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15.1: Radioactivity
Q15.1.1
Define radioactivity.
S15.1.1
Radioactivity is the spontaneous emission of particles and electromagnetic radiation from nuclei of unstable atoms.
Q15.1.2
Give an example of a radioactive element. How do you know if it is radioactive?
Q15.1.3
How many protons and neutrons are in each isotope?
1. $_{5}^{11}\textrm{B}$
2. $_{13}^{27}\textrm{Al}$
3. $_{}^{56}\textrm{Fe}$
4. $_{}^{224}\textrm{Rn}$
S15.1.3
1. 5 protons; 6 neutrons
2. 13 protons; 14 neutrons
3. 26 protons; 30 neutrons
4. 86 protons; 138 neutrons
Q15.1.4
How many protons and neutrons are in each isotope?
1. $_{1}^{2}\textrm{H}$
2. $_{48}^{112}\textrm{Cd}$
3. $_{}^{252}\textrm{Es}$
4. $_{}^{40}\textrm{K}$
Q15.1.5
Describe an alpha particle. What nucleus is it equivalent to?
S15.1.5
An alpha particle is a collection of two protons and two neutrons and is equivalent to a helium nucleus.
Q15.1.6
Describe a beta particle. What subatomic particle is it equivalent to?
Q15.1.7
What are gamma rays?
S15.1.7
Gamma rays are high-energy electromagnetic radiation given off in radioactive decay.
Q15.1.8
Why is it inappropriate to refer to gamma rays as “gamma particles”?
Q15.1.9
Plutonium has an atomic number of 94. Write the nuclear equation for the alpha particle emission of plutonium-244. What is the daughter isotope?
S15.1.9
$_{94}^{244}\textrm{Pu}\rightarrow \: _{92}^{240}\textrm{U}+\: _{2}^{4}\textrm{He}$
daughter isotope: $_{}^{240}\textrm{U}$
Q15.1.10
Francium has an atomic number of 87. Write the nuclear equation for the alpha particle emission of francium-212. What is the daughter isotope?
S15.1.11
$_{50}^{121}\textrm{Sn}\rightarrow \: _{51}^{121}\textrm{Sb}+\: _{-1}^{0}\textrm{e}$
daughter isotope: $_{}^{121}\textrm{Sb}$
Q15.1.11
Tin has an atomic number of 50. Write the nuclear equation for the beta particle emission of tin-121. What is the daughter isotope?
Q15.1.12
Technetium has an atomic number of 43. Write the nuclear equation for the beta particle emission of technetium-99. What is the daughter isotope?
Q15.1.13
Energies of gamma rays are typically expressed in units of megaelectron volts (MeV), where 1 MeV = 1.602 × 10−13 J. Using the data provided in the text, calculate the energy in megaelectron volts of the gamma ray emitted when radon-222 decays.
0.51 MeV
Q15.1.14
The gamma ray emitted when oxygen-19 gives off a beta particle is 0.197 MeV. What is its energy in joules? (See Exercise 13 for the definition of a megaelectron volt.)
Q15.1.15
Which penetrates matter more deeply—alpha particles or beta particles? Suggest ways to protect yourself against both particles.
S15.1.15
Beta particles penetrate more. A thick wall of inert matter is sufficient to block both particles.
Q15.1.16
Which penetrates matter more deeply—alpha particles or gamma rays? Suggest ways to protect yourself against both emissions.
Q15.1.17
Define nuclear fission.
S15.1.17
Nuclear fission is the breaking down of large nuclei into smaller nuclei, usually with the release of excess neutrons.
Q15.1.18
What general characteristic is typically necessary for a nucleus to undergo spontaneous fission?
Q15.2.1
Do all isotopes have a half-life? Explain your answer.
S15.2.1
Only radioactive isotopes have a half-life.
Q15.2.2
Which is more radioactive—an isotope with a long half-life or an isotope with a short half-life?
Q15.2.3
How long does it take for 1.00 g of palladium-103 to decay to 0.125 g if its half-life is 17.0 d?
51.0 d
Q15.2.4
How long does it take for 2.00 g of niobium-94 to decay to 0.0625 g if its half-life is 20,000 y?
Q15.2.5
It took 75 y for 10.0 g of a radioactive isotope to decay to 1.25 g. What is the half-life of this isotope?
25 y
Q15.2.6
It took 49.2 s for 3.000 g of a radioactive isotope to decay to 0.1875 g. What is the half-life of this isotope?
Q15.2.7
The half-live of americium-241 is 432 y. If 0.0002 g of americium-241 is present in a smoke detector at the date of manufacture, what mass of americium-241 is present after 100.0 y? After 1,000.0 y?
S15.2.7
0.000170 g; 0.0000402 g
Q15.2.8
If the half-life of tritium (hydrogen-3) is 12.3 y, how much of a 0.00444 g sample of tritium is present after 5.0 y? After 250.0 y?
Q15.2.9
Explain why the amount left after 1,000.0 y in Exercise 7 is not one-tenth of the amount present after 100.0 y, despite the fact that the amount of time elapsed is 10 times as long.
S15.2.9
Radioactive decay is an exponential process, not a linear process.
Q15.2.10
Explain why the amount left after 250.0 y in Exercise 8 is not one-fiftieth of the amount present after 5.0 y, despite the fact that the amount of time elapsed is 50 times as long.
Q15.2.11
An artifact containing carbon-14 contains 8.4 × 10−9 g of carbon-14 in it. If the age of the artifact is 10,670 y, how much carbon-14 did it have originally? The half-life of carbon-14 is 5,730 y.
3.1 × 10−8 g
Q15.2.12
Carbon-11 is a radioactive isotope used in positron emission tomography (PET) scans for medical diagnosis. Positron emission is another, though rare, type of radioactivity. The half-life of carbon-11 is 20.3 min. If 4.23 × 10−6 g of carbon-11 is left in the body after 4.00 h, what mass of carbon-11 was present initially?
Define rad.
S15.3.1
a unit of radioactive exposure equal to 0.01 J of energy per gram of tissue
Define rem.
Q15.3.3
How does a becquerel differ from a curie?
S15.3.3
A becquerel is 1 decay/s, whereas a curie is 3.7 × 1010 decays/s.
Define curie.
Q15.3.5
A sample of radon gas has an activity of 140.0 mCi. If the half-life of radon is 1,500 y, how long before the activity of the sample is 8.75 mCi?
6.0 × 103 y
Q15.3.6
A sample of curium has an activity of 1,600 Bq. If the half-life of curium is 24.0 s, how long before its activity is 25.0 Bq?
Q15.3.7
If a radioactive sample has an activity of 65 µCi, how many disintegrations per second are occurring?
S15.3.7
2.41 × 106 disintegrations per second
Q15.3.8
If a radioactive sample has an activity of 7.55 × 105 Bq, how many disintegrations per second are occurring?
Q15.3.9
A sample of fluorine-20 has an activity of 2.44 mCi. If its half-life is 11.0 s, what is its activity after 50.0 s?
0.104 mCi
Q15.3.10
Strontium-90 has a half-life of 28.1 y. If 66.7 Bq of pure strontium-90 were allowed to decay for 15.0 y, what would the activity of the remaining strontium-90 be?
Q15.3.11
How long does it take 100.0 mCi of fluorine-20 to decay to 10.0 mCi if its half-life is 11.0 s?
Q15.3.12
Technetium-99 is used in medicine as a source of radiation. A typical dose is 25 mCi. How long does it take for the activity to reduce to 0.100 mCi? The half-life of 99Tc is 210,000 y.
Q15.3.13
Describe how a radiation exposure in rems is determined.
S15.3.13
by using a film badge, which is exposed by the radiation, or a Geiger counter
Q15.3.14
Which contributes more to the rems of exposure—alpha or beta particles? Why?
Q15.3.15
Use Table 15.4.2 to determine which sources of radiation exposure are inescapable and which can be avoided. What percentage of radiation is unavoidable?
S15.3.14
Radioactive atoms in the body, most terrestrial sources, cosmic sources, and nuclear energy sources are likely unavoidable, which is about 27% of the total exposure. If exposure to radon gas is added, the total unavoidable exposure increases to 82%.
Q15.3.16
Name two isotopes that contribute to the radioactivity in our bodies.
Q15.3.17
Explain how a film badge works to detect radiation.
S15.3.17
Film is exposed by the radiation. The more radiation film is subjected to, the more exposed it becomes.
Q15.3.18
Explain how a Geiger counter works to detect radiation.
15.4: Uses of Radioactive Isotopes
Q15.4.1
Define tracer and give an example of how tracers work.
S15.4.1
A tracer is a radioactive isotope that can be detected far from its original source to trace the path of certain chemicals. Hydrogen-3 can be used to trace the path of water underground.
Q15.4.2
Name two isotopes that have been used as tracers.
Q15.4.3
Explain how radioactive dating works.
S15.4.3
If the initial amount of a radioactive isotope is known, then by measuring the amount of the isotope remaining, a person can calculate how old that object is since it took up the isotope.
Q15.4.4
Name two isotopes that have been used in radioactive dating.
Q15.4.5
The current disintegration rate for carbon-14 is 14.0 Bq. A sample of burnt wood discovered in an archeological excavation is found to have a carbon-14 disintegration rate of 3.5 Bq. If the half-life of carbon-14 is 5,730 y, approximately how old is the wood sample?
11,500 y
Q15.4.6
A small asteroid crashes to Earth. After chemical analysis, it is found to contain 1 g of technetium-99 to every 3 g of ruthenium-99, its daughter isotope. If the half-life of technetium-99 is 210,000 y, approximately how old is the asteroid?
Q15.4.7
What is a positive aspect of the irradiation of food?
S15.4.7
increased shelf life (answers will vary)
Q15.4.8
What is a negative aspect of the irradiation of food?
Q15.4.9
Describe how iodine-131 is used to both diagnose and treat thyroid problems.
S15.4.9
The thyroid gland absorbs most of the iodine, allowing it to be imaged for diagnostic purposes or preferentially irradiated for treatment purposes.
Q15.4.10
List at least five organs that can be imaged using radioactive isotopes.
Q15.4.11
Which radioactive emissions can be used therapeutically?
gamma rays
Q15.4.12
Which isotope is used in therapeutics primarily for its gamma ray emissions?
15.5: Nuclear Energy
Q15.5.1
According to Einstein’s equation, the conversion of 1.00 g of matter into energy generates how much energy?
9.00 × 1013 J
Q15.5.2
How much matter needs to be converted to energy to supply 400 kJ of energy, the approximate energy of 1 mol of C–H bonds? What conclusion does this suggest about energy changes of chemical reactions?
Q15.5.3
In the spontaneous fission of lead-208, the following reaction occurs: $_{}^{208}\textrm{Pb}\rightarrow \: _{ }^{129}\textrm{I}\: +\: _{ }^{76}\textrm{Cu}\: +\: 3_{ }^{1}\textrm{n}$ For every mole of lead-208 that decays, 0.1002 g of mass is lost. How much energy is given off per mole of lead-208 reacted?
9.02 × 1012 J
Q15.5.4
In the spontaneous fission of radium-226, the following reaction occurs: $_{}^{226}\textrm{Ra}\rightarrow \: _{ }^{156}\textrm{Pm}\: +\: _{ }^{68}\textrm{Co}\: +\: 2_{ }^{1}\textrm{n}$ For every mole of radium-226 that decays, 0.1330 g of mass is lost. How much energy is given off per mole of radium-226 reacted?
Q15.5.5
Recalculate the amount of energy from Exercise 3 in terms of the number of grams of lead-208 reacted.
4.34 × 1010 J/g
Q15.5.6
Recalculate the amount of energy from Exercise 4 in terms of the number of grams of radium-226 reacted.
Q15.5.7
What is the energy change of this fission reaction? Masses in grams are provided. $\underset{241.0569}{_{ }^{241}\textrm{Pu}}\rightarrow \: \underset{139.9106}{_{ }^{140}\textrm{Ba}}\: +\: \underset{89.9077}{_{ }^{90}\textrm{Sr}}\: +\: \underset{11\times 1.0087}{11_{ }^{1}\textrm{n}}$
−1.28 × 1013 J
Q15.5.8
What is the energy change of this fission reaction? Masses in grams are provided. $\underset{247.0704}{_{ }^{247}\textrm{Cm}}\rightarrow \: \underset{106.9099}{_{ }^{107}\textrm{Ru}}\: +\: \underset{130.9085}{_{ }^{131}\textrm{Te}}\: +\: \underset{9\times 1.0087}{9_{ }^{1}\textrm{n}}$
Q15.5.9
The two rarer isotopes of hydrogen—deuterium and tritium—can also be fused to make helium by the following reaction: $^{2}H+^{3}H\rightarrow ^{4}\textrm{He}+^{1}n$ In the course of this reaction, 0.01888 g of mass is lost. How much energy is given off in the reaction of 1 mol of deuterium and tritium?
1.70 × 1012 J
Q15.5.10
A process called helium burning is thought to occur inside older stars, forming carbon: $3^{4}He\rightarrow ^{12}C$ If the reaction proceeds with 0.00781 g of mass lost on a molar basis, how much energy is given off?
Q15.5.11
Briefly describe how a nuclear reactor generates electricity.
S15.5.11
A nuclear reactor controls a nuclear reaction to produce energy in usable amounts. The energy produced generates steam, which is used to turn a turbine that generates electricity for general use.
Q15.5.12
Briefly describe the difference between how a nuclear reactor works and how a nuclear bomb works.
Q15.5.13
What is a chain reaction?
S15.5.13
a process that generates more reaction pathways for each previous reaction
Q15.5.14
Why must uranium be enriched to supply nuclear energy?
Additional Exercises
QExtra.1
Given that many elements are metals, suggest why it would be unsafe to have radioactive materials in contact with acids.
SExtra.1
Acids can dissolve many metals; a spilled acid can lead to contamination.
QExtra.2
Many alpha-emitting radioactive substances are relatively safe to handle, but inhaling radioactive dust can be very dangerous. Why?
QExtra.3
Uranium can be separated from its daughter isotope thorium by dissolving a sample in acid and adding sodium iodide, which precipitates thorium(III) iodide:
$Th^{3+}(aq)+3I^{-}(aq)\rightarrow ThI_{3}(s)$
If 0.567 g of Th3+ were dissolved in solution, how many milliliters of 0.500 M NaI(aq) would have to be added to precipitate all the thorium?
14.7 mL
QExtra.4
Thorium oxide can be dissolved in acidic solution: $ThO_{2}(s)+4H^{+}\rightarrow Th^{4+}(aq)+2H_{2}O(l)$
How many milliliters of 1.55 M HCl(aq) are needed to dissolve 10.65 g of ThO2?
QExtra.5
Radioactive strontium is dangerous because it can chemically replace calcium in the human body. The bones are particularly susceptible to radiation damage. Write the nuclear equation for the beta emission of strontium-90.
SExtra.5
$_{38}^{90}\textrm{Sr}\rightarrow _{39}^{90}\textrm{Y}+_{-1}^{ 0}\textrm{e}$
QExtra.6
Write the nuclear equation for the beta emission of iodine-131, the isotope used to diagnose and treat thyroid problems.
QExtra.7
A common uranium compound is uranyl nitrate hexahydrate [UO2(NO3)2 ⋅ 6H2O]. What is the formula mass of this compound?
502.15 g/mol
QExtra.8
Plutonium forms three oxides: PuO, PuO2, and Pu2O3. What are the formula masses of these three compounds?
QExtra.9
A banana contains 600 mg of potassium, 0.0117% of which is radioactive potassium-40. If 1 g of potassium-40 has an activity of 2.626 × 105 Bq, what is the activity of a banana?
about 18 Bq
QExtra.10
Smoke detectors typically contain about 0.25 mg of americium-241 as part of the smoke detection mechanism. If the activity of 1 g of americium-241 is 1.26 × 1011 Bq, what is the activity of americium-241 in the smoke detector?
QExtra.11
Uranium hexafluoride (UF6) reacts with water to make uranyl fluoride (UO2F2) and HF. Balance the following reaction: $UF_{6}+H_{2}O\rightarrow UO_{2}F_{2}+HF$
SExtra.11
$UF_{6}+2H_{2}O\rightarrow UO_{2}F_{2}+4HF$
QExtra.12
The cyclopentadienyl anion (C5H5) is an organic ion that can make ionic compounds with positive ions of radioactive elements, such as Np3+. Balance the following reaction:
$NpCl_{3}+Be(C_{2}H_{5})_{2}\rightarrow Np(C_{2}H_{5})_{3}+BeCl_{2}$
QExtra.13
If the half-life of hydrogen-3 is 12.3 y, how much time does it take for 99.0% of a sample of hydrogen-3 to decay?
81.7 y
QExtra.14
If the half-life of carbon-14 is 5,730 y, how long does it take for 10.0% of a sample of carbon-14 to decay?
QExtra.15
Although bismuth is generally considered stable, its only natural isotope, bismuth-209, is estimated to have a half-life of 1.9 × 1019 y. If the universe is estimated to have a lifetime of 1.38 × 1010 y, what percentage of bismuth-209 has decayed over the lifetime of the universe? (Hint: Be prepared to use a lot of decimal places.)
SExtra.15
about 0.000000005%
QExtra.16
The most common isotope of uranium (uranium-238) has a half-life of 4.5 × 109 y. If the universe is estimated to have a lifetime of 1.38 × 1010 y, what percentage of uranium-238 has decayed over the lifetime of the universe?
QExtra.17
Refer to Table 15.4.1, and separate the sources of radioactive exposure into voluntary and involuntary sources. What percentage of radioactive exposure is involuntary?
SExtra.17
Radioactive atoms in the body, terrestrial sources, and cosmic sources are truly involuntary, which is about 27% of the total. Radon exposure, medical sources, consumer products, and even nuclear energy sources can be avoided.
QExtra.18
With reference to Table 15.4.1, suggest ways that a practical person can minimize exposure to radioactivity. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/15%3A_Nuclear_Chemistry/15.E%3A_Nuclear_Chemistry_%28Exercises%29.txt |
Organic chemistry is the study of the chemistry of carbon compounds. Why focus on carbon? Carbon has properties that give its chemistry unparalleled complexity. It forms four covalent bonds, which give it great flexibility in bonding. It makes fairly strong bonds with itself (a characteristic called catenation), allowing for the formation of large molecules; it also forms fairly strong bonds with other elements, allowing for the possibility of a wide variety of substances. No other element demonstrates the versatility of carbon when it comes to making compounds. So an entire field of chemistry is devoted to the study of the compounds and reactivity of one element.
Because of the potential for complexity, chemists have defined a rather rigorous system to describe the chemistry of carbon. We will introduce some of that system in this chapter. Should you continue your study of chemistry beyond this text, you will find a much larger world of organic chemistry than we can cover in a single chapter.
• 16.1: Prelude to Organic Chemistry
Photosynthesis involves a whole series of reactions with many chemicals, enzymes, breaking and making chemical bonds, the transfer of electrons and H+ ions, and other chemical processes. The elucidation of the actual steps of photosynthesis—a process still unduplicated artificially—is a major achievement of modern chemistry.
• 16.2: Hydrocarbons
The simplest organic compounds are hydrocarbons and are composed of carbon and hydrogen. Hydrocarbons can be aliphatic or aromatic; aliphatic hydrocarbons are divided into alkanes, alkenes, and alkynes. The combustion of hydrocarbons is a primary source of energy for our society.
• 16.3: Branched Hydrocarbons
A unique name can be given to branched hydrocarbons. A unique structure can be drawn for the name of a hydrocarbon.
• 16.4: Alkyl Halides and Alcohols
The presence of a halogen atom (F, Cl, Br, or I; X is also used to represent any halogen atom) is one of the simplest functional groups. Organic compounds that contain a halogen atom are called alkyl halides.
• 16.5: Other Oxygen-Containing Functional Groups
Aldehydes, ketones, carboxylic acids, esters, and ethers have oxygen-containing functional groups.
• 16.6: Other Functional Groups
There are some common, and important, functional groups that contain elements other than oxygen. In this section, we will consider three of them: amine, amide, and thiol functional groups.
• 16.7: Polymers
Polymers are long molecules composed of chains of units called monomers. Several important biological polymers include proteins, starch, cellulose, and DNA.
• 16.E: Organic Chemistry (Exercises)
These are exercises and select solutions to accompany Chapter 16 of the "Beginning Chemistry" Textmap formulated around the Ball et al. textbook.
Thumbnail: DNA Double Helix (Public Domain; National Human Genome Research Institute via Wikipedia).
16: Organic Chemistry
All life on Earth is ultimately based on photosynthesis. Photosynthesis is the process by which plants absorb CO2 and H2O from their environment and, in the presence of sunlight, convert those substances into a simple sugar (glucose) and ultimately starches and other building blocks of life. The net photosynthesis chemical reaction is as follows:
$6CO_{2}+6H_{2}O \overset{light}{\rightarrow} C_{6}H_{12}O_{6}+6O_{2}\nonumber$
Oxygen is also a product of photosynthesis. Most forms of animal life (including people) depend on oxygen to breathe, which makes plants indispensable. Virtually all food sources come from plants, eaten either directly (as fruits, vegetables, or grains) or indirectly (as feedstock for meat animals such as cattle, poultry, pigs, sheep, goats, and the like). Plants are absolutely necessary for life to exist.
The net reaction for photosynthesis is misleadingly simple. A series of reactions, called light-dependent reactions, start by the absorption of light by pigments (not just chlorophyll, as commonly misunderstood) in plant cells. This is followed by a series of light-independent reactions, so named not because they happen in the dark, but because they do not directly involve light. However, light-independent reactions involve the products of reactions stimulated by light, so they ultimately depend on light. The whole series of reactions involves many chemicals, enzymes, breaking and making chemical bonds, the transfer of electrons and H+ ions, and other chemical processes. The elucidation of the actual steps of photosynthesis—a process still unduplicated artificially—is a major achievement of modern chemistry. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/16%3A_Organic_Chemistry/16.01%3A_Prelude_to_Organic_Chemistry.txt |
Learning Objectives
• Identify alkanes, alkenes, alkynes, and aromatic compounds.
• List some properties of hydrocarbons.
The simplest organic compounds are those composed of only two elements: carbon and hydrogen. These compounds are called hydrocarbons. Hydrocarbons themselves are separated into two types: aliphatic hydrocarbons and aromatic hydrocarbons.
Aliphatic hydrocarbons are hydrocarbons based on chains of C atoms. There are three types of aliphatic hydrocarbons. Alkanes are aliphatic hydrocarbons with only single covalent bonds. Alkenes are hydrocarbons that contain at least one C–C double bond, and Alkynes are hydrocarbons that contain a C–C triple bond. Occasionally, we find an aliphatic hydrocarbon with a ring of C atoms; these hydrocarbons are called cycloalkanes (or cycloalkenes or cycloalkynes).
Aromatic hydrocarbons have a special six-carbon ring called a benzene ring. Electrons in the benzene ring have special energetic properties that give benzene physical and chemical properties that are markedly different from alkanes. Originally, the term aromatic was used to describe this class of compounds because they were particularly fragrant. However, in modern chemistry the term aromatic denotes the presence of a six-membered ring that imparts different and unique properties to a molecule.
The simplest alkanes have their C atoms bonded in a straight chain; these are called normal alkanes. They are named according to the number of C atoms in the chain. The smallest alkane is methane:
The next-largest alkane has two C atoms that are covalently bonded to each other. For each C atom to make four covalent bonds, each C atom must be bonded to three H atoms. The resulting molecule, whose formula is C2H6, is ethane:
Propane has a backbone of three C atoms surrounded by H atoms. You should be able to verify that the molecular formula for propane is C3H8:
The diagrams representing alkanes are called structural formulas because they show the structure of the molecule. As molecules get larger, structural formulas become more and more complex. One way around this is to use a condensed structural formula, which lists the formula of each C atom in the backbone of the molecule. For example, the condensed structural formula for ethane is CH3CH3, while for propane it is CH3CH2CH3. Table $1$ - The First 10 Alkanes, gives the molecular formulas, the condensed structural formulas, and the names of the first 10 alkanes.
Table $1$ The First 10 Alkanes
Molecular Formula Condensed Structural Formula Name
CH4 CH4 methane
C2H6 CH3CH3 ethane
C3H8 CH3CH2CH3 propane
C4H10 CH3CH2CH2CH3 butane
C5H12 CH3CH2CH2CH2CH3 pentane
C6H14 CH3(CH2)4CH3 hexane
C7H16 CH3(CH2)5CH3 heptane
C8H18 CH3(CH2)6CH3 octane
C9H20 CH3(CH2)7CH3 nonane
C10H22 CH3(CH2)8CH3 decane
Because alkanes have the maximum number of H atoms possible according to the rules of covalent bonds, alkanes are also referred to as saturated hydrocarbons.
Alkenes have a C–C double bond. Because they have less than the maximum number of H atoms possible, they are unsaturated hydrocarbons. The smallest alkene—ethene—has two C atoms and is also known by its common name ethylene:
The next largest alkene—propene—has three C atoms with a C–C double bond between two of the C atoms. It is also known as propylene:
What do you notice about the names of alkanes and alkenes? The names of alkenes are the same as their corresponding alkanes except that the ending is -ene, rather than -ane. Using a stem to indicate the number of C atoms in a molecule and an ending to represent the type of organic compound is common in organic chemistry, as we shall see.
With the introduction of the next alkene, butene, we begin to see a major issue with organic molecules: choices. With four C atoms, the C–C double bond can go between the first and second C atoms or between the second and third C atoms:
2 structural formulas for butene, with the first butene having the double bond on the first and second carbon from the left and the latter having its double bond on the second and third carbon from the left.
(A double bond between the third and fourth C atoms is the same as having it between the first and second C atoms, only flipped over.) The rules of naming in organic chemistry require that these two substances have different names. The first molecule is named 1-butene, while the second molecule is named 2-butene. The number at the beginning of the name indicates where the double bond originates. The lowest possible number is used to number a feature in a molecule; hence, calling the second molecule 3-butene would be incorrect. Numbers are common parts of organic chemical names because they indicate which C atom in a chain contains a distinguishing feature.
The compounds 1-butene and 2-butene have different physical and chemical properties, even though they have the same molecular formula—C4H8. Different molecules with the same molecular formula are called isomers. Isomers are common in organic chemistry and contribute to its complexity.
Example $1$
Based on the names for the butene molecules, propose a name for this molecule.
Solution
With five C atoms, we will use the pent- stem, and with a C–C double bond, this is an alkene, so this molecule is a pentene. In numbering the C atoms, we use the number 2 because it is the lower possible label. So this molecule is named 2-pentene.
Exercise $1$
Based on the names for the butene molecules, propose a name for this molecule.
A structural formula of a six carbon molecule with a double bond on the third and fourth carbon from the left. There are twelve hydrogen atoms in total.
Answer
3-hexene
Alkynes, with a C–C triple bond, are named similarly to alkenes except their names end in -yne. The smallest alkyne is ethyne, which is also known as acetylene:
Propyne has the structure
Structural formula showing three carbon molecules with a triple bond present between the first and second carbon atom. The appropriate number of hydrogen atoms is attached to each carbon atom.
With butyne, we need to start numbering the position of the triple bond, just as we did with alkenes:
Two structural formula of butyne. One butyne has a triple bond between the first and second carbon atom, while two butyne has the triple bond between the second and third carbon atom.
Aromatic compounds contain the benzene unit. Benzene itself is composed of six C atoms in a ring, with alternating single and double C–C bonds:
The six carbons are arranged in a hexagon pattern with one hydrogen atom emerging outwards from each carbon atom. The presence of a double bond is alternated between every other carbon atom.
The alternating single and double C–C bonds give the benzene ring a special stability, and it does not react like an alkene as might be suspected. Benzene has the molecular formula C6H6; in larger aromatic compounds, a different atom replaces one or more of the H atoms.
As fundamental as hydrocarbons are to organic chemistry, their properties and chemical reactions are rather mundane. Most hydrocarbons are nonpolar because of the close electronegativities of the C and H atoms. As such, they dissolve only sparingly in H2O and other polar solvents. Small hydrocarbons, such as methane and ethane, are gases at room temperature, while larger hydrocarbons, such as hexane and octane, are liquids. Even larger hydrocarbons are solids at room temperature and have a soft, waxy consistency.
Hydrocarbons are rather unreactive, but they do participate in some classic chemical reactions. One common reaction is substitution with a halogen atom by combining a hydrocarbon with an elemental halogen. Light is sometimes used to promote the reaction, such as this one between methane and chlorine:
$CH_{4}+Cl_{2}\overset{light}{\rightarrow} CH_{3}Cl+HCl\nonumber$
Halogens can also react with alkenes and alkynes, but the reaction is different. In these cases, the halogen reacts with the C–C double or triple bond and inserts itself onto each C atom involved in the multiple bonds. This reaction is called an addition reaction. One example is
The reaction conditions are usually mild; in many cases, the halogen reacts spontaneously with an alkene or an alkyne.
Hydrogen can also be added across a multiple bond; this reaction is called a hydrogenation reaction. In this case, however, the reaction conditions may not be mild; high pressures of H2 gas may be necessary. A platinum or palladium catalyst is usually employed to get the reaction to proceed at a reasonable pace:
$CH_{2}=CH_{2}+H_{2}\overset{metal\: catalyst}{\rightarrow} CH_{3}CH_{3}\nonumber$
By far the most common reaction of hydrocarbons is combustion, which is the combination of a hydrocarbon with O2 to make CO2 and H2O. The combustion of hydrocarbons is accompanied by a release of energy and is a primary source of energy production in our society (Figure $2$ - Combustion). The combustion reaction for gasoline, for example, which can be represented by C8H18, is as follows:
$2C^{8}H_{18}+25O_{2}\rightarrow 16CO_{2}+18H_{2}O+\sim 5060kJ\nonumber$
Key Takeaways
• The simplest organic compounds are hydrocarbons and are composed of carbon and hydrogen.
• Hydrocarbons can be aliphatic or aromatic; aliphatic hydrocarbons are divided into alkanes, alkenes, and alkynes.
• The combustion of hydrocarbons is a primary source of energy for our society.
Exercise $2$
1. Define hydrocarbon. What are the two general types of hydrocarbons?
2. What are the three different types of aliphatic hydrocarbons? How are they defined?
3. Indicate whether each molecule is an aliphatic or an aromatic hydrocarbon; if aliphatic, identify the molecule as an alkane, an alkene, or an alkyne.
4. Indicate whether each molecule is an aliphatic or an aromatic hydrocarbon; if aliphatic, identify the molecule as an alkane, an alkene, or an alkyne.
5. Indicate whether each molecule is an aliphatic or an aromatic hydrocarbon; if aliphatic, identify the molecule as an alkane, an alkene, or an alkyne.
6. Indicate whether each molecule is an aliphatic or an aromatic hydrocarbon; if aliphatic, identify the molecule as an alkane, an alkene, or an alkyne.
7. Name and draw the structural formulas for the four smallest alkanes.
8. Name and draw the structural formulas for the four smallest alkenes.
9. What does the term aromatic imply about an organic molecule?
10. What does the term normal imply when used for alkanes?
11. Explain why the name 1-propene is incorrect. What is the proper name for this molecule?
12. Explain why the name 3-butene is incorrect. What is the proper name for this molecule?
13. Name and draw the structural formula of each isomer of pentene.
14. Name and draw the structural formula of each isomer of hexyne.
15. Write a chemical equation for the reaction between methane and bromine.
16. Write a chemical equation for the reaction between ethane and chlorine.
17. Draw the structure of the product of the reaction of bromine with propene.
18. Draw the structure of the product of the reaction of chlorine with 2-butene.
19. Draw the structure of the product of the reaction of hydrogen with 1-butene.
20. Draw the structure of the product of the reaction of hydrogen with 1-butene.
21. Write the balanced chemical equation for the combustion of heptane.
22. Write the balanced chemical equation for the combustion of nonane.
Answers
1. an organic compound composed of only carbon and hydrogen; aliphatic hydrocarbons and aromatic hydrocarbons
2.
1. aliphatic; alkane
2. aromatic
3. aliphatic; alkene
3.
1. aliphatic; alkane
2. aliphatic; alkene
3. aromatic
4.
5.
6. Aromatic means that the molecule has a benzene ring.
7.
8. The 1 is not necessary. The name of the compound is simply propene.
9.
10.
11. CH4 + Br2 → CH3Br + HBr
12.
13.
14.
15. C7H16 + 11O2 → 7CO2 + 8H2O | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/16%3A_Organic_Chemistry/16.02%3A_Hydrocarbons.txt |
Learning Objectives
• Name a branched hydrocarbon from its structure.
