Br}$, and $\ce{I}$ will exhibit multiple oxidation numbers in compounds.
Summary
• 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.
• Six rules for determining oxidation numbers are listed.
• Examples of oxidation number determinations are provided. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/22%3A_Oxidation-Reduction_Reactions/22.06%3A_Assigning_Oxidation_Numbers.txt |
Zinc is an important component of many kinds of batteries. This metal is mined as zinc compounds, one of which is zinc carbonate. To obtain the pure metal, the ore must go through the following chemical processes:
1. High temperatures and hot air blasts are used to roast the ore.
$\ce{ZnCO_3} \left( s \right) + \text{heat} \rightarrow \ce{ZnO} \left( s \right) + \ce{CO_2} \left( g \right)\nonumber$
2. Then the $\ce{ZnO}$ is treated with carbon.
\begin{align*} \ce{ZnO} \left( s \right) + \ce{C} \left( s \right) + \text{heat} &\rightarrow \ce{Zn} \left( g \right) + \ce{CO} \left( g \right) \ \ce{ZnO} \left( s \right) + \ce{CO} \left( g \right) + \text{heat} &\rightarrow \ce{Zn} \left( g \right) + \ce{CO_2} \left( g \right) \end{align*}\nonumber
The result is the pure metal, which can then be fabricated into a variety of products.
Changes in Oxidation Number in Redox Reactions
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 $\left( \ce{Cu^0} \right)$.
A loss of negatively-charged electrons corresponds to an increase in oxidation number, while a gain of electrons corresponds to a decrease in oxidation number. Therefore, the element or ion that is oxidized undergoes an increase in oxidation number. The element or ion that is reduced undergoes a decrease in oxidation number. The table below summarizes the processes of oxidation and reduction.
Processes of Oxidation and Reduction
Table $1$: Processes of Oxidation and Reduction
Oxidation Reduction
Complete loss of electrons (ionic reaction). Complete gain of electrons (ionic reaction).
Gain of oxygen. Loss of oxygen.
Loss of hydrogen in a molecular compound. Gain of hydrogen in a molecular compound.
Increase in oxidation number. Decrease in oxidation number.
Example $1$
Use changes in oxidation number to determine which atoms are oxidized and which atoms are reduced in the following reaction. Identify the oxidizing and reducing agent.
$\ce{Fe_2O_3} \left( s \right) + 3 \ce{CO} \left( g \right) \rightarrow 2 \ce{Fe} \left( s \right) + 3 \ce{CO_2} \left( g \right)\nonumber$
Step 1: Plan the problem.
Use the oxidation number rules to assign oxidation numbers to each atom in the balanced equation. Coefficients do not affect oxidation numbers. The oxidized atom increases in oxidation number and the reduced atom decreases in oxidation number.
Step 2: Solve.
$\overset{+3}{\ce{Fe_2}} \overset{-2}{\ce{O_3}} \left( s \right) + 3 \overset{+2}{\ce{C}} \overset{-2}{\ce{O}} \left( g \right) \rightarrow 2 \overset{0}{\ce{Fe}} \left( s \right) + 3 \overset{+4}{\ce{C}} \overset{-2}{\ce{O_2}} \left( g \right)\nonumber$
The element carbon is oxidized because its oxidation number increases from $+2$ to $+4$. The iron (III) ion within the $\ce{Fe_2O_3}$ is reduced because its oxidation number decreases from $+3$ to $0$. The carbon monoxide $\left( \ce{CO} \right)$ is the reducing agent since it contains the element that is oxidized. The $\ce{Fe^{3+}}$ ion is the oxidizing agent since it is reduced in the reaction. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/22%3A_Oxidation-Reduction_Reactions/22.07%3A_Changes_in_Oxidation_Number_in_Redox_Reactions.txt |
The reaction of copper wire with nitric acid produces a colorful mix of products that include copper (II) nitrate, nitrogen dioxide, and water. Copper salts are blue in solution, reflecting the rather unique arrangements of electrons in the $d$ orbital as the copper ionizes from metallic copper.
Identifying Reaction Types
A redox reaction must involve a change in oxidation number for two of the elements involved in the reaction. The oxidized element increases in oxidation number, while the reduced element decreases in oxidation number.
Single-replacement reactions are redox reactions because two different elements appear as free elements (oxidation number of zero) on one side of the equation and as part of a compound on the other side. Therefore, oxidation numbers must change.
$\ce{Zn} + 2 \ce{HCl} \rightarrow \ce{ZnCl_2} + \ce{H_2}\nonumber$
$\ce{Zn}$ is oxidized from $\ce{Zn^0}$ to $\ce{Zn^{2+}}$ and the $\ce{H}$ is reduced from $\ce{H^+}$ to $\ce{H^0}$.
Combustion reactions are redox reactions because elemental oxygen $\left( \ce{O_2} \right)$ acts as the oxidizing agent and is itself reduced.
$\ce{CH_4} + 2 \ce{O_2} \rightarrow \ce{CO_2} + 2 \ce{H_2O}\nonumber$
Most combination and decomposition reactions are redox reactions, since elements are usually transformed into compounds and vice-versa. The thermite reaction involves ferric oxide and metallic aluminum:
$\ce{Fe_2O_3} + 2 \ce{Al} \rightarrow \ce{Al_2O_3} + 2 \ce{Fe}\nonumber$
We see that the iron is reduced and the aluminum oxidized during the course of the reaction.
So what types of reactions are not redox reactions? Double-replacement reactions, such as the one below, are not redox reactions because ions are simply recombined without any transfer of electrons.
$\overset{+1}{\ce{Na_2}} \overset{+6}{\ce{S}} \overset{-2}{\ce{O_4}} \left( aq \right) + \overset{+2}{\ce{Ba}} ( \overset{+5}{\ce{N}} \overset{-2}{\ce{O_3}} ) \left( aq \right) \rightarrow 2 \overset{+1}{\ce{Na}} \overset{+5}{\ce{N}} \overset{-2}{\ce{O_3}} \left( aq \right) + \overset{+2}{\ce{Ba}} \overset{+6}{\ce{S}} \overset{-2}{\ce{O_4}} \left( s \right)\nonumber$
Note that the oxidation numbers for each element remain unchanged in the reaction.
Acid-base reactions involve a transfer of a hydrogen ion instead of an electron. Acid-base reactions, like the one below, are also not redox reactions.
$\overset{+1}{\ce{H}} \overset{-1}{\ce{F}} \left( aq \right) + \overset{-3}{\ce{N}} \overset{+1}{\ce{H_3}} \left( aq \right) \rightarrow \overset{-3}{\ce{N}} \overset{+1}{\ce{H_4^+}} \left( aq \right) + \overset{-1}{\ce{F^-}}\nonumber$
Again, the transfer of an $\ce{H^+}$ ion leaves the oxidation numbers unaffected. In summary, redox reactions can always be recognized by a change in oxidation number of two of the atoms in the reaction. Any reaction in which no oxidation numbers change is not a redox reaction.
Summary
• A redox reaction must involve a change in oxidation number for two of the elements involved in the reaction.
• The oxidized element increases in oxidation number, while the reduced element decreases in oxidation number.
• Single-replacement reactions and combustion reactions are redox reactions.
• Most combination and decomposition reactions are redox reactions.
• Double-replacement reactions and acid-base reactions are not redox reactions.
22.09: Balancing Redox Reactions- Oxidation Number Change Method
Sulfuric acid is produced in extremely large quantities in the United States (about 40 million tons/year). This material is manufactured by oxidizing sulfur to form sulfur trioxide. The $\ce{SO_3}$ is then dissolved in water to form the sulfuric acid. Most of the sulfuric acid produced is used in fertilizers. This acid is also found in lead-acid car batteries.
Balancing Redox Reactions: Oxidation-Number-Change Method
One way to balance redox reactions is by keeping track electron transfer, by using the oxidation numbers of each of the atoms. For the oxidation-number-change method, start with the unbalanced skeleton equation. The example below is for the reaction of iron (II) oxide with carbon monoxide. This reaction is one that takes place in a blast furnace during the processing of iron ore into metallic iron:
$\ce{Fe_2O_3} \left( s \right) + \ce{CO} \left( g \right) \rightarrow \ce{Fe} \left( s \right) + \ce{CO_2} \left( g \right)\nonumber$
$\overset{+3}{\ce{Fe_2}} \overset{-2}{\ce{O_3}} \left( s \right) + \overset{+2}{\ce{C}} \overset{-2}{\ce{O}} \left( g \right) \rightarrow \overset{0}{\ce{Fe}} \left( s \right) + \overset{+4}{\ce{C}} \overset{-2}{\ce{O_2}} \left( g \right)\nonumber$
The carbon atom's oxidation number increases by 2, while the iron atom's oxidation number decreases by 3. As written, the number of electrons lost does not equal the number of electrons gained. In a balanced redox equation, these must be equal. So, the increase in oxidation number of one atom must be made equal to the decrease in oxidation number of the other.
Step 4: Use coefficients to make the total increase in oxidation number equal to the total decrease in oxidation number. In this case, the least common multiple of 2 and 3 is 6. So the oxidation-number increase should be multiplied by 3, while the oxidation-number decrease should be multiplied by 2. The coefficient is also applied to the formulas in the equation. So a 3 is placed in front of the $\ce{CO}$ and in front of the $\ce{CO_2}$. A 2 is placed in front of the $\ce{Fe}$ on the right side of the equation. The $\ce{Fe_2O_3}$ does not require a coefficient because the subscript of 2 after the $\ce{Fe}$ indicates that there are already two iron atoms.
Step 5: Check the balancing for both atoms and charge. Occasionally, a coefficient may need to be placed in front of a molecular formula that was not involved in the redox process. In the current example, the equation is now balanced.
$\ce{Fe_2O_3} \left( s \right) + 3 \ce{CO} \left( g \right) \rightarrow 2 \ce{Fe} \left( s \right) + 3 \ce{CO_2} \left( g \right)\nonumber$ | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/22%3A_Oxidation-Reduction_Reactions/22.08%3A_Identifying_Reaction_Types.txt |
The picture below shows one of the two Thunder Dolphin amusement ride trains. This train has an orange stripe, while its companion has a yellow stripe. Pigments of these colors are often made with a dichromate salt (usually sodium or potassium dichromate). These brightly colored compounds serve as strong oxidizing agents in chemical reactions.
Balancing Redox Equations: Half-Reaction Method
Another method for balancing redox reactions uses half-reactions. Recall that a half-reaction is either the oxidation or reduction that occurs, treated separately. The half-reaction method works better than the oxidation-number method when the substances in the reaction are in aqueous solution. The aqueous solution is typically either acidic or basic, so hydrogen ions or hydroxide ions are present.
In general, the half-reactions are first balanced by atoms separately. Electrons are included in the half-reactions. These are then balanced so that the number of electrons lost is equal to the number of electrons gained. Finally, the two half-reactions are added back together. The example is the oxidation of $\ce{Fe^{2+}}$ ions to $\ce{Fe^{3+}}$ ions by dichromate $\left( \ce{Cr_2O_7^{2-}} \right)$ in acidic solution. The dichromate ions are reduced to $\ce{Cr^{3-}}$ ions.
Step 1: Write the unbalanced ionic equation.
$\ce{Fe^{2+}} \left( aq \right) + \ce{Cr_2O_7^{2-}} \left( aq \right) \rightarrow \ce{Fe^{3+}} \left( aq \right) + \ce{Cr^{3+}} \left( aq \right)\nonumber$
Notice that the equation is far from balanced, as there are no oxygen atoms on the right side. This will be resolved by the balancing method.
Step 2: Write separate half-reactions for the oxidation and the reduction processes. Determine the oxidation numbers first, if necessary.
\begin{align*} &\text{Oxidation:} \: \ce{Fe^{2+}} \left( aq \right) \rightarrow \ce{Fe^{3+}} \left( aq \right) \ &\text{Reduction:} \: \overset{+6}{\ce{Cr_2}} \ce{O_7^{2-}} \left( aq \right) \rightarrow \ce{Cr^{3+}} \left( aq \right) \end{align*}\nonumber
Step 3: Balance the atoms in the half-reactions other than hydrogen and oxygen. In the oxidation half-reaction above, the iron atoms are already balanced. The reduction half-reaction needs to be balanced with the chromium atoms.
$\ce{Cr_2O_7^{2-}} \left( aq \right) \rightarrow 2 \ce{Cr^{3+}} \left( aq \right)\nonumber$
Step 4: Balance oxygen atoms by adding water molecules to the appropriate side of the equation. For the reduction half-reaction above, seven $\ce{H_2O}$ molecules will be added to the product side.
$\ce{Cr_2O_7^{2-}} \left( aq \right) \rightarrow 2 \ce{Cr^{3+}} \left( aq \right) + 7 \ce{H_2O} \left( l \right)\nonumber$
Now the hydrogen atoms need to be balanced. In an acidic medium, add hydrogen ions to balance. In this example, fourteen $\ce{H^+}$ ions will be added to the reactant side.
$14 \ce{H^+} \left( aq \right) + \ce{Cr_2O_7^{2-}} \left( aq \right) \rightarrow 2 \ce{Cr^{3+}} \left( aq \right) + 7 \ce{H_2O} \left( l \right)\nonumber$
Step 5: Balance the charges by adding electrons to each half-reaction. For the oxidation half-reaction, the electrons will need to be added to the product side. For the reduction half-reaction, the electrons will be added to the reactant side. By adding one electron to the product side of the oxidation half-reaction, there is a $2+$ total charge on both sides.
$\ce{Fe^{2+}} \left( aq \right) \rightarrow \ce{Fe^{3+}} \left( aq \right) + \ce{e^-}\nonumber$
There is a total charge of $12+$ on the reactant side of the reduction half-reaction $\left( 14 - 2 \right)$. The product side has a total charge of $6+$ due to the two chromium ions $\left( 2 \times 3 \right)$. To balance the charge, six electrons need to be added to the reactant side.
$6 \ce{e^-} + 14 \ce{H^+} \left( aq \right) + \ce{Cr_2O_7^{2-}} \left( aq \right) \rightarrow 2 \ce{Cr^{3+}} \left( aq \right) + 7 \ce{H_2O} \left( l \right)\nonumber$
Now equalize the electrons by multiplying everything in one or both equations by a coefficient. In this example, the oxidation half-reaction will be multiplied by six.
$6 \ce{Fe^{2+}} \left( aq \right) \rightarrow 6 \ce{Fe^{3+}} \left( aq \right) + 6 \ce{e^-}\nonumber$
Step 6: Add the two half-reactions together. The electrons must cancel. Balance any remaining substances by inspection. If necessary, cancel out $\ce{H_2O}$ or $\ce{H^+}$ that appear on both sides.
\begin{align*} 6 \ce{Fe^{2+}} \left( aq \right) &\rightarrow 6 \ce{Fe^{3+}} \left( aq \right) + \cancel{ 6 \ce{e^-}} \ \cancel{6 \ce{e^-}} + 14 \ce{H^+} \left( aq \right) + \ce{Cr_2O_7^{2-}} \left( aq \right) &\rightarrow 2 \ce{Cr^{3+}} \left( aq \right) + 7 \ce{H_2O} \left( l \right) \ \hline 14 \ce{H^+} \left( aq \right) + 6 \ce{Fe^{2+}} \left( aq \right) + \ce{Cr_2O_7^{2-}} \left( aq \right) &\rightarrow 6 \ce{Fe^{3+}} \left( aq \right) + 2 \ce{Cr^{3+}} \left( aq \right) + 7 \ce{H_2O} \left( l \right) \end{align*}\nonumber
Step 7: Check the balancing. In the above equation, there are $14 \: \ce{H}$, $6 \: \ce{Fe}$, $2 \: \ce{Cr}$, and $7 \: \ce{O}$ on both sides. The net charge is $24+$ on both sides. The equation is balanced.
22.11: Half-Reaction Method in Basic Solution
Cyanide is a very toxic material. Generated mainly by industrial manufacturing processes, this anion can cause neurological effects and damage to sensitive tissues such as the thyroid gland.
Treatment with chlorine gas in basic solution effectively destroys any cyanide present by converting it to harmless nitrogen gas. The reaction is as follows:
$2 \ce{NaCN} + 5 \ce{Cl_2} + 12 \ce{NaOH} \rightarrow \ce{N_2} + 2 \ce{Na_2CO_3} + 10 \ce{NaCl} + 6 \ce{H_2O}\nonumber$
Half-Reaction Method in Basic Solution
For reactions that occur in basic solution rather than acidic solution, the steps to balance the reaction are primarily the same. However, after finishing step 6, add an equal number of $\ce{OH^-}$ ions to both sides of the equation. Combine the $\ce{H^+}$ and $\ce{OH^-}$ to make $\ce{H_2O}$ and cancel out any water molecules that appear on both sides. Using the example of the oxidation of $\ce{Fe^{2+}}$ ions by dichromate $\left( \ce{Cr_2O_7^{2-}} \right)$, we would get the following three steps:
1. Add the hydroxide ions:
$14 \ce{OH^-} \left( aq \right) + 14 \ce{H^+} \left( aq \right) + 6 \ce{Fe^{2+}} \left( aq \right) + \ce{Cr_2O_7^{2-}} \left( aq \right) \rightarrow 6 \ce{Fe^{3+}} \left( aq \right) + 2 \ce{Cr^{3+}} \left( aq \right) + 7 \ce{H_2O} \left( l \right) + 14 \ce{OH^-} \left( aq \right)\nonumber$
2. Combine the hydrogen ions and hydroxide ions to make water:
$14 \ce{H_2O} \left( l \right) + 6 \ce{Fe{2+}} \left( aq \right) + \ce{Cr_2O_7^{2-}} \left( aq \right) \rightarrow 6 \ce{Fe^{3+}} \left( aq \right) + 2 \ce{Cr^{3+}} \left( aq \right) + 7 \ce{H_2O} \left( l \right) + 14 \ce{OH^-} \left( aq \right)\nonumber$
3. Cancel out seven water molecules from both sides to get the final equation:
$7 \ce{H_2O} \left( l \right) + 6 \ce{Fe^{2+}} \left( aq \right) + \ce{Cr_2O_7^{2-}} \left( aq \right) \rightarrow 6 \ce{Fe^{3+}} \left( aq \right) + 2 \ce{Cr^{3+}} \left( aq \right) + 14 \ce{OH^-} \left( aq \right)\nonumber$
The equation is still balanced by atoms and by charge, but the presence of hydroxide ions rather than hydrogen ions means that the reaction takes place in basic solution. Typically, most redox reactions will actually only proceed in one type of solution or the other. The oxidation of $\ce{Fe^{2+}}$ by $\ce{Cr_2O_7^{2-}}$ does not occur in basic solution, and was only balanced this way to demonstrate the method.
In summary, the choice of which balancing method to use depends on the kind of reaction. The oxidation-number method works best if the oxidized and reduced species appear only once on each side of the equation, and if no acids or bases are present. The half-reaction method is more versatile and works well for reactions involving ions in aqueous solution. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/22%3A_Oxidation-Reduction_Reactions/22.10%3A_Balancing_Redox_Reactions-_Half-Reaction_Method.txt |
Gold and silver are widely used metals for making jewelry. One of the reasons these metals are employed for this purpose is that they are very unreactive. They do not react in contact with most other metals, so they are more likely to stay in tact under challenging conditions. No one wants their favorite piece of jewelry to fall apart!
Direct Redox Reactions
When a strip of zinc metal is placed into a blue solution of copper (II) sulfate (figure below), a reaction immediately begins as the zinc strip begins to darken. If left in the solution for a longer period of time, the zinc will gradually decay due to oxidation to zinc ions. At the same time, the copper (II) ions from the solution are reduced to copper metal (see second figure below), which causes the blue copper (II) sulfate solution to become colorless.
The process that occurs in this redox reaction is shown below as two separate half-reactions, which can then be combined into the full redox reaction.
$\begin{array}{ll} \text{Oxidation:} & \ce{Zn} \left( s \right) \rightarrow \ce{Zn^{2+}} \left( aq \right) + 2 \ce{e^-} \ \text{Reduction:} & \ce{Cu^{2+}} \left( aq \right) + 2 \ce{e^-} \rightarrow \ce{Cu} \left( s \right) \ \hline \text{Full Reaction:} & \ce{Zn} \left( s \right) + \ce{Cu^{2+}} \left( aq \right) \rightarrow \ce{Zn^{2+}} \left( aq \right) + \ce{Cu} \left( s \right) \end{array}\nonumber$
Why does this reaction occur spontaneously? The activity series is a list of elements in descending order of reactivity. An element that is higher in the activity series is capable of displacing an element that is lower on the series in a single-replacement reaction. This series also lists elements in order of ease of oxidation. The elements at the top are the easiest to oxidize, while those at the bottom are the most difficult to oxidize. The table below shows the activity series together with each element's oxidation half-reaction.
Activity Series of Metals
Table $1$: Activity Series of Metals
Element Oxidation Half-Reaction
Most active or most easily oxidized. Lithium $\ce{Li} \left( s \right) \rightarrow \ce{Li^+} \left( aq \right) + \ce{e^-}$
Potassium $\ce{K} \left( s \right) \rightarrow \ce{K^+} \left( aq \right) + \ce{e^-}$
Barium $\ce{Ba} \left( s \right) \rightarrow \ce{Ba^{2+}} \left( aq \right) + 2 \ce{e^-}$
Calcium $\ce{Ca} \left( s \right) \rightarrow \ce{Ca^{2+}} \left( aq \right) + 2 \ce{e^-}$
Sodium $\ce{Na} \left( s \right) \rightarrow \ce{Na^+} \left( aq \right) + \ce{e^-}$
Magnesium $\ce{Mg} \left( s \right) \rightarrow \ce{Mg^{2+}} \left( aq \right) + 2 \ce{e^-}$
Aluminum $\ce{Al} \left( s \right) \rightarrow \ce{Al^{3+}} \left( aq \right) + 3 \ce{e^-}$
Zinc $\ce{Zn} \left( s \right) \rightarrow \ce{Zn^{2+}} \left( aq \right) + 2 \ce{e^-}$
Iron $\ce{Fe} \left( s \right) \rightarrow \ce{Fe^{2+}} \left( aq \right) + 2 \ce{e^-}$
Nickel $\ce{Ni} \left( s \right) \rightarrow \ce{Ni^{2+}} \left( aq \right) + 2 \ce{e^-}$
Tin $\ce{Sn} \left( s \right) \rightarrow \ce{Sn^{2+}} \left( aq \right) + 2 \ce{e^-}$
Lead $\ce{Pb} \left( s \right) \rightarrow \ce{Pb^{2+}} \left( aq \right) + 2 \ce{e^-}$
Hydrogen $\ce{H_2} \left( g \right) \rightarrow 2 \ce{H^+} \left( aq \right) + 2 \ce{e^-}$
Copper $\ce{Cu} \left( s \right) \rightarrow \ce{Cu^{2+}} \left( aq \right) + 2 \ce{e^-}$
Mercury $\ce{Hg} \left( l \right) \rightarrow \ce{Hg^{2+}} \left( aq \right) + 2 \ce{e^-}$
Silver $\ce{Ag} \left( s \right) \rightarrow \ce{Ag^+} \left( aq \right) + \ce{e^-}$
Platinum $\ce{Pt} \left( s \right) \rightarrow \ce{Pt^{2+}} \left( aq \right) + 2 \ce{e^-}$
Least active or most difficult to oxidize. Gold $\ce{Au} \left( s \right) \rightarrow \ce{Au^{3+}} \left( aq \right) + 3 \ce{e^-}$
Notice that zinc is listed above copper on the activity series. which means that zinc is more easily oxidized than copper. That is why copper (II) ions can act as an oxidizing agent when put into contact with zinc metal. Ions of any metal that is below zinc, such as lead or silver, would oxidize the zinc in a similar reaction. These types of reactions are called direct redox reactions because the electrons flow directly from the atoms of one metal to the cations of the other metal. However, no reaction will occur if a strip of copper metal is placed into a solution of zinc ions, because the zinc ions are not able to oxidize the copper. In other words, such a reaction is nonspontaneous.
Summary
• The activity series of metals is given.
• Parameters for spontaneous reactions between metals are described. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/23%3A_Electrochemistry/23.01%3A_Direct_Redox_Reactions.txt |
Metal sculptures 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 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.
The reaction of zinc metal with copper (II) ions is called a direct redox process or reaction. The electrons that are transferred in the reaction go directly from the \(\ce{Zn}\) atoms on the surface of the strip to the \(\ce{Cu^{2+}}\) ions in the area of the solution right next to the zinc strip. Electricity, on the other hand, requires the passage of electrons through a conducting medium, such as a wire, in order to do work. This work could be lighting a light bulb or powering a refrigerator or heating a house. When the redox reaction is direct, those electrons cannot be made to do work. Instead, we must separate the oxidation process from the reduction process and force the electrons to move from one place to another in between. That is the key to the structure of the electrochemical cell. 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.
Summary
• Electrochemistry is a branch of chemistry that deals with the interconversion of chemical energy and electrical energy.
• An electrochemical cell is any device that converts chemical energy into electrical energy, or electrical energy into chemical energy.
• An electrochemical reaction requires:
• A solution where redox reactions can occur.
• A conductor for electron transfer.
• A salt bridge for ions to move through.
23.03: Voltaic Cells
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.
Voltaic Cells
A voltaic cell is an electrochemical cell that uses a spontaneous redox reaction to produce electrical energy.
The voltaic cell (see figure above) consists of two separate compartments. A half-cell is one part of a voltaic cell in which either the oxidation or reduction half-reaction takes place. The left half-cell is a strip of zinc metal in a solution of zinc sulfate. The right half-cell is a strip of copper metal in a solution of copper (II) sulfate. The strips of metal are called electrodes. An electrode is a conductor in a circuit that is used to carry electrons to a nonmetallic part of the circuit. The nonmetallic part of the circuit is the electrolyte solution in which the electrodes are placed. A metal wire connects the two electrodes. A switch opens or closes the circuit. A porous membrane is placed between the two half-cells to complete the circuit.
The various electrochemical processes that occur in a voltaic cell occur simultaneously. It is easiest to describe them in the following steps, using the above zinc-copper cell as an example.
1. Zinc atoms from the zinc electrode are oxidized to zinc ions. This happens because zinc is higher than copper on the activity series, and so is more easily oxidized.
$\ce{Zn} \left( s \right) \rightarrow \ce{Zn^{2+}} \left( aq \right) + 2 \ce{e^-}\nonumber$
The electrode at which oxidation occurs is called the anode. The zinc anode gradually diminishes as the cell operates, due to the loss of zinc metal. The zinc ion concentration in the half-cell increases. Because of the production of electrons at the anode, it is labeled as the negative electrode.
2. The electrons that are generated at the zinc anode travel through the external wire and register a reading on the voltmeter. They continue to the copper electrode.
3. Electrons enter the copper electrode where they combine with the copper (II) ions in the solution, reducing them to copper metal.
$\ce{Cu^{2+}} \left( aq \right) + 2 \ce{e^-} \rightarrow \ce{Cu} \left( s \right)\nonumber$
The electrode at which reduction occurs is called the cathode. The cathode gradually increases in mass because of the production of copper metal. The concentration of copper (II) ions in the half-cell solution decreases. The cathode is the positive electrode.
4. Ions move through the membrane to maintain electrical neutrality in the cell. In the cell illustrated above, sulfate ions will move from the copper side to the zinc side to compensate for the decrease in $\ce{Cu^{2+}}$ and the increase in $\ce{Zn^{2+}}$.
The two half-reactions can again be summed to provide the overall redox reaction occurring in the voltaic cell:
$\ce{Zn} \left( s \right) + \ce{Cu^{2+}} \left( aq \right) \rightarrow \ce{Zn^{2+}} \left( aq \right) + \ce{Cu} \left( s \right)\nonumber$
Summary
• A voltaic cell is an electrochemical cell that uses a spontaneous redox reaction to produce electrical energy.
• The structure of a voltaic cell includes two half-cells; each being one part of a voltaic cell in which either the oxidation or reduction half-reaction takes place, and each containing an electrode.
• The electrode at which oxidation occurs is called the anode.
• The electrode at which reduction occurs is called the cathode.
• The reactions producing electron flow are given. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/23%3A_Electrochemistry/23.02%3A_Electrochemical_Reaction.txt |
The voltmeter doesn't measure volts directly; it measures electric current flow. However, current flow and voltage can be directly related to one another. The first meters were called galvanometers, and they used basic laws of electricity to determine voltage. They were heavy and hard to work with, but got the job done. The first multimeters were developed in the 1920s, but true portability had to wait until printed circuits and transistors replaced the cumbersome wires and vacuum tubes.
Electrical Potential
Electrical potential is a measurement of the ability of a voltaic cell to produce an electric current. Electrical potential is typically measured in volts $\left( \text{V} \right)$. The voltage that is produced by a given voltaic cell is the electrical potential difference between the two half-cells. It is not possible to measure the electrical potential of an isolated half-cell. For example, if only a zinc half-cell were constructed, no complete redox reaction could occur, and so no electrical potential could be measured. It is only when another half-cell is combined with the zinc half-cell that an electrical potential difference—or voltage—can be measured.
The electrical potential of a cell results from a competition for electrons. In a zinc-copper voltaic cell, it is the copper (II) ions that will be reduced to copper metal. That is because the $\ce{Cu^{2+}}$ ions have a greater attraction for electrons than the $\ce{Zn^{2+}}$ ions in the other half-cell. Instead, the zinc metal is oxidized. The reduction potential is a measure of the tendency of a given half-reaction to occur as a reduction in an electrochemical cell. In a given voltaic cell, the half-cell that has the greater reduction potential is the one in which reduction will occur. In the half-cell with the lower reduction potential, oxidation will occur. The cell potential $\left( E_\text{cell} \right)$ is the difference in reduction potential between the two half-cells in an electrochemical cell.
Summary
• Electrical potential is a measurement of the ability of a voltaic cell to produce an electric current.
• The reduction potential is a measure of the tendency of a given half-reaction to occur as a reduction in an electrochemical cell.
• The cell potential $\left( E_\text{cell} \right)$ is the difference in reduction potential between the two half-cells in an electrochemical cell.
23.05: Standard Hydrogen Electrode
It's a human tendency to compare ourselves to someone else. Can I run faster than you? Am I taller than my dad?—these are relative comparisons that don't provide much useful data. When we use a standard for our comparisons, one thing is much more comparable to another. One meter is the same distance everywhere in the world, so a 100 meter track in one country is exactly the same distance as a 100 meter track in another country—this is a universal basis for the comparison of a running time, for example.
Standard Hydrogen Electrode
The activity series allows us to predict the relative reactivity of different materials when used in oxidation-reduction processes. We also know we can create electric current by a combination of chemical processes. But how do we predict the expected amount of current that will flow through the system? We measure this flow as voltage (an electromotive force or potential difference).
In order to do this, we need some way of comparing the extent of electron flow in various chemical systems. The best way to do this is to have a baseline that we use—a standard that everything can be measured against. For determination of half-reaction current flows and voltages, we use the standard hydrogen electrode. The figure below illustrates this electrode. A platinum wire conducts the electricity through the circuit. The wire is immersed in a $1.0 \: \text{M}$ strong acid solution and $\ce{H_2}$ gas is bubbled in at a pressure of one atmosphere and a temperature of $25^\text{o} \text{C}$. The half-reaction at this electrode is $\ce{H_2} \rightarrow 2 \ce{H^+} + 2 \ce{e^-}$.
Under these conditions, the potential for the hydrogen reduction is defined as exactly zero. We call this $E^0$, the standard reduction potential.
We can then use this system to measure the potentials of other electrodes in the half-cell. A metal and one of its salts (sulfate is often used) is in the second half-cell. We will use zinc as our example (see figure below).
As we observe the reaction, we notice that the mass of solid zinc decreases during the course of the reaction. This suggests that the reaction occurring in that half-cell is:
$\ce{Zn} \left( s \right) \rightarrow \ce{Zn^{2+}} \left( aq \right) + 2 \ce{e^-}\nonumber$
So, we have the following process occurring in the cell:
\begin{align*} &\ce{Zn} \left( s \right) \rightarrow \ce{Zn^{2+}} \left( aq \right) + 2 \ce{e^-} \: \left( \text{anode - oxidation} \right) \ &2 \ce{H^+} \left( aq \right) + 2 \ce{e^-} \rightarrow \ce{H_2} \left( g \right) \: \left( \text{cathode - reduction} \right) \end{align*}\nonumber
The measured cell voltage is 0.76 volts (abbreviated as $\text{V}$).
We define the standard emf (electromotive force) of the cell as:
\begin{align*} E^0_\text{cell} &= E^0_\text{cathode} - E^0_\text{anode} \ 0.76 \: \text{V} &= 0 - E^0_{\ce{Zn} \: \text{cell}} \ E^0_{\ce{Zn} \: \text{cell}} &= -0.76 \: \text{V} \end{align*}\nonumber
We can do the same determination with a copper cell (figure below).
As we run the reaction, we see the mass of the copper increases, so we write the half-reaction:
$\ce{Cu^{2+}} + 2 \ce{e^-} \rightarrow \ce{Cu}\nonumber$
This makes the copper electrode the cathode. We now have the two half-reactions:
\begin{align*} &\ce{H_2} \rightarrow 2 \ce{H^+} + 2 \ce{e^-} \: \left( \text{anode} \right) \ &\ce{Cu^{2+}} + 2 \ce{e^-} \rightarrow \ce{Cu} \: \left( \text{cathode} \right) \end{align*}\nonumber
We determine the $E^0$ for the system to be $0.34 \: \text{V}$.
Again,
\begin{align*} E^0_\text{cell} &= E^0_\text{cathode} - E^0_\text{anode} \ 0.34 \: \text{V} &= E^0_\text{copper} - 0 \end{align*}\nonumber
So, copper potential $= 0.34 \: \text{V}$.
Now, we want to build a system in which both zinc and copper are involved. We know from the activity series that zinc will be oxidized and copper reduced, so we can use the values at hand:
$E^0_\text{cell} = 0.34 \: \text{V} \: \left( \text{copper} \right) - \left( -0.76 \: \text{V} \: \text{zinc} \right) = 1.10 \: \text{volts for the cell}\nonumber$
Summary
• The standard hydrogen electrode is used for determination of half-reaction current flows and voltages.
• The structure of the standard hydrogen electrode is described.
• Examples of using the standard hydrogen electrode to determine reduction potentials are given. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/23%3A_Electrochemistry/23.04%3A_Electrical_Potential.txt |
When exposed to moisture, steel will begin to rust fairly quickly. This creates a significant problem for items like nails that are exposed to the atmosphere. The nails can be protected by being coated with zinc metal, to make a galvanized nail. The zinc is more likely to oxidize than the iron in the steel, so it prevents rust from developing on the nail.
Calculating Standard Cell Potentials
In order to function, any electrochemical cell must consist of two half-cells.The table below can be used to determine the reactions that will occur and the standard cell potential for any combination of two half-cells, without actually constructing the cell. The half-cell with the higher reduction potential according to the table will undergo reduction within the cell. The half-cell with the lower reduction potential will undergo oxidation within the cell. If those specifications are followed, the overall cell potential will be a positive value. The cell potential must be positive in order for redox reaction of the cell to be spontaneous in the reverse direction.
Half-Reaction $E^0 \: \left( \text{V} \right)$
Table $1$: Standard Reduction Potentials at $25^\text{o} \text{C}$
$\ce{F_2} + 2 \ce{e^-} \rightarrow \ce{F^-}$ +2.87
$\ce{PbO_2} + 4 \ce{H^+} + \ce{SO_4^{2-}} + 2 \ce{e^-} \rightarrow \ce{PbSO_4} + 2 \ce{H_2O}$ +1.70
$\ce{MnO_4^-} + 8 \ce{H^+} + 5 \ce{e^-} \rightarrow \ce{Mn^{2+}} + 4 \ce{H_2O}$ +1.51
$\ce{Au^{3+}} + 3 \ce{e^-} \rightarrow \ce{Au}$ +1.50
$\ce{Cl_2} + 2 \ce{e^-} \rightarrow 2 \ce{Cl^-}$ +1.36
$\ce{Cr_2O_7^{2-}} + 14 \ce{H^+} + 6 \ce{e^-} \rightarrow 2 \ce{Cr^{3+}} + 7 \ce{H_2O}$ +1.33
$\ce{O_2} + 4 \ce{H^+} + 4 \ce{e^-} \rightarrow 2 \ce{H_2O}$ +1.23
$\ce{Br_2} + 2 \ce{e^-} \rightarrow 2 \ce{Br^-}$ +1.07
$\ce{NO_3^-} + 4 \ce{H^+} + 3 \ce{e^-} \rightarrow \ce{NO} + 2 \ce{H_2O}$ +0.96
$2 \ce{Hg^{2+}} + 2 \ce{e^-} \rightarrow \ce{Hg_2^{2+}}$ +0.92
$\ce{Hg^{2+}} + 2 \ce{e^-} \rightarrow \ce{Hg}$ +0.85
$\ce{Ag^+} + \ce{e^-} \rightarrow \ce{Ag}$ +0.80
$\ce{Fe^{3+}} + \ce{e^-} \rightarrow \ce{Fe^{2+}}$ +0.77
$\ce{I_2} + 2 \ce{e^-} \rightarrow 2 \ce{I^-}$ +0.53
$\ce{Cu^+} + \ce{e^-} \rightarrow \ce{Cu}$ +0.52
$\ce{O_2} + 2 \ce{H_2O} + 4 \ce{e^-} \rightarrow 4 \ce{OH^-}$ +0.40
$\ce{Cu^{2+}} + 2 \ce{e^-} \rightarrow \ce{Cu}$ +0.34
$\ce{Sn^{4+}} + 2 \ce{e^-} \rightarrow \ce{Sn^{2+}}$ +0.13
$2 \ce{H^+} + 2 \ce{e^-} \rightarrow \ce{H_2}$ 0.00
$\ce{Pb^{2+}} + 2 \ce{e^-} \rightarrow \ce{Pb}$ -0.13
$\ce{Sn^{2+}} + 2 \ce{e^-} \rightarrow \ce{Sn}$ -0.14
$\ce{Ni^{2+}} + 2 \ce{e^-} \rightarrow \ce{Ni}$ -0.25
$\ce{Co^{2+}} + 2 \ce{e^-} \rightarrow \ce{Co}$ -0.28
$\ce{PbSO_4} + 2 \ce{e^-} \rightarrow \ce{Pb} + \ce{SO_4^{2-}}$ -0.31
$\ce{Cd^{2+}} + 2 \ce{e^-} \rightarrow \ce{Cd}$ -0.40
$\ce{Fe^{2+}} + 2 \ce{e^-} \rightarrow \ce{Fe}$ -0.44
$\ce{Cr^{3+}} + 3 \ce{e^-} \rightarrow \ce{Cr}$ -0.74
$\ce{Zn^{2+}} + 2 \ce{e^-} \rightarrow \ce{Zn}$ -0.76
$2 \ce{H_2O} + 2 \ce{e^-} \rightarrow \ce{H_2} + 2 \ce{OH^-}$ -0.83
$\ce{Mn^{2+}} + 2 \ce{e^-} \rightarrow \ce{Mn}$ -1.18
$\ce{Al^{3+}} + 3 \ce{e^-} \rightarrow \ce{Al}$ -1.66
$\ce{Be^{2+}} + 2 \ce{e^-} \rightarrow \ce{Be}$ -1.70
$\ce{Mg^{2+}} + 2 \ce{e^-} \rightarrow \ce{Mg}$ -2.37
$\ce{Na^+} + \ce{2^-} \rightarrow \ce{Na}$ -2.71
$\ce{Ca^{2+}} + 2 \ce{e^-} \rightarrow \ce{Ca}$ -2.87
$\ce{Sr^{2+}} + 2 \ce{e^-} \rightarrow \ce{Sr}$ -2.89
$\ce{Ba^{2+}} + 2 \ce{e^-} \rightarrow \ce{Ba}$ -2.90
$\ce{Rb^+} + \ce{e^-} \rightarrow \ce{Rb}$ -2.92
$\ce{K^+} + \ce{e^-} \rightarrow \ce{K}$ -2.92
$\ce{Cs^+} + \ce{e^-} \rightarrow \ce{Cs}$ -2.92
$\ce{Li^+} + \ce{e^-} \rightarrow \ce{Li}$ -3.05
Example $1$
Calculate the standard cell potential of a voltaic cell that uses the $\ce{Ag}$/$\ce{Ag^+}$ and $\ce{Sn}$/$\ce{Sn^{2+}}$ half-cell reactions. Write the balanced equation for the overall cell reaction that occurs. Identify the anode and the cathode.
Known
• $E^0_\ce{Ag} = +0.80 \: \text{V}$
• $E^0_\ce{Sn} = -0.14 \: \text{V}$
Unknown
The silver half-cell will undergo reduction because its standard reduction potential is higher. The tin half-cell will undergo oxidation. The overall cell potential can be calculated by using the equation $E^0_\text{cell} = E^0_\text{red} - E^0_\text{oxid}$.
Step 2: Solve.
\begin{align*} \text{oxidation} \: \left( \text{anode} \right) &: \ce{Sn} \left( s \right) \rightarrow \ce{Sn^{2+}} \left( aq \right) + 2 \ce{e^-} \ \text{reduction} \: \left( \text{cathode} \right) &: \ce{Ag^+} \left( aq \right) + \ce{e^-} \rightarrow \ce{Ag} \left( s \right) \end{align*}\nonumber
Before adding the two reactions together, the number of electrons lost in the oxidation must equal the number of electrons gained in the reduction. The silver half-cell reaction must be multiplied by two. After doing that and adding to the tin half-cell reaction, the overall equation is obtained.
$\text{overall equation} \: \: \: \: \: \: \ce{Sn} \left( s \right) + 2 \ce{Ag^+} \left( aq \right) \rightarrow \ce{Sn^{2+}} \left( aq \right) + 2 \ce{Ag} \left( s \right)\nonumber$
The cell potential is calculated.
$E^0_\text{cell} = E^0_\text{red} - E^0_\text{oxid} = +0.80 - \left( -0.14 \: \text{V} \right) = +0.94 \: \text{V}\nonumber$
Step 3: Think about your result.
The standard cell potential is positive, so the reaction is spontaneous as written. Tin is oxidized at the anode, while silver ion is reduced at the cathode. Note that the voltage for the silver ion reduction is not doubled, even though the reduction half-reaction had to be doubled to balance the overall redox equation.
Oxidizing and Reducing Agents
A substance which is capable of being reduced very easily is a strong oxidizing agent. Conversely, a substance which is capable of being oxidized very easily is a strong reducing agent. According to the standard cell potential table, fluorine $\left( \ce{F_2} \right)$ is the strongest oxidizing agent. It will oxidize any substance below on the table. For example, fluorine will oxidize gold metal, according to the following reaction:
$3 \ce{F_2} \left( g \right) + 2 \ce{Au} \left( s \right) \rightarrow 6 \ce{F^-} \left( aq \right) + 2 \ce{Au^{3+}} \left( aq \right)\nonumber$
Lithium metal $\left( \ce{Li} \right)$ is the strongest reducing agent. It is capable of reducing any substance above on the table. For example, lithium will reduce water according to the following reaction:
$2 \ce{Li} \left( s \right) + 2 \ce{H_2O} \left( l \right) \rightarrow 2 \ce{Li^+} \left( aq \right) 2 \ce{OH^-} \left( aq \right) + \ce{H_2} \left( g \right)\nonumber$
Using the table above will allow you to predict whether reactions will occur or not. For example, nickel metal is capable of reducing copper (II) ions, but is not capable of reducing zinc ions. This is because nickel $\left( \ce{Ni} \right)$ is below $\ce{Cu^{2+}}$, but is above $\ce{Zn^{2+}}$ in the table.
Summary
• Standard cell potential calculations are described.
• Guidelines for making predictions of reaction possibilities using standard cell potentials are given. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/23%3A_Electrochemistry/23.06%3A_Calculating_Standard_Cell_Potentials.txt |
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 that separated the disks. There were no voltmeters at the time (and no insight that the electric current was due to electron flow). So, 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 lucky that we have voltmeters today to measure electric current, instead of having to test the voltage out for ourselves! Two variations on the basic voltaic cell are the dry cell and the lead storage battery.
Batteries: Dry Cells
Many common batteries, such as those used in a flashlight or remote control, are voltaic dry cells. These batteries are called dry cells because the electrolyte is a paste. They are relatively inexpensive, but do not last a long time and are not rechargeable.
In the zinc-carbon dry cell, the anode is a zinc container, while the cathode is a carbon rod through the center of the cell. The paste is made of manganese (IV) oxide $\left( \ce{MnO_2} \right)$, ammonium chloride $\left( \ce{NH_4Cl} \right)$, and zinc chloride $\left( \ce{ZnCl_2} \right)$ in water. The half-reactions for this dry cell are:
Anode (oxidation):
$\ce{Zn} \left( s \right) \rightarrow \ce{Zn^{2+}} \left( aq \right) + 2 \ce{e^-}\nonumber$
Cathode (reduction):
$2 \ce{MnO_2} \left( s \right) + 2 \ce{NH_4^+} \left( aq \right) + 2 \ce{e^-} \rightarrow \ce{Mn_2O_3} \left( s \right) + 2 \ce{NH_3} \left( aq \right) + \ce{H_2O} \left( l \right)\nonumber$
The paste prevents the contents of the dry cell from freely mixing, so a salt bridge is not needed. The carbon rod is a conductor only and does not undergo reduction. The voltage produced by a fresh dry cell is $1.5 \: \text{V}$, but decreases during use.
An alkaline battery is a variation on the zinc-carbon dry cell. The alkaline battery has no carbon rod and uses a paste of zinc metal and potassium hydroxide, instead of a solid metal anode. The cathode half-reaction is the same, but the anode half-reaction is different.
Anode (oxidation):
$\ce{Zn} \left( s \right) + 2 \ce{OH^-} \left( aq \right) \rightarrow \ce{Zn(OH)_2} \left( s \right) + 2 \ce{e^-}\nonumber$
Advantages of the alkaline battery are that it has a longer shelf life, and the voltage does not decrease during use.
Lead Storage Batteries
A battery is a group of electrochemical cells combined together as a source of direct electric current at a constant voltage. Dry cells are not true batteries since they are only one cell. The lead storage battery is commonly used as the power source in cars and other vehicles. It consists of six identical cells joined together, each of which has a lead anode and a cathode made of lead (IV) oxide $\left( \ce{PbO_2} \right)$ packed on a metal plate.