• Draw the structural formula of a branched hydrocarbon from its name.
Not all hydrocarbons are straight chains. Many hydrocarbons have branches of C atoms attached to a chain. These branched alkanes are isomers of straight-chain alkanes having the same number of C atoms. However, they are different compounds with different physical and chemical properties. As such, they need different names. How do we name branched hydrocarbons?
There are a series of rules for naming branched alkanes (and, ultimately, for all organic compounds). These rules make up the system of nomenclature for naming organic molecules. Worldwide, the International Union of Pure and Applied Chemistry (IUPAC) has developed the system of nomenclature for organic compounds. So these rules are sometimes called the IUPAC rules of nomenclature. By learning and applying these rules, you can name any organic compound when given its structure or determine the unique structure of a molecule from its name. You have already learned the basics of nomenclature—the names of the first 10 normal hydrocarbons. Here, we will add some steps to the procedure so you can name branched hydrocarbons.
First, given the structure of an alkane, identify the longest continuous chain of C atoms. Note that the longest chain may not be drawn in a straight line. The longest chain determines the parent name of the hydrocarbon. For example, in the molecule
the longest chain of carbons has six C atoms. Therefore, it will be named as a hexane. However, in the molecule
the longest chain of C atoms is not six, but seven, as shown. So this molecule will be named as a heptane.
The next step is to identify the branches, or substituents, on the main chain. The names of the substituents, or alkyl groups, are derived from the names of the parent hydrocarbons; however, rather than having the ending -ane, the substituent name has the ending -yl. Table \(1\) - Substituent Names, lists the substituent names for the five smallest substituents.
Table \(1\) Substituent Names
Substituent Formula Number of C Atoms Name of Substituent
CH3 1 methyl-
CH3CH2 2 ethyl-
CH3CH2CH2 3 propyl-
CH3CH2CH2CH2 4 butyl-
CH3CH2CH2CH2CH2 5 pentyl-
and so forth and so forth and so forth
In naming the branched hydrocarbon, the name of the substituent is combined with the parent name of the hydrocarbon without spaces. However, there is likely one more step. The longest chain of the hydrocarbon must be numbered, and the numerical position of the substituent must be included to account for possible isomers. As with double and triple bonds, the main chain is numbered to give the substituent the lowest possible number. For example, in this alkane
the longest chain is five C atoms long, so it is a pentane. There is a one-carbon substituent on the third C atom, so there is a methyl group at position 3. We indicate the position using the number, which is followed by a hyphen, the substituent name, and the parent hydrocarbon name—in this case, 3-methylpentane. That name is specific to that particular hydrocarbon and no other molecule. Organic chemistry nomenclature is very specific!
It is common to write the structural formula of a hydrocarbon without the H atoms, for clarity. So we can also represent 3-methylpentane as
where it is understood that any unwritten covalent bonds are bonds with H atoms. With this understanding, we recognize that the structural formula for 3-methylpentane refers to a molecule with the formula of C6H14.
Example \(1\)
Name this molecule.
Solution
The longest continuous carbon chain has seven C atoms, so this molecule will be named as a heptane. There is a two-carbon substituent on the main chain, which is an ethyl group. To give the substituent the lowest numbering, we number the chain from the right side and see that the substituent is on the third C atom. So this hydrocarbon is 3-ethylheptane.
Exercise \(1\)
Name this molecule.
Answer
2-methylpentane
Branched hydrocarbons may have more than one substituent. If the substituents are different, then give each substituent a number (using the smallest possible numbers) and list the substituents in alphabetical order, with the numbers separated by hyphens and with no spaces in the name. So the molecule
is 3-ethyl-2-methylpentane.
If the substituents are the same, then use the name of the substituent only once, but use more than one number, separated by a comma. Also, put a numerical prefix before the substituent name that indicates the number of substituents of that type. The numerical prefixes are listed in Table \(2\) - Numerical Prefixes to Use for Multiple Substituents. The number of the position values must agree with the numerical prefix before the substituent.
Table \(2\): Numerical Prefixes to Use for Multiple Substituents
Number of Same Substituent Numerical Prefix
2 di-
3 tri-
4 tetra-
5 penta-
and so forth and so forth
Consider this molecule:
The longest chain has four C atoms, so it is a butane. There are two substituents, each of which consists of a single C atom; they are methyl groups. The methyl groups are on the second and third C atoms in the chain (no matter which end the numbering starts from), so we would name this molecule 2,3-dimethylbutane. Note the comma between the numbers, the hyphen between the numbers and the substituent name, and the presence of the prefix di- before the methyl. Other molecules—even with larger numbers of substituents—can be named similarly.
Example \(2\)
Name this molecule.
Solution
The longest chain has seven C atoms, so we name this molecule as a heptane. We find two one-carbon substituents on the second C atom and a two-carbon substituent on the third C atom. So this molecule is named 3-ethyl-2,2-dimethylheptane.
Exercise \(2\)
Name this molecule.
Answer
4,4,5-tripropyloctane
Alkenes and alkynes are named in a similar fashion. The biggest difference is that when identifying the longest carbon chain, it must contain the C–C double or triple bond. Furthermore, when numbering the main chain, the double or triple bond gets the lowest possible number. This means that there may be longer or higher-numbered substituents than may be allowed if the molecule were an alkane. For example, this molecule
is 2,4-dimethyl-3-heptene (note the number and the hyphens that indicate the position of the double bond).
Example \(3\)
Name this molecule.
Solution
The longest chain that contains the C–C triple bond has six C atoms, so this is a hexyne molecule. The triple bond starts at the third C atom, so this is a 3-hexyne. Finally, there are two methyl groups on the chain; to give them the lowest possible number, we number the chain from the left side, giving the methyl groups the second position. So the name of this molecule is 2,2-dimethyl-3-hexyne.
Exercise \(3\)
Name this molecule.
Answer
2,3,4-trimethyl-2-pentene
Once you master naming hydrocarbons from their given structures, it is rather easy to draw a structure from a given name. Just draw the parent chain with the correct number of C atoms (putting the double or triple bond in the right position, as necessary) and add the substituents in the proper positions. If you start by drawing the C atom backbone, you can go back and complete the structure by adding H atoms to give each C atom four covalent bonds. From the name 2,3-dimethyl-4-propyl-2-heptene, we start by drawing the seven-carbon parent chain with a double bond starting at the third carbon:
We add two one-carbon substituents to this structure on the second and third C atoms:
We finish the carbon backbone by adding a three-carbon propyl group to the fourth C atom in the parent chain:
If we so choose, we can add H atoms to each C atom to give each carbon four covalent bonds, being careful to note that the C atoms in the double bond already have an additional covalent bond. (How many H atoms do you think are required? There will need to be 24 H atoms to complete the molecule.)
Example \(4\)
Draw the carbon backbone for 2,3,4-trimethylpentane.
Solution
First, we draw the five-carbon backbone that represents the pentane chain:
According to the name, there are three one-carbon methyl groups attached to the second, the third, and the fourth C atoms in the chain. We finish the carbon backbone by putting the three methyl groups on the pentane main chain:
Exercise \(4\)
Draw the carbon backbone for 3-ethyl-6,7-dimethyl-2-octene.
Answer
Naming substituted benzene molecules is straightforward. If there is only one substituent, the substituent is named as a side chain on a benzene molecule, like this:
If there are two or more substituents on a benzene molecule, the relative positions must be numbered, just as an aliphatic chain of C atoms is numbered. The substituent that is first alphabetically is assigned position 1, and the ring is numbered in a circle to give the other substituents the lowest possible number(s).
If a benzene ring is treated as a substituent, it is given the name phenyl-. The following molecule is 3-phenylpentane:
where the H atoms have been omitted for clarity.
Summary
A unique name can be given to branched hydrocarbons. A unique structure can be drawn for the name of a hydrocarbon.
Exercise \(5\)
1. How does a branched hydrocarbon differ from a normal hydrocarbon?
2. How does a substituent get its unique name?
3. Name this molecule.
4. Name this molecule.
5. Name this molecule.
6. Name this molecule.
7. Name this molecule.
8. Name this molecule.
9. Name this molecule.
10. Name this molecule.
11. Name this molecule.
12. Name this molecule.
13. Draw the carbon backbone for each molecule.
1. 3,4-diethyloctane
2. 2,2-dimethyl-4-propylnonane
14. Draw the carbon backbone for each molecule.
1. 3-ethyl-4-methyl-3-heptene
2. 3,3-diethyl-1-pentyne
15. Draw the carbon backbone for each molecule.
1. 4-ethyl-4-propyl-2-octyne
2. 5-butyl-2,2-dimethyldecane
16. Draw the carbon backbone for each molecule.
1. 3,4-diethyl-1-hexyne
2. 4-propyl-3-ethyl-2-methyloctane
17. The name 2-ethylhexane is incorrect. Draw the carbon backbone and write the correct name for this molecule.
18. The name 3-butyl-7-methyloctane is incorrect. Draw the carbon backbone and write the correct name for this molecule.
Answers
1. A branched hydrocarbon does not have all of its C atoms in a single row.
2.
3. 3-methyl-2-hexene
4.
5. 4,4-dimethyl-1-pentene
6.
7. 2,4-dimethyl-2-pentene
8.
9. 3,4-diethyloctane
10.
11. 1-bromo-4-chlorobenzene
12.
13.
14. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/16%3A_Organic_Chemistry/16.03%3A_Branched_Hydrocarbons.txt |
Learning Objectives
• Define functional group.
• Identify and name a simple alkyl halide.
• Identify and name a simple alcohol.
• Predict the product(s) of an elimination reaction of an alkyl halide or an alcohol.
A functional group is any collection of atoms and/or bonds with certain characteristic chemical reactions. We have already seen two functional groups: the C–C double bond and the C–C triple bond. They undergo certain characteristic chemical reactions—for example, the addition of a halogen across the multiple bond.
The presence of a halogen atom (F, Cl, Br, or I; also, X is used to represent any halogen atom) is one of the simplest functional groups. Organic compounds that contain a halogen atom are called alkyl halides. We have already seen some examples of alkyl halides when the addition of halogens across double and triple bonds was introduced in Section 16.3 - "Branched Hydrocarbons;" the products of these reactions were alkyl halides.
A simple alkyl halide can be named like an ionic salt, first by stating the name of the parent alkane as a substituent group (with the -yl suffix) and then the name of the halogen as if it were the anion. So CH3Cl has the common name of methyl chloride, while CH3CH2Br is ethyl bromide and CH3CH2CH2I is propyl iodide. However, this system is not ideal for more complicated alkyl halides.
The systematic way of naming alkyl halides is to name the halogen as a substituent, just like an alkyl group, and use numbers to indicate the position of the halogen atom on the main chain. The name of the halogen as a substituent comes from the stem of the element's name plus the ending -o, so the substituent names are fluoro-, chloro-, bromo- and iodo-. If there is more than one of a certain halogen, we use numerical prefixes to indicate the number of each kind, just as with alkyl groups. For example, this molecule
is 2-bromobutane, while this molecule
is 2,3-dichloropentane. If alkyl groups are present, the substituents are listed alphabetically. Numerical prefixes are ignored when determining the alphabetical ordering of substituent groups.
Example \(1\)
Name this molecule.
Solution
The longest carbon chain has five C atoms, so the molecule is a pentane. There are two chlorine substituents located on the second and third C atoms, with a one-carbon methyl group on the third C atom as well. The correct name for this molecule is 2,3-dichloro-3-methylpentane.
Exercise \(1\)
Name this molecule.
Answer
1,1,2-tribromopropane
Another simple functional group is the covalently bonded OH group. This is the alcohol functional group. It is not the hydroxide ion; rather than being present as a negatively charged species, in organic chemistry it is a covalently bonded functional group.
Like alkyl halides, alcohols have a common naming system and a more formal system. The common system is similar to that of alkyl halides: name the alkyl group attached to the OH group, ending with the suffix -yl, and add the word alcohol as a second word. So CH3OH is methyl alcohol; CH3CH2OH is ethyl alcohol, and CH3CH2CH2OH is propyl alcohol.
As with alkyl halides, though, this system is limited (although for smaller alcohols, it is very common in everyday usage). The formal system of naming uses the name of the hydrocarbon containing the OH group and having the correct number of C atoms, dropping the final -e of the name and appending the suffix -ol. Thus CH3OH is methanol and CH3CH2OH is ethanol. For larger alcohol molecules, we use a number to indicate the position of the OH group on the longest carbon chain, similar to the number needed for alkenes and alkynes. Again, the carbon chain is numbered to give the OH group the lowest number, no matter how large the other numbers are. So CH3CH2CH2OH is 1-propanol, while CH3CHOHCH3 is 2-propanol. (A common component in many medicine cabinets, 2‑propanol is also known as isopropanol or isopropyl alcohol (Figure \(1\) - Isopropyl Alcohol).
Another acceptable way of naming an alcohol—especially a more complicated molecule—is to name the OH group as the hydroxy substituent and give it a numerical position like an alkyl group or a halogen atom. Thus, 2-propanol would be called 2-hydroxypropane by this convention.
Example \(2\)
Name this molecule as an alcohol and as a substituted alkane.
Solution
The longest carbon chain containing the OH group has four C atoms, so the parent hydrocarbon is butane. Because the OH group is on the first C atom, it is 1-butanol. There is a methyl group on the second C atom, as well as a Cl atom, so the formal name for this alcohol is 2-chloro-2-methyl-1-butanol. If naming the alcohol group as a substituent, it would be 2-chloro-1-hydroxy-2-methylbutane.
Exercise \(2\)
Name this molecule as an alcohol and as a substituted alkane.
Answer
2,2,2-trichloroethanol; 2,2,2-trichloro-1-hydroxyethane
Most alkyl halides are insoluble in H2O. Smaller alcohols, however, are very soluble in H2O because these molecules can engage in hydrogen bonding with H2O molecules. For larger molecules, however, the polar OH group is overwhelmed by the nonpolar alkyl part of the molecule. While methanol is soluble in H2O in all proportions, only about 2.6 g of pentanol will dissolve in 100 g of H2O. Larger alcohols have an even lower solubility in H2O.
One reaction common to alcohols and alkyl halides is elimination, the removal of the functional group (either X or OH) and an H atom from an adjacent carbon. The general reaction can be written as follows:
where Z represents either the X or the OH group. The biggest difference between elimination in alkyl halides and elimination in alcohols is the identity of the catalyst: for alkyl halides, the catalyst is a strong base; for alcohols, the catalyst is a strong acid. For compounds in which there are H atoms on more than one adjacent carbon, a mixture of products results.
Example \(3\)
Predict the organic product(s) of this reaction.
Solution
Under these conditions, an HOH (otherwise known as H2O) molecule will be eliminated, and an alkene will be formed. It does not matter which adjacent carbon loses the H atom; in either case the product will be
which is propene.
Exercise \(3\)
Predict the organic product(s) of this reaction.
Answer
1-butene and 2-butene
Key Takeaways
• Alkyl halides have a halogen atom as a functional group.
• Alcohols have an OH group as a functional group.
• Nomenclature rules allow us to name alkyl halides and alcohols.
• In an elimination reaction, a double bond is formed as an HX or an HOH molecule is removed.
Exercise \(4\)
1. Define functional group and give two examples.
2. What is elimination? How does it differ for alkyl halides and alcohols?
3. Name this molecule.
4. Name this molecule.
5. Name this molecule.
6. Name this molecule.
7. Name this molecule.
8. Name this molecule.
9. Name this molecule.
10. Name this molecule.
11. Predict the product(s) of this elimination reaction.
12. Predict the product(s) of this elimination reaction.
13. Predict the product(s) of this elimination reaction.
14. Predict the product(s) of this elimination reaction.
Answers
1. a group of atoms with a certain reactivity; halogen atoms and alcohol groups (answers will vary).
2.
3. 2-bromobutane
4.
5. 2-chloro-3-fluoro-4-methylheptane
6.
7. 2-methyl-2-propanol
8.
9. 4-octanol
10.
11. 2-pentene
12.
13. 2-hexene and 3-hexene
1. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/16%3A_Organic_Chemistry/16.04%3A_Alkyl_Halides_and_Alcohols.txt |
Learning Objective
• Identify the aldehyde, ketone, acid, ester, and ether functional groups.
There are other functional groups that contain O atoms. Before we introduce them, we define the carbonyl group, which is formed when an O atom and a C atom are joined by a double bond:
The other two bonds on the C atom are attached to other atoms. It is the identities of these other atoms that define what specific type of compound an organic molecule is.
If one bond of the carbonyl group is made to an H atom, then the molecule is classified as an aldehyde (If there are two H atoms, there is only 1 C atom). When naming aldehydes, the main chain of C atoms must include the carbon in the carbonyl group, which is numbered as position 1 in the carbon chain. The parent name of the hydrocarbon is used, but the suffix -al is appended. (Do not confuse -al with -ol, which is the suffix used for alcohols.) So we have
Methanal has a common name with which you may be familiar: formaldehyde. The main thing to note about aldehydes is that the carbonyl group is at the end of a carbon chain.
A carbonyl group in the middle of a carbon chain implies that both remaining bonds of the carbonyl group are made to C atoms. This type of molecule is called a ketone. Despite the fact that aldehydes and ketones have the same carbonyl group, they have different chemical and physical properties and are properly grouped as two different types of compounds. The smallest ketone has three C atoms in it. When naming a ketone, we take the name of the parent hydrocarbon and change the suffix to -one:
The common name for propanone is acetone. With larger ketones, we must use a number to indicate the position of the carbonyl group, much like a number is used with alkenes and alkynes:
There is another way to name ketones: name the alkyl groups that are attached to the carbonyl group and add the word ketone to the name. So propanone can also be called dimethyl ketone, while 2-butanone is called methyl ethyl ketone.
Example \(1\)
Draw the structure of 2-pentanone.
Solution
This molecule has five C atoms in a chain, with the carbonyl group on the second C atom. Its structure is as follows:
The combination of a carbonyl functional group and an OH group makes the carboxyl group.
Molecules with a carboxyl group are called carboxylic acids. As with aldehydes, the functional group in carboxylic acids is at the end of a carbon chain. Also as with aldehydes, the C atom in the functional group is counted as one of the C atoms that defines the parent hydrocarbon name. To name carboxylic acids, the parent name of the hydrocarbon is used, but the suffix -oic acid is added:
Methanoic acid and ethanoic acid are also called formic acid and acetic acid, respectively. Formic acid is the compound that makes certain ant bites sting, while acetic acid is the active substance in vinegar.
How acidic are carboxylic acids? It turns out that they are not very acidic. No carboxylic acid is on the list of strong acids (Table \(1\)). This means that all carboxylic acids are weak acids. A 1 M solution of formic acid is only about 1.3% dissociated into H+ ions and formate ions, while a similar solution of acetic acid is ionized by about only 0.4%. Some carboxylic acids are stronger—for example, trichloroacetic acid is about 45% dissociated in aqueous solution. But no carboxylic acid approaches the 100% dissociation amount required by the definition of a strong acid.
As their name suggests, however, carboxylic acids do act like acids in the presence of bases. The H atom in the carboxyl group comes off as the H+ ion, leaving a carboxylate anion:
Carboxylate ions are named from the acid name: the -oic acid is replaced with -oate to name the ion.
Example \(2\):
Complete the chemical reaction. Can you name the carboxylate ion formed?
Solution
The OH ion removes the H atom that is part of the carboxyl group:
The carboxylate ion, which has the condensed structural formula CH3CO2, is the ethanoate ion, but it is commonly called the acetate ion.
Exercise \(1\)
Complete the chemical reaction. Can you name the carboxylate ion formed?
Answer
The anion is the methanoate ion, which is commonly called the formate ion.
One reaction to consider is that of a carboxylic acid and an alcohol. When combined under the proper conditions, a water molecule will be removed, and the remaining pieces will combine to form a new functional group—the ester functional group:
Note how the acid molecule contributes one alkyl side (represented by R), while the alcohol contributes the other side (represented by R′). Esters are named using the alkyl group name from the alcohol plus the carboxylate name from the acid. For example, the molecule
is called methyl propanoate.
Chemistry is Everywhere: Esters, Fragrances, and Flavorings
Esters are very interesting compounds, in part because many have very pleasant odors and flavors. (Remember, never taste anything in the chemistry lab!) Many esters occur naturally and contribute to the odor of flowers and the taste of fruits. Other esters are synthesized industrially and are added to food products to improve their smell or taste; it is likely that if you eat a product whose ingredients include artificial flavorings, those flavorings are esters. Here are some esters and their uses, thanks to their odors, flavors, or both:
Table with four columns and seven rows. The first column on the left is labeled Ester and underneath in the rows has different esters. The second column is labeled Tastes/Smells Like and underneath in the rows has different tastes and smells. The third column is labeled Ester and underneath in the rows has different esters. The last and fourth column is labeled Tastes/Smells Like and underneath in the rows has different tastes and smells.
Ester Tastes/Smells Like Ester Tastes/Smells Like
allyl hexanoate pineapple isobutyl formate raspberry
benzyl acetate pear isobutyl acetate pear
butyl butanoate pineapple methyl phenylacetate honey
ethyl butanoate banana nonyl caprylate orange
ethyl hexanoate pineapple pentyl acetate apple
ethyl heptanoate apricot propyl ethanoate pear
ethyl pentanoate apple propyl isobutyrate rum
Finally, the ether functional group is an O atom that is bonded to two organic groups: R-O-R′
The two R groups may be the same or different. Naming ethers is like the alternate way of naming ketones. In this case, the R groups are named sequentially, and the word ether is appended. The molecule CH3OCH3 is dimethyl ether, while CH3OCH2CH3 is methyl ethyl ether. Diethyl ether, another ether, was once used as an anesthetic, but its flammability and toxicity caused it to fall out of favor. Smaller ether molecules that are liquids at room temperature are common solvents for organic chemical reactions.
Key Takeaway
• Aldehydes, ketones, carboxylic acids, esters, and ethers have oxygen-containing functional groups.
Exercise \(2\)
1. Name a similarity between the functional groups found in aldehydes and ketones. Can you name a difference between them?
2. Explain how a carboxylic acid is used to make an ester.
3. Name each molecule.
4. Name each molecule.
5. Name each molecule.
6. Name each molecule.
7. Name the molecule.
8. Name the molecule.
9. Give an alternate but acceptable name to the molecule in Exercise 3b.
10. Give an alternate but acceptable name to the molecule in Exercise 4b.
11. Complete this chemical reaction. KOH." style="width: 550px; height: 82px;" width="550px" height="82px" data-cke-saved-src="/@api/deki/files/96632/Ex_11.png" src="/@api/deki/files/96632/Ex_11.png" data-quail-id="209">
12. Complete this chemical reaction.
13. The drug known as aspirin has this molecular structure: Identify the functional group(s) in this molecule.
14. The drug known as naproxen sodium is the sodium salt of this molecule: (The extra H atoms are omitted for clarity.) Identify the functional group(s) in this molecule.
15. Identify the ester made by reacting these molecules.
16. Identify the ester made by reacting these molecules.
Answers
1. They both have a carbonyl group, but an aldehyde has the carbonyl group at the end of a carbon chain, and a ketone has the carbonyl group in the middle.
2.
1. propanal
2. 2-butanone
3.
1. 3-methylbutanoic acid
2. ethyl propionate
4.
5. ethyl propyl ether
6.
7. ethyl methyl ketone
8.
9. H2O + KCH3CH2CO2
10.
11. acid, ester, and aromatic (benzene ring)
12.
13. propyl propionate | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/16%3A_Organic_Chemistry/16.05%3A_Other_Oxygen-Containing_Functional_Groups.txt |
Learning Objective
• Identify the amine, amide, and thiol functional groups.
There are some common and important functional groups that contain elements other than oxygen. In this section, we will consider three of them.
Nitrogen-Containing Compounds
An amine is an organic derivative of ammonia (NH3). In amines, one or more of the H atoms in NH3 is substituted with an organic group. A primary amine has one H atom substituted with an R group:
A secondary amine has two H atoms substituted with R groups:
A tertiary amine has all three H atoms substituted with R groups:
Naming simple amines is straightforward: name the R groups as substituents and then add the suffix -amine, using numerical suffixes on the substituent names as necessary. This amine
is diethylamine (the H atoms on the C atoms are omitted for clarity), while this amine
is ethyldipropylamine.
Example \(1\)
Name this amine.
Solution
This amine has a methyl group, an ethyl group, and a propyl group. Listing the names in alphabetical order, this amine is ethylmethylpropylamine.
Exercise \(1\)
Name this amine.
Answer
triethylamine
As with NH3, the N atom in amines can accept a proton onto the lone electron pair on the N atom. That is, amines act as Brønsted-Lowry bases (i.e., proton acceptors):
The amine becomes an ion, the organic counterpart of the ammonium (NH4+) ion.
All amines are weak bases. The weakness of amines is about the same as that of carboxylic acids. N-containing organic compounds are very common in nature, and they all act as weak bases. Some of these compounds have rather complicated structures. Figure \(1\) - Some Naturally Occurring N-Containing Compounds, shows some N-containing substances that you may recognize.
Nitrogen-containing compounds occur frequently in nature. Here are some that you might encounter in the course of your everyday life.
An amide functional group is a combination of an amine group and a carbonyl group:
Amides are actually formed by bringing together an amine-containing molecule and a carboxylic acid-containing molecule. A molecule of H2O is lost, much like when an ester forms:
The bond between the N of the amine group and the C of the carbonyl group is called an amide bond. Amide bonds are particularly important in biological molecules called proteins, which are composed of strings of amino acids—molecules that have an amine group and a carboxylic acid group within them. The amine group on one amino acid reacts with the carboxylic acid group of another amino acid, making a chain held together by amide bonds. We will consider proteins later in this chapter.
Example \(2\)
Draw the structure of the amide formed by the combination of ethylamine and butanoic acid.
Solution
The structures of ethylamine and butanoic acid are as follows:
When they come together to make an amide, an H2O molecule is lost, and the N of the amine group bonds to the C of the carboxyl group. The resulting molecule is as follows:
Exercise \(2\)
Draw the structure of the amide formed by the combination of methylamine and formic acid.
Answer
Sulfur-Containing Compounds
Sulfur is below oxygen on the periodic table, and it occasionally shows some similar chemistry. One similarity is that an S atom can take the place of an O atom in an alcohol, to make a molecule that looks like this:
R-SH
The sulfur analog of an alcohol is called a thiol. The formal way of naming a thiol is similar to that of alcohols, except that instead of using the suffix-ol, you use -thiol as the suffix. The following illustrates thiol nomenclature:
An older system uses the word mercaptan in naming simple thiols, much like the word alcohol is used with small alcohols. These thiols can also be named like this:
Many thiols have strong, objectionable odors; indeed, the spray from skunks is composed of thiols and is detectable by the human nose at concentrations less than 10 ppb. Because natural gas is odorless, thiols are intentionally added to natural gas—at very low levels, of course—so that gas leaks can be more easily detected. Not all thiols have objectionable odors; this thiol is responsible for the odor of grapefruit:
One amino acid that is a thiol is cysteine:
Cysteine plays an important role in protein structure. If two cysteine amino acids in a protein chain approach each other, they can be oxidized, and a S–S bond (also known as a disulfide bond) is formed:
R–SH + HS–R → R–S–S–R
where the R group is the rest of the cysteine molecule. The disulfide bond is strong enough to fix the position of the two cysteine groups, thus imposing a structure on the protein. Hair is composed of about 5% cysteine, and the breaking and remaking of disulfide bonds between cysteine units is the primary mechanism behind straightening and curling hair (hair "perms").
Food and Drink Application: Amino Acids, Essential and Otherwise
The text mentioned cysteine, an amino acid. Amino acids are the fundamental building blocks of proteins, a major biological component. Proteins are a necessary part of the diet; meat, eggs, and certain plant foods such as beans and soy are good sources of protein and amino acids.
All life on Earth—from the lowliest single-celled organism to humans to blue whales—relies on proteins for life, so all life on Earth is dependent on amino acids. The human body contains 20 different amino acids (curiously, other organisms may have a different number of amino acids). However, not all of them must be obtained from the diet. The body can synthesize 12 amino acids. The other 8 must be obtained from the diet. These 8 amino acids are called the essential amino acids. Daily requirements range from 4 mg per kilogram of body weight for tryptophan to 40 mg per kilogram of body weight for leucine. Infants and children need a greater mass per kg of body weight to support their growing bodies; also, the number of amino acids that are considered essential for infants and children is greater than for adults due to the greater protein synthesis associated with growth.
Because of the existence of essential amino acids, a diet that is properly balanced in protein is necessary. Rice and beans, a very popular food dish in Latin cuisines, actually provides all of the essential amino acids in one dish; without one component, the dish would be nutritionally incomplete. Corn (maize) is the most-grown grain crop in the world, but an over-reliance on it as a primary food source deprives people of lysine and tryptophan, which are two essential amino acids. (Indeed, it is now widely accepted that the disappearance of certain native American groups was largely due to the overuse of corn as the staple food.) People on restricted diets, whether out of necessity or by choice (e.g., vegetarians), may be missing the proper amount of an essential amino acid. It is important to vary the diet when possible to ensure ingestion of a wide range of protein sources.
Key Takeaway
• Other functional groups include amine, amide, and thiol functional groups.
Exercise \(3\)
1. What are the structure and name of the smallest amine?
2. What are the structure and name of the smallest thiol?
3. Identify each compound as a primary, secondary, or tertiary amine.
4. Identify each compound as a primary, secondary, or tertiary amine.
5. Write the chemical reaction between each amine in Exercise 3 and HCl.
6. Write the chemical reaction between each amine in Exercise 4 and HNO3.
7. Name each amine.
8. Name each amine.
9. A peptide is a short chain of amino acids connected by amide bonds. How many amide bonds are present in this peptide?
10. How many amide bonds are present in this peptide? (See Exercise 9 for the definition of a peptide.)
11. Draw the backbone structure of the amide formed by reacting propylamine with propanoic acid.
12. Draw the backbone structure of the amide formed by reacting hexylamine with ethanoic acid.
13. Name each thiol using the -thiol suffix.
1. C4H9–SH
14. Name each thiol in Exercise 13 with the mercaptan label.
15. One component of skunk spray is 3-methyl-1-butanethiol. Draw its structure. (The 1 indicates the position of the S atom.)
16. An S–S bond can be fairly easily broken into proteins, yielding two lone cysteine units in a protein chain. Is this process an oxidation or a reduction? Explain your answer.
Answers
1. CH3NH2; methylamine
2.
1. primary
2. tertiary
3. secondary
3.
1. C3H3CO2HSHNH2 + HCl → C3H3CO2HSHNH3Cl
2. (C6H11)(C2H5)(CH3)N + HCl → (C6H11)(C2H5)(CH3)NHCl
3. (C2H5)(CH3)NH + HCl → (C2H5)(CH3)NH2Cl
4.
1. ethylmethylamine
2. phenylamine
5.
6. two
7.
8.
1. cyclohexanethiol
2. butanethiol
9. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/16%3A_Organic_Chemistry/16.06%3A_Other_Functional_Groups.txt |
Learning Objectives
• Define the terms monomer and polymer.
• Draw the structure of a polymer from its monomer.
Among other applications, organic chemistry has had a huge impact on the development of modern materials called polymers. Many objects in daily life are composed of polymers; curiously, so are several important biological materials. Consider a molecule with a double bond, such as ethylene:
Imagine the bond between the carbons opening up and attacking another ethylene molecule:
The double bond between the two carbons is depicted with electron pushing arrows attacking another ethylene.
Now imagine further that the second ethylene molecule's double bond opens up and attacks a third ethylene molecule, which also opens up its double bond and attacks a fourth ethylene molecule, and so forth. The end result is a long, virtually endless molecule:
This long, almost nonstop molecule is called a polymer (from the Greek meaning "many parts"). The original part, ethylene, is called the monomer (meaning "one part"). The process of making a polymer is called polymerization. A polymer is an example of a macromolecule, the name given to a large molecule.
Simple polymers are named after their monomers. The ethylene polymer is formally called poly(ethylene), although in common use, the names are used without parentheses: polyethylene. Because adding one monomer to another forms this polymer, polyethylene is an example of a type of polymer called addition polymers. Figure \(1\) - Some Monomers and Their Addition Polymers, lists some addition polymers and their monomers. One of them, poly(ethylene oxide), results not from the opening of a double bond but the opening of a ring in the monomer; the concept of bonding with other monomers, however, is the same.
Example \(1\)
Draw the polymer that results from the polymerization of tetrafluoroethylene.
Solution
In the case of this monomer, the double bond opens up and joins to other monomers, just as with ethylene. The polymer that is made has this structure:
Exercise \(1\)
Draw the polymer that results from the polymerization of vinyl chloride.
Answer
Another type of polymer is the condensation polymer, which is a polymer made when two different monomers react together and release some other small molecule as a product. We have already seen an example of this, in the formation of an amide bond:
Here, H2O is released when the ends of the molecules react to form a polymer.
Related to condensation polymers are the copolymers, polymers made from more than one type of monomer. For example, ethylene and propylene can be combined into a polymer that is a mixture of the two monomers. A common form of synthetic rubber called styrene butadiene rubber (SBR) is made from two monomers: styrene and butadiene:
The physical and chemical properties of polymers vary widely, based on their monomers, structures, and additives. Among the other properties that can be modified based on these factors include: solubility in H2O and other solvents, melting point, flammability, color, hardness, transparency, film thickness, wet-ability, surface friction, mold-ability, particle size...the list goes on.
The uses of polymers are almost too numerous to consider. Anything that you might describe as "plastic" is likely a polymer. Polymers are used to make everything from toothbrushes to computer cases to automobile parts. Many epoxy-based adhesives are condensation polymers that adhere strongly to other surfaces. Polyurethane paints and coatings are polymers, as are the polyester fabrics used to make clothing. Nylon, Dacron, and Mylar are polymers (in fact, both Dacron and Mylar are forms of polyethylene terephthalate [PET]). The product known as Saran Wrap was originally constructed from Saran, a name for poly(vinylidene chloride), which was relatively impervious to oxygen and could be used as a barrier to help keep food fresh. (It has since been replaced with polyethylene, which is not as impervious to atmospheric oxygen.) Poly(vinyl chloride) is the third-most produced polymer [after poly(ethylene) and poly(propylene)] and is used to make everything from plastic tubing to automobile engine parts, water pipes to toys, flooring to waterbeds and pools.
All the polymers we have considered so far are based on a backbone of (largely) carbon. There is another class of polymers based on a backbone of Si and O atoms; these polymers are called silicones. The Si atoms have organic groups attached to them, so these polymers are still organic. One example of a silicone is as follows:
Silicones are used to make oils and lubricants; they are used as sealants for glass objects (such as aquariums) and films for waterproofing objects. Solid silicones are heat resistant and rubbery and are used to make cookware and electrical insulation.
Some very important biological materials are polymers. Of the three major food groups, polymers are represented in two: proteins and carbohydrates. Proteins are polymers of amino acids, which are monomers that have an amine functional group and a carboxylic acid functional group. These two groups react to make a condensation polymer, forming an amide bond:
Proteins are formed when hundreds or even thousands of amino acids form amide bonds to make polymers. Proteins play a crucial role in living organisms.
A carbohydrate is a compound that has the general formula Cn(H2O)n. Many carbohydrates are relatively small molecules, such as glucose:
Linking hundreds of glucose molecules together makes a relatively common material known as starch:
Starch is an important source of energy in the human diet. Note how individual glucose units are joined together. They can also be joined together in another way, like this:
This polymer is known as cellulose. Cellulose is a major component in the cell walls of plants. Curiously, despite the similarity in the building blocks, some animals (such as humans) cannot digest cellulose; those animals that can digest cellulose typically rely on symbiotic bacteria in the digestive tract for the actual digestion. Animals do not have the proper enzymes to break apart the glucose units in cellulose, so it passes through the digestive tract and is considered dietary fiber.
DNA
Deoxyribonucleic acid (DNA) and ribonucleic acid (RNA) are also polymers, composed of long, three-part chains consisting of phosphate groups, sugars with 5 C atoms (ribose or deoxyribose), and N-containing rings referred to as bases. Each combination of the three parts is called a nucleotide; DNA and RNA are essentially polymers of nucleotides that have rather complicated, but intriguing structures (Figure \(2\)). DNA is the fundamental material in chromosomes and is directly responsible for heredity, while RNA is an essential substance in protein synthesis.
Key Takeaways
• Polymers are long molecules composed of chains of units called monomers.
• Several important biological polymers include proteins, starch, cellulose, and DNA.
Exercise \(2\)
1. Explain the relationship between a monomer and a polymer.
2. Must a monomer have a double bond to make a polymer? Give an example to illustrate your answer.
3. Draw the polymer made from this monomer.
4. Draw the polymer made from this monomer.
5. What is the difference between an addition polymer and a condensation polymer?
6. What is the difference between a condensation polymer and a copolymer?
7. List three properties of polymers that vary widely with composition.
8. List three uses of polymers.
9. Draw the silicone made from this monomer.
10. Draw the silicone made from this monomer.
11. Explain how starch is a polymer.
12. What is the difference between starch and cellulose?
13. Explain how protein is a polymer.
14. What are the parts that compose DNA?
Answers
1. A polymer is many monomers bonded together.
2.
3.
4. In an addition polymer, no small molecule is given off as a product; whereas in a condensation polymer, small parts of each monomer come off as a small molecule.
5.
6. solubility in H2O and other solvents, melting point, flammability, color, hardness, transparency, film thickness, wetability, surface friction, moldability, and particle size (answers will vary)
7.
8.
9. Starch is composed of many glucose monomer units.
10.
11. Proteins are polymers of amino acids, which act as the monomers. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/16%3A_Organic_Chemistry/16.07%3A_Polymers.txt |
Additional Exercises
1. Cycloalkanes are named based on the number of C atoms in them, just like regular alkanes, but with the prefix cyclo- on the name. What are the names of the three smallest cycloalkanes?
2. Cycloalkenes are named similarly to cycloalkanes (see Exercise 1). What are the names of the cycloalkenes with five, six, and seven C atoms?
3. Draw the carbon backbone of all noncyclic alkanes with only four C atoms.
4. Draw the carbon backbone of all noncyclic alkanes with only five C atoms.
5. Cyclic alkanes can also have substituent groups on the ring. Draw the carbon backbone of all cyclic alkanes with only four C atoms.
6. Cyclic alkanes can also have substituent groups on the ring. Draw the carbon backbone of all cyclic alkanes with only five C atoms.
7. Draw and name all possible isomers of pentene.
8. Draw and name all possible normal (that is, straight-chain) isomers of heptyne.
9. Polyunsaturated alkenes have more than one C–C double bond. Draw the carbon backbone of all possible noncyclic polyunsaturated alkenes with four C atoms and two double bonds. What are the complete molecular formulas for each possible molecule?
10. Draw the carbon backbone of all possible five-carbon cyclic alkenes with two double bonds, assuming no substituents on the ring.
11. If a hydrocarbon is combined with enough halogen, all the H atoms will eventually be substituted with that halogen atom. Write the balanced chemical reaction between ethane and excess chlorine.
12. If a hydrocarbon is combined with enough halogen, all the H atoms will eventually be substituted with that halogen atom. Write the balanced chemical reaction between butane and excess bromine.
13. Molecules with multiple double bonds can also participate in addition reactions. Draw the structure of the product when butadiene, CH2=CH–CH=CH2, reacts with chlorine.
14. Molecules with multiple double bonds can also participate in addition reactions. Draw the structure of the product when allene, CH2=C=CH2, reacts with bromine.
15. What is the maximum number of methyl groups that can be on a propane backbone before the molecule cannot be named as a propane compound?
16. Explain why cycloethane cannot exist as a real molecule.
17. In the gasoline industry, what is called isooctane is actually 2,2,4-trimethylpentane. Draw the structure of isooctane.
18. Isooctane (see Exercise 17) is an isomer of what straight-chain alkane?
19. The actual name for the explosive TNT is 2,4,6-trinitrotoluene. If the structure of TNT is propose the structure of the parent compound toluene.
20. Phenol is hydroxybenzene, the simplest aromatic alcohol. Picric acid is an explosive derivative of phenol whose formal name is 2,4,6-trinitrophenol. With reference to Exercise 19, draw the structure of picric acid.
21. Draw the structures of all possible straight-chain isomers of bromopentane.
22. Draw the structures of all the possible isomers of butanol. Include branched isomers.
23. What is the final product of the double elimination of HCl from 1,1-dichloroethane?
24. Draw the structure of the final product of the double elimination of 1,3-dibromopropane.
25. Draw the structure and name of the alcohol whose double elimination would yield the same product as in Exercise 23. Name the molecule as a hydroxyl-substituted compound.
26. Draw the structure and name of the alcohol whose double elimination would yield the same product as in Exercise 24. Name the molecule as a hydroxyl-substituted compound.
27. Draw the smallest molecule that can have a separate aldehyde and carboxylic acid group.
28. Name the functional group(s) in urea, a molecule with the following structure:
29. Ethyl acetate is a common ingredient in nail-polish remover because it is a good solvent. Draw the structure of ethyl acetate.
30. A lactone is an ester that has its ester functional group in a ring. Draw the structure of the smallest possible lactone (which is called acetolactone, which might give you a hint about its structure).
31. Draw the structure of diethyl ether, once used as an anesthetic.
32. The smallest cyclic ether is called an epoxide. Draw its structure.
33. The odor of fish is caused by the release of small amine molecules, which vaporize easily and are detected by the nose. Lemon juice contains acids that react with the amines and make them not as easily vaporized, which is one reason why adding lemon juice to seafood is so popular. Write the chemical reaction of HCl with trimethylamine, an amine that is given off by seafood.
34. Putrescine and cadaverine are molecules with two amine groups on the opposite ends of a butane backbone and a pentane backbone, respectively. They are both emitted by rotting corpses. Draw their structures and determine their molecular formulas.
35. With four monomers, draw two possible structures of a copolymer composed of ethylene and propylene.
36. With four monomers, draw two possible structures of a copolymer composed of ethylene and styrene.
37. Draw the silicone that can be made from this monomer:
38. One of the ingredients in the original Silly Putty was a silicone polymer with two methyl groups on each Si atom. Draw this silicone.
Answers
1. cyclopropane, cyclobutane, and cyclopentane
2.
3.
4.
5.
6. Both molecular formulas are C4H6.
7.
8. C2H6 + 6Cl2 → C2Cl6 + 6HCl
9.
10.
11. two
12.
13.
14.
15.
16. ethyne
17.
18. The names are 1,2-dihydroxyethane and 1,1-dihydroxyethane, respectively.
19.
20.
21.
22.
23. (CH3)3N + HCl → (CH3)3NHCl
24.
25. (answers will vary)
• | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/16%3A_Organic_Chemistry/16.E%3A_Organic_Chemistry_%28Exercises%29.txt |
Chemistry is the study of matter and the ways in which different forms of matter combine with each other. You study chemistry because it helps you to understand the world around you. Everything you touch or taste or smell is a chemical, and the interactions of these chemicals with each other define our universe. Chemistry forms the fundamental basis for biology and medicine. From the structure of proteins and nucleic acids, to the design, synthesis and manufacture of drugs, chemistry allows you an insight into how things work. Chapter One in this text will introduce you to matter, atoms and their structure. You will learn the basics of scientific measurement and you will gain an appreciation of the scale of chemistry; from the tiniest atom to the incredibly large numbers dealt with in the “mole concept” (Chapter 4). Chapter One lays the foundation on which we will build our understanding.
Thumbnail: Two small test tubes held in spring clamps. (CC BY-SA 3.0 Unported; Amitchell125 via Wikipedia)
01: Measurements and Atomic Structure
Chemistry is the branch of science dealing with the structure, composition, properties, and the reactive characteristics of matter. Matter is anything that has mass and occupies space. Thus, chemistry is the study of literally everything around us – the liquids that we drink, the gasses we breathe, the composition of everything from the plastic case on your phone to the earth beneath your feet. Moreover, chemistry is the study of the transformation of matter. Crude oil is transformed into more useful petroleum products such as gasoline and kerosene by the process of refining. Some of these products are further transformed into plastics. Crude metal ores are transformed into metals, that can then be fashioned into everything from foil to automobiles. Potential drugs are identified from natural sources, isolated and then prepared in the laboratory. Their structures are systematically modified to produce the pharmaceuticals that have led to vast advances in modern medicine. Chemistry is at the center of all of these processes and chemists are the people that study the nature of matter and learn to design, predict and control these chemical transformations. Within the branches of chemistry you will find several apparent subdivisions. Inorganic chemistry, historically, focused on minerals and metals found in the earth, while organic chemistry dealt with carbon-containing compounds that were first identified in living things. Biochemistry is an outgrowth of the application of organic chemistry to biology and relates to the chemical basis for living things. In the later chapters of this text we will explore organic and biochemistry in a bit more detail and you will notice examples of organic compounds scattered throughout the text. Today, the lines between the various fields have blurred significantly and a contemporary chemist is expected to have a broad background in all of these areas.
In this chapter we will discuss some of the properties of matter, how chemists measure those properties and we will introduce some of the vocabulary that is used throughout chemistry and the other physical sciences.
Let’s begin with matter. Matter is defined as any substance that has mass. It’s important to distinguish here between weight and mass. Weight is the result of the pull of gravity on an object. On the Moon, an object will weigh less than the same object on Earth because the pull of gravity is less on the Moon. The mass of an object, however, is an inherent property of that object and does not change, regardless of location, gravitational pull, or whatever. It is a property that is solely dependent on the quantity of matter within the object.
Contemporary theories suggests that matter is composed of atoms. Atoms themselves are constructed from neutrons, protons and electrons, along with an ever-increasing array of other subatomic particles. We will focus on the neutron, a particle having no charge, the proton, which carries a positive charge, and the electron, which has a negative charge. Atoms are incredibly small. To give you an idea of the size of an atom, a single copper penny contains approximately 28,000,000,000,000,000,000,000 atoms (that’s 28 sextillion). Because atoms and subatomic particles are so small, their mass is not readily measured using pounds, ounces, grams or any other scale that we would use on larger objects. Instead, the mass of atoms and subatomic particles is measured using atomic mass units (abbreviated amu). The atomic mass unit is based on a scale that relates the mass of different types of atoms to each other (using the most common form of the element carbon as a standard). The amu scale gives us a convenient means to describe the masses of individual atoms and to do quantitative measurements concerning atoms and their reactions. Within an atom, the neutron and proton both have a mass of one amu; the electron has a much smaller mass (about 0.0005 amu).
Figure 1.2: Atoms are incredible small. Atoms are incredibly small. To give you an idea of the size of an atom, a single copper penny contains approximately 28,000,000,000,000,000,000,000 atoms (that’s 28 sextillion).
Atomic theory places the neutron and the proton in the center of the atom in the nucleus. In an atom, the nucleus is very small, very dense, carries a positive charge (from the protons) and contains virtually all of the mass of the atom. Electrons are placed in a diffuse cloud surrounding the nucleus. The electron cloud carries a net negative charge (from the charge on the electrons) and in a neutral atom there are always as many electrons in this cloud as there are protons in the nucleus (the positive charges in the nucleus are balanced by the negative charges of the electrons, making the atom neutral).
An atom is characterized by the number of neutrons, protons and electrons that it possesses. Today, we recognize at least 116 different types of atoms, each type having a different number of protons in its nucleus. These different types of atoms are called elements. The neutral element hydrogen (the lightest element) will always have one proton in its nucleus and one electron in the cloud surrounding the nucleus. The element helium will always have two protons in its nucleus. It is the number of protons in the nucleus of an atom that defines the identity of an element. Elements can, however, have differing numbers of neutrons in their nucleus. For example, stable helium nuclei exists that contain one, or two neutrons (but they all have two protons). These different types of helium atoms have different masses (3 or 4 amu) and they are called isotopes. For any given isotope, the sum of the numbers of protons and neutrons in the nucleus is called the mass number. All elements exist as a collection of isotopes, and the mass of an element that we use in chemistry, the atomic mass, is the average of the masses of these isotopes. For helium, there is approximately one isotope of Helium-3 for every million isotopes of Helium-4, hence the average atomic mass is very close to 4 (4.002602).
As different elements were discovered and named, abbreviations of their names were developed to allow for a convenient chemical shorthand. The abbreviation for an element is called its chemical symbol. A chemical symbol consists of one or two letters, and the relationship between the symbol and the name of the element is generally apparent. Thus helium has the chemical symbol He, nitrogen is N, and lithium is Li. Sometimes the symbol is less apparent but is decipherable; magnesium is Mg, strontium is Sr, and manganese is Mn. Symbols for elements that have been known since ancient times, however, are often based on Latin or Greek names and appear somewhat obscure from their modern English names. For example, copper is Cu (from cuprum), silver is Ag (from argentum), gold is Au (from aurum), and iron is (Fe from ferrum). Throughout your study of chemistry, you will routinely use chemical symbols and it is important that you begin the process of learning the names and chemical symbols for the common elements. By the time you complete General Chemistry, you will find that you are adept at naming and identifying virtually all of the 116 known elements. Table 1.1 contains a starter list of common elements that you should begin learning now!
Table 1.1: Names and Chemical Symbols for Common Elements
Element Chemical Symbol Element Chemical Symbol
Hydrogen H Phosphorus P
Helium He Sulfur S
Lithium Li Chlorine Cl
Beryllium Be Argon Ar
Boron B Potassium K
Carbon C Calcium Ca
Nitrogen N Iron Fe
Oxygen O Copper Cu
Fluorine F Zinc Zn
Neon Ne Bromine Br
Sodium Na Silver Ag
Magnesium Mg Iodine I
Aluminum Al Gold Au
Silicon Si Lead Pb
The chemical symbol for an element is often combined with information regarding the number of protons and neutrons in a particular isotope of that atom to give the atomic symbol. To write an atomic symbol, you begin with the chemical symbol, then write the atomic number for the element (the number of protons in the nucleus) as a subscript, preceding the chemical symbol. Directly above this, as a superscript, now write the mass number for the isotope, that is, the total number of protons and neutrons in the nucleus. Thus, for helium, the atomic number is 2 and there are two neutrons in the nucleus for the most common isotope making the atomic symbol ${\displaystyle {}_{2}^{4}{\text{He}}}$. In the definition of the atomic mass unit, the “most common isotope of carbon”, ${\displaystyle {}_{6}^{12}{\text{C}}}$, is defined as having a mass of exactly 12 amu and the atomic masses of the remaining elements are based on their masses relative to this isotope. Chlorine (chemical symbol Cl) consists of two major isotopes, one with 18 neutrons (the most common, comprising 75.77% of natural chlorine atoms) and one with 20 neutrons (the remaining 24.23%). The atomic number of chlorine is 17 (it has 17 protons in its nucleus), therefore the chemical symbols for the two isotopes are ${\displaystyle {}_{17}^{35}{\text{Cl}}}$ and ${\displaystyle {}_{17}^{37}{\text{Cl}}}$.
When data is available regarding the natural abundance of various isotopes of an element, it is simple to calculate the average atomic mass. In the example above, ${\displaystyle {}_{17}^{35}{\text{Cl}}}$ was the most common isotope with an abundance of 75.77% and ${\displaystyle {}_{17}^{37}{\text{Cl}}}$ had an abundance of the remaining 24.23%. To calculate the average mass, first convert the percentages into fractions; that is, simply divide them by 100. Now, chlorine-35 represents a fraction of natural chlorine of 0.7577 and has a mass of 35 (the mass number). Multiplying these, we get (0.7577 × 35) = 26.51. To this, we need to add the fraction representing chlorine-37, or (0.2423 × 37) = 8.965; adding, (26.51 + 8.965) = 35.48, which is the weighted average atomic mass for chlorine. Whenever we do mass calculations involving elements or compounds (combinations of elements), we always need to use average atomic masses. | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/01%3A_Measurements_and_Atomic_Structure/1.1%3A_Why_Study_Chemistry.txt |
The number of protons in the nucleus of an element is called the atomic number of that element. Chemists typically place elements in order of increasing atomic numbers in a special arrangement that is called the periodic table.
The periodic table is not simply a grid of elements arranged numerically. In the periodic table, the elements are arranged in horizontal rows called periods (numbered in blue) and vertically into columns called groups. These groups are numbered by two, somewhat conflicting, schemes. In the simplest presentation, favored by the International Union of Pure and Applied Chemistry (IUPAC), the groups are simply numbered 1-18. The convention in much of the world, however, is to number the first two groups 1A and 2A, the last six groups 3A-8A; the middle ten groups are then numbered 1B-8B (but not in that order!). While the IUPAC numbering appears much simpler, in this text we will use the current USA nomenclature (1A-8A). The reason for this choice will become more apparent in Chapter 3 when we discuss “valence” and electron configuration in more detail. The actual layout of the periodic table is based on the grouping of the elements according to chemical properties. For example, elements in each Group of the periodic table (each vertical column) will share many of the same chemical properties. As we discuss the properties of elements and the ways they combine with other elements, the reasons for this particular arrangement of the periodic table will become more obvious.
As you can see, each element in the periodic table is represented by a box containing the chemical symbol, the atomic number (the number of protons in the nucleus) and the atomic mass of the element. Remember that the atomic mass is the weighted average of the masses of all of the natural isotopes of the particular element.
Periodic tables are often colored, or shaded, to distinguish groups of elements that have similar properties or chemical reactivity. The broadest classification is into metals, metalloids (or semi-metals) and nonmetals. The elements in Groups 1A – 8A are called the representative elements and the elements in Groups 3 - 15 are called the transition metals. the metallic elements are shown in purple. Metals are solids (except for mercury), can conduct electricity and are usually malleable (can be rolled into sheets) and are ductile (can be drawn into wires). Metals are usually separated into the main group metals (the elements colored purple in Groups 1A - 5A) and the transition metals (in Groups 3 – 15). Nonmetals (yellow in the Figure) do not conduct electricity (with the exception of carbon in the form of graphite) and have a variety of physical states (some are solids, some liquids and some gasses). Two important subclasses of nonmetals are the halogens (Group 7A) and the inert gasses (or noble gasses; Group 8A). At the border between metals and nonmetals lie the elements boron, silicon, germanium, astatine, antimony and tellurium. These elements share physical properties of metals and nonmetals and are called metalloids, or semi-metals. The common semiconductors silicon and germanium are in this group and it is their unique electrical properties that make transistors and other solid-state devices possible. Later in this book we will see that the position of elements in the periodic table also correlates with their chemical reactivity. | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/01%3A_Measurements_and_Atomic_Structure/1.2%3A_Organization_of_the_Elements_-_The_Periodic_Table.txt |
In Section 1.1, we stated that a single copper penny contains approximately 28,000,000,000,000,000,000,000 atoms. This is a huge number. If we were to measure the diameter of an atom of hydrogen, it would be about 0.00000000000026 inches across. This is an incredibly small number. Chemists routinely use very large and very small numbers in calculations. In order to allow us to use this range of numbers efficiently, chemists will generally express numbers using exponential, or scientific notation. In scientific notation, a number n is shown as the product of that number and 10, raised to some exponent x; that is, (n × 10x). The number 102 is equal to 100. If we multiply 2 × 102, that is equivalent to multiplying 2 × 100, or 200. Thus 200 can be written in scientific notation as 2 × 102. When we convert a number to scientific notation, we begin by writing a the first (non-zero) digit in the number. If the number contains more than one digit, we write a decimal point, followed by all of the remaining digits. Next we inspect the number to see what power of 10 this decimal must be multiplied by to give the original number. Operationally, what you are doing is moving decimal places. Take the number of atoms in a penny, 28,000,000,000,000,000,000,000. We would begin by writing 2.8. To get the power of 10 that we need, we begin with the last digit in the number and count the number of places that we must move to the left to reach our new decimal point. In this example, we must move 22 places to the left. The number is therefore the product of 2.8 and 1022, and the number is written in scientific notation as 2.8 × 1022.
Let’s look at a very small number; for example, 0.00000000000026 inches, the diameter of a hydrogen atom. We want to place our decimal point between the two and the six. To do this, we have to move the decimal point in our number to the right thirteen places. When you are converting a number to scientific notation and you move the decimal point to the right, the power of 10 must have a negative exponent. Thus our number would be written $2.6 \times 10^{-13}$ inches. A series of numbers in decimal format and in scientific notation are shown in Table $1$ below.
Table $1$: Examples of Numbers in Decimal Format and in Scientific Notation
Decimal Format Scientific Notation
274 2.74 × 102
0.0035 3.5 × 10–3
60221415 6.0221415 × 107
0.125 1.25 × 10–1
402.5 4.025 × 102
0.0002001 2.001 × 10–4
10,000 1 × 104
Exercise $1$
Convert the following numbers into scientific notation:
1. 93,000,000
2. 708,010
3. 0.000248
4. 800.0
Exercise $2$
Convert the following numbers from scientific notation into decimal format:
1. 6.02 × 104
2. 6.00 × 10-4
3. 4.68 × 10-2
4. 9.3 × 107
1.4: SI and Metric Units
Within the sciences, we use the system of weights and measures that are defined by the International System of Units which are generally referred to as SI Units. At the heart of the SI system is a short list of base units defined in an absolute way without referring to any other units. The base units that we will use in this text, and later in General Chemistry include the meter (m) for distance, the kilogram (kg) for mass and the second (s) for time. The volume of a substance is a derived unit based on the meter, and a cubic meter (m3) is defined as the volume of a cube that is exactly 1 meter on all edges.
Because most laboratory work that takes place in chemistry is on a relatively small scale, the mass of a kilogram (about 2.2 pounds) is too large to be convenient and the gram is generally utilized, where a gram (g) is defined as 1/1000 kilograms. Likewise, a volume of one cubic meter is too large to be practical in the laboratory and it is common to use the cubic centimeter to describe volume. A cubic centimeter is a cube that is 1/100 meter on each edge, a teaspoon holds approximately 5 cubic centimeters. For liquids and gasses, chemists will usually describe volume using the liter, where a liter (L) is defined as 1000 cubic centimeters.
SI base units are typically represented using the abbreviation for the unit itself, preceded by a metric prefix, where the metric prefix represents the power of 10 that the base unit is multiplied by. The set of common metric prefixes are shown in Table \(1\).
Table \(1\): Common Metric Prefixes
Factor Name Symbol
10-12 pico p
10-9 nano n
10-6 micro µ
10-3 milli m
10-2 centi c
10-1 deci d
1 none
103 kilo k
106 mega M
109 giga G
Using this Table as a reference, we see the metric symbol “c” represents the factor 10-2; thus writing “cm” is equivalent to writing (10–2 × m). Likewise, we could describe 1/1000 of a meter as mm, where the metric symbol “m” represents the factor 10-3. The set of metric prefixes and their symbols that are shown in Table 1.3 are widely used in chemistry and it is important that you memorize them and become adept at relating the prefix (and its’ symbol) to the corresponding factor of 10. | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/01%3A_Measurements_and_Atomic_Structure/1.3%3A_Scientific_Notation.txt |
Because chemists often deal with measurements that are both very small (as in the size of an atom) and very large (as in numbers of atoms), it is often necessary to convert between units of metric measurement. For example, a mass measured in grams may be more convenient to work with if it was expressed in mg (10–3 × g). Converting between metric units is an exercise in unit analysis (also called dimensional analysis). Unit analysis is a form of proportional reasoning where a given measurement can be multiplied by a known proportion or ratio to give a result having a different unit, or dimension. For example, if you had a sample of a substance with a mass of 0.0034 grams and you wished to express that mass in mg you could use the following unit analysis:
The given quantity in this example is the mass of 0.0034 grams. The quantity that you want to find is the mass in mg, and the known proportion or ratio is given by the definition of the metric prefix, that is one mg is equal to 10-3 grams. Expressing this as a proportion or ratio, you could say there is one mg per 10-3 grams, or:
Looking at this expression, the numerator, 1 mg, is equivalent to saying 1 × 10–3 g, which is identical to the value in the denominator. This ratio, therefore has a numeric value of one (anything divided by itself is one, by definition). Algebraically, we know that we are allowed to multiply any number by one and that number will be unchanged. If, however, the number has units, and we multiply it by a ratio containing units, the units in the number will multiply and divide by the units of the ratio, giving the original number (remember you are multiplying by one) but with different units. In the present case, if we multiply the given by the known ratio, the unit “g” will cancel, leaving “mg” as the only remaining unit. The original number in grams has therefore been converted to milligrams, the units that you wanted to find.
The method that we used to solve this problem can be generalized as: given × known ratio = find. The given is a numerical quantity (with its units), the known ratio is based on the metric prefixes and is set up so that the units in the denominator of the ratio match the units of given and the units in the numerator match those in find. When these are multiplied, the number from given will now have the units of find. In the ratio used in the example, “g” (the units of given) appear in the denominator and “mg” (the units of find) appear in the numerator.
As an example of a case where the units of the known ratio must be inverted, if you wanted to convert 1.3 × 107 µg into grams, the given would be 1.3 × 107 µg, the find would be grams and the known ratio would be based on the definition of µg as one µg per 10-6 grams. This ratio must be expressed in the solution with µg (the units of given) in the denominator and g (the units of find) in the numerator.
Note that instead of “one µg per 10-6 grams”, we must invert the known ratio and state it as either “10-6 grams per 1 µg” so that the units of given (µg) will cancel. We can do this inversion because the ratio still has a numeric value of one. Simple ratios like these can also be used to convert English measurements in to their metric equivalents. The ratio relating inches to meters is .
Exercise $1$
Convert the following metric measurements into the indicated units:
1. 9.3 × 10-4 g into ng
2. 278 g into mg
Exercise $2$
Convert the following metric measurements into the indicated units:
1. 2,057 grams - as kg
2. 1.25 × 10-7 meters - as µm
3. 6.58 × 104 meters - as km
4. 2.78 × 10-1 grams - as mg
In the examples we have done thus far, we have been able to write a known ratio based on the definition of the appropriate metric prefix. But what if we wanted to take a number that was expressed in milligrams and convert it to a number with the units of nanograms? In a case like this, we need to use two known ratios in sequence; the first with the units of given (mg) in the denominator and the second with the units of find (ng) in the numerator. For example, if we were given 0.00602 mg and asked to find ng, we could set up a ratio based on grams per mg. If we solved the problem at this point, we would have a result with the units of grams. To get a final answer in terms of ng, we would need to multiply this intermediate result (the new given) by a ratio based on nanograms per gram.
In the first two terms, the units of “mg” cancel and in the second two terms, “g” cancels leaving only “ng”, the units of find. One of the reassuring pleasures of doing these types of problems is that, if you set up your problem and the units cancel, leaving only the units of find, you know you have set up the problem correctly! All you have to do is to do the sequential calculations and you know your answer is correct!
Exercise $3$
Convert the following metric measurements into the indicated units:
1. 2,057 mg - into kg
2. 1.25 × 10-7 km - into µm
3. 9.3 × 10-4 pg - into ng
4. 6.5 × 104 mm - into km | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/01%3A_Measurements_and_Atomic_Structure/1.5%3A_Unit_Conversion_with_the_Metric_System.txt |
Experimental work in scientific laboratories will generally involve measurement. Whenever we make a measurement, we always strive to make our value as accurate as possible.
The distance is more than 50 mm, the last numbered digit shown before the second arrow. If we look very carefully, we see that the second arrow is about half-way between the fourth and fifth division following the 50 mm mark. The measurement is therefore greater than 54 mm and less than 55 mm and is about 54.5 mm. The last digit in our measurement is estimated, but the first two digits are exact. In any measurement, like this, the last digit that you report is always an estimated digit. If we were to say that the measurement was 54 mm, that would be incorrect, because we know it’s larger. If we were to say the measurement was 54.5567 mm, that would be nonsense because our scale does not show that degree of accuracy. In a measurement in science, the estimated digit is called the least significant digit, and the total number of exact digits plus the estimated digit is called the number of significant figures in the measurement. Thus the measurement in the figure, 54.5, has three significant figures (3 SF). By adhering to this rule, we can look at any measured value and immediately know the accuracy of the measurement that was done. In order to properly interpret the number of significant figures in a measurement, we have to know how to interpret measurements containing zeros. For example, an object is found to have a mass of 602 mg. The last digit (the 2) is estimated and the first two digits are exact. The measurement therefore is accurate to three significant figures. We could also express this measurement in grams using the metric conversion ratio , making the measurement 0.00602 g. We now have three additional digits in our number (called leading zeros), but is our number any more accurate? No; in a measurement, leading zeros (zeros that appear before the number) are never significant.