The cathode and anode are both immersed in an aqueous solution of sulfuric acid, which acts as the electrolyte. The cell reactions are:
Anode (oxidation):
$\ce{Pb} \left( s \right) + \ce{SO_4^{2-}} \left( aq \right) \rightarrow \ce{PbSO_4} \left( s \right) + 2 \ce{e^-}\nonumber$
Cathode (reduction):
$\ce{PbO_2} \left( s \right) + 4 \ce{H^+} \left( aq \right) + \ce{SO_4^{2-}} \left( aq \right) + 2 \ce{e^-} \rightarrow \ce{PbSO_4} \left( s \right) + 2 \ce{H_2O} \left( l \right)\nonumber$
Overall:
$\ce{Pb} \left( s \right) + \ce{PbO_2} \left( s \right) + 4 \ce{H^+} \left( aq \right) + 2 \ce{SO_4^{2-}} \left( aq \right) \rightarrow 2 \ce{PbSO_4} \left( s \right) + 2 \ce{H_2O} \left( l \right)\nonumber$
Each cell in a lead storage battery produces $2 \: \text{V}$, so a total of $12 \: \text{V}$ is generated by the entire battery. This is used to start a car or power other electrical systems.
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 nonspontaneous, 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 recharge is not $100\%$ efficient, because some of the lead (II) sulfate falls from the electrodes and collects on the bottom of the cells.
Summary
• Construction of a dry cell and a battery are given.
• Chemical reactions for both types are described. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/23%3A_Electrochemistry/23.07%3A_Batteries.txt |
In 1989, two scientists announced that they had achieved "cold fusion", the process of fusing together elements at essentially room temperature to achieve energy production. The hypothesis was that the fusion would produce more energy than was required to cause the process to occur. Their process involved the electrolysis of heavy water (water molecules containing some deuterium instead of normal hydrogen) on a palladium electrode. The experiments could not be reproduced and their scientific reputations were pretty well shot. However, in more recent years, both industry and government researchers are taking another look at this process. The device illustrated above is part of a government project, and NASA is completing some studies on the topic as well. Cold fusion may not be so "cold" after all.
Electrolytic Cells
A voltaic cell uses a spontaneous redox reaction to generate an electric current. It is also possible to do the opposite. When an external source of direct current is applied to an electrochemical cell, a reaction that is normally nonspontaneous can be made to proceed. Electrolysis is the process in which electrical energy is used to cause a nonspontaneous chemical reaction to occur. Electrolysis is responsible for the appearance of many everyday objects, such as gold-plated or silver-plated jewelry, and chrome-plated car bumpers.
An electrolytic cell is the apparatus used for carrying out an electrolysis reaction. In an electrolytic cell, electric current is applied to provide a source of electrons for driving the reaction in a nonspontaneous direction. In a voltaic cell, the reaction goes in a direction that releases electrons spontaneously. In an electrolytic cell, the input of electrons from an external source forces the reaction to go in the opposite direction.
The spontaneous direction for the reaction between $\ce{Zn}$ and $\ce{Cu}$ is for the $\ce{Zn}$ metal to be oxidized to $\ce{Zn^{2+}}$ ions, while the $\ce{Cu^{2+}}$ ions are reduced to $\ce{Cu}$ metal. This makes the zinc electrode the anode and the copper electrode the cathode. When the same half-cells are connected to a battery via the external wire, the reaction is forced to run in the opposite direction. The zinc electrode is now the cathode and the copper electrode is the anode.
$\begin{array}{lll} \text{oxidation (anode)} & \ce{Cu} \left( s \right) \rightarrow \ce{Cu^{2+}} \left( aq \right) + 2 \ce{e^-} & E^0 = -0.34 \: \text{V} \ \text{reduction (cathode)} & \ce{Zn^{2+}} \left( aq \right) + 2 \ce{e^-} \rightarrow \ce{Zn} \left( s \right) & E^0 = -0.76 \: \text{V} \ \hline \text{overall reaction} & \ce{Cu} \left( s \right) + \ce{Zn^{2+}} \left( aq \right) \rightarrow \ce{Cu^{2+}} \left( aq \right) + \ce{Zn} \left( s \right) & E^0_\text{cell} = -1.10 \: \text{V} \end{array}\nonumber$
The standard cell potential is negative, indicating a nonspontaneous reaction. The battery must be capable of delivering at least $1.10 \: \text{V}$ of direct current in order for the reaction to occur. Another difference between a voltaic cell and an electrolytic cell is the signs of the electrodes. In a voltaic cell, the anode is negative and the cathode is positive. In an electrolytic cell, the anode is negative and the cathode is positive. In an electrolytic cell, the anode is positive because it is connected to the positive terminal of the battery. The cathode is therefore negative. Electrons still flow through the cell from the anode to the cathode.
Summary
• The function of an electrolytic cell is described.
• Reactions illustrating electrolysis are given. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/23%3A_Electrochemistry/23.08%3A_Electrolytic_Cells.txt |
With fossil fuels becoming more expensive and less available, scientists are looking for other energy sources. Hydrogen has long been considered an ideal source, since it does not pollute the air when it burns. The problem has been finding ways to generate hydrogen economically. One new approach that is being studied is photoelectrolysis—the generation of electricity using photovoltaic cells to split water molecules. This technique is still in the research stage, but appears to be a very promising source of power in the future.
Electrolysis of Water
The electrolysis of water produces hydrogen and oxygen gases. The electrolytic cell consists of a pair of platinum electrodes immersed in water to which a small amount of an electrolyte such as $\ce{H_2SO_4}$ has been added. The electrolyte is necessary because pure water will not carry enough charge due to the lack of ions. At the anode, water is oxidized to oxygen gas and hydrogen ions. At the cathode, water is reduced to hydrogen gas and hydroxide ions.
$\begin{array}{lll} \text{oxidation (anode):} & 2 \ce{H_2O} \left( l \right) \rightarrow \ce{O_2} \left( g \right) + 4 \ce{H^+} \left( aq \right) + 4 \ce{e^-} & E^0 = -1.23 \: \text{V} \ \text{reduction (cathode):} & 2 \ce{H_2O} \left( l \right) + 2 \ce{e^-} \rightarrow \ce{H_2} \left( g \right) + 2 \ce{OH^-} \left( aq \right) & E^0 = -0.83 \: \text{V} \ \hline \text{overall reaction:} & 2 \ce{H_2O} \left( l \right) \rightarrow \ce{O_2} \left( g \right) + 2 \ce{H_2} \left( g \right) & E^0_\text{cell} = -2.06 \: \text{V} \end{array}\nonumber$
In order to obtain the overall reaction, the reduction half-reaction was multiplied by two to equalize the electrons. The hydrogen ion and hydroxide ions produced in each reaction combine to form water. The $\ce{H_2SO_4}$ is not consumed in the reaction.
23.10: Electrolysis of Molten Salts and Electrolysis of Brine
Production of $\ce{NaOH}$ is an important industrial process. Three different methods are employed, all of which involve the use of electricity. When calculating the price of sodium hydroxide, a company has to charge in order to make a profit; the cost of electricity has to be factored in. To make a metric ton of $\ce{NaOH}$, between $3300$-$5000 \: \text{kWh}$ (kilowatt hours) are required. Compare that with the power needed to run an average house. You could power a home for 6 - 10 months with the same amount of electricity.
Electrolysis of Molten Sodium Chloride
Molten (liquid) sodium chloride can be electrolyzed to produce sodium metal and chlorine gas. The electrolytic cell used in the process is called a Down's cell (see figure below).
In a Down's cell, the liquid sodium ions are reduced at the cathode to liquid sodium metal. At the anode, liquid chlorine ions are oxidized to chlorine gas. The reactions and cell potentials are shown below:
$\begin{array}{lll} \text{oxidation (anode):} & 2 \ce{Cl^-} \left( l \right) \rightarrow \ce{Cl_2} \left( g \right) + 2 \ce{e^-} & E^0 = -1.36 \: \text{V} \ \text{reduction (cathode):} & \ce{Na^+} \left( l \right) + \ce{e^-} \rightarrow \ce{Na} \left( l \right) & E^0 = -2.71 \: \text{V} \ \hline \text{overall reaction:} & 2 \ce{Na^+} \left( l \right) + 2 \ce{Cl^-} \left( l \right) \rightarrow 2 \ce{Na} \left( l \right) + \ce{Cl_2} \left( g \right) & E^0_\text{cell} = -4.07 \: \text{V} \end{array}\nonumber$
The battery must supply over 4 volts to carry out this electrolysis. This reaction is a major source of production of chlorine gas and is the only way to obtain pure sodium metal. Chlorine gas is widely used in cleaning, disinfecting, and in swimming pools.
Electrolysis of Aqueous Sodium Chloride
It may be logical to assume that the electrolysis of aqueous sodium chloride, called brine, would yield the same result through the same reactions as the process in molten $\ce{NaCl}$. However, the reduction reaction that occurs at the cathode does not produce sodium metal, instead, the water is reduced. This is because the reduction potential for water is only $-0.83 \: \text{V}$ compared to $-2.71 \: \text{V}$ for the reduction of sodium ions. This makes the reduction of water preferable because its reduction potential is less negative. Chlorine gas is still produced at the anode, just as in the electrolysis of molten $\ce{NaCl}$.
$\begin{array}{lll} \text{oxidation (anode):} & 2 \ce{Cl^-} \left( aq \right) \rightarrow \ce{Cl_2} \left( g \right) + 2 \ce{e^-} & E^0 = -1.36 \: \text{V} \ \text{reduction (cathode):} & 2 \ce{H_2O} \left( l \right) + 2 \ce{e^-} \rightarrow \ce{H_2} \left( g \right) + 2 \ce{OH^-} \left( aq \right) & E^0 = -0.83 \: \text{V} \ \hline \text{overall reaction} & 2 \ce{Cl^-} \left( aq \right) + 2 \ce{H_2O} \left( l \right) \rightarrow \ce{Cl_2} \left( g \right) + \ce{H_2} \left( g \right) + 2 \ce{OH^-} \left( aq \right) & E^0_\text{cell} = -2.19 \: \text{V} \end{array}\nonumber$
Since hydroxide ions are also a product of the net reaction, the important chemical sodium hydroxide $\left( \ce{NaOH} \right)$ is obtained from evaporation of the aqueous solution at the end of the hydrolysis.
Summary
• The reactions involving the electrolysis of molten $\ce{NaCl}$ are described.
• The reactions involving the electrolysis of brine are described.
23.11: Electroplating
The astrolabe was a device used to study the motions of planets and to do surveying; most were made of brass. Persian mystics also used astrolabes for following the stars and making astrological predictions.
Electroplating
Many decorative objects, like jewelry, are manufactured with the aid of an electrolytic process. Electroplating is a process in which a metal ion is reduced in an electrolytic cell and the solid metal is deposited onto a surface. The figure below shows a cell in which copper metal is to be plated onto a second metal.
The cell consists of a solution of copper sulfate and a strip of copper which acts as the anode. The metal $\left( \ce{Me} \right)$ is the cathode. The anode is connected to the positive electrode of a battery, while the metal is connected to the negative electrode.
When the circuit is closed, copper metal from the anode is oxidized, allowing copper ions to enter the solution.
$\text{anode:} \: \: \: \ce{Cu^0} \left( s \right) \rightarrow \ce{Cu^{2+}} \left( aq \right) + 2 \ce{e^-}\nonumber$
Meanwhile, copper ions from the solution are reduced to copper metal on the surface of the cathode (the second metal):
$\text{cathode:} \: \: \: \ce{Cu^{2+}} \left( aq \right) + 2 \ce{e^-} \rightarrow \ce{Cu^0} \left( s \right)\nonumber$
The concentration of copper ions in the solution is effectively constant. This is because the electroplating process transfers metal from the anode to the cathode of the cell. Other metals commonly plated onto objects include chromium, gold, silver, and platinum. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/23%3A_Electrochemistry/23.09%3A_Electrolysis_of_Water.txt |
Figure $3$: The radioactive danger symbol. (By Clkr-Free-Vector-Images(opens in new window); under Pixabay license)
If you visit the nuclear medicine department of a large hospital, you are very likely to see the symbol shown above. This sign means that radioactive materials are present, and special safety precautions need to be taken. These materials are used for the diagnosis and treatment of many diseases. The people using these materials are specially trained to handle them safely. Radioactive materials can be dangerous and should be respected, but need not be feared.
Discovery of Radioactivity
John Dalton first proposed his atomic theory in an 1804 lecture to the Royal Institution, a prestigious British scientific society. In this talk, he put forth the idea that all atoms of an element were identical, and that atoms were indestructible. In a little over 100 years, both of these ideas were shown to be incorrect.
In 1919, studies on atomic weights led Francis Aston (1877-1925) to the conclusion that some elements with different atomic weights were actually the same element in different isotope forms. Aston used a mass spectrograph to separate isotopes of different elements. He won the Nobel Prize in Chemistry for this work in 1922.
Natural Radioactivity
In 1895, Wilhelm Röntgen produced great interest around what caused the phenomenon of x-rays. One researcher of this phenomenon was Henri Becquerel. Becquerel studied the fluorescent properties of uranium salts, believing they had something to do with x-rays. He soon learned that uranium could expose a photographic plate without an external input of energy (thought to be needed to produce uranium fluorescence). Becquerel also showed that the uranium salt emissions could be affected by a magnetic field, which was not true for x-rays.
Pierre and Marie Curie studied the properties of uranium salts with the express purpose of identifying the details of these emissions. They were the first to coin the term "radioactivity", meaning the spontaneous emission of radiation in the form of particles or high energy photons, resulting from a nuclear reaction. The major contributions to the work came from Marie, who showed that the amount of radioactivity present was due to the amount of a specific element, and not due to some chemical reaction. She discovered the element polonium and named it after her native Poland. Madame Curie shared the 1903 Nobel Prize in Physics with her husband Pierre and Henri Becquerel. She won the Nobel Prize in Chemistry in 1911.
Ernest Rutherford, a later researcher, was able to demonstrate three different types of radioactive emissions. These emission types differed in terms of mass, charge, and their ability to penetrate materials. He designated them simply as alpha $\left( \alpha \right)$ emissions, beta $\left( \beta \right)$ emissions, and gamma $\left( \gamma \right)$ emissions.
Radioactivity involves the spontaneous emission of material and/or energy from the nucleus of an atom. The most common radioactive atoms have high atomic numbers and contain a large excess of neutrons. Some typical radioactive elements are technetium (atomic number 43), promethium (atomic number 61), and all elements atomic number 84 (polonium) and higher. There are four primary types of emission, either involving the release of a particle from the nucleus or the release of energy. In many instances, both energy and a particle are produced by the radioactive event.
It should be noted that some elements considered to be stable do have radioactive isotopes. Carbon-14 is radioactive, but this isotope is only a small fraction of the total amount of carbon in existence (about one part per trillion carbon atoms). Hydrogen also has a radioactive isotope known as hydrogen-3 or tritium, again a very small fraction of the total elemental hydrogen present.
Summary
• Francis Aston discovered that some elements with different atomic weights were actually the same element in different isotope forms.
• Henri Becquerel showed that uranium salt emissions could be affected by a magnetic field, which was not true for x-rays.
• Marie and Pierre Curie were the first to coin the term radioactivity, meaning the spontaneous emission of radiation in the form of particles or high energy photons, resulting from a nuclear reaction.
• Ernest Rutherford, a later researcher, was able to demonstrate three different types of radioactive emissions: alpha $\left( \alpha \right)$ emissions, beta $\left( \beta \right)$ emissions, and gamma $\left( \gamma \right)$ emissions. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/24%3A_Nuclear_Chemistry/24.01%3A_Discovery_of_Radioactivity.txt |
Food irradiation is a sensitive subject for many people. The practice involves exposing food to ionizing radiation in order to kill harmful bacteria (such as salmonella) that cause sickness. The food is essentially unchanged and does not lose any nutritive value. Parasites and insect pests are easily destroyed by this process, while bacteria take longer to kill. Viruses are not affected by the radiation treatment. But don't worry—the food is not radioactive, and you will not glow in the dark if you eat it.
Nuclear Decay Processes
Radioactive decay involves the emission of a particle and/or energy as one atom changes into another. In most instances, the atom changes its identity to become a new element. There are four different types of emissions that occur.
Alpha Emission
Alpha $\left( \alpha \right)$ decay involves the release of helium ions from the nucleus of an atom. This ion consists of two protons and two neutrons and has a $2+$ charge. Release of an $\alpha$-particle produces a new atom that has an atomic number two less than the original atom and an atomic weight that is four less. A typical alpha decay reaction is the conversion of uranium-238 to thorium:
$\ce{^{238}_{92}U} \rightarrow \ce{^{234}_{90}Th} + \ce{^4_2 \alpha}^+\nonumber$
We see a decrease of two in the atomic number (uranium to thorium) and a decrease of four in the atomic weight (238 to 234). Usually the emission is not written with atomic number and weight indicated since it is a common particle whose properties should be memorized. Quite often the alpha emission is accompanied by gamma $\left( \gamma \right)$ radiation, a form of energy release. Many of the largest elements in the periodic table are alpha-emitters.
Beta Emission
Beta $\left( \beta \right)$ decay is a more complicated process. Unlike the $\alpha$-emission, which simply expels a particle, the $\beta$-emission involves the transformation of a neutron in the nucleus to a proton and an electron. The electron is then ejected from the nucleus. In the process, the atomic number increases by one while the atomic weight stays the same. As is the case with $\alpha$-emissions, $\beta$-emissions are often accompanied by $\gamma$-radiation.
A typical beta decay process involves carbon-14, often used in radioactive dating techniques. The reaction forms nitrogen-14 and an electron:
$\ce{^{14}_6C} \rightarrow \ce{^{14}_7N} + \ce{^0_{-1}e}\nonumber$
Again, the beta emission is usually simply indicated by the Greek letter $\beta$; memorization of the process is necessary in order to follow nuclear calculations in which the Greek letter $\beta$ appears without further notation.
Positron Emission
A positron is a positive electron (a form of antimatter). This rare type of emission occurs when a proton is converted to a neutron and a positron in the nucleus, with ejection of the positron. The atomic number will decrease by one, while the atomic weight does not change. A positron is often designated by $\beta^+$.
Carbon-11 emits a positron to become boron-11:
$\ce{^{11}_6C} \rightarrow \ce{^{11}_5B} + \ce{^0_{+1} \beta}\nonumber$
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.
$\ce{^{106}_{47}Ag} + \ce{^0_{-1}e} \rightarrow \ce{^{106}_{46}Pd}\nonumber$
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 Emission
Gamma $\left( \gamma \right)$ radiation is simply energy. It may be released by itself or, more commonly, in association with other radiation events. There is no change of atomic number or atomic weight in a simple $\gamma$-emission. Often, an isotope may produce $\gamma$-radiation as a result of a transition in a metastable isotope. This type of isotope may just "settle", with a shifting of particles in the nucleus. The composition of the atom is not altered, but the nucleus could be considered more "comfortable" after the shift. This shift increases the stability of the isotope from the energetically unstable (or "metastable") isotope to a more stable form of the nucleus.
Summary
• Radioactive decay involves the emission of a particle and/or energy as one atom changes into another.
• Alpha $\left( \alpha \right)$ decay involves the release of helium ions from the nucleus of an atom.
• Beta $\left( \beta \right)$ decay involves the transformation of a neutron in the nucleus to a proton and an electron.
• A positron is a positive electron (a form of antimatter). This rare type of emission occurs when a proton is converted to a neutron and a positron in the nucleus, with ejection of the positron.
• Gamma $\left( \gamma \right)$ radiation is simply energy. It may be released by itself or, more commonly, in association with other radiation events. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/24%3A_Nuclear_Chemistry/24.02%3A_Nuclear_Decay_Processes.txt |
Uranium is worth more on the open market than it was several years ago. So, people are out there looking for more uranium. However, uranium-hunters are not waving Geiger counters around to find this material—all the surface uranium has been found. Instead, they must study geology. Deep holes are drilled to explore the underground geology of a site, and the rocks obtained are chemically analyzed. No more clicking—just modern-day geology and chemistry.
Detection of Radioactivity
Radioactivity is determined by measuring the number of decay processes per unit time. Perhaps the easiest way is simply to determine the number of counts/minute, with each count measuring a single decay process, such as the emission of an $\alpha$-particle. A particular isotope may have an activity of 5,000 counts/minute $\left( \text{cpm} \right)$ while another isotope might only have $250 \: \text{cpm}$. The amount of activity gives a rough indication of the amount of the radioisotope present—the higher the activity, the more radioactive isotope in the sample.
Units of Measurement
The curie $\left( \text{Ci} \right)$ is one measure of the rate of decay (named after Pierre and Marie Curie). One curie is equivalent to $3.7 \times 10^{10}$ disintegrations per second. Since this is obviously a large and unwieldy number, radiation is often expressed in millicuries or microcuries (still very large numbers). Another measure is the becquerel $\left( \text{Bq} \right)$, named after Henri Becquerel. The becquerel is defined as an activity of one disintegration/second. Both of these units are concerned with the disintegration rate of the radioactive isotope, and give no indication of dosage to the target material.
Personal Dosimeters
Measurement of exposure to radioactivity is important for anyone who deals with radioactive materials on a regular basis. Perhaps the simplest device is a personal dosimeter—a film badge that will fog up when exposed to radiation. The amount of fogging is proportional to the amount of radiation present. These devices are not very sensitive to low levels of radiation. More sensitive systems use crystals that respond in some way to radioactivity by registering the number of emissions in a given time. These systems tend to be more sensitive and more reliable than film badges.
Geiger Counters
A Geiger counter provides a sensitive means of detecting radioactivity. A tube is filled with an inert gas, which will conduct electricity when radiation enters it. When a charged particle comes into the tube, it changes the electrical potential between the anode and the cathode. This change in potential in the tube produces a change in voltage in the electrical circuit and registers as a count. Geiger counters are fairly inexpensive and reliable, so they are useful in a wide range of applications. More complicated types of counters are also available, but are generally used in sophisticated experiments.
Summary
• The curie $\left( \text{Ci} \right)$ is one measure of the rate of radioactive decay; one curie is equivalent to $3.7 \times 10^{10}$ disintegrations per second.
• Another measure of radioactive decay is the becquerel $\left( \text{Bq} \right)$, defined as an activity of one disintegration/second.
• The dosimeter is a device to measure radioactivity—a film badge that will fog up when exposed to radiation.
• The Geiger counter provides a sensitive means of detecting radioactivity. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/24%3A_Nuclear_Chemistry/24.03%3A_Detection_of_Radioactivity.txt |
Uranium isotopes produce plutonium-239 as a decay product. The plutonium can be used in nuclear weapons and is a power source for nuclear reactors, which generate electricity. This isotope has a half-life of 24,100 years, causing concern in regions where radioactive plutonium has accumulated and is stored. At some storage sites, the waste is slowly leaking into the groundwater and contaminating nearby rivers. The 24,100 year half-life means that it will be present in the earth for a very long time.
Half-Life
Radioactive materials lose some of their activity each time a decay event occurs. This loss of activity can be estimated by determining the half-life of an isotope. The half-life is defined as the period of time needed for one half of a given quantity of a substance to undergo a change. For a radioisotope, every time a decay event occurs, a count is detected on the Geiger counter, or other measuring device. A specific isotope might have a total count of $30,000 \: \text{cpm}$. In one hour, the count could be $15,000 \: \text{cpm}$ (half of the original count). So, the half-life of that isotope is one hour. Some isotopes have long half-lives—the half-life of $\ce{U}$-234 is 245,000 years. Other isotopes have shorter half-lives. $\ce{I}$-131, used in thyroid scans, has a half-life of 8.02 days.
Half-Life Calculations
Information on the half-life of an isotope can be used to calculate how much radioactivity of that isotope will be present after a certain period of time. There is a formula that allows calculation at any time after the initial count, but we are just going to look at loss of activity after different half-lives. The isotope $\ce{I}$-125 is used in certain laboratory procedures and has a half-life of 59.4 days. If the initial activity of a sample of $\ce{I}$-125 is $32,000 \: \text{cpm}$, how much activity will be present in 178.2 days? Begin by determining how many half-lives are represented by 178.2 days:
$\frac{178.2 \: \text{days}}{59.4 \: \text{days/half-life}} = 3 \: \text{half-lives}\nonumber$
Then, simply count activity:
\begin{align*} \text{initial activity} \: \left( t_0 \right) &= 32,000 \: \text{cpm} \ \text{after one half-life} \: &= 16,000 \: \text{cpm} \ \text{after two half-lives} \: &= 8,000 \: \text{cpm} \ \text{after three half-lives} &= 4,000 \: \text{cpm} \end{align*}\nonumber
Be sure to keep in mind that the initial count is at time zero $\left( t_0 \right)$, and to subtract from that count at the first half-life. The second half-life has an activity of half the previous count (not the initial count).
For the more mathematically inclined, the following formula can be used to calculate the amount of radioactivity remaining after a given time:
$N_t = N_0 \times \left( 0.5 \right)^\text{number of half-lives}\nonumber$
where $N_t =$ activity at time $t$ and $N_0 =$ initial activity at time $t = 0$.
If we have an initial activity of $42,000 \: \text{cpm}$, what will the activity be after four half-lives?
\begin{align*} N_t &= N_0 \left( 0.5 \right)^4 \ &= \left( 42,000 \right) \left( 0.5 \right) \left( 0.5 \right) \left( 0.5 \right) \left( 0.5 \right) \ &= 2625 \: \text{cpm} \end{align*}\nonumber
The graph above illustrates a typical decay curve for a radioactive material. The activity decreases by one-half during each succeeding half-life. Half-lives of different elements vary considerably, as shown in the table below:
Table $1$: Isotope Half-Lives
Isotope Decay Mode Half-Life
Cobalt-60 beta 5.3 years
Neptunium-237 alpha 2.1 million years
Polonium-214 alpha 0.00016 seconds
Radium-224 alpha 3.7 days
Tritium ($\ce{H}$-3) beta 12 years
We have talked about the activity and decay of individual isotopes. In the real world, there is a decay chain that takes place until a stable end-product is produced. For $\ce{U}$-238, the chain is a long one, with a mix of isotopes having very different half-lives. The end of the chain resides in lead, a stable element that does not decay further.
Summary
• Radioactive materials lose some of their activity each time a decay event occurs.
• Radioactive loss of activity can be estimated by determining the half-life of an isotope.
• Half-life is defined as the period of time needed for one half of a given quantity of a substance to undergo a change.
• Half-lives of different elements vary considerably.
24.05: Background Radiation
Sitting in a hot bath or spa has always been a great prescription for dealing with sore muscles. People used to believe that it was even more beneficial to immerse themselves in radioactive hot springs, and drink water containing radioactive materials (some still do). In the early 1900s, people spent millions of dollars on treatments and "radioactive water" with the belief that all of their ailments would cease. Radioactivity in the water was usually due to radon gas that leaked up from deep underground, formed by decay of other radioisotopes.
Background Radiation
We are all exposed to a small amount of radiation in our daily lives. This background radiation comes from naturally occurring sources and from human-produced radiation. Exposure to x-rays and nuclear medicine isotopes, ground sources, and cosmic radiation account for almost half of the background exposure of the average American. Radon gas, formed from the decay of uranium and thorium isotopes, is responsible for a little over half the total amount of background radiation. See the table below for background sources.
Sources of Background Radiation
Table \(1\): Sources of Background Radiation
radon \(54\%\)
consumer products \(3\%\)
nuclear medicine \(4\%\)
cosmic radiation \(8\%\)
terrestrial \(8\%\)
internal \(11\%\)
x-rays \(11\%\)
other \(1\%\)
The Problem of Radon
Small amounts of uranium and thorium are found in the soil of a large number of areas in the U.S. When radioactive isotopes of these elements decay, radon is one of the products formed. Radon is a colorless, odorless gas and is chemically inert, as it is one of the noble gases.
Radon is also radioactive and can easily be inhaled into the lungs. Over time, this internal radon exposure can lead to the development of lung cancer. The incidence of lung cancer in smokers exposed to radon is much higher than in non-smokers exposed to radon, since smoking has already produced some lung damage, and the radon simply makes the damage worse. Radon exposure is highest in homes lacking good air circulation to move the gas out of the residence. There are a number of inexpensive approaches to decreasing your exposure to radon. A good start is to test your living area for radon with a radon test kit.
Summary
• Background radiation is defined.
• Sources of background radiation are listed. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/24%3A_Nuclear_Chemistry/24.04%3A_Half-Life.txt |
Nuclear fission was first discovered by two German scientists, Fritz Strassman and Otto Hahn, in the 1930s. They began their work by bombarding uranium with neutrons, hoping to create larger elements. Instead, they were very surprised to find $\ce{Ba}$-141, a much smaller element. They immediately contacted a fellow scientist in the field, Lise Meitner, who carried out calculations to demonstrate that fission had taken place.
Nuclear Fission
Radioactive decay by the release of alpha or beta particles is not the only way new isotopes are formed. When a neutron collides with a nucleus, the nucleus splits into two isotopes, each of which is roughly half the mass of the original atom. A small amount of mass is "left over" and released as energy, as predicted by Einstein's famous equation $E = mc^2$, that relates mass and energy. This process is known as nuclear fission. The neutron must be a "slow" neutron, traveling at a speed that is approximately that of the molecules of a gas at the same temperature in the system producing the neutrons. High-speed ("fast") neutrons will not result in nuclear fission.
The example above illustrates the basic nuclear fission process. A neutron (generally produced by some controlled process, not usually a natural event) collides with an atom of $\ce{U}$-235. Momentarily, a $\ce{U}$-236 atom forms, which then splits into two smaller atoms ($\ce{Kr}$-93 and $\ce{Ba}$-141) in the diagram. This process results in the release of three new neutrons, which can then initiate fission reactions with more atoms. We will see later how this propagation of neutrons can be employed in a reactor for the generation of electricity.
An extended version of this process can be seen in the figure below. Not every collision of a neutron with $\ce{U}$-235 results in a fission reaction. A neutron from the initial fission process may strike an atom of $\ce{U}$-238, which does not continue the process. Another neutron may not collide with a nucleus, and be lost in the environment. However, a third neutron produced from the initial collision can collide with more $\ce{U}$-235, and continue the chain reaction to produce more neutrons.
Typical nuclear fission reactions balance in terms of mass. The total mass of the reactants is equal to the total mass of the products:
$\ce{^{235}_{92}U} + \ce{^1_0n} \rightarrow \ce{^{92}_{36}Kr} + \ce{^{142}_{56}Ba} + 2 \ce{^1_0n} + \text{energy}\nonumber$
There are a total of 236 mass units on the left of the equation and 236 mass units on the right. In the same manner, we see 92 protons on the left and 92 on the right. The energy that is released is the binding energy that holds the nucleus together.
Another set of fission products from $\ce{U}$-235 can be seen in the following reaction:
$\ce{^{235}_{92}U} + \ce{^1_0n} \rightarrow \ce{^{95}_{42}Mo} + \ce{^{139}_{57}La} + 2 \ce{^1_0n} + \text{energy}\nonumber$
Again, we see that the total number of mass units and of protons is equal on both sides of the equation.
Summary
• Nuclear fission is the process of a neutron colliding with a nucleus. The nucleus splits into two isotopes, each of which is roughly half the mass of the original atom. A small amount of mass is "left over" and released as energy.
• Examples of nuclear fission processes are illustrated.
24.07: Nuclear Power Generation
On Wednesday, March 28, 1979, the residents of Middleton, PA woke up to a very scary situation. A nuclear power plant near the town had experienced a series of malfunctions that led to the release of some radioactive gases into the atmosphere, along with a partial meltdown of the reactor core. Fortunately, follow-up studies showed that there were no health effects on workers or the general public. A thorough investigation was conducted that led to significant improvements in safety and operation of these nuclear power plants. One of the two reactors was shut down completely, while the other remained in operation until its permanent deactivation in 2014.
Nuclear Power Generation
The generation of electricity is critical for operation of businesses, health care delivery, schools, homes, and other areas requiring the use of electrical power. According to 2011 statistics, coal is used for \(42\%\) of the total power generated, with natural gas being employed for another \(25\%\). Nuclear power plants are employed in about \(19\%\) of the cases, with renewable energy sources supplying the last \(13\%\). All of these fuels are used to heat water to generate steam. The steam then turns a turbine to generate electricity.
The diagram below shows the layout of a typical nuclear power plant. The radioactive rods are in the red container along with water, which is heated to steam. The energy for this heat comes from fission reactions of uranium. The steam passes through the turbine and causes the turbine to spin, generating electricity. As the steam condenses, it is run through a cooling tower to lower its temperature. The water then recirculates through the reactor core to be used again.
The control rods play an important role in the modulation of the nuclear chain reaction (usually a collision of a neutron with uranium). Each collision produces more neutrons than were present initially. If left unsupervised, the reaction would soon get out of control. Rods are commonly made of boron or a number of metals and metal alloys. The purpose of the control rods is to absorb neutrons to regulate the rate of the chain reaction, so that the water does not overheat and destroy the reactor.
Nuclear power is also used to propel ships. The turbine can be connected to a propeller system. The rotating turbine shaft will turn the propeller to move the ship.
Summary
• The importance of nuclear power in generating electricity is described.
• The operation of a nuclear power plant is described. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/24%3A_Nuclear_Chemistry/24.06%3A_Nuclear_Fission_Processes.txt |
A number of reactions take place in the sun that cannot be duplicated on Earth. Some of these reactions involve the formation of large elements from smaller ones. So far, we have only been able to observe formation of very small elements here on Earth. The reaction sequence observed appears to be the following: hydrogen-1 atoms collide to form the larger hydrogen isotopes, hydrogen-2 (deuterium) and hydrogen-3 (tritium). In the process, positrons and gamma rays are formed. The positrons will collide with any available electrons and annihilate, producing more gamma rays. In the process, tremendous amounts of energy are produced to keep us warm and continue supplying reactions.
Nuclear Fusion
In contrast to nuclear fission, which results in smaller isotopes being formed from larger ones, the goal of nuclear fusion is to produce larger materials from the collision of smaller atoms. The forcing of the smaller atoms together results in tighter packing, and the release of energy. As seen in the figure below, energy is released in the formation of the larger atom, helium $\left( \ce{He} \right)$ from the fusion of hydrogen-2 and hydrogen-3, as well as from the expulsion of a neutron.
This release of energy is what drives research on fusion reactors today. If such a reaction could be accomplished efficiently on Earth, it could provide a clean source of nuclear energy. Unlike fission reactions, nuclear fusion does not produce radioactive products that represent hazards to living systems.
Nuclear fusion reactions in the laboratory have been extraordinarily difficult to achieve. Extremely high temperatures (i.e., millions of degrees) are required. Methods must be developed to force the atoms together and hold them together long enough to react. The neutrons released during the fusion reactions can interact with atoms in the reactor, and convert them to radioactive materials. There has been some success in the field of nuclear fusion reactions, but the journey to feasible fusion power is still a long and uncertain one.
Summary
• Nuclear fusion produces larger materials from the collision of smaller atoms. The forcing of the smaller atoms together results in tighter packing, and the release of energy.
• Examples of nuclear fusion reactions are given.
24.09: Penetrating Ability of Emissions
Designated containers can store radioisotopes that will be administered to patients as part of a medical treatment. Isotopes are stored in lead containers that block the radiation from escaping and exposing people. These lead containers are called a "pig", a term long used for referring to castings of lead or other metals. The origin of the term is not clear, but was probably first applied to these containers by the people who made them.
Penetrating Ability of Emissions
The various emissions differ considerably in their ability to go through matter, known as their penetrating ability. The $\alpha$-particle has the least penetrating power since it is the largest and slowest emission. It can be blocked by a sheet of paper or a human hand. Beta particles are more penetrating than alpha particles, but can be stopped by a thin sheet of aluminum. Of the three basic types of emissions, gamma particles are the most penetrating. A thick lead shield is required to stop gamma emissions. Positrons represent a special case in that they annihilate when they come in contact with electrons. The collision of a positron and an electron results in the formation of two gamma emissions that go 180 degrees away from each other.
Blocking of alpha particles can easily be accomplished by as little as $10 \: \text{mm}$ plastic or paper. Beta emissions represent a somewhat different situation. The negative charge on a beta particle has the potential for activating the element being used to block the radiation. Lead and tungsten are large atoms with many protons and neutrons in their nuclei. While the beta electron may be blocked, the target material could become irradiated in the process.
High-density materials are much more effective protection against gamma emissions than low-density ones. Gamma rays are usually blocked effectively by lead shielding. The thickness of the shielding will determine the effectiveness of the protection offered by the lead.
Summary
• The penetrating ability of a radioactive emission is its ability to go through matter.
• The relative penetrating abilities of radioactive emissions are described. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/24%3A_Nuclear_Chemistry/24.08%3A_Nuclear_Fusion.txt |
Bacterial contamination in our food often makes the news. There are many bacteria present on raw food, especially raw meat. Campylobacter, salmonella, and other microorganisms can be present, even after cooking, if the meat has not been sufficiently exposed to the heat. Ionizing radiation can be used to disrupt the DNA-RNA-protein synthesis cycle that allows the bacteria to reproduce. Cobalt-60 is a common radiation source, as is cesium-137. But...just to be safe...order that burger well-done!
Effects of Radiation
In order to better understand how cellular radiation damage occurs, we need to take a quick review of how the cell functions. DNA in the nucleus is responsible for protein synthesis and for regulation of many cellular functions. In the process of protein synthesis, DNA partially unfolds to produce messenger RNA (mRNA). The mRNA leaves the nucleus and interacts with ribosomes, transfer RNA, amino acids, and other cellular constituents in the cytoplasm. Through a complex series of reactions, proteins are produced to carry out a number of specialized processes within the organism. Anything that disturbs this flow of reactions can produce damage to the cell.
The major effect of ionizing radiation on the cell is the disruption of the DNA strand. With the DNA structure damaged, the cell cannot reproduce in its normal fashion. Protein synthesis is affected, as are a number of processes necessary for proper cell function. One common effect is the generation of cancer cells. These cells have an abnormal structure due to the damaged DNA. In addition, they usually grow rapidly since the normal control processes regulating cell growth have been changed by the altered composition of the DNA. Tissue damage is also common in people with severe exposure to radiation.
Effects of Radiation on Humans
We can see two general types of effects when humans are exposed to radiation. Low-level exposure can lead to development of cancer. The regulatory processes regulating cell growth are disrupted, leading to uncontrolled growth of abnormal cells. Acute exposure can produce nausea, weakness, skin burns, and internal tissue damage. Cancer patients receiving radiation therapy experience these symptoms; the radiation is targeted to a specific site in the body so that the damage is primarily to the cancer cells, and the patient is able to recover from the exposure.
Summary
• Anything that affects DNA replication and protein synthesis can damage a cell.
• Effects of ionizing radiation on protein synthesis are listed.
• The impact of ionizing radiation on human health is discussed.
24.11: Radioisotopes in Medical Diagnosis and Treatment
The molecule picture below is thyroxine, a compound produced by the thyroid gland. This molecule regulates how the body uses energy. In a condition known as hypothyroidism, the thyroid makes less thyroxine than normal. A person with this disease feels tired all the time and often puts on weight. Hypothyroidism is treated with thyroid hormone supplements.
Radioisotopes in Medical Diagnosis and Treatment
Radioisotopes are widely used to diagnose disease and as effective treatment tools. For diagnosis, the isotope is administered, and then located in the body using a scanner of some sort. The decay product (often gamma emission) can be located, and the intensity measured. The amount of isotope taken up by the body can then give information as to the extent of the medical problem.
An isotope of iodine (\(\ce{I}\)-131) is used in both the diagnosis and treatment of thyroid cancer. The thyroid will normally absorb iodine to produce the iodine-containing thyroid hormones. An overactive thyroid gland will absorb the radioactive material, which can then destroy excess thyroid tissue or any cancer of the thyroid. The material is sometimes used to image cancers in other parts of the body.
Technetium-99m is perhaps the most widely used radioisotope in diagnosis and treatment (the "m" stands for metastable). This isotope decays to \(\ce{Tc}\)-99 and a gamma emission of low intensity, making the radiation damage fairly negligible. The half-life is about six hours, so it will remain in the body for some time. \(\ce{Tc}\)-99m can be used to look at cardiac damage. The isotope flows in the bloodstream; if there is less blood flow in the heart, there will be less isotope concentrated in the heart muscle. Similar information can be obtained for blood flow in the brain.
Isotopes can be very useful in scans to locate cancer cells. The patient in the above image has multiple tumors that have spread (metastasized) from the main tumor. A radioisotope has been attached to antibodies that bind to specific cancer cells. The very dark spots in the armpits, neck, and groin represent areas where tumor cells exist.
Many other examples could be presented. There are presently over 25 different isotopes in use for diagnosis and treatment. A very partial list can be seen in the table below.
Radioisotopes Employed in Diagnosis and/or Treatment
Table \(1\): Radioisotopes Employed in Diagnosis and/or Treatment
Isotope Half-Life Application
\(\ce{Cr}\)-51 28 days Label red blood cells.
\(\ce{Fe}\)-59 446 days Study iron metabolism in spleen.
\(\ce{Xe}\)-133 5 days Study lung function.
\(\ce{Ho}\)-166 26 hours Cancer treatment. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/24%3A_Nuclear_Chemistry/24.10%3A_Effects_of_Radiation.txt |
Does your heart beat faster when you are scared? Do you have a tender feeling in your heart for that "special person"? Sorry to disappoint you, but that is actually simply a response generated by the brain. Science tells us that the seat of emotion is in the brain. Using PET scans and other techniques, scientists look at specific areas of the brain that process and store information dealing with strong emotions. They haven't localized the love site, but they're working on it.
PET Scans
One of the more interesting and useful approaches to radioisotope use in medicine is positron emission topography (PET), often referred to as a PET scan. This technique is especially useful in studying the processes in the brain. Many compounds do not enter the brain because of what is called the "blood-brain barrier", a filtering system to block material from being transported into brain tissue. This mechanism serves to protect the brain from a wide variety of harmful substances.
In order to get a good picture of what is happening in the brain, radiolabels are attached to different compounds that will enter the brain. Since the brain uses about $25\%$ of the glucose found in the body, this molecule is often labeled with a positron emitter such as $\ce{F}$-18 (half-life of 109.8 minutes) to study brain function in general. Other labels are attached to specific compounds that will localize in certain areas of the brain to look at specific structures.
The PET scanner detects gamma emissions from the collision of a positron with an electron (see figure below). As the positron is released from the nucleus of the atom, it will collide with an electron. This meeting of matter (electron) with antimatter (positron) results in annihilation of both particles, and the release of two gamma emissions that are $180^\text{o}$ apart from one another. The apparatus detects these gamma rays and stores the data in a computer. From this information, a detailed picture of the brain can be developed.
One useful application of PET scanning is in the diagnosis of Alzheimer's disease. This debilitating memory loss condition primarily occurs in elderly individuals. A protein known as beta-amyloid gradually forms deposits in the brain called plaque. Memory loss and impaired movement are the result of the plaque growth.
The compound known as Pittsburgh compound B is often used to identify areas of plaque in the brain. The radiolabel is $\ce{C}$-11 (half-life of 20.38 minutes).
The label attaches to plaque and can be observed using PET scans.
The computer translates the amount of isotope into a color scale, with red indicating a high level of radioactivity and yellow indicating somewhat less activity. We can see from the scans that the cognitively healthy individual shows the presence of very little plaque in the brain (see figure above). The individual with Alzheimer's demonstrates high concentrations of beta-amyloid in numerous areas of the brain.
Other studies have been done looking at brain function in drug addicts. One of the theories about drug addiction involves the amount of dopamine action in the brain (a chemical that is a part of the system to transport nerve impulses). Studies of dopamine action have been helpful in understanding addictive processes.
The figure above shows the binding of chemicals that attach to dopamine receptors. The non-addicted individuals have large numbers of receptors for dopamine. The addicted persons show less binding to these receptors, indicating that fewer receptors are present. Since dopamine is somehow linked with the sense of pleasure, these data may help to bring a better understanding to the biochemical processes in drug addiction.
Summary
• Positron Emission Topography (PET) scans make use of radioisotopes in medicine.
• As an example, PET scans can be used to detect Alzheimer's Disease and to help scientists learn about addiction. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/24%3A_Nuclear_Chemistry/24.12%3A_PET_Scans.txt |
Organic chemistry is a very vast and complex subject. There are millions of known organic compounds—far more than the number of inorganic compounds. The reason for this lies within the uniqueness of carbon's structure and bonding capabilities. Carbon has four valence electrons, and therefore makes four separate covalent bonds in compounds. Carbon has the ability to bond to itself repeatedly, making long chains of carbon atoms, as well as ringed structures. These bonds can be single, double, or triple covalent bonds. Carbon readily makes covalent bonds with other elements—primarily hydrogen, oxygen, nitrogen, halogens, and several other nonmetals.
25: Organic Chemistry
How many carbon-containing molecules are there? The current estimate is around 20 million different known organic compounds. Why the uncertainty? Every day, scientists are coming up with new compounds. Some of these materials are of interest for a research project, while others are destined to be developed for a commercial market. As soon as we think we know how many organic compounds exist, more are discovered and our number quickly becomes out of date.
Organic Chemistry
At one time in history, it was thought that only living things were capable of synthesizing the carbon-containing compounds present in cells. For that reason, the term organic was applied to those compounds. Eventually it was proved that carbon-containing compounds could be synthesized from inorganic substances, but the term "organic" has remained. Currently, organic compounds are defined as covalently bonded compounds containing carbon, excluding carbonates and oxides. By this definition, compounds such as carbon dioxide $\left( \ce{CO_2} \right)$ and sodium carbonate $\left( \ce{Na_2CO_3} \right)$ are considered to be inorganic. Organic chemistry is the study of all organic compounds.
Organic chemistry is a very vast and complex subject. There are millions of known organic compounds—far more than the number of inorganic compounds. The reason lies within the uniqueness of carbon's structure and bonding capabilities. Carbon has four valence electrons, and therefore makes four separate covalent bonds in compounds. Carbon has the ability to bond to itself repeatedly, making long chains of carbon atoms, as well as ringed structures. These bonds can be single, double, or triple covalent bonds. Carbon readily makes covalent bonds with other elements—primarily hydrogen, oxygen, nitrogen, halogens, and several other nonmetals. The figures below show ball-and-stick models of two of the many organic compounds.
The related field of biochemistry overlaps to some extent with organic chemistry. Biochemistry is the study of the chemistry of living systems. Many biochemical compounds are considered to be organic chemicals. Both of the molecules shown above are biochemical materials in terms of their use in the body, but organic chemicals in terms of their structure and chemical reactivity.
Summary
• Organic compounds are defined as covalently bonded compounds containing carbon, excluding carbonates and oxides. (By this definition, compounds such as carbon dioxide $\left( \ce{CO_2} \right)$ and sodium carbonate $\left( \ce{Na_2CO_3} \right)$ are considered to be inorganic.)
• Organic chemistry is the study of all organic compounds.
• Biochemistry is the study of the chemistry of living systems.
25.02: Straight-Chain Alkanes
As our country looks at the prospect of oil shortages in the future, we are searching for alternative transportation fuel sources. One very viable possibility is propane gas. Power and acceleration capacities for propane-powered vehicles are comparable to gasoline-powered vehicles, and fuel efficiency is greater. Propane has a higher octane rating than regular gasoline, leading to much longer engine life. When properly structured, propane engines can produce lower amounts of air pollution. We are seeing a growing use of propane in buses, trucks, and police cars. Maybe your next car will burn propane.