Let’s consider another measurement; we are told that a distance is 1700 m. The first thing to notice is that this number does not have a decimal point. What this tells us is that the estimated digit in this number is the 7, and that this number only has two significant figures. The last two zeros in this measurement are called trailing zeros; in numbers without a decimal point, trailing zeros are never significant. If, however, the distance was reported as 1700.00 m, the presence of the decimal point would imply that the last zero was the estimated digit (zeros can be estimated too) and this number would have six significant figures. Stated as a rule, in a number containing a decimal point, trailing zeros are always significant. These simple rules for interpreting zeros in measurements are collected below:
Rules for Handling Zeroes when interpreting Significant Figures
In numbers with a decimal point
• leading zeroes are never significant
• trailing zeroes are always significant
In numbers without a decimal point
• trailing zeroes are never significant
In all numbers
• zeroes which appear between non-zero digits are always significant
Applying these rules to some examples:
• 117.880 m contains six significant figures; the number has a decimal point, so the trailing zero is significant.
• 0.002240 g contains four significant figures; the number has a decimal point so the trailing zero is significant, but the leading zeros are not.
• 1,000,100 contains five significant figures; the number does not have a decimal point, so the trailing zeros are not significant. The zeros between the first and fifth digits, however, are significant.
• 6.022 × 1023 contains four significant figures. In scientific notation, all of the significant figures in a measurement are shown before the exponent. (Remember this when you are converting measurements into scientific notation.)
Exercise $1$
Determine the number of significant figures in each of the following numbers:
1. 2,057,000
2. 1.250600
3. 9.300 × 10-4
4. 6.05 × 104 | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/01%3A_Measurements_and_Atomic_Structure/1.6%3A_Significant_Figures.txt |
As we learned in Section 1.1, modern atomic theory places protons and neutrons in the nucleus of an atom and electrons are placed in a diffuse cloud surrounding this nucleus. As chemists and physicists began examining the structure of atoms, however, it became apparent that all of the electrons in atoms were not equivalent. The electrons were not randomly placed in one massive “cloud”, rather they seemed to be arranged in distinct energy levels and energy was required to move electrons from a lower to a higher energy level. A mathematical model of atomic structure was developed in the early nineteenth century that defined these energy levels as quantum levels, and today this description is generally referred to as quantum mechanics.
According to the quantum model of the atom, electron for the known elements can reside in seven different quantum levels, denoted by the principal quantum number n, where n has a value of one to seven. As the quantum number increases, the average energy of the electrons having that quantum number also increases. Each of the seven rows in the periodic table corresponds to a different quantum number. The first row (n = 1) can only accommodate two electrons. Thus an element in the first row of the periodic table can have no more than two electrons (hydrogen has one, and helium has two). The second row (n = 2) can accommodate eight electrons and an element in the second row of the periodic table will have two electrons in the first level (it is full) and up to eight electrons in the second level.
Quantum theory also tells us that the electrons in a given energy level are not all equivalent. Within an energy level electrons reside within sublevels (or subshells). The sublevels for any given level are identified by the letters, s, p, d and f and the total number of sublevels is also given by the quantum number, n. The s sublevel can accommodate two electrons, the p holds six, the d holds 10 and the f can hold 14. The elements in the first row in the periodic table (n = 1) have electrons only in the 1s sublevel (n = 1, therefore there can only be one sublevel). The single electron in hydrogen would be identified as 1s1 and the two electrons in helium would be identified as 1s2. Fluorine, in the second row of the periodic table (n = 2), has an atomic number of nine and therefore has nine electrons. The electrons in fluorine are arranged, two in the first level (1s2), two in the 2s suborbital (2s2) and five 2p suborbital (2p5). If we were to write the electron configuration for fluorine, we would write it as 1s2 2s2 2p5. Each of the sublevels in an atom is also associated with an orbital, where an orbital is simply a region of space where the electron is likely to be found.
1.8: Filling Orbitals with Electrons
As stated above, an s sublevel can accommodate two electrons, the p accommodates six, there can be 10 in the d sublevel and 14 in the f. Although there are two electrons in the s sublevel, these electrons are not identical; they differ in the quantum property known as spin. As a simple device to illustrate this, the electrons within a suborbital are often represented as arrows pointing up or down, graphically representing opposite spin axes (↑ and ↓). Electrons are added to sublevels according to Hund’s rules which state that every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied, and all electrons in singly occupied orbitals have the same spin. When a subshell is doubly occupied, the electrons have opposite spins.
For example, carbon has a filled 1s sublevel, a filled 2s sublevel and two electrons in the 2p sublevel (2p2).
The electron configuration for fluorine is 1s2 2s2 2p5.
This sequence continues nicely until the third period; it turns out the 3d orbitals are slightly higher in energy than the 4s orbital, therefore the 4s fills with two electrons, and then the next 10 electrons are placed in the 3d orbital. This is a general trend in the periodic table, and the order of filling can be easily predicted by the scheme where you simply follow the arrows on the diagonal to determine the next orbital to fill.
One of the shortcuts that is often used when writing electron configuration is to show “core” electrons simply as the inert gas from the preceding period. For example, fluorine is in the second period (n = 2). That means that the orbitals associated with the first period are already filled, just like they are in the inert gas, helium (He). Therefore, instead of writing the configuration for fluorine as we did above, we can replace the 1s2 with the “helium core”.
Calcium is in the fourth period and in Group 2. That means that the first three quantum levels are filled (n = 1, 2 and 3) just like they are in argon.
Exercise \(1\): Electron Configurations
1. Write the complete electron configurations for the elements beryllium and carbon.
2. Identify the elements corresponding to the following electron configurations:
• 1s2 2s1 and 1s2 2s2 2p6. | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/01%3A_Measurements_and_Atomic_Structure/1.7%3A_Atomic_Structure_and_Electron_Configuration.txt |
• Matter is defined as any substance that has mass. Matter is composed of atoms. that are constructed primarily from neutrons, protons and electrons. Neutrons have no charge, protons, carry a positive charge, and electrons, have a negative charge.
• The mass of atoms and subatomic particles is measured using atomic mass units (abbreviated amu); protons and neutrons have a mass of one amu, and the mass of an electron is negligible.
• The neutron and the proton are in the center of the atom in the nucleus. Virtually all of the mass of the atom resides in the nucleus. Electrons are placed in a diffuse cloud surrounding the nucleus.
• The electron cloud carries a net negative charge and in a neutral atom there are always as many electrons in this cloud as there are protons in the nucleus.
• The identity of an atom is defined by the number of protons in its nucleus; each unique type of atom is called an element. Elements with the same number of protons, but differing numbers of neutrons in their nucleus are called isotopes. The atomic mass of an element is the weighted average of the masses each of these isotopes.
• Each element is referred to using its chemical symbol, which is an abbreviation of its name (many symbols are based on Latin or Greek names).
• The atomic symbol for an element consists of the chemical symbol with the atomic number for the element as a subscript, preceding the chemical symbol, and directly above this, a superscript showing the mass number for the particular isotope of the element.
• The average atomic mass for an element can be calculated as the sum of the fraction of each isotope within the natural abundance, multiplied by the mass number of that isotope; or, average atomic mass = f1M1 + f2M2 + f3M3
• The number of protons in the nucleus of an element is called the atomic number of that element. Elements are typically arranged in order of increasing atomic numbers in the periodic table. In the periodic table, horizontal rows are called periods and vertical columns are called groups.
• Typically in the sciences, very large or very small numbers are shown using scientific notation (exponential notation) where a number n is shown as the product of that number and 10, raised to some exponent x; that is, (n × 10x).
• In the SI (or metric) system, the unit for distance is the meter (m), kilogram (kg) is used for mass and second (s) for time. The volume of a substance is a derived unit based on the meter, and a cubic meter (m3) is defined as the volume of a cube that is exactly 1 meter on all edges. Typically, in the laboratory, mass is expressed in grams (g) (1/1000 of a kilogram) and the cubic centimeter (cc) is to describe volume. A cubic centimeter is a cube that is 1/100 meter on each edge. For liquids and gasses, volume is usually described using the liter, where a liter (L) is defined as 1000 cubic centimeters.
• SI base units are typically represented using the abbreviation for the unit itself, preceded by a metric prefix, where the metric prefix represents the power of 10 that the base unit is multiplied by.
• When converting between metric units, a simple algorithm involves taking a given measurement and multiplying it by a known proportion or ratio to give a result having the metric unit, or dimension, that you were trying to find.
• In a measurement in science, the last digit that is reported is estimated, and this digit is called the least significant digit; this, along with the total number of exact digits plus the estimated digit is called the number of significant figures in the measurement. When identifying the number of significant figures in a measurement, all leading zeros are excluded. Zeros that are surrounded by non-zero digits are included, and, for numbers with a decimal point, trailing zeros are also included. If a number does not have a decimal point, trailing zeros are not included. A number written in scientific notation includes all significant digits in n; (n × 10x).
• According to the quantum model of the atom, electrons reside in seven different quantum levels, denoted by the principal quantum number n, where n has a value of one to seven, corresponding to the seven rows in the periodic table. The first row (n = 1) can accommodate two electrons; the second row (n = 2) can accommodate eight electrons; the third row (n = 3), eighteen, up to a maximum of 2n2 for the known elements.
• Quantum theory also tells us that the electrons in a given energy level reside within sublevels (or subshells). The sublevels for any given level are identified by the letters, s, p, d and f and the quantum number for the level, written as 1s2 2s2 2p5, etc. Each of the sublevels is also associated with an orbital, where an orbital is simply a region of space where the electron is likely to be found.
• When adding electrons to sublevels, Hund’s rules state that every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied, and all electrons in singly occupied orbitals have the same spin (shown using “up and down” arrows). Electrons are added in order of increasing energy of the sublevel, not necessarily in numeric order. | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/01%3A_Measurements_and_Atomic_Structure/1.S%3A_Measurements_and_Atomic_Structure_%28Summary%29.txt |
In chapter one, we defined matter as anything that has mass and occupies space. In this chapter, we will look more closely at matter and break it into sub-classes including pure substances and mixtures. We will look at the physical state that matter can adopt (solids, liquids, and gasses) And we will learned how to characterized these using intensive properties such as density. Finally, we will look at changes in these properties and define them in terms of simple physical changes (such as share of physical state) And chemical changes, where one or more substances combine to make new substances.
• 2.1: Pure Substances and Mixtures
A compound is a substance that results from the combination of two or more elements in a constant ratio. The chemical formula of which tells us the elements that are present and the ratio of the elements to each other. In a molecule, the atoms are not only bonded together in a constant ratio, but they are bonded in a specific geometric arrangement as well.
• 2.2: The States of Matter
At low temperature, water exists as a solid (ice). As the temperature increases, water exists as a liquid, and at high temperature, as water vapor, a gas.These three forms of water represent the three states of matter: solids, liquids and gases. States of matter are examples of physical properties of a substance. Other physical properties include appearance (shiny, dull, smooth, rough), odor, electrical conductivity, thermal conductivity, hardness and density, to name just a few.
• 2.3: Density, Proportion and Dimensional Analysis
The physical state of a substance at under a defined set of conditions (like temperature and pressure) is an intensive property of a substance. An intensive property is defined as a property that is inherent to the substance and is not dependent on the sample size. Density, the mass-to-volume ratio of a substance, is another example of an intensive property.
• 2.4: Chemical and Physical Properties and Changes
Changes in outward appearances that do not alter the chemical nature of the substance and make no new substance are called physical changes. A chemical change has occurred when the chemical composition of a material changes and a new substance is produced. Chemical properties are simply the set of chemical changes that are possible for that substance.
• 2.5: Conservation of Mass
The law of mass conservation states that there is no detectable change in the total mass of materials when they react chemically to form new materials.
• 2.S: The Physical and Chemical Properties of Matter (Summary)
02: The Physical and Chemical Properties of Matter
In Chapter 1, we learned that atoms are composed of electrons, protons and neutrons and that the number of protons in the nucleus of an atom (the atomic number) defines the identity of that element. For example, an atom with six protons in its nucleus is a carbon atom; seven protons makes it nitrogen; eight protons makes it oxygen, and so on. The periodic table organizes these elements by atomic number and there are currently over 118 known elements.
Because there are clearly more than 118 different types of substances in the world around us, we can see that most substances that we encounter are not pure elements, but are composed of different elements combined together. In chemistry, we refer to these as compounds, which we define as a substance that results from the combination of two or more elements in a constant ratio. For example, water is a compound composed of two hydrogen atoms bonded to one oxygen atom. We can show the ratio of hydrogen to oxygen in this compound by using subscripts on the chemical symbols for each element. Thus, water (two hydrogens and one oxygen) can be written as H2O. This shorthand notation for water is called a chemical formula. For any compound, the chemical formula tells us the elements that are present and the ratio of the elements to each other. Later we will see that water is a member of a special sub-type of compound, called a molecular compound. In a molecule, the atoms are not only bonded together in a constant ratio, but they are bonded in a specific geometric arrangement as well. In the following chapter, we will look more closely at how elements are bonded together in compounds, but first we will examine some of the properties of chemical substances.
When we speak of a pure substance, we are speaking of something that contains only one kind of matter. This can either be one single element or one single compound, but every sample of this substance that you examine must contain exactly the same thing with a fixed, definite set of properties. If we take two or more pure substances and mix them together, we refer to this as a mixture. Mixtures can always be separated again into component pure substances,because bonding among the atoms of the constituent substances does not occur in a mixture. Whereas a compound may have very different properties from the elements that compose it, in mixtures the substances keep their individual properties. For example sodium is a soft shiny metal and chlorine is a pungent green gas. These two elements can 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 types, based on the uniformity of their composition. 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 one can vary the amounts of each component. 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 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. | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/02%3A_The_Physical_and_Chemical_Properties_of_Matter/2.1%3A_Pure_Substances_and_Mixtures.txt |
As described in Section 2.1, a molecule of water is composed of two atoms of hydrogen bonded to one atom of oxygen(H2O). All water molecules are exactly the same (same ratio of elements, same geometric bonding pattern), but we encounter water in three different forms in the world around us. At low temperature, water exists as a solid (ice). As the temperature increases, water exists as a liquid, and at high temperature, as water vapor, a gas. These three forms of water represent the three states of matter: solids, liquids and gases. States of matter are examples of physical properties of a substance. Other physical properties include appearance (shiny, dull, smooth, rough), odor, electrical conductivity, thermal conductivity, hardness and density, to name just a few. We will discuss density in more detail in the next section, but first let’s examine the states of matter and how they differ on an atomic level.
If ice, liquid water and water vapor all consist of identical molecules, then what accounts for the difference in their properties? So far, we have talked about molecules as if they were standing still, but in fact, they are always moving. In chemistry, we often explain the states of matter in terms of the kinetic molecular theory (KMT). The word kinetic refers to motion and the kinetic molecular theory suggests that atoms and molecules are always in motion. The energy associated with this motion is termed kinetic energy. The amount of kinetic energy that a particle has is a direct function of temperature, and it is the kinetic energy of the water molecules under different conditions that determines the different properties of the three states of water.
Atoms and molecules move in different ways under different conditions because of the forces attracting them to each other, called intermolecular forces. Intermolecular forces is a general term describing the fact that all atoms, and molecules share a certain inherent attraction for each other. These attractive forces are much weaker than the bonds that hold molecules together, but in a large cluster of atoms or molecules the sum of all of these attractive forces can be quite significant.
Now, consider a group of molecules or atoms clustered together and held in place by these attractive forces. At low temperature, the molecules or atoms will remain stuck together in a lump of defined shape and structure, like water in the form of an ice cube. This is referred to as the solid phase. At the atomic level, the molecules or atoms in a solid are closely packed, and although they are still all rapidly moving, their movements are so small that they can be thought of as vibrating about a fixed position. As an analogy here, think of a handful of small magnets stuck together in a solid mass. Solids and liquids are the most tightly packed states of matter. Because of the intermolecular forces, solids have a defined shape, which is independent of the container in which they are placed. As energy is added to the system, usually in the form of heat, the individual molecules or atoms acquire enough energy to overcome some of the attractive intermolecular forces between them so that neighboring particles are free to move past or slide over one another. This state of matter is called the liquid phase. As in a solid, in a liquid, the attractive forces are strong enough to hold the molecules or atoms close together so they are not easily compressed and have a definite volume. Unlike in a solid, however, the particles will flow (slide over each other) so that they can assume the shape of their container.
Finally, if enough energy is put into the system, the individual molecules or atoms acquire enough energy to totally break all of the attractive forces between them and they are free to separate and rapidly move throughout the entire volume of their container. This is called the gas phase and atoms or molecules in the gas phase will totally fill whatever container they occupy, taking on the shape and volume of their container. Because there is so much space between the particles in a gas, a gas is highly compressible, which means that the molecules can be forced closer together to fit in a much smaller space. We are all familiar with cylinders of compressed gas, where the compressibility of gasses is exploited to allow a large amount of gas to be transported in a very small space.
Returning to our example of water, at low temperature, water exists as the solid, ice. As the solid is warmed, the water molecules acquire enough energy to overcome the strongest of the attractive forces between them and the ice melts to form liquid water. This transition from the solid phase to the liquid phase happens at a fixed temperature for each substance called the melting point. The melting point of a solid is one of the physical properties of that solid. If we remove energy from the liquid molecules they will slow down enough for the attractive forces to take hold again and a solid will form. The temperature that this happens is called the freezing point and is the same temperature as the melting point.
As more energy is put into the system, the water heats up, the molecules begin moving faster and faster until there is finally enough energy in the system to totally overcome the attractive forces. When this happens, the water molecules are free to fly away from each other, fill whatever container they are occupying and become a gas. The transition from the liquid phase to the gas phase happens at a fixed temperature for each substance and is called the boiling point. Like the melting point, the boiling point is another physical property of a liquid.
Phase transitions for a typical substance can be shown using simple diagram showing the physical states, separated by transitions for melting and boiling points. For example, if you are told that a pure substance is 15˚ C above its boiling point, you can use the diagram to plot the temperature relative to the boiling point. Because you are above the boiling point, the substance will exist in the gas phase.
There are, however, some exceptions to the rules for changes of state that we have just established,. For example, ice is a solid and the molecules in the interior are held together tightly by intermolecular forces. Surface molecules, however, are exposed and they have the opportunity to absorb energy from the environment (think of a patch of snow on a bright sunny day). If some of these surface molecules absorb enough energy, they can break the attractive forces that are holding them and escape as a gas (water vapor) without ever going through the liquid phase. The transition from a solid directly into a gas is called sublimation. The reverse process, a direct transition from a gas to a solid, is called deposition. Perhaps the most common example of a solid that does not melt, but only sublimes, is dry ice (solid carbon dioxide;CO2). This property of dry ice is what makes it a good refrigerant for shipping perishables. It is quite cold, keeping things well frozen, but does not melt into a messy liquid as it warms during shipment.
Just like surface molecules in solids can move directly into the gas phase, surface molecules in liquids also absorb energy from the environment and move into the gas phase, even though the liquid itself is below the boiling point. This is the process of vaporization (evaporation). The reverse process, a transition from a gas to a liquid, is called condensation. Liquid substances undergo vaporization and the space above any liquid has molecules of that substance in the gas state. This is called the vapor pressure of the liquid, and vapor pressure (at a given temperature) is another of the physical properties of liquid substances.
Summarizing what we know about the different states of matter:
In a gas:
• the molecules or atoms are highly separated, making a gas highly compressible,
• attractive forces between the particles are minimal, allowing the gas to take on the shape and volume of its container.
In a liquid:
• the molecules or atoms are closely spaced, making a liquid much less compressible than a gas,
• attractive forces between the particles are intermediate, allowing the molecules or atoms to move past, or slide over one another,
• liquids have a definite volume, but will take on the shape of their container.
In a solid:
• the attractive forces are strong, keeping the atoms or molecules in relatively fixed positions,
• the neighboring atoms or molecules are close together, making the solid not compressible and giving it a definite shape that is independent of the shape and size of its container. | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/02%3A_The_Physical_and_Chemical_Properties_of_Matter/2.2%3A_The_States_of_Matter.txt |
In the previous section, we have learned about the states of matter. The physical state of a substance at under a defined set of conditions (like temperature and pressure) is an intensive property of a substance. An intensive property is defined as a property that is inherent to the substance and is not dependent on the sample size. Density, the mass-to-volume ratio of a substance, is another example of an intensive property.
If you picked up equal sized samples of aluminum and gold, you would immediately notice that one was much heavier than the other. The atomic mass of gold is over seven times greater than the atomic mass of aluminum, so although the two samples are the same size, the lump of gold is significantly more massive than the equally sized lump of aluminum. We can say that gold is more dense than aluminum.
Making this a quantitative measurement, one cubic centimeter of gold has a mass of 19.3 grams (remember that a cubic centimeter is the volume of a cube that is exactly one cm on each side, and it has the units of cm3. We defined density as the mass-to-volume ratio of a substance. For gold, the mass is 19.3 grams and the volume is 1 cm3. The mass-to-volume ratio of gold is and the density (d) of gold is written as d = 19.3 g/cm3.
Returning to our block of aluminum;experimentally, one cubic centimeter of aluminum has a mass of 2.70 grams. The mass-to-volume ratio of aluminum is , and the density of aluminum is therefore 2.70 g/cm3, about 7 times less than that of gold.
Density is a physical property that can be measured for all substances, solids, liquids and gasses. For solids and liquids, density is often reported using the units of g/cm3. Densities of gasses, which are significantly lower than the densities of solids and liquids, are often given using units grams/liter (g/L, remembering from SI and Metric Units" data-cke-saved-href="/Bookshelves/Introductory_Chemistry/Book:_Introductory_Chemistry_Online_(Young)/01:_Measurements_and_Atomic_Structure/1.4:_SI_and_Metric_Units" href="/Bookshelves/Introductory_Chemistry/Book:_Introductory_Chemistry_Online_(Young)/01:_Measurements_and_Atomic_Structure/1.4:_SI_and_Metric_Units" data-quail-id="28">Section 1.4 that a liter is defined as 1000 cm3).
The definition of density that we used previously was the mass-to-volume ratio of a substance. This is also stated as “mass per unit volume”. The word per in this context implies that a mathematical relationship exists between mass and volume. In this case, the relationship is the ratio of mass-to-volume. Whenever two factors can be related by a ratio or fraction we can use unit analysis to solve problems relating those factors. For density, the ratio is mass-to-volume. If a sample of iron has a mass of 23.4 grams and a volume of 3.00 cm3, the density of iron can be calculated as:
$d=\frac{mass}{volume} \nonumber$
$d=\frac{23.4g}{3.00cm^{3}}=7.80\: g/cm^{3} \nonumber$
In this calculation, our two experimental numbers are 23.4 and 3.00. Each of these numbers has three significant figures (remember, the trailing zeros in 3.00 are significant because the number has a decimal point). Our answer must therefore also be accurate to three significant figures, or 7.80.
2.4: Chemical and Physical Properties and Changes
A lump of gold can be hammered into a very thin sheet of gold foil (it is the most malleable of all of the elements). Nonetheless, the gold in the foil sheet is still just elemental gold;nothing has changed except the physical appearance of the sample. The same is true if you take any solid pure substance and melt it, or convert it to a gas. The atomic or molecular structure of the substance has not changed, it simply has a different physical appearance. Changes in outward appearances that do not alter the chemical nature of the substance and make no new substance are called physical changes. Pure carbon, in the form of a briquette, can be smashed to a fine power without changing the fact that it is still just elemental carbon (thus, this is a physical change), but if pure carbon is heated in the presence of oxygen, something else happens. The carbon slowly disappears (often in flames) and the carbon atoms now appear as a compound with oxygen with the formula CO2. Carbon dioxide is a totally different substance than either the carbon or the oxygen that we started with. For example, carbon is a black solid and carbon dioxide is a colorless gas. You know that a chemical change has occurred when the chemical composition of the material changes and a new substance is produced.
Just like we defined a set of physical properties for substances, we can also define a set of chemical properties. Chemical properties are simply the set of chemical changes that are possible for that substance. For the element magnesium (Mg), we could say that chemical properties include:
• the reaction with oxygen to form MgO
• the reaction with hydrochloric acid to form MgCl2 and hydrogen gas (H2)
• the reaction with solid carbon dioxide (dry ice) to form MgO and carbon
Chemical changes can almost always be detected with one of our physical senses. Thus, when magnesium reacts with oxygen (burns in air) a bright white flame is produced, heat is evolved and the shiny metallic magnesium is converted to a crumbly white powder MgO. In the reaction with hydrochloric acid (the molecule HCl dissolved in water), the solid metallic magnesium disappears, bubbles of hydrogen gas (H2) are evolved, heat is produced, and a clear solution containing MgCl2 is formed. In the reaction with solid carbon dioxide (dry ice), a bright white flame is produced, heat is evolved and the shiny metallic magnesium is converted to a crumbly white powder and solid carbon. In general, when you are trying to identify a chemical change, look for evidence of heat or light, the evolution of a gas, a change in color or the formation of new solid products from otherwise clear solutions. | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/02%3A_The_Physical_and_Chemical_Properties_of_Matter/2.3%3A_Density_Proportion_and_Dimensional_Analysis.txt |
When substances undergo chemical changes their physical state is usually dramatically altered. Despite this dramatic change, however, no matter is lost or created. We can show this with the reaction of magnesium metal with oxygen to form magnesium oxide. Instead of burning the magnesium metal openly in the air, if you were to seal the magnesium and air together in a glass vessel, weigh it, heat it to promote reaction, and then weigh the vessel again, you would find that there was no change in total mass. The mass of the product, magnesium oxide, would exactly equal the masses of the substances that reacted (oxygen gas and magnesium metal).
This is analogous to an experiment performed by the French chemist, Lavoisier, in the 1770s in which he heated metallic tin (Sn) with air in a closed vessel. This, and other experiments of the time, provided the data that led to the law of mass conservation. Formally, the law states, there is no detectable change in the total mass of materials when they react chemically to form new materials.
Basically, what the conservation law says is that whenever a chemical change occurs, the total mass of the substances reacting must equal the total mass of the substances that are produced. Sometimes this is stated as mass is conserved or mass is neither created nor destroyed in a chemical reaction. For example, when charcoal is burned in oxygen, the mass of the (charcoal + oxygen) must equal the mass of the (carbon dioxide, water vapor and ash) that is produced. The conservation of mass is one of the fundamental principles on which modern chemistry is based.
2.S: The Physical and Chemical Properties of Matter (Summary)
• A compound is defined as a substance that results from the combination of two or more elements in a constant ratio. In a compound such as water, we show the ratio of the elements (hydrogen and oxygen) by using subscripts on the chemical symbols for each element. Thus, water (two hydrogens and one oxygen) is written using the chemical formula H2O. In a molecule, the atoms are not only bonded together in a constant ratio, but they are also bonded in a specific geometric arrangement.
• A pure substance contains only one kind of matter;it can be a single element or a single chemical compound. Two or more pure substances mixed together constitute a mixture;you can always separate a mixture by simple physical means.
• A heterogeneous mixture is not uniform and different samples of the mixture will have a different compositions. A homogeneous mixture, is uniform and 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.
• Any pure substance, under appropriate conditions, can exist in three different states: solids, liquids and gases. States of matter are examples of physical properties of a substance. Other physical properties include appearance (shiny, dull, smooth, rough), odor, electrical conductivity, thermal conductivity, hardness and density, etc.
• Solids have both a definite shape and volume. Liquids have a definite volume, but take on the shape of their container. Gasses have neither a definite shape nor volume, and both of these are defined by the shape and volume of their container.
• The kinetic molecular theory (KMT) is generally used to explain physical states of matter. The KMT suggests that atoms and molecules are always in motion and are loosely bound to each other by attractive called intermolecular forces. In a solid, the kinetic energy (energy of motion) associated with the atoms or molecules is insufficient to break these forces and the particles are essentially fixed in place, adjacent to each other. In a liquid, there is enough kinetic energy to break some of the attractive forces, allowing the particles to “slip and slide” next to each other, but there is not enough energy to allow them to escape. In a gas, there is sufficient kinetic energy to totally overcome the forces and the particles have no interactions with each other.
• A change of state from a solid to a liquid occurs at a defined temperature (which) called the melting point (or freezing point);this temperature is a unique physical property of the substance. The transition from a liquid to a gas, likewise, occurs at the boiling point. A direct transition from a solid to a gas is called sublimation.
• An intensive property is defined as a property that is inherent to the substance and is not dependent on the sample size. Density, the ratio of mass-to-volume for a substance, is a classic example of an intensive property.
• Density is calculated by taking the mass of a sample of a substance, and dividing that by the volume of that sample. Density for solids is typically expressed using units of grams per cubic centimeter (g cm-3);liquids as grams per milliliter (g mL-1) and gasses as grams per liter (g L-1), although any mixture of mass and volume units may be used. Remember, a mL has the same volume as a cm3, and a L is simply 1000 mL.
• Physical changes are changes in outward appearances that do not alter the chemical nature of the substance and produce no new substance. When a chemical change occurs, a new substance is produced. Just like physical properties describe the appearance or intensive properties of a substance, chemical properties describe the set of chemical changes that are possible for that substance.
• The law of mass conservation (conservation of mass) simply states, that there is no detectable change in the total mass of materials when they react chemically (undergo a chemical change) to form new substances. | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/02%3A_The_Physical_and_Chemical_Properties_of_Matter/2.5%3A_Conservation_of_Mass.txt |
When elements combine to make compounds, they form strong chemical bonds to each other so that the compounds now behaves as a single new substance. In this chapter, we will see that there are two ways that atoms bond to each other: either by transferring electrons or by sharing them. These two types of bonding are called ionic and covalent. We will learn to draw Lewis structure for atoms and molecules and see how we can use these to show (and predict) bonding patterns in compounds. Finally, we will learn how to systematically name and classify compounds using simple rules based on the type of bonds used to construct them.
03: Chemical Bonding and Nomenclature
If we take two or more atoms and bond them together chemically so that they now behave as a single substance, we have made a chemical compound. We will see that the process of bonding actually involves either the sharing, or the net transfer, of electrons from one atom to another. The two types of bonding are covalent, for the sharing of electrons between atoms, and ionic, for the net transfer of electrons between atoms. Covalent or ionic bonding will determine the type of compound that will be formed.
In Chapter 1, we used atomic theory to describe the structure of the fluorine atom. We said that neutral fluorine has nine protons in its nucleus (an atomic number of 9), nine electrons surrounding the nucleus (to make it neutral), and the most common isotope has ten neutrons in its nucleus, for a mass number of 19. Further, we said that the nine electrons exist in two energy levels; the first energy level contains two electrons and is written 1s2. The second energy level contains seven electrons, distributed as 2s2 2p5. The outermost electron level in any atom is referred to as the valence shell. For the representative elements (remember, this includes all of the elements except for the transition metals), the number of electrons in the valence shell corresponds to the Group number of the element in the periodic table. Group 1A elements will have one valence electron, Group 6A elements will have six valence electrons, and so on. Fluorine is a Group 7A element and has seven valence electrons. We can show the electron configuration for fluorine using a Lewis diagram (or electron-dot structure), named after the American chemist G. N. Lewis, who proposed the concepts of electron shells and valence electrons. In a Lewis diagram, the electrons in the valence shell are shown as small “dots” surrounding the atomic symbol for the element.
$\cdot\underset{..}{\overset{..}F}\cdot \nonumber$
When more than four electrons are present in the valence shell, they are shown as pairs when writing the Lewis diagram (but never more than pairs). Lewis diagrams for the atoms in the second period are shown below:
$\cdot Li\; \; \; \; \cdot Be\: \cdot \; \; \; \; \cdot \overset{.}B\cdot \; \; \; \; \cdot \underset{.}{\overset{.}C}\cdot \nonumber$
$\cdot\underset{.}{\overset{..}N}\cdot\; \; \; \; :\underset{.}{\overset{..}O}\cdot\; \; \; \; \cdot\underset{..}{\overset{..}F}\cdot\; \; \; \; :\underset{..}{\overset{..}Ne}: \nonumber$
As you look at the dot-structures please understand that it makes no difference where you place the electrons, or the electron pairs, around the symbol, as long as pairs are shown whenever there are four or more valence electrons.
If you examine the Lewis diagram for neon (Ne) above, you will see that the valence shell is filled; that is, there are eight electrons in the valence shell. Elements in Group 8A of the periodic table are called noble gasses; they are very stable and do not routinely combine with other elements to form compounds (although today, many compounds containing noble gasses are known). Modern bonding theory tells us that this stability arises because the valence shell in the noble gasses is completely filled. When the valence shell is not full, theory suggests that atoms will transfer or share electrons with other atoms in order to achieve a filled valence shell… that is, the electron configuration of the noble gasses. Chemical bonding can then be viewed as a quest by atoms to acquire (or lose) enough electrons so that their valence shells are filled, that is, to achieve a “noble gas configuration”. This is often referred to as the “octet rule”; the desire for elements to obtain eight electrons in the valence shell (except of course for helium where the noble gas configuration is two valence electrons).
Atoms can achieve a noble gas configuration by two methods; the transfer of electrons from their valence shells to another atom, or by sharing electrons with another atom. If you examine the Lewis diagram for lithium (Li), you will see that it has only one valence electron. If lithium was to transfer this electron to another atom, it would be left with two electrons in the 1s-orbital (denoted as 1s2). This is the same electron configuration as helium (He), and so by losing this electron, lithium has achieved a noble gas configuration. Because electrons carry a negative charge, the loss of this electron leaves lithium with a single positive charge. This is the lithium cation and it is shown as Li+.
Returning to fluorine (F), in order to achieve the 2s2 2p6 configuration of neon (Ne), fluorine needs to gain one valence electron. Because fluorine has gained one electron, it now has one negative charge. This is the fluoride anion and it is shown as F-. The transfer of electrons in order to achieve a noble gas configuration is the process known as ionic bonding, and this will be covered in more detail later in this chapter.
Exercise $1$
1. Sodium and chlorine are both third-period elements. Draw Lewis diagrams for each of these elements.
2. What number of electrons would chlorine have to gain in order to achieve a “noble gas configuration”? What would be the charge on chlorine?
3. What number of electrons would Na have to lose to obtain the noble gas configuration of Ne with eight valence electrons? What charge would Na have? | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/03%3A_Chemical_Bonding_and_Nomenclature/3.01%3A_Compounds_Lewis_Diagrams_and_Ionic_Bonds.txt |
A second method by which atoms can achieve a filled valence shell is by sharing valence electrons with another atom. Thus fluorine, with one unpaired valence electron, can share that electron with an unshared electron on another fluorine to form the compound, F2 in which the two shared electrons form a chemical bond holding the two fluorine atoms together. When you do this, each fluorine now has the equivalent of eight electrons in its valence shell; three unshared pairs and one pair that is shared between the two atoms. Note that when you are counting electrons, the electrons that are shared in the covalent bond are counted for each atom, individually. A chemical bond formed by sharing electrons between atoms is called a covalent bond. When two or more atoms are bonded together utilizing covalent bonds, the compound is referred to as a molecule.
There is a simple method, given below, that we can use to construct Lewis diagrams for diatomic and for polyatomic molecules:
• Begin by adding up all of the valence electrons in the molecule. For F2, each fluorine has seven, giving a total of 14 valence electrons.
• Next, draw your central atom. For a diatomic molecule like F2, both atoms are the same, but if several different atoms are present, the central atom will be to the left (or lower) in the periodic table.
• Next, draw the other atoms around the central atom, placing two electrons between the atoms to form a covalent bond.
• Distribute the remaining valence electrons, as pairs, around each of the outer atoms, so that they all are surrounded by eight electrons.
• Place any remaining electrons on the central atom.
• If the central atom is not surrounded by an octet of electrons, construct multiple bonds with the outer atoms until all atoms have a complete octet.
• If there are an odd number of valence electrons in the molecule, leave the remaining single electron on the central atom.
Let’s apply these rules for the Lewis diagram for chlorine gas, Cl2. There are 14 valence electrons in the molecule. Both atoms are the same, so we draw them next to each other and place two electrons between them to form the covalent bond. Of the twelve remain electrons, we now place six around one chlorine (to give an octet) and then place the other six around the other chlorine (our central atom). Checking, we see that each atom is surrounded by an octet of valence electrons, and so our structure is complete.
$:\underset{..}{\overset{..}{Cl}}\cdot \cdot \underset{..}{\overset{..}{Cl}}: \nonumber$
All of the Group 7A elements (the halogens), have valence shells with seven electrons and all of the common halogens exist in nature as diatomic molecules; fluorine, F2; chlorine, Cl2; bromine, Br2 and iodine I2 (astatine, the halogen in the sixth period, is a short-lived radioactive element and its chemical properties are poorly understood). Nitrogen and oxygen, Group 5A and 6A elements, respectively, also exists in nature as diatomic molecules (N2 and O2). Let’s consider oxygen; oxygen has six valence electrons (a Group 6A element). Following the logic that we used for chlorine, we draw the two atoms and place one pair of electrons between them, leaving 10 valence electrons. We place three pairs on one oxygen atom, and the remaining two pairs on the second (our central atom). Because we only have six valence electrons surrounding the second oxygen atom, we must move one pair from the other oxygen and form a second covalent bond (a double bond) between the two atoms. Doing this, each atom now has an octet of valence electrons.
$:\underset{..}{O}::\overset{..}{O}: \nonumber$
Nitrogen has five valence electrons. Sharing one on each atom gives the first intermediate where each nitrogen is surrounded by six electrons (not enough!). Sharing another pair, each nitrogen is surrounded by seven electrons, and finally, sharing the third, we get a structure where each nitrogen is surrounded by eight electrons; a noble gas configuration (or the “octet rule”). Nitrogen is a very stable molecule and relatively unreactive, being held together by a strong triple covalent bond.
$:N\vdots \vdots N: \nonumber$
As we have constructed Lewis diagrams, thus far, we have strived to achieve an octet of electrons around every element. In nature, however, there are many exceptions to the “octet rule”. Elements in the first row of the periodic table (hydrogen and helium) can only accommodate two valence electrons. Elements below the second row in the periodic table can accommodate, 10, 12 or even 14 valence electrons (we will see an example of this in the next section). Finally, in many cases molecules exist with single unpaired electrons. A classic example of this is oxygen gas (O2). We have previously drawn the Lewis diagram for oxygen with an oxygen-oxygen double bond. Physical measurements on oxygen, however, suggest that this picture of bonding is not quite accurate. The magnetic properties of oxygen, O2, are most consistent with a structure having two unpaired electrons in the configuration shown below:
$:\underset{.}{\overset{..}{O}}\cdot \cdot \underset{.}{\overset{..}{O}}: \nonumber$
In this Lewis diagram, each oxygen atom is surrounded by seven electrons (not eight). This electronic configuration may explain why oxygen is such a reactive molecule (reacting with iron, for example, to form rust); the unpaired electrons on the oxygen molecule are readily available to interact with electrons on other elements to form new chemical compounds.
Another notable exception to the “octet rule” is the molecule NO (nitrogen monoxide). Combining one nitrogen (Group 5A) with one oxygen (Group 6A) gives a molecule with eleven valence electrons. There is no way to arrange eleven electrons without leaving one electron unpaired. Nitric oxide is an extremely reactive molecule (by virtue of its unshared electron) and has been found to play a central role is biochemistry as a reactive, short-lived molecule involved in cellular communication.
$\cdot \underset{.}{\overset{..}{N}}\cdot \cdot \underset{.}{\overset{..}{O}}: \nonumber$
As useful as Lewis diagrams can be, chemists tire of drawing little dots and, for a shorthand representation of a covalent bond, a short line (a line-bond) is often drawn between the two elements. Whenever you see atoms connected by a line-bond, you are expected to understand that this represents two shared electrons in a covalent bond. Further, the unshared pairs of electrons on the bonded atoms are sometimes shown, and sometimes they are omitted. If unshared pairs ore omitted, the chemist reading the structure is assumed to understand that they are present.
$\underset{..}{\overset{..}{F}}\cdot \cdot \underset{..}{\overset{..}{F}}:\; or\; :\underset{..}{\overset{..}{F}}-\underset{..}{\overset{..}{F}}:\; or\; F-F \nonumber$
$:N\vdots \vdots N:\; or\; :N\equiv N:\; or\; N\equiv N \nonumber$ | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/03%3A_Chemical_Bonding_and_Nomenclature/3.02%3A_Covalent_Bonding.txt |
As mentioned in Section 3.1, elements can also transfer electrons to another element in order to achieve the noble gas configuration. Consider sodium. Sodium (Na) is in Group 3A. Its first energy level is completely filled (1s2 ), the second energy level is also filled (2s2 2p6) and there is a single electron in the third energy level (3s1). Energetically, the easiest way for sodium to achieve an octet of valence electrons is by transferring its valence electron to an acceptor atom. This will leave the sodium atom with the same electron configuration as neon (1s22s2 2p6), and will satisfy the “octet rule”. Because sodium loses one electron (with its negative charge) the sodium atom must now have a positive charge. Atoms, or covalently bound groups of atoms with a positive charge are called cations, and the sodium cation is written as Na+.
If the acceptor atom in the example above was chlorine, the third valence shell would now be filled, matching the electron configuration of argon. Because the chlorine atom has accepted an electron, with its negative charge, the chlorine atom must now have a negative charge. Atoms, or covalently bound groups of atoms with a negative charge are called anions, and the chlorine anion is written as Cl-. Although electron transfer has occurred between the two atoms in this example, there is no direct bond holding the sodium cation and the chlorine anion together, other than the simple electrostatic attraction between the two charged atoms. This is the real difference between ionic and covalently bound atoms; covalent molecules are held together in a specific geometry which is dictated by the electrons that they share. Ionic compounds are held together by simple electrostatic attraction and, unless these atoms are present in organized crystals, there is no defined geometric order to this attraction. Ionic compounds are often referred to as salts.
Exercise \(1\)
1. Hydrogen and oxygen react to form water, H2O. Draw a Lewis diagram for water using the line-bond shorthand.
2. Draw the Lewis diagram for the molecules, hydrogen chloride, BrCl, and hydrogen cyanide (HCN).
3.04: Identifying Molecular and Ionic Compounds
The tendency for two or more elements to combine and form a molecule that is stabilized by covalent bonds (a molecular compound) can be predicted simply by the location of the various elements on the periodic table. In Chapter 1, we divided the elements in the periodic table into (seemingly) arbitrary groupings; the metals, the non-metals, the semi-metals, and so on. These groupings are not arbitrary, but are largely based on physical properties and on the tendency of the various elements to bond with other elements by forming either an ionic or a covalent bond. As a general rule of thumb, compounds that involve a metal binding with either a non-metal or a semi-metal will display ionic bonding. Compounds that are composed of only non-metals or semi-metals with non-metals will display covalent bonding and will be classified as molecular compounds. Thus, the compound formed from sodium and chlorine will be ionic (a metal and a non-metal). Nitrogen monoxide (NO) will be a covalently bound molecule (two non-metals), silicon dioxide (SiO2) will be a covalently bound molecule (a semi-metal and a non-metal) and MgCl2 will be ionic (a metal and a non-metal). Later in this chapter we will see that many covalent compounds have bonds that are highly polarized with greater electron density around one atom than the other. These compounds are often described as having “ionic character” and these types of covalent bonds can often be readily broken to form sets of ions.
Exercise \(1\)
1. Determine whether each of the following compounds is likely to exist as a molecule, or as an ionic compound:
1. Hydrogen fluoride; HF
2. Silicon tetrachloride; SiCl4
3. Elemental sulfur as S8
4. Disodium dioxide; Na2O
5. PF3
6. Be3N2
7. AlP
8. CBr4 | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/03%3A_Chemical_Bonding_and_Nomenclature/3.03%3A_Lewis_Representation_of_Ionic_Compounds.txt |
The compound NaOH has wide industrial use and is the active ingredient in drain cleaners. Based on the discussion in the previous section, we would expect NaOH to be an ionic compound because it contains sodium, a Group 1A metal. Hydrogen and oxygen, however, are nonmetals, and we would expect these to bond together covalently. This compound, called sodium hydroxide, is an example of an ionic compound formed between a metal ion (sodium) and a polyatomic ion (HO). Charged groups of atoms, like HO, that are bonded together covalently are called polyatomic ions. Within an ionic compound a polyatomic behaves as a single unit forming salts with other cations or anions.
Using the rules described in Section 3.2, we can draw a Lewis diagram for HO. Oxygen has six valence electrons and hydrogen has one, for a total of seven. The central atom in our structure will be hydrogen (it is to the left of oxygen in the periodic table). Next, because this is a polyatomic ion with a single negative charge, we add the extra electron to the central atom, pair the electrons and then draw the two atoms bonded together. Next, the six remaining electrons are distributed around the oxygen to form an octet. Finally, the polyatomic ion is enclosed in brackets with the charge as a superscript to show that the ion behaves as a single unit.
Polyatomic ions are very common in chemistry. It is essential that you memorize these and be able to correlate the name, the composition and the charge for each of them, as they will be discussed freely throughout the remainder of the course and you will be expected to know these in General Chemistry.
Example \(1\):
Construct a Lewis diagram for the polyatomic ion CO32.
Solution
Oxygen has six valence electrons and carbon has four; therefore in CO32 there will be a total of 22 valence electrons, plus two additional electrons from the 2- charge. The central atom in our structure will be carbon (it is to the left of oxygen in the periodic table). Next, we draw the carbon (our central atom) with its’ four electrons and add the additional two electrons from the charge. The three oxygens are placed around the carbon and the electrons are arranged to form the three covalent bonds. Next, the 18 remaining electrons are distributed around the oxygens so that they all have a full octet. The carbon, however, is only surrounded by six electrons. To remedy this, we move one electron pair in to form a double bond to one of the oxygen atoms. Finally, the polyatomic ion is enclosed in brackets with the charge as a superscript.
We need to understand that the process of placing electrons into a particular bond in a compound is an artificial aspect of building Lewis diagrams. In fact, the electrons are added to the polyatomic ion, but it is impossible to know exactly where they went.
Exercise \(1\)
1. Draw Lewis diagrams for NO2- and NH4+.
3.06: Resonance
In the previous section, we constructed a Lewis diagram for the carbonate anion. Our final structure showed two carbon-oxygen single bonds and one carbon-oxygen double bond. The structure that we drew is shown below, along with two other possible representations for the carbonate anion. These structures differ only in the position of the carbon-oxygen double bond.
So which of these is correct? Actually, they all are! These are all “proper” Lewis diagrams for a covalent structure having constant geometry, and the diagrams differ only in the manner that we have arbitrarily arranged the electrons. These Lewis diagrams are called resonance forms. For the carbonate anion, there are three equivalent resonance forms that can be drawn. It is important to note that the electrons are not “hopping” between the atoms, but that the electrons are spread evenly between the carbon and all three oxygens and that each carbon-oxygen bond has a bond-order of 1.33 (one and one-third covalent bonds). A structure such as this is called the resonance hybrid, and although it most clearly represents the actual bonding in the compound, it is often difficult to understand the nature of the bonding when structures are represented as resonance hybrids. A full discussion of resonance is beyond the scope of an introductory text and, for structures such as the carbonate anion, we will accept any of the proper resonance forms shown above. | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/03%3A_Chemical_Bonding_and_Nomenclature/3.05%3A_Polyatomic_Ions.txt |
If we were to construct a Lewis diagram for molecular hydrogen (H2), we would pair the single valence electrons on each atom to make a single covalent bond. Each hydrogen would now have two electrons in its valence shell, identical to helium. The mathematical equations chemists use to describe covalent bonding can be solved to predict the regions of space surrounding the molecule in which these electrons are likely to be found. A particularly useful application of these calculations generates a molecular surface that is color coded to show electron density surrounding the molecule. This type of molecular surface is called an electrostatic potential map. When this type of calculation is done for molecules consisting of two (or more) different atoms, the results can be strikingly different. Consider the molecule HF. Hydrogen, with one valance electron, can share that electron with fluorine (with seven valence electrons) to form a single covalent bond.
In this electrostatic potential map, blue is used to indicate low electron density (a relative positive charge) and red indicates high electron density (a relative negative charge); the colors light blue, green, yellow and orange indicate the increasing charge gradient. The molecule HF is clearly very polar, meaning that a significant difference in electron density exists across the length of the molecule. The electrostatic potential map for HF contrasts significantly with that for H2, where the charge was quite symmetrical (a uniform green color). Hydrogen fluoride (HF) can be described as a very polar molecule, while hydrogen (H2) is nonpolar.
The origin of the polarization of the HF covalent bond has to do with electronegativity, an inherent property of all atoms. Within the periodic table, there is a trend for atoms to attract electrons towards themselves when they are bonded to another atom (as in HF). Atoms that tend to strongly attract electrons have a high electronegativity, relative to atoms that have a relatively low tendency to attract electrons towards themselves. The modern electronegativity scale was devised by Linus Pauling in 1932 and, in the Pauling scale, atoms in the periodic table vary in electronegativity from a low of 0.8 for cesium to a maximum of 4.0 for fluorine.
In the molecule HF, the electronegativity of the hydrogen is 2.2 and fluorine is 4.0. This difference leads to the profound polarization of the HF covalent bond which is apparent in the electrostatic potential map.
Polarizations of covalent bonds also occur in more complex molecules. In water, the oxygen has an electronegativity of 3.5; hydrogen is 2.2. Because of this, each of the H-O bonds is polarized with greater electron density towards the oxygen. Within the molecule, H2O, the effect of this polarization becomes apparent in the electrostatic potential map, as shown in Figure 3.18. The end of the molecule with the oxygen has a high electron density and the hydrogen ends are electron deficient. We will see in later chapters that the polarization of water, caused by the difference in electronegativities, gives water the special properties that allows it to dissolve ionic compounds, and basically support life as we know it. Within organic chemistry (the study of carbon-containing molecular compounds), you will appreciate that the relative reactivity of organic molecules with each other is largely dependent on the polarization of covalent bonds in these molecules.
3.08: Exceptions to the Octet Rule
Returning briefly to classical Lewis diagrams. Consider the diagram shown below for the molecule SF4. In constructing this diagram, six valence electrons are placed around the sulfur and seven valence electrons are placed around each fluorine. As we attempt to pair these to form covalent bonds, we note that there are “too many” electrons on the sulfur! We clearly cannot form a covalent bond using three electrons, so we split one pair, move the single electrons into bonding position and form bonds with the remaining two fluorines. The “extra pair” of electrons just sits there on the sulfur and does not participate directly in the bonding.
This is an example of valence expansion. In general, elements below the second period in the periodic table (S, Se, Te, etc.) will commonly have 10 – 12 electrons in their valence shells. As in SF4, these electrons are not directly involved in the formation of covalent bonds, but they affect the overall reactivity of the particular molecule.
In general, all molecular compounds containing elements that appear below the second row in the periodic table are capable of valence expansion and you need to be very careful when you are drawing Lewis diagrams for these compounds. As we saw in Section 3.1 for the molecule nitrogen oxide (NO), stable molecules also exist in which atoms are not surrounded by an octet of electrons. Another example of this is the molecule BF3, which is shown in the following example.
Example \(1\): Boron Trifluoride
Construct a Lewis diagram for the molecule BF3.
Solution
Boron has three valence electrons and fluorine has seven. The central atom in our structure will be boron (it is to the left of fluorine in the periodic table). Next, we draw the boron (our central atom) with its’ three electrons and place the three fluorines around the boron with the electrons arranged to form the three covalent bonds. Each of the fluorines have a full octet. The boron, however, is only surrounded by six electrons. Because of this, the boron in BF3 is a powerful electron acceptor and forms strong complexes with electrons from other compounds. In Chapter 8 we will see that this property is called Lewis acidity and BF3 is a very powerful Lewis acid. | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/03%3A_Chemical_Bonding_and_Nomenclature/3.07%3A_Electronegativity_and_the_Polar_Covalent_Bond.txt |
In an ionic compound the total number of charges on the cations must equal the total number of charges on the anions; that is, the compound must be neutral. In Section 3.2 we described how a sodium atom can donate an electron to another atom in order to form an ion with a full octet of electrons in its outermost electron shell (the same electron configuration as Ne). The charge on the sodium atom, when this happens, is now 1+, because it has eleven protons in its nucleus, but is only surrounded by 10 electrons. Lithium, likewise, can lose one electron to form Li1+ and be left with the same electron configuration as He. In fact, all of the Group 1A metals can lose a single electron to form 1+ ions. Elements in Group 2A can each lose two electrons to form 2+ ions and achieve a noble gas configuration. In fact, the group that a main-group element is associated with in the periodic table will dictate the valence (or charge) of its corresponding ion. Metals in Groups 1A, 2A and 3A will form ions with 1+, 2+ and 3+ charges, respectively.
Main-group nonmetals can easily achieve an octet of valence electrons by accepting electrons from other elements. Thus Group 5A elements can accept three electrons to form 3- ions, Group 6A elements accept two electrons to form 2- ions and Group 7A elements (the halogens) accept one electron to form 1- ions. For example, oxygen (Group 6A) needs to accept two electrons to achieve the electron configuration of neon. This gives oxygen a total of 10 electrons, but it only has eight protons in its nucleus (its atomic number is 8), therefore, the oxygen ion has a net charge of 2- (O2-).
To write a formula for an ionic compound composed of main group elements (or containing polyatomic ions) you need to adjust the ratio of anions and cations so that the resulting molecule is electrically neutral. For example, consider an ionic compound containing sodium and chlorine. Lithium is a Group 1A element and will form a 1+ ion; fluorine is a Group 7A element and will form a 1- ion. Neutrality is achieved when one lithium is paired with one fluorine, or LiF. For a compound composed of calcium and chlorine, the Group 2A calcium will form a 2+ ion while chlorine forms a 1- ion. To achieve neutrality, there must be two chlorines for every calcium, and the formula must be as CaCl2. Aluminum (Group 3A) will form a 3+ ion. If this was paired with oxygen (Group 6A) which forms a 2- ion neutrality would only be achieved if two Al3+ ions (for a total of six positive charges) were paired with three O2- ions (a total of six negative charges).
Consider a compound consisting of sodium and the polyatomic ion sulfate (SO42-). Sodium (Group 1A) yields a 1+ cation and so there must be two sodiums in the compound for every sulfate (which has a 2- charge), or Na2SO4. For a compound containing calcium (Group 2A) and nitrate (NO3-), two nitrate anions must be present for every calcium 2+ cation. In a compound containing multiple copies of a polyatomic ion, the entire ion is enclosed in parenthesis with a subscript to indicate the number of units. Thus the compound from calcium and nitrate would be written as Ca(NO3)2. | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/03%3A_Chemical_Bonding_and_Nomenclature/3.09%3A_Common_Valence_States_and_Ionic_Compounds.txt |
The simplest ionic compounds consist of a single type of cation associated with a single type of anion. Nomenclature for these compounds is trivial; the cation is named first, followed by the anion. If the anion is a single element, the suffix ide is added to the root name of the element.
When you are constructing names for ionic compounds, you do not use “multipliers” to indicate how many cations or anions are present in the compound. For example NaI is named sodium iodide; Na2S is named sodium sulfide; CaCl2 is named calcium chloride. The chemist reading the name is assumed to have sufficient knowledge to pair the elements properly based on their common valence states. There are exceptions to this simple nomenclature, however. Many transition metals exist as more than one type of cation. Thus, iron exists as Fe2+ and Fe3+ cations (they are referred to as “oxidation states”, and will be covered in detail in Chapter 5). When you are naming an ionic compound containing iron, it is necessary to indicate which oxidation state the metal has. For metals, the oxidation state is the same as the charge. Thus Fe2+ in a compound with chloride would have a formula FeCl2 and would be named iron (II) chloride, with the oxidation state (the charge on the iron) appearing as a Roman numeral in parenthesis after the cation. The cation Fe3+ paired with oxygen would have the formula Fe2O3 and would have the name iron (III) oxide.
The procedure for naming ionic compounds contain polyatomic ions is identical to that described above for simple ions. Thus, CaCO3 is named calcium carbonate; Na2SO4 is named sodium sulfate; (NH4)2HPO4 (a compound with two polyatomic ions) is named ammonium hydrogen phosphate; and Pb2+ paired with SO42-, PbSO4 is named lead (II) sulfate.
Example \(1\):
Write a correct chemical formula for each of the following ionic compounds:
1. Calcium bromide
2. Aluminum oxide
3. Copper (II) chloride
4. Iron (III) oxide
Solution
1. Calcium is 2+, bromide is 1-; CaBr2.
2. Aluminum is 3+, oxide is 2-; Al2O3.
3. From the oxidation state that is given, copper is 2+, chloride is 1-; CuCl2.
4. From the oxidation state, iron is 3+, oxide is 2-; Fe2O3.
Example \(1\):
Write a proper chemical name for each of the following ionic compounds:
1. Li2S
2. CaO
3. NiCl2
4. FeO
Solution
1. We don’t use multipliers, so this is simply lithium sulfide.
2. This is simply calcium oxide.
3. We don’t have to specify an oxidation state for nickel, so this is nickel chloride.
4. We must specify that iron is 2+ in this compound; iron (II) oxide.
Exercise \(1\)
Write a correct chemical formula for each of the following ionic compounds:
1. Sodium phosphide
2. Iron (II) nitrite
3. Calcium hydrogen phosphate
4. Chromium (III) oxide
Exercise \(1\)
Write a proper chemical name for each of the following ionic compounds:
1. NaBr
2. CuCl2
3. Fe(NO3)3
4. (NH4)3PO4
3.11: Nomenclature of Molecular Compounds
The nomenclature of simple binary molecular compounds (covalently bonded compounds consisting of only two elements) is slightly more complicated than the nomenclature of ionic compounds because multipliers must be used to indicate the ratio of the elements in the molecule; the multiplier mono is only used for the second element in a compound.
Further, when you are naming a molecular compound, you must also decide which element should be listed first. In general, elements appearing to the left or lower in the periodic table are listed first in the name. Once you have decided on the order, the second element is named using the element root and ide, just like in ionic compounds. Thus, for CCl4, carbon is to the left of chlorine (Group 4A vs. Group 5A), so it is listed first. There are four chlorines, so the multiplier tetra is used, and the name is carbon tetrachloride. Compounds containing hydrogen are generally an exception, and the hydrogen is listed as the first element in the name. Thus, H2S would be named using the multiplier di to indicate that there are two hydrogens and mono to indicate that there is only one sulfur, or, dihydrogen monosulfide.
For the molecule SO2; they are both Group 6A elements, but sulfur is lower in the periodic table (Row 3 vs. Row 2) so it is first in the name. There are two oxygens, so the multiplier is di and the name is sulfur dioxide.
For the molecule NO; nitrogen is to the left of oxygen (Group 6A vs. Group 5A) so it is first in the name. There is one oxygen, so the multiplier is mono and, following the rules, the name would be “nitrogen monooxide”. In this case, however, the second “o” in the name is dropped (to allow for easier pronunciation ) and the name is shortened to nitrogen monoxide. Distinguish this from another oxide of nitrogen, N2O4. Again nitrogen is first and needs the multiplier di. There are four oxygens, so the multiplier is tetra, but once again the multiplier is shortened (again, the “a” is dropped) and the name is dinitrogen tetroxide.
Example \(1\):
Write a correct chemical formula for each of the following molecular compounds:
1. Chlorine monofluoride
2. Dihydrogen monosulfile
3. Carbon tetrabromide
4. Bromine
1. ClF
2. H2S
3. CBr4
4. Br2
Example \(1\):
Write a proper chemical name for each of the following molecular compounds:
1. IF
2. PCl3
3. I2
4. N2F2
Solution
1. Iodine monofluoride
2. Phosphorus trichloride
3. Iodine
4. Dinitrogen difluoride | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/03%3A_Chemical_Bonding_and_Nomenclature/3.10%3A_Nomenclature_of_Ionic_Compounds.txt |
• In covalent bonding, electrons are shared between atoms. In ionic bonding, electrons are transferred from one atom to another. Compounds that are formed using only covalent bonds are termed molecular compounds.
• The outermost electron level in any atom is referred to as the valence shell. The electron configuration of the valence shell of an atom can be shown graphically using a Lewis diagram (or electron-dot structure). The arrangement of “dots” around the chemical symbol for the element are shown singly up through four electrons, and then paired until eight electrons are present.
• To form ions from individual elements, electrons are added or subtracted from the valence shell in order to completely fill the shell with eight electrons (the octet rule). The charge on the ion reflects the electrons added or removed. Representative elements through Group IIIA will lose electrons to form cations, while those in groups IVAVIIA will gain electrons and form anions.
• A covalent bond is constructed in a Lewis diagram by pairing a set of unpaired electrons from two different atoms. For the purposes of the “octet rule”, a pair of shared electrons is counted as two electrons for each atom. Multiple covalent bonds (double bonds and triple bonds) are used, if necessary, to give each bonded atom a full octet (except, of course, for helium and hydrogen). When two or more atoms are bonded together utilizing covalent bonds, the compound is referred to as a molecule.
• As a rule of thumb, ionic bonds will be formed whenever the compound contains a metal. Covalent bonding will be observed in compounds containing only semimetals or nonmetals.
• Groups of covalently bonded semimetals or nonmetals which are charged are called polyatomic ions. Common examples include sulfate dianion, nitrate anion, phosphate trianion, etc. These polyatomic ions are commonly paired with metals forming ionic compounds.
• Many (but not all) polyatomic ions can be drawn in two or more equivalent Lewis representations. These are called resonance forms of the ion. The actual electronic structure of the ion is a combination of these Lewis structures and is called the resonance hybrid.
• The electronegativity of an element is a measure of the tendency of that element to attract electrons towards itself. Electronegativities range from 0.6 to 4.0, with fluorine as the most electronegative element (a value of 4.0). The general trend in the periodic table is for electronegativity to increase from the lower left-hand corner (Fr) to the upper right-hand corner (F).
• Covalent bonds formed between atoms with different electronegativity will be polarized with the greatest electron density localized around the most electronegative atom. The effect of electronegativity on electron distribution within a molecule can be shown using a computer-calculated electrostatic potential map where colors are used to represent electron density.
• Elements in periods 3 – 7 can accommodate more that eight electrons in there valence shells. This phenomena is called valence shell expansion and molecules involving these elements may have 10 – 14 valence electrons in properly drawn Lewis diagrams. Exceptions to the “octet rule” also exist where the valence shell contains less than eight electrons, or contains unpaired electrons.
• When naming simple, binary ionic compounds, the cation is named first using the name of the element, followed by the anion, where the suffix ide is added to the root name of the element. Multipliers are not used. For transition metals in which the metal can assume a variety of oxidation states (different positive charges), the charge of the metal ion is shown in the name using Roman numerals, in parenthesis, following the name of the element (i.e., iron (III) chloride).
• When naming simple binary molecular compounds (compounds containing only covalent bonds) the least electronegative element is (generally) named first, followed by the second element, where the suffix ide is again added to the root name of the element. In molecular compounds multipliers are used to indicate the number of each atom present (mono-, di-, tri-, tetra-, etc.) with the exception that mono is not used for the first element in the compound. | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/03%3A_Chemical_Bonding_and_Nomenclature/3.S%3A_Chemical_Bonding_and_Nomenclature_%28Summary%29.txt |
In the preceding chapters, you should have gained an appreciation of the scale of chemistry, with regard to the physical size and mass of individual atoms and compounds. Because a typical samples of a substance (such as a copper penny) contains so many atoms, chemistry have defined a unit by which we can easily county these large numbers; this units is called a mole. However, we will see that a mole is more than a just a large number; a mole is also directly related to the atomic weight of atoms and compounds and the mole concept will give us the tools to consider chemical reactions in a quantitative manner. This chapter serves as an introduction to the concepts and we will continues to build upon this foundation when we consider stoichiometry in the following chapters.
Thumbnail: A photograph of Scalopus aquaticus. The eastern American mole (Scalopus aquaticus linnacus) showing the large forelimbs used to excavate tunnels. (CC BY-SA 3.0; Kenneth Catania, Vanderbilt University).
04: The Mole and Measurement in Chemistry
One of the important concepts to grasp in chemistry is scale. Atoms and molecules are very small. A single atom of hydrogen has a mass of about 1.67 x 10-24 grams (that’s 0.00000000000000000000000167 grams). One cubic centimeter of water (one mL) contains about 3.3 x 1022 water molecules (that’s 33 sextillion molecules). Because chemists routinely use numbers that are both incredibly small and incredibly large, unique units of measurement have been developed to simplify working with these numbers. As we learned in Chapter 1, the atomic mass unit (amu) helps us talk about the mass of atoms on a scale appropriate to atoms (one grams is about 600 sextillion amu). In this chapter, we will introduce the concept of a mole to help us talk about numbers of atoms on a scale appropriate to the size of a sample we could work with in a laboratory (for example, in grams). The mole is defined as the number of atoms contained in exactly 12 grams of carbon-12 (the isotope ${$/extract_itex]displaystyle {}_{6}^{12}{\text{C}}} ). Chemist have measured this number and it turns out to be 6.0221415 x 1023. We can think of the term mole as a number, just like the word dozen represents the number 12. We will use the mole to represent this very large number (a chemist’s dozen) and we will see that there is a special relationship between a mole of a pure substance and the mass of the substance measured in amu. The origin of the mole concept is generally attributed to the Italian chemical physicist, Amadeo Avogadro. In 1811, Avogadro published an important article that drew a distinction between atoms and molecules (although these terms were not in use at the time). As part of his explanation of the behavior of gasses, he suggested that equal volumes of all gases at the same temperature and pressure contained the same number of molecules. This statement is referred to as Avogadro’s Hypothesis and today we commonly refer to the number of things in a mole, (6.0221415 x 1023) as Avogadro’s number (this is rounded to 6.02 x 1023 for most calculations). Because a mole can be thought of as a number, you can convert any number to moles by dividing that number by 6.02 x 1023. For example, at the time of this writing, the national debt of the United States is about 7.9 trillion dollars (7.9 x 1012 dollars). This could be expressed as moles of dollars as shown below: \[\left ( 7.9\times 10^{12}dollars \right )\left ( \frac{1\: mol\: dollars}{6.02\times 10^{23}dollars} \right )=1.3\times 10^{-11}\: mol\: dollars \nonumber$
The ratio of any number to the number of things in a mole is often referred to as a mole fraction.