Straight-Chain Alkanes
Hydrocarbons
A hydrocarbon is an organic compound that is made up of only carbon and hydrogen. A hydrocarbon is the simplest kind of organic molecule, and is the basis for all other more complex organic compounds. Hydrocarbons can be divided into two broad categories: aliphatic and aromatic. Aliphatic hydrocarbons are hydrocarbons that do not contain the benzene group or a benzene ring. Aromatic hydrocarbons contain one or more benzene rings. In this section, we will focus on the aliphatic hydrocarbons.
Alkanes
An alkane is a hydrocarbon in which there are only single covalent bonds. The simplest alkane is methane, with the molecular formula $\ce{CH_4}$. The carbon is the central atom and makes four single covalent bonds to hydrogen atoms.
The next simplest alkane is called ethane $\left( \ce{C_2H_6} \right)$, and consists of two carbon atoms with a single covalent bond between them. Each carbon is then able to bond to three hydrogen atoms. The alkane series progresses from there, increasing the length of the carbon chain by one carbon at a time. Structural formulas for ethane, propane $\left( \ce{C_3H_8} \right)$, and butane $\left( \ce{H_4H_{10}} \right)$ are shown below.
These alkanes are called straight-chain alkanes because the carbon atoms are connected in one continuous chain with no branches. Naming and writing structural and molecular formulas for the straight-chain alkanes is straightforward. The name of each alkane consists of a prefix that specifies the number of carbon atoms and the ending -ane. The molecular formula follows the pattern of $\ce{C_{n}H_{2n_2}}$ where $n$ is the number of carbons in the chain. The table below lists the first ten members of the alkane series.
First Ten Members of the Alkane Series
Table $1$: First Ten Members of the Alkane Series
Name Molecular Formula Condensed Structural Formula Boiling Point $\left( ^\text{o} \text{C} \right)$
Methane $\ce{CH_4}$ $\ce{CH_4}$ -161.0
Ethane $\ce{C_2H_6}$ $\ce{CH_3CH_3}$ -88.5
Propane $\ce{C_3H_8}$ $\ce{CH_3CH_2CH_3}$ -42.0
Butane $\ce{C_4H_{10}}$ $\ce{CH_3CH_2CH_2CH_3}$ 0.5
Pentane $\ce{C_5H_{12}}$ $\ce{CH_3CH_2CH_2CH_2CH_3}$ 36.0
Hexane $\ce{C_6H_{14}}$ $\ce{CH_3CH_2CH_2CH_2CH_2CH_3}$ 68.7
Heptane $\ce{C_7H_{16}}$ $\ce{CH_3CH_2CH_2CH_2CH_2CH_2CH_3}$ 98.5
Octane $\ce{C_8H_{18}}$ $\ce{CH_3CH_2CH_2CH_2CH_2CH_2CH_2CH_3}$ 125.6
Nonane $\ce{C_9H_{20}}$ $\ce{CH_3CH_2CH_2CH_2CH_2CH_2CH_2CH_2CH_3}$ 150.7
Decane $\ce{C_{10}H_{22}}$ $\ce{CH_3CH_2CH_2CH_2CH_2CH_2CH_2CH_2CH_2CH_3}$ 174.1
Note that the table shows a variation of a structural formula called a condensed structural formula. In this formula, the covalent bonds are understood to exist between each carbon and the hydrogens associated with it, as well as between carbon atoms. This table also shows that the boiling points of the alkanes steadily increase as the length of the carbon chain increases. This is due to an increase in the strength of the intermolecular attractive forces and is a general feature of other organic molecules as well.
Summary
• A hydrocarbon is an organic compound that is made up of only carbon and hydrogen; it is the simplest kind of organic molecule.
• Aliphatic hydrocarbons are hydrocarbons that do not contain the benzene group or a benzene ring.
• Aromatic hydrocarbons contain one or more benzene rings.
• An alkane is a hydrocarbon in which there are only single covalent bonds. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/25%3A_Organic_Chemistry/25.01%3A_Organic_Chemistry.txt |
Tracing your family tree can be both fun and exciting. In order to do this correctly, it helps to know the exact names of your family members. Just a first, middle, or last name is not enough. A traceable family tree is one in which all relatives are carefully and precisely identified. After all, you would prefer that great-great-great-uncle to be royalty, and not a horse thief!
Branched Alkanes
Beginning with butane, there is an alternate structure possible that is not a straight chain. The structural formula below shows a structure with a three-carbon chain that has a \(\ce{-CH_3}\) group attached to the middle carbon.
The name of this molecule is 2-methylpropane. The molecular formula is still \(\ce{C_4H_{10}}\), which is the same formula as butane. A structural isomer is one of multiple molecules that have the same molecular formula, but different structural formulas. Butane and 2-methylpropane are structural isomers.
2-methylpropane is an example of a type of alkane called a branched alkane. The IUPAC system of nomenclature for branched alkanes follows a set of steps which will be applied to the example molecule below.
1. Find the longest continuous chain of carbon atoms in the molecule. This is called the parent chain. In the example, the longest chain is eight carbon atoms, and so the parent hydrocarbon is octane.
2. Number the carbon atoms in the parent chain. To do this, start at the end that will give the smallest numbers possible to the carbon atoms where the branches originate. In the example above, the branches are on carbons 3 and 5 when the chain is numbered left-to-right. If it were to be numbered right-to-left, the branches would be on carbons 4 and 6, so the left-to-right order is preferable.
3. The atoms attached to the parent chain (branches) are called substituents. A substituent that is a hydrocarbon is called an alkyl group. The names of alkyl groups use the same prefixes as the alkanes, but with a -yl suffix. So a 1-carbon alkyl group is a methyl group, a 2-carbon alkyl group is an ethyl group, and so on. The substituents are named by placing the number from the parent carbon chain in front of the name of the substituent. In the current example, we have 3-methyl and 4-ethyl substituents.
4. Use a prefix to indicate the appearance of more than one of the same substituent in the structural formula. Two of the same group is di-, three is tri-, four is tetra-, etc. For example, if methyl groups were attached to both carbons 2 and 3, that part of the name would be 2,3-dimethyl-. This rule does not apply to the structure pictured above.
5. Multiple different substituents are listed in alphabetical order. Ignore any of the prefixes from rule 4. In the current example, the 5-ethyl- comes before the 3-methyl-.
6. Commas are used to separate multiple numbers. Hyphens come between the number and the name of the substituent. The parent name comes immediately after the last substituent. There are no blank spaces in the name.
The correct name for the above structure, according to the IUPAC system, is 5-ethyl-3-methyloctane.
Summary
• A structural isomer is one of multiple molecules that have the same molecular formula, but different structural formulas.
• Nomenclature rules for branched hydrocarbons are given.
25.04: Alkenes and Alkynes
One of the most effective ways to cut metal is with an oxy-acetylene torch. Very high temperatures are obtained when acetylene burns in oxygen. Mixed 1:1 with oxygen, a temperature of over $3000^\text{o} \text{C}$ can be achieved. The amount of energy released is high—the net heat of combustion is $1300 \: \text{kJ/mol}$. Safety precautions need to be observed, as the gas is very explosive. The oxy-acetylene torch is one of the top tools for welding and cutting.
Alkenes and Alkynes
Alkenes
An alkene is a hydrocarbon with one or more carbon-carbon double covalent bonds. The simplest alkene is composed of two carbon atoms and is called ethene (shown below). Each carbon is bonded to two hydrogen atoms, in addition to the double bond between them.
The hybridization of each carbon atom is $sp^2$ with trigonal planar geometry. All the atoms of the molecule lay in one plane. Like the alkane series, the names of alkenes are based on the number of atoms in the parent chain. Naming follows the same rules as for alkanes, with the addition of using a number to indicate the location of the double bond. Propene $\left( \ce{C_3H_6} \right)$ has three carbons total, while butene $\left( \ce{C_4H_8} \right)$ has four. The general formula for alkenes with one double bond is $\ce{C_{n}H_{2n}}$. Alkenes are called unsaturated hydrocarbons. An unsaturated hydrocarbon is a hydrocarbon that contains less than the maximum number of hydrogen atoms that can possibly bond with the number of carbon atoms present.
The location of the carbon-carbon double bond can vary. The 4-carbon alkene generic name is butene. Since the double bond can be located in more than one place, we have 1-butene and 2-butene:
Molecules with multiple double bonds are also quite common. The formula below shows a four-carbon chain with double bonds between carbons 1 and 2, and between carbons 3 and 4. This molecule is called 1,3-butadiene.
Alkynes
An alkyne is a hydrocarbon with one or more carbon-carbon triple covalent bonds. The simplest alkyne consists of two carbon atoms and is called ethyne (common name: acetylene).
The ethyne molecule is linear, with $sp$ hybridization for each carbon atom. The general formula of alkynes with one triple bond is $\ce{C_{n}H_{2n-2}}$. Alkynes are also unsaturated hydrocarbons. Other alkynes exist, such as 2-pentyne:
Summary
• An alkene is a hydrocarbon with one or more carbon-carbon double covalent bonds.
• An alkyne is a hydrocarbon with one or more carbon-carbon triple covalent bonds.
• An unsaturated hydrocarbon is a hydrocarbon that contains less than the maximum number of hydrogen atoms that can possibly bond with the number of carbon atoms present.
• Structures of alkene and alkyne are given. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/25%3A_Organic_Chemistry/25.03%3A_Branched_Alkanes.txt |
As we delve into the complexities of organic chemistry, we will see how molecular shape affects reactions. One common reaction for alkenes is the addition of hydrogen across the double bond to form the corresponding alkane. Because of the geometry of the reaction, the different 2-butene shapes have different heats of reaction. These differences are important, both from a theoretical standpoint, as well as from the point of view of industrial applications. Greater energy requirements mean a higher cost and a more expensive product.
Isomers
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.
Structural Isomers
A structural isomer is one in which two or more organic compounds have the same molecular formulas, but different structures. The two pentane molecules below differ only in the location of the methyl group.
Alkenes can also demonstrate structural isomerism. In alkenes, there are multiple structural isomers based on where in the chain the double bond occurs. The condensed structural formulas of 1-butene and 2-butene show this.
The number in the name of the alkene refers to the lowest numbered carbon in the chain that is part of the double bond.
Geometric 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
• Isomers are molecules with the same molecular formulas, but different arrangements of atoms.
• A structural isomer is one in which two or more organic compounds have the same molecular formulas, but different structures.
• Geometric isomers are isomers in which the order of atom bonding is the same, but the arrangement of atoms in space is different.
• Examples of alkane and alkene isomers are given. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/25%3A_Organic_Chemistry/25.05%3A_Isomers.txt |
Although cyclohexane can be isolated from petroleum products, a major source of this chemical is the hydrogenation of benzene. Much of the cyclohexane produced is used to manufacture intermediates for the production of nylon. Items such as nylon balloons no doubt had their start in a chemical plant, where hydrogen gas and benzene were reacted at high temperatures to form cyclohexane. This cycloalkane then undergoes nitration to begin the process of forming the long strands of nylon that can be made into balloons, ropes, clothing, and many other useful products.
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 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 left is called the chair conformation, while the one on the right is called the boat 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.
Summary
• 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.)
• 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.
• Names and structures of typical cyclic hydrocarbons are given. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/25%3A_Organic_Chemistry/25.06%3A_Cyclic_Hydrocarbons.txt |
Friedrich Kekulé was a German chemist in the 1800s. Supposedly, he was thinking about the structure of benzene one night as he fell asleep. While asleep, he dreamed of a snake eating its own tail. He used this idea to propose the cyclic structure of benzene. Whether or not he actually had this dream has been debated ever since. Whatever really happened, the tale persists today.
Aromatic Hydrocarbons
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 that 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. Benzene rings are common in a great number of natural substances and biomolecules. The figure below shows the structural formulas for vanilla and naphthalene. Naphthalene is a chemical which is commonly used in mothballs.
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 one another 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
• Benzene is the parent compound of the large family of organic compounds known as aromatic compounds.
• 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.
• Nomenclature for benzene compounds is dictated. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/25%3A_Organic_Chemistry/25.07%3A_Aromatic_Hydrocarbons.txt |
A class of simple alkyl halides called chlorofluorocarbons (CFCs) was once very widely used in aerosol sprays and as refrigerants. One such example is \(\ce{CCl_3F}\), or trichlorofluoromethane. Unfortunately, CFCs are harmful to the ozone layer of our upper atmosphere. Ozone is critical in limiting the amount of damaging ultraviolet radiation that reaches the Earth. CFCs react with the ozone and damage it, leaving the earth less protected.
Beginning in the late 1970s, ozone depletion was recognized as a significant environmental issue. The most dramatic decrease in ozone occurs seasonally over the continent of Antarctica. The size and duration of the ozone hole steadily increased, with the largest hole recorded in 2006. Fortunately, most countries have recognized the danger of CFCs and dramatically curtailed their use in recent years. It is hoped that ozone depletion will slow, and that the ozone layer may eventually be restored to its earlier levels.
Alkyl Halides
An alkyl halide is an organic compound in which one or more halogen atoms are substituted for one or more hydrogen atoms in a hydrocarbon. 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. 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.
Uses of Alkyl Halides
Alkyl halides are often used as synthetic intermediates in the laboratory. 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 has led to a widespread decrease in applications for these materials. 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.
Summary
• An alkyl halide is an organic compound in which one or more halogen atoms are substituted for one or more hydrogen atoms in a hydrocarbon.
• Alkyl halides are often used as synthetic intermediates in the laboratory. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/25%3A_Organic_Chemistry/25.08%3A_Alkyl_Halides.txt |
Water freezes at $0^\text{o} \text{C}$, which creates problems for cars in the winter. The water in the engine will freeze and crack the engine block. To prevent this, antifreeze is added to lower the freezing point of the liquid. The most common antifreeze is an alcohol known as propylene glycol. It has largely replaced the much more toxic ethylene glycol. Methanol can also be used as an antifreeze, mainly in windshield wiper fluid.
Alcohols
An alcohol is an organic compound that contains one or more hydroxyl $\left( \ce{-OH} \right)$ groups. The general formula for alcohols is $\ce{R-OH}$. Do not confuse alcohols with inorganic bases that contain the hydroxide ion $\left( \ce{OH^-} \right)$. The $\ce{-OH}$ group in an alcohol is covalently bonded to a carbon atom and does not ionize in solution. The steps for naming alcohols are listed below.
1. Name the parent compound by finding the longest continuous carbon atom chain that also contains the hydroxyl group. If there is one hydroxyl group in the molecule, change the final -e in the name of the alkane to -ol. If there is more than one hydroxyl group, use the full name of the alkane and add a suffix to indicate the number of hydroxyl groups. For example, two hydroxyl groups is -diol, three is -triol, etc.
2. Number the carbon chain in a way that makes the sum of the hydroxyl numbers as low as possible.
3. Add the numerical prefix into the name before the name of the alcohol.
4. Separate numbers with commas and separate numbers from names or prefixes with a hyphen. There are no spaces in the name.
Following are three examples of alcohols and their IUPAC names.
Aliphatic alcohols can be classified according to the number of $\ce{R}$ groups attached to the carbon with the hydroxyl group. If one $\ce{R}$ group is attached to that carbon, the alcohol is a primary alcohol. If two $\ce{R}$ groups are attached, the alcohol is a secondary alcohol. If three $\ce{R}$ groups are attached, the alcohol is a tertiary alcohol. Shown below is an example of each. The primary alcohol is 1-propanol, the secondary alcohol is 2-butanol, and the tertiary alcohol is 2-methyl-2-propanol.
Properties of Alcohols
The smallest and lightest alcohols (methanol, ethanol, propanol) are completely soluble in water in all proportions. In a solution, the hydroxyl groups of alcohol molecules and the water molecules form hydrogen bonds with each other, resulting in complete miscibility. However, as the length of the carbon chain increases, the solubility decreases. The solubility of 1-butanol is $7.4 \: \text{g}$ per $100 \: \text{g}$ of water, while that of 1-pentanol is $2.7 \: \text{g}$ per $100 \: \text{g}$ of water, and 1-octanol is $0.06 \: \text{g}$ per $100 \: \text{g}$ of water. The carbon chain portion of the larger alcohol molecule is nonpolar, and leads to the decreased solubility of the overall compound.
The presence of hydrogen bonds in alcohols also explains the relatively high boiling points of alcohols compared to alkanes of similar molar mass (see table below).
Compound Formula Molar Mass $\left( \text{g/mol} \right)$ Boiling Point $\left( ^\text{o} \text{C} \right)$
Table $1$: Boiling Point Comparison of Alkanes and Alcohols
ethane $\ce{CH_3CH_3}$ 32 -88
methanol $\ce{CH_3OH}$ 30 64.7
propane $\ce{CH_3CH_2CH_3}$ 44 -42.1
ethanol $\ce{CH_3CH_2OH}$ 46 78.3
Only weak London dispersion forces hold molecules of nonpolar alkanes together in the liquid phase. Consequently, less energy is required to break these molecules away from the surface of the liquid and turn them into a vapor. The stronger hydrogen bonding between alcohol molecules means that more energy is required to convert the liquid to vapor, and boiling points are therefore high.
Summary
• An alcohol is an organic compound that contains one or more hydroxyl $\left( \ce{-OH} \right)$ groups.
• The lightest alcohols (methanol, ethanol, propanol) are completely soluble in water in all proportions; as the length of the carbon chain increases, the solubility decreases.
• The presence of hydrogen bonds in alcohols also explains the relatively high boiling points of alcohols compared to alkanes of similar molar mass. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/25%3A_Organic_Chemistry/25.09%3A_Alcohols.txt |
Race car drivers are always looking for that "edge" that will make their car (legally) faster than their competitors' cars. One way to get more speed is to burn the gasoline in the car engine more efficiently. Methyl-t-butyl ether (MTBE) has been used for this purpose, but is being discontinued due to concerns about the contamination of drinking water by leaking fuel tanks that contain this compound.
Ethers
An ether is an organic compound in which two hydrocarbon groups are bonded to the same atom of oxygen. An ether is represented by the general formula \(\ce{R-O-R'}\). The \(\ce{R'}\) in the formula means that the hydrocarbon group can be the same as \(\ce{R}\), or it can be different. The steps for naming ethers are listed below.
1. The parent compound is given by the word ether, which appears at the end of the name.
2. The names of each alkyl group come before the word ether. If the two alkyl groups are the same, the prefix di- is used. If the two alkyl groups are different, they are listed in alphabetical order.
3. Spaces are left between the names of the alkyl groups (if different) and before the word ether.
Shown below are two examples of ethers with their IUPAC names.
Properties of Ethers
Like alcohols, ethers are also quite soluble in water. The lone-pair electrons on the oxygen atom of the ether can form a hydrogen bond with the hydrogen atoms of water molecules. As with alcohols, the solubility is greater for ethers that have shorter length \(\ce{R}\) groups. The boiling points of ethers are much lower than the boiling points of alcohols. Ether molecules do not have hydrogen atoms that are covalently bonded to a highly electronegative atom, and so ether molecules cannot form hydrogen bonds with each other. The weaker intermolecular force acting between ether molecules results in boiling points which are much closer to alkanes of similar molar mass than to alcohols.
The anesthetic effects of ethers were first discovered in the 1840s. Diethyl ether was used as a general anesthetic for patients undergoing surgery for many years. However, ethers are very flammable and have undesirable side effects, such as nausea and vomiting. Safer alternatives to ethers are now used in anesthesia, and the primary use of ethers today is as a solvent for other organic compounds.
Summary
• An ether is an organic compound in which two hydrocarbon groups are bonded to the same atom of oxygen.
• IUPAC naming rules for ethers are dictated.
• Ethers are quite soluble in water.
• The primary use of ethers today is as a solvent for other organic compounds.
25.11: Aldehydes and Ketones
There's nothing like the smell of a fresh cinnamon roll; the taste is even better. What is that delicious taste owed to? This flavoring comes from the bark of a tree (actually, several different kinds of trees). One of the major compounds responsible for the taste and odor of cinnamon is cinnamaldehyde. Cinnamon has been widely used through centuries to treat a number of different disorders. In ancient times, doctors believed it could cure snakebite poisoning, freckles, and the common cold. Today, there are several research studies being carried out on the health benefits of cinnamon. So, enjoy that cinnamon roll—it just might have a health benefit or two.
Aldehydes and Ketones
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. 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 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 so 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.
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 by safer alternatives for preservation purposes. 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 once used in most nail polish removers, but has largely been replaced by other solvents.
Summary
• Aldehydes and ketones are two related categories of organic compounds that both contain the carbonyl group.
• 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.
• Aldehydes and ketones generally have lower boiling points than those of alcohols. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/25%3A_Organic_Chemistry/25.10%3A_Ethers.txt |
Vinegar (dilute acetic acid) is available in every grocery store in the country. The large one-gallon jugs can be found in households everywhere. One magazine published an article listing 150 ways to use vinegar, from getting dirt off of your computer to cleaning window blinds. A mixture of vinegar and baking soda is an excellent way to clean sink drains. You can also clean your carpets and your piano keys. Not bad for something that may have struck you as merely an ingredient in salad dressing!
Carboxylic Acids
Organic acids such as acetic acid all contain a functional group called a carboxyl group.
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:
1. Name the parent compound by finding the longest continuous chain that contains the carboxyl group.
2. Change the -e at the end of the name of the alkane to -oic acid.
Properties of Carboxylic Acids
Carboxylic acids are all weak acids. In aqueous solution, the $\ce{O-H}$ bond of the hydroxyl group can break, yielding a negative carboxylate ion and the hydrogen ion.
The smaller members of the aliphatic carboxylic acid series are colorless, volatile liquids with strong odors. Ethanoic acid is commonly known as acetic acid; common household vinegar is a $5\%$ solution of acetic acid. Larger carboxylic acids are solids with low melting points. There are numerous aromatic carboxylic acids, which are all crystalline solids. Carboxylic acids can form intermolecular hydrogen bonds, and thus have relatively high melting and boiling points compared to other organic compounds that cannot hydrogen bond. Carboxylic acids with shorter carbon chains are very soluble in water, while those with longer carbon chains are less soluble.
Many carboxylic acids occur naturally in plants and animals. Citrus fruits such as oranges and lemons contain citric acid.
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.
Summary
• Organic acids, such as acetic acid, all contain a functional group called a carboxyl group.
• 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.
25.13: Esters
Perfumes contain ingredients from a number of sources. Musk is obtained from animals, but the vast majority of perfume components are obtained from plants. Approximately 2,000 plant species have been used as sources for perfume materials. The needed chemicals are extracted using solvent extraction or distillation. The oils are diluted with ethanol to varying degrees, depending on the price of the finished product—the less ethanol present (meaning there is a higher percentage of the active scent ingredients), the more expensive the perfume.
Esters
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. The steps for naming esters, along with two examples, are shown below.
1. Identify and name the alkyl group $\left( \ce{R'} \right)$ that has replaced the hydrogen of the hydroxyl group. This is the first part of the ester name.
2. Name the carboxylic acid portion of the molecule $\left( \ce{R-COO} \right)$, but change the ending of the name from -oic acid to -oate. This is the second part of the ester name.
3. Leave a space between the alkyl group name and the name of the carboxylic acid derivative.
Esters have lower boiling points than the carboxylic acids from which they were derived because they cannot form hydrogen bonds with each other. This is because there are no hydrogen atoms bonded to oxygen atoms, as in the acid. Esters with shorter carbon chains are soluble in water, while those with longer chains are less soluble.
Esters are very commonly found in plants, and are responsible for many distinct odors and flavors. For example, methyl salicylate has the odor and flavor of oil of wintergreen, while propyl ethanoate has that of a pear.
25.14: Amines
For many people, sleeping problems have to do with an amine that the brain makes called melatonin. This compound affects the sleep-wake cycle and is affected by sunlight. During the winter, the daily cycle of melatonin production may be affected by less sunlight hours. If this cycle is changed, the person may have trouble sleeping. Melatonin supplements are available in pharmacies and health food stores, and may be of help for those with sleeping problems.
Amines
An amine is an organic compound that can be considered to be a derivative of ammonia $\left( \ce{NH_3} \right)$. The general structure of an amine can be abbreviated as $\ce{R-NH_2}$, where $\ce{R}$ is a carbon chain. However, similar to alcohols, amines can be primary, secondary, or tertiary.
The nitrogen atom of a primary amine is bonded to two hydrogen atoms and one carbon. The nitrogen atom of a secondary amine is bonded to one hydrogen and two carbons. The nitrogen atom of a tertiary amine is bonded to three carbon atoms. Amines are typically named by a common system, rather than by IUPAC guidelines. The common system for naming amines, along with several examples, is shown below.
1. Name the alkyl groups that are attached to the nitrogen atom of the amine. If there is more than one different alkyl group, put them in alphabetical order. If there are two or three of the same alkyl group, use the di- or tri- prefix.
2. Follow the alkyl group name with the suffix -amine, with no spaces.
Properties of Amines
Amines are weak bases due to the presence of a lone pair of electrons on the nitrogen atom. This lone pair can attract the hydrogen atom from a water molecule, causing the bond between it and the oxygen atom to break. The resultant products are the conjugate acid of the amine and the hydroxide ion.
Amines are capable of hydrogen bonding, though their boiling points are generally a bit lower than the corresponding alcohol. 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. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/25%3A_Organic_Chemistry/25.12%3A_Carboxylic_Acids.txt |
Organic compounds can be used for a wide variety of applications. One organic compound is halothane, an alkyl halide. This material is used as a general anesthetic for surgical procedures. When it was first developed in 1956, it was very popular, but has had limited use more recently because of toxicity issues—specifically liver damage.
Substitution Reactions
A substitution reaction is a reaction in which one or more atoms replace another atom or group of atoms in a molecule. 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:
$\ce{CH_4} \left( g \right) + \ce{Cl_2} \left( g \right) \overset{\text{UV light}}{\rightarrow} \ce{CH_3Cl} \left( g \right) + \ce{HCl} \left( g \right)\nonumber$
The reaction produces chloromethane and hydrogen chloride. When the mixture is allowed to react for longer periods of time further, substitution reactions may occur.
$\ce{CH_3Cl} \left( g \right) + \ce{Cl_2} \left( g \right) \overset{\text{UV light}}{\rightarrow} \ce{CH_2Cl_2} \left( g \right) + \ce{HCl} \left( g \right)\nonumber$
The product above is dichloromethane. Further substitution produces 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 hydrofluoric acid $\left( \ce{HF} \right)$, because the fluorine atom bonds preferentially to the carbon.
$\ce{CCl_4} \left( g \right) + \ce{HF} \left( g \right) \overset{\ce{FeBr_3}}{\rightarrow} \ce{CCl_3F} \left( g \right) + \ce{HCl} \left( g \right)\nonumber$
The fluorine atom substitutes for a chlorine atom in the reaction.
Benzene is a fairly stable and unreactive molecule due to the electron delocalization around the six-membered ring. The treatment of benzene with a halogen in the presence of an iron catalyst causes the substitution of a halogen atom for a hydrogen atom. The resulting molecule is called an aryl halide.
$\ce{C_6H_6} \left( l \right) + \ce{Br_2} \left( l \right) \overset{\ce{FeBr_3}}{\rightarrow} \ce{C_6H_5Br} \left( l \right) + \ce{HBr} \left( g \right)\nonumber$
Alkyl groups can be introduced onto a benzene ring by the reaction of benzene with an alkyl halide, using aluminum chloride as the catalyst. In the reaction below, benzene reacts with chloroethane to produce ethylbenzene.
$\ce{C_6H_6} \left( l \right) + \ce{CH_3CH_2Cl} \left( g \right) \overset{\ce{AlCl_3}}{\rightarrow} \ce{C_6H_5CH_2CH_3} \left( l \right) + \ce{HCl} \left( g \right)\nonumber$
The reaction of an alkyl halide with an inorganic hydroxide base at elevated temperature produces an alcohol. The molecular reaction to produce methanol from iodomethane and sodium hydroxide is shown below.
$\ce{CH_3I} \left( l \right) + \ce{NaOH} \left( aq \right) \overset{100^\text{o} \text{C}}{\rightarrow} \ce{CH_3OH} \left( l \right) + \ce{NaI} \left( aq \right)\nonumber$
Summary
• A substitution reaction is a reaction in which one or more atoms replace another atom or group of atoms in a molecule.
• Examples of substitution reactions are given.
25.16: Addition Reactions
There is some debate these days about the benefits of using margarine versus butter on your toast (or pancakes, or muffin). Margarine is less expensive than butter, and is lower in fat and cholesterol. Margarine is made from vegetable fats using hydrogenation to reduce the double bonds in the fatty acids. Hydrogen gas is bubbled through the liquid oil and reacts with the carbon-carbon double bonds present in the long-chain fatty acids. The product is less likely to spoil than butter.
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 be thought of as adding a molecule across the double bond or triple bond of an alkene or alkyne. Addition reactions are useful ways to introduce a new functional group into an organic molecule.
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. The reaction is typically performed with the use of a platinum catalyst. Ethene reacts with hydrogen to form ethane.
$\ce{CH_2=CH_2} \left( g \right) + \ce{H_2} \left( g \right) \overset{\ce{Pt}}{\rightarrow} \ce{CH_3CH_3} \left( g \right)\nonumber$
Alkyl halides can be produced from an alkene by the addition of either the elemental halogen or the hydrogen halide. When the reactant is the halogen, the product is a disubstituted alkyl halide, as in the addition of bromine to ethene:
$\ce{CH_2=CH_2} \left( g \right) + \ce{Br_2} \left( l \right) \rightarrow \ce{CH_2BrCH_2Br} \left( g \right) \: \text{(1,2-dibromoethane)}\nonumber$
The addition of bromine to an unknown organic compound is indeed a test for saturation in the compound. Bromine has a distinctive brownish-orange color, while most bromoalkanes are colorless. When bromine is slowly added to the compound, the orange color will fade if it undergoes the addition reaction to the hydrocarbon. If the orange color remains, then the original compound was already saturated, and no reaction occurred.
A monosubstituted alkyl halide can be produced by the addition of a hydrogen halide to an alkene. Shown below is the formation of chloroethane.
$\ce{CH_2=CH_2} \left( g \right) + \ce{HCl} \left( g \right) \rightarrow \ce{CH_3CH_2Cl} \left( g \right)\nonumber$
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.
$\ce{CH_2=CH_2} + \ce{H_2O} \rightarrow \ce{CH_3CH_2OH}\nonumber$
Under modest reaction conditions, benzene resists addition reactions because adding a molecule across a double bond in a benzene ring disrupts the ring of delocalized electrons; this greatly destabilizes the molecule. However, under conditions of high temperature and pressure, and with an appropriate catalyst, benzene will slowly react with three molecules of hydrogen to produce cyclohexane.
$\ce{C_6H_6} + 3 \ce{H_2} \overset{\ce{Pt}}{\rightarrow} \ce{C_6H_{12}}\nonumber$
Summary
• An addition reaction is a reaction in which an atom or molecule is added to an unsaturated molecule, making a single product.
• Hydrogenation is a reaction that occurs when molecular hydrogen is added to an alkene to produce an alkane; the reaction is typically performed with the use of a platinum catalyst.
• A hydration reaction is a reaction in which water is added to an alkene. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/25%3A_Organic_Chemistry/25.15%3A_Substitution_Reactions.txt |
Benzoic acid is widely used as a food preservative, either as the carboxylic acid or as the sodium benzoate salt. This compound is most effective when added to acidic foods such as fruit juices and soft drinks. The major industrial source of benzoic acid is the partial oxidation of toluene with oxygen. The process is inexpensive and environmentally benign.
Oxidation Reactions
Oxidation can be defined as the addition of oxygen to a molecule, or the removal of hydrogen from a molecule. 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:
$\text{oxidation:} \: \: \ce{CH_3CH_3} \overset{\ce{-H_2}}{\rightarrow} \ce{CH_2=CH_2} \overset{\ce{-H_2}}{\rightarrow} \ce{CH \equiv CH}\nonumber$
The reactions are reversible, and so an alkyne can be reduced first to an alkene, and then to an alkane:
$\text{reduction:} \: \: \ce{CH \equiv CH} \overset{\ce{_H_2}}{\rightarrow} \ce{CH_2=CH_2} \overset{\ce{+H_2}}{\rightarrow} \ce{CH_3CH_3}\nonumber$
The alkane is the most reduced form of a hydrocarbon, while the alkyne is the most oxidized form.
Oxidation reactions in organic chemistry often involve the addition of oxygen to a compound, which changes the particular functional group of that compound. The following sequence shows how methane can be oxidized first to methanol, then to methanal, then to methanoic acid, and finally to carbon dioxide.
$\begin{array}{ccccccccc} \ce{CH_4} & \overset{\text{gain of oxygen}}{\longrightarrow} & \ce{CH_3OH} & \overset{\text{loss of hydrogen}}{\longrightarrow} & \ce{CH_2O} & \overset{\text{gain of oxygen}}{\longrightarrow} & \ce{HCOOH} & \overset{\text{loss of hydrogen}}{\longrightarrow} & \ce{CO_2} \ \text{methane} & & \text{methanol} & & \text{methanal} & & \text{methanoic acid} & & \text{carbon dioxide} \end{array}\nonumber$
Each step in the process is either a gain of oxygen or a loss of hydrogen. Each step also releases energy, which explains why the complete combustion of alkanes to carbon dioxide is an extremely exothermic reaction.
The oxidation of an alcohol can produce either an aldehyde or a ketone. Ethanol can be oxidized in the laboratory through a heating process combined with the addition of an oxidizing agent such as the dichromate ion, which catalyzes the reaction in an acidic solution. The reaction produces the aldehyde ethanal (acetaldehyde).
$\ce{CH_3CH_2OH} \underset{\ce{H^+}}{\overset{\ce{Cr_2O_7^{2-}}}{\rightarrow}} \ce{CH_3CHO}\nonumber$
When the alcohol to be oxidized is a secondary alcohol, the oxidation product is a ketone rather than an aldehyde. The oxidation of the simplest secondary alcohol, 2-propanol, yields propanone.
$\ce{CH_3CHOCH_3} \underset{\ce{H^+}}{\overset{\ce{Cr_2O_7^{2-}}}{\rightarrow}} \ce{CH_3COCH_3}\nonumber$
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.
When a primary alcohol is oxidized to an aldehyde, the reaction is difficult to stop because the aldehyde is easily oxidized further to the corresponding carboxylic acid. The oxidation of ethanal produces ethanoic (acetic) acid.
$\ce{CH_3CHO} \underset{\ce{H^+}}{\overset{\ce{Cr_2O_7^{2-}}}{\rightarrow}} \ce{CH_3COOH}\nonumber$
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, and so the ketone cannot be converted to a carboxylic acid.
Summary
• Oxidation can be defined as the addition of oxygen to a molecule, or the removal of hydrogen from a molecule.
• Primary alcohols can be oxidized to aldehydes, and then further oxidized to carboxylic acids.
• When the alcohol to be oxidized is a secondary alcohol, the oxidation product is a ketone rather than an aldehyde.
• Tertiary alcohols cannot be oxidized. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/25%3A_Organic_Chemistry/25.17%3A_Oxidation_Reactions.txt |
Vegetable oils are being explored for a variety of uses in which they could replace petroleum products. One such application is in the field of lubricants. Every moving part in machinery (such as engine pistons) needs lubrication to decrease friction and prolong the life of the equipment. Petroleum products serve this purpose now, but are not good for the environment. New techniques for making specialized esters from vegetable oil are being explored, with the purpose of making the compounds more stable and more useful as lubricants.
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.
This reaction forms a molecule called a dipeptide and the carbon-nitrogen covalent bond is called a peptide bond. When repeated numerous times, a lengthy molecule called a protein is eventually produced.
Esterification
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 is 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.
The esterification reaction is reversible. When an ester is heated in the presence of a strong base such as sodium hydroxide, the ester breaks down. The products are an alcohol and the conjugate base of the carboxylic acid as a salt.
$\begin{array}{ccccccc} \ce{CH_3COOCH_2CH_3} & + & \ce{NaOH} & \rightarrow & \ce{CH_3COO^- Na^+} & + & \ce{CH_3CH_2OH} \ \text{ethyl ethanoate} & & & & \text{sodium acetate} & & \text{ethanol} \end{array}\nonumber$
The sodium hydroxide is not acting as a catalyst, but is consumed in the reaction.
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.
$\begin{array}{ccccccc} \ce{C_{17}H_{35}COOC_2H_5} & + & \ce{NaOH} & \rightarrow & \ce{C_{17}H_{35}COO^- Na^+} & + & \ce{C_2H_5OH} \ \text{ethyl stearate} & & & & \text{sodium stearate (soap)} & & \end{array}\nonumber$
Summary
• A condensation reaction is a reaction in which two molecules combine to form a single molecule.
• An esterification is a condensation reaction in which an ester is formed from an alcohol and a carboxylic acid.
• Saponification describes the alkaline hydrolysis reaction of an ester. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/25%3A_Organic_Chemistry/25.18%3A_Condensation_Reactions.txt |
We reap the benefits of using Styrofoam containers, but don't often consider where they end up. Styrofoam materials do not break down quickly under exposure to the elements. When buried in a landfill, styrofoam will remain intact for a long time. The good news is that there is not a lot of this pollutant found in landfills (maybe about $0.5\%$ by weight of the total mass of garbage). There is no good way to recycle Styrofoam at present, but in the future, a creative scientist may change that.
Polymerization - Addition Polymers
Polymers are very different from 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 in a repeating pattern. The small molecules which make up the polymer are called monomers. Polymers generally form either from an addition reaction or a condensation reaction.
Addition Polymers
An addition polymer is a polymer formed by chain addition reactions between monomers that contain a double bond. Molecules of ethene can polymerize with each other under the right conditions to form the polymer called polyethylene.
$n \ce{CH_2=CH_2} \rightarrow \ce{-(CH_2CH_2)}_n-\nonumber$
The letter $n$ stands for the number of monomers that are joined in repeated fashion to make the polymer, and can have a value in the hundreds or even thousands.
The reactions above show the basic steps to form an addition polymer:
1. Initiation - a free radical initiator $\left( \ce{X}^* \right)$ attacks the carbon-carbon double bond (first step above). The initiator can be something like hydrogen peroxide. This material can easily split to form two species with a free electron attached to each: $\ce{H-O-O-H} \rightarrow 2 \ce{H-O} \cdot$. This free radical attacks a carbon-carbon double bond. One of the pi electrons forms a single bond with the initiator while the other pi electron forms a new free radical on the carbon atom.
2. Propagation - the new free radical compound interacts with another alkane, continuing the process of chain growth (second step above).
3. Termination occurs whenever two free radicals come in contact with one another (not shown). The two free electrons form a covalent bond and the free radical on each molecule no longer exists.
Polyethylene can have different properties depending on the length of the polymer chains, and on how efficiently they pack together. Some common products made from different forms of polyethylene include plastic bottles, plastic bags, and harder plastic objects such as milk crates.
Several other kinds of unsaturated monomers can be polymerized, and are components in common household products. Polypropylene is stiffer than polyethylene, and is in plastic utensils and some other types of containers.
Polystyrene is used in insulation and in molded items such as coffee cups.
Polyvinyl chloride (PVC) is extensively used for plumbing pipes.
Polyisoprene is a polymer of isoprene and is better known as rubber. It is produced naturally by rubber trees, but several variants have been developed which demonstrate improvements on the properties of natural rubber.
Summary
• A polymer is a large molecule formed of many smaller molecules covalently bonded in a repeating pattern; they are the primary components of all sorts of plastics and related compounds.
• The small molecules which make up polymers are called monomers.
• Polymers generally form either from an addition reaction or a condensation reaction.
• An addition polymer is a polymer formed by chain addition reactions between monomers that contain a double bond.
• The basic steps to form an addition polymer are: (1) initiation, (2) propagation, (3) termination
25.20: Polymerization - Condensation Polymers
Animal intestines and silk were used for all guitar strings for centuries, until modern technology and changes in musical taste brought about significant changes. There are two major types of guitar strings in use today. Steel strings (first developed around 1900) are found on acoustic and electric guitars. They have a bright, crisp sound that lends itself well to diverse music such as jazz, rock 'n' roll, and bluegrass. Nylon strings are a more recent development. During World War II, the silk and animal products needed to manufacture steel guitar strings were not available. Nylon quickly proved to be a more-than-adequate substitute. Now nylon strings are found on all classical guitars. Their sound is somewhat softer than the steel strings, making the tone quality well-suited for the classical genre of music.
Polymerization - Condensation Polymers
A condensation polymer is a polymer formed by condensation reactions. Monomers of condensation polymers must contain two functional groups so that each monomer can link up with two other monomers. One type of condensation polymer is called a polyamide. An amide is characterized by the functional group shown below wherein the carbon of a carbonyl group is bonded to the nitrogen of an amine.
One pair of monomers that can form a polyamide is that of adipic acid and hexanediamine. Adipic acid is a carboxylic acid with two carboxyl groups on either end of the molecule. Hexanediamine has amino groups on either end of a six-carbon chain. When these molecules react with each other, a molecule of water is eliminated, classifying it as a condensation reaction (see figure below).
The polymer that results from the repetition of the condensation reaction is a polyamide called nylon-66. Nylon-66 was first invented in 1935 and has been used in all sorts of products. Polyamides, including Nylon-66, are commonly found in fibers and clothing, cooking utensils, fishing line, and carpeting—among many other applications.
Polyester is another common type of condensation polymer. Recall that esters are formed from the reaction of an alcohol with a carboxylic acid. When both the acid and alcohol have two functional groups, the ester is capable of being polymerized. One such polyester is called polyethylene terephthalate (PET) and is formed from the reaction of ethylene glycol with terephthalic acid. The structure of PET is shown below.
PET is used in tires, photographic film, food packaging, and clothing. Polyester fabric is used in permanent-press clothing. Its resistance to wrinkling comes from the cross-linking of the polymer strands.
Summary
• A condensation polymer is a polymer formed by condensation reactions.
• Polyamides and polyesters are common types of condensation polymers. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/25%3A_Organic_Chemistry/25.19%3A_Polymerization_-_Addition_Polymers.txt |
The brain is a marvelous organ. And it's a hungry one, too. The major fuel for the brain is glucose—a carbohydrate. The average adult brain represents about \(2\%\) of our body's weight, but uses \(25\%\) of the glucose in the body. Moreover, specific areas of the brain use glucose at different rates. If you are concentrating hard (taking a test, for example), certain parts of the brain need a lot of extra glucose, while other parts of the brain only use their normal amount.
Monosaccharides
Some foods that are high in carbohydrates include bread, pasta, and potatoes. Because carbohydrates are easily digested, athletes often rely on carbohydrate-rich foods to enable a high level of performance.
The term carbohydrate comes from the fact that the majority contain carbon, hydrogen, and oxygen in a ratio of 1:2:1, making for an empirical formula of \(\ce{CH_2O}\). This is somewhat misleading, because the molecules are not actually hydrates of carbon at all. Carbohydrates are monomers and polymers of aldehydes and ketones that have multiple hydroxyl groups attached.
Carbohydrates are the most abundant source of energy found in most foods. The simplest carbohydrates, also called simple sugars, are plentiful in fruits. A monosaccharide is a carbohydrate consisting of one sugar unit. Common examples of simple sugars or monosaccharides are glucose and fructose. Both of these monosaccharides are referred to as hexoses, since they have six carbons. Glucose is abundant in many plant sources, and makes up sweeteners such as corn sugar and grape sugar. Fructose is found in many fruits, as well as in honey. These sugars are structural isomers of one another, with the difference being that glucose contains an aldehyde functional group, whereas fructose contains a ketone functional group.
Glucose and fructose are both very soluble in water. In aqueous solution, the predominant forms are not the straight-chain structure shown above. Rather, they adopt a cyclic structure (see figure below). Glucose is six membered ring, while fructose is a five-membered ring. Both rings contain an oxygen atom.
Another important group of monosaccharides are the pentoses, containing five carbons in the chain. Ribose and deoxyribose are two pentoses that are components of the structures of DNA and RNA.
Summary
• Carbohydrates are monomers and polymers of aldehydes and ketones that have multiple hydroxyl groups attached.
• A monosaccharide is a carbohydrate consisting of one sugar unit.
• Common examples of simple sugars or monosaccharides are glucose and fructose.
• Another important group of monosaccharides are the pentoses, containing five carbons in the chain; DNA and RNA are partly comprised of pentoses.
26.02: Disaccharides
Milk is one of the basic foods needed for good nutrition, especially for growing children. It contains vitamins and minerals necessary for healthy development. Unfortunately, milk and other dairy products also contain lactose, a carbohydrate that can make some people very ill. Lactose intolerance is a condition in which the lactose in milk cannot be digested well in the small intestine. The undigested lactose then moves into the large intestine where bacteria attack it, forming large amounts of gas. Symptoms of lactose intolerance include bloating, cramps, nausea, and vomiting. Often, in the case of children, the individual will outgrow this problem. Avoidance of foods containing lactose is recommended for people who show signs of lactose intolerance. Since dairy products can provide many vital nutrients, tablets can be taken that provide the needed digestive materials in the small intestine. Lactose-free milk is also readily available.
Disaccharides
The simple sugars form the foundation of more complex carbohydrates. The cyclic forms of two sugars can be linked together by means of a condensation reaction. The figure below shows how a glucose molecule and a fructose molecule combine to form a sucrose molecule. A hydrogen atom from one molecule and a hydroxyl group from the other molecule are eliminated as water, with a resulting covalent bond linking the two sugars together at that point.
Sucrose, commonly known as table sugar, is an example of a disaccharide. A disaccharide is a carbohydrate formed by the joining of two monosaccharides. Other common disaccharides include lactose and maltose. Lactose, a component of milk, is formed from glucose and galactose, while maltose is formed from two glucose molecules.
During digestion, these disaccharides are hydrolyzed in the small intestine to form the component monosaccharides, which are then absorbed across the intestinal wall and into the bloodstream to be transported to the cells.
Summary
• A disaccharide is a carbohydrate formed by the joining of two monosaccharides.
• Common disaccharides include sucrose, lactose, and maltose.
26.03: Polysaccharides
As the weather warms up, the runners come out. Not just the casual joggers, but also the really serious runners who actually enjoy running all 26.2 miles of a marathon. Prior to these races (and a lot of shorter ones), you hear a lot about carbo-loading. This practice involves eating a lot of starch in the days prior to the race. The starch is converted to glucose, which is normally used for biochemical energy. Excess glucose is stored as glycogen in liver and muscle tissue to be used when needed. If there is a lot of glycogen available, the muscles will have more biochemical energy to draw on when needed for the long run. Meanwhile, the rest of us will sit at the sidewalk restaurant, eating our spaghetti, and enjoying watching other people work hard.
Polysaccharides
Many simple sugars can combine by repeated condensation reactions until a very large molecule is formed. A polysaccharide is a complex carbohydrate polymer formed from the linkage of many monosaccharide monomers. One of the best known polysaccharides is starch, the main form of energy storage in plants. Starch is a staple in most human diets. Foods such as corn, potatoes, rice, and wheat have high starch contents. Starch is made of glucose monomers and occurs in both straight-chain and branched forms. Amylose is the straight-chain form, and consists of hundreds of linked glucose molecules. The branched form of starch is called amylopectin. In the small intestine, starch is hydrolyzed to form glucose. The glucose can then be converted to biochemical energy or stored for later use.
Glycogen is an even more highly branched polysaccharide of glucose monomers that serves the function of energy storage in animals. Glycogen is made and stored primarily in the cells of the liver and muscles.
Cellulose is another polymer of glucose, consisting of anywhere from hundreds to over ten thousand monomers. It is the structural component of the cell walls of green plants and is the single most common organic molecule on Earth. Roughly \(33\%\) of all plant matter is cellulose. The linkage structure in cellulose is different than that of starch, and cellulose is indigestible except by a few microorganisms that live in the digestive tracts of cattle and termites. The figure below shows a triple strand of cellulose. There is no branching and the fibers adopt a very stiff rod-like structure with numerous hydrogen bonds between the fibers adding to its strength. Cellulose is the main component of paper, cardboard, and textiles made from cotton, linen, and other plant fibers.
Summary
• A polysaccharide is a complex carbohydrate polymer formed from the linkage of many monosaccharide monomers.
• One of the best known polysaccharides is starch, the main form of energy storage in plants.
• Glycogen is an even more highly branched polysaccharide of glucose monomers that serves the function of storing energy in animals.
• Cellulose is another polymer of glucose; it is the structural component of the cell walls of green plants. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/26%3A_Biochemistry/26.01%3A_Monosaccharides.txt |
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, and 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 shown any advantage obtained by the athletes themselves; they are 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 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.
The amino acid and carboxyl group 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. Some side chains consist of nonpolar aliphatic or aromatic hydrocarbons. Other side chains are polar, while some are acidic or basic.
The table below lists the names of the 20 naturally occurring amino acids along with a three-letter abbreviation which is used to describe sequences of linked amino acids.
Table $1$: Amino acids and Abbreviations
Amino acids and Abbreviations
Amino Acid Abbreviation Amino Acid Abbreviation
Alanine Ala Leucine Leu
Arginine Arg Lysine Lys
Asparagine Asp Methionine Met
Aspartic acid Asp Phenylalanine Phe
Cysteine Cys Proline Pro
Glutamine Gln Serine Ser
Glutamic acid Glu Threonine Thr
Glycine Gly Tryptophan Trp
Histidine His Tyrosine Tyr
Isoleucine Ile Valine Val
Another more recent set of abbreviations employs only one letter. Leucine would be designated by L, serine by S, tyrosine by Y. The advantage of this system comes when listing the amino acid sequence of a protein that may contain over 100 amino acids in its chain.
Summary
• 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.
• The nature of an amino acid's side chain accounts for the variability in chemical and physical properties of amino acids.
• There are 20 amino acids formed and used by living organisms.
26.05: Peptides
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. Merrifield 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 and a free carboxyl group on the other end. 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 Gyl-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.
Summary
• 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.
• 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. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/26%3A_Biochemistry/26.04%3A_Amino_Acids.txt |
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, however, 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 important parts of the human 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 the protein. The secondary structure is a highly regular sub-structure of the protein. The two most common types of protein secondary structure are the alpha helix and the beta sheet. An alpha helix consists of amino acids that adopt a spiral shape. A beta sheet 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 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.
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. Hemoglobin is a very large protein found in red blood cells, whose function is to bind and carry oxygen throughout the bloodstream. As pictured below, 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. 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.
Summary
• 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 primary structure is the amino acid sequence of the protein (unique to each protein).
• The secondary structure is a highly regular sub-structure of the protein; the two most common types being the alpha helix and the beta sheet.
• The tertiary structure is the overall three-dimensional structure of the protein.
• The quaternary structure of a protein refers to the specific interaction and orientation of the subunits of that protein. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/26%3A_Biochemistry/26.06%3A_Proteins.txt |
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.
$\ce{H_2NCONH_2} \left( aq \right) + \ce{H_2O} \left( l \right) \overset{\text{urease}}{\rightarrow} 2 \ce{NH_3} \left( g \right) + \ce{CO_2} \left( g \right)\nonumber$
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
An enzyme is a protein that acts as a biological catalyst. Recall that a catalyst is a substance that increases the rate of a chemical reaction without itself being consumed in the reaction. Cellular processes consist of many chemical reactions that must occur quickly in order for the cell to function properly. Enzymes catalyze most of the chemical reactions that occur in a cell. A substrate is the molecule or molecules on which the enzyme acts. In the urease catalyzed reaction above, urea is the substrate. The figure below diagrams a typical enzymatic reaction.
The first step in the reaction is that the substrate binds to a specific part of the enzyme molecule. 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 binds 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.
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.
The 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. Cyanide is a potent poison which acts as a competitive inhibitor. It binds to the active site of the enzyme cytochrome c oxidase and interrupts cellular respiration. The binding of the cyanide to the enzyme is irreversible and the affected organism dies quickly.
Non-competitive Inhibition
A non-competitive inhibitor does not bind at the active site. It attaches at some other site on the enzyme, and changes the shape of the protein. This shift in three-dimensional structure alters the shape of the active site so that the substrate will no longer fit in the site properly (see figure below).
Cofactors
Some enzymes require the presence of a non-protein molecule called a cofactor on order to function properly. Cofactors can be inorganic metal ions or small organic molecules. Many vitamins, such as B vitamins, act as cofactors. Some metal ions which function as cofactors for various enzymes include zinc, magnesium, potassium, and iron.
Summary
• An enzyme is a protein that acts as a biological catalyst.
• A substrate is the molecule or molecules on which the enzyme acts.
• The active site is the place on an enzyme where the substrate binds.
• Each enzyme catalyzes only one chemical reaction with only one substrate.
• An inhibitor is a molecule which interferes with the function of an enzyme, either by slowing or stopping the chemical reaction.
• A competitive inhibitor is a molecule that binds to the active site of an enzyme without reacting, thus preventing the substrate from binding.
• A non-competitive inhibitor does not bind at the active site; it attaches at some other site on the enzyme.
• Some enzymes require the presence of a non-protein molecule called a cofactor on order to function properly. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/26%3A_Biochemistry/26.07%3A_Enzymes.txt |
There is a lot of interest these days in 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 triglyceride (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, eat a large pizza, and expect these pills to keep you healthy. It's important to eat healthfully and get exercise.
Triglycerides
A lipid is a member of a class of water-insoluble compounds that includes oils, fats, and waxes. Oils and fats are based on the same general structure, but fats are solids at room temperature, while oils are liquids. Butter is an example of a fat and is derived from animals. Some oils include olive oil and canola oil, which are obtained from plants. Lipids are an essential part of a healthy diet, though excess dietary fat can be harmful. Lipids store energy in the body and are also needed to keep cell membranes healthy.
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 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 body. 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.
Fats may be either saturated or unsaturated. A saturated fat is a fat that consists of triglycerides whose carbon chains consist entirely of carbon-carbon single bonds. Therefore, the carbon chains are saturated with the maximum number of hydrogen atoms possible. An unsaturated fat is a fat that consists of triglycerides whose carbon chains contain one or more carbon-carbon double bonds. A fat with one double bond is called monounsaturated, while a fat with multiple double bonds is called polyunsaturated (see figure below).
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. The figure below shows the percentages of fat types in some common foods.
Summary
• A lipid is a member of a class of water-insoluble compounds that includes oils, fats, and waxes.
• One type of lipid is called a triglyceride, an ester derived from glycerol combined with three fatty acid molecules.
• A saturated fat is a fat that consists of triglycerides whose carbon chains consist entirely of carbon-carbon single bonds.
• An unsaturated fat is a fat that consists of triglycerides whose carbon chains contain one or more carbon-carbon double bonds. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/26%3A_Biochemistry/26.08%3A_Triglycerides.txt |
If you were to go to the dentist to get a tooth pulled, you would not want to feel any pain. The dentist would inject an anesthetic into your gum to numb it. 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.
Summary
• A phospholipid is a lipid that contains a phosphate group.
• A phospholipid consists of a hydrophilic (water-loving) head and hydrophobic (water-fearing) tail.
• 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.
• Phospholipid bilayers are critical components of cell membranes.
26.10: Waxes
From the 1700s up to the late 1900s, whalers searched the ocean for the sperm whale. Whaling was a dangerous occupation—the prey averaged about fifty feet in length. But when successful, the search was worth it. One large whale could produce up to 500 gallons of oil, valuable for making candles, ointments, cosmetic creams, and industrial lubricants. In 1988, sperm whales (and other whale species) were placed under international protection because their numbers were diminishing rapidly. Today, various vegetable oils are used in place of whale oils.
Waxes
Another category of lipid molecule is waxes. Waxes are esters of long-chain fatty acids and long-chain alcohols. Waxes are soft solids with generally low melting points and are insoluble in water. The figure below shows the structure of cetyl palmitate, a natural wax present in sperm whales.
One of the best known natural waxes is beeswax, though many other animals and plants synthesize waxes naturally. Waxes can be found on leaves of plants and on the skin, hair, or feathers of animals, where they function to keep these structures pliable and waterproof. Humans take advantage of the protective properties of natural and synthetic waxes in such applications as floor polish and car wax. Other common waxes include jojoba, carnauba, and wool wax, which is also known as lanolin.
Summary
• Waxes are esters of long-chain fatty acids and long-chain alcohols.
• Humans take advantage of the protective properties of natural and synthetic waxes in such applications as floor polish and car wax. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/26%3A_Biochemistry/26.09%3A_Phospholipids.txt |
Cancer treatment is a complex and challenging effort. Cancer cells grow without the usual controls that act on normal cells. One approach to treating cancer is to alter the structure of the DNA in order to slow down or stop the growth of the abnormal cells. Compounds that structurally resemble the normal building blocks of DNA have been shown to be very effective in stopping some forms of cancer from spreading throughout the body.
Nucleic Acids
The Swiss biochemist Friedrich Miescher first discovered nitrogen-containing compounds in the nuclei of cells in 1869. The term nucleic acid was used to describe these molecules because of their discovery within the cell nucleus, and because of the presence of phosphate groups and their relationship to phosphoric acid. A nucleic acid is a large biopolymer consisting of many nucleotides. The two primary nucleic acids which are found in cells are deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). DNA is the carrier of genetic information and is ultimately responsible for how cells produce proteins in order to carry out all the functions necessary for life. RNA is a related molecule that is involved in the mechanism by which the information stored in DNA is eventually converted into protein molecules.
The basic components of nucleic acids are nucleotides. A nucleotide is a molecule that contains a five-carbon sugar, a phosphate group, and a nitrogen-containing base. The five-carbon sugar is either ribose, in the case of RNA, or deoxyribose, in the case of DNA. The only difference between the two molecule is the presence of a hydroxyl group attached to one member of the carbon ring in RNA. In DNA, that same carbon atom is attached only to a hydrogen atom (see figure below). Note that in drawing the structure of organic molecules, the single hydrogen atoms are not shown in the structure, but are understood to be attached at each carbon point unless another molecule is shown.
The nucleotides form the backbone of RNA and DNA. Each nucleotide consists of a base, a pentose (either ribose or deoxyribose) and phosphate groups (see figure below). Three of the bases in RNA and DNA are identical (adenine, cytosine, and guanine). Thymine is found in DNA, while uracil is found in RNA.
Summary
• A nucleic acid is a large biopolymer consisting of many nucleotides.
• The two primary nucleic acids which are found in cells are deoxyribonucleic acid (DNA) and ribonucleic acid (RNA).
• A nucleotide is a molecule that contains a five-carbon sugar, a phosphate group, and a nitrogenous base.
26.12: DNA and RNA
Linus Pauling was one of the greatest scientists of the twentieth century. Pauling was a two-time Nobel Prize winner (in chemistry in 1954, and the peace prize in 1962). However, he didn't always come in first. In the 1950s, there was a great deal of interest in the structure of DNA. Pauling spent some time on this puzzle, although he was primarily interested in proteins. He proposed a DNA structure where the bases were on the outside and the phosphate groups were on the inside. This idea turned out to be incorrect, but it certainly did not take away from his outstanding scientific reputation.
DNA and RNA
The three parts of a DNA nucleotide are assembled as shown in the figure below.
Every DNA and RNA polymer consists of multiple nucleotides strung together into extremely long chains. The only variation in each nucleotide is the identity of the nitrogenous base. The figure above shows one example of a nitrogenous base, called adenine. There are only five different nitrogenous bases found in all nucleic acids. The four bases of DNA are adenine, thymine, cytosine, and guanine, abbreviated A, T, C, and G respectively. In RNA, the base thymine is not found and is instead replaced by a different base called uracil, abbreviated U. The other three bases are present in both DNA and RNA.
The specific structure of DNA proved elusive to scientists for many years. In 1953, James Watson and Francis Crick proposed that the structure of DNA consists of two side-by-side polynucleotide chains wrapped into the shape of a double helix. One aspect of this structure is that each nitrogenous base on one of the DNA strands must be paired up with another base on the opposite strand. The figure below illustrates the base pairing. Each adenine base is always paired with a thymine, while each cytosine is paired with a guanine. The bases fit together perfectly from one strand to the other and are also held together by hydrogen bonds. The A-T pairing contains two hydrogen bonds, while the C-G pairing contains three hydrogen bonds. The ends of each strand are labeled either with 3' or 5', based on a numbering of the deoxyribose sugar ring.
The double helical structure of DNA is shown in the figure below.
Summary
• Every DNA and RNA polymer consists of multiple nucleotides strung together into extremely long chains.
• The only variation in each nucleotide is the identity of the nitrogenous base.
• The four bases of DNA are adenine, thymine, cytosine, and guanine, abbreviated A, T, C, and G respectively. In RNA, the base thymine is not found and is instead replaced by a different base called uracil, abbreviated U; the other three bases are present in both DNA and RNA.
• James Watson and Francis Crick proposed that the structure of DNA consists of two side-by-side polynucleotide chains wrapped into the shape of a double helix. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/26%3A_Biochemistry/26.11%3A_Nucleic_Acids.txt |
The Human Genome Project started in 1990 with the lofty goal of sequencing the complete set of human DNA. This project was completed in April 2003, ahead of schedule and under the budget set for it (a rare occurrence for a government project). With this knowledge, we can now identify genetic disorders quickly and personalize treatment for many diseases. However, much work still remains to fully understand the connections between specific DNA sequences and specific diseases.
The Genetic Code
Each particular organism contains many protein molecules that are specific to that organism. The particular base sequence of DNA is responsible for the production of all of the different proteins that are present in each and every living thing that has ever inhabited the Earth. How does that work? Cells use the unique sequence of DNA bases to decide which proteins to synthesize. A gene is a segment of DNA that carries a code for making a specific polypeptide chain. The cell essentially decodes the DNA in order to make whatever peptides and proteins are needed by that organism.
The genetic code works as a series of three-letter codes. Each sequence of three letters, called a triplet, corresponds to one of the twenty common amino acids. The triplets are read by the cell, one after the other, in the process of protein synthesis. The table below shows all of the possible triplets and the amino acids that result from each three-letter code.
Table \(1\): DNA Triplet Codes for Amino Acids
DNA Triplet Codes for Amino Acids
AAA Lys GAA Glu TAA Stop CAA Gln
AAG Lys GAG Glu TAG Stop CAG Gln
AAT Asn GAT Asp TAT Tyr CAT His
AAC Asn GAC Asp TAC Tyr CAC His
AGA Arg GGA Gly TGA Stop CGA Arg
AGG Arg GGG Gly TGG Trp CGG Arg
AGT Ser GGT Gly TGT Cys CGT Arg
AGC Ser GGC Gly TGC Cys CGC Arg
ATA Ile GTA Val TTA Leu CTA Leu
ATG Met GTG Val TTG Leu CTG Leu
ATT Ile GTT Val TTT Phe CTT Leu
ATC Ile GTC Val TTC Phe CTC Leu
ACA Thr GCA Ala TCA Ser CCA Pro
ACG Thr GCG Ala TCG Ser CCG Pro
ACT Thr GCT Ala TCT Ser CCT Pro
ACC Thr GCC Ala TCC Ser CCC Pro
The DNA code word GCA corresponds to the amino acid arginine, while the DNA code word TCG corresponds to the amino acid serine. Most amino acids are represented by more than one possible triplet code, but each triplet code yields only one particular amino acid. Three of the DNA code words (TAA, TAG, and TGA) are stop or termination code words. The translation of a DNA base sequence begins with a start code word and runs until a stop code word is reached.
Even with only four different bases, the number of possible nucleotide sequences in a DNA chain is virtually limitless. The particular DNA sequence of an organism constitutes the genetic blueprint for that organism. This genetic blueprint is found in the nucleus of each cell of the organism, and is passed on from parents to offspring. The incredible diversity of life on Earth stems from the differences in the genetic code of every living thing.
Summary
• The particular base sequence of DNA is responsible for the production of all of the different proteins that are present in each and every living thing that has ever inhabited the Earth.
• A gene is a segment of DNA that carries a code for making a specific polypeptide chain.
• The genetic code works as a series of three-letter codes.
26.14: Protein Synthesis
The assembly line is an American invention that was developed around 1901 to mass-produce cars. Prior to that time, teams of workers would build a car together. With the advent of the assembly line, cars could be produced much more quickly, and at lower cost. The assembly line idea quickly spread to other products. Being able to line up parts in order, and have a smooth process for putting those parts together, means that an item can be produced quickly and reproducibly, coming out the same way every time.
Protein Synthesis
The process of protein synthesis is summarized in the diagram below. DNA produces an RNA template, which then directs the amino acids to be introduced into the growing protein chain in the proper sequence. A specific transfer-RNA (tRNA) attaches to each specific amino acid and brings the amino acid to the RNA for incorporation.
The first step in the process is transcription—the unfolding of DNA and the production of a messenger-RNA (mRNA) strand. This step takes place in the nucleus of the cell.
The DNA uncoils and provides the pattern for the formation of a single strand of mRNA. After production of the RNA, the DNA refolds into the original double helix. The mRNA is exported to the cytoplasm (outside the nucleus) for further processing.
Amino acids will link with specific tRNA molecules for proper placement in the protein chain. The tRNA is a small coiled molecule that accepts an amino acid on one end and matches up to a specific three-base portion of the mRNA on the other end. The tRNA interacts with the mRNA so as to put the amino acid in the proper sequence for the developing protein. After adding the amino acid to the sequence, the tRNA is then cleaved from the amino acid and recycled for further use in the process.
The process of amino acid assembly takes place in the ribosome. This structure consists of two subunits containing ribosomal RNA that enclose the mRNA, and catalyze the formation of the amide linkages in the growing protein in a process known as translation. When protein synthesis is complete, the two subunits dissociate and release the completed protein chain.
The process of protein synthesis is fairly fast. Amino acids are added to the growing peptide chain at a rate of about 3-5 amino acids per second. A small protein (100-200 amino acids) can be produced in a minute or less.
Summary
• The first step of protein synthesis is transcription—the unfolding of DNA and the production of a messenger-RNA (mRNA) strand.
• In the second step of protein synthesis—translation—tRNA and mRNA interact to code amino acids into growing polypeptide chains. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/26%3A_Biochemistry/26.13%3A_Genetic_Code.txt |
What is chemistry? Simply put, chemistry is the study of the interactions of matter with other matter and with energy. This seems straightforward enough. However, the definition of chemistry includes a wide range of topics that must be understood to gain a mastery of the topic or even take additional courses in chemistry. In this book, we will lay the foundations of chemistry in a topic-by-topic fashion to provide you with the background you need to successfully understand chemistry.
• 1.1: Chemistry - The Central Science
Chemistry is the study of matter and how it behaves. The scientific method is the general process by which we learn about the natural universe.
• 1.2: States of Matter
A state of matter is one of the distinct forms that different phases of matter take on. Three states of matter are readily observable in everyday life: solid, liquid, and gas. Each state has characteristic properties that are dictate by the nature of the matter and the conditions on which it exists.
• 1.3: Classification of Matter
Matter can be described with both physical properties and chemical properties. Matter can be identified as an element, a compound, or a mixture.
• 1.4: Chemical Elements and Symbols
An element is a substance that cannot be broken down into simpler chemical substances. There are about 90 naturally occurring elements known on Earth. Using technology, scientists have been able to create nearly 30 additional elements that do not occur in nature. Today, chemistry recognizes 118 elements—some of which were created an atom at a time.
• 1.5: Chemical Reactions - Examples of Chemical Change
Chemical reactions are the processes by which chemicals interact to form new chemicals with different compositions. Simply stated, a chemical reaction is the process where reactants are transformed into products. How chemicals react is dictated by the chemical properties of the element or compound- the ways in which a compound or element undergoes changes in composition.
• 1.6: Physical Quantities - Units and Scientific Notation
To understand chemistry, we need a clear understanding of the units chemists work with and the rules they follow for expressing numbers.
• 1.7: Measuring Mass, Length, and Volume
The SI base units specifies certain units for various types of quantities, based on seven fundamental units. We will use most of the fundamental units in chemistry. This section addresses the measurements: mass, length, and volume
• 1.8: Measurement and Significant Figures
Significant figures properly report the number of measured and estimated digits in a measurement. There are rules for applying significant figures in calculations.
• 1.9: Rounding Off Numbers
Before dealing with the specifics of the rules for determining the significant figures in a calculated result, we need to be able to round numbers correctly.
• 1.10: Problem Solving - Unit Conversions and Estimating Answers
A unit can be converted to another unit of the same type with a conversion factor. Using this method, called dimensional analysis or the factor label method, is an essential skill that will be applied in many chapters of this textbook.
• 1.11: Temperature, Heat, and Energy
Heat and temperature are a closely related topics, although the difference between the two are often conflated. The key difference is that heat deals with the transfer of thermal energy, whereas temperature is a metric of molecular kinetic energy. Heat is a process and temperature is a property.
• 1.12: Density and Specific Gravity
Knowledge of density is important in the characterization and separation of materials. Information about density allows us to make predictions about the behavior of matter.
01: Matter and Measurements
Learning Objectives
• Define chemistry in relation to other sciences.
• Use physical and chemical properties, including phase, to describe matter.
Chemistry is the study of matter—what it consists of, what its properties are, and how it changes. Being able to describe the ingredients in a cake and how they change when the cake is baked is called chemistry. Matter is anything that has mass and takes up space—that is, anything that is physically real. Some things are easily identified as matter—this book, for example. Others are not so obvious. Because we move so easily through air, we sometimes forget that it, too, is matter.
Chemistry is one branch of science. Science is the process by which we learn about the natural universe by observing, testing, and then generating models that explain our observations. Because the physical universe is so vast, there are many different branches of science (Figure \(1\)). Thus, chemistry is the study of matter, biology is the study of living things, and geology is the study of rocks and the earth. Mathematics is the language of science, and we will use it to communicate some of the ideas of chemistry.
Although we divide science into different fields, there is much overlap among them. For example, some biologists and chemists work in both fields so much that their work is called biochemistry. Similarly, geology and chemistry overlap in the field called geochemistry. Figure \(1\) shows how many of the individual fields of science are related; there are many other fields of science in addition to the ones listed here.
How do scientists work? Generally, they follow a process called the scientific method. The scientific method is an organized procedure for learning answers to questions and making explanations for observations. The steps of the scientific method may not be as clear-cut in real life as described here, but most scientific work follows this general outline.
1. Propose a hypothesis. A scientist generates a testable idea, or hypothesis, to try to answer a question or explain an observation about how the natural universe works. Some people use the word theory in place of hypothesis, but the word hypothesis is the proper word in science. For scientific applications, the word theory is a general statement that describes a large set of observations and data. A theory represents the highest level of scientific understanding.
2. Test the hypothesis. A scientist evaluates the hypothesis by devising and carrying out experiments to test it. If the hypothesis passes the test, it may be a proper answer to the question. If the hypothesis does not pass the test, it may not be a good answer.
3. Refine the hypothesis if necessary. Depending on the results of experiments, a scientist may want to modify the hypothesis and then test it again. Sometimes the results show the original hypothesis to be completely wrong, in which case a scientist will have to devise a new hypothesis.
Not all scientific investigations are simple enough to be separated into these three discrete steps. But these steps represent the general method by which scientists learn about our natural universe.
Physical and Chemical Properties and Changes
The properties that chemists use to describe matter fall into two general categories. Physical properties are characteristics that describe matter. They include characteristics such as size, shape, color, and mass. These characteristics can be observed or measured without changing the identity of the matter in question. Chemical properties are characteristics that describe how matter changes its chemical structure or composition. An example of a chemical property is flammability—a material’s ability to burn—because burning (also known as combustion) changes the chemical composition of a material. The observation of chemical properties involves a chemical change of the matter in question, resulting in matter with a different identity and different physical and chemical properties. Part of understanding matter is being able to describe it. One way chemists describe matter is to assign different kinds of properties to different categories.
We know from our experience with water that substances can change from one phase to another if the conditions are right. Typically, varying the temperature of a substance (and, less commonly, the pressure exerted on it) can cause a physical process in which a substance changes from one phase to another. Examples are summarized in Table \(1\).
Table \(1\): Physical Changes
Change Name
solid to liquid melting, fusion
solid to gas sublimation
liquid to gas boiling, evaporation
liquid to solid solidification, freezing
gas to liquid condensation
gas to solid deposition
The difference between a physical reaction and a chemical reaction is composition. In a chemical reaction, there is a change in the composition of the substances in question; in a physical change there is a difference in the appearance, smell, or simple display of a sample of matter without a change in composition. Although we call them physical "reactions," no reaction is actually occurring. In order for a reaction to take place, there must be a change in the elemental composition of the substance in question. Thus, we shall simply refer to "physical reactions" as physical changes from now on.
Example \(1\)
Classify each of the following changes as physical or chemical:
1. condensation of steam
2. burning of gasoline
3. souring of milk
4. dissolving of sugar in water
5. melting of gold
Solution
The best way to determine whether a change is physical or chemical is to determine if the substance identity has changed or not.
1. Steam is water in the gas phase. When it condenses, changes from gas to liquid, it remains water. Therefore, this is a physical change.
2. When gasoline burns, it changes into different substances. This is a chemical change.
3. When milk becomes sour, several chemical changes have occured; proteins denature, sugars break down, and acid is produced. The milk is now made up of different substances.
4. Sugar dissolves easily into water but it still remains sugar. If the water is evaporated, the sugar will recrystalize as a solid again. Dissolving is a physical change, the substance does not change identity.
5. Melting solid gold into liquid gold is a physical change much like melting solid water (an ice cube) into liquid water is a physical change.
Exercise \(1\)
Classify each of the following changes as physical or chemical:
1. coal burning
2. ice melting
3. mixing chocolate syrup with milk
4. explosion of a firecracker
5. magnetizing of a screwdriver
Answer a:
chemical
Answer b:
physical
Answer c:
physical
Answer d:
chemical
Answer e:
physical
Key Takeaways
• Chemistry is the study of matter and how it behaves.
• The scientific method is the general process by which we learn about the natural universe.
• When a substance changes from one state to another, this is a physical change.
• If a substance undergoes a change into one or more different substances, this is a chemical change. | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/01%3A_Matter_and_Measurements/1.01%3A_Chemistry_-_The_Central_Science.txt |
Learning Objectives
• Describe the different physical states of matter; solid, liquid, and gas.
• Understand how external conditions can affect the states of matter.
Water can take many forms. At low temperatures (below $0^\text{o} \text{C}$), it is a solid. When at "normal" temperatures (between $0^\text{o} \text{C}$ and $100^\text{o} \text{C}$), it is a liquid. While at temperatures above $100^\text{o} \text{C}$, water is a gas (steam). The state the water is in depends upon the temperature. Each state (solid, liquid, and gas) has its own unique set of physical properties.
Matter typically exists in one of three states: solid, liquid, or gas and these different states of matter have different properties (Table $1$):
• 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. Forces between atoms or molecules are not strong enough to hold them together.
• 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. Forces between atoms or molecules are strong enough to keep the molecules relatively close together but not strong enough to prevent them from moving past 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. Forces between atoms or molecules are strong enough to keep the molecules together and to prevent them from moving past one another.
The state a given substance exhibits is a physical property. Some substances exist as gases at room temperature (oxygen and carbon dioxide), while 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. Figure $2$ shows the differences among solids, liquids, and gases at the molecular level. A solid has definite volume and shape, a liquid has a definite volume but no definite shape, and a gas has neither a definite volume nor shape (Table $1$).
These three descriptions each imply that the matter has certain physical properties when in these states. A solid has a definite shape and a definite volume. Liquids ordinarily have a definite volume but not a definite shape; they take the shape of their containers. Gases have neither a definite shape nor a definite volume, and they expand to fill their containers. We encounter matter in each phase every day; in fact, we regularly encounter water in all three phases: ice (solid), water (liquid), and steam (gas).
Table $1$: Characteristics of each Phase of Matter
Gas Liquid Solid
Shape
no definite shape (takes the shape of its container)
no definite shape (takes the shape of its container)
definite shape (rigid)
Volume
particles move in random motion with little or no attraction to each other
has definite volume
definite volume
Mobility
particles move in random motion with little or no attraction to each other
particles are free to move over each other, but are still attracted to each other
particles vibrate around fixed axes
Compressibility
highly compressible
weakly compressible
weakly compressible
Adding energy to matter gives its atoms or molecules the ability to resist some of the forces holding them together. For example, heating ice to its melting point gives its molecules enough energy to move. The ice melts and becomes liquid water. Similarly, heating liquid water to its boiling point gives its molecules enough energy to pull apart from one another so they no longer have contact. The liquid water vaporizes and becomes water vapor.
State of Matter Depends on the External Conditions
The temperature of the melting and boiling points depend on the identity of the substance and the atmospheric pressure. Each substance has its own boiling and melting points that depend on the properties of the substance. As an example, the values for water are given in Table $2$. Note how the boiling point of water varies greatly with pressure.
Table $1$: Boiling point of water as a function of pressure
Altitude (ft) Pressure (atm) Boiling Point (oC)
-500 1.05 100.5
0 1.00 100
4000 0.892 96
7000 0.797 93
Example $1$
Isopropyl alcohol is a colorless, flammable chemical compound with a strong odor. Its melting point is –89°C and it boiling point is 82.5°C. Is isopropyl alcohol a solid, liquid or gas at room temperature (25°C)?
Solution
Room temperature (25°C) is above the melting point of isopropyl alcohol (–89°C), but lower than its boiling point (82.5°C), therefore, it is a liquid at room temperature.
Exercise $1$
Freon-12 is used as a refrigerant and aerosol spray propellant. Its melting point is –157.7°C and it boiling point is –29.8 °C. Is Freon-12 a solid, liquid or gas at room temperature (25°C).
Answer
Freon-12 is a gas at room temperature
Key Takeaways
• Matter exists in different physical states.
• Changes in conditions such as temperature and pressure can allow matter to change state.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/01%3A_Matter_and_Measurements/1.02%3A_States_of_Matter.txt |
Learning Objectives
• Categorize different types of matter as a pure substances or mixtures.
• Explain the difference between an element and a compound.
• Explain the difference between a homogeneous mixture and a heterogeneous mixture.
One useful way of organizing our understanding of matter is to think of a hierarchy that extends down from the most general and complex substances to the simplest and most fundamental (Figure \(1\)). At the top of this hierarchy is two broad categories into which all matter can be classified: pure substances and mixtures. A pure substance is a form of matter that has a constant composition (meaning that all samples of this substance have uniform composition) and properties that are constant throughout the sample (meaning that there is only one set of properties such as melting point, color, boiling point, etc. throughout the matter). If we take two or more pure substances and physically mix them together, we refer to this as a mixture. A mixture does not have constant composition or properties throughout.
Elements and compounds are both examples of pure substances. A substance that cannot be broken down into chemically simpler components is an element. Oxygen, O, and hydrogen, H, are each examples of elements. A pure substance that can be broken down into chemically simpler components (because it made up of more than one element) is a compound. For example, the compound water, H2O, is formed when hydrogen and oxygen chemically combine in a fixed ratio of 2 hydrogen atoms for every 1 oxygen atom.
Compounds may have different chemical and physical properties from the individual elements that they are composed of. Mixtures, on the other hand, are physical blends of two or more components, each of which retains its own identity and properties. Mixtures can always be separated again into the component pure substances, because bonding among the atoms of the constituent substances does not occur. For example sodium is a soft shiny metal and chlorine is a pungent green gas. These two elements chemically combine to form the compound, sodium chloride (table salt) which is a white, crystalline solid having none of the properties of either sodium or chlorine. If, however, you mixed table salt with ground pepper, you would still be able to see the individual grains of each of them and, if you were patient, you could take tweezers and carefully separate them back into pure salt and pure pepper.
Mixtures fall into two categories, based on the uniformity of their composition (Figure \(1\)). The first, called a heterogeneous mixture, is distinguished by the fact that different samples of the mixture may have a different composition. For example, if you open a container of mixed nuts and pull out a series of small samples and examine them, the exact ratio of peanuts-to-almonds in the samples will always be slightly different, no matter how carefully you mix them. Common examples of heterogeneous mixtures include dirt, gravel, and vegetable soup.
In a homogeneous mixture, on the other hand, any sample that you examine will have exactly the same composition as any other sample. Within chemistry, the most common type of homogeneous mixture is a solution which is one substance dissolved completely within another. Think of a solution of pure sugar dissolved in pure water. Any sample of the solution that you examine will have exactly the same ratio of sugar-to-water, which means that it is a homogeneous mixture. Even in a homogeneous mixture, the properties of the components are generally recognizable. Thus, sugar-water tastes sweet (like sugar) and is wet (like water). Unlike a compound, which has a fixed, definite ratio, in a mixture the amounts of each component can vary. For example, when you add a little sugar to one cup of tea and a lot of sugar to another, each cup will contain a homogeneous mixture of tea and sugar but they will have a different ratio of sugar-to-tea and a different taste. If you add so much sugar that some does not dissolve and stays on the bottom, however, the mixture is no longer homogeneous, it is heterogeneous; you could easily separate the two components (Figure \(1\)).
Example \(1\)
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. I. Tea is a solution of compounds in water, so it is not chemically pure. It is usually separated from tea leaves by filtration.
II. Because the composition of the solution is uniform throughout, it is a homogeneous mixture.
2. I. Orange juice contains particles of solid (pulp) as well as liquid; it is not chemically pure.
II. Because its composition is not uniform throughout, orange juice is a heterogeneous mixture.
3. I. 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.
II. The regions of different composition indicate that a compact disc is a heterogeneous mixture.
4. I. Aluminum oxide is a single, chemically pure compound.
5. I. Selenium is one of the known elements.
Exercise \(1\)
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 \(2\)
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 different materials, so it is a heterogeneous mixture.
3. Water is a substance; more specifically, because water is composed of a fixed ratio hydrogen and oxygen atoms, it is a compound.
4. Oxygen, a substance, is an element.
Exercise \(2\)
How would a chemist categorize each example of matter?
1. coffee
2. hydrogen
3. an egg
Answer a:
Coffee, assuming it is filtered, is a variety of substances dissolved in water, therefore, it is a homogeneous mixture, or a solution.
Answer b:
Hydrogen is a known element.
Answer c:
An egg is composed of many different substances with different compositions between the yolk and white, it is a heterogeneous mixture.
Summary
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.
Contributors and Attributions
This page was constructed from content via the following contributor(s) and edited (topically or extensively) by the LibreTexts development team to meet platform style, presentation, and quality:
• Henry Agnew (UC Davis)
• Paul R. Young, Professor of Chemistry, University of Illinois at Chicago, Wiki: AskTheNerd; PRYaskthenerd.com - pyounguic.edu; ChemistryOnline.com | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/01%3A_Matter_and_Measurements/1.03%3A_Classification_of_Matter.txt |
Learning Objectives
• Identify names and symbols of common chemical elements.
• Represent a chemical compound with a chemical formula.
As described in the previous section, an element is a pure substance that cannot be broken down into simpler chemical substances. There are about 90 naturally occurring elements known on Earth. Using technology, scientists have been able to create nearly 30 additional elements that do not occur in nature. Today, chemistry recognizes 118 elements—some of which were created an atom at a time. Figure $1$ shows some of the chemical elements.
Elemental Names and Symbols
Each element has a name. Some of these names date from antiquity, while others are quite new. Today, the names for new elements are proposed by their discoverers but must be approved by the International Union of Pure and Applied Chemistry, an international organization that makes recommendations concerning all kinds of chemical terminology.
Today, new elements are usually named after famous scientists.
The names of the elements can be cumbersome to write in full, especially when combined to form the names of compounds. Therefore, each element name is abbreviated as a one- or two-letter chemical symbol. By convention, the first letter of a chemical symbol is a capital letter, while the second letter (if there is one) is a lowercase letter. The first letter of the symbol is usually the first letter of the element’s name, while the second letter is some other letter from the name. Some elements have symbols that derive from earlier, mostly Latin names, so the symbols may not contain any letters from the English name. Table $1$ lists the names and symbols of some of the most familiar elements.
Table $1$: Element Names and Symbols
Element Name Element Symbol Element Name Element Symbol
aluminum Al magnesium Mg
argon Ar manganese Mn
arsenic As mercury Hg*
barium Ba neon Ne
bismuth Bi nickel Ni
boron B nitrogen N
bromine Br oxygen O
calcium Ca phosphorus P
carbon C platinum Pt
chlorine Cl potassium K*
chromium Cr silicon Si
copper Cu* silver Ag*
fluorine F sodium Na*
gold Au* strontium Sr
helium He sulfur S
hydrogen H tin Sn*
iron Fe tungsten W
iodine I uranium U
lead Pb* zinc Zn
lithium Li zirconium Zr
*The symbol comes from the Latin name of element. The symbol for tungsten comes from its German name—wolfram.
Element names in languages other than English are often close to their Latin names. For example, gold is oro in Spanish and or in French (close to the Latin aurum), tin is estaño in Spanish (compare to stannum), lead is plomo in Spanish and plomb in French (compare to plumbum), silver is argent in French (compare to argentum), and iron is fer in French and hierro in Spanish (compare to ferrum). The closeness is even more apparent in pronunciation than in spelling.
Elements in Nature and the Human Body
The elements vary widely in abundance. In the universe as a whole, the most common element is hydrogen (about 90% of atoms), followed by helium (most of the remaining 10%). All other elements are present in relatively minuscule amounts, as far as we can detect. On the planet Earth, however, the situation is rather different (Table $2$). Oxygen makes up 46.1% of the mass of Earth’s crust (the relatively thin layer of rock forming Earth’s surface), mostly in combination with other elements, while silicon makes up 28.2%. Hydrogen, the most abundant element in the universe, makes up only 0.14% of Earth’s crust.
Table $2$: Elemental Composition of Earth and the Human Body
Earth’s Crust Human Body
Element Percentage Element Percentage
oxygen 46.1 oxygen 61
silicon 28.2 carbon 23
aluminum 8.23 hydrogen 10
iron 5.53 nitrogen 2.6
calcium 4.15 calcium 1.4
sodium 2.36 phosphorus 1.1
magnesium 2.33 sulfur 0.20
potassium 2.09 potassium 0.20
titanium 0.565 sodium 0.14
hydrogen 0.14 chlorine 0.12
phosphorus 0.105 magnesium 0.027
all others 0.174 silicon 0.026
Source: D. R. Lide, ed. CRC Handbook of Chemistry and Physics, 89th ed. (Boca Raton, FL: CRC Press, 2008–9), 14–17.
Table $2$ also lists the relative abundances of elements in the human body. If you compare both compositions, you will find disparities between the percentage of each element in the human body and on Earth. Oxygen has the highest percentage in both cases, but carbon, the element with the second highest percentage in the body, is relatively rare on Earth and does not even appear as a separate entry; carbon is part of the 0.174% representing “other” elements.
How does the human body concentrate so many apparently rare elements? The relative amounts of elements in the body have less to do with their abundances on Earth than with their availability in a form we can assimilate. We obtain oxygen from the air we breathe and the water we drink. We also obtain hydrogen from water. On the other hand, although carbon is present in the atmosphere as carbon dioxide, and about 80% of the atmosphere is nitrogen, we obtain those two elements from the food we eat, not the air we breathe.
Looking Closer: The Phosphorous Bottleneck
There is an element that we need more of in our bodies than is proportionately present in Earth’s crust, and this element is not easily accessible. Phosphorus makes up 1.1% of the human body but only 0.105% of Earth’s crust. We need phosphorus for our bones and teeth, and it is a crucial component of all living cells. Unlike carbon, which can be obtained from carbon dioxide, there is no phosphorus compound present in our surroundings that can serve as a convenient source. Phosphorus, then, is nature’s bottleneck. Its availability limits the amount of life our planet can sustain.
Higher forms of life, such as humans, can obtain phosphorus by selecting a proper diet (plenty of protein); but lower forms of life, such as algae, must absorb it from the environment. When phosphorus-containing detergents were introduced in the 1950s, wastewater from normal household activities greatly increased the amount of phosphorus available to algae and other plant life. Lakes receiving this wastewater experienced sudden increases in growth of algae. When the algae died, concentrations of bacteria that ate the dead algae increased. Because of the large bacterial concentrations, the oxygen content of the water dropped, causing fish to die in large numbers. This process, called eutrophication, is considered a negative environmental impact.
Today, many detergents are made without phosphorus so the detrimental effects of eutrophication are minimized. You may even see statements to that effect on detergent boxes. It can be sobering to realize how much impact a single element can have on life—or the ease with which human activity can affect the environment.
Example $1$
Write the chemical symbol for each element without consulting the above tables.
1. bromine
2. boron
3. carbon
4. calcium
5. gold
Strategy: The symbol for some of the more common elements is the first one or two letters of the element name. Test yourself to see if you know the symbol, then check your answer in the above tables. You will learn the element symbols as you practice.
1. Br
2. B
3. C
4. Ca
5. Au
Exercise $1$
Write the chemical symbol for each element without consulting the above tables.
1. manganese
2. magnesium
3. neon
4. nitrogen
5. silver
Answer a
Mn
Answer b
Mg
Answer c
Ne
Answer d
N
Answer e
Ag
Example $2$
What element is represented by each chemical symbol?
1. Na
2. Hg
3. P
4. K
5. I
1. sodium
2. mercury
3. phosphorus
4. potassium
5. iodine
Exercise $2$
What element is represented by each chemical symbol?
1. Pb
2. Sn
3. U
4. O
5. F
Answer a
lead
Answer b
tin
Answer c
uranium
Answer d
oxygen
Answer e
fluorine
Chemical Formulas
A chemical formula is an expression that shows each of the elements in a compound and the relative proportions of those elements. Water is composed of hydrogen and oxygen in a 2:1 ratio and its chemical formula is $\ce{H_2O}$. Sulfuric acid is one of the most widely produced chemicals in the United States and is composed of the elements hydrogen, sulfur, and oxygen; the chemical formula for sulfuric acid is $\ce{H_2SO_4}$. Sucrose (table sugar) consists of carbon, hydrogen, and oxygen in a 12:22:11 ratio. The chemical formula of these are:
\underbrace{\color{red} {\ce{H2}} \color{blue} {\ce{O}}}_{\begin{aligned} & \color{red} {2\, \text{H atoms}} \ & \color{blue} {1\, \text{O atom}} \end{aligned}} \quad \underbrace{\color{red} \ce{H2} \color{green} \ce{S} \color{blue} \ce{O4}}_{\begin{aligned} & \color{red} 2\, \text{H atoms} \ \color{green} &1\, \text{S atom} \ & \color{blue} 4\, \text{O atoms} \end{aligned}} \quad \underbrace{\color{red} \ce{C12} \color{black} \ce{H22} \color{blue} \ce{O11}}_{\begin{aligned} & \color{red} 12\, \text{H atoms} \ & \color{black} 22\, \text{C atoms} \ & \color{blue} 11\, \text{O atoms} \end{aligned}}
Notice that the oxygen and sulfur in water and sulfuric acid, respectively, do not have a "1" subscripts - this is assumed.
Sometimes certain groups of atoms are bonded together within the chemical and act as a single unit. Polyatomic ions will be discussed later and are enclosed in parenthesis followed by a subscript if more than one of the same ion exist in a chemical formula. For example, the formula $\ce{Ca3(PO4)2}$ represents a compound with:
• 3 $\ce{Ca}$ atoms and
• 2 $\ce{PO_4^{3-}}$ polyatomic ions
To count the total number of atoms for formulas with polyatomic ions enclosed in parenthesis, use the subscript as a multiplier for each atom or number of atoms.
$\ce{Ca_3(PO_4)} \color{red} \ce{_2} \nonumber$
and decomposing this to elements gives
• 3 $\ce{Ca}$ atoms
• $\color{red} 2 \color{black} \times 1$ $\ce{P}$ atoms
• $\color{red} 2 \color{black} \times 4$ $\ce{O}$ atoms
That is, 3 $\ce{Ca}$ atoms, 2 $\ce{P}$ atoms, and 8 $\ce{O}$ atoms
Chemical formula can be used in chemical equations. For example, the reaction of hydrogen gas ($\ce{H2}$) burning with oxygen gas ($\ce{O2}$) to form water ($\ce{H2O}$) is written as:
$\ce{2H_2 + O_2 \rightarrow 2H_2O} \nonumber$
Exercise $3$
Identify the elements in each of the following chemical formulas and what is the ratio of different elements in the chemical formulas:
1. $\ce{NaOH}$
2. $\ce{NaCl}$
3. $\ce{CaCl2}$
4. $\ce{CH3COOH}$
Answer a
Sodium $\ce{Na}$, oxygen $\ce{O}$, and hydrogen $\ce{H}$ are present. This is sodium hydroxide and is also known as lye or caustic soda.
This is a 1:1:1 ratio of sodium, oxygen, and hydrogen, respectively.
Answer b
Sodium $\ce{Na}$ and chlorine $\ce{O}$ are present. This is sodium chloride and is also known as table salt.
This is a 1:1 ratio of sodium and chlorine, respectively.
Answer c
Calcium $\ce{Ca}$ and Chlorine $\ce{Cl}$ are present. This is calcium chloride and is a different type of salt than sodium chloride.
This is a 1:2 ratio of calcium and chlorine, respectively.
Answer d
Carbon $\ce{C}$, Oxygen $\ce{O}$, and Hydrogen $\ce{H}$ are present. This is acetic acid and is also known as vinegar.
This is a 2:2:4 (or 1:1:2) ratio of carbon, oxygen, and hydrogen, respectively.
Key Takeaways
• All matter is composed of elements, which are represented by one- or two-letter symbols.