Exercise $1$
1. It is estimated there are 7 x 1022 stars in the universe. How many moles of stars is this?
2. It is estimated there are 7.5 x 1018 grains of sand on the earth. How many moles of sand grains is this?
3. You have 0.0555 moles of jelly donuts. What number of donuts would that be?
4. You drink a small bottle of drinking water that contains 13 moles of water. What is the number of molecules of water you drank? | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/04%3A_The_Mole_and_Measurement_in_Chemistry/4.1%3A_Measurement_and_Scale_-_The_Mole_Concept.txt |
As we described in Section 4.1, in chemistry, the term mole can be used to describe a particular number. The number of things in a mole is large, very large (6.0221415 x 1023). We are all familiar with common copy-machine paper that comes in 500 sheet reams. If you stacked up 6.02 x 1023 sheets of this paper, the pile would reach from the earth to the moon 80 billion times! The mole is a huge number, and by appreciating this, you can also gain an understanding of how small molecules and atoms really are.
Chemists work simultaneously on the level of individual atoms, and on the level of samples large enough to work with in the laboratory. In order to go back and forth between these two scales, they often need to know how many atoms or molecules there are in the sample they’re working with. The concept that allows us to bridge these two scales is molar mass. Molar mass is defined as the mass in grams of one mole of a substance. The units of molar mass are grams per mole, abbreviated as g/mol.
The mass of a single isotope of any given element (the isotopic atomic mass) is a value relating the mass of that isotope to the mass of the isotope carbon-12 (${\displaystyle {}_{6}^{12}{\text{C}}}$); a carbon atom with six proton and six neutrons in its’ nucleus, surrounded by six electrons. The atomic mass of an element is the relative average of all of the naturally occurring isotopes of that element and atomic mass is the number that appears in the periodic table. We have defined a mole based on the isotopic atomic mass of carbon-12. By definition, the molar mass of carbon-12 is numerically the same, and is therefore exactly 12 grams. Generalizing this definition, the molar mass of any substance in grams per mole is numerically equal to the mass of that substance expressed in atomic mass units. For example, the atomic mass of an oxygen atom is 16.00 amu; that means the molar mass of an oxygen atom is 16.00 g/mol. Further, if you have 16.00 grams of oxygen atoms, you know from the definition of a mole that your sample contains 6.022 x 1023 oxygen atoms.
The concept of molar mass can also be applied to compounds. For a molecule (for example, nitrogen, N2) the mass of molecule is the sum of the atomic masses of the two nitrogen atoms. For nitrogen, the mass of the N2 molecule is simply (14.01 + 14.01) = 28.02 amu. This is referred to as the molecular mass and the molecular mass of any molecule is simply the sum of the atomic masses of all of the elements in that molecule. The molar mass of the N2 molecule is therefore 28.02 g/mol. For compounds that are not molecular (ionic compounds), it is improper to use the term “molecular mass” and “formula mass” is generally substituted. This is because there are no individual molecules in ionic compounds. However when talking about a mole of an ionic compound we will still use the term molar mass. Thus, the formula mass of calcium hydrogen carbonate is 117.10 amu and the molar mass of calcium hydrogen carbonate is 117.10 grams per mole (g/mol).
Exercise $1$
Find the molar mass of each of the following compounds:
• Sand - silicon dioxide (SiO2)
• Draino - sodium hydroxide (NaOH)
• Nutrasweet - Aspartame (C14H18N2O5)
• Bone phosphate - calcium phosphate Ca3(PO4)2 | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/04%3A_The_Mole_and_Measurement_in_Chemistry/4.2%3A_Molar_Mass.txt |
As described in the previous section, molar mass is expressed as “grams per mole”. The word per in this context implies a mathematical relationship between grams and mole. Think of this as a ratio. The fact that a per relationship, ratio, exists between grams and moles implies that you can use dimensional analysis to interconvert between the two. For example, if we wanted to know the mass of 0.50 mole of molecular hydrogen (H2) we could set up the following equations:
The known molar mass of H2 is:
$\left ( \frac{2.06g\: H_{2}}{1\: mol\: H_{2}} \right ) \nonumber$
We are given that we have 0.50 moles of H2 and we want to find the number of grams of H2 that this represents. To perform the dimensional analysis, we arrange the known and the given so that the units cancel, leaving only the units of the item we want to find.
$(0.5mol\: H_{2})\times \left ( \frac{2.06g\: H_{2}}{1\: mol\: H_{2}} \right )=x\, g\: H_{2}=1.0g\: H_{2} \nonumber$
Exercise $1$
1. Determine the mass of 0.752 mol of H2 gas.
2. How many moles of molecular hydrogen are present in 6.022 grams of H2?
3. If you have 22.414 grams of Cl2, how many moles of molecular chlorine do you have?
We can also use what is often called a per relationship (really just a ratio) to convert between number of moles and the number to things (as in 6.02 x 1023 things per mole). For example, if we wanted to know how many molecules of H2 are there in 3.42 moles of H2 gas we could set up the following equations:
The known ratio of molecules per mole is :
$\left ( \frac{6.02\times 10^{23}molecules\: H_{2}}{1\: mol\: H_{2}} \right ) \nonumber$
We are given that we have 3.42 moles of H2 and we want to find the number of molecules of H2 that this represents. To perform the dimensional analysis, we arrange the known and the given so that the units cancel, leaving only the units of the item we want to find.
$(3.42mol\: H_{2})\times \left ( \frac{6.02\times 10^{23}molecules\: H_{2}}{1\: mol\: H_{2}} \right )=x\: molecules\: H_{2}=2.06\times 10^{24}\: molecules\: H_{2} \nonumber$
And finally, we can combine these two operations and use the per relationships to convert between mass and the number of atoms or molecules. For example, if we wanted to know how many molecules of H2 are there in 6.022 grams of H2 gas we could set up the following series of equations:
The known molar mass of H2 is
$\left ( \frac{2.016gH_{2}}{1molH_{2}} \right ) \nonumber$
The known ratio of molecules per mole is ${\displaystyle \left({\frac {6.{\text{02 }}\times {\text{ 10}}^{\text{23}}{\text{ }}molecules{\text{ H}}_{\text{2}}}{1{\text{ }}mol{\text{ H}}_{\text{2}}}}\right)}$
$\left ( \frac{6.02\times 10^{23}molecules\: H_{2}}{1\: mol\: H_{2}} \right ) \nonumber$
We are given that we have 6.022 grams of H2 and we want to find the number of molecules of H2 that this represents. As always, to perform the dimensional analysis, we arrange the known ratios and the given so that the units cancel, leaving only the units of the item we want to find.
$(6.022gH_{2})\times \left ( \frac{1molH_{2}}{2.016gH_{2}}\right )\times \left ( \frac{6.02\times 10^{23}molecules\: H_{2}}{1\: mol\: H_{2}} \right )=x\; molecules\; H_{2}=1.80\times 10^{24} \nonumber$
Exercise $1$
1. A sample of molecular chlorine is found to contain 1.0 x 1020 molecules of Cl2. What is the mass (in grams) of this sample?
2. How many moles of sand, silicon dioxide (SiO2), and how many molecules of sand are found in 1.00 pound (454g) of sand?
3. You add 2.64 x 1023 molecules of sodium hydroxide (Drano™; NaOH), to your drain. How many moles are this and how many grams? | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/04%3A_The_Mole_and_Measurement_in_Chemistry/4.3%3A_Mole-Mass_Conversions.txt |
We are all familiar with the term percentage. We take an exam that is worth 100 points, we get 96 correct and we describe our score as 96%. We arrive at 96% by first taking our score and dividing it by the total number of points to get the fraction that we got correct. To convert the fraction to a percentage, we multiply the fraction by 100.
$\left ( \frac{96\; points}{100\; points\; total} \right )\times 100=96\% \nonumber$
Applying this concept to molecules, we could describe HCl as consisting of one atom of hydrogen and one atom of chlorine. Likewise, we could use mole nomenclature and say that one mole of the molecule consists of one mole of hydrogen and one mole of chlorine, and that one mole of HCl has a mass of 36.46 grams. These descriptions, however, tell us nothing about how much of this mass is attributable to the hydrogen and how much comes from the chlorine. In order to do this, we need to speak of the percentage composition of a molecule, that is, what percent of the total mass arises from each element. These calculations are simple and involve taking the atomic mass of the element in question and dividing by the molar mass of the molecule.
For HCl, the fraction of hydrogen in HCl is given by the molar mass of hydrogen divided by the molar mass of HCl:
$\frac{1.008\frac{g}{mol}}{36.46\frac{g}{mol}} \nonumber$
and the percentage of hydrogen in HCl is obtained by multiplying the fraction by 100:
$0.02765\times 100=2.765\% \nonumber$
Combining the steps, the percentage of chlorine in HCl can be calculated by dividing the molar mass of chlorine by the molar mass of HCl and multiplying by 100:
$\frac{1.008\frac{g}{mol}}{36.46\frac{g}{mol}}=0.02765 \nonumber$
Exercise $1$
1. Find the percentage of fluorine in calcium fluoride (CaF2).
These types of problems can also be presented as mass calculations. For example, determine the mass of calcium in 423.6 grams of CaF2. Collecting our known, given and find values:
The known molar mass of Ca is
$\frac{40.08g\; Ca}{1mol\; Ca} \nonumber$
The known molar mass of CaF2 is
$\frac{78.08g\; CaF_{2}}{1mol\; Cl_{2}} \nonumber$
We are given that we have a sample of CaF2 with a mass of 423.6 grams and we want to find the mass of Ca in this sample. We could find the mass of Ca if we knew the fraction of Ca in CaF2. We could then multiply this fraction by the known mass of CaF2 to obtain the mass of calcium in the sample. The fraction of Ca in CaF2 is the ratio of the two known molar masses (the percentage, before you multiply by 100). As always, to perform the dimensional analysis, we arrange the known and the given so that the units cancel, leaving only the units of the item we want to find.
$(423.6g\; CaF_{2})\times \left ( \frac{1mol}{78.08g\; CaF_{2}} \right )\times \left ( \frac{40.08g\; Ca}{1mol\; Ca}\right )=x\: g\; Ca=217.4\: g\; Ca \nonumber$
Note that in this example, you want to use the fraction of Ca in CaF2, not the percentage; if you used percentage, you would have to divide your answer by 100 to get the proper number of grams of Ca in the sample.
Exercise $1$
1. A sample of CaF2 is known to contain 18.00 grams of calcium; what mass of fluorine is contained in this sample?
2. Carbon dioxide is a green house gas produced in combustion. What is the percentage of oxygen in CO2?
3. Barium sulfate is use when x-raying the gastrointestinal track. Determine the mass of barium in 523 grams of BaSO4.
4. Sulfuric acid, H2SO4 is the most used chemical in industrial processes. If a sample of sulfuric acid contained 3.67 g of hydrogen how many grams of sulfur would it contain? | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/04%3A_The_Mole_and_Measurement_in_Chemistry/4.4%3A_Percentage_Composition.txt |
In Chapter 2, we introduced the concept of a chemical compound as a substance that results from the combination of two or more atoms, in such a way that the atoms are bonded together in a constant ratio. We represented that ratio using the symbols for the atoms in the molecule, with subscripts to indicate the fixed ratios of the various atoms. The result is a molecular formula, and in Chapter 3, we used molecular formulas to devise chemical names for both molecular and ionic compounds. As we have seen in this chapter, molecular formulas can be used to directly calculate the molar mass of a compound.
Many of the methods, however, that chemists use in the laboratory to determine the composition of compounds do not give the molecular formula of the compound directly, but instead simply yield the lowest whole-number ratio of the elements in the compound. A formula such as this is called an empirical formula. For example, the molecular formula for glucose is C6H12O6, but the simplest whole-number ratio of the elements in glucose is CH2O; if you multiply each element in (CH2O) by six, you obtain the molecular formula for glucose. An empirical formula cannot be converted into a molecular formula unless you know the molar mass of the compound. For example, the empirical formula for acetic acid (the acidic component in vinegar) is identical to that for glucose (CH2O). If you analyzed these two compounds and determined only an empirical formula, you could not identify which compound you had.
Conversion of an empirical formula into a molecular formula requires that you know the molar mass of the compound in question. Knowing this, you can calculate a molecular formula based on the fact that an empirical formula can always be multiplied by an integer n to yield a molecular formula. Thus, some value of n, multiplying each element in CH2O will yield the molecular formula of acetic acid. The value of n can be determined as follows:
$n\times (CH_{2}O),\: where\; n=\frac{molar\; mass\; of\; the\; compound}{molar\; mass\; of\; the\; empirical\; formula} \nonumber$
For acetic acid, the molar mass is 60.05 g/mol, and the molar mass of the empirical formula CH2O is 30.02 g/mol. The value of the integer n for acetic acid is therefore,
$n=\frac{60.05\; g/mol}{30.02\; g/mol}=2 \nonumber$
And the molecular formula is C2H4O2.
Note that n must be an integer and that your calculation should always yield a whole number (or very close to one).
Exercise $1$
1. A compound is determined to have a molar mass of 58.12 g/mol and an empirical formula of C2H5; determine the molecular formula for this compound.
2. Benzene is an intermediate in the production of many important chemicals used in the manufacture of plastics, drugs, dyes, detergents and insecticides. Benzene has an empirical formula of CH. It has a molar mass of 78.11 g/mol.What is the molecular formula?
4.S: The Mole and Measurements in Chemistry (Summary)
• The mole is defined as the number of atoms contained in exactly 12 grams of carbon-12 (the isotope ). There are 6.0221415 x 1023 particles in a mole. Remember, a mole is just a number (like dozen) and you can have a mole of anything.
• The concept of a mole is based on Avogadro’s Hypothesis (equal volumes of all gases at the same temperature and pressure contained the same number of molecules) and the number of particles in a mole (6.0221415 x 1023) is commonly referred to as Avogadro’s number (typically rounded to 6.02 x 1023 for most calculations).
• Because atomic masses, and the number of particles in a mole, are both based on the isotopic atomic mass of the isotope carbon-12, the mass of any substance expressed in atomic mass units is numerically equal to the molar mass of the substance in grams per mole. Thus, exactly 12 grams of carbon-12 contains exactly a mole of carbon atoms; likewise, 31.9988 grams of O2 contains 6.02214 x 1023 oxygen molecules (note, six significant figures), etc.
• To convert the number of moles of a substance into the mass of a substance, you simply need to multiply (moles x molar mass).
• To convert the mass of a substance into the number of moles, you simply need to divide the mass by the molar mass.
• To convert the number of moles of a substance into the number of particles of that substance, you simply need to multiply (moles x Avogadro’s number).
• The percentage composition of a compound, simply tells us what percent of the total mass arises from each element in the compound. To do the calculation, simply take the atomic mass of the element in question and divide it by the molar mass of the molecule.
• The empirical formula for a compound is the lowest whole-number ratio of the elements in that compound. For example, the molecular formula for glucose is C6H12O6, but the simplest whole-number ratio of the elements in glucose is CH2O. | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/04%3A_The_Mole_and_Measurement_in_Chemistry/4.5%3A_Empirical_and_Molecular_Formulas.txt |
In Chapter 2, we learned that chemical changes result in the transformation of one chemical substance into a different substance having a new set of chemical and physical properties. The transformation of one substance into another is called a chemical reaction and is described using a chemical equation. In this chapter we will learn how to write and balance simple chemical equations. We will learn the basic types of chemical reactions and we will learn how to predict the products that are likely to be formed when these reactions occur. We will examine a special type of chemical reaction in which one of the products has low solubility in water and precipitates from solution. Understanding the basic rules of solubility is simple and again will allow us to predict when this type of reaction is likely to be observed. Finally, we will address the energetics of chemical reactions, laying a fundamental background for the study of reaction rates and equilibrium later in the course.
Thumbnail: When a zinc rod is inserted into a beaker that contains an aqueous solution of copper(II) sulfate, a spontaneous redox reaction occurs: the zinc electrode dissolves to give Zn2+(aq) ions, while Cu2+(aq) ions are simultaneously reduced to metallic copper. The reaction occurs so rapidly that the copper is deposited as very fine particles that appear black, rather than the usual reddish color of copper.
05: Chemical Reactions
In Chapter 2, we classified changes in our environment utilizing the concepts of physical and chemical changes. We said that a physical change alters the appearance of a substance without changing its molecular structure. Ice melts, water evaporates and mountains are slowly weathered into dust. All of these change the characteristics of substances, but they do not alter its basic structure. A chemical change, however, results in the transformation of one molecular substance into another. Gasoline burns, reacting with oxygen in the atmosphere, generating light, heat, and converting the carbon-based molecules into carbon dioxide gas and water vapor. When substances combine like this and undergo chemical changes, we say that a chemical reaction has occurred. Some chemical reactions are quite evident, like the burning of gasoline, and involve the production of heat or light. In other types of chemical reactions, gases are evolved, color changes occur and clear solutions become cloudy, with the ultimate formation of an insoluble substance (a precipitate). Chemical changes can also be quite obscure and their occurrence can only be detected by sophisticated chemical analysis.
Sometimes chemical changes occur spontaneously, others require the input of energy (heat) in order to occur. Chemical reactions can occur rapidly, like the explosive reaction of sodium metal in the presence of water, and others occur very slowly, like the rusting of iron or the tarnish that slowly develops on some metal surfaces exposed to air. In this chapter we will learn to represent chemical reactions using chemical equations. We will learn to balance these equations, explore types of reactions and learn to predict products from simple reactions. Central to all of this is the concept of the chemical equation.
5.2: Chemical Equations
The processes that occur during a chemical change can be represented using a chemical equation. In a chemical equation, the chemical formulas for the substance or substances that undergo the chemical reaction (the reactants) and the formulas for the new substance or substances that are formed (the products) are both shown, and are linked by an arrow. The arrow in a chemical equation has the properties of an “equals sign” in mathematics, and because of this, in a chemical equation, there must be the same number and types of atoms on each side of the arrow.
ReactantsProducts
As an example of a chemical reaction, consider the reaction between solid carbon and oxygen gas to form carbon dioxide gas. This chemical equation for this reaction can be written as shown below.
C (s) + O2 (g) → CO2 (g)
In this equation, we have used (s) and (g) to represent the physical state of the reactants at the time of the reaction (solid and gas). Other abbreviations that are often used include (l) for liquid and (aq) to indicate that the reactant or product is dissolved in aqueous solution.
As we inspect this equation we see that there is one carbon atom on each side of the arrow and that there are two oxygen atoms on each side. An equation in which there are the same number and types of atoms on both sides of the arrow is referred to as balanced. As you write chemical equations, it is important to remember those elements that naturally occur as diatomic molecules (Table 1.1). Remember that when you include these as reactants or products, remember to indicate that they are diatomic by using the subscript “2”.
Table 1.1 Common Diatomic Elements
Element Chemical Formula
Hydrogen H2
Oxygen O2
Nitrogen N2
Fluorine F2
Chlorine Cl2
Bromine Br2
Iodine I2
Exercise \(1\)
1. Write a chemical equation for the reaction of solid iron with solid sulfur to form solid iron(II) sulfide.
2. Write a chemical equation for the reaction of solid carbon with solid magnesium oxide to form carbon monoxide gas and magnesium metal. | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/05%3A_Chemical_Reactions/5.1%3A_Chemical_Changes_and_Chemical_Reactions.txt |
In another example of a chemical reaction, sodium metal reacts with chlorine gas to form solid sodium chloride. An equation describing this process is shown below.
Na (s) + Cl2 (g) → NaCl (s)
Inspection of this equation, however, shows that, while there is one sodium atom on each side of the arrow, there are two chlorine atoms in the reactants and only one in the products. This equation is not balanced, and is therefore not a valid chemical equation. In order to balance this equation, we must insert coefficients (not subscripts) in front of the appropriate reactants or products so that the same number and types of atoms appear on both sides of the equation. Because chlorine is diatomic, there are two chlorines in the reactants and there must also be two chlorines in the products. In order to accomplish this, we place the coefficient “2” in front of the product, NaCl. Now we are balanced for chlorine, but there are two atoms of sodium in the products and only one shown in the reactants. To resolve this, we need to place the coefficient “2” in front of the sodium in the reactant, to give the equation shown below.
2 Na (s) + Cl2(g) → 2 NaCl (s)
In this equation, there are two sodiums in the reactants, two sodiums in the products, two chlorines in the reactants and two chlorines in the products; the equation is now balanced.
There are many different strategies that people use in order to balance chemical equations. The simplest methods, where you examine and modify coefficients in some systematic order, is generally called “balancing by inspection”. These methods are generally useful for most simple chemical equations, although mathematical algorithms are often necessary for highly complex reactions. One version of the “inspection” method that we will use in this section can be called the “odd-even” approach. Looking at the first equation that we wrote for the sodium-chlorine reaction, we note that there are an odd number of chlorines in the products and an even number of chlorines in the reactants. The first thing we did in balancing this equation was to insert the multiplier “2” in front of the product (NaCl) so that there were now an even number of chlorines on both sides of the equation. Once we did that, we simply had to balance the other element (Na) which was “odd” on both sides, and the equation was easily balanced. When you are using this approach with more complicated equations, it is often useful to begin by balancing the most complex molecule in the equation first (the one with the most atoms), and focus on the element in this compound that is present in the greatest amount.
Another example where the “odd-even” approach works well is the decomposition of hydrogen peroxide to yield water and oxygen gas, as shown below.
H2O2 (aq) → O2 (g) + H2O (l)
As we inspect this equation, we note that there are an even number of oxygen atoms in the reactants and an odd number of oxygens in the products. Specifically, water has only one oxygen (in the products) and the number of oxygen atoms in the products can be made “even” by inserting the coefficient “2” in front of H2O. Doing this (shown below) we note that we now have four hydrogens in the products and only two in the reactants.
H2O2 (aq) → O2 (g) + 2 H2O (l)
Balancing the hydrogens by inserting “2” in front of H2O2 in the reactants gives us an equation with four hydrogens on both sides on four oxygens on both sides; the equation is now balanced.
2 H2O2 (aq) → O2 (g) + 2 H2O (l)
Exercise \(1\)
Write a balanced chemical equation for the reactions given below:
1. When hydrogen gas reacts is combined with oxygen gas and the mixture ignited with a spark, water is formed in a violent reaction.
2. Lead (IV) oxide reacts with HCl to give lead (II) chloride, chlorine gas and water.
3. Solid potassium chlorate decomposes on heating to form solid KCl and oxygen gas.
4. An aqueous solution of barium chloride reacts with an aqueous solution of sodium sulfate to form solid barium sulfate and a solution of sodium chloride.
5. Hydrogen reacts with nitrogen to give ammonia, according to the equation shown below; balance this equation.
_____H2 (g) + _____N2 (g) → _____NH3 (g)
6. Zinc metal reacts with aqueous HCl to give hydrogen gas and zinc chloride, according to the equation shown below; balance this equation.
_____Zn (s) + _____HCl (aq) → _____H2 (g) + _____ZnCl2 (aq)
7. Iron(III) oxide reacts with chlorine gas to give iron(III) chloride and oxygen gas, according to the equation shown below; balance this equation.
_____Fe2O3 (s) + _____Cl2 (g) → _____FeCl3 (s) + _____O2 (g)
8. Sodium metal reacts with ammonia to give sodium amide and hydrogen gas, according to the equation shown below; balance this equation.
_____Na (s) + _____NH3 (l) → _____H2 (g) + _____NaNH2 (s)
9. Ethane reacts with oxygen gas to give carbon dioxide and water vapor, according to the equation shown below; balance this equation.
_____C2H6 (g) + _____O2 (g) → _____CO2 (g) + _____H2O (g) | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/05%3A_Chemical_Reactions/5.3%3A_Balancing_Chemical_Equations.txt |
The reactions we have examined in the previous sections can be classified into a few simple types. Organizing reactions in this way is useful because it will assist us in predicting the products of unknown reactions. There are many different classifications of chemical reactions, but here we will focus on the following types: synthesis, decomposition, single replacement and double replacement. In addition, we will see that some of these reactions involve changes in the oxidation numbers of the reactants and products; these will be referred to as oxidation-reduction, or “redox” reactions.
The first type of reaction we will consider is a synthesis reaction (also called a combination reaction). In a synthesis reaction, elements or compounds undergo reaction and combine to form a single new substance. The reaction of sodium metal with chlorine gas to give sodium chloride is an example of a synthesis reaction where both reactants are elements.
2 Na (s) + Cl2 (g) → 2 NaCl (s)
In this reaction, sodium metal and chlorine gas have combined to yield (synthesize) the more complex molecule, sodium chloride. Another example of a synthesis reaction, where compounds are involved as reactants, is the reaction of the organic molecule ethene (a carbon-based molecule – covered in more depth in Chapter 14) with HBr to yield the organic molecule bromoethane.
In this example, two molecular compounds (the organic compound, ethene, and hydrogen bromide) have combined to yield (synthesize) the new molecule, bromoethane. In a similar manner, synthesis reaction can also involve elements reacting with compounds.
In decomposition reactions, a single compound will break down to form two or more new substances. The substances formed can be elements, compounds, or a mixture of both elements and compounds. Two simple examples of decomposition reactions are shown below.
Cu2S (s) → 2 Cu (s) + S (s)
CaCO3 (s) → CaO (s) + CO2 (g)
In a single-replacement reaction (also called a single-displacement reaction) an element and a compound will react so that their elements are switched. In other words, an element will typically displace another element from within a compound. As a general rule, metals will replace metals in compounds and non-metals will typically replace non-metals. An example of a single replacement reaction is shown below.
Zn (s) + CuCl2 (s) → ZnCl2 (s) + Cu (s)
In this example, elemental zinc has displaced the metal, copper, from copper(II) chloride to form zinc chloride and elemental copper. In the reactants, zinc was elemental and in the products, it is present within the compound, zinc chloride. Likewise, copper was present in a compound in the reactants and is elemental in the products.
In another example, iron metal will react with an aqueous solution of copper sulfate to give copper metal and iron(II) sulfate.
Fe (s) + CuSO4 (aq) → Cu (s) + FeSO4 (aq)
In this reaction, elemental iron replaces copper in a compound with sulfate anion and elemental copper metal is formed; metal replaces metal. The tendency of metals to replace other metals in single-replacement reactions is often referred to as an activity series. In the activity series shown below, any metal will replace any other metal which is to the right of itself in the series. Thus, iron will replace copper, as shown above, but copper metal would not replace iron from iron sulfate (iron is more active than copper). The activity series is useful in predicting whether a given single-replacement reaction will occur or not. Note that hydrogen is also included in the table, although it is not generally considered a “metal”. Within this table, metals which occur before hydrogen will react with acids to form hydrogen gas and metal salts. Copper, silver, mercury and gold are less active than hydrogen and will not liberate hydrogen from simple acids. The metals in Groups I & II of the periodic table (Li, Na, K, Rb, Cs, Ca, Sr and Ba) are so reactive that they will directly react with water to liberate hydrogen gas and form metal hydroxides. These are generally referred to as “active metals”.
The activity series is useful in predicting whether a given single-replacement reaction will occur or not. Note that hydrogen is also included in the table, although it is not generally considered a “metal”. Within this table, metals which occur before hydrogen will react with acids to form hydrogen gas and metal salts. Copper, silver, mercury and gold are less active than hydrogen and will not liberate hydrogen from simple acids. The metals in Groups I & II of the periodic table (Li, Na, K, Rb, Cs, Ca, Sr and Ba) are so reactive that they will directly react with water to liberate hydrogen gas and form metal hydroxides. These are generally referred to as “active metals”.
A double-replacement reaction (or double-displacement) two ionic compounds in aqueous solution switch anions and form two new compounds. In order for a chemical reaction to occur in a double-replacement reaction, one of the new compounds that is formed must be insoluble in water, forming a solid precipitate (or a gas, to be covered in Chapter 6). If both of the new compounds which are formed are water-soluble, then no reaction has occurred. In the reactants, there were two cations and two anions in solution, and in the products there are the same two cations and two anions, in the same solution; nothing has happened. An example of a double-replacement reaction is shown below.
BaCl2 (aq) + Na2SO4 (aq) → BaSO4 (s) + 2 NaCl (aq)
In this reaction, solid barium sulfate is formed as a precipitate. This is a chemical change and this is a valid chemical reaction. In order to predict whether a double-replacement reaction will occur or not you need to understand rules for predicting solubility of ionic compounds. These rules will be covered in a later section, but are not related to the activity series discussed above.
Table 5.2 Activity Series for Common Metals
Li > K > Ba > Sr > Ca >
Na > Mg > Al > Mn > Zn >
Fe > Cd > Co > Ni > Sn >
Pb > H > Cu > Ag > Hg >
Au | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/05%3A_Chemical_Reactions/5.4%3A_Classifying_Chemical_Reactions.txt |
The oxidation number of an element represents the total number of electrons which have been removed (a positive oxidation state) or added (a negative oxidation state) to get the element into its present state. The term oxidation describes the loss of electrons by an element and an increase in oxidation state; the term reduction describes the gain of electrons and a decrease in oxidation state. Oxidation numbers for elements in compounds can be calculated using a simple set of rules, which are reproduced below in Table 5.3.
Table 5.3 Rules for Assigning Oxidation Numbers
Table 5.3 Rules for Assigning Oxidation Numbers
1. The oxidation number of an element in the free state is zero.
2. A monoatomic ion will have an oxidation number that is equal to its charge.
3. In compounds with metals, hydrogen will be –1, otherwise it will always be +1.
4. Oxygen, within a compound, will generally have an oxidation number of –2.
5. Halogens will be –1, except in compounds with oxygen.
6. Sulfur will generally be –2, except in compounds with oxygen.
7. In a molecular compound, the most electronegative element is assigned a negative oxidation number
In many chemical reactions, the oxidation number of elements change. Consider the synthesis reaction shown below. In the reactants, carbon and oxygen are both elements and their oxidation numbers are zero (Rule 1). In the product, oxygen will have an oxidation number of –2 (Rule 4), therefore, carbon in CO2 must have an oxidation number of +4 in order to balance the four negative charges on the oxygens. During this reaction, the oxidation number of carbon has changed from zero in the reactants to +4 in the products and the oxidation number of oxygen has changed from zero to –2. This is an example of a redox reaction; a chemical reaction in which the oxidation numbers of elements change on going from reactants to products.
C (s) + O2 (g) → CO2 (g)
In a redox reaction, the element that “loses electrons” is said to be oxidized and will have an increase in its oxidation number. In the example above, the oxidation number of carbon increases from zero to +4; it has “lost electrons” and has been oxidized. The element that “gains electrons” in a redox reaction is said to be reduced and will have a decrease in its oxidation number. In the reaction above, the oxidation number of oxygen has decreased from zero to –2; it has “gained electrons” and has been reduced.
Exercise \(1\)
Arsonic and nitric acids react to form nitrogen monoxide, arsenic acid and water according to the equation shown below. Is this an example of a redox reaction?
2 HNO3 (aq) + 3 H3AsO3 (aq) → 2 NO (g) + 3 H3AsO4 (aq) + H2O (l)
Exercise \(1\)
For each of the reactions given below, calculate the oxidation number of each of the elements in the reactants and the products and determine if the reaction involves oxidation-reduction. If it is a redox reaction, identify the elements that have been oxidized and reduced.