• Chemical compounds are represented by formulas using element symbols and numerical subscripts to represent the ratio of each element in the compound.
• Anonymous
1.05: Chemical Reactions- Examples of Chemical Change
Chemical reactions are the processes by which chemicals interact to form new chemicals with different compositions. Simply stated, a chemical reaction is the process where reactants are transformed into products. How chemicals react is dictated by the chemical properties of the element or compound- the ways in which a compound or element undergoes changes in composition.
Chemical reactions are constantly occurring in the world around us; everything from the rusting of an iron fence to the metabolic pathways of a human cell are all examples of chemical reactions. Chemistry is an attempt to classify and better understand these reactions. One key reaction in modern civilization is combustion
A chemical reaction is typically represented by a chemical equation, which represents the change from reactants to products. The left hand side of the equation represents the reactants, while the right hand side represents the products. A typical chemical reaction is written with stoichiometric coefficients, which show the relative amounts of products and reactants involved in the reaction. Each compound is followed by a parenthetical note of the compound’s physical state: $(l)$ for liquid, $(s)$ for solid, $(g)$ for gas. The symbol $(aq)$ is also commonly used in order to represent an aqueous solution, in which compounds are dissolved in water.
Butane is a gas at room temperature and atmospheric pressure and is highly flammable, colorless, easily liquefied gas (under light pressure). Butane can be used for gasoline blending, as a fuel gas, fragrance extraction solvent, either alone or in a mixture with propane, and as a feedstock for the manufacture of ethylene and butadiene, a key ingredient of synthetic rubber. The chemical formula of butane if $\ce{C4H10}$. When oxygen is plentiful, butane burns to form carbon dioxide and water vapor as observed in modern lighters (Figure $1$).
This reaction in words is:
$\text{butane} + \text{oxygen} \rightarrow \text{carbon dioxide} + \text{water}$
and the corresponding chemical equation for this reaction is:
$\ce{2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(g)}$
In the above chemical equation, $\ce{C4H10}$ and $\ce{O2}$ are the reactants that reacted to form the products: $\ce{CO2}$ and $\ce{H2O}$,
Writing Chemical Equations
To write an accurate chemical equation, two things must occur:
1. Each product and reactant must be written using its chemical formula
2. Coefficients are used in front of the chemical formulas to reflect the ratio species (discussed further in a later chapter)
3. Adding the phase of each chemical in parentheses (although this is often dropped for convenience)
Key Takeaways
• Chemical reactions are written to represent chemical changes.
• Chemical formulas are used to represent the reactants (starting substance) and the products (the new substances). | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/01%3A_Matter_and_Measurements/1.04%3A_Chemical_Elements_and_Symbols.txt |
Learning Objectives
• Express quantities properly using a number and a unit.
• Recognize the different measurement systems used in chemistry.
• Describe how prefixes are used in the metric system and identify how the prefixes milli-, centi-, and kilo- compare to the base unit.
• Understand when and how to use scientific notation to represent measurements.
Measurements
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 using a variety of devices or measuring tools, many of which are similar to those used in everyday life. Rulers are used to measure length, balances (scales) are used to measure mass (weight), and graduated cylinders or pipettes are used to measure volume. Measurements made using these devices are expressed 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 “5.2 kilometers,” we know that the quantity has been expressed in units of kilometers and that the number of kilometers is 5.2.
$\color{red} \underbrace{5.2}_{\text{number}} \color{blue} \underbrace{\text{kilometers}}_{\text{unit}} \nonumber$
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—for example, 12 miles, 12 kilometers, 12 furlongs, or 12 yards.
Without units, a number can be meaningless, confusing, or possibly life threatening. Suppose a doctor prescribes phenobarbital to control a patient’s seizures and states a dosage of “100” without specifying units. Not only will this be confusing to the medical professional giving the dose, but the consequences can be dire: 100 mg given three times per day can be effective as an anticonvulsant, but a single dose of 100 g is more than 10 times the lethal amount.
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.
Example $1$
Identify the number and the unit in each quantity.
1. one dozen eggs
2. 2.54 centimeters
3. a box of pencils
4. 88 meters per second
Answers
1. The number is one, and the unit is dozen.
2. The number is 2.54, and the unit is centimeter.
3. The number 1 is implied because the quantity is only a box. The unit is box of pencils.
4. The number is 88, and the unit is meters per second. Note that in this case the unit is actually a combination of two units: meters and seconds.
Exercise $2$
Identify the number and the unit in each quantity.
1. 99 bottles of soda
2. 60 miles per hour
3. 32 fluid ounces
4. 98.6 degrees Fahrenheit
Answer a
The number is 99, and the unit is bottles of soda.
Answer b
The number is 60, and the unit is miles per hour.
Answer c
The number 32, and the unit is fluid ounces
Answer d
The number is 98.6, and the unit is degrees Fahrenheit
The International System of Units
How long is a yard? It depends on whom you ask and when you asked the question. Today we have a standard definition of the yard, which you can see marked on every football field. If you move the ball ten yards, you get a first down and it doesn't matter whether you are playing in Los Angeles, Dallas, or Green Bay. But at one time that yard was arbitrarily defined as the distance from the tip of the king's nose to the end of his outstretched hand. Of course, the problem there is simple: new king, new distance (and then you have to remark all those football fields).
SI Base Units
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 seven base units 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.
The Metric System is Not Ubiquitously Adopted
The map below shows the adoption of the SI units in countries around the world. The United States has legally adopted the metric system for measurements, but does not use it in everyday practice. Great Britain and much of Canada use a combination of metric and imperial units.
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. Commonly used prefixes are bolded.
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 $\left( \text{pm} \right)=10^{-12} \: \text{m}$
Just as expressing a quantity without a unit is meaningless, so is using the incorrect format for units and prefixes. As you may have noticed, most metric abbreviations are lowercase. We use "$\text{m}$" for meter and not "$\text{M}$", which you will see later to represent solution concentration, something very different from length. However, when it comes to volume, the base unit "liter" is abbreviated as "$\text{L}$" and not "$\text{l}$". So we would write 3.5 milliliters 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
Solutions to example explaining unit abbreviations.
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 $2$
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
Scientific Notation
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. In scientific notation, these numbers are expressed in the form
$N \times 10^n$
where N is a number 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. 7.86
4. 12,378
5. 0.00032
6. 61.06700
7. 2002.080
8. 0.01020
Solution
Solutions to example explaining how numbers look in scientific notation.
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 This is usually expressed simply as 7.86. (Recall that 100 = 1.) $7.86 \times 10^0$
d Because the decimal point was moved four places to the left, n = 4. $1.2378 \times 10^4$
e Because the decimal point was moved four places to the right, n = −4. $3.2 \times 10^{−4}$
f Because the decimal point was moved one place to the left, n = 1.
$6.106700 \times 10^1$
g Because the decimal point was moved three places to the left, n = 3. $2.002080 \times 10^3$
h Because the decimal point was moved two places to the right, n = -2. $1.020 \times 10^{−2}$
Scientific Notation: Addition and Subtraction
Before numbers expressed in scientific notation can be added or subtracted, they must be converted to a form in which all the exponents have the same value. The appropriate operation is then carried out on the values of N. Example $2$ illustrates how to do this.
Example $2$: Expressing Sums and Differences in Scientific Notation
Carry out the appropriate operation and then express the answer in scientific notation.
1. $(1.36 \times 10^2) + (4.73 \times 10^3) \nonumber$
2. $(6.923 \times 10^{−3}) − (8.756 \times 10^{−4}) \nonumber$
Solution
Solution to example explaining how sums and differences look in scientific notation.
Explanation Answer
a
Both exponents must have the same value, so these numbers are converted to either
$(1.36 \times 10^2) + (47.3 \times 10^2) = (1.36 + 47.3) \times 10^2 = 48.66 × 10^2$
or
$(0.136 \times 10^3) + (4.73 \times 10^3) = (0.136 + 4.73) \times 10^3) = 4.87 \times 10^3$.
Choosing either alternative gives the same answer, reported to two decimal places:
In converting 48.66 × 102 to scientific notation, $n$ has become more positive by 1 because the value of $N$ has decreased.
$4.87 \times 10^3$
b
Converting the exponents to the same value gives either
$(6.923 \times 10^{-3}) − (0.8756 \times 10^{-3}) = (6.923 − 0.8756) \times 10^{−3}$
or
$(69.23 \times 10^{-4}) − (8.756 \times 10^{-4}) = (69.23 − 8.756) \times 10^{−4} = 60.474 \times 10^{−4}$.
In converting 60.474 × 10-4 to scientific notation, $n$ has become more positive by 1 because the value of $N$ has decreased.
$6.047 \times 10^{−3}$
Scientific Notation: Multiplication and Division
When multiplying numbers expressed in scientific notation, we multiply the values of $N$ and add together the values of $n$. Conversely, when dividing, we divide $N$ in the dividend (the number being divided) by $N$ in the divisor (the number by which we are dividing) and then subtract n in the divisor from n in the dividend. In contrast to addition and subtraction, the exponents do not have to be the same in multiplication and division. Examples of problems involving multiplication and division are shown in Example $3$.
Example $3$: Expressing Products and Quotients in Scientific Notation
Perform the appropriate operation and express your answer in scientific notation.
1. $[ (6.022 \times 10^{23})(6.42 \times 10^{−2}) \nonumber$
2. $\dfrac{ 1.67 \times 10^{-24} }{ 9.12 \times 10 ^{-28} } \nonumber$
3. $\dfrac{ (6.63 \times 10^{−34})(6.0 \times 10) }{ 8.52 \times 10^{−2}} \nonumber$
Solution
Solutions to example explaining how products and quotients look in scientific notation.
Explanation Answer
a
In multiplication, we add the exponents:
$(6.022 \times 10^{23})(6.42 \times 10^{−2})= (6.022)(6.42) \times 10^{[23 + (−2)]} = 38.7 \times 10^{21} \nonumber$
In converting $38.7 \times 10^{21}$ to scientific notation, $n$ has become more positive by 1 because the value of $N$ has decreased.
$3.87 \times 10^{22}$
b
In division, we subtract the exponents:
${1.67 \times 10^{−24} \over 9.12 \times 10^{−28}} = {1.67 \over 9.12} \times 10^{[−24 − (−28)]} = 0.183 \times 10^4 \nonumber$
In converting $0.183 \times 10^4$ to scientific notation, $n$ has become more negative by 1 because the value of $N$ has increased.
$1.83 \times 10^3$
c
This problem has both multiplication and division:
${(6.63 \times 10^{−34})(6.0 \times 10) \over (8.52 \times 10^{−2})} = {39.78 \over 8.52} \times 10^{[−34 + 1 − (−2)]} \nonumber$
$4.7\times 10^{-31}$
Key Takeaways
• Identifying a quantity properly requires both a number and a unit.
• 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. | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/01%3A_Matter_and_Measurements/1.06%3A_Physical_Quantities_-_Units_and_Scientific_Notation.txt |
Learning Objective
• Describe the units (and abbreviations) that go with various quantities.
• Derive new units by combining numerical prefixes with units.
Mass vs. Weight
One of the many interesting things about travel in outer space is the idea of weightlessness. If something is not fastened down, it will float in mid-air. Early astronauts learned that weightlessness had bad effects on bone structure. If there was no pressure on the legs, those bones would begin to lose mass. Weight provided by gravity is needed to maintain healthy bones. Specially designed equipment is now a part of every space mission so the astronauts can maintain good body fitness.
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 $1 \: \text{g}$.
The Difference Between Mass and Weight
The mass of a body is a measure of its inertial property or how much matter it contains. The weight of a body is a measure of the force exerted on it by gravity or the force needed to support it. Gravity on earth gives a body a downward acceleration of about 9.8 m/s2. In common parlance, weight is often used as a synonym for mass in weights and measures. For instance, the verb “to weigh” means “to determine the mass of” or “to have a mass of.” The incorrect use of weight in place of mass should be phased out, and the term mass used when mass is meant. The SI unit of mass is the kilogram (kg). In science and technology, the weight of a body in a particular reference frame is defined as the force that gives the body an acceleration equal to the local acceleration of free fall in that reference frame. Thus, the SI unit of the quantity weight defined in this way (force) is the newton (N).
Length
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
In addition to the fundamental units, SI also allows for derived units based on a fundamental unit or units. There are many derived units used in science. For example, the derived unit for area comes from the idea that area is defined as width times height. Because both width and height are lengths, they both have the fundamental unit of meter, so the unit of area is meter × meter, or meter2 (m2). This is sometimes spoken as "square meters." A unit with a prefix can also be used to derive a unit for area, so we can also have cm2, mm2, or km2 as acceptable units for area.
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. One 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 $1 \: \text{dm}^3$. A smaller unit of volume that is commonly used is the milliliter ($\text{mL}$ - note the capital $\text{L}$ which is a standard practice). 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 $\left( \text{cm}^3 \right)$. There are $1000 \: \text{mL}$ in $1 \: \text{L}$, which is the same as saying that there are $1000 \: \text{cm}^3$ in $1 \: \text{dm}^3$.
Another definition of a liter is one-tenth of a meter cubed. Because one-tenth of a meter is 10 cm, then a liter is equal to 1,000 cm3 (Figure $3$). Because 1 L equals 1,000 mL, we conclude that 1 mL equals 1 cm3; thus, these units are interchangeable.
Units not only are multiplied together but also can be divided. For example, if you are traveling at one meter for every second of time elapsed, your velocity is 1 meter per second, or 1 m/s. The word per implies division, so velocity is determined by dividing a distance quantity by a time quantity. Other units for velocity include kilometers per hour (km/h) or even micrometers per nanosecond (μm/ns). Later, we will see other derived units that can be expressed as fractions.
Key Takeaways
• Mass is a measure of the amount of matter that an object contains. Mass is independent of location.
• Weight is a measure of force that is equal to the gravitational pull on an object. Weight depends on location.
• Units can be multiplied and divided to generate new units for quantities like the liter for volume. | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/01%3A_Matter_and_Measurements/1.07%3A_Measuring_Mass_Length_and_Volume.txt |
Learning Objectives
• Identify the number of significant figures in a reported value.
• Apply the concept of significant figures to report a measurement with the proper number of digits.
Scientists have established certain conventions for communicating the degree of precision of a measurement, which is dependent on the measuring device used (See Figure $1$). Imagine, for example, that you are using a meterstick to measure the width of a table. The centimeters (cm) marked on the meterstick, tell you how many centimeters wide the table is. Many metersticks also have markings for millimeters (mm), so we can measure the table to the nearest millimeter. The measurement made using millimeters is more precise, it is closer to the actual length of the table. Most metersticks do not have any smaller (or more precise) markings indicated, so you cannot report the measured width of the table any more precise than to the nearest millimeter.
The concept of significant figures takes the limitation of measuring devices into account. The significant figures of a measured quantity are defined as all the digits known with certainty (those indicated by the markings on the measuring device) and the first uncertain, or estimated, digit (one digit past the smallest marking on the measuring device). It makes no sense to estimate and report any digits after the first uncertain one, so it is the last significant digit reported in a measurement. Zeros are used when needed to place the significant figures in their correct positions. Thus, zeros are sometimes counted as significant figures but are sometimes only used as placeholders (see the rules for significant figures below for more details).
“Sig figs” is a common abbreviation for significant figures.
Rules for Determination of Significant Figures
Consider the earlier example of measuring the width of a table with a meterstick. If the table is measured and reported as being 1,357 mm wide, the number 1,357 has four significant figures. The 1 (thousands place), the 3 (hundreds place), and the 5 (tens place) are certain; the 7 (ones place) is assumed to have been estimated. It would make no sense to report such a measurement as 1,357.0 (five Sig Figs) or 1,357.00 (six Sig Figs) because that would suggest the measuring device was able to determine the width to the nearest tenth or hundredth of a millimeter, when in fact it shows only tens of millimeters and therefore the ones place was estimated.
On the other hand, if a measurement is reported as 150 mm, the 1 (hundreds) and the 5 (tens) are known to be significant, but how do we know whether the zero is or is not significant? The measuring device could have had marks indicating every 100 mm or marks indicating every 10 mm. How can you determine if the zero is significant (the estimated digit), or if the 5 is significant and the zero a value placeholder?
The rules for deciding which digits in a measurement are significant are as follows:
Rule 1: All nonzero digits in a measurement are significant.
• 237 has three significant figures.
• 1.897 has four significant figures.
Rule 2: Zeros that appear between other nonzero digits (i.e., "middle zeros") are always significant.
• 39,004 has five significant figures.
• 5.02 has three significant figures.
Rule 3: Zeros that appear in front of all of the nonzero digits are called leading zeros. Leading zeros are never significant.
• 0.008 has one significant figure.
• 0.000416 has three significant figures.
Rule 4: Zeros that appear after all nonzero digits are called trailing zeros. A number with trailing zeros that lacks a decimal point may or may not be significant.
• 1400 is ambiguous.
• $1.4 \times 10^3$ has two significant figures.
• $1.40 \times 10^3$ three significant figures.
• $1.400 \times 10^3$ has four significant figures.
Rule 5: Trailing zeros in a number with a decimal point are significant. This is true whether the zeros occur before or after the decimal point.
• 620.0 has four significant figures.
• 19.000 has five significant figures.
It needs to be emphasized that just because a certain digit is not significant does not mean that it is not important or that it can be left out. Though the zero in a measurement of 140 may not be significant, the value cannot simply be reported as 14. An insignificant zero functions as a placeholder for the decimal point. When numbers are written in scientific notation, this becomes more apparent. The measurement 140 can be written as $1.4 \times 10^2$, with two significant figures in the coefficient or as $1.40 \times 10^3$, with three significant figures. A number less than one, such as 0.000416, can be written in scientific notation as $4.16 \times 10^{-4}$, which has 3 significant figures. In some cases, scientific notation is the only way to correctly indicate the correct number of significant figures. In order to report a value of 15,000,00 with four significant figures, it would need to be written as $1.500 \times 10^7$.
Exact Quantities
When numbers are known exactly, the significant figure rules do not apply. This occurs when objects are counted rather than measured. For example, a carton of eggs has 12 eggs. The actual value cannot be 11.8 eggs, since we count eggs in whole number quantities. So the 12 is an exact quantity. Exact quantities are considered to have an infinite number of significant figures; the importance of this concept will be seen later when we begin looking at how significant figures are dealt with during calculations. Numbers in many conversion factors, especially for simple unit conversions, are also exact quantities and have infinite significant figures. There are exactly 100 centimeters in 1 meter and exactly 60 seconds in 1 minute. Those values are definitions and are not the result of a measurement.
Example $2$
Give the number of significant figures in each. Identify the rule for each.
1. 5.87
2. 0.031
3. 52.90
4. 00.2001
5. 500
6. 6 atoms
Solution
Solution to example explaining how many significant figures are in each number.
Explanation Answer
a. All three numbers are significant (rule 1). 5.87 , three significant figures
b. The leading zeros are not significant (rule 3). The 3 and the 1 are significant (rule 1) 0.031, two significant figures
c. The 5, the 2 and the 9 are significant (rule 1). The trailing zero is also significant (rule 5). 52.90, four significant figures
d. The leading zeros are not significant (rule 3). The 2 and the 1 are significant (rule 1) and the middle zeros are also significant (rule 2). 00.2001, four significant figures
e. The number is ambiguous. It could have one, two or three significant figures. 500, ambiguous
f. The 6 is a counting number. A counting number is an exact number. 6, infinite
Exercise $2$
Give the number of significant figures in each.
1. 36.7 m
2. 0.006606 s
3. 2,002 kg
4. 306,490,000 people
5. 3,800 g
Answer a:
three significant figures.
Answer b:
four significant figures.
Answer c:
four significant figures.
Answer d:
Infinite (Exact number)
Answer e:
Ambiguous, could be two, three or four significant figure.
Summary
Uncertainty exists in all measurements. The degree of uncertainty is affected in part by the quality of the measuring tool. Significant figures give an indication of the certainty of a measurement. Rules allow decisions to be made about how many digits to use in any given situation. | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/01%3A_Matter_and_Measurements/1.08%3A_Measurement_and_Significant_Figures.txt |
Learning Objectives
• Use significant figures correctly in arithmetical operations.
Rounding
Before dealing with the specifics of the rules for determining the significant figures in a calculated result, we need to be able to round numbers correctly. To round a number, first decide how many significant figures the number should have. Once you know that, round to that many digits, starting from the left. If the number immediately to the right of the last significant digit is less than 5, it is dropped and the value of the last significant digit remains the same. If the number immediately to the right of the last significant digit is greater than or equal to 5, the last significant digit is increased by 1.
Consider the measurement $207.518 \: \text{m}$. Right now, the measurement contains six significant figures. How would we successively round it to fewer and fewer significant figures? Follow the process as outlined in Table $1$.
Number of Significant Figures Rounded Value Reasoning
Table $1$: Rounding examples
6 207.518 All digits are significant
5 207.52 8 rounds the 1 up to 2
4 207.5 2 is dropped
3 208 5 rounds the 7 up to 8
2 210 8 is replaced by a 0 and rounds the 0 up to 1
1 200 1 is replaced by a 0
Notice that the more rounding that is done, the less reliable the figure is. An approximate value may be sufficient for some purposes, but scientific work requires a much higher level of detail.
It is important to be aware of significant figures when you are mathematically manipulating numbers. For example, dividing 125 by 307 on a calculator gives 0.4071661238… to an infinite number of digits. But do the digits in this answer have any practical meaning, especially when you are starting with numbers that have only three significant figures each? When performing mathematical operations, there are two rules for limiting the number of significant figures in an answer—one rule is for addition and subtraction, and one rule is for multiplication and division.
In operations involving significant figures, the answer is reported in such a way that it reflects the reliability of the least precise operation. An answer is no more precise than the least precise number used to get the answer.
Rules for Calculations With Measured Numbers
How are significant figures handled in calculations? It depends on what type of calculation is being performed.
Multiplication and Division
For multiplication or division, the rule is to count the number of significant figures in each number being multiplied or divided and then limit the significant figures in the answer to the lowest count. An example is as follows:
The final answer, limited to four significant figures, is 4,094. The first digit dropped is 1, so we do not round up.
Scientific notation provides a way of communicating significant figures without ambiguity. You simply include all the significant figures in the leading number. For example, the number 450 has two significant figures and would be written in scientific notation as 4.5 × 102, whereas 450.0 has four significant figures and would be written as 4.500 × 102. In scientific notation, all significant figures are listed explicitly.
Example $1$
Write the answer for each expression using scientific notation with the appropriate number of significant figures.
1. 23.096 × 90.300
2. 125 × 9.000
Solution
a
Solution to (a) in example question.
Explanation Answer
The calculator answer is 2,085.5688, but we need to round it to five significant figures. Because the first digit to be dropped (in the tenths place) is greater than 5, we round up to 2,085.6. $2.0856 \times 10^3$
b
Solution to (b) in example question.
Explanation Answer
The calculator gives 1,125 as the answer, but we limit it to three significant figures. $1.13 \times 10^3$
Addition and Subtraction
If the calculation is an addition or a subtraction, the rule is as follows: limit the reported answer to the rightmost column that all numbers have significant figures in common. For example, if you were to add 1.2 and 4.71, we note that the first number stops its significant figures in the tenths column, while the second number stops its significant figures in the hundredths column. We therefore limit our answer to the tenths column.
We drop the last digit—the 1—because it is not significant to the final answer.
The dropping of positions in sums and differences brings up the topic of rounding. Although there are several conventions, in this text we will adopt the following rule: the final answer should be rounded up if the first dropped digit is 5 or greater and rounded down if the first dropped digit is less than 5.
Example $2$
1. 13.77 + 908.226
2. 1,027 + 611 + 363.06
Solution
a
Solution to (a) in example question.
Explanation Answer
The calculator answer is 921.996, but because 13.77 has its farthest-right significant figure in the hundredths place we need to round the final answer to the hundredths position. Because the first digit to be dropped (in the thousandths place) is greater than 5, we round up to 922.00 $922.00 = 9.2200 \times 10^2$
b
Solution to (b) in example question.
Explanation Answer
The calculator gives 2,001.06 as the answer, but because 611 and 1027 has its farthest-right significant figure in the ones place, the final answer must be limited to the ones position. $2,001.06 = 2.001 \times 10^3$
Exercise $2$
Write the answer for each expression using scientific notation with the appropriate number of significant figures.
1. 217 ÷ 903
2. 13.77 + 908.226 + 515
3. 255.0 − 99
4. 0.00666 × 321
Answer a:
$0.240 = 2.40 \times 10^{-1}$
Answer b:
$1,437 = 1.437 \times 10^3$
Answer c:
$156 = 1.56 \times 10^2$
Answer d:
$2.14 = 2.14 \times 10^0$
Remember that calculators do not understand significant figures. You are the one who must apply the rules of significant figures to a result from your calculator
Example $3$
1. 2(1.008g) + 15.99 g
2. 137.3 s + 2(35.45 s)
3. ${118.7 g \over 2} - 35.5 g$
Solution
a
Solution to (a) in example question.
Explanation Answer
2(1.008g) + 15.99 g =
Perform multiplication first.
2 (1.008g 4 sig figs) = 2.016 g 4 sig figs
The number with the least number of significant figures is 1.008g; the number 2 is an exact number and therefore has infinite number of significant figures.
Then, perform the addition.
2.016 g thousandths place + 15.99 g hundredths place (least precise) = 18.006 g
Round the final answer.
Round the final answer to the hundredths place since 15.99 has its farthest right significant figure in the hundredths place (least precise).
18.01 g (rounding up)
b
Solution to (b) in example question.
Explanation Answer
137.3 s + 2 (35.45 s) =
Perform multiplication first.
2 (35.45 s 4 sig figs) = 70.90 s 4 sig figs
The number with the least number of significant figures is 35.45; the number 2 is an exact number and therefore has infinite number of significant figures.
Then, perform the addition.
137.3 s tenths place (least precise) + 70.90 s hundredths place = 208.20 s
Round the final answer.
Round the final answer to the tenths place based on 137.3 s.
208.2 s
c
Solution to (c) in example question.
Explanation Answer
${118.7 g \over 2} - 35.5 g$ =
Perform division first.
${118.7g \over 2}$ 4 sig figs = 59.35 g 4 sig figs
The number with the least number of significant figures is 118.7g; the number 2 is an exact number and therefore has infinite number of significant figures.
Perform subtraction next.
59.35 g hundredths place − 35.5 g tenths place (least precise) = 23.85 g
Round the final answer.
Round the final answer to the tenths place based on 35.5 g.
23.9 g (rounding up)
Exercise $3$
Complete the calculations and report your answers using the correct number of significant figures.
1. 5(1.008s) - 10.66 s
2. 99.0 cm+ 2(5.56 cm)
Answer a
-5.62 s
Answer b
110.2 cm
Summary
• Rounding
• If the number to be dropped is greater than or equal to 5, increase the number to its left by 1. e.g. 2.9699 rounded to three significant figures is 2.97
• If the number to be dropped is less than 5, there is no change. e.g. 4.00443 rounded to four sig. figs. is 4.004
• The rule in multiplication and division is that the final answer should have the same number of significant figures as there are in the number with the fewest significant figures.
• The rule in addition and subtraction is that the answer is given the same number of decimal places as the term with the fewest decimal places.
Contributions & Attributions
This page was constructed from content via the following contributor(s) and edited (topically or extensively) by the LibreTexts development team to meet platform style, presentation, and quality:
• Henry Agnew (UC Davis) | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/01%3A_Matter_and_Measurements/1.09%3A_Rounding_Off_Numbers.txt |
Learning Objectives
• Convert a value reported in one unit to a corresponding value in a different unit.
The ability to convert from one unit to another is an important skill. For example, a nurse with 50 mg aspirin tablets who must administer 0.2 g of aspirin to a patient needs to know that 0.2 g equals 200 mg, so 4 tablets are needed. Fortunately, there is a simple way to convert from one unit to another.
Conversion Factors
If you learned the SI units and prefixes described previously, then you know that 1 cm is 1/100th of a meter.
$1\; \rm{cm} = \dfrac{1}{100} \; \rm{m} \nonumber$
or
$100\; \rm{cm} = 1\; \rm{m} \nonumber$
Suppose we divide both sides of the equation by 1 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:
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. A fraction that has equivalent quantities in the numerator and the denominator but expressed in different units is called a conversion factor.
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:
$\dfrac{3.55}{1} \times \dfrac{100 \; \rm{cm}}{1} = 355 \; \rm{cm} \label{Ex1}$
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:
$\text{quantity (in old units)} \times \text{conversion factor} = \text{quantity (in new units)} \nonumber$
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 (Equation \ref{Ex1}), 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.
Significant Figures in Conversions
How do conversion factors affect the determination of significant figures? Numbers in conversion factors based on prefix changes, such as kilograms to grams, are not considered in the determination of significant figures in a calculation because the numbers in such conversion factors are exact. Exact numbers are defined or counted numbers, not measured numbers, and can be considered as having an infinite number of significant figures. (In other words, 1 kg is exactly 1,000 g, by the definition of kilo-.) Counted numbers are also exact. If there are 16 students in a classroom, the number 16 is exact. In contrast, conversion factors that come from measurements (such as density, as we will see shortly) or are approximations have a limited number of significant figures and should be considered in determining the significant figures of the final answer.
Example $1$
1. The average volume of blood in an adult male is 4.7 L. What is this volume in milliliters?
2. A hummingbird can flap its wings once in 18 ms. How many seconds are in 18 ms?
Solution
1. We start with what we are given, 4.7 L. We want to change the unit from liters to milliliters. There are 1,000 mL in 1 L. From this relationship, we can construct two conversion factors:
$\dfrac{1\; \rm{L}}{1,000\; \rm{mL}} \; \text{ or } \; \dfrac{1,000 \; \rm{mL}}{1\; \rm{L}} \nonumber$
We use the conversion factor that will cancel out the original unit, liters, and introduce the unit we are converting to, which is milliliters. The conversion factor that does this is the one on the right.
$4.7 \cancel{\rm{L}} \times \dfrac{1,000 \; \rm{mL}}{1\; \cancel{\rm{L}}} = 4,700\; \rm{mL} \nonumber$
Because the numbers in the conversion factor are exact, we do not consider them when determining the number of significant figures in the final answer. Thus, we report two significant figures in the final answer.
1. We can construct two conversion factors from the relationships between milliseconds and seconds:
$\dfrac{1,000 \; \rm{ms}}{1\; \rm{s}} \; \text{ or } \; \dfrac{1\; \rm{s}}{1,000 \; \rm{ms}} \nonumber$
To convert 18 ms to seconds, we choose the conversion factor that will cancel out milliseconds and introduce seconds. The conversion factor on the right is the appropriate one. We set up the conversion as follows:
$18 \; \cancel{\rm{ms}} \times \dfrac{1\; \rm{s}}{1,000 \; \cancel{\rm{ms}}} = 0.018\; \rm{s} \nonumber$
The conversion factor’s numerical values do not affect our determination of the number of significant figures in the final answer.
Exercise $1$
Perform each conversion.
1. 101,000 ns to seconds
2. 32.08 kg to grams
Answer a
$101,000 \cancel{\rm{ns}} \times \dfrac{1\; \rm{s}}{1,000,000,000\; \cancel{\rm{ns}}} = 0.000101\; \rm{s} \nonumber$
Answer b
$32.08 \cancel{\rm{kg}} \times \dfrac{1,000 \; \rm{g}}{1\; \cancel{\rm{kg}}} = 32,080\; \rm{g} \nonumber$
Conversion Factors From Different Units
Conversion factors can also be constructed for converting between different kinds of units. For example, density can be used to convert between the mass and the volume of a substance. Consider mercury, which is a liquid at room temperature and has a density of 13.6 g/mL. The density tells us that 13.6 g of mercury have a volume of 1 mL. We can write that relationship as follows:
13.6 g mercury = 1 mL mercury
This relationship can be used to construct two conversion factors:
$\mathrm{\dfrac{13.6\:g}{1\:mL}\:and\:\dfrac{1\:mL}{13.6\:g}} \nonumber$
Which one do we use? It depends, as usual, on the units we need to cancel and introduce. For example, suppose we want to know the mass of 16 mL of mercury. We would use the conversion factor that has milliliters on the bottom (so that the milliliter unit cancels) and grams on top so that our final answer has a unit of mass:
\begin{align*} \mathrm{16\:\cancel{mL}\times\dfrac{13.6\:g}{1\:\cancel{mL}}} &= \mathrm{217.6\:g} \[4pt] &\approx \mathrm{220\:g} \end{align*} \nonumber
In the last step, we limit our final answer to two significant figures because the volume quantity has only two significant figures; the 1 in the volume unit is considered an exact number, so it does not affect the number of significant figures. The other conversion factor would be useful if we were given a mass and asked to find volume, as the following example illustrates.
Density can be used as a conversion factor between mass and volume.
Example $2$: Mercury Thermometer
A mercury thermometer for measuring a patient’s temperature contains 0.750 g of mercury. What is the volume of this mass of mercury?
Solution
Because we are starting with grams, we want to use the conversion factor that has grams in the denominator. The gram unit will cancel algebraically, and milliliters will be introduced in the numerator.
\begin{align*} 0.750 \; \cancel{\rm{g}} \times \dfrac{1\; \rm{mL}}{13.6 \; \cancel{\rm{g}}} &= 0.055147 \ldots \; \rm{mL} \[4pt] &\approx 0.0551\; \rm{mL} \end{align*} \nonumber
We have limited the final answer to three significant figures.
Exercise $2$
What is the volume of 100.0 g of air if its density is 1.3 g/L?
Answer
$100.0 \cancel{\rm{g}} \times \dfrac{1\; \rm{L}}{1.3\; \cancel{\rm{g}}} = 76.92307692\; \rm{L} ≈ 77 L \nonumber$
Because the density (1.3 g/L) has only 2 significant figures, we are rounding off the final answer to 2 significant figures.
Problem Solving With 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. You can either memorize the relationship between kilometers and millimeters, or you can do the conversion in two steps. Most people prefer to convert in steps.
To do a stepwise conversion, we first convert the given amount to the base unit. In this example, the base unit is meters. We know that there are 1,000 m in 1 km:
$54.7\; \cancel{\rm{km}} \times \dfrac{1,000 \; \rm{m}}{1\; \cancel{\rm{km}}} = 54,700\; \rm{m} \nonumber$
Then we take the result (54,700 m) and convert it to millimeters, remembering that there are $1,000\; \rm{mm}$ for every $1\; \rm{m}$:
\begin{align*} 54,700 \; \cancel{\rm{m}} \times \dfrac{1,000 \; \rm{mm}}{1\; \cancel{\rm{m}}} &= 54,700,000 \; \rm{mm} \[4pt] &= 5.47 \times 10^7\; \rm{mm} \end{align*} \nonumber
We have expressed the final answer in scientific notation.
As a shortcut, both steps in the conversion can be combined into a single, multistep expression:
Concept Map
Calculation
\begin{align*} 54.7\; \cancel{\rm{km}} \times \dfrac{1,000 \; \cancel{\rm{m}}}{1\; \cancel{\rm{km}}} \times \dfrac{1,000 \; \rm{mm}}{1\; \cancel{\rm{m}}} &= 54,700,000 \; \rm{mm} \[4pt] &= 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 canceling out until only the unit needed in the answer is left.
Either method—one step at a time or all the steps together—is acceptable. If you do all the steps together, the restriction for the proper number of significant figures should be done after the last step. As long as the math is performed correctly, you should get the same answer no matter which method you use.
Example $3$
Convert 58.2 ms to megaseconds in one multistep calculation.
Solution
First, convert the given unit (ms) to the base unit—in this case, seconds—and then convert seconds to the final unit, megaseconds:
Calculation
\begin{align*} 58.2 \; \cancel{\rm{ms}} \times \dfrac{\cancel{1 \rm{s}}}{1,000\; \cancel{\rm{ms}}} \times \dfrac{1\; \rm{Ms}}{1,000,000\; \cancel{ \rm{s}}} &=0.0000000582\; \rm{Ms} \[4pt] &= 5.82 \times 10^{-8}\; \rm{Ms} \end{align*} \nonumber
Neither conversion factor affects the number of significant figures in the final answer.
Exercise $3$
Convert 43.007 mg to kilograms in one multistep calculation.
Answer
\begin{align*} 43.007 \; \cancel{\rm{mg}} \times \dfrac{\cancel{1 \rm{g}}}{1,000\; \cancel{\rm{mg}}} \times \dfrac{1\; \rm{kg}}{1,000\; \cancel{ \rm{g}}} &=0.000043007\; \rm{kg} \[4pt] &= 4.3007 \times 10^{-5}\; \rm{kg} \end{align*} \nonumber.
Neither conversion factor affects the number of significant figures in the final answer.
Career Focus: Pharmacist
A pharmacist dispenses drugs that have been prescribed by a doctor. Although that may sound straightforward, pharmacists in the United States must hold a doctorate in pharmacy and be licensed by the state in which they work. Most pharmacy programs require four years of education in a specialty pharmacy school.
Pharmacists must know a lot of chemistry and biology so they can understand the effects that drugs (which are chemicals, after all) have on the body. Pharmacists can advise physicians on the selection, dosage, interactions, and side effects of drugs. They can also advise patients on the proper use of their medications, including when and how to take specific drugs properly. Pharmacists can be found in drugstores, hospitals, and other medical facilities.
Curiously, an outdated name for pharmacist is chemist, which was used when pharmacists formerly did a lot of drug preparation, or compounding. In modern times, pharmacists rarely compound their own drugs, but their knowledge of the sciences, including chemistry, helps them provide valuable services in support of everyone’s health.
Key Takeaway
• A unit can be converted to another unit of the same type with a conversion factor.
Concept Review Exercises
1. How do you determine which quantity in a conversion factor goes in the denominator of the fraction?
2. State the guidelines for determining significant figures when using a conversion factor.
3. Write a concept map (a plan) for how you would convert $1.0 \times 10^{12}$ nanoliters (nL) to kiloliters (kL).
Answers
1. The unit you want to cancel from the numerator goes in the denominator of the conversion factor.
2. Exact numbers that appear in many conversion factors do not affect the number of significant figures; otherwise, the normal rules of multiplication and division for significant figures apply.
3. Concept Map: Convert the given (nanoliters, nL) to liters; then convert liters to kiloliters. | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/01%3A_Matter_and_Measurements/1.10%3A_Problem_Solving_-_Unit_Conversions_and_Estimating_Answers.txt |
Learning Objectives
• To identify the different between temperature and heat
• To recognize the different scales used to measuring temperature
Temperature
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 Temperature 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 Temperature 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 Temperature 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 kelvin.
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:
$T \,\text{(in °C)} + 273.15 = T \, \text{(in K)} \tag{3.10.1} \label{3.10.1}$
$T \, \text{ (in K)} − 273.15 = T \; \text{(in °C)} \tag{3.10.2} \label{3.10.2}$
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: where the coefficient for degrees Fahrenheit is exact. (Some calculators have a function that allows you to convert directly between °F and °C.) There is only one temperature for which the numerical value is the same on both the Fahrenheit and Celsius scales: −40°C = −40°F. The relationship between the scales are as follows:
$°C = \dfrac{5}{9} \times (°F-32) \tag{3.10.3} \label{3.10.3}$
$°F = \dfrac{9}{5} \times (°C)+32 \tag{3.10.4} \label{3.10.4}$
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{3.10.3}:
\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{3.10.1}:
\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 K, 9980 °F
Answer (b)
2807 K, 5084 °F
Answer (c)
-195.79 K, -320.42 °F
Heat
While the concept of temperature may seem familiar to you, many people confuse temperature with heat. as discussed above, 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.
When scientists speak of heat, they are referring to energy that is transferred from an object with a higher temperature to an object with a lower temperature as a result of the temperature difference. Heat will "flow" from the hot object to the cold object until both end up at the same temperature. When you cook with a metal pot, you witness energy being transferred in the form of heat. Initially, only the stove element is hot – the pot and the food inside the pot are cold. As a result, heat moves from the hot stove element to the cold pot. After a while, enough heat is transferred from the stove to the pot, raising the temperature of the pot and all of its contents.
Energy
Just like matter, energy is a term that we are all familiar with and use on a daily basis. Before you go on a long hike, you eat an energy bar; every month, the energy bill is paid; on TV, politicians argue about the energy crisis. But what is energy? If you stop to think about it, energy is very complicated. When you plug a lamp into an electric socket, you see energy in the form of light, but when you plug a heating pad into that same socket, you only feel warmth. Without energy, we couldn't turn on lights, we couldn't brush our teeth, we couldn't make our lunch, and we couldn't travel to school. In fact, without energy, we couldn't even wake up because our bodies require energy to function. We use energy for every single thing that we do, whether we're awake or asleep.
Energy is measured in one of two common units: the calorie and the joule. The joule $\left( \text{J} \right)$ is the SI unit of energy. The calorie is familiar because it is commonly used when referring to the amount of energy contained within food. A calorie $\left( \text{cal} \right)$ is the quantity of heat required to raise the temperature of 1 gram of water by $1^\text{o} \text{C}$. For example, raising the temperature of $100 \: \text{g}$ of water from $20^\text{o} \text{C}$ to $22^\text{o} \text{C}$ would require $100 \times 2 = 200 \: \text{cal}$.
Calories contained within food are actually kilocalories $\left( \text{kcal} \right)$. In other words, if a certain snack contains 85 food calories, it actually contains $85 \: \text{kcal}$ or $85,000 \: \text{cal}$. In order to make the distinction, the dietary calorie is written with a capital C.
$1 \: \text{kilocalorie} = 1 \: \text{Calorie} = 1000 \: \text{calories}$
To say that the snack "contains" 85 Calories means that $85 \: \text{kcal}$ of energy are released when that snack is processed by your body.
Heat changes in chemical reactions are typically measured in joules rather than calories. The conversion between a joule and a calorie is shown below.
$1 \: \text{J} = 0.2390 \: \text{cal or} \: 1 \: \text{cal} = 4.184 \: \text{J}$
We can calculate the amount of heat released in kilojoules when a 400. Calorie hamburger is digested.
$400 \: \text{Cal} = 400 \: \text{kcal} \times \dfrac{4.184 \: \text{kJ}}{1 \: \text{kcal}} = 1.67 \times 10^3 \: \text{kJ}$
Heat Capacity and Specific Heat
If a swimming pool and wading pool, both full of water at the same temperature, were subjected to the same input of heat energy, the wading pool would certainly rise in temperature more quickly than the swimming pool. Because of its much larger mass, the swimming pool of water has a larger "heat capacity" than the wading pool. Similarly, different substances respond to heat in different ways. If a metal chair sits in the bright sun on a hot day, it may become quite hot to the touch. An equal mass of water in the same sun will not become nearly as hot.
We would say that water has a high heat capacity (the amount of heat required to raise the temperature of an object by $1^\text{o} \text{C}$). Water is very resistant to changes in temperature, while metals in general are not. The specific heat of a substance is the amount of energy required to raise the temperature of 1 gram of the substance by $1^\text{o} \text{C}$. The symbol for specific heat is $c_p$, with the $p$ subscript referring to the fact that specific heats are measured at constant pressure. The units for specific heat can either be joules per gram per degree $\left( \text{J/g}^\text{o} \text{C} \right)$ or calories per gram per degree $\left( \text{cal/g}^\text{o} \text{C} \right)$ (Table $1$). This text will use $\text{J/g}^\text{o} \text{C}$ for specific heat.
$\text{specific heat}= \dfrac{\text{heat}}{\text{mass} \times \text{cal/g}^\text{o} \text{C}}$
Notice that water has a very high specific heat compared to most other substances.
Table $1$: Specific Heat Capacities
Substance Specific Heat Capacity
at 25oC in J/g oC
Substance Specific Heat Capacity
at 25oC in J/g oC
$\ce{H2}$ gas 14.267 steam @ 100oC 2.010
$\ce{He}$ gas 5.300 vegetable oil 2.000
$\ce{H2O(l)}$ 4.184 sodium 1.23
lithium
3.56
air 1.020
ethyl alcohol
2.460
magnesium 1.020
ethylene glycol
2.200
aluminum 0.900
ice @ 0oC
2.010
Concrete 0.880
steam @ 100oC
2.010
glass 0.840
Water is commonly used as a coolant for machinery because it is able to absorb large quantities of heat (see table above). Coastal climates are much more moderate than inland climates because of the presence of the ocean. Water in lakes or oceans absorbs heat from the air on hot days and releases it back into the air on cool days.
Example $1$: Measuring Heat
A flask containing $\mathrm{8.0 \times 10^2\; g}$ of water is heated, and the temperature of the water increases from 21 °C to 85 °C. How much heat did the water absorb?
Solution
To answer this question, consider these factors:
• the specific heat of the substance being heated (in this case, water)
• the amount of substance being heated (in this case, 800 g)
• the magnitude of the temperature change (in this case, from 21 °C to 85 °C).
The specific heat of water is 4.184 J/g °C, so to heat 1 g of water by 1 °C requires 4.184 J. We note that since 4.184 J is required to heat 1 g of water by 1 °C, we will need 800 times as much to heat 800 g of water by 1 °C. Finally, we observe that since 4.184 J are required to heat 1 g of water by 1 °C, we will need 64 times as much to heat it by 64 °C (that is, from 21 °C to 85 °C).
This can be summarized using the equation:
\begin{align*} q&=c×m×ΔT \[4pt] &=c×m×(T_\ce{final}−T_\ce{initial}) \[4pt] &=\mathrm{(4.184\:J/\cancel{g}°C)×(800\:\cancel{g})×(85−21)°C}\[4pt] &=\mathrm{(4.184\:J/\cancel{g}°\cancel{C})×(800\:\cancel{g})×(64)°\cancel{C}}\[4pt] &=\mathrm{210,000\: J(=210\: kJ)} \end{align*}
Because the temperature increased, the water absorbed heat and $q$ is positive
Exercise $1$
How much heat, in joules, must be added to a $\mathrm{5.00 \times 10^2 \;g}$ iron skillet to increase its temperature from 25 °C to 250 °C? The specific heat of iron is 0.451 J/g °C.
Answer
$\mathrm{5.05 \times 10^4\; J}$
Summary
• Three different scales are commonly used to measure temperature: Fahrenheit (expressed as °F), Celsius (°C), and Kelvin (K).
• Heat capacity is the amount of heat required to raise the temperature of an object by $1^\text{o} \text{C}$).
• The specific heat of a substance is the amount of energy required to raise the temperature of 1 gram of the substance by $1^\text{o} \text{C}$.
Contributions & Attributions
This page was constructed from content via the following contributor(s) and edited (topically or extensively) by the LibreTexts development team to meet platform style, presentation, and quality: | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/01%3A_Matter_and_Measurements/1.11%3A_Temperature_Heat_and_Energy.txt |
Learning Objectives
• Define density and specific gravity.
• Perform calculations involving both density and specific gravity.
After trees are cut, logging companies often move these materials down a river to a sawmill where they can be shaped into building materials or other products. The logs float on the water because they are less dense than the water they are in. Knowledge of density is important in the characterization and separation of materials. Information about density allows us to make predictions about the behavior of matter.
Density
A golf ball and a table tennis ball are about the same size. However, the golf ball is much heavier than the table tennis ball. Now imagine a similar size ball made out of lead. That would be very heavy indeed! What are we comparing? By comparing the mass of an object relative to its size, we are studying a property called density. Density is the ratio of the mass of an object to its volume.
\begin{align} \text{density} &= \dfrac{\text{mass}}{\text{volume}} \label{eq1} \[4pt] D &= \dfrac{m}{V} \label{eq2} \end{align}
Density is usually a measured property of a substance, so its numerical value affects the significant figures in a calculation. Notice that density is defined in terms of two dissimilar units, mass and volume. That means that density overall has derived units, just like velocity. Common units for density include g/mL, g/cm3, g/L, kg/L, and even kg/m3. Densities for some common substances are listed in Table $1$.
Table $1$: Densities of Some Common Substances
Liquids and Solids Density at $20^\text{o} \text{C} \: \left( \text{g/mL} \right)$ Gases Density at $20^\text{o} \text{C} \: \left( \text{g/L} \right)$
Ethanol 0.79 Hydrogen 0.084
Ice $\left( 0^\text{o} \text{C} \right)$ 0.917 Helium 0.166
Corn oil 0.922 Air 1.20
Water 0.998 Oxygen 1.33
Water $\left( 4^\text{o} \text{C} \right)$ 1.000 Carbon dioxide 1.83
Corn syrup 1.36 Radon 9.23
Aluminum 2.70
Copper 8.92
Lead 11.35
Mercury 13.6
Gold 19.3
The SI units of density are kilograms per cubic meter $\left( \text{kg/m}^3 \right)$, since the $\text{kg}$ and the $\text{m}$ are the SI units for mass and length respectively. In everyday usage in a laboratory, this unit is awkwardly large. Most solids and liquids have densities that are conveniently expressed in grams per cubic centimeter $\left( \text{g/cm}^3 \right)$. Since a cubic centimeter is equal to a milliliter, density units can also be expressed as $\text{g/mL}$. Gases are much less dense than solids and liquids, so their densities are often reported in $\text{g/L}$. Water has a density of $1.0 \: \text{g/mL}$.
Because of how it is defined, density can act as a conversion factor for switching between units of mass and volume. For example, suppose you have a sample of aluminum that has a volume of 7.88 cm3. How can you determine what mass of aluminum you have without measuring it? You can use the volume to calculate it. If you multiply the given volume by the known density (from Table $1$), the volume units will cancel and leave you with mass units, telling you the mass of the sample:
Start with Equation \ref{eq1} $\text{density} = \dfrac{m}{V} \nonumber$
and insert the relavant numbers
$\dfrac{2.7g}{cm^3} = \dfrac{m}{7.88 \, cm^3} \nonumber$
Cross multiplying both sides (right numerator x left denominator = left numerator x right denominator), we get the following expression with answer and appropriate unit.
$7.88\cancel{cm^3}\times \dfrac{2.7\,g}{\cancel{cm^3}}= 21 g \text{ of aluminum} \nonumber$
Since most materials expand as temperature increases, the density of a substance is temperature dependent and usually decreases as temperature increases. You known that ice floats in water and it can be seen from the table that ice is less dense. Alternatively, corn syrup, being denser, would sink if placed in water.
Example $1$
An $18.2 \: \text{g}$ sample of zinc metal has a volume of $2.55 \: \text{cm}^3$. Calculate the density of zinc.
Solution
Step 1: List the known quantities and plan the problem.
Known
• Mass $= 18.2 \: \text{g}$
• Volume $= 2.55 \: \text{cm}^3$
Unknown
• Density $= ? \: \text{g/cm}^3$
Use Equation \ref{eq1} to solve the problem.
Step 2: Calculate
$D = \frac{m}{V} = \frac{18.2 \: \text{g}}{2.55 \: \text{cm}^3} = 7.14 \: \text{g/cm}^3$
Step 3: Think about your result.
If $1 \: \text{cm}^3$ of zinc has a mass of about 7 grams, then 2 and a half $\text{cm}^3$ will have a mass about 2 and a half times as great. Metals are expected to have a density greater than that of water and zinc's density falls within the range of the other metals listed above.
Since density values are known for many substances, density can be used to determine an unknown mass or an unknown volume. Dimensional analysis will be used to ensure that units cancel appropriately.
Example $2$
1. What is he mass of $2.49 \: \text{cm}^3$ of aluminum?
2. What is the volume of $50.0 \: \text{g}$ of aluminum?
Solution
Step 1: List the known quantities and plan the problem.
Known
• Density $= 2.70 \: \text{g/cm}^3$
• 1. Volume $= 2.49 \: \text{cm}^3$
• 2. Mass $= 50.0 \: \text{g}$
Unknown
• 1. Mass $= ? \: \text{g}$
• 2. Volume $= ? \: \text{cm}^3$
Use the equation for density, $D = \frac{m}{V}$, and dimensional analysis to solve each problem.
Step 2: Calculate
$1. \: \: 2.49 \: \text{cm}^3 \times \frac{2.70 \: \text{g}}{1 \: \text{cm}^3} = 6.72 \: \text{g}$
$2. \: \: 50.0 \: \text{g} \times \frac{1 \: \text{cm}^3}{2.70 \: \text{g}} = 18.5 \: \text{cm}^3$
In problem 1, the mass is equal to the density multiplied by the volume. In problem 2, the volume is equal to the mass divided by the density.
Step 3: Think about your results.
Because a mass of $1 \: \text{cm}^3$ of aluminum is $2.70 \: \text{g}$, the mass of about $2.5 \: \text{cm}^3$ should be about 2.5 times larger. The $50 \: \text{g}$ of aluminum is substantially more than its density, so that amount should occupy a relatively large volume.
Specific Gravity
Specific gravity is the ratio of the density (mass of a unit volume) of a substance to the density of a given reference material, often a liquid.
$\text{specific gravity} = \dfrac{\text{Density of a substance}(\cancel{g/mL})}{\text{Density of the water at the same temperature}(\cancel{g/mL})}$
If a substance's relative density is less than one then it is less dense than water and similarly, if greater than 1 then it is denser than water. If the relative density is exactly 1 then the densities are equal. For example, an ice cube, with a relative density of about 0.91, will float on water and a substance with a relative density greater than 1 will sink.
A hydrometer is an instrument used for measuring the specific density of liquids based on the concept of buoyancy (Figure $2$). A hydrometer usually consists of a sealed hollow glass tube with a wider bottom portion for buoyancy, a ballast such as lead or mercury for stability, and a narrow stem with graduations for measuring. The liquid to test is poured into a tall container, often a graduated cylinder, and the hydrometer is gently lowered into the liquid until it floats freely. The point at which the surface of the liquid touches the stem of the hydrometer correlates to relative density. Hydrometers can contain any number of scales along the stem corresponding to properties correlating to the density.
Summary
• Density is the ratio of the mass of an object to its volume.
• Gases are less dense than either solids or liquids.
• Both liquid and solid materials can have a variety of densities.
• For liquids and gases, the temperature will affect the density to some extent. | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/01%3A_Matter_and_Measurements/1.12%3A_Density_and_Specific_Gravity.txt |
• 2.1: Atomic Theory and the Structure of Atoms
Atoms are the ultimate building blocks of all matter. The modern atomic theory establishes the concepts of atoms and how they compose matter.
• 2.2: Elements and Atomic Number
Atoms are composed of three main subatomic particles: protons, neutrons, and electrons. Protons and neutrons are grouped together in the nucleus of an atom, while electrons orbit about the nucleus.
• 2.3: Isotopes and Atomic Weight
Elements can be identified by their atomic number and mass number. Isotopes are atoms of the same element that have different masses. Atoms have a mass that is based largely on the number of protons and neutrons in their nucleus.
• 2.4: The Periodic Table
The chemical elements are arranged in a chart called the periodic table. Some characteristics of the elements are related to their position on the periodic table.
• 2.5: Some Characteristics of Different Groups
The periodic table is useful for understanding atomic properties that show periodic trends. Periodic trends are specific patterns that are present in the periodic table that illustrate different aspects of a certain element, including its size and its electronic properties. Periodic trends, arising from the arrangement of the periodic table, provide chemists with an invaluable tool to quickly predict an element's properties.
• 2.6: Electronic Structure of Atoms
Electrons in atoms have quantized energies. The state of electrons in atoms is described by four quantum numbers. Electrons are organized into shells and subshells about the nucleus of an atom.
• 2.7: Electron Configurations
There are a set of general rules that are used to figure out the electron configuration of an atomic species: Aufbau Principle, Hund's Rule and the Pauli-Exclusion Principle.
• 2.8: Electron Configurations and the Periodic Table
The arrangement of electrons in atoms is responsible for the shape of the periodic table. Electron configurations can be predicted by the position of an atom on the periodic table
• 2.9: Electron-Dot Symbols
A Lewis electron dot diagram (or electron dot diagram or a Lewis diagram or a Lewis structure) is a representation of the valence electrons of an atom that uses dots around the symbol of the element.
02: Atoms and the Periodic Table
Learning Objectives
• State the modern atomic theory.
• Describe how atoms are constructed.
The smallest piece of an element that maintains the identity of that element is called an atom. Individual atoms are extremely small. It would take about fifty million atoms in a row to make a line that is 1 cm long. The period at the end of a printed sentence has several million atoms in it. Atoms are so small that it is difficult to believe that all matter is made from atoms-but it is.
The modern atomic theory, proposed about 1803 by the English chemist John Dalton, 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. With that size, it takes over 18 million of these atoms, lined up side by side, to equal the width of your little 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 a 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. 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.
• In chemical reactions, atoms are combined, separated, or rearranged.
Dalton's atomic theory has been largely accepted by the scientific community, with the exception of three changes. We know now that (1) an atom can be further subdivided, (2) all atoms of an element are not identical in mass, and (3) using nuclear fission and fusion techniques, we can create or destroy atoms by changing them into other atoms.
These concepts form the basis of chemistry. Although the word atom comes from a Greek word that means "indivisible," we understand now that atoms themselves are composed of smaller parts called subatomic particles. The first part to be discovered was the electron, a tiny subatomic particle with a negative charge. It is often represented as e, with the right superscript showing the negative charge. Later, two larger particles were discovered. The proton, a subatomic particle with a positive charge. is a more massive (but still tiny) subatomic particle with a positive charge, represented as p+. The neutron is a subatomic particle with about the same mass as a proton but no charge. It is represented as either n or n0. We now know that all atoms of all elements are composed of electrons, protons, and (with one exception) neutrons. Table $1$ summarizes the properties of these three subatomic particles.
Table $1$: Properties of the Three Subatomic Particles
Name Symbol Mass (approx.; g) Mass (approx.; amu) Charge
Proton p+ 1.673 × 10−24 1.0073 +1
Neutron n, n0 1.675 × 10−24 1.0087 none
Electron e 9.109 × 10−28 5.486 × 10−4 –1
Atoms and subatomic particles are so small that it doesn't quite make sense to measure their masses in grams. A more useful unit to measure atomic mass is the atomic mass unit ($\text{amu}$), where 1 amu = 1.660539 × 10−24 g or one-twelfth of the mass of a carbon-12 atom. As you can see in the table above, the mass of 1 proton and 1 neutron are each 1 amu in this system. Carbon-12 contains six protons and six neutrons and is assigned a mass of exactly 12 amu.
How are these subatomic particles arranged in atoms? They are not arranged at random. Experiments by Ernest Rutherford in England in the 1910's pointed to a nuclear model with atoms that has the protons and neutrons in a central nucleus with the electrons in orbit about the nucleus. The relatively massive protons and neutrons are collected in the center of an atom, in a region called the nucleus of the atom (plural nuclei). The electrons are outside the nucleus and spend their time orbiting in space about the nucleus. (Figure $2$). Because protons and neutrons are so massive compared to electrons, Table $1$, nearly all of the mass of an atom is contained in the nucleus.
Note Atoms In Action
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 demonstrate in Video $1$, control of individual atoms can be use used 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.
Key Takeaways
• Chemistry is based on the modern atomic theory, which states that all matter is composed of atoms.
• Atoms themselves are composed of protons, neutrons, and electrons.
• Each element has its own atomic number, which is equal to the number of protons in its nucleus. | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/02%3A_Atoms_and_the_Periodic_Table/2.01%3A_Atomic_Theory_and_the_Structure_of_Atoms.txt |
Learning Objectives
• Use the atomic number and mass number to describe an atom.
The modern atomic theory states that atoms of one element are the same, while atoms of different elements are different. What makes atoms of different elements different? The fundamental characteristic that all atoms of the same element share is the number of protons. All atoms of hydrogen have one and only one proton in the nucleus; all atoms of iron have 26 protons in the nucleus. This number of protons is so important to the identity of an atom that it is called the atomic number (Z). The number of protons in an atom is the atomic number of the element. Thus, hydrogen has an atomic number of 1, while iron has an atomic number of 26. Each element has its own characteristic atomic number.
Atoms are electrically neutral, meaning that the overall electric charge is zero. This is because the number of protons (positive charge) equals the number of electrons (negative charge). Therefore, the atomic number also provides the number of electrons. For example, helium has Z = 2, which tells us that there are 2 protons in the nucleus and 2 electrons outside of the nucleus. Sometimes atoms will gain or lose electrons resulting in a difference in the number of protons and electrons, which means the charge is no longer zero. Atoms that have a charge are called ions and will be discussed further in later chapters.
\[atomic\ number = Z = p^+ = e^{–1}\]
As we learned previously, protons and neutrons, which are found in the nucleus of an atom, each have a mass of ~1 amu. Because an electron has negligible mass relative to that of a proton or a neutron, the majority of an atom's mass is in the nucleus. The mass number (A) is defined as the total number of protons (\(p^+\)) and neutrons (\(n\)) in an atom:
\[mass\ number = A = p^+ + n\]
Atoms of the same element always have the same number of protons, same Z, but often have different numbers of neutrons, therefore, different mass numbers. These atoms are called isotopes and will be discussed in more detail in the next chapter.
Example \(1\)
1. The most common carbon atoms have six protons and six neutrons in their nuclei. What are the atomic number and the mass number of these carbon atoms?
2. An isotope of uranium has an atomic number of 92 and a mass number of 235. What are the number of protons and neutrons in the nucleus of this atom?
Solution
1. If a carbon atom has six protons in its nucleus, its atomic number is 6. If it also has six neutrons in the nucleus, then the mass number is 6 + 6, or 12.
2. If the atomic number of uranium is 92, then that is the number of protons in the nucleus. Because the mass number is 235, then the number of neutrons in the nucleus is 235 − 92, or 143.
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
Example \(2\):
1. What is the symbol for an isotope of uranium that has an atomic number of 92 and a mass number of 235?
2. How many protons and neutrons are in \(\ce{_{26}^{56}Fe}\)
Solution
1. The symbol for this isotope is \(\ce{_{92}^{235}U}\)
2. This iron atom has 26 protons and 56 − 26 = 30 neutrons.
Exercise \(2\)
How many protons are in \(\ce{_{11}^{23} Na}\)
Answer
11 protons
Key Takeaways
• Atoms themselves are composed of protons, neutrons, and electrons.
• Each element has its own atomic number, which is equal to the number of protons in its nucleus.
• Isotopes of an element contain different numbers of neutrons.
• Elements are represented by an atomic symbol.
Contributors
• Lisa Sharpe Elles, University of Kansas | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/02%3A_Atoms_and_the_Periodic_Table/2.02%3A_Elements_and_Atomic_Number.txt |
Learning Objectives
• Explain how isotopes differ from one another.
• Calculate the atomic mass of an element from the masses and relative percentages of the isotopes of the element.
Isotopes
As introduced previously, atoms of a specific element are distinguished from other elements by their atomic number, (the number of protons). Atoms of the same element always have the same number of protons, however, the number of neutrons can vary. Isotopes are atoms of the same element that contain different numbers of neutrons. This difference in neutron amount affects the mass number (A) but not the atomic number (Z). In a chemical laboratory, isotopes of an element appear and react the same. For this reason, it is difficult to distinguish between different isotopes. In contrast, nuclear scientists can identify and separate different types of atomic nuclei. The technology required for this process is more sophisticated that what could be found in a typical chemical laboratory.
Figure $1$ shows an easy way to represent isotopes with a nuclear symbol, which includes the atomic or element symbol (represented by $X$), the mass number, $A$, and the atomic number, $Z$. Thus, for the isotope of carbon that has 6 protons and 6 neutrons, the symbol is:
$\ce{_{6}^{12}C} \nonumber$
where $C$ is the symbol for carbon, 6 represents the atomic number, and 12 represents the mass number.
It is also common to state the mass number after the name of an element to indicate a particular isotope. Carbon-12 represents an isotope of carbon with 6 protons and 6 neutrons, while uranium-238 is an isotope of uranium that has 92 protons and 146 neutrons.
Most elements on the periodic table have at least two stable isotopes. For example, in addition to $\ce{^{12}C}$, a typical sample of carbon contains 1.11% $\ce{_6^{13}C}$, with 7 neutrons and 6 protons, and a trace of $\ce{_6^{14}C}$, with 8 neutrons and 6 protons. The nucleus of $\ce{_6^{14}C}$ is not stable, however, but undergoes a slow radioactive decay that is the basis of the carbon-14 dating technique used in archeology. Many elements other than carbon have more than one stable isotope; tin, for example, has 10 isotopes. There are about twenty elements that exist in only one isotopic form (sodium and fluorine are examples of these).
An important series of isotopes is found with hydrogen atoms. Most hydrogen atoms have a nucleus with only a single proton. About 1 in 10,000 hydrogen nuclei, however, also has a neutron; this particular isotope is called deuterium. An extremely rare hydrogen isotope, tritium, has 1 proton and 2 neutrons in its nucleus. Figure $2$ compares the three isotopes of hydrogen.
There are currently over 3,500 isotopes known for all the elements. When scientists discuss individual isotopes, they need an efficient way to specify the number of neutrons in any particular nucleus. A/Z and symbol-mass formats can be used to display periodic table information. When viewing either of these two notations, isotopic differences can be obtained.
The discovery of isotopes required a minor change in Dalton’s atomic theory. Dalton thought that all atoms of the same element were exactly the same.
Look at the A/Z formats for the three isotopes of hydrogen in Table $1$. Note how the atomic number (bottom value) remains the same while the atomic masses (top number) are varied. All isotopes of a particular element will vary in neutrons and mass. This variance in mass will be visible in the symbol-mass format of same isotopes as well.
Table $1$
Common Name A/Z formats symbol-mass format Expanded Name
Hydrogen $\mathrm{^{1}_{1}H}$ $\text{H-1}$ hydrogen-1
Deuterium $\mathrm{^{2}_{1}H}$ $\text{H-2}$ hydrogen-2
Tritium $\mathrm{^{3}_{1}H}$ $\text{H-3}$ hydrogen-3
Both A/Z or symbol-mass formats can be utilized to determine the amount of subatomic particles (protons, neutrons, and electrons) contained inside an isotope. When given either format, these mass values should be used to calculate the number of neutrons in the nucleus.
Atomic Weight
Since most naturally occurring elements samples are mixtures of isotopes, it is useful to use an average weight of an element. The atomic mass of an element is the weighted mass of all the naturally presented isotopes (on earth). To determine the most abundant isotopic form of an element, compare given isotopes to the weighted average on the periodic table. For example, the three hydrogen isotopes in Figure $2$ are H-1, H-2, and H-3. The atomic mass or weighted average of hydrogen is around 1.008 amu ( look again to the periodic table). Of the three hydrogen isotopes, H-1 is closest in mass to the weighted average; therefore, it is the most abundant. The other two isotopes of hydrogen are quite rare, but are very exciting in the world of nuclear science.
You can calculate the atomic mass (or average mass) of an element provided you know the relative abundances (the fraction of an element that is a given isotope), the element's naturally occurring isotopes, and the masses of those different isotopes. We can calculate this by the following equation:
$\text{Atomic mass} = \left( \%_1 \right) \left( \text{mass}_1 \right) + \left( \%_2 \right) \left( \text{mass}_2 \right) + \cdots \label{eq1}$
Averages like Equation 1 are known as weighted averages. An element's atomic mass can be calculated provided the relative abundances of the element's naturally occurring isotopes and the masses of those isotopes are known. If all the abundances are not provided, it is safe to assume that all numbers should add up to 100%.
For example, Boron has two naturally occurring isotopes. In a sample of boron, $20\%$ of the atoms are $\text{B-10}$, which is an isotope of boron with 5 neutrons and mass of $10 \: \text{amu}$. The other $80\%$ of the atoms are $\text{B-11}$, which is an isotope of boron with 6 neutrons and a mass of $11 \: \text{amu}$. How do we calculate the atomic mass of boron?
Boron has two isotopes so we will use the Equation \ref{eq1} and substitute the relative abundances and atomic masses of Boron into Equation \ref{eq1}:
\begin{align} \text{Atomic mass} &= \left( 0.20 \right) \left( 10 \right) + \left( 0.80 \right) \left( 11 \right) \nonumber \ &= 10.8 \: \text{amu}\nonumber \end{align}\nonumber
The mass of an average boron atom, and thus boron's atomic mass, is $10.8 \: \text{amu}$.
Example $1$: Atomic Weight of Neon
Neon has three naturally occurring isotopes. In a sample of neon, $90.92\%$ of the atoms are $\ce{Ne}$-20, which is an isotope of neon with 10 neutrons and a mass of $19.99 \: \text{amu}$. Another $0.3\%$ of the atoms are $\ce{Ne}$-21, which is an isotope of neon with 11 neutrons and a mass of $20.99 \: \text{amu}$. The final $8.85\%$ of the atoms are $\ce{Ne}$-22, which is an isotope of neon with 12 neutrons and a mass of $21.99 \: \text{amu}$. What is the atomic mass of neon?
Solution
Neon has three isotopes. We will use the equation:
$\text{Atomic mass} = \left( \%_1 \right) \left( \text{mass}_1 \right) + \left( \%_2 \right) \left( \text{mass}_2 \right) + \cdots \nonumber$
Substitute these into the equation, and we get:
\begin{align*} \text{Atomic mass} &= \left( 0.9092 \right) \left( 19.99 \right) + \left( 0.003 \right) \left( 20.99 \right) + \left( 0.0885 \right) \left( 21.99 \right) \ &= 20.17 \: \text{amu} \end{align*}
The mass of an average neon atom is $20.17 \: \text{amu}$
Exercise $1$
Magnesium has the three isotopes listed in the following table:
Table showing the 3 isotopes of magnesium, the exact mass of each, and the percent abundance of each.
Isotope Exact Mass (amu) Percent Abundance (%)
24Mg 23.98504 78.70
25Mg 24.98584 10.13
26Mg 25.98259 11.17
Use these data to calculate the atomic mass of magnesium.
Answer
24.31 amu
Applications of Isotopes
During the Manhattan project, the majority of federal funding was dedicated the separation of uranium isotopes. The two most common isotopes of uranium are U-238 and U-235. About 99.3% of uranium is of the U-238 variety, this form is not fissionable and will not work in a nuclear weapon or reaction. The remaining 0.7% is U-235 which is fissionable, but first had to be separated from U-238. This separation process is called enrichment. During World War II, a nuclear facility was built in Oak Ridge, Tennessee to accomplish this project. At the time, the enrichment process only produced enough U-235 for one nuclear weapon. This fuel was placed inside the smaller of the two atomic bombs (Little Boy) dropped over Japan.
Uranium is a natural element that can be found in several different countries. Countries that do not have natural uranium supplies would need to obtain it from one of the countries below. Most nuclear reactors that provide energy rely on U-235 as a source of fuel. Fortunately, reactors only need 2-5% U-235 for the production of megawatts or even gigawatts of power. If the purification process exceeds this level, than it is likely a country is focusing on making nuclear weapons. For example, Manhattan Project scientists enriched U-235 up to 90% in order to produce the Little Boy weapon.
Abbreviations like HEU (highly enriched uranium) and LEU (low-enriched uranium) are used frequently by nuclear scientists and groups. HEU is defined as being over 20% pure U-235 and would not be used in most commercial nuclear reactors. This type of material is used to fuel larger submarines and aircraft carriers. If the purification of U-235 reaches 90%, then the HEU is further classified as being weapons grade material. This type of U-235 could be used to make a nuclear weapon (fission or even fusion based). As for LEU, its U-235 level would be below this 20% mark. LEU would be used for commercial nuclear reactors and smaller, nuclear powered submarines. LEU is not pure enough to be used in a conventional nuclear weapon, but could be used in a dirty bomb. This type of weapon uses conventional explosives like dynamite to spread nuclear material. Unlike a nuclear weapon, dirty bombs are not powerful enough to affect large groups of buildings or people. Unfortunately, the spread of nuclear material would cause massive chaos for a community and would result in casualties.
Summary
• The isotopes of an element have different masses and are identified by their mass numbers.
• An element's atomic mass is the weighted average of the masses of the isotopes of an element
• An element's atomic mass can be calculated provided the relative abundances of the element's naturally occurring isotopes and the masses of those isotopes are known. If all the abundances are not provided, it is safe to assume that all numbers should add up to 100%.
Concept Review Exercises
1. Why is the atomic number so important to the identity of an atom?
2. What is the relationship between the number of protons and the number of electrons in an atom?
3. How do isotopes of an element differ from each other?
4. What is the mass number of an element?
Answers
1. The atomic number defines the identity of an element. It describes the number of protons in the nucleus.
2. In an electrically neutral atom, the number of protons equals the number of electrons.
3. Isotopes of an element have the same number of protons but have different numbers of neutrons in their nuclei.
4. The mass number is the sum of the numbers of protons and neutrons in the nucleus of an atom. | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/02%3A_Atoms_and_the_Periodic_Table/2.03%3A_Isotopes_and_Atomic_Weight.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.
In the 19th century, many previously unknown elements were discovered, and scientists noted that certain sets of elements had similar chemical properties. For example, chlorine, bromine, and iodine react with other elements (such as sodium) to make similar compounds. Likewise, lithium, sodium, and potassium react with other elements (such as oxygen) to make similar compounds. Why is this so?
In 1864, Julius Lothar Meyer, a German chemist, organized the elements by atomic mass and grouped them according to their chemical properties. Later that decade, Dmitri Mendeleev, a Russian chemist, organized all the known elements according to similar properties. He left gaps in his table for what he thought were undiscovered elements, and he made some bold predictions regarding the properties of those undiscovered elements. When elements were later discovered whose properties closely matched Mendeleev’s predictions, his version of the table gained favor in the scientific community. Because certain properties of the elements repeat on a regular basis throughout the table (that is, they are periodic), it became known as the periodic table.
Mendeleev had to list some elements out of the order of their atomic masses to group them with other elements that had similar properties.
The periodic table is one of the cornerstones of chemistry because it organizes all the known elements on the basis of their chemical properties. A modern version is shown in Figure \(1\). Most periodic tables provide additional data (such as atomic mass) in a box that contains each element’s symbol. The elements are listed in order of atomic number.
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).
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.
Metals, Nonmetals, and Metalloids
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 semimetal, 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 semimetals (or metalloids). Elements adjacent to the bold zigzag line in the right-hand portion of the periodic table have semimetal properties.
Example \(1\)
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. The atomic number of selenium is 34, which places it in period 4 and group 16. In Figure \(2\), selenium lies above and to the right of the diagonal line marking the boundary between metals and nonmetals, so it should be a nonmetal. Note, however, that because selenium is close to the metal-nonmetal dividing line, it would not be surprising if selenium were similar to a semimetal in some of its properties.
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 \(1\)
Based on its location in the periodic table, do you expect indium (In) to be a nonmetal, a metal, or a semimetal?
Answer
metal
Representative, Transition, and Inner-transition
Another way to categorize the elements of the periodic table is shown in Figure \(3\). The first two columns on the left (groups 1 and 2) and the last six columns on the right (groups 13-19) are called the main group or representative elements. The ten-column block between these columns (groups 3-12) 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.
To Your Health: Transition Metals in the Body
Most of the elemental composition of the human body consists of main group elements. The most abundant non-main group element is iron, at 0.006 percentage by mass. Because iron has relatively massive atoms, it would appear even lower on a list organized in terms of percent by atoms rather than percent by mass.
Iron is a transition metal and the chemistry of iron makes it a key component in the proper functioning of red blood cells.
Red blood cells are cells that transport oxygen from the lungs to cells of the body and then transport carbon dioxide from the cells to the lungs. Without red blood cells, animal respiration as we know it would not exist. The critical part of the red blood cell is a protein called hemoglobin. Hemoglobin combines with oxygen and carbon dioxide, transporting these gases from one location to another in the body. Hemoglobin is a relatively large molecule, with a mass of about 65,000 u.
The crucial atom in the hemoglobin protein is iron. Each hemoglobin molecule has four iron atoms, which act as binding sites for oxygen. It is the presence of this particular transition metal in your red blood cells that allows you to use the oxygen you inhale.
Other transition metals have important functions in the body, despite being present in low amounts. Zinc is needed for the body’s immune system to function properly, as well as for protein synthesis and tissue and cell growth. Copper is also needed for several proteins to function properly in the body. Manganese is needed for the body to metabolize oxygen properly. Cobalt is a necessary component of vitamin B-12, a vital nutrient. These last three metals are present in the body in very small quantities. However, even these small quantities are required for the body to function properly.
Contributions & Attributions
This page was constructed from content via the following contributor(s) and edited (topically or extensively) by the LibreTexts development team to meet platform style, presentation, and quality: | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/02%3A_Atoms_and_the_Periodic_Table/2.04%3A_The_Periodic_Table.txt |
Learning Objectives
• Describe how some characteristics of elements relate to their positions on the periodic table.
The periodic table is useful for understanding atomic properties that show periodic trends. Periodic trends are specific patterns that are present in the periodic table that illustrate different aspects of a certain element, including its size and its electronic properties. Major periodic trends include atomic radius, melting point, among many other properties as we will discuss. Periodic trends, arising from the arrangement of the periodic table, provide chemists with an invaluable tool to quickly predict an element's properties. These trends exist because of the similar atomic structure of the elements within their respective group families or periods, and because of the periodic nature of the elements.
One important atomic property is the atomic radius, which is a measure of the atomic size, usually the distance from the nucleus to the out electron shell. However, since this boundary is not well-defined, there are multiple definitions of atomic radius. Irrespective of the definition used, a clear periodic trend can be observed when atomic radius is plotted vs. atomic number (Figure \(1\)). The radii generally decrease along each row of the table and increase down each group. The radius increases sharply between the noble gas at the end of each period and the alkali metal at the beginning of the next period. The largest atoms are found in the lower left corner of the periodic table and the smallest are found in the upper right corner
The melting point is a metric of the energy required to transform the solid phase of a substance into a liquid. Generally, the stronger the bond between the atoms of an element, the more energy required to break that bond. The melting points exhibit comparable, albeit more complex, periodic trends as observed in the atomic radii (Figure \(2\)). Key feature of these trends are:
• Metals generally possess a high melting point.
• Most non-metals possess low melting points.
• The non-metal carbon possesses the highest melting point of all the elements. The semi-metal boron also possesses a high melting point.
The trends of the atomic radii and melting points (and other chemical and physical properties of the elements) can be explained by the electron shell theory of the atom discussed in the following sections.
Element Characteristics By Group
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. For example, the elements of Group 1 are known as the alkali metals, Group 2 are the alkaline earth metals, Group 17 are the halogens, and Group 18 are the noble gases.
• Group 1: The Alkali Metals lithium (Li), sodium (Na), potassium (K), rubidium (Rb), cesium (Cs), and francium (Fr) are soft, shiny, and highly reactive metals. 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. Hydrogen is unique in that it is generally placed in Group 1, but it is not a metal.
• Group 2: The Alkaline Earth Metals – beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra) are shiny, slivery-white, somewhat reactive metals. 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 – fluorine (F), chlorine (Cl), bromine (Br), iodine (I), and astatine (At) are nonmetals. 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).
• Group 18: The Noble Gases – helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn) are gases at room temperature and pressure. Because the noble gases are composed of only single atoms, they are called monatomic. 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.
To Your Health: Radon
Radon is an invisible, odorless noble gas that is slowly released from the ground, particularly from rocks and soils whose uranium content is high. Because it is a noble gas, radon is not chemically reactive. Unfortunately, it is radioactive, and increased exposure to it has been correlated with an increased lung cancer risk.
Because radon comes from the ground, we cannot avoid it entirely. Moreover, because it is denser than air, radon tends to accumulate in basements, which if improperly ventilated can be hazardous to a building’s inhabitants. Fortunately, specialized ventilation minimizes the amount of radon that might collect. Special fan-and-vent systems are available that draw air from below the basement floor, before it can enter the living space, and vent it above the roof of a house.
After smoking, radon is thought to be the second-biggest preventable cause of lung cancer in the United States. The American Cancer Society estimates that 10% of all lung cancers are related to radon exposure. There is uncertainty regarding what levels of exposure cause cancer, as well as what the exact causal agent might be (either radon or one of its breakdown products, many of which are also radioactive and, unlike radon, not gases). The US Environmental Protection Agency recommends testing every floor below the third floor for radon levels to guard against long-term health effects.
Example \(1\): Groups
Provide the family/group names and period numbers (horizontal values) of each element.
1. Li
2. Ar
3. Ra
Solution:
1. Lithium is an alkali metal. It is located in period two.
2. Argon is a noble gas. It is located in period three.
3. Radium is an alkaline metal. It is located in period seven.
Example \(2\): Classification of Elements
Provide elemental names for the following combinations:
1. The alkali metal in period three.
2. The halogen in period two
3. A metalloid in period four
4. A transition metal in period three
Solution:
1. Sodium
2. Fluorine
3. Germanium or Arsenic
4. There are no transition metals in period three (gotcha!)
Key Takeaways
• The chemical elements are arranged in a chart called the periodic table.
• Some characteristics of the elements are related to their position on the periodic table.
• The number of valence electrons of an element can be determined by the group (vertical column) number in the Periodic Table. Elements with the same number of valence electrons have similar chemical properties. | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/02%3A_Atoms_and_the_Periodic_Table/2.05%3A_Some_Characteristics_of_Different_Groups.txt |
Learning Objectives
• Describe how electrons are grouped within atoms into shells, subshells, and orbitals using quantum numbers.
You now know that the periodic table is arranged in groups and periods (columns and rows) based on chemical and physical properties of the different elements. The first element, hydrogen (Z=1) has one proton and one electron and as you move right across the rows, each subsequent element has one additional proton and electron. You may have asked yourself, why are periodic trends observed across the rows and down the groups? Or, why do the rows have different numbers of elements, giving the table a unique shape?
These questions can be answered by learning more about the electrons in atoms. Although we have discussed the general arrangement of subatomic particles in atoms, we have said little about how electrons occupy the space around the nucleus. Do they move around the nucleus at random, or do they exist in some ordered arrangement?
In 1913, the Danish scientist Niels Bohr suggested that the electron in a hydrogen atom could not have any random energy, having only certain fixed values of energy that were indexed by the number n (now called a quantum number). Bohr suggested that the energy of the electron in hydrogen was quantized because it was in a specific orbit; much like the steps on a staircase does not have half or quarter stairs or the keys on a piano don't have notes in between, there are no energy levels in between each orbit. Figure \(1\) shows a model of the hydrogen atom based on Bohr's ideas.
Bohr's ideas were useful, but were applicable only to the hydrogen atom. However, later researchers generalized Bohr's ideas into a new theory called quantum mechanics, which explains the behavior of electrons as if they were acting as a wave, not as particles. Quantum mechanics predicts two major things: quantized energies for electrons of all atoms (not just hydrogen) and an organization of electrons within atoms. Electrons are no longer thought of as being randomly distributed around a nucleus or restricted to certain orbits (in that regard, Bohr was wrong). Instead, electrons are collected into groups (shells) and subgroups (subshells) that explain much about the chemical behavior of the atom.
In the quantum-mechanical model of an atom, which is the modern and currently accepted model, the location of electrons in the atom are described by four quantum numbers, not just the one predicted by Bohr. Much like your home address can be used to locate you in a specific state, city, street, and house number, the first three quantum numbers identify approximately where electrons are in an atom. The fourth quantum number describes the electron and whether it is spin up or down (clockwise or counterclockwise). The theory and mathematics behind these four quantum numbers are well beyond the scope of this textbook, however, it is useful to learn some of the basics in order to understand how atoms behave and interact with (react) with other atoms.
Electron Arrangements: Shells, Subshells, and Orbitals
Electrons are organized according to their energies into sets called shells (labeled by the principle quantum number, n). Generally the higher the energy of a shell, the farther it is (on average) from the nucleus. Shells do not have specific, fixed distances from the nucleus, but an electron in a higher-energy shell will spend more time farther from the nucleus than does an electron in a lower-energy shell.
Shells are further divided into subsets of electrons called subshells, labeled by type as s, p, d, or f. The first shell has only one subshell, s. The second shell has two subshells, s and p; the third shell has three subshells, s, p, and d, and the fourth shell has four subshells, s, p, d, and f. Within each subshell, electrons are arranged into different numbers of orbitals, an s subshell is made up of one s orbital, a p subshell has two p orbitals, a d subshell, five d orbitals, and an f subshell, seven f orbitals. Each orbital has a different shape and orientation around the nucleus (Figure \(1\), however, rather than representing an orbit, as the name suggests, orbitals define a boundary for the region of space where a given electron is most likely to be found. Lastly, a single orbital can hold up to two electrons each with a different spin.
It is important to note that according to quantum theory, there are specific allowed combinations of quantum numbers and others that are not allowed. For example, shell two can only have two subshells, s with one orbital and p with 3 orbitals, therefore, this shell can hold a maximum of eight electrons (four orbitals times two electrons each). It takes practice to learn the allowed combinations as shown in Table \(1\ but it is helpful to visualize the atom as a sphere with the nucleus in the center. Close to the nucleus, there is a smaller amount of space for electrons – a smaller shell. As the number of electrons increases, the shells that hold the electrons get larger and thus further away from the nucleus.
Table \(1\): Shells and Subshells
Shell Number of Subshells Names of Subshells Number of Orbitals (per Subshell) Number of Electrons (per Subshell) Total Electrons (per Shell)
1 1 1s 1 2 2
2 2 2s and 2p 1, 3 2, 6 8
3 3 3s, 3p, and 3d 1, 3, 5 2, 6, 10 18
4 4 4s, 4p, 4d, and 4f 1, 3, 5, 7 2, 6, 10, 14 32
All of this information about the shell, subshell, and orbital is put together to make up the "address" for an electron and all of the addresses for all the electrons in an atom make up the electron configuration, which is described more later. | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/02%3A_Atoms_and_the_Periodic_Table/2.06%3A_Electronic_Structure_of_Atoms.txt |
Learning Objectives
• Describe how electrons are arranged in an atom using electron configurations.
Previously we discussed the concept of electron shells, subshells, orbitals, and electron spin. It is the arrangement of electrons into shells and subshells that most concerns us here, so we will focus on that.
We use numbers to indicate which shell an electron is in. The first shell, closest to the nucleus and with the lowest-energy electrons, is shell 1. This first shell has only one subshell, which is labeled 1s and can hold a maximum of 2 electrons. We combine the shell and subshell labels when referring to the organization of electrons about a nucleus and use a superscript to indicate how many electrons are in a subshell. Thus, because a hydrogen atom has its single electron in the s subshell of the first shell, we use 1s1 (spoken as “one-ess-one”) to describe the electron arrangement or distribution of electrons in hydrogen. This structure is called an electron configuration and is unique to hydrogen.
Helium atoms have 2 electrons. Both electrons fit into the 1s subshell because s subshells contain one s orbital which can hold up to 2 electrons; therefore, the electron configuration for helium atoms is 1s2 (spoken as “one-ess-two”).
The 1s subshell can hold a maximum of 2 electrons, so the electron configuration for a lithium atom, which has three electrons, cannot be 1s3. Two of the lithium electrons can fit into the 1s subshell, but the third electron must go into the second shell and the lower energy orbital, which is the 2s orbital. Therefore, we write the electron configuration of a lithium atom as 1s22s1 (spoken as “one-ess-two two-ess-one”).
The shell diagram for a lithium atom (Figure $1$). The shell closest to the nucleus (first shell) has 2 dots representing the 2 electrons in 1s, while the outermost shell (2s) has 1 electron.
There are a set of general rules that are used to figure out the electron configuration of an atomic species: Aufbau Principle, Hund's Rule and the Pauli-Exclusion Principle.
• Rule 1 (Aufbau Principle): Electrons occupy the lowest-energy orbitals (closest to the nucleus) possible, starting with 1s, then 2s, 2p, and continuing on to higher energy (further away from the nucleus). Shells increase in energy in order from 1 to 2 to 3, and so on. Within these shells, an s subshell is the lowest energy followed by p, then d, then f.
• Rule 2 (Hund's Rule): When electrons occupy degenerate orbitals (i.e. same shell and subshell), they must first singly occupy (half-fill) each empty orbital in a subshell before double occupying (completely filling) them. Furthermore, the most stable configuration results when the spins are parallel (i.e. all spin up or all spin down). For example, all three p orbitals in a p subshell will have one electron before a single p orbital contains two electrons.
• Rule 3 (Pauli-Exclusion Principle): Each electron is described with a unique set of four quantum numbers (a unique address). Therefore, if two electrons occupy the same orbital, they must have different spins. This is the reason all orbitals can hold a maximum of two electrons.
Continuing on the periodic table to the next largest atom, beryllium, with 4 electrons, the electron configuration is 1s22s2. Now that the 2s subshell is filled, electrons in larger atoms, starting with boron, begin filling the 2p subshell, which can hold a maximum of six electrons. The next six elements progressively fill up the 2p subshell:
• B: 1s22s22p1
• C: 1s22s22p2
• N: 1s22s22p3
• O: 1s22s22p4
• F: 1s22s22p5
• Ne: 1s22s22p6
At the end of the period the element neon, has filled the 2s, and 2p subshells, which completes the second shell. Now atoms with more electrons now must begin the third shell starting with the 3s subshell. The first two subshells of the third shell are filled in order—for example, the electron configuration of aluminum, with 13 electrons, is 1s22s22p63s23p1. However, a curious thing happens after the 3p subshell is filled: the 4s subshell begins to fill before the 3d subshell does. In fact, the exact ordering of subshells becomes more complicated at this point (after argon, with its 18 electrons), so we will not consider the electron configurations of larger atoms.
Table $1$: Electron Configurations of the First 20 Elements
Atomic Number Element Symbol Outermost Shell Electron Configuration Noble Gas Configuration
1 H 1 1s 1 1s 1
2 He 1 1s 2 1s 2
3 Li 2 1s 2 2s 1 [He] 2s 1
4 Be 2 1s 2 2s 2 [He] 2s 2
5 B 2 1s 2 2s 2 2p1 [He] 2s 2 2p1
6 C 2 1s 2 2s 2 2p2 [He] 2s 2 2p2
7 N 2 1s 2 2s 2 2p3 [He] 2s 2 2p3
8 O 2 1s 2 2s 2 2p4 [He] 2s 2 2p4
9 F 2 1s 2 2s 2 2p5 [He] 2s 2 2p5
10 Ne 2 1s 2 2s 2 2p6 [He] 2s 2 2p6
11 Na 3 1s 2 2s 2 2p6 3s 1 [Ne] 3s 1
12 Mg 3 1s 2 2s 2 2p6 3s 2 [Ne] 3s 2
13 Al 3 1s 2 2s 2 2p6 3s 2 3p1 [Ne] 3s 2 3p1
14 Si 3 1s 2 2s 2 2p6 3s 2 3p2 [Ne]3s 2 3p2
15 P 3 1s 2 2s 2 2p6 3s 2 3p3 [Ne] 3s 2 3p3
16 S 3 1s 2 2s 2 2p6 3s 2 3p4 [Ne] 3s 2 3p4
17 Cl 3 1s 2 2s 2 2p6 3s 2 3p5 [Ne] 3s 2 3p5
18 Ar 3 1s 2 2s 2 2p6 3s 2 3p6 [Ne] 3s 2 3p6
19 K 4 1s 2 2s 2 2p6 3s 2 3p6 4s 1 [Ar] 4s 1
20 Ca 4 1s 2 2s 2 2p6 3s 2 3p6 4s 2 [Ar] 4s 2
Noble Gas Configuration
The electron configuration of sodium is $1s^2 2s^2 2p^6 3s^1$ (Table $1$). The first ten electrons of the sodium atom are the inner-shell electrons and the configuration of just those ten electrons is exactly the same as the configuration of the element neon $\left( Z=10 \right)$. This provides the basis for a shorthand notation for electron configurations called the noble gas configuration, which atom consists of the elemental symbol of the last noble gas prior to that atom, followed by the configuration of the remaining electrons. So for sodium, we make the substitution of $\left[ \ce{Ne} \right]$ for the $1s^2 2s^2 2p^6$ part of the configuration. Sodium's noble gas configuration becomes $\left[ \ce{Ne} \right] 3s^1$.
The electron filling diagram shown below in Figure $2$ is useful in remembering the order for electrons to occupy shells and subshells. Although it is much easier to use the periodic table as a guide for electron filling as you will see in the next section.
Example $1$: Electronic Configuration of Phosphorus Atoms
Using Figure $2$ as your guide, write the electron configuration of a neutral phosphorus atom. The atomic number of P is 15.
Solution
A neutral phosphorus atom has 15 electrons. Two electrons can go into the 1s subshell, 2 can go into the 2s subshell, and 6 can go into the 2p subshell. That leaves 5 electrons. Of those 5 electrons, 2 can go into the 3s subshell, and the remaining 3 electrons can go into the 3p subshell. Thus, the electron configuration of neutral phosphorus atoms is 1s22s22p63s23p3.
Exercise $1$: Electronic Configuration of Chlorine Atoms
Using Figure $2$ as your guide, write the electron configuration of a neutral chlorine atom. The atomic number of Cl is 17.
Answer
A neutral chlorine atom has 17 electrons. Two electrons can go into the 1s subshell, 2 can go into the 2s subshell, and 6 can go into the 2p subshell. That leaves 7 electrons. Of those 7 electrons, 2 can go into the 3s subshell, and the remaining 5 electrons can go into the 3p subshell. Thus, the electron configuration of neutral chlorine atoms is 1s22s22p63s23p5.
Orbital Diagrams
An orbital diagram is the more visual way to represent the arrangement of all the electrons in a particular atom. In an orbital diagram, the individual orbitals are shown as squares and orbitals within a sublevel are drawn next to each other horizontally. Each sublevel is labeled by its shell and sublevel. Electrons are indicated by arrows inside of the squares. An arrow pointing upwards indicates one spin direction, while a downward pointing arrow indicates the other direction. The orbital filling diagrams for hydrogen, helium, and lithium are shown in the figure below.