1. Cu2S → 2 Cu + S
Reactants: Cu _____ S _____
Products: Cu _____ S _____
Element oxidized: __________ Element Reduced __________
2. CaCO3 → CaO + CO2
Reactants: Ca _____ C _____ O _____
Products: Ca _____ C _____ O _____
Element oxidized: __________ Element Reduced __________
3. Fe2O3 + 3 H2 → 2 Fe + 3 H2O
Reactants: Fe _____ O _____ H _____
Products: Fe _____ O _____ H _____
Element oxidized: __________ Element Reduced __________
4. AgNO3 + NaCl → AgCl (s) + NaNO3
Reactants: Ag _____ N _____ O _____ Na _____ Cl _____
Products: Ag _____ N _____ O _____ Na _____ Cl _____
Element oxidized: __________ Element Reduced __________ | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/05%3A_Chemical_Reactions/5.5%3A_Oxidation_and_Reduction_Reactions.txt |
Part of the lure of chemistry is that things don’t always work out the way you expect. You plan a reaction, anticipate the products and, quite often, the results astound you! The exercise, then, is trying to figure out what was formed, why, and whether your observation leads to other useful generalizations. The first step in this process of discovery is anticipating or predicting the products which are likely to be formed in a given chemical reaction. The guidelines we describe here will accurately predict the products of most classes of simple chemical reactions. As your experience in chemistry grows, however, you will begin to appreciate the unexpected!
In simple synthesis reactions involving reaction of elements, such as aluminum metal reacting with chlorine gas, the product will be a simple compound containing both elements. In this case, it is easiest to consider the common charges that the elements adopt as ions and build your product accordingly. Aluminum is a Group III element and will typically form a +3 ion. Chlorine, being Group VII, will accept one electron and form a monoanion. Putting these predictions together, the product is likely to be AlCl3. In fact, if aluminum metal and chlorine gas are allowed to react, solid AlCl3 is the predominant product.
2 Al (s) + 3 Cl2 (g) → 2 AlCl3 (s)
The synthesis reaction involving the non-metals hydrogen gas and bromine can be approached similarly. The product will contain both elements. Hydrogen, Group I, has one valence electron and will form one covalent bond. Bromine, Group VII, has seven valence electrons and will form one covalent bond. The likely product is therefore HBr, with one covalent bond between the hydrogen and the bromine.
H2 (g) + Br2 (g) < → 2 HBr (g)
For a single-replacement reaction, recall that (in general) metals will replace metals and non-metals will replace non-metals. For the reaction between lead(IV) chloride and fluorine gas, the fluorine will replace the chlorine, leading to a compound between lead and fluorine and the production of elemental chlorine. The lead can be viewed as a “spectator” in the reaction and the product is likely to be lead(IV) fluoride. The complete equation is shown below.
PbCl4 (s) + 2 F2 (g) → PbF4 (s) + 2 Cl2 (g)
In single-replacement reactions in which metals (or carbon or hydrogen) are expected to replace metals, first you should check the activity series to see if any reaction is anticipated. Remember that metals can only replace metals that are less active than themselves (to the right in the Table). If the reaction is predicted to occur, use the same general guidelines that we used above. For example, solid iron reacting with aqueous sulfuric acid (H2SO4). In this reaction the question is whether iron will displace hydrogen and form hydrogen gas. Consulting the activity series, we see that hydrogen is to the right of iron, meaning that the reaction is expected to occur. Next, we reason that iron will replace hydrogen, leading to the formation of iron sulfate, where the sulfate is the “spectator” ion. The formation of hydrogen gas requires a change in oxidation number in the hydrogen of +1 to zero. Two hydrogen atoms must therefore be reduced (a decrease in oxidation number) and the two electrons required for the reduction must come from the iron. The charge on the iron is therefore most likely to be +2 (it starts off at zero and donates two electrons to the hydrogens). The final product is therefore most likely iron(II) sulfate. The complete equation is shown below.
Fe (s) + H2SO4 (aq) → FeSO4 (aq) + H2 (g)
Decomposition reactions are the most difficult to predict, but there are some general trends that are useful. For example, most metal carbonates will decompose on heating to yield the metal oxide and carbon dioxide.
NiCO3 (s) → NiO (s) + CO2 (g)
Metal hydrogen carbonates also decompose on heating to give the metal carbonate, carbon dioxide and water.
2 NaHCO3 (s) → Na2CO3 (s) + H2O (g) + CO2 (g)
Finally, many oxygen-containing compounds will decompose on heating to yield oxygen gas and “other compounds”. Identifying these compounds and building an understanding of why and how they are formed is one of the challenges of chemistry. Some examples:
H2O2 (aq) → O2 (g) + H2O (l)
2 HgO (s) → O2 (g) + 2 Hg (l)
2 KClO3 (s) → 3 O2 (g) + 2 KCl (s)
The potential products in double-replacement reactions are simple to predict; the anions and cations simply exchange. Remember, however, that one of the products must precipitate, otherwise no chemical reaction has occurred. For the reaction between lead(II) nitrate and potassium iodide, the products are predicted to be lead(II) iodide and potassium nitrate. No redox occurs, and the product, lead iodide, precipitates from the solution as a bright yellow solid. The question of how do you predict this type of solubility trend is addressed in the next section.
Pb(NO3)2 (aq) + 2 KI(aq)< → PbI2 (s) + 2 KNO3 (aq) | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/05%3A_Chemical_Reactions/5.6%3A_Predicting_Products_from_Chemical_Reactions.txt |
The solubility of many simple ionic compounds can be predicted by applying the set of rules shown below.
1. Salts of the alkali metal ions and the ammonium ion, Li+, Na+, K+, and NH4+ are almost always soluble.
2. Virtually all metal nitrates and metal acetates are soluble.
3. Metal halides are generally soluble, except for salts of Ag+, Pb2+, Cu+ and Hg+.
4. Metal sulfates are generally soluble, except for salts of Ba2+, Pb2+ and Ca2+.
5. With exception of the alkali metal ions and ammonium (Rule 1), the following salts are generally insoluble: metal carbonates (CO32-), metal phosphates (PO43-) and metal chromates (CrO42-).
6. Metal hydroxides and metal sulfides are generally insoluble, except for those covered by Rule 1 and Ca2+, Sr2+ and Ba2+.
Applying these rules to the reaction between lead nitrate and potassium iodide, the reactants are both soluble (Rule 1 and Rule 2). In the products, potassium nitrate will be soluble (Rule 2) and lead iodide will be insoluble, based on Rule 3.
Exercise \(1\)
Mixing each of the following salt solutions results in the formation of a precipate. In each case, identify the insoluble salt.
1. NaCl + Pb(NO3)2
2. Fe(C2H3O2)3 + KOH
3. Ca(NO3)2 + K2SO4
4. Li2S + CuSO4
5. Co(C2H3O2)2 + LiOH
Exercise \(2\)
For each of the ionic compounds given below, determine whether or not the compound will be soluble in water, according to the trends given above.
1. AgNO3 soluble insoluble
2. MgCl2 soluble insoluble
3. Na2SO4 soluble insoluble
4. AgCl soluble insoluble
5. Ba(NO3)2 soluble insoluble
6. PbI2 soluble insoluble
7. Mg(NO3)2 soluble insoluble
8. BaSO4 soluble insoluble
9. FeCl3 soluble insoluble
10. Pb(CH3COO)2 soluble insoluble
5.8: The Energetics of Chemical Reactions
Many of the chemical reactions that we have discussed in this chapter occur with the generation of significant amounts of light and heat. A prime example is the synthesis reaction between zinc and sulfur, described by the equation shown below.
Zn (s) + S (s) → ZnS (s)
Initially, both elements are present as fine powders. They are mixed (carefully) and the mixture is stable sitting of the laboratory bench. When the mixture is touched with a heated metal rod, however, a violent reaction occurs (the reaction is termed exothermic; producing heat) and zinc sulfide is formed as the product. The reaction is obviously favorable, so why does it need heat to start the reaction? This concept can be understood by considering a reaction coordinate diagram for a simple one-step reaction.
In a reaction coordinate diagram, the y-axis corresponds to energy. The initial and final “energy wells” represent the ground-state energies of the reactants and products, and the path connecting them describes the energy changes that occur in the course of the reaction.
Looking at the reaction coordinate diagram for the zinc sulfide reaction, the reactants sit at an initial energy level that is characteristic and unique for each element or compound. Likewise, zinc sulfide sits at a lower overall energy level; that means that the conversion from the elements to the compound is favorable and that heat is liberated during the reaction. If the energy level of the products was higher than the reactants, the reaction would be unfavorable and heat would be absorbed during the reaction (the reaction is said to be endothermic; consuming heat).
Why, then, does the zinc sulfide reaction need energy input before the reaction begins? The answer lies in the curved path that connects the reactants and products in the reaction coordinate diagram. In order for the zinc-sulfur mixture to react, enough energy must be put into the system so that the energy level of the reactants equals the highest hill in the diagram. Once that point is reached, the reactants can “tumble down” the energy hill and form the more stable products (with the evolution of all of the excess energy as heat, light, etc.).
The top of the energy hill in a reaction coordinate diagram is called the transition state (or activated complex). In modern chemical theory, the transition state is the energy maximum corresponding to the processes of bond-making and bond-breaking. The energy required to go from the reactants to the transition state is the activation energy. Reactions that occur with little requirement for heat simply have a small activation energy. The magnitude of the activation energy controls the rate of a reaction and the difference in energy between the reactants and the products controls the equilibrium distribution of the products and reactants in a reversible reaction. We will return to these concepts when we address reaction rates and equilibrium later in the book. | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/05%3A_Chemical_Reactions/5.7%3A_Predicting_Solubility_Trends.txt |
• The processes that occur during a chemical change can be represented using a chemical equation. In a chemical equation, the chemical formulas for the substance or substances that undergo the chemical reaction (the reactants) and the formulas for the new substance or substances that are formed (the products) are both shown, and are linked by an arrow. The arrow in a chemical equation has the properties of an “equals sign” in mathematics, and because of this, in a chemical equation, there must be the same number and types of atoms on each side of the arrow.
• A chemical equation in which the same number and types of atoms appear on each side of the arrow is called balanced. In order to balance an equation, insert coefficients in front of the appropriate reactants or products until there are the same number and types of atoms on both sides of the arrow.
• In a synthesis reaction', elements or compounds undergo reaction and combine to form a single new substance.
• In a decomposition' reaction, a single compound will break down to form two or more new substances. The substances formed can be elements, compounds, or a mixture of both.
• In a single-replacement' (single-displacement) reaction, an element and a compound will react so that their elements are switched. As a general rule, metals will replace metals in compounds and non-metals will typically replace non-metals.
• A double-replacement' (double-displacement) reaction, two ionic compounds in aqueous solution switch anions and form two new compounds. In order for a chemical reaction to occur, one of the new compounds that is formed must be insoluble in water, forming a solid precipitate or a gas.
• The oxidation number' of an element represents the total number of electrons which have been removed (a positive oxidation state) or added (a negative oxidation state) to get the element into its present state. The term oxidation describes the loss of electrons by an element and an increase in oxidation state; the term reduction describes the gain of electrons and a decrease in oxidation state.
• A chemical reaction in which oxidation numbers undergo a change is called a redox reaction. In a redox reaction, the element that “loses electrons” is said to be oxidized and will have an increase in its oxidation number. The element that “gains electrons” in a redox reaction is said to be reduced and will have a decrease in its oxidation number.
• In a simple synthesis reaction involving reaction of elements, the product will be a compound containing both elements. Write the product considering the common charges that the elements adopt as ions or the number of bonds that the elements typically form in molecules.
• In a simple single-replacement reaction, (in general) metals (including carbon and hydrogen) will tend to replace metals and that non-metals will replace non-metals.
• In a double-replacement reaction, the anions and cations simply of the two compounds simply exchange. In order for a reaction to occur, one of the products must precipitate, otherwise no chemical reaction has occurred. Changes in oxidation numbers do not occur in double-replacement reactions.
• Solubility trends can be predicted using a simple set of rules shown in Table 5.5; you should review there rules, remembering to apply them in order.
• The energy required to initiate a chemical reaction is called the activation energy. The greater the activation energy, the slower, or less favorable a reaction will be. The magnitude of the activation energy is directly linked to the rate of a chemical reaction. | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/05%3A_Chemical_Reactions/5.S%3A_Chemical_Reactions_%28Summary%29.txt |
With the background that we have built in this course, thus far, we can now approach a more quantitative view of chemistry. While the notion of chemistry and math (together in the same room) may make you want to scream, we will see in this chapter that concepts such as chemical stoichiometry and mass balance are not overwhelming and can be approached using the same problem-solving algorithms that we have mastered in previous chapters. Once the concept of stoichiometric balance has been mastered, we will finally tackle limiting reactants. Limiting reactant problems may appear challenging, but we will see it is the same calculation that we do routinely… we simply have to do the calculations twice.
• 6.1: An Introduction to Stoichiometry
Stoichiometry… what a wonderful word! It sounds so complex and so chemical. In fact, it’s a fairly simple concept; stoichiometry is the relationship between the molar masses of chemical reactants and products in a given chemical reaction.
• 6.2: Molar Stoichiometry in Chemical Equations
• 6.3: Mass Calculations
The methods described in the previous section allow us to express reactants and products in terms of moles, but what if we wanted to know how many grams of a reactant would be required to produce a given number of grams of a certain product? This logical extension is, of course, trivial!
• 6.4: Percentage Yield
Stoichiometric calculations will give you a theoretical yield for a reaction; the yield that you should obtain assuming that the reaction proceeds with 100% efficiency and that no material is lost in handling. The amount of material that you isolate from a given reaction is called the actual yield and it is always less than the theoretical yield. The percentage of the theoretical yield that you actually isolate is called the percentage yield.
• 6.5: Limiting Reactants
You may have noticed that, in many of the problems in this chapter, we stated that one reactant reacted with an excess of a second reactant. In all of these cases, the theoretical yield of product is determined by the limiting reactant in the reaction, and some of the excess reactant is left over.
• 6.S: Quantitative Relationships in Chemistry (Summary)
06: Quantitative Relationships in Chemistry
Stoichiometry… what a wonderful word! It sounds so complex and so chemical. In fact, it’s a fairly simple concept; stoichiometry is the relationship between the molar masses of chemical reactants and products in a given chemical reaction. In Chapter 5 we learned to balance chemical equations by inserting numerical coefficients in front of reactants or products so that there were the same number and types of atoms on each side of the equation. Thus, in the reaction between sodium metal and chlorine gas, the balanced chemical equation is:
$\ce{2 Na (s) + Cl_2 (g) \rightarrow 2 NaCl (s)} \nonumber$
This equation tells us that two atoms of sodium react with one molecule of chlorine gas to give two sodium chlorides. The coefficients in front of the sodium and the sodium chloride are called the stoichiometric coefficients for this reaction. If we were to totally react a single molecule of chlorine gas in this reaction, this stoichiometry tells us that two atoms of sodium (atomic mass 22.99) having a total mass of 45.98 amu, would react with one molecule of chlorine gas (having a mass 70.90 amu) to give two sodium chlorides (formula mass 137.94), for a total of 275.9 amu of product. Because chemists usually don’t speak of chemical reactions in terms of individual atoms or molecules, it is much more common to describe the stoichiometry of a reaction in terms of grams of reactants and products, or more conveniently in terms of moles. Molar stoichiometry is simply the expression of the coefficients of a reaction in terms of moles of reactants and products. | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/06%3A_Quantitative_Relationships_in_Chemistry/6.1%3A_An_Introduction_to_Stoichiometry.txt |
If we were to describe the reaction of sodium metal with chlorine gas in molar terms, we would say that two moles of sodium metal combine with one mole of Cl2 to give two moles of sodium chloride. In terms of mass, two moles of sodium, having a total mass of 45.98 grams, would react with one mole of chlorine gas (a mass 70.90 grams) to give two moles of sodium chloride, for a total of 275.9 grams of product. Likewise, the molar stoichiometry for the decomposition of hydrogen peroxide (H2O2) to form oxygen and water, can be described simply as two moles of H2O2 decompose to form one mole of oxygen gas and two moles of water.
$2 H_2O_2 (aq) \rightarrow O_2 (g) + 2 H_2O (l) \nonumber$
The stoichiometric coefficients for this reaction gives us the key information about the relationship between molar quantities of reactants and products, but in the real world, we will not always be working with exactly two moles of hydrogen peroxide. What if you want to know how much oxygen gas will be formed when 0.28 moles of H2O2 decompose? One way to solve this type of problem is to utilize a tool that we will call a reaction pathway. The reaction pathway is a kind of simple map of the stoichiometry of a reaction, which uses arrows to show the relationship between reactants and products.
Using the given-find-ratio algorithm that we introduced back in Chapter 1, if we were given mol reactant and we wanted to find mol product, we could set up a simple equation as follows:
$(\text{mol product})=\cancel{(\text{mol reactant})} \left ( \dfrac{\text{mol product}}{ \cancel{\text{mol reactant}}}\right ) \nonumber$
The units mol reactant cancel to give the solution in mol product. If we were given mol product and we wanted to find mol reactant, we would set up the equation as follows in order for the units to properly cancel:
$(\text{mol reactant})=\cancel{(\text{mol product})} \left ( \dfrac{\text{mol reactant}}{\cancel{\text{mol product}}}\right ) \nonumber$
This basic approach can be used to solve for any molar (mass, or gas) conversion based on a balanced chemical equation, as long as you are careful to set up the ratios so that the units cancel, giving you the desired solution with the proper units. As an example, return to the question of the decomposition of H2O2. If 0.28 moles of H2O2 decompose, according to the equation given below, how many moles of oxygen gas (O2) will be formed?
$\left ( \dfrac{1\; mol\; O_{2}}{2\; mol\; H_{2}O_{2}}\right ) \nonumber$
We set up the problem to solve for mol product; the general equation is:
$(molproduct)=(molreactant)\left ( \frac{molproduct}{molreactant}\right ) \nonumber$
The stoichiometric mole ratio is set up so that mol reactant will cancel, giving a solution in mol product. Substituting,
$(x\, molO_{2})=(0.28\, mol\; H_{2}O_{2})\times \left ( \frac{1\; mol\; O_{2}}{2\; mol\; H_{2}O_{2}}\right )=0.14\, mol\; O_{2} \nonumber$
Thus, the decomposition of 0.28 mol of H2O2 will produce 0.14 mol of the product, oxygen gas (O2).
Exercise $1$
1. Iron (III) oxide reacts with hydrogen gas to form elemental iron and water, according to the balanced equation shown below. How many moles of iron will be formed from the reduction of excess iron (III) oxide by 0.58 moles of hydrogen gas?
2. When an impure sample containing an unknown amount of Fe2O3 is reacted with excess hydrogen gas, 0.16 moles of solid Fe are formed. How many moles of Fe2O3 were in the original sample?
Exercise $2$
Ammonia is produced industrially from nitrogen and hydrogen according to the equation:
$N_2 + 3 H_2 \rightarrow 2 NH_3 \nonumber$
1. If you are given 6.2 moles of nitrogen how many mole of ammonia could you produce?
2. How many moles of hydrogen would you need to fully react with 6.2 moles of nitrogen?
3. If you wished to produce 11 moles of ammonia how many moles of nitrogen would you need to start with? | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/06%3A_Quantitative_Relationships_in_Chemistry/6.2%3A_Molar_Stoichiometry_in_Chemical_Equations.txt |
The methods described in the previous section allow us to express reactants and products in terms of moles, but what if we wanted to know how many grams of a reactant would be required to produce a given number of grams of a certain product? This logical extension is, of course, trivial! In Chapter 4, we learned to express molar quantities in terms of the masses of reactants or products. For example, the reduction of iron (III) oxide by hydrogen gas, produces metallic iron and water. If we were to ask how many grams of elemental iron will be formed by the reduction of 1.0 grams of iron (III) oxide, we would simply use the molar stoichiometry to determine the number of moles of iron that would be produced, and then convert moles into grams using the known molar mass. For example, one gram of Fe2O3 can be converted into mol Fe2O3 by remembering that moles of a substance is equivalent to grams of that substance divided by the molar mass of that substance:
$moles=\left ( \frac{grams}{molar\; mass} \right )=\left ( \frac{grams}{grams/mol} \right )=(grams)\times (mol/grams) \nonumber$
Using this approach, the mass of a reactant can be inserted into our reaction pathway as the ratio of mass-to-molar mass. This is shown here for the reduction of 1.0 gram of Fe2O3.
$Given:\left ( \frac{1.0g\: Fe_{2}O_{3}}{159.70\frac{g\: Fe_{2}O_{3}}{mol\: Fe_{2}O_{3}}} \right )\; \; Find: x\: mol\: Fe \nonumber$
We set up the problem to solve for mol product; the general equation is:
$(molproduct)=(molreactant)\times \left ( \frac{molproduct}{molreactant} \right ) \nonumber$
The stoichiometric mole ratio is set up so that mol reactant will cancel, giving a solution in mol product. Substituting,
$x\: mol\: Fe=\left ( \frac{1.0g\: Fe_{2}O_{3}}{159.70\frac{g\: Fe_{2}O_{3}}{mol\: Fe_{2}O_{3}}} \right )\times \left ( \frac{2\: mol\: Fe}{1\: mol\: Fe_{2}O_{3}} \right ) \nonumber$
It is often simpler to express the ratio (mass)/(molar mass) as shown below,
$\left ( \frac{1.0g\: Fe_{2}O_{3}}{159.70\frac{g\: Fe_{2}O_{3}}{mol\: Fe_{2}O_{3}}} \right )=(1.0g\: Fe_{2}O_{3})\times \left ( \frac{1\: mol\: Fe_{2}O_{3}}{159.70\: g\: Fe_{2}O_{3}} \right ) \nonumber$
Doing this, and rearranging,
$x\: mol\: Fe=(1.0g\: Fe_{2}O_{3})\times \left ( \frac{1\: mol\: Fe_{2}O_{3}}{159.70\: g\: Fe_{2}O_{3}} \right )\times \left ( \frac{2\: mol\: Fe}{1\: mol\: Fe_{2}O_{3}} \right )=0.013\: mol \nonumber$
That is, the reduction of 1.0 grams of Fe2O3 by excess hydrogen gas will produce 0.013 moles of elemental iron. All of these calculations are good to two significant figures based on the mass of iron (III) oxide in the original problem (1.0 grams). Note that we have two conversion factors (ratios) in this solution; one from mass to molar mass and the second, the stoichiometric mole ratio from the balanced chemical equation. Knowing that we have 0.013 moles of Fe, we could now convert that into grams by knowing that one mole of Fe has a mass of 55.85 grams; the yield would be 0.70 grams.
We could also modify our basic set-up so that we could find the number of grams of iron directly.
Here we have simply substituted the quantity (moles molar mass) to get mass of iron that would be produced. Again, we set up the problem to solve for mol product;
$(molproduct)=(molreactant)\times \left ( \frac{molproduct}{molreactant} \right ) \nonumber$
In place of mol product and mol reactant, we use the expressions for mass and molar mass, as shown in the scheme above. The stoichiometric mole ratio is set up so that mol reactant (the given) will cancel, giving a solution in mol product. Substituting,
$x\: mol\: Fe\left ( \frac{55.85g\: Fe}{mol\: Fe} \right )=(1.0g\: Fe_{2}O_{3})\times \left ( \frac{1\: mol\: Fe_{2}O_{3}}{159.70\: g\: Fe_{2}O_{3}} \right )\times \left ( \frac{2\: mol\: Fe}{1\: mol\: Fe_{2}O_{3}} \right ) \nonumber$
Rearranging and canceling units,
$x\: g\: Fe=\left ( \frac{55.85g\: Fe}{mol\: Fe} \right ) \left ( \frac{1\: g\: Fe_{2}O_{3}\times mol\: Fe_{2}O_{3}}{159.70\: g\: Fe_{2}O_{3}} \right )\times \left ( \frac{2\: mol\: Fe}{1\: mol\: Fe_{2}O_{3}} \right )=0.70g \nonumber$
Exercise $1$
Aqueous solutions of silver nitrate and sodium chloride react in a double-replacement reaction to form a precipitate of silver chloride, according to the balanced equation shown below.
AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq)
If 3.06 grams of solid AgCl are recovered from the reaction mixture, what mass of AgNO3 was present in the reactants?
Exercise $2$
Aluminum and chlorine gas react to form aluminum chloride according to the balanced equation shown in below.
2 Al (s) + 3 Cl2 (g) → 2 AlCl3 (s)
If 17.467 grams of chlorine gas are allowed to react with excess Al, what mass of solid aluminum chloride will be formed?
Exercise $3$
Ammonia, NH3, is also used in cleaning solutions around the house and is produced from nitrogen and hydrogen according to the equation:
N2 + 3 H2 → 2 NH3
1. If you have 6.2 moles of nitrogen what mass of ammonia could you hope to produce?
2. If you have 6.2 grams of nitrogen how many grams of hydrogen would you need? | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/06%3A_Quantitative_Relationships_in_Chemistry/6.3%3A_Mass_Calculations.txt |
When we use stoichiometric calculations to predict quantities in a reaction, our results are based on the assumption that everything happens in reality exactly as described by the chemical equation. Unfortunately, that is not always the case. When you work in the laboratory, things often go wrong. When you are weighing out reactants and transferring the materials to reaction vessels, some material will often remain on the spatula or in the weighing vessel. As you collect product, some may spill, or in a vigorous reaction, material may escape from the reaction vessel. Stoichiometric calculations will give you a theoretical yield for a reaction; the yield that you should obtain assuming that the reaction proceeds with 100% efficiency and that no material is lost in handling. The amount of material that you isolate from a given reaction is called the actual yield and it is always less than the theoretical yield. The percentage of the theoretical yield that you actually isolate is called the percentage yield.
Consider the reaction between silver nitrate and sodium chloride to form solid silver chloride. If we react 10.00 grams silver nitrate with excess sodium chloride, we would predict that we would obtain:
We set up the problem to solve for mol product; the general equation is:
$(molproduct)=(molreactant)\times \left ( \frac{molproduct}{molreactant} \right ) \nonumber$
For mol product and mol reactant, we use the expressions for (mass)/(molar mass), as shown in the scheme above. The stoichiometric mole ratio is set up so that mol reactant will cancel, giving a solution in mol product. Substituting,
$(x\: g\: AgCl)\left ( \frac{1\: mol\: AgCl}{143.32\: g\: AgCl} \right )=(10.00\: g\: AgNO_{3})\left ( \frac{1\: mol\: AgNO_{3}}{169.88\: g\: AgNO_{3}} \right )\times \left ( \frac{1\: mol\: AgCl}{1\: mol\: AgNO_{3}} \right ) \nonumber$
$(x\: g\: AgCl)=(10.00\: g\: AgNO_{3})\left ( \frac{1\: mol\: AgNO_{3}}{169.88\: g\: AgNO_{3}} \right )\times \left ( \frac{1\: mol\: AgCl}{1\: mol\: AgNO_{3}} \right )\times \left ( \frac{143.32\: g\: AgCl}{1\: mol\: AgCl} \right )=8.440\: g \nonumber$
This mass, calculated from the masses of starting materials and the stoichiometry of the equation is the theoretical yield. Because the silver chloride is a precipitate from an aqueous solution, however, we must filter it, dry it, transfer it to our balance and weigh it, before we can measure how much product we obtain as our actual yield. As we filter it, a small amount of solid is likely to remain stuck to the sides of the flask. When it is dry and we transfer it to the balance, some solid will remain on the filter paper, some on the spatula, and (most likely) your lab partner will sneeze at an inopportune moment and blow some of it all over the desktop. Considering all of this, it is very unlikely that we will end up with an actual yield of 8.044 grams of solid AgCl.
Let’s assume that we have done all of these operations (including the sneeze) and when we weigh our solid AgCl, we actually obtain 7.98 grams of solid. We know that we obtain 7.98 grams of product (our actual yield) and we calculated that we should obtain 8.440 grams of product (the theoretical yield). The percentage of the theoretical yield that we obtain is called the percentage yield, and is calculated as (actual yield)/(theoretical yield) 100. In the present case:
$\left ( \frac{7.98g}{8.440g} \right )\times 100=94.5\% \nonumber$
The percentage yield of solid AgCl that we obtained in this reaction is therefore 94.5% (not bad, actually, considering your lab partner). The concept of percentage yield is generally applied to all experimental work in chemistry.
Exercise $1$
Powdered zinc and solid sulfur combine explosively to form zinc sulfide. You and your lab partner carefully mix 0.010 mole of solid zinc powder with exactly 0.010 mole of powdered sulfur in a small porcelain crucible. Knowing your lab partner, you allow your instructor to ignite the mixture. The explosion forms a cloud of ZnS, scatters some all over the ground, and leaves a crusty pile of product in the crucible. You transfer this and determine that 0.35 grams of solid product has been recovered. Calculate the percentage yield.
Exercise $2$
The Harber process is used making ammonia from nitrogen and hydrogen according to the equation shown below. The yield of the reaction, however, is not 100%.
N2 + 3 H2 → 2 NH3
1. Suppose you end up with 6.2 moles of ammonia, but the reaction stoichiometry predicts that you should have 170.0 grams of ammonia. What is the percent yield for this reaction?
2. If you started with 6.2 grams of nitrogen and you produce 6.2 grams of ammonia what would be the percent yield? | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/06%3A_Quantitative_Relationships_in_Chemistry/6.4%3A_Percentage_Yield.txt |
Gloves will typically come is right- and left-handed models. In order to make a pair of gloves, you need one that is designed to fit each hand. If you had a box containing 50 left-handed gloves and another box containing 40 right-handed gloves, you could make 40 proper pairs and you would have ten left-handed gloves left over. The number of pairs of gloves that you could assemble is limited by the glove in the smallest number (the right-handed glove). The other glove in this example, the left, is present in excess.
The same sort of logic applies to chemical reactions in which there are two or more reactants. In Example 6.4, we carefully weighed out 0.010 mole of solid zinc and solid sulfur in order to react them to form 0.010 mole of the product, ZnS. If instead, we had reacted 0.010 mole of Zn with 0.020 moles of sulfur, how much ZnS would have (theoretically) formed? The answer is still 0.010 mole of ZnS. What happens to the leftover sulfur? It just sits there! When the reaction is complete, there is (theoretically) 0.010 mole of ZnS mixed with the remaining 0.010 mole of sulfur. 10 atoms of Zn reacting with 20 atoms of S yield 10 molecules of ZnS with 10 atoms of S remaining.
You may have noticed that, in many of the problems in this chapter, we stated that one reactant reacted with an excess of a second reactant. In all of these cases, the theoretical yield of product is determined by the limiting reactant in the reaction, and some of the excess reactant is left over. If aluminum and chlorine gas, a diatomic gas, react to form aluminum chloride according to the equation shown below,
2 Al (s) + 3 Cl2 (g) → 2 AlCl3 (s)
and there are 2.0 moles of aluminum and 14 moles of chlorine present, two moles of aluminum chloride are formed and 11 moles of chlorine gas remain in excess. Our stoichiometry is:
$\left ( \frac{3\: mol\: Cl_{2}}{2\: mol\: AlCl_{3}} \right ) \nonumber$
If three moles of chlorine gas are used in the reaction, (14 – 3) = 11 moles of chlorine must remain. When a problem is presented and one reactant is labeled as excess, the theoretical yield of product is equal to the moles of the limiting reagent, adjusted for the stoichiometry of the reaction.
Although it would be easier if reactants were routinely labeled as “limiting” or “excess”, more commonly problems are written in such a way that it is not always trivial to identify the limiting reactant. For example, if you were told that 6.0 grams of aluminum was reacted with 3.8 grams of chlorine gas and you were asked to calculate the mass of AlCl3 that would be formed, there would be no simple way to identify the limiting- and excess reactants. In a case like this what you want to do is to simply solve the problem twice. First you would calculate the number of moles of aluminum in 6.0 grams, and then calculate how many moles of AlCl3 could be formed. Next, you calculate how many moles of chlorine are present in 3.8 grams of chlorine gas and again, calculate how many moles of AlCl3 could be formed. Whichever reagent produces the smallest number of moles of product must be limiting and the other reagent must be in excess.
Find: x moles of AlCl3
We set up the problem to solve for mol product for each reactant. The general equation is:
$(molproduct)=(molreactant)\times \left ( \frac{molproduct}{molreactant} \right ) \nonumber$
The solutions for both reactants are:
$x\: mol\: AlCl_{3}=(6.0\: g\: Al)\left ( \frac{1\: mol\: Al}{26.98\: g\: Al} \right )\left ( \frac{1\: mol\: AlCl}{1\: mol\: Al} \right ) \nonumber$
$x\: mol\: AlCl_{3}=(3.8\: g\: Cl_{2})\left ( \frac{1\: mol\: Cl_{2}}{70.90\: g\: Cl_{2}} \right )\left ( \frac{2\: mol\: AlCl}{3\: mol\: Cl_{2}} \right ) \nonumber$
Solving these equations, we see that, beginning with 6.0 grams of aluminum, 0.22 moles of AlCl3 can be formed, and that, beginning with 3.8 grams of chlorine, 0.036 moles of AlCl3 can be formed.