According to the Aufbau Principle, sublevels and orbitals are filled with electrons in order of increasing energy. Since the $s$ sublevel consists of just one orbital, the second electron simply pairs up with the first electron as in helium. The next element is lithium and necessitates the use of the next available sublevel, the $2s$.
The orbital diagram for carbon is shown in Figure $10$. There are two $2p$ electrons for carbon and each occupies its own $2p$ orbital (Hund's Rule).
Oxygen has four $2p$ electrons. After each $2p$ orbital has one electron in it, the fourth electron can be placed in the first $2p$ orbital with a spin opposite that of the other electron in that orbital.
If you keep your papers in manila folders, you can pick up a folder and see how much it weighs. If you want to know how many different papers (articles, bank records, or whatever else you keep in a folder), you have to take everything out and count. A computer directory, on the other hand, tells you exactly how much you have in each file. We can get the same information on atoms. If we use an orbital filling diagram, we have to count arrows. When we look at electron configuration data, we simply add up the numbers.
Example $3$: Carbon Atoms
Draw the orbital filling diagram for carbon and write its electron configuration.
Known
• Atomic number of carbon, Z=6
Use the order of fill diagram to draw an orbital filling diagram with a total of six electrons. Follow Hund's rule. Write the electron configuration.
Orbital filling diagram for carbon.
Electron configuration 1s22s22p2
Step 3: Think about your result.
Following the 2s sublevel is the 2p, and p sublevels always consist of three orbitals. All three orbitals need to be drawn even if one or more is unoccupied. According to Hund's rule, the sixth electron enters the second of those p orbitals and has the same spin as the fifth electron. | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/02%3A_Atoms_and_the_Periodic_Table/2.07%3A_Electron_Configurations.txt |
Learning Objectives
• Relate the electron configurations of the elements to the shape of the periodic table.
• Determine the expected electron configuration of an element by its place on the periodic table.
Remember, that the periodic table as a tool for organizing the known chemical elements(Figure \(1\)). The elements are listed by atomic number (the number of protons in the nucleus), and elements with similar chemical properties are grouped together in columns. Why does the periodic table have the structure it does? The answer is rather simple, if you understand electron configurations: the shape of the periodic table mimics the filling of the subshells with electrons.
The shape of the periodic table mimics the filling of the subshells with electrons.
Let us start with H and He. Their electron configurations are 1s1 and 1s2, respectively; with He, the n = 1 shell is filled. These two elements make up the first row of the periodic table (Figure \(1\))
The next two electrons, for Li and Be, would go into the 2s subshell. Figure \(2\) shows that these two elements are adjacent on the periodic table.
For the next six elements, the 2p subshell is being occupied with electrons. On the right side of the periodic table, these six elements (B through Ne) are grouped together (Figure \(3\)).
The next subshell to be filled is the 3s subshell. The elements when this subshell is being filled, Na and Mg, are back on the left side of the periodic table (Figure \(4\)).
Next, the 3p subshell is filled with the next six elements (Figure \(5\)).
Instead of filling the 3d subshell next, electrons go into the 4s subshell (Figure \(6\)).
After the 4s subshell is filled, the 3d subshell is filled with up to 10 electrons. This explains the section of 10 elements in the middle of the periodic table (Figure \(7\)).
And so forth. As we go across the columns of the periodic table, the overall shape of the table outlines how the electrons are occupying the shells and subshells.
The first two columns on the left side of the periodic table are where the s subshells are being occupied. Because of this, the first two rows of the periodic table are labeled the s block. Similarly, the p block are the right-most six columns of the periodic table, the d block is the middle 10 columns of the periodic table, while the f block is the 14-column section that is normally depicted as detached from the main body of the periodic table. It could be part of the main body, but then the periodic table would be rather long and cumbersome. Figure \(8\) shows the blocks of the periodic table.
The electrons in the highest-numbered shell, plus any electrons in the last unfilled subshell, are called valence electrons; the highest-numbered shell is called the valence shell. (The inner electrons are called core electrons.) The valence electrons largely control the chemistry of an atom. If we look at just the valence shell's electron configuration, we find that in each column, the valence shell's electron configuration is the same. For example, take the elements in the first column of the periodic table: H, Li, Na, K, Rb, and Cs. Their electron configurations (abbreviated for the larger atoms) are as follows, with the valence shell electron configuration highlighted:
Table shows first column of the periodic table and their electron configurations.
H: 1s1
Li: 1s22s1
Na: [Ne]3s1
K: [Ar]4s1
Rb: [Kr]5s1
Cs: [Xe]6s1
They all have a similar electron configuration in their valence shells: a single s electron. Because much of the chemistry of an element is influenced by valence electrons, we would expect that these elements would have similar chemistry-and they do. The organization of electrons in atoms explains not only the shape of the periodic table but also the fact that elements in the same column of the periodic table have similar chemistry.
The same concept applies to the other columns of the periodic table. Elements in each column have the same valence shell electron configurations, and the elements have some similar chemical properties. This is strictly true for all elements in the s and p blocks. In the d and f blocks, because there are exceptions to the order of filling of subshells with electrons, similar valence shells are not absolute in these blocks. However, many similarities do exist in these blocks, so a similarity in chemical properties is expected.
Similarity of valence shell electron configuration implies that we can determine the electron configuration of an atom solely by its position on the periodic table. Consider Se, as shown in Figure \(9\). It is in the fourth column of the p block. This means that its electron configuration should end in a p4 electron configuration. Indeed, the electron configuration of Se is [Ar]4s23d104p4, as expected.
Example \(1\)
From the element's position on the periodic table, predict the valence shell electron configuration for each atom (Figure \(10\)).
1. Ca
2. Sn
Solution
1. Ca is located in the second column of the s block. We would expect that its electron configuration should end with s2. Calcium's electron configuration is [Ar]4s2.
2. Sn is located in the second column of the p block, so we expect that its electron configuration would end in p2. Tin's electron configuration is [Kr]5s24d105p2.
Exercise \(1\)
From the element's position on the periodic table, predict the valence shell electron configuration for each atom. Figure \(10\).
1. Ti
2. Cl
Answer a
[Ar]4s23d2
Answer b
[Ne]3s23p5
Example \(2\): Aluminum
Write the electron configuration of neutral aluminum atom. The atomic number of Al is 13.
Solution
Aluminum has 13 electrons.
Start at Period 1 of the periodic table, Figure \(2\). Place two electrons in the 1s subshell (1s2).
Proceed to Period 2 (left to right direction). Place the next two electrons in the 2s subshell (2s2) and the next six electrons in the 2p subshell (2p6).
Proceed to Period 3 (left to right direction). Place the next two electrons in the 3s subshell (3s2) and the last one electron in the 3p subshell (3p1).
The electron configuration of Aluminum is 1s22s22p63s23p1
Exercise \(2\)
Using Figure \(2\) as your guide, write the electron configuration of the atom that has 20 electrons
Answer
Start at Period 1 of Figure \(2\). Place two electrons in the 1s subshell (1s2).
Proceed to Period 2 (left to right direction). Place the next two electrons in the 2s subshell (2s2) and the next six electrons in the 2p subshell (2p6).
Proceed to Period 3 (left to right direction). Place the next two electrons in the 3s subshell (3s2) and the next six electron in the 3p subshell (3p6).
Proceed to Period 4. Place the remaining two electrons in the 4s subshell (4s2).
The electron configuration is 1s22s22p63s23p64s2
Summary
The arrangement of electrons in atoms is responsible for the shape of the periodic table. Electron configurations can be predicted by the position of an atom on the periodic table | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/02%3A_Atoms_and_the_Periodic_Table/2.08%3A_Electron_Configurations_and_the_Periodic_Table.txt |
Learning Objective
• Draw a Lewis electron dot symbol for a given atom.
In almost all cases, chemical bonds are formed by interactions of valence electrons in atoms. To facilitate our understanding of how valence electrons interact, a simple way of representing those valence electrons would be useful.
A Lewis electron-dot symbol (or electron-dot symbol or a Lewis symbol) is a representation of the valence electrons of an atom that uses dots around the symbol of the element. The number of dots equals the number of valence electrons in the atom. These dots are arranged to the right and left and above and below the symbol, with no more than two dots on a side. (It does not matter what order the positions are used.) For example, the electron-dot symbol for hydrogen is simply
$\mathbf{H}\mathbf{\cdot}$
Because the side is not important, the electron-dot symbol could also be drawn as follows:
$\mathbf{\dot{H}}\; \; or\; \mathbf{\cdot}\mathbf{H}\; \; \; or\; \; \; \mathbf{\underset{.}H}$
The electron-dot symbol for helium, with two valence electrons, is as follows:
$\mathbf{He}\mathbf{:}$
By putting the two electrons together on the same side, we emphasize the fact that these two electrons are both in the 1s subshell; this is the common convention we will adopt, although there will be exceptions later. The next atom, lithium, has an electron configuration of 1s22s1, so it has only one electron in its valence shell. Its electron dot diagram resembles that of hydrogen, except the symbol for lithium is used:
$\mathbf{Li}\mathbf{\cdot}$
Beryllium has two valence electrons in its 2s shell, so its eelectron-dot symbol is like that of helium:
$\mathbf{Be}\mathbf{:}$
The next atom is boron. Its valence electron shell is 2s22p1, so it has three valence electrons. The third electron will go on another side of the symbol:
$\mathbf{\dot{Be}}\mathbf{:}$
Again, it does not matter on which sides of the symbol the electron dots are positioned.
For carbon, there are four valence electrons, two in the 2s subshell and two in the 2p subshell. As usual, we will draw two dots together on one side, to represent the 2s electrons. However, conventionally, we draw the dots for the two p electrons on different sides. As such, the electron-dot symbol for carbon is as follows:
$\mathbf{\cdot \dot{C}}\mathbf{:}$
With N, which has three p electrons, we put a single dot on each of the three remaining sides:
$\mathbf{\cdot}\mathbf{\dot{\underset{.}N}}\mathbf{:}$
For oxygen, which has four p electrons, we now have to start doubling up on the dots on one other side of the symbol. When doubling up electrons, make sure that a side has no more than two electrons.
$\mathbf{\cdot}\mathbf{\ddot{\underset{.}O}}\mathbf{:}$
Fluorine and neon have seven and eight dots, respectively:
$\mathbf{:}\mathbf{\ddot{\underset{.}F}}\mathbf{:}$
$\mathbf{:}\mathbf{\ddot{\underset{.\: .}Ne}}\mathbf{:}$
With the next element, sodium, the process starts over with a single electron because sodium has a single electron in its highest-numbered shell, the n = 3 shell. By going through the periodic table, we see that the electron-dot symbol of atoms will never have more than eight dots around the atomic symbol.
Example $1$:
What is the electron-dot symbol 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 electron-dot symbol for each element?
1. phosphorus
2. argon
Answer
$\mathbf{\cdot }\mathbf{\dot{\underset{.}P}}\mathbf{:}\nonumber$
$\mathbf{:}\mathbf{\ddot{\underset{.\, .}Ar}}\mathbf{:}\nonumber$
Summary
• Lewis electron-dot symbols use dots to represent valence electrons around an atomic symbol. | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/02%3A_Atoms_and_the_Periodic_Table/2.09%3A_Electron-Dot_Symbols.txt |
When you think of bonding, you may not think of ions or molecules. Like most of us, you probably think of bonding between people. Like people, molecules bond — and some bonds are stronger than others. It's hard to break up a mother and baby, or a molecule made up of one oxygen and two hydrogen atoms! A chemical bond is a force of attraction between atoms or ions. Bonds form when atoms share or transfer valence electrons. Valence electrons are the electrons in the outer energy level of an atom that may be involved in chemical interactions. Valence electrons are the basis of all chemical bonds.
• 3.1: Ions
Atoms can gain or lose electrons to form positively charged or negatively charged species called ions.
• 3.2: Ions and the Octet Rule
Atoms tend to gain or lose electrons to achieve an octet (8 valence electrons). Electron configurations can be used to show how many electrons are needed to complete an octet and form an ion.
• 3.3: Ions of Some Common Elements
The periodic table can be used to determine ion charge.
• 3.4: Periodic Properties and Ion Formation
The energy changes associated with ion formation are called ionization energy and electron affinity. Ionization energy is the energy required to remove an electron and electron affinity is the energy released when an atom gains an electron. This is a more quantitative way to determine which atoms will form anions and which atoms will form cations.
• 3.5: Naming Monoatomic Ions
Cations and anions formed from one element (monoatomic ions) are assigned names based on nomenclature rules used by chemists. Cations are named using the element name plus the word "ion". In cases where an atom forms ions of variable charge, Roman numerals are used to specify each ion. Anions are named using the element name with the ending changed to "ide".
• 3.6: Polyatomic Ions
• 3.7: Ionic Bonds
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.
• 3.8: Formulas of Ionic Compounds
Although ionic compounds are made up of ions with charges, the overall charge must be zero, because matter is electrically neutral. Therefore, proper chemical formulas for ionic compounds can be determined by balancing the total positive charge with the total negative charge.
• 3.9: Naming Ionic Compounds
Each ionic compound has its own unique name that comes from the names of the ions that make up the formula unit. By convention, the cation is always named first followed by the anion. Because the ratio of ions in a compound is determined by the charge on each ion, no prefixes are needed to indicate how many of each ion are present. When a compound contains an ion with variable charge, a Roman numeral is used to specify the charge.
• 3.10: Some Properties of Ionic Compounds
Ionic compounds are composed of cations and anions that are strongly attracted to each other. Hence, ionic solids have very high melting points and are extremely hard. When dissolved in water, the ions separate from each other, allowing them to form electrolyte solutions.
• 3.11: H⁺ and OH⁻ Ions - An Introduction to Acids and Bases
Hydrogen ions (H+) and hydroxide ions (OH–) are two important ions discussed later in this text as acidic and basic ions respectively. Both of these ions can form compounds that you will see in organic and biological chemistry.
Thumbnail: The crystal structure of sodium chloride, NaCl, a typical ionic compound. The purple spheres represent sodium cations, \(\ce{Na^{+}}\), and the green spheres represent chloride anions, \(\ce{Cl^{−}}\). (Public Domain; Benjah-bmm27 via Wikipedia)
03: Ionic Compounds
Learning Objectives
• Describe how an ion is formed.
• Distinguish the difference between the two types of ions.
Ions
As introduced previously, atoms contain a nucleus with neutrons and positively charged protons, surrounded by negatively charged electrons. In an atom, the total number of electrons, negative charge, equals the total number of protons, positive charge, and therefore, atoms are electrically neutral or uncharged. If an atom loses or gains electrons, it will become a positively or negatively charged particle, called an ion. The loss of one or more electrons results in more protons than electrons and an overall positively charged ion, called a cation. For example, a sodium atom with one less electron is a cation, Na+, with a +1 charge (Figure \(1\)).
When an atom gains one or more electrons, it becomes a negatively charged anion, because there are more electrons than protons. When chlorine gains one electron it forms a chloride ion, Cl, with a –1 charge (Figures \(2\))
The names for positive and negative ions are pronounced CAT-eye-ons (cations) and ANN-eye-ons (anions), respectively.
Note Naming Ions
Cations are named using the element name plus "ion" to indicate it is charged. Anions are named by changing the element name ending to "ide". For example, a magnesium ion is formed when neutral magnesium loses electrons and a fluoride ion is formed when neutral fluorine gains electrons.
Example \(1\)
A calcium (Ca) atom loses two electrons and a sulfur (S) atom gains two electrons. Determine if the resulting ions are cations or anions? Write the ion symbols for each.
Solution
When calcium (Z = 20) gains two electrons the resulting ion will have 18 electrons and 20 protons and therefore a charge of +2 (there are two more positive protons than negative electrons), it is a cation. The symbol for a calcium ion is Ca2+.
When sulfur (Z = 16) gains two electrons the resulting ion will have 18 electrons and 16 protons and therefore a charge of –2 (there are two more negative electrons than positive protons), it is an anion. The symbol for a sulfide ion is S2.
Key Takeaways
• Ions are formed when atoms gain or lose electrons.
• Ions can be positively charged (cations) or negatively charged (anions).
Contributors
• Lisa Sharpe Elles, University of Kansas | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/03%3A_Ionic_Compounds/3.01%3A_Ions.txt |
Learning Objectives
• Use the octet rule and electron configurations to determine if an atom will gain or lose electrons forming anions or cations.
Ions are formed when an atom, usually on the left side of the periodic table, reacts with and transfers one or more electrons to another atom, usually on the right side of the periodic table. These electrons are usually lost from or gained into the valence shell, or outermost energy level (shell). Why do some atoms lose electrons and others gain electrons? How can we predict the number of electrons lost or gained? Which ions are stable and which ions do not form at all? These questions are best answered by looking at electron configurations and considering what is called the octet rule, which states that atoms gain or lose electrons to form a stable, noble gas configuration, i.e., a filled subshell containing eight electrons. Therefore, it is useful to take a closer look at electron configurations to further illustrate ion formation and electron transfer between atoms.
The electron configuration for sodium shows that there are ten core electrons and one valence electron in the third energy level. When sodium loses the single valence electron, forming the cation Na+, the electron configuration is now identical to that of neon, a stable noble gas with eight valence electrons.
$\begin{array}{lcl} \ce{Na} & \rightarrow & \ce{Na^+} + \ce{e^-} \ 1s^2 \: 2s^2 \: 2p^6 \: 3s^1 & & 1s^2 \: 2s^2 \: 2p^6 \end{array}$
Chlorine also has ten core electrons and valence electrons in the third energy level. However, chlorine has seven valence electrons, one less than the noble gas argon, which has eight valence electrons. Thus, chlorine will gain one electron, forming the anion, Cl, and achieving a stable noble gas configuration.
$\begin{array}{lcl} \ce{Cl} + \ce{e^-} & \rightarrow & \ce{Cl^-} \ 1s^2 \: 2s^2 \:2p^6 \: 3s^2 \: 3p^5 & & 1s^2 \: 2s^2 \: 2p^6 \: 3s^2 \: 3p^6 \end{array}$
The octet rule and the periodic table can be used to predict what ions will form; main group elements on the left side of the periodic table (metals in groups 1, 2, and 13) tend to lose electrons (form cations) to achieve the same electron configuration as the noble gas just before them in the table. The number of electrons the atom will lose depends on what group the atom is in, i.e., how many valence electrons it has. Main group elements on the right side of the periodic table (nonmetals in groups 15-17) will gain electrons to achieve the same electron configuration as the noble gas just after them in the table. Again, the number of electrons the atom will gain depends on the number of valence electrons it has and how many are needed to reach the filled subshell, eight electrons.
Note Violation of the Octet Rule
It is not impossible to violate the octet rule. Consider sodium: 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 Na+ ion. We could remove another electron by adding even more energy to the ion, to make the Na2+ 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 Na+ 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.
Example $1$
Write the electron configuration of aluminum atom (Z = 13) and underline the valence electrons. How many electrons are gained/lost to form an aluminum ion? Write the symbol and the electron configuration for an aluminum ion.
Solution
The electron configuration of Al atom is 1s22s22p63s23p1. Aluminum has three valence electrons in the third energy level, (3s23p1). The cation, Al3+, is formed when these three valence electrons are lost, leaving the configuration for the noble gas neon, 1s22s22p6.
Exercise $1$
Write the electron configuration of oxygen atom (Z = 8) and underline the valence electrons. How many electrons are gained/lost to form an oxide ion? Write the symbol and electron configuration for oxide ion.
Answer
The electron configuration of O atom is 1s22s22p4. Oxygen has six valence electrons in the second energy level, (2s22p4). The anion O2 is formed when two electrons are gained in the valence shell. The resulting electron configuration, 1s22s22p6, which is also identical to the configuration for the noble gas neon.
3.03: Ions of Some Common Elements
Learning Objectives
• Become familiar with the charge of some common ions.
• Use the periodic table to predict ion charge.
In many cases, elements that belong to the same group (vertical column) on the periodic table form ions with the same charge because they have the same number of valence electrons. Thus, as introduced previously, the periodic table becomes a tool for remembering the charges on many ions. For example, all ions made from alkali metals, the first column on the periodic table, have a +1 charge. Ions made from alkaline earth metals, the second group on the periodic table, have a +2 charge. On the other side of the periodic table, the next-to-last column, the halogens, form ions having a −1 charge. Figure \(1\) shows how the charge on many ions can be predicted by the location of an element on the periodic table.
Some elements, especially transition metals, can form ions with variable charges. Figure \(1\) shows the characteristic charges for some of these ions. Notice that there is no simple pattern for transition metal ions (or for the larger main group elements) as there is with the main group ions. This is because the transition metals have electrons in d subshell and do not follow the octet rule. In order for an element such as iron (Fe) to achieve the same noble gas configuration of argon (Ar), it would need to lose 6 electrons in the 3d subshell and 2 electrons in the 4s subshell. An iron ion with a charge of +8 is not very likely, therefore, the octet rule is not applicable to transition elements.
Note
For a multiply-charged ion, the correct convention is to write the charge number first followed by the sign. For example, the barium cation is written Ba2+, not Ba+2.
Example \(1\)
Which of these ions is not likely to form?
1. Mg+
2. K+
Solution
(a) Mg is in Group 2A and has two valence electrons. It achieves octet by losing two electrons to form Mg2+ cation. Losing only one electron to form Mg+ does not make an octet, hence, Mg+ is not likely to form.
Exercise \(2\)
Which of these ions is not likely to form?
1. S3
2. N3
Answer
(a) S is in Group 6A and has six valence electrons. It achieves octet by gaining two electrons to form S2 anion. Gaining three electrons to form S3does not make it octet, hence, S3 is not likely to form. | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/03%3A_Ionic_Compounds/3.02%3A_Ions_and_the_Octet_Rule.txt |
Learning Objectives
• Describe ionization energy and electron affinity and how this relates to ion formation trends.
We have seen that elements often gain or lose enough electrons to achieve the valence electron configuration of the nearest noble gas. Why is this so? In this section, we develop a more quantitative approach to predicting such reactions by examining periodic trends in the energy changes that accompany ion formation.
Ionization Energy
Because atoms do not spontaneously lose electrons, energy is required to remove an electron from an atom to form a cation. Chemists define the ionization energy ($IE$) of an element as the amount of energy needed to remove an electron from the gaseous atom $A$ in its ground state. $IE$ is therefore the energy required for the reaction:
$A(g) \rightarrow A^+(g) + e^- \;\;\ \text{energy required=IE } \label{3.4.1}$
Because an input of energy is required, the ionization energy is always positive ($IE > 0$) for the reaction as written in Equation $1$. Larger values of ($IE$) mean that the electron is more tightly bound to the atom and thus, harder to remove. Typical units for ionization energies are kilojoules/mole (kJ/mol) or electron volts (eV):
$1\; eV / atom = 96.49\; kJ/mol$
As you move across a row on the periodic table, the ionization energies generally tend to increase. This means that elements on the left side of the periodic table lose electrons more easily (requires less energy to remove an electron) than those on the right side of the periodic table. The trend in ionization energy can be explained by considering the trend in atomic radius (explained in an earlier chapter). As you move from left to right on the periodic table, the atomic size decreases and the electrostatic interactions between the nucleus and valence electrons increases, which increases the energy required to remove electrons, thus the ($IE$) increases.
If you look closely at the trends in ionization energy, you will notice that there are some "exceptions" where you see a decrease in energy rather than an increase. In periods 1 and 2, you can see this decrease in ($IE$) between groups 2 and 3 and again between groups 5 and 6. These variations in the trend can be further explained by looking closely at the electron configurations of the atoms in question. As you move from magnesium to aluminum, one electron is added to the 3p subshell. This 3p electron is slightly further from the nucleus (higher in energy) and is therefore, more easy to remove compared to the 3s electrons. The decrease between phosphorous and sulfur occurs because the added electron in sulfur is the first to be paired in the p subshell. These two electrons in the same p orbital repel each other, making the sulfur atom slightly less stable than would otherwise be expected, as is true of all the group 16 elements. This electron is easier to remove because it will lead to more stability.
Example $1$: Lowest First Ionization Energy
Use their locations in the periodic table to predict which element has the lowest first ionization energy: Ca, K, Mg, Na, Rb, or Sr.
Given: six elements
Asked for: element with lowest first ionization energy
Strategy:
Locate the elements in the periodic table. Based on trends in ionization energies across a row and down a column, identify the element with the lowest first ionization energy.
Solution:
These six elements form a rectangle in the two far-left columns of the periodic table. Because we know that ionization energies increase from left to right in a row and from bottom to top of a column, we can predict that the element at the bottom left of the rectangle will have the lowest first ionization energy: Rb.
Exercise $1$: Highest First Ionization Energy
Use their locations in the periodic table to predict which element has the highest first ionization energy: As, Bi, Ge, Pb, Sb, or Sn.
Answer
$\ce{As}$
Electron Affinity
The electron affinity ($EA$) of an element $A$ is defined as the energy change that occurs when an electron is added to a gaseous atom or ion:
$A(g)+e^- \rightarrow A^-(g) \;\;\; \text{energy change=}EA \label{7.5.1}$
Unlike ionization energies, which are always positive for a neutral atom because energy is required to remove an electron, electron affinities can be negative (energy is released when an electron is added), positive (energy must be added to the system to produce an anion), or zero (the process is energetically neutral). This sign convention is consistent with a negative value corresponded to the energy change for an exothermic process, which is one in which heat is released (Figure $2$).
The chlorine atom has the most negative electron affinity of any element, which means that more energy is released when an electron is added to a gaseous chlorine atom than to an atom of any other element:
$\ce{ Cl(g) + e^- \rightarrow Cl^- (g)} \;\;\; EA=-346\; kJ/mol \label{7.5.2}$
In contrast, beryllium does not form a stable anion, so its effective electron affinity is
$\ce{ Be(g) + e^- \rightarrow Be^- (g)} \;\;\; EA \ge 0 \label{7.5.3}$
In general, electron affinities of the main-group elements become less negative as we proceed down a column. This is because as the energy level increases, the extra electrons enter orbitals that are increasingly far from the nucleus, and it is less easy to gain extra electrons forming anions.
Example $2$: Contrasting Electron Affinities of Sb, Se, and Te
Based on their positions in the periodic table, which of Sb, Se, or Te would you predict to have the most negative electron affinity?
Given: three elements
Asked for: element with most negative electron affinity
Strategy:
1. Locate the elements in the periodic table. Use the trends in electron affinities going down a column for elements in the same group. Similarly, use the trends in electron affinities from left to right for elements in the same row.
2. Place the elements in order, listing the element with the most negative electron affinity first.
Solution:
We know that electron affinities become less negative going down a column (except for the anomalously low electron affinities of the elements of the second row), so we can predict that the electron affinity of Se is more negative than that of Te. We also know that electron affinities become more negative from left to right across a row, and that the group 15 elements tend to have values that are less negative than expected. Because Sb is located to the left of Te and belongs to group 15, we predict that the electron affinity of Te is more negative than that of Sb. The overall order is Se < Te < Sb, so Se has the most negative electron affinity among the three elements.
Exercise $2$: Contrasting Electron Affinities of Rb, Sr, and Xe
Based on their positions in the periodic table, which of Rb, Sr, or Xe would you predict to most likely form a gaseous anion?
Answer
Rb | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/03%3A_Ionic_Compounds/3.04%3A_Periodic_Properties_and_Ion_Formation.txt |
Learning Objectives
• Name monoatomic ions using the defined nomenclature rules.
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 Cations
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. Iron, for example, can form two cations, each of which, when combined with the same anion, makes a different ionic 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, 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.
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 \(1\): Names of Some Cations
Element Charge Symbol Common System Name Stock System Name
chromium 2+ Cr2+ chromous ion chromium(II) ion
3+ Cr3+ chromic ion chromium(III) ion
copper 1+ Cu+ cuprous ion copper(I) ion
2+ Cu2+ cupric ion copper(II) ion
iron 2+ Fe2+ ferrous ion iron(II) ion
3+ Fe3+ ferric ion iron(III) ion
lead 2+ Pb2+ plumbous ion lead(II) ion
4+ Pb3+ plumbic ion lead(IV) ion
tin 2+ Sn2+ stannous ion tin(II) ion
4+ Sn4+ stannic ion tin(IV) ion
Naming Anions
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.
Table \(2\): Some Monatomic Anions
Element Charge Symbol Name
fluorine 1– F fluoride ion
chlorine 1– Cl chloride ion
bromine 1– Br bromide ion
iodine 1– I iodide ion
oxygen 2– O2− oxide ion
sulfur 2– S2− sulfide ion
phosphorous 3– P3− phosphide ion
nitrogen 3– N3− nitride ion
Example \(1\)
Name each ion.
1. Ca2+
2. S2−
3. SO32
4. NH4+
5. Cu+
Answer a
the calcium ion
Answer b
the sulfide ion (from Table \(2\) )
Answer c
the sulfite ion
Answer d
the ammonium ion
Answer e
the copper(I) ion or the cuprous ion (copper can form cations with either a 1+ or 2+ charge, so we have to specify which charge this ion has
Exercise \(1\)
Name each ion.
1. Fe2+
2. Fe3+
3. SO42
4. Ba2+
5. HCO3
Answer a
the iron (II) or ferrous ion
Answer b
the iron (III) or ferric ion
Answer c
the sulfate ion
Answer d
the barium ion
Answer e
the bicarbonate ion or hydrogen carbonate ion
Example \(2\)
Write the formula for each ion.
1. the bromide ion
2. the phosphate ion
3. the cupric ion
4. the magnesium ion
Answer a
Br
Answer b
PO43
Answer c
Cu2+
Answer d
Mg2+
Exercise \(2\)
Write the formula for each ion.
1. the fluoride ion
2. the carbonate ion
3. the ferrous ion
4. the potassium ion
Answer a
F
Answer b
CO32-
Answer c
Fe2+
Answer d
K+
Note Chemistry Is Everywhere: Salt
The element sodium (part [a] in the accompanying figure) is a very reactive metal; given the opportunity, it will react with the sweat on your hands and form sodium hydroxide, which is a very corrosive substance. The element chlorine (part [b] in the accompanying figure) is a pale yellow, corrosive gas that should not be inhaled due to its poisonous nature. Bring these two hazardous substances together, however, and they react to make the ionic compound sodium chloride (part [c] in the accompanying figure), known simply as salt.
Salt is necessary for life. Na+ ions are one of the main ions in the human body and are necessary to regulate the fluid balance in the body. Cl ions are necessary for proper nerve function and respiration. Both of these ions are supplied by salt. The taste of salt is one of the fundamental tastes; salt is probably the most ancient flavoring known, and one of the few rocks we eat.
The health effects of too much salt are still under debate, although a 2010 report by the US Department of Agriculture concluded that "excessive sodium intake…raises blood pressure, a well-accepted and extraordinarily common risk factor for stroke, coronary heart disease, and kidney disease."US Department of Agriculture Committee for Nutrition Policy and Promotion, Report of the Dietary Guidelines Advisory Committee on the Dietary Guidelines for Americans, accessed January 5, 2010. It is clear that most people ingest more salt than their bodies need, and most nutritionists recommend curbing salt intake. Curiously, people who suffer from low salt (called hyponatria) do so not because they ingest too little salt but because they drink too much water. Endurance athletes and others involved in extended strenuous exercise need to watch their water intake so their body's salt content is not diluted to dangerous levels. | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/03%3A_Ionic_Compounds/3.05%3A_Naming_Monoatomic_Ions.txt |
Learning Objectives
• To identify and name polyatomic ions.
Some ions consist of groups of atoms covalently bonded together and have an overall electric charge. Because these ions contain more than one atom, they are called polyatomic ions. The structures, names, and formulas of some polyatomic ions are found in the below figure and table.
Polyatomic ions have defined formulas, names, and charges that cannot be modified in any way. Table \(1\) lists the ion names and ion formulas of the most common polyatomic ions. For example, \(\ce{NO_3^{−}}\) is the nitrate ion; it has one nitrogen atom and three oxygen atoms and an overall −1 charge.
Table \(1\): Common Polyatomic Ion Names and Formulas
Ion Name Ion Formula
hyddronium ion H3O+
ammonium ion NH4+
hydroxide ion OH
cyanide ion CN
carbonate ion CO32
bicarbonate or hydrogen carbonate HCO3
acetate ion C2H3O2 or CH3CO2
nitrate ion NO3
nitrite ion NO2
sulfate ion SO42
sulfite ion SO32
phosphate ion PO43
phosphite ion PO33
Note that only two polyatomic ions in this table are cations, hydronium ion (H3O+) and ammonium ion (NH4+), the remaining polyatomic ions are all negatively-charged and, therefore, are classified as anions. However, only two of these, the hydroxide ion and the cyanide ion, are named using the "-ide" suffix that is typically indicative of negatively-charged ions. The remaining polyatomic anions, which all contain oxygen, in combination with another non-metal, exist as part of a series in which the number of oxygens within the polyatomic unit can vary. A single suffix, "-ide," is insufficient for distinguishing the names of the anions in a related polyatomic series. Therefore, "-ate" and "-ite" suffixes are employed, in order to denote that the corresponding polyatomic ions are part of a series. Additionally, these suffixes also indicate the relative number of oxygens that are contained within the polyatomic ions. Note that all of the polyatomic ions whose names end in "-ate" contain one more oxygen than those polyatomic anions whose names end in "-ite." Unfortunately, much like the common system for naming transition metals, these suffixes only indicate the relative number of oxygens that are contained within the polyatomic ions. For example, both the nitrate ion, symbolized as NO3, and the sulfate ion, symbolized as SO42, share an "-ate" suffix, however, the former contains three oxygens, and the latter contains four. Additionally, both the nitrate ion and the sulfite ion contain three oxygens, but these polyatomic ions do not share a common suffix. Unfortunately, the relative nature of these suffixes mandates that the ion formula/ion name combinations of the polyatomic ions must simply be memorized.
3.07: Ionic Bonds
Learning Objectives
• Define an ionic bond, ionic compound, and electrostatic force
• Recognize the complexity of three-dimensional ionic bond interactions involved in ionic compounds
Oppositely charged particles attract each other. This attractive force is often referred to as an electrostatic attraction. An ionic bond is the electrostatic electrostatic attraction that holds ions together in an ionic compound. The strength of the ionic bond is directly dependent upon the quantity of the charges and inversely dependent on the distance between the charged particles. A cation with a 2+ charge will make a stronger ionic bond than a cation with a 1+ charge. A larger ion makes a weaker ionic bond because of the greater distance between its electrons and the nucleus of the oppositely charged ion.
We will use sodium chloride as an example to demonstrate the nature of the ionic bond and how it forms. As you know, sodium is a metal and loses its one valence electron to become a cation. Chlorine is a nonmetal and gains one electron in becoming an anion. However, electrons cannot be simply "lost" to nowhere in particular. A more accurate way to describe what is happening is that a single electron is transferred from the sodium atom to the chlorine atom as shown below.
The ionic bond is the attraction of the \(\ce{Na^{+}}\) ion for the \(\ce{Cl^{-}}\) ion. It is conventional to show show the cation without dots around the symbol to emphasize that the original energy level that contained the valence electron is now empty. The anion is now shown with a complete octet of electrons.
Ionic compounds are held together by attractive electrostatic interactions between cations and anions. In contrast to the simplified electron transfer depicted above for sodium and chlorine, the cations and anions in ionic compounds are arranged in space to form an extended three-dimensional array that maximizes the number of attractive electrostatic interactions and minimizes the number of repulsive electrostatic interactions (Figure \(1\)). In other words, each ion is attracted to many of its neighboring ions forming a sodium chloride crystal or ionic solid. | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/03%3A_Ionic_Compounds/3.06%3A_Polyatomic_Ions.txt |
Learning Objectives
• Write the chemical formula for a simple ionic compound based on the charges of the cations and anions in the compound.
In every ionic compound, the total number of positive charges of the cations equals the total number of negative charges of the anions. Thus, ionic compounds are electrically neutral overall, even though they contain positive and negative ions. We can use this observation to help us write the formula of an ionic compound. The formula of an ionic compound must have a ratio of ions such that the numbers of positive and negative charges are equal.
Ionic Compounds Formed From Monoatomic Ions
Consider an Na atom in the presence of a Cl atom. The two atoms have these Lewis electron dot diagrams and electron configurations:
$\mathbf{Na\, \cdot }\; \; \; \; \; \; \; \; \; \; \mathbf{\cdot }\mathbf{\ddot{\underset{.\: .}Cl}}\mathbf{\: :}\nonumber$
$\left [ Ne \right ]3s^{1}\; \; \; \; \left [ Ne \right ]3s^{2}3p^{5}\nonumber$
For the Na atom to obtain an octet, it must lose an electron; for the Cl atom to obtain an octet, it must gain an electron. An electron transfers from the Na atom to the Cl atom:
$\mathbf{Na\, \cdot }\curvearrowright \mathbf{\cdot }\mathbf{\ddot{\underset{.\: .}Cl}}\mathbf{\: :}\nonumber$
resulting in two ions—the Na+ ion and the Cl ion:
$\mathbf{Na\, \cdot }^{+}\; \; \; \; \; \; \; \; \mathbf{:}\mathbf{\ddot{\underset{.\: .}Cl}}\mathbf{\: :}^{-}\nonumber$
$\left [ Ne \right ]\; \; \; \; \; \left [ Ne \right ]3s^{2}3p^{6}\nonumber$
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:
$\mathbf{Na\, \cdot }^{+}\; + \; \mathbf{:}\mathbf{\ddot{\underset{.\: .}Cl}}\mathbf{\: :}^{-}\rightarrow Na^{+}Cl^{-}\; \; or\; \; NaCl\nonumber$
where we have written the final formula (the formula for sodium chloride) as per the convention for ionic compounds, without listing the charges explicitly. As explained previously, the attraction between oppositely charged ions is called an ionic bond.
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:
$\mathbf{Mg\,}^{2+}\; + \; \mathbf{:}\mathbf{\ddot{\underset{.\: .}O}}\mathbf{\: :}^{2-}\; \; \; \; \; Mg^{2+}O^{2-}\; or\; MgO\nonumber$
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 form an overall neutrally charged ionic compound, which we write as Na2O. The need for the number of electrons lost to be equal to the number of electrons gained explains why ionic compounds have the ratio of cations to anions that they do. This is also required by the law of conservation of matter.
Example $1$
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. We need two Cl atoms to accept the two electrons from one Ca atom. The transfer process is as follows:
The oppositely charged ions attract one another to make CaCl2.
Exercise $1$
With arrows, illustrate the transfer of electrons to form potassium sulfide from K atoms and S atoms.
Answer
For compounds in which the ratio of ions is not as obvious, the subscripts in the formula can be obtained by crossing charges: use the absolute value of the charge on one ion as the subscript for the other ion. This method is shown schematically in Figure 3.3.2.
When crossing charges, it is sometimes necessary to reduce the subscripts to their simplest ratio to write the empirical formula. Consider, for example, the compound formed by Pb4+ and O2−. Using the absolute values of the charges on the ions as subscripts gives the formula Pb2O4. This simplifies to its correct empirical formula PbO2. The empirical formula has one Pb4+ ion and two O2− ions.
Example $2$: Predicting the Formula of an Ionic Compound
The gemstone sapphire (Figure $2$) is mostly a compound of aluminum and oxygen that contains aluminum cations, Al3+, and oxygen anions, O2−. What is the formula of this compound?
Solution
Because the ionic compound must be electrically neutral, it must have the same number of positive and negative charges. Two aluminum ions, each with a charge of 3+, would give us six positive charges, and three oxide ions, each with a charge of 2−, would give us six negative charges. The formula would be Al2O3.
Exercise $2$
Predict the formula of the ionic compound formed between the sodium cation, Na+, and the sulfide anion, S2−.
Answer
Na2S
Ionic Compounds Formed From Polyatomic Ions
Many ionic compounds contain polyatomic ions as the cation, the anion, or both. As with simple ionic compounds, these compounds must also be electrically neutral, so their formulas can be predicted by treating the polyatomic ions as discrete units. We use parentheses in a formula to indicate a group of atoms that behave as a unit. For example, the formula for calcium phosphate, one of the minerals in our bones, is Ca3(PO4)2. This formula indicates that there are three calcium ions (Ca2+) for every two phosphate $\left(\ce{PO4^{3-}}\right)$ groups. The $\ce{PO4^{3-}}$ groups are discrete units, each consisting of one phosphorus atom and four oxygen atoms, and having an overall charge of 3−. The compound is electrically neutral, and its formula shows a total count of three Ca, two P, and eight O atoms.
Example $3$: Predicting the Formula of a Compound with a Polyatomic Anion
Baking powder contains calcium dihydrogen phosphate, an ionic compound composed of the ions Ca2+ and $\ce{H2PO4-}$. What is the formula of this compound?
Solution
The positive and negative charges must balance, and this ionic compound must be electrically neutral. Thus, we must have two negative charges to balance the 2+ charge of the calcium ion. This requires a ratio of one Ca2+ ion to two $\ce{H2PO4-}$ ions. We designate this by enclosing the formula for the dihydrogen phosphate ion in parentheses and adding a subscript 2. The formula is Ca(H2PO4)2.
Exercise $3$
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:
Mg2+ and CO32- = MgCO3
Answer b:
Al3+ and C2H3O2- = Al(C2H3O2)3
Formula Unit
Ionic compounds exist as alternating positive and negative ions in regular, three-dimensional arrays called crystals (Figure $3$). As you can see, there are no individual $\ce{NaCl}$ “particles” in the array; instead, there is a continuous lattice of alternating sodium and chloride ions. However, we can use the ratio of sodium ions to chloride ions, expressed in the lowest possible whole numbers or simplest formula, as a way of describing the compound. In the case of sodium chloride, the ratio of sodium ions to chloride ions, expressed in lowest whole numbers, is 1:1, so we use $\ce{NaCl}$ (one $\ce{Na}$ symbol and one $\ce{Cl}$ symbol) to represent the compound. Thus, $\ce{NaCl}$ is the chemical formula for sodium chloride, which is a concise way of describing the relative number of different ions in the compound. A macroscopic sample is composed of myriads of NaCl pairs; each individual pair called a formula unit. Although it is convenient to think that $\ce{NaCl}$ crystals are composed of individual $\ce{NaCl}$ units, Figure $3$ shows that no single ion is exclusively associated with any other single ion. Each ion is surrounded by ions of opposite charge. | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/03%3A_Ionic_Compounds/3.08%3A_Formulas_of_Ionic_Compounds.txt |
Learning Objectives
• Write the names for ionic compounds by recognizing and naming the ions in the formula unit.
Ionic compounds are named using the formula unit and by following some important conventions. First, the name of the cation is written first followed by the name of the anion. Because most metals form cations and most nonmetals form anions, formulas typically list the metal first and then the nonmetal. Second, charges are not included in the name (or the formula). Remember that in an ionic compound, the component species are ions, not neutral atoms, even though the formula does not contain charges. The proper formula for an ionic compound will show how many of each ion is needed to balance the total positive and negative charges; the name does not need to include indication of this ratio.
There are two main types of ionic compound with different naming rules for each; Type I: compounds containing cations of main group elements and Type II: compounds containing cations of variable charge (generally transition metals). Below we will look at examples of each type to learn the rules for naming.
Type I Ionic Compounds
Cations of main group elements do not have variable charges and are the simply named by placing the name of the cation first, followed by the name of the anion, and dropping the word ion from both parts. For example, what is the name of the compound whose formula is \(\ce{Ba(NO3)2}\)?
The compound’s name does not need to indicate that there are two nitrate ions for every barium ion. You must determine the ratio of ions in the formula unit by balancing the positive and negative charges.
Type II Ionic Compounds
Some metals can form cations with variable charges. When naming 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 \(\ce{FeCl2}\) and \(\ce{FeCl3}\). In the first compound, the iron ion has a 2+ charge because there are two \(\ce{Cl^{−}}\) 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 \(\ce{Cl^{−}}\) 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. If we were to use the stems and suffixes of the common system, the names would be ferrous chloride and ferric chloride, respectively.
Ionic Compounds in Your Cabinets
Every day you encounter and use a large number of ionic compounds. Some of these compounds, where they are found, and what they are used for are listed in Table \(1\). Look at the label or ingredients list on the various products that you use during the next few days, and see if you run into any of those in this table, or find other ionic compounds that you could now name or write as a formula.
Table \(1\): Everyday Ionic Compounds
Ionic Compound Name Use
NaCl sodium chloride ordinary table salt
KI potassium iodide added to “iodized” salt for thyroid health
NaF sodium fluoride ingredient in toothpaste
NaHCO3 sodium bicarbonate baking soda; used in cooking (and in antacids)
Na2CO3 sodium carbonate washing soda; used in cleaning agents
NaOCl sodium hypochlorite active ingredient in household bleach
CaCO3 calcium carbonate ingredient in antacids
Mg(OH)2 magnesium hydroxide ingredient in antacids
Al(OH)3 aluminum hydroxide ingredient in antacids
NaOH sodium hydroxide lye; used as drain cleaner
K3PO4 potassium phosphate food additive (many purposes)
MgSO4 magnesium sulfate added to purified water
Na2HPO4 sodium hydrogen phosphate anti-caking agent; used in powdered products
Na2SO3 sodium sulfite preservative
As you practice naming compounds, use Figure \(1\) as a guide.
Example \(3\)
Name each ionic compound, using both Stock and common systems if necessary.
1. Ca3(PO4)2
2. (NH4)2Cr2O7
3. KCl
4. CuCl
5. SnF2
Answer a
calcium phosphate
Answer b
ammonium dichromate (the prefix di- is part of the name of the anion)
Answer c
potassium chloride
Answer d
copper(I) chloride or cuprous chloride
Answer e
tin(II) fluoride or stannous fluoride
Exercise \(3\)
Name each ionic compound, using both Stock and common systems if necessary.
1. ZnBr2
2. Fe(NO3)3
3. Al2O3
4. CuF2
5. AgF
Answer a
zinc bromide
Answer b
iron (III) nitrate or ferric nitrate
Answer c
aluminum oxide
Answer d
copper (II) fluoride or cupric fluoride
Answer e
silver fluoride | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/03%3A_Ionic_Compounds/3.09%3A_Naming_Ionic_Compounds.txt |
Learning Objectives
• Describe the basic physical properties of ionic compounds.
The figure below shows just a few examples of the color and brilliance of naturally occurring ionic crystals. The regular and orderly arrangement of ions in the crystal lattice is responsible for the various shapes of these crystals, while transition metal ions give rise to the colors.
Melting Points
Because of the many simultaneous attractions between cations and anions that occur, ionic crystal lattices are very strong. The process of melting an ionic compound requires the addition of large amounts of energy in order to break all of the ionic bonds in the crystal. For example, sodium chloride has a melting temperature of about 800 oC. As a comparison, the molecular compound water melts at 0 °C.
Shattering
Ionic compounds are generally hard, but brittle. Why? It takes a large amount of mechanical force, such as striking a crystal with a hammer, to force one layer of ions to shift relative to its neighbor. However, when that happens, it brings ions of the same charge next to each other (see below). The repulsive forces between like-charged ions cause the crystal to shatter. When an ionic crystal breaks, it tends to do so along smooth planes because of the regular arrangement of the ions.
Conductivity
Another characteristic property of ionic compounds is their electrical conductivity. The figure below shows three experiments in which two electrodes that are connected to a light bulb are placed in beakers containing three different substances.
In the first beaker, distilled water does not conduct a current because water is a molecular compound. In the second beaker, solid sodium chloride also does not conduct a current. Despite being ionic and thus composed of charged particles, the solid crystal lattice does not allow the ions to move between the electrodes. Mobile charged particles are required for the circuit to be complete and the light bulb to light up. In the third beaker, the NaCl has been dissolved into the distilled water. Now the crystal lattice has been broken apart and the individual positive and negative ions can move.
Example \(1\)
Write the dissociation equation of solid NaCl in water.
Solution
NaCl(s) → Na+(aq) + Cl(aq)
Exercise \(1\)
Write the dissociation equation of solid NH4NO3 in water.
Answer
NH4NO3(s) → NH4+(aq) + NO3(aq)
Key Takeaways
• Ionic compounds have high melting points.
• Ionic compounds are hard and brittle.
• Ionic compounds dissociate into ions when dissolved in water.
• Solutions of ionic compounds and melted ionic compounds conduct electricity, but solid materials do not.
• An ionic compound can be identified by its chemical formula: metal + nonmetal or polyatomic ions.
Contributors and Attributions
• CK-12 Foundation by Sharon Bewick, Richard Parsons, Therese Forsythe, Shonna Robinson, and Jean Dupon.
3.11: H and OH Ions - An Introduction to Acids and Bases
Learning Objectives
• To identify H+ as an acid and OH as a base.
You may have some idea, from your day-to-day experiences, of some chemical substances that are considered acids or bases. Generally, a compound that is sour is recognized as being an acid. Some familiar acids may include citric acid, which is a molecule found in citrus fruits such as lemons and limes, and acetic acid, the molecule that makes up vinegar. Bases, on the other hand, include chemicals such as sodium bicarbonate (baking soda), which is bitter-tasting, and sodium hydroxide, which is found in cleaning substances and is not recommended for ingestion.
More details about acids, bases, and the reactions they undergo will be discussed later in this text. However, because you have just learned about ions and ionic compounds, it is worth pointing out two important ions that are used to identify and distinguish if a substance is acidic or basic; hydrogen ions, \(\ce{H^{+}}\), and hydroxide ions, \(\ce{OH^{−}}\). In fact, one definition of acids and bases states that an acid will produce \(\ce{H^{+}}\) when dissolved in water and a base will produce a \(\ce{OH^{−}}\) when dissolved in water.
Ionic compounds that are basic are easily recognized because the hydroxide ion is part of the formula and name. Some common examples are sodium hydroxide, \(\ce{NaOH}\), and calcium hydroxide, \(\ce{Ca(OH)2}\). When dissolved in water, sodium hydroxide will split into its constituent ions, sodium ions (\(\ce{Na+}\)) and hydroxide ions (\(\ce{OH^{−}}\)) in a 1:1 ratio. However, according to it's chemical formula, calcium hydroxide will produce two hydroxide ions for every one calcium ion, a 1:2 ratio.
Compounds that produce hydrogen ions in water contain one or more hydrogen ions in the chemical formula and usually have special names to help recognize them as acids. Hydrochloric acid (\(\ce{HCl}\)), nitric acid (\(\ce{HNO3}\)), and carbonic acid (\(\ce{H2CO3}\)) are all acids. \(\ce{HCl}\) and \(\ce{HNO3}\) each dissolve in water producing a \(\ce{H^{+}}\) and an anion (chloride and nitrate respectively). As indicated by the formula for carbonic acid there are two \(\ce{H^{+}}\) for every one carbonate ion. The below table lists some common acids. See if you can determine the ratio of \(\ce{H^{+}}\) to anions produced when these compounds are dissolved in water.
Table \(1\) Common Acids and Their Anions
Acid Name Acid Formula Anion Name Anion Formula
acetic acid \(\ce{CH3COOH}\) acetate ion \(\ce{CH3COO^{-}}\)
carbonic acid \(\ce{H2CO3}\) bicarbonate ion \(\ce{HCO3^{-}}\)
carbonate ion \(\ce{CO3^{2-}}\)
hydrobromic acid \(\ce{HBr}\) bromide ion \(\ce{Br^{-}}\)
nitric acid \(\ce{HNO3}\) nitrate ion \(\ce{NO3^{-}}\)
nitrous acid \(\ce{HNO2}\) nitrous ion \(\ce{NO2^{-}}\)
phosphoric acid \(\ce{H3PO4}\) dihydrogen phosphate ion \(\ce{H2PO4^{-}}\)
hydrogen phosphate ion \(\ce{HPO4^{2-}}\)
phosphate ion \(\ce{PO4^{3-}}\)
sulfuric acid \(\ce{H2SO4}\) hydrogen sulfate ion \(\ce{HSO4^{-}}\)
sulfate ion \(\ce{SO4^{2-}}\)
sulfurous acid \(\ce{H2SO3}\) hydrogen sulfite ion \(\ce{HSO3^{-}}\)
sulfite ion \(\ce{SO3^{2-}}\) | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/03%3A_Ionic_Compounds/3.10%3A_Some_Properties_of_Ionic_Compounds.txt |
• 4.1: Covalent Bonds
You have already seen examples of substances that contain covalent bonds. One substance mentioned previously was water (H₂O). You can tell from its formula that it is not an ionic compound; it is not composed of a metal and a nonmetal. Consequently, its properties are different from those of ionic compounds. A covalent bond is formed between two atoms by sharing electrons.
• 4.2: Covalent Bonds and the Periodic Table
Atoms share electrons and form covalent bonds to satisfy the octet rule. The periodic table and trends in valence electrons can be used to determine the number of bonds an atom is most likely to form.
• 4.3: Multiple Covalent Bonds
Some molecules must have multiple covalent bonds between atoms to satisfy the octet rule.
• 4.4: Coordinate Covalent Bonds
Each of the covalent bonds that we have looked at so far has involved each of the atoms that are bonding contributing one of the electrons to the shared pair. There is an alternate type of covalent bond in which one of the atoms provided both of the electrons in a shared pair.
• 4.5: Characteristics of Molecular Compounds
Ionic compounds and molecular compounds have very different physical properties.
• 4.6: Molecular Formulas and Lewis Structures
Molecules can be represented using formulas, which give information about the number and type of atoms bonded together. Different types of structural formulas show the bonds between atoms and sometimes give information about molecular shape as well.
• 4.7: Drawing Lewis Structures
Molecules can be represented using Lewis structures, which show how electrons are arranged around the atoms in a molecule as bonded pairs of electrons (bonds) and lone pairs of electrons. These structures are useful for explaining and depicting molecular shapes and chemical reactivity.
• 4.8: The Shapes of Molecules
According to valence-shell electron-pair repulsion theory, VSEPR, the electron groups minimize repulsion by getting as far apart as possible from each other. Therefore, the approximate shape of a molecule can be predicted from the number of electron groups and the number of surrounding atoms.
• 4.9: Polar Covalent Bonds and Electronegativity
Covalent bonds between different atoms have different bond lengths. Covalent bonds can be polar or nonpolar, depending on the electronegativity difference between the atoms involved.
• 4.10: Polar Molecules
The molecular polarity of a diatomic molecule is determined by the bond polarity. The polarity of molecules with more than one bond must be determined by first identifying the molecular structure and then the bond polarity. If the dipole moments point in a similar direction, there is a net molecular dipole. If the dipole moments point in opposite directions, they cancel out and there is no net dipole.
• 4.11: Naming Binary Molecular Compounds
The chemical formula of a simple covalent compound can be determined from its name. The name of a simple covalent compound can be determined from its chemical formula.
Thumbnail: Covalently bonded hydrogen and carbon in a w:molecule of methane. (CC BY-SA 2.5; DynaBlast via Wikipedia)
04: Molecular Compounds
Learning Objectives
• Describe how covalent bonds form using the octet rule.
Formation of Covalent Bonds
Nonmetal atoms frequently form covalent bonds with other nonmetal atoms. Covalent bonds are formed between two atoms when both have similar tendencies to attract electrons to themselves (i.e., when both atoms have identical or fairly similar ionization energies and electron affinities). For example, two hydrogen atoms bond covalently to form an H2 molecule; each hydrogen atom in the H2 molecule has two electrons stabilizing it, giving each atom the same number of valence electrons as the noble gas He.
The bond in a hydrogen molecule, measured as the distance between the two nuclei, is about 7.4 × 10−11 m, or 74 picometers (pm; 1 pm = 1 × 10−12 m). This particular bond length represents the lowest potential energy state of two hydrogen atoms and is a balance between several forces: the attractions between oppositely charged electrons and nuclei, the repulsion between two negatively charged electrons, and the repulsion between two positively charged nuclei.
A plot of the potential energy of the system as a function of the internuclear distance (Figure \(2\)) shows that energy decreases as two hydrogen atoms move toward each other. Starting on the far right, we have two separate hydrogen atoms with a particular potential energy, indicated by the red line. Along the x-axis is the distance between the two atoms. As the two atoms approach each other (moving left along the x-axis), their valence orbitals (1s) begin to overlap. The single electrons on each hydrogen atom then interact with both atomic nuclei, occupying the space around both atoms. The strong attraction of each shared electron to both nuclei stabilizes the system, and the potential energy decreases as the bond distance decreases. If the atoms continue to approach each other, the positive charges in the two nuclei begin to repel each other, and the potential energy increases.
Lewis Structures
Chemists frequently use Lewis structures to represent covalent bonding in molecular substances. For example, the Lewis symbols of two separate hydrogen atoms are as follows:
The Lewis structures 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. For simplicity’s sake, it is common 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:
You will notice that 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.
Diatomic Molecules
Hydrogen (\(\ce{H2}\)) and fluorine (\(\ce{F2}\)) are both described as diatomic molecules. These elements and others (see Table \(1\)) exist naturally as molecules rather than as individual atoms. It is important to note that the names of these elements represent molecules and not individual atoms. When describing a single atom rather than a molecule, the word atom is used.
Table \(1\): Elements That Exist as Diatomic Molecules
Hydrogen (\(\ce{H2}\)) Oxygen (\(\ce{O2}\)) Nitrogen (\(\ce{N2}\)) Fluorine (\(\ce{F2}\))
Chlorine (\(\ce{Cl2}\)) Bromine (\(\ce{Br2}\)) Iodine (\(\ce{I2}\)) | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/04%3A_Molecular_Compounds/4.01%3A_Covalent_Bonds.txt |
Learning Objectives
• Predict the number of covalent bonds formed based on the elements involved and their position on the periodic table.
• Describe the important exceptions to the octet rule.
Diatomic molecules such as hydrogen (\(\ce{H2}\)), chlorine (\(\ce{Cl2}\)), fluorine (\(\ce{F2}\)), etc. containing covalent bonds between two of the same type of atom are only a few examples of the vast number of molecules that can form. Two different atoms can also share electrons and form covalent bonds. For example, water, (\(\ce{H2O}\)), has two covalent bonds between a single oxygen atom and two hydrogen atoms. Ammonia, (\(\ce{NH3}\), is a central nitrogen atom bonded to three hydrogen atoms. Methane, (\(\ce{CH4}\), is a single carbon atom covalently bonded to four hydrogen atoms. In these examples the central atoms form different numbers of bonds to hydrogen atoms in order to complete their valence subshell and form octets.
How Many Covalent Bonds Are Formed?
The number of bonds that an atom can form can often be predicted from the number of electrons needed to reach an octet (eight valence electrons); this is especially true of the nonmetals of the second period of the periodic table (C, N, O, and F). For example, each atom of a group 14 element has four electrons in its outermost shell and therefore requires four more electrons to reach an octet. These four electrons can be gained by forming four covalent bonds, as illustrated here for carbon in CCl4 (carbon tetrachloride) and silicon in SiH4 (silane). Because hydrogen only needs two electrons to fill its valence shell, it is an exception to the octet rule and only needs to form one bond. The transition elements and inner transition elements also do not follow the octet rule since they have d and f electrons involved in their valence shells.
Group 15 elements such as nitrogen have five valence electrons in the atomic Lewis symbol: one lone pair and three unpaired electrons. To obtain an octet, these atoms form three covalent bonds, as in NH3 (ammonia). Oxygen and other atoms in group 16 obtain an octet by forming two covalent bonds:
The number of electrons required to obtain an octet determines the number of covalent bonds an atom can form. This is summarized in the table below. In each case, the sum of the number of bonds and the number of lone pairs is 4, which is equivalent to eight (octet) electrons.
Table showing 4 different atoms, each of their number of bonds, and each of their number of lone pairs.
Atom (Group number) Number of Bonds Number of Lone Pairs
Carbon (Group 14) 4 0
Nitrogen (Group 15) 3 1
Oxygen (Group 16) 2 2
Fluorine (Group 17) 1 3
Because hydrogen only needs two electrons to fill its valence shell, it follows the duet rule. Hydrogen only needs to form one bond to complete a duet of electrons. This is the reason why H is always a terminal atom and never a central atom.
Example \(1\)
Examine the Lewis structure of OF2 below. Count the number of bonds formed by each element. Based on the element's location in the periodic table, does it correspond to the expected number of bonds shown in Table 4.1? Does the Lewis structure below follow the octet rule?
Solution
Yes. F (group 7A) forms one bond and O (group 6A) forms 2 bonds. Each atom is surrounded by 8 electrons. This structure satisfies the octet rule.
Exercise \(1\)
Examine the Lewis structure of NCl3 below. Count the number of bonds formed by each element. Based on the element's location in the periodic table, does it correspond to the expected number of bonds shown in Table 4.1? Does the Lewis structure below follow the octet rule?
Answer
Both Cl and N form the expected number of bonds. Cl (group 7A) has one bond and 3 lone pairs. The central atom N (group 5A) has 3 bonds and one lone pair. Yes, the Lewis structure of NCl3 follows the octet rule.
Octet Rule Exceptions
As important and useful as the octet rule is in chemical bonding, there are many covalent molecules with central atoms that do not have eight electrons in their Lewis structures. This does not mean that the octet rule is useless—quite the contrary. As with many rules, there are exceptions, or violations. These molecules fall into three categories:
• Odd-electron molecules have an odd number of valence electrons, and therefore have an unpaired electron.
• Electron-deficient (diminished octet) molecules have a central atom that has fewer electrons than needed for a noble gas configuration.
• Expanded octet (hypervalent) molecules have a central atom that has more electrons than needed for a noble gas configuration.
Odd-electron molecules
Although they are few, some stable compounds, often called free radicals, 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 \(\ce{NO}\), \(\ce{NO2}\), and \(\ce{ClO2}\). The Lewis electron dot diagram for \(\ce{NO}\), a compound produced in internal combustion engines when oxygen and nitrogen react at high temperatures, is as follows:
As you can see, the nitrogen and oxygen share four electrons between them. The oxygen atom has an octet of electrons but the nitrogen atom has only seven valence electrons, two electrons in the double bond, one lone pair, and one additional lone electron. Although \(\ce{NO}\) is a stable compound, it is very chemically reactive, as are most other odd-electron compounds.
Electron-deficient molecules
These stable compounds have less than eight electrons around an atom in the molecule, i.e. they have less than an octet. 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 \(\ce{B}\) atom. A well-known example is \(\ce{BF3}\):
Expanded Octet Molecules
Elements in the second period of the periodic table (n = 2) can accommodate only eight electrons in their valence shell orbitals because they have only four valence orbitals (one 2s and three 2p orbitals). Elements in the third and higher periods (n ≥ 3) have more than four valence orbitals and can share more than four pairs of electrons with other atoms because they have empty d orbitals in the same shell. Molecules formed from these elements have expanded octets and are sometimes called hypervalent molecules. Phosphorous pentachloride shares five pairs of electrons for a total of ten electrons in the valence shell.
In some expanded octet molecules, such as IF5 and XeF4, some of the electrons in the outer shell of the central atom are lone pairs:
Example \(2\)
Identify each violation to the octet rule by drawing a Lewis electron dot diagram.
1. \(\ce{ClO}\)
2. \(\ce{SF6}\)
Solution
1. 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:
1. In \(\ce{SF6}\), the central \(\ce{S}\) 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 \(2\): Xenon Difluoride
Identify the violation to the octet rule in \(\ce{XeF2}\) by drawing a Lewis electron dot diagram.
Answer
The Xe atom has an expanded valence shell with more than eight electrons around it.
Concept Review Exercises
1. How is a covalent bond formed between two atoms?
2. How does covalent bonding allow atoms in group 6A to satisfy the octet rule?
Answers
1. Covalent bonds are formed by two atoms sharing electrons.
2. The atoms in group 6A make two covalent bonds. | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/04%3A_Molecular_Compounds/4.02%3A_Covalent_Bonds_and_the_Periodic_Table.txt |
Learning Objectives
• Identify when a multiple bond (double or triple) is needed to complete an octet.
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.
In addition to nitrogen and oxygen, carbon will also commonly form multiple bonds to complete valence octets. Additional examples involving these three atoms are shown below.
A double bond is formed between the carbon and oxygen atoms in CH2O (formaldehyde) and between the two carbon atoms in C2H4 (ethylene):
A triple bond is formed between carbon and oxygen in carbon monoxide (CO) and between carbon and nitrogen in the cyanide ion (CN):
Example \(1\): Writing Lewis Structures
NASA’s Cassini-Huygens mission detected a large cloud of toxic hydrogen cyanide (HCN) on Titan, one of Saturn’s moons. Titan also contains ethane (H3CCH3), acetylene (HCCH), and ammonia (NH3). What are the Lewis structures of these molecules?
Solution
Calculate the number of valence electrons.
• HCN: (1 × 1) + (4 × 1) + (5 × 1) = 10
• H3CCH3: (1 × 3) + (2 × 4) + (1 × 3) = 14
• HCCH: (1 × 1) + (2 × 4) + (1 × 1) = 10
• NH3: (5 × 1) + (3 × 1) = 8
Draw a skeleton and connect the atoms with single bonds. Remember that H is never a central atom:
Where needed, distribute electrons to the terminal atoms:
• HCN: six electrons placed on N
• H3CCH3: no electrons remain
• HCCH: no terminal atoms capable of accepting electrons
• NH3: no terminal atoms capable of accepting electrons
Where needed, place remaining electrons on the central atom:
• HCN: no electrons remain
• H3CCH3: no electrons remain
• HCCH: four electrons placed on carbon
• NH3: two electrons placed on nitrogen
Where needed, rearrange electrons to form multiple bonds in order to obtain an octet on each atom:
• HCN: form two more C–N bonds
• H3CCH3: all atoms have the correct number of electrons
• HCCH: form a triple bond between the two carbon atoms
• NH3: all atoms have the correct number of electrons
Exercise \(1\)
Both carbon monoxide, CO, and carbon dioxide, CO2, are products of the combustion of fossil fuels. Both of these gases also cause problems: CO is toxic and CO2 has been implicated in global climate change. What are the Lewis structures of these two molecules?
Answer
4.04: Coordinate Covalent Bonds
Learning Objectives
• Describe the difference between a covalent bond and a coordinate covalent bond.
• Identify molecules that form coordinate covalent bonds.
Remember when you were younger, and were told to share your favorite toy with your brother, sister, or friend? You probably didn't want to share, but did anyway. It likely turned out that you had more fun playing with the toy together than if you had kept it to yourself. Atoms often share electrons with other atoms that have nothing to contribute to the situation forming a coordinate covalent bond; the end result is a new structure.
Coordinate Covalent Bonds
Each of the covalent bonds that we have looked at so far has involved each of the atoms that are bonding contributing one of the electrons to the shared pair. There is an alternate type of covalent bond in which one of the atoms provides both of the electrons in a shared pair. Carbon monoxide, \(\ce{CO}\), is a toxic gas that is released as a byproduct during the burning of fossil fuels. The bonding between the \(\ce{C}\) atom and the \(\ce{O}\) atom can be thought of in the following procession:
At this point, a double bond has formed between the two atoms, with each atom providing one of the electrons to each bond. The oxygen atom now has a stable octet of electrons, but the carbon atom only has six electrons and is unstable. This situation is resolved if the oxygen atom contributes one of its lone pairs in order to make a third bond with the carbon atom.
The carbon monoxide molecule is correctly represented by a triple covalent bond between the carbon and oxygen atoms. One of the bonds formed is a coordinate covalent bond, a covalent bond in which one of the atoms contributes both of the electrons in the shared pair. Once formed, a coordinate covalent bond is the same as any other covalent bond. It is not as if the two conventional bonds in the \(\ce{CO}\) molecule are stronger or different in any other way than the coordinate covalent bond.
Electron-deficient molecules, like \(\ce{BF3}\), are very reactive and will often combine with other molecules forming coordinate covalent bonds. The central boron atom in \(\ce{BF3}\) does not have eight electrons, and is therefore very reactive. It can readily combine with a molecule containing a central atom with a lone pair of electrons. For example, \(\ce{NH3}\) reacts with \(\ce{BF3}\) because the lone pair on nitrogen can be shared with the boron atom:
Summary
• Coordinate covalent bonds can form when one atom provides a lone pair of electrons to the bond.
• Coordinate covalent bonds are as strong as other covalent bonds.
Review
1. Where does the third covalent bond in the CO molecule come from?
2. Why is the incorrect structure for CO above wrong?
3. Are coordinate covalent bonds stronger or weaker than regular covalent bonds?
4.05: Characteristics of Molecular Compounds
Learning Objectives
• Compare the properties of ionic and molecular compounds.
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 (discussed further in a later chapter) between molecules is far less than the energy required to break the ionic bonds in a crystalline ionic compound. Since molecular compounds are composed of neutral molecules, their electrical conductivity is generally quite poor, whether in the solid or liquid state. Ionic compounds do not conduct electricity in the solid state because of their rigid structure, but conduct well when either molten or dissolved into a solution. The water solubility of molecular compounds is variable and depends primarily on the type of intermolecular forces involved. Substances that exhibit hydrogen bonding or dipole-dipole forces are generally water soluble, whereas those that exhibit only London dispersion forces are generally insoluble. Most, but not all, ionic compounds are quite soluble in water. 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
In summary, covalent compounds are softer, have lower boiling and melting points, are more flammable, are less soluble in water and do not conduct electricity compared to ionic compounds. The individual melting and boiling points, solubility and other physical properties of molecular compounds depend on molecular polarity. | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/04%3A_Molecular_Compounds/4.03%3A_Multiple_Covalent_Bonds.txt |
Learning Objectives
• Understand the different ways to represent molecules.
There are many "universal languages" in the world. Musicians of every culture recognize music embodied in a series of notes on a staff. This passage from a Bach cello suite could be played by any trained musician from any country, because there is an agreement as to what the symbols on the page mean. In the same way, molecules are represented using symbols and a language that all chemists agree upon.
Molecular Formulas
A molecular formula is a chemical formula of a molecular compound that shows the kinds and numbers of atoms present in a molecule of the compound. Ammonia is a compound of nitrogen and hydrogen as shown below:
Note from the example that there are some standard rules to follow in writing molecular formulas. The number of atoms of each kind is indicated by a subscript following the atom. If there is only one atom, no number is written. If there is more than one atom of a specific kind, the number is written as a subscript following the atom. We would not write \(\ce{N_3H}\) for ammonia, because that would mean that there are three nitrogen atoms and one hydrogen atom in the molecule, which is incorrect.
Although it is useful for describing a molecule, the molecular formula does not tell us anything about the shape of the molecule, where the different atoms are, or what kinds of bonds are formed. Structural formulas are much more useful to communicate more detailed information about a molecule because they show which atoms are bonded to one another and, in some cases, the approximate arrangement of the atoms in space. Knowing the structural formula of a compound enables chemists to create a three-dimensional model, which provides information about how that compound will behave physically and chemically.
Figure \(3\) shows some of the different ways to portray the structure of a slightly more complex molecule: methanol. These representations differ greatly in their information content. For example, the molecular formula for methanol (Figure \(\PageIndex{3a}\)) gives only the number of each kind of atom; writing methanol as CH4O tells nothing about its structure. In contrast, the structural formula (Figure \(\PageIndex{3b}\)) indicates how the atoms are connected, but it makes methanol look as if it is planar (which it is not). Both the ball-and-stick model (part (c) in Figure \(3\)) and the perspective drawing (Figure \(\PageIndex{3d}\)) show the three-dimensional structure of the molecule. The latter (also called a wedge-and-dash representation) is the easiest way to sketch the structure of a molecule in three dimensions. It shows which atoms are above and below the plane of the paper by using wedges and dashes, respectively; the central atom is always assumed to be in the plane of the paper. The space-filling model (part (e) in Figure \(3\)) illustrates the approximate relative sizes of the atoms in the molecule, but it does not show the bonds between the atoms. In addition, in a space-filling model, atoms at the “front” of the molecule may obscure atoms at the “back.”
Although a structural formula, a ball-and-stick model, a perspective drawing, and a space-filling model provide a significant amount of information about the structure of a molecule, each requires time and effort. Consequently, chemists often use a condensed structural formula (part (f) in Figure \(3\)), which omits the lines representing bonds between atoms and simply lists the atoms bonded to a given atom next to it. Multiple groups attached to the same atom are shown in parentheses, followed by a subscript that indicates the number of such groups. For example, the condensed structural formula for methanol is CH3OH, which indicates that the molecule contains a CH3 unit that looks like a fragment of methane (CH4). Methanol can therefore be viewed either as a methane molecule in which one hydrogen atom has been replaced by an –OH group or as a water molecule in which one hydrogen atom has been replaced by a –CH3 fragment. Because of their ease of use and information content, we use condensed structural formulas for molecules throughout this text. Ball-and-stick models are used when needed to illustrate the three-dimensional structure of molecules, and space-filling models are used only when it is necessary to visualize the relative sizes of atoms or molecules to understand an important point.
Example \(1\): Molecular Formulas
Write the molecular formula for each compound. The condensed structural formula is given.
1. Sulfur monochloride (also called disulfur dichloride) is a vile-smelling, corrosive yellow liquid used in the production of synthetic rubber. Its condensed structural formula is ClSSCl.
2. Ethylene glycol is the major ingredient in antifreeze. Its condensed structural formula is HOCH2CH2OH.
3. Trimethylamine is one of the substances responsible for the smell of spoiled fish. Its condensed structural formula is (CH3)3N.
Given: condensed structural formula
Asked for: molecular formula
Strategy:
1. Identify every element in the condensed structural formula and then determine whether the compound is organic or inorganic.
2. As appropriate, use either organic or inorganic convention to list the elements. Then add appropriate subscripts to indicate the number of atoms of each element present in the molecular formula.
Solution:
The molecular formula lists the elements in the molecule and the number of atoms of each.
1. A Each molecule of sulfur monochloride has two sulfur atoms and two chlorine atoms. Because it does not contain mostly carbon and hydrogen, it is an inorganic compound. B Sulfur lies to the left of chlorine in the periodic table, so it is written first in the formula. Adding subscripts gives the molecular formula S2Cl2.
2. A Counting the atoms in ethylene glycol, we get six hydrogen atoms, two carbon atoms, and two oxygen atoms per molecule. The compound consists mostly of carbon and hydrogen atoms, so it is organic. B As with all organic compounds, C and H are written first in the molecular formula. Adding appropriate subscripts gives the molecular formula C2H6O2.
3. A The condensed structural formula shows that trimethylamine contains three CH3 units, so we have one nitrogen atom, three carbon atoms, and nine hydrogen atoms per molecule. Because trimethylamine contains mostly carbon and hydrogen, it is an organic compound. B According to the convention for organic compounds, C and H are written first, giving the molecular formula C3H9N.
Exercise \(1\): Molecular Formulas
Write the molecular formula for each molecule.
1. Chloroform, which was one of the first anesthetics and was used in many cough syrups until recently, contains one carbon atom, one hydrogen atom, and three chlorine atoms. Its condensed structural formula is \(\ce{CHCl3}\).
2. Hydrazine is used as a propellant in the attitude jets of the space shuttle. Its condensed structural formula is \(\ce{H2NNH2}\).
3. Putrescine is a pungent-smelling compound first isolated from extracts of rotting meat. Its condensed structural formula is H2NCH2CH2CH2CH2NH2. This is often written as \(\ce{H2N(CH2)4NH2}\) to indicate that there are four CH2 fragments linked together.
Answer a
CHCl3
Answer b
N2H4
Answer c | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/04%3A_Molecular_Compounds/4.06%3A_Molecular_Formulas_and_Lewis_Structures.txt |
Learning Objectives
• Draw Lewis structures for molecules.
For very simple molecules and molecular ions, we can write the Lewis structures by merely pairing up the unpaired electrons on the constituent atoms. See these examples:
For more complicated molecules, it is helpful to follow the step-by-step procedure outlined here:
1. 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.
2. 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.
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.
Now let’s apply this procedure to some particular compounds.
The central atom is usually the least electronegative element in the molecule or ion; hydrogen and the halogens are usually terminal.
The \(H_2O\) Molecule
1. Because H atoms are almost always terminal, the arrangement within the molecule must be HOH.
2. 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.
3. Placing one bonding pair of electrons between the O atom and each H atom gives H:O:H, with 4 electrons left over.
4. Each H atom has a full valence shell of 2 electrons.
5. Adding the remaining 4 electrons to the oxygen (as two lone pairs) gives the following structure:
Because this structure gives oxygen an octet and each hydrogen two electrons, we do not need to use step 6.
The \(OCl^−\) Ion
1. With only two atoms in the molecule, there is no central atom.
2. Oxygen (group 16) has 6 valence electrons, and chlorine (group 17) has 7 valence electrons; we must add one more for the negative charge on the ion, giving a total of 14 valence electrons.
3. Placing a bonding pair of electrons between O and Cl gives O:Cl, with 12 electrons left over.
4. If we place six electrons (as three lone pairs) on each atom, we obtain the following structure:
Each atom now has an octet of electrons, so steps 5 and 6 are not needed. The Lewis electron structure is drawn within brackets as is customary for a molecular ion, with the overall charge indicated outside the brackets, and the bonding pair of electrons is indicated by a solid line. OCl is the hypochlorite ion, the active ingredient in chlorine laundry bleach and swimming pool disinfectant.
The \(CH_2O\) Molecule
1. Because carbon is less electronegative than oxygen and hydrogen is normally terminal, C must be the central atom. One possible arrangement is as follows:
2. 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.
3. 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. 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. There are no electrons left to place on the central atom.
6. 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:
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.
Example \(1\)
Write the Lewis electron structure for each species.
1. NCl3
2. S22
3. NOCl
Given: chemical species
Asked for: Lewis electron structures
Strategy:
Use the six-step procedure to write the Lewis electron structure for each species.
Solution:
1. Nitrogen is less electronegative than chlorine, and halogen atoms are usually terminal, so nitrogen is the central atom. The nitrogen atom (group 15) has 5 valence electrons and each chlorine atom (group 17) has 7 valence electrons, for a total of 26 valence electrons. Using 2 electrons for each N–Cl bond and adding three lone pairs to each Cl account for (3 × 2) + (3 × 2 × 3) = 24 electrons. Rule 5 leads us to place the remaining 2 electrons on the central N:
Nitrogen trichloride is an unstable oily liquid once used to bleach flour; this use is now prohibited in the United States.
2. In a diatomic molecule or ion, we do not need to worry about a central atom. Each sulfur atom (group 16) contains 6 valence electrons, and we need to add 2 electrons for the −2 charge, giving a total of 14 valence electrons. Using 2 electrons for the S–S bond, we arrange the remaining 12 electrons as three lone pairs on each sulfur, giving each S atom an octet of electrons:
3. Because nitrogen is less electronegative than oxygen or chlorine, it is the central atom. The N atom (group 15) has 5 valence electrons, the O atom (group 16) has 6 valence electrons, and the Cl atom (group 17) has 7 valence electrons, giving a total of 18 valence electrons. Placing one bonding pair of electrons between each pair of bonded atoms uses 4 electrons and gives the following:
Adding three lone pairs each to oxygen and to chlorine uses 12 more electrons, leaving 2 electrons to place as a lone pair on nitrogen:
Because this Lewis structure has only 6 electrons around the central nitrogen, a lone pair of electrons on a terminal atom must be used to form a bonding pair. We could use a lone pair on either O or Cl. Because we have seen many structures in which O forms a double bond but none with a double bond to Cl, it is reasonable to select a lone pair from O to give the following:
All atoms now have octet configurations. This is the Lewis electron structure of nitrosyl chloride, a highly corrosive, reddish-orange gas.
Exercise \(1\)
Write Lewis electron structures for CO2 and SCl2, a vile-smelling, unstable red liquid that is used in the manufacture of rubber.
1.
2. | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/04%3A_Molecular_Compounds/4.07%3A_Drawing_Lewis_Structures.txt |
Learning Objectives
• Predict the structures of small molecules using valence shell electron pair repulsion (VSEPR) theory.
The Lewis electron-pair approach can be used to predict the number and types of bonds between the atoms in a substance, and it indicates which atoms have lone pairs of electrons. This approach gives no information about the actual arrangement of atoms in space, however. We continue our discussion of structure and bonding by introducing the valence-shell electron-pair repulsion (VSEPR) model (pronounced “vesper”), which can be used to predict the shapes of many molecules and polyatomic ions. Keep in mind, however, that the VSEPR model, like any model, is a limited representation of reality; the model provides no information about bond lengths or the presence of multiple bonds.
The VSEPR Model
The VSEPR model can predict the structure of nearly any molecule or polyatomic ion in which the central atom is a nonmetal, as well as the structures of many molecules and polyatomic ions with a central metal atom. The premise of the VSEPR theory is that electron pairs located in bonds and lone pairs repel each other and will therefore adopt the geometry that places electron pairs as far apart from each other as possible. This theory is very simplistic and does not account for the subtleties of orbital interactions that influence molecular shapes; however, the simple VSEPR counting procedure accurately predicts the three-dimensional structures of a large number of compounds, which cannot be predicted using the Lewis electron-pair approach.
We can use the VSEPR model to predict the geometry of most polyatomic molecules and ions by focusing only on the number of electron pairs around the central atom, ignoring all other valence electrons present. According to this model, valence electrons in the Lewis structure form electron groups (regions of electron density), which may consist of a single bond, a double bond, a triple bond, a lone pair of electrons, or even a single unpaired electron, which in the VSEPR model is counted as a lone pair. Because electrons repel each other electrostatically, the most stable arrangement of electron groups (i.e., the one with the lowest energy) is the one that minimizes repulsions. Groups are positioned around the central atom in a way that produces the molecular structure with the lowest energy, as illustrated in Figure \(1\).
It is important to note that electron group geometry around a central atom is not the same thing as its molecular structure. Electron group geometries describe all regions where electrons are located, bonds as well as lone pairs. Molecular structure describes the location of the atoms alone, not including the lone pair electrons.
We differentiate between these two situations by naming the geometry that includes all electron pairs the electron group geometry. The structure that includes only the placement of the atoms in the molecule is called the molecular structure (or molecular shape). The electron group geometries will be the same as the molecular structures when there are no lone electron pairs around the central atom, but they will be different when there are lone pairs present on the central atom.
Predicting Electron Group Geometry and Molecular Structure
The following procedure uses VSEPR theory to determine electron group geometry and molecular structures (molecular shape):
1. Draw the Lewis structure of the molecule or polyatomic ion.
2. Count the number of electron groups or regions of electron density (lone pairs and bonds) around the central atom. A single, double, or triple bond counts as one electron group.
3. Determine the electron group geometry by placing the groups as far apart as possible.
4. Determine the molecular structure (looking at the bonded groups only).
Table \(1\) summarizes the shapes of molecules based on the number of electron groups and surrounding atoms.
Table \(1\): Summary of Electron Group Geometries and Molecular Structures
Number of Electron Groups on Central Atom Number of Bonding Groups Number of Lone Pairs Electron Group Geometry Molecular Structure
2 2 0 linear linear
3 3 0 trigonal planar trigonal planar
3 2 1 trigonal planar bent 120°
4 4 0 tetrahedral tetrahedral
4 3 1 tetrahedral trigonal pyramidal
4 2 2 tetrahedral bent 109°
Two Electron Groups
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, which is 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 structure is linear. Examples include BeH2 and CO2:
Three Electron Groups
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 molecular structure is called bent 120° or angular.
Four Electron Groups
A molecule with four electron groups about the central atom orients the four groups in the direction of a tetrahedron with bond angles of approximately 109.5°. If there are four atoms attached to these electron groups, then the molecular structure 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 109° 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.
Shapes of Molecules with Double or Triple Bonds
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.
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.
For \(\ce{PCl3}\), 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 \(\ce{NOF}\) 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 \(\ce{CH2Cl2}\)?
Answer
Tetrahedral
Exercise \(2\)
Ethylene (\(\ce{C2H4}\)) 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.
Molecules With Multiple Central Atoms
The VSEPR model can be used to predict the structure of somewhat more complex molecules with more than one central atom by using VSEPR as described above for each central atom individually. We will demonstrate with methyl isocyanate (CH3–N=C=O), a volatile and highly toxic molecule that is used to produce the pesticide Sevin.
Start by looking at the electron groups around the first carbon atom at the left, which is connected to three H atoms and one N atom by single bonds. There are four groups or electrons or four bonds around the carbon. We can therefore predict the CH3–N portion of the molecule to be roughly tetrahedral, similar to methane:
The nitrogen atom is connected to one carbon by a single bond and to the other carbon by a double bond, producing a total of three bonds, C–N=C. For nitrogen to have an octet of electrons, it must also have a lone pair:
Because multiple bonds are not shown in the VSEPR model, the nitrogen is effectively surrounded by three electron groups. Thus according to the VSEPR model, the C–N=C fragment should be bent with an angle ~120°.
The carbon in the –N=C=O fragment is doubly bonded to both nitrogen and oxygen, which in the VSEPR model gives carbon a total of two electron pairs. The N=C=O angle should therefore be 180°, or linear. The three fragments combine to give the following structure:
Certain patterns are seen in the structures of moderately complex molecules. For example, carbon atoms with four bonds (such as the carbon on the left in methyl isocyanate) are generally tetrahedral. Similarly, the carbon atom on the right has two double bonds that are similar to those in CO2, so its geometry, like that of CO2, is linear. Recognizing similarities to simpler molecules will help you predict the molecular geometries of more complex molecules.
Example \(3\)
Use the VSEPR model to predict the molecular geometry of propyne (H3C–C≡CH), a gas with some anesthetic properties.
Given: chemical compound
Asked for: molecular geometry
Strategy:
Count the number of electron groups around each carbon, recognizing that in the VSEPR model, a multiple bond counts as a single group. Use Figure \(3\) to determine the molecular geometry around each carbon atom and then deduce the structure of the molecule as a whole.
Solution:
Because the carbon atom on the left is bonded to four other atoms, we know that it is approximately tetrahedral. The next two carbon atoms share a triple bond, and each has an additional single bond. Because a multiple bond is counted as a single bond in the VSEPR model, each carbon atom behaves as if it had two electron groups. This means that both of these carbons are linear, with C–C≡C and C≡C–H angles of 180°.
Exercise \(3\)
Predict the geometry of allene (H2C=C=CH2), a compound with narcotic properties that is used to make more complex organic molecules.
Answer
The terminal carbon atoms are trigonal planar, the central carbon is linear, and the C–C–C angle is 180°. | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/04%3A_Molecular_Compounds/4.08%3A_The_Shapes_of_Molecules.txt |
Learning Objectives
• Describe electronegativity and polarity.
• Use electronegativity values to predict bond polarity.
Our discussions of bonding thus far have focused on two types, ionic and covalent. In ionic bonds, like \(NaCl\), electrons are transferred; the 3s electron is stripped from the Na atom and is incorporated into the electronic structure of the Cl atom, and the compound is most accurately described as consisting of individual \(Na^+\) and \(Cl^-\) ions. In covalent bonding, unpaired electrons from individual atoms are shared in order to fill the valence shell of each atom. When a covalent bond is formed between the same type of atoms, such as \(Cl_2\), the electrons are shared equally between the two. However, when a covalent bond is formed between different types of atoms, the electrons are not necessarily shared equally. In these compounds their bond character falls between the two extremes: transferred and shared equally.
Bond Polarity
As demonstrated below, bond polarity is a useful concept for describing the sharing of electrons between atoms, within a covalent bond:
• A nonpolar covalent bond (Figure \(\PageIndex{1a}\))is one in which the electrons are shared equally between two atoms.
• A polar covalent bond (Figure \(\PageIndex{1b}\)) is one in which one atom has a greater attraction for the electrons than the other atom.
• If the relative attraction of an atom for electrons is great enough, then the bond is an ionic bond (Figure \(\PageIndex{1c}\)).
Electron density in a polar bond is distributed unevenly and is greater around the atom that attracts the electrons more than the other. For example, the electrons in the H–Cl bond of a hydrogen chloride molecule spend more time near the chlorine atom than near the hydrogen atom. Note that the shaded area around Cl in Figure \(\PageIndex{1b}\) is much larger than it is around H. This imbalance in electron density results in a buildup of partial negative charge (designated as δ−) on one side of the bond (Cl) and a partial positive charge (designated δ+) on the other side of the bond (H).
Any covalent bond between atoms of different elements is a polar bond, but the degree of polarity varies widely. Some bonds between different elements are only minimally polar, while others are strongly polar. Ionic bonds can be considered the ultimate in polarity, with electrons being transferred rather than shared. To judge the relative polarity of a covalent bond, chemists use electronegativity, which is a relative measure of how strongly an atom attracts electrons when it forms a covalent bond.
Electronegativity
Because the tendency of an element to gain or lose electrons is so important in determining its chemistry, various methods have been developed to quantitatively describe this tendency. The most important method uses a measurement called electronegativity, defined as the relative ability of an atom to attract electrons to itself in a chemical compound. Elements with high electronegativities tend to acquire electrons in chemical reactions and are found in the upper right corner of the periodic table. Elements with low electronegativities tend to lose electrons in chemical reactions and are found in the lower left corner of the periodic table.
Unlike ionization energy or electron affinity, the electronegativity of an atom is not a simple, fixed property that can be directly measured in a single experiment. In fact, an atom’s electronegativity should depend to some extent on its chemical environment because the properties of an atom are influenced by its neighbors in a chemical compound. Nevertheless, when different methods for measuring the electronegativity of an atom are compared, they all tend to assign similar relative values to a given element. For example, all scales predict that fluorine has the highest electronegativity and cesium the lowest of the stable elements, which suggests that all the methods are measuring the same fundamental property.
Electronegativity is a function of:
1. the atom's ionization energy (how strongly the atom holds on to its own electrons) and
2. the atom's electron affinity (how strongly the atom attracts other electrons).
Both of these are properties of the isolated atom. An element will be highly electronegative if it has a large (negative) electron affinity and a high ionization energy (always positive for neutral atoms). Thus, it will attract electrons from other atoms and resist having its own electrons attracted away.
Electronegativity is defined as the ability of an atom in a particular molecule to attract electrons to itself. The greater the value, the greater the attractiveness for electrons.
The Pauling Electronegativity Scale
The original electronegativity scale, developed in the 1930s by Linus Pauling (1901– 1994) was based on measurements of the strengths of covalent bonds between different elements. Pauling arbitrarily set the electronegativity of fluorine at 4.0 (although today it has been refined to 3.98), thereby creating a scale in which all elements have values between 0 and 4.0.
Periodic variations (trends) in Pauling’s electronegativity values are illustrated in Figures \(2\) and \(3\). If we ignore the inert gases and elements for which no stable isotopes are known, we see that fluorine is the most electronegative element and cesium is the least electronegative nonradioactive element. Because electronegativities generally increase diagonally from the lower left to the upper right of the periodic table, elements lying on diagonal lines running from upper left to lower right tend to have comparable values (e.g., O and Cl, and N, S, and Br).
The polarity of a covalent bond can be judged by determining the difference in the electronegativities of the two atoms making the bond. The greater the difference in electronegativities, the greater the imbalance of electron sharing in the bond. Although there are no hard and fast rules, the general rule, (see Figure \(5\)), is if the difference in electronegativities is less than about 0.4, the bond is considered nonpolar; if the difference is greater than 0.4, the bond is considered polar. If the difference in electronegativities is large enough (generally greater than about 1.8), the resulting compound is considered ionic rather than covalent. An electronegativity difference of zero, of course, indicates a nonpolar covalent bond.
Example \(1\)
Describe the electronegativity difference between each pair of atoms and the resulting polarity (or bond type).
1. C and H
2. H and H
3. Na and Cl
4. O and H
Solution
1. Carbon has an electronegativity of 2.5, while the value for hydrogen is 2.1. The difference is 0.4, which is rather small. The C–H bond is therefore considered nonpolar.
2. Both hydrogen atoms have the same electronegativity value—2.1. The difference is zero, so the bond is nonpolar.
3. Sodium’s electronegativity is 0.9, while chlorine’s is 3.0. The difference is 2.1, which is rather high, and so sodium and chlorine form an ionic compound.
4. With 2.1 for hydrogen and 3.5 for oxygen, the electronegativity difference is 1.4. We would expect a very polar bond. The sharing of electrons between O and H is unequal with the electrons more strongly drawn towards O.
Exercise \(1\)
Describe the electronegativity (EN) difference between each pair of atoms and the resulting polarity (or bond type).
1. C and O
2. K and Br
3. N and N
4. Cs and F
Answer a:
The EN difference is 1.0 , hence polar. The sharing of electrons between C and O is unequal with the electrons more strongly drawn towards O.
Answer b:
The EN difference is greater than 1.8, hence ionic.
Answer c:
Identical atoms have zero EN difference, hence nonpolar.
Answer d:
The EN difference is greater than 1.8, hence ionic.
Looking Closer: Linus Pauling
Arguably the most influential chemist of the 20th century, Linus Pauling (1901–94) is the only person to have won two individual (that is, unshared) Nobel Prizes. In the 1930s, Pauling used new mathematical theories to enunciate some fundamental principles of the chemical bond. His 1939 book The Nature of the Chemical Bond is one of the most significant books ever published in chemistry.
By 1935, Pauling’s interest turned to biological molecules, and he was awarded the 1954 Nobel Prize in Chemistry for his work on protein structure. (He was very close to discovering the double helix structure of DNA when James Watson and James Crick announced their own discovery of its structure in 1953.) He was later awarded the 1962 Nobel Peace Prize for his efforts to ban the testing of nuclear weapons.
Linus Pauling was one of the most influential chemists of the 20th century.
In his later years, Pauling became convinced that large doses of vitamin C would prevent disease, including the common cold. Most clinical research failed to show a connection, but Pauling continued to take large doses daily. He died in 1994, having spent a lifetime establishing a scientific legacy that few will ever equal. | textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/04%3A_Molecular_Compounds/4.09%3A_Polar_Covalent_Bonds_and_Electronegativity.txt |
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