Keeping score, 6.0 grams of Al yields 0.22 moles of AlCl3, and 3.8 grams of chlorine gas yields 0.036 moles of AlCl3. The lowest yield comes from the chlorine gas, therefore it must be limiting and aluminum must be in excess. The reaction in the problem will therefore produce 0.036 moles of product, which is equivalent to:
$0.036\: mol\: AlCl_{3}(133.33\: g/mol)=4.8\: g\: AlCl_{3} \nonumber$
Exercise $1$
Lead (IV) chloride reacts with fluorine gas to give lead (IV) fluoride and Cl2. If 0.023 moles of fluorine gas reacts with 5.3 grams of lead (IV) chloride, what mass of lead (IV) fluoride will be formed?
Although “limiting reactant problems” may be tedious, they are not difficult. When you are faced with a limiting reactant problem, just remember, you do the simple molar yield calculations twice, one for each reactant. The reactant that yields the lowest molar quantity is your limiting reagent and the molar value you calculate determines the theoretical yield in the problem.
Exercise $2$
Ammonia, which is the active ingredient in “smelling salts”, is prepared from nitrogen and hydrogen according to the equation shown below.
N2 + 3 H2 → 2 NH3
1. If you mix 5.0 mol of nitrogen and 10.0 moles of hydrogen how many moles of ammonia would you produce? Which reactant is in excess?
2. If you have 6.2 grams of nitrogen and you react it with 6.2 grams of hydrogen how many grams of ammonia would you produce? Which reactant is the limiting reactant?
6.S: Quantitative Relationships in Chemistry (Summary)
• Stoichiometry is the relationship between the masses of chemical reactants and products in a given chemical reaction. The coefficients placed in a chemical equation in order to balance it are called the stoichiometric coefficients. Molar stoichiometry is simply the expression of the coefficients of a reaction in terms of moles of reactants and products.
• In order to find the number of moles of a product that is produced in a chemical reaction when you are given moles of reactant, simply multiply the moles of reactant by the stoichiometric ratio relating that reactant and the desired product; i.e., $(molreactant)\times \left ( \frac{molproduct}{molreactant} \right ) \nonumber$
• In order to find the number of moles of a product that are produced in a chemical reaction when you are given mass of reactant, simply divide the mass of reactant by the molar mass (to get moles reactant) and then multiply by the stoichiometric ratio relating that reactant and the desired product; i.e., $\left ( \frac{grams}{grams/mol} \right )\times \left ( \frac{molproduct}{molreactant} \right ) \nonumber$
• Always remember, mass divided by molar mass equals moles; $\left ( \frac{grams}{grams/mol} \right )=mol \nonumber$
• The mass or the number of moles that you calculate for a product based on reaction stoichiometry is called the theoretical yield for the reaction. The amount of material that you actually isolate from a given reaction is called the actual yield and it is always less than the theoretical yield. The ratio of the actual and theoretical yields, expressed as a percentage is called the percentage yield.
• If a reaction requires more than one reactant and if you are given the mass, or the number of moles of each reactant, you must approach the calculation as a limiting reactant problem. To solve a limiting reactant problem, simply perform the standard mass calculation for each reactant, noting the mass (or number of moles) of product formed in each calculation. The reactant that yields the smallest amount or product from these calculations is called the limiting reactant. Reactants that yield larger amounts of products in these calculations are called excess reactants. The theoretical yield in the reaction will be based solely on the calculated amount for the limiting reactant.
• If a reactant in a chemical reaction is said to be “in excess”, you assume that you have unlimited amount of the reactant, and that it will never be the limiting reactant. | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/06%3A_Quantitative_Relationships_in_Chemistry/6.5%3A_Limiting_Reactants.txt |
Water is the most remarkable solvent! The O—H bonds in water are polarized due to the differences in electronegativity between hydrogen and oxygen. When this uneven charge distribution is coupled with the fact that water has a “bent” molecular geometry, the two covalent bond dipoles combine to form a molecular dipole (shown in the electrostatic potential map on the right. This molecular dipole allows water to surround and stabilize ions in solution, making water a powerful solvent for the dissolution of polar and ionic compounds. If we know the amount of solute that we have dissolved in a given volume of solvent, we can define the term molarity, as the number of moles of solute in each liter of solution. Finally, by combining the concept of molarity with what we have learned about simple stoichiometric calculations, we can now approach quantitative chemical calculations in solution.
07: Aqueous Solutions
Water is an amazing solvent, and has remarkable physical and chemical properties that make it the essential ingredient to life as we know it. The special properties of water come from the fact that the elements hydrogen and oxygen have differing electronegativities. In Chapter 3 we learned that covalent bonds formed between atoms of differing electronegativity are polarized. Because electronegativity is a measure of how strongly a given atom attracts electrons to itself, the atom in the covalent bond with the highest electronegativity will tend to draw the bonding electrons towards itself, resulting in a bond that is electron-rich on one end and electron-poor on the other. Covalent bonds that are polarized are said to have a dipole, where the term dipole moment refers to the direction and magnitude of the charge separation.
Consider water. The electronegativities of hydrogen and oxygen are 2.20 and 3.44, respectively. That means that in each covalent bond, the electrons will be attracted towards the oxygen, leaving the hydrogen electron-poor. In Chapter 3, we used a calculated electrostatic potential map to visualize the electron density around molecules. The map for water is shown to the left and is colored using red to indicate a high electron density and blue to show electron-poor regions. Because electrons carry a negative charge, this also means that the red regions of the molecule are anionic (negative) and that the blue regions are cationic (positive).
7.2: Molecular Dipoles
Electrostatic potential maps are useful because they clearly show the electron distribution around covalent bonds within molecules. They must be calculated, however, using sophisticated computer programs, and then rendered in color for visualization. Because of this, the polarization of covalent bonds is typically shown using a special arrow (a dipole arrow) to indicate the direction in which the bond is polarized. A dipole arrow is crossed at the beginning (as in a plus sign) and points in the direction of the greatest electron density. Thus for hydrogen fluoride, the electronegativities are 2.20 and 3.98 for the hydrogen and fluorine, respectively. We would predict that the H—F bond would be polarized with the greatest electron density towards the fluorine.
A molecule such as water, with two covalent bonds, will have two local dipoles, each oriented along the covalent bonds, as shown below. Because water is asymmetric (it has a bend structure) both of these local dipoles point in the same direction, generating a molecular dipole, in which the entire molecule has a charge imbalance, with the “oxygen end” being anionic and the “hydrogen end” being cationic.
Molecules with local dipoles do not necessarily possess a molecular dipole. Consider the molecule boron trihydride (BH3). The BH3 molecule is planar with all three hydrogens spaced evenly surrounding the boron (trigonal planar). The electronegativities of boron and hydrogen are 2.04 and 2.20, respectively. The bonds in BH3 will therefore be somewhat polarized, with the local dipoles oriented towards the hydrogen atoms, as shown below. But because the molecule is symmetrical, the three dipole arrows cancel and, as a molecule, BH3 has no net molecular dipole.
Exercise \(1\)
For each of the molecules of NH3 and CO2 indicate whether a molecular dipole exists. If a dipole does exist, use a dipole arrow to indicate the direction of the molecular dipole.
7.3: Dissolution of Ionic Compounds
A simple ionic compound, such as sodium chloride (NaCl) consists of a sodium cation and a chloride anion. Because these are oppositely charge ions, they are strongly attracted to each other. This attraction is non-specific and the sodium cation would also be strongly attracted to any anion. When an ionic compound dissolves in water, the individual cations and anions are completely surrounded by water molecules, but these water molecules are not randomly oriented. A sodium cation in water will be surrounded by water molecules oriented so that the negative end of the molecular dipole is in contact with the sodium cation. Likewise, the waters surrounding the chloride anion are oriented so that the positive end of the molecular dipole contacts the anion.When arranged like this, the charged poles of the water molecules neutralize, and thus stabilize the charges on the ions.
The ability of water to interact with and stabilize charge particles goes well beyond the water molecules that actually touch the ion. Surrounding the inner water shell is another shell of waters that will orient themselves so that their dipoles bind to the exposed dipoles from the inner shell. As the subsequent layers of water surround each other, the positive charge from the cation is dispersed or spread out over the whole group of interacting molecules. The cluster then becomes effectively neutral allowing the charged ion to exist free in solution, removed from its counter-ion (the chloride). The dynamic collection of water molecules surrounding an ion in solution is referred to as the solvation shell and it is the ability of water to solvate and stabilize ions that makes water such an important solvent, both in chemistry and in biology.
In addition to ionic compounds, water will also dissolve and stabilize most molecules that are polar, that is, if they possess a molecular dipole. The organic compound, methyl propionate, contains a highly polar carbon-oxygen double bond. The electrostatic potential map in the figure clearly shows the resulting molecular dipole and methyl propionate is quite soluble in water; 6.2 grams of methyl propionate will dissolve in 100 mL of water. The organic molecule propane, does not possess a significant molecular dipole and is only very slightly soluble in water. | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/07%3A_Aqueous_Solutions/7.1%3A_Hydrogen_Bonding_and_the_Properties_of_Water.txt |
As described in the previous section, sodium chloride is quite soluble in water. At 25 ˚C (about room temperature), 359 grams of sodium chloride will dissolve in one liter of water. If you were to add more sodium chloride to the solution, it would not dissolve, because a given volume of water can only dissolve, disperse and stabilize a fixed amount of solute (the stuff that dissolves). This amount is different for every compound and it depends on the structure of the particular compound and how that structure interacts with the solvation shell. When a substance is dissolved in water to the point that no more will go into solution, we say the solution is saturated. For most compounds, heating the solution will allow more of the substance to dissolve, hence it is important to note the temperature when you are speaking of the solubility of a particular compound.
If we had a saturated solution of sodium chloride at 25 ˚C, we could quote the concentration as 359 grams/L, but because we know the molar mass of sodium chloride (58.44 grams/mole), we could also express our concentration as:
$\left ( \frac{(359\, g)\times \frac{1\, mole}{58.44\, g}}{1\, L} \right )=6.14\, moles/L \nonumber$
In chemistry, the units of moles/L are called molarity, with the abbreviation M. Thus we could say that our saturated solution of sodium chloride was 6.14 molar, or 6.14 M.
The advantage of expressing concentrations in terms of molarity is that these solutions can now be used in chemical reactions of known stoichiometry because any volume of the solution corresponds directly to a known number of moles of a particular compound. For example, the molar mass of potassium bromide is 119.0 g/mole. If we dissolved 119.0 grams of KBr in 1.000 L of water, the concentration would be 1.000 mole/L, or 1.000 M. If we now took half o this solution (0.500 L) we know that we would also have 0.500 moles of KBr.
We can determine the concentration of a solution using the problem-solving algorithm we introduced back in Chapter 1. For example, if you wan to find the molarity of a solution containing 42.8 grams of KBr in 1.00 L of water, you would identify the given and 42.8 g, your ratio is the molar mass (119 g/mole) and you want to find molarity (or moles/L). Remembering to set the equation up so that the units of given appear in the denominator of the ratio, the number of moles is:
$42.8\, g)\times \left ( \frac{1\, mole}{119\, g} \right )=0.360\, moles \nonumber$
and, the molarity is:
$\left ( \frac{0.360\, mole}{1.00\, L} \right )=0.360\, moles/L\; or\; 0.360\, M \nonumber$
When you become comfortable with the simple two-step method, you can combine steps and simply divide your given mass by the given volume to get the result directly. Thus, if you had 1.73 grams of KBr in 0.0230 L of water, your concentration would be:
$\left ( \frac{(1.73\, g)\times \frac{1\, mole}{119\, g}}{0.0230\, L} \right )=6.32\, moles/L\; or\; 0.632\, M \nonumber$
We can also solve these problems backwards, that is, convert molarity into mass. For example; determine the number of grams of KBr that are present in 72.5 mL of a 1.05 M solution of KBr. Here we are given a volume of 0.0725 L and our ratio is the molarity, or (1.05 moles/L). We first solve for moles,
$0.0725\, L\times \left ( \frac{1.05\, mole}{1.00\, L} \right )=0.0761\, moles \nonumber$
and then convert to mass using:
$0.0761\, moles\times \left ( \frac{119\, grams}{1\, mole} \right )=9.06\, grams\, of\, KBr \nonumber$
Exercise $1$
A sample of 12.7 grams of sodium sulfate (Na2SO4) is dissolved in 672 mL of distilled water.
1. What is the molar concentration of sodium sulfate in the solution?
2. What is the concentration of sodium ion in the solution?
Exercise $1$
Calculate the mass of sodium chloride required to make 125.0 mL of a 0.470 M NaCl solution. If you dissolve 5.8g of NaCl in water and then dilute to a total of 100.0 mL, what will be the molar concentration of the resulting sodium chloride solution? | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/07%3A_Aqueous_Solutions/7.4%3A_Concentration_and_Molarity.txt |
As we learned in Chapter 5, double replacement reactions involve the reaction between ionic compounds in solution and, in the course of the reaction, the ions in the two reacting compounds are “switched” (they replace each other). As an example, silver nitrate and sodium chloride react to form sodium nitrate and the insoluble compound, silver chloride.
$\ce{AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq)} \nonumber$
Because these reactions occur in aqueous solution, we can use the concept of molarity to directly calculate the number of moles of products that will be formed, and hence the mass of precipitates. In the reaction shown above, if we mixed 123 mL of a 1.00 M solution of NaCl with 72.5 mL of a 2.71 M solution of AgNO3, we could calculate the moles (and hence, the mass) of AgCl that will be formed as follows:
First, we must examine the reaction stoichiometry. In this reaction, one mole of AgNO3 reacts with one mole of NaCl to give one mole of AgCl. Because our ratios are one, we don’t need to include them in the equation. Next, we need to calculate the number of moles of each reactant:
$0.123L\times \left ( \frac{1.00\: mole}{1.00\: L} \right )=0.123\: moles\: NaCl \nonumber$
$0.0725L\times \left ( \frac{2.71\: mole}{1.00\: L} \right )=0.196\: moles\: AgNO_{3} \nonumber$
Because this is a limiting reactant problem, we need to recall that the moles of product that can be formed will equal the smaller of the number of moles of the two reactants. In this case, NaCl is limiting and AgNO3 is in excess. Because our stoichiometry is one-to-one, we will therefore form 0.123 moles of AgCl. Finally, we can convert this to mass using the molar mass of AgCl:
$0.0725L\times \left ( \frac{2.71\: mole}{1.00\: L} \right )=0.196\: moles\: AgNO_{3} \nonumber$
In a reaction where the stoichiometry is not one-to-one, you simply need to include the stoichiometric ratio in you equations. Thus, for the reaction between lead (II) nitrate and potassium iodide, two moles of potassium iodide are required for every mole of lead (II) iodide that is formed.
$\ce{Pb(NO3)2 (aq) + 2 KI (aq) → PbI2 (s) + 2 KNO3 (aq)} \nonumber$ For example: 1.78 grams of lead (II) nitrate are dissolved in 17.0 mL of water and then mixed with 25.0 mL of 2.5 M potassium iodide solution. What mass of lead (II) iodide will be formed and what will be the final concentration of potassium nitrate in the solution? Again, we need to look at this as a limiting reactant problem and first calculate the number of moles of each reactant: $1.78\: g\times \left ( \frac{1.00\: mole}{331.2\: g} \right )=5.37\times 10^{-3}\: moles\: Pb(NO_{3})_{2} \nonumber$ $0.0025\: L\times \left ( \frac{2.50\: mole}{1.00\: L} \right )=6.25\times 10^{-3}\: moles\: KI \nonumber$ The stoichiometry of this reaction is given by the ratios: $\left ( \frac{1\: mole\: PbI_{2}}{2\: mole\: KI} \right )\; and\; \left ( \frac{1\: mole\: PbI_{2}}{1\: mole\: Pb(NO_{3})_{2}} \right ) \nonumber$ so the number of moles of product that would be formed from each reactant is calculated as:
$\left ( \frac{1\: mole\: PbI_{2}}{1\: mole\: Pb(NO_{3})_{2}} \right ) \nonumber$
$6.25\times 10^{-3}\: moles\: KI\times \left ( \frac{1\: mole\: PbI_{2}}{2\: moles\: KI} \right )=3.12\times 10^{-3}\: moles\: PbI_{2} \nonumber$
Potassium iodide produces the smaller amount of PbI2 and hence, is limiting and lead (II) nitrate is in excess. The mass of lead (II) iodide that will be produced is then calculated from the number of moles and the molar mass:
$3.12\times 10^{-3}\: moles\: \times \left ( \frac{461\: grams}{1\: mole} \right )=1.44\: grams\: PbI_{2} \nonumber$
To determine the concentration of potassium nitrate in the final solution, we need to note that two moles of potassium nitrate are formed for every mole of PbI2, or a stoichiometric ratio of $\left ( \frac{2\: moles\: KNO_{3}}{1\: mole\: PbI_{2}} \right ) \nonumber$
Our final volume is (17.0 + 25.0) = 42.0 mL, and the concentration of potassium nitrate is calculated as:
$\frac{3.12\times 10^{-3}\: moles\:PbI_{2}\times \left ( \frac{2\: moles\: KNO_{3}}{1\: mole\: PbI_{2}} \right )}{0.0420\: L}=0.148\; moles\; KNO_{3}/L\; or\; 0.148\; M \nonumber$
Exercise $1$
1. A sample of 12.7 grams of sodium sulfate (Na2SO4) is dissolved in 672 mL of distilled water.
1. What is the molar concentration of sodium sulfate in the solution?
2. What is the concentration of sodium ion in the solution?
2. How many moles of sodium sulfate must be added to an aqueous solution that contains 2.0 moles of barium chloride in order to precipitate 0.50 moles of barium sulfate?
3. If 1.0 g of NaN3 reacts with 25 mL of 0.20 M NaNO3 according to the reaction shown below, how many moles of N2(g) are produced?
$5 NaN3(s) + NaNO3(aq) → 3 Na2O(s) + 8 N2(g) \nonumber$ | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/07%3A_Aqueous_Solutions/7.5%3A_Solution_Stoichiometry.txt |
In the laboratory, a chemist will often prepare solutions of known concentration beginning with a standard stock solution. A stock solution is generally concentrated and, of course, the molar concentration of the solute must be known. To perform a reaction, a measured amount of this stock solution will be withdrawn and added to another reactant, or will be diluted into a larger volume for some other use. The calculations involved in these dilutions are trivial and simply involve calculating the number of moles transferred and dividing this by the final volume. For example, 15.0 mL of a stock solution of 1.00 M hydrochloric acid (HCl) is withdrawn and diluted into 75 mL of distilled water; what is the final concentration of hydrochloric acid?
First, the number of moles of HCl is calculated from the volume added and the concentration of the stock solution:
$0.0150L\times \left ( \frac{1.00\: moles}{1\: L} \right )=0.0150\; moles\; HCl \nonumber$
We have diluted this number of moles into (15.0 + 75.0) = 90.0 mL, therefore the final concentration of HCl is given by:
$\left ( \frac{0.0150\; moles\; HCl}{0.0900\: L} \right )=(0.167\; moles\; HCl/L)\; or\; 0.167\; M \nonumber$
An even simpler way to approach these problems is to multiply the initial concentration of the stock solution by the ratio of the aliquot (the amount withdrawn from the stock solution) to the final volume, using the equation below:
$(stock\; concentration)\times \left ( \frac{volume\; of\; the\; aliquot}{final\; volume} \right )=final\; concentration \nonumber$
Using this method on the previous problem,
$(1.00\; M)\times \left ( \frac{15.0\; ml}{90.0\; ml} \right )=0.167\; M \nonumber$
Note that we did not have to convert our volumes (15.0 and 90.0 mL) into L when we use this approach because the units of volume cancel in the equation. If the units that are given for the aliquot and the final volume are different, a metric conversion ratio may be required. For example, 10.0 µL of a 1.76 M solution of HNO3 (nitric acid) are diluted into 10.0 mL of distilled water; what is the final concentration of nitric acid?
In this problem, we need to convert µL and mL into a common unit. We can do this using the ratios,
$\left ( \frac{10^{-6}L}{1\mu L} \right )\; and\; \left ( \frac{10^{-3}L}{1\: mL} \right ) \nonumber$
We need to multiply each of our volumes by the appropriate factor to get our volumes in terms of liters, and then simply multiply by the initial concentration. Thus,
$1.76\: M\times \left \{ \frac{10.0\mu L\left ( \frac{10^{-6}L}{1\mu L}\right )}{10.0\: mL\left ( \frac{10^{-3}L}{1\: mL}\right )} \right \}=1.76\times 10^{-3}M \nonumber$
The final volume in this problem is actually (1.00×10-2 L) + (1.00×10-5 L) = 1.001×10-2 L, but because our calculation is only accurate to three significant figures, the volume of the aliquot is not significant and the final volume has been rounded.
The standard method we have used here can also be adapted to the type of problem in which you need to find the volume of a stock solution that must be diluted to a certain volume in order to produce a solution of a given concentration. For example, what volume of 0.029 M CaCl2 must be diluted to exactly 0.500 L in order to give a solution that is 50.0 µM?
In order to solve this problem in the simplest of terms, we should re-examine the above equation:
$(stock\; concentration)\times \left ( \frac{volume\; of\; the\; aliquot}{final\; volume} \right )=final\; concentration \nonumber$
This equation can be re-written as:
$\left ( \frac{volume\; of\; the\; aliquot}{final\; volume} \right )=\left ( \frac{final\; concentration}{stock\; concentration} \right ) \nonumber$
Or
$\left ( \frac{V}{V_{f}} \right )=\left ( \frac{C_{f}}{C_{i}} \right ) \nonumber$
where Ci and Cf are the stock and final concentrations, respectively, V is the volume of the aliquot and Vf is the final volume of the solution. Stated another way, this is simply a set of ratios; aliquot to final volume, and final concentration to initial concentration (operationally, these ratios will always be “small value/larger value”). Working with this set of ratios, we can directly solve this type of problem as follows:
First, we need to convert our final concentration (50.0 µM) into M, to match the units of our stock solution. The metric multiplier for µ is 10-6, making our final concentration 50.0×10-6 M, or more properly, 5.00×10-5 M. Our equation is therefore:
$\left ( \frac{V}{0.500L} \right )=\left ( \frac{5.00\times 10^{-5}M}{0.029M} \right ) \nonumber$
$V=\left ( \frac{5.00\times 10^{-5}M}{0.029M} \right )\times 0.500L \nonumber$
The volume of the aliquot, V, is 8.62×10-4 L, or using the conversion factor
$\left ( \frac{10^{3}mL}{1L} \right ) \nonumber$
the required volume is 0.86 mL (there are only two significant figures in the concentration of the stock solution, 0.029 M).
Dilution problems can be solved directly using the above equation, or, as you become more comfortable with the math, using the initial and final ratios like we did in this problem (remember, the numbers in the two ratios are “smaller/larger”).
Exercise $1$
1. A 1.50 mL aliquot of a 0.177 M solution of sulfuric acid (H2SO4) is diluted into 10.0 mL of distilled water, to give solution A. A 10.0 mL aliquot of A is then diluted into 50.0 mL of distilled water, to give solution B. Finally, 10.0 mL of B is diluted into 900.0 mL of distilled water to give solution C. Additional distilled water is then added to C to give a final volume of 1.0000 L. What is the final concentration of sulfuric acid in solution C?
2. A solution was prepared by mixing 250 mL of 0.547 M NaOH with 50.0 mL of 1.62 M NaOH and then diluting to a final volume of 1.50 L. What is the molarity of Na+ in this solution? To what final volume should 75.00 mL of 0.889 M HCl(aq) be diluted to prepare 0.800 M HCl(aq)?
7.S: Aqueous Solutions (Summary)
• Covalent bonds formed between atoms of differing electronegativity are polarized, resulting in a bond that is electron-rich on one end and electron-poor on the other. Covalent bonds that are polarized are said to have a dipole, where the term dipole moment refers to the direction and magnitude of the charge separation.
• If a molecule is asymmetric (such as a molecule with a bend structure) local dipoles along covalent bonds can combine, generating a molecular dipole, in which the entire molecule has an imbalance with regard to electron distribution. This can be shown with an dipole arrow (with a positive end) indicating the direction of the charge separation in the molecule.
• If a molecule is symmetrical (such as BH3, which is trigonal planar), the individual dipoles associated with the covalent bonds cancel, leaving a molecule with no molecular dipole.
• Water has a significant molecular dipole, allowing it to strongly interact with other polar molecules and with individual ions from ionic compounds. Because of this, water is able to break the electrostatic attraction between ions in compounds and to move the ions into solution. In solution, cations will be surrounded by a solvation shell where the water molecules are oriented so that the negative end of the water molecule interacts with the cation. Likewise, the cationic end of water will surround and solvate anions.
• Molarity is simply defined as the number of moles of a solute dissolved in one liter of solvent, or (moles/L). The abbreviation for molarity is the uppercase M.
• You should remember that concentration multiplied by volume gives the number of moles of solute; (moles/LL=moles.
• When you are given the amount of solute in grams, remember, mass divided by molar mass gives moles. Dividing this by volume (in liters) gives molarity; $\frac{\left ( \frac{grams}{grams/mole} \right )}{L}=molarity \nonumber$
• In a standard solution, we simply know the molarity of the solute(s). Because concentration (the molarity) multiplied by volume gives us moles, we can calculate the number of moles in given volume and use this value in standard stoichiometric calculations.
• A sample of a solution of known volume is called an aliquot. When an aliquot of a solution is diluted into a larger volume, the final concentration can be calculated as: $\left ( \frac{volume\; of\; the\; aliquot}{final\; volume} \right )=\left ( \frac{final\; concentration}{stock\; concentration} \right ) \nonumber$ or $\left ( \frac{V}{V_{f}} \right )=\left ( \frac{C_{f}}{C_{i}} \right ) \nonumber$ where Ci and Cf are the stock and final concentrations, respectively, V is the volume of the aliquot and Vf is the final volume of the solution. This relationship is also often stated as V1C1 = V2C2, where the subscripts refer to the initial and final concentrations and volumes. | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/07%3A_Aqueous_Solutions/7.6%3A_Dilution_of_Concentrated_Solutions.txt |
We have all heard of acids… acid indigestion, the acid content of vinegars and cheap wines, vats of sulfuric acid and arch villains. In the last chapter we learned about the molecular dipole in water and how water molecules form hydrogen bonds among themselves, and to other polar molecules. In this chapter we will see that this hydrogen bonding can be strong enough to actually break the O—H bonds in water and in other hydrogen bonded molecules, transferring the proton to water to form the hydronium ion, H3O+. In its simplest sense, the formation of the hydronium ion is the “acidity” that we are all familiar with. We will see how we measure this acidity (the pH scale). We will get an introduction to the concept of equilibrium and you will finally come to appreciate the utility of the log and anti-log buttons on your scientific calculator!
08: Acids Bases and pH
In Chapter 7, we explored the unique properties of water that allow it to serve as a powerful solvent with the ability to dissolve both ionic compounds, as well as polar molecular compounds. We attributed this to the ability of water molecules to align themselves so that the polarized hydrogen-oxygen bonds could stabilize cations, anions, and virtually any compound that also contained a significantly polarized covalent bond. By this logic, it is not at all surprising that water can also react strongly with itself, and indeed water exists as a vast network of molecules aligned so that their positive and negative dipoles interact with each other. A bond that is formed from a hydrogen atom, which is part of a polar covalent bond (such as the O—H bond) to another, more electronegative atom (that has at least one unshared pair of electrons in its valence shell) is called a hydrogen bond. Recall that oxygen has two unshared pairs in its valence shell, and the hydrogen-oxygen interaction in water is the classic example of a hydrogen bond. Hydrogen bonds are weak, relative to covalent bonds. The energy required to break the O—H covalent bond (the bond dissociation energy) is about 111 kcal/mole, or in more proper SI units, 464 kJ/mole. The energy required to break an O—H••••O hydrogen bond is about 5 kcal/mole (21 kJ/mole), or less than 5% of the energy of a “real” covalent bond. Even though hydrogen bonds are relatively weak, if you consider that every water molecule is participating in a least four hydrogen bonds, the total energy of hydrogen bonding interactions can rapidly become significant. Hydrogen bonding is generally used to explain the high boiling point of water (100 ˚C). For many compounds which do not possess highly polarized bonds, boiling points parallel the molar mass of the compound. Methane, CH4, has a molar mass of 16 and a boiling point of –164 ˚C. Water, with a molar mass of 18, has a boiling point of +100 ˚C. Although these two compounds have similar molar masses, a significant amount of energy must be put into the polar molecule, water, in order to move into the gas phase, relative to the non-polar methane. The extra energy that is required is necessary to break down the hydrogen bonding network.
Hydrogen bonding is also important is DNA. According to the Watson-Crick model, the double helix of DNA is assembled and stabilized by hydrogen pairing between matching “bases”. The hydrogen bonds are formed between the oxygen atoms (red) and the adjacent N—H bonds, and between the central nitrogen (blue) and the adjacent N—H bond. It is suggested that the precise alignment of these hydrogen bonds contributes to stability of the double helix and ensures the proper alignment of the corresponding base pairs. | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/08%3A_Acids_Bases_and_pH/8.1%3A_Hydrogen_Bonding.txt |
In general, acids can be thought of as molecular compounds containing at least one hydrogen which is covalently bonded to a more electronegative atom. As an example, consider the compound hydrogen fluoride (HF). As we discussed in Chapter 7, the electronegativities of hydrogen and fluorine are 2.1 and 4.0, respectively, and the hydrogen-fluoride covalent bond is very highly polarized.
Because of this, when hydrogen fluoride is dissolved in water, water molecules orient themselves around HF so that the water dipoles interact with, and stabilize, the highly polarized H—F bond. Important to this stabilization is the hydrogen bond that is formed between the hydrogen of HF and the oxygen of an adjacent water. This hydrogen bond not only stabilizes the HF molecular dipole, but it also weakens the H—F covalent bond. As a result of this weakening, the H—F bond stretches (the bond length increases) and then fully breaks. The hydrogen that was hydrogen-bonded to the water molecule now becomes fully bonded to the oxygen, forming the species H3O+ (the hydronium ion) and the fluorine now exists as a fluoride anion. This is known as the process of acid dissociation.
The chemical equation describing the acid dissociation reaction of HF is given in Equation 8.2a. The products of the reaction, fluoride anion and the hydronium ion, are oppositely charged ions, and it is reasonable to assume that they will be attracted to each other. If they do come in contact, it is also reasonable to suggest that the process of hydrogen transfer that we described above can also happen in reverse. That is, H3O+ can hydrogen bond to the fluoride ion and the hydrogen can be transferred back, to form HF and water. The chemical equation describing this process is shown in Equation 8.2b. In fact, these two reaction do occur simultaneously (and very rapidly) in solution. When we speak of a set of forward- and back-reactions that occur together on a very fast time-scale, we describe the set of reactions as an equilibrium and we use a special double arrow in the chemical reaction to show this (Equation 8.2c). Equation 8.2c can be said to represent the equilibrium dissociation of HF in water.
HF (aq) + H2O(l) → H3O+ (aq) + F (aq) Eq. 8.2a
HF (aq) + H2O(l)←H3O+ (aq) + F (aq) Eq. 8.2b
HF (aq) + H2O(l) ⇄ H3O+ (aq) + F (aq) Eq. 8.2c
For any equilibrium, an equilibrium constant can be written that describes whether the products or the reactants will be the predominant species in solution. We will address this fully in Chapter 10, but according to the Law of Mass Action, the equilibrium constant, K for this reaction, is simply given by the ratio of the activities of the products and reactants. To simplify the calculations, the activities of the solutes are approximated by the concentration in units of molarity. Note that any solid or liquid reactants or products, or solvents (such as water) are considered to be pure substances, and so have an activity that is equal to 1. Thus for the ionization of HF
$K_{a}=\frac{a_{H_3O^+}·a_{OH^-}}{a_{HF}·a_{H_2O}} \approx \frac{[H_{3}O^{+}][F^{-}]}{[HF](1)}=\frac{[H_{3}O^{+}][F^{-}]}{[HF]} \nonumber$
When you are dealing with acids, the equilibrium constant is generally called an acid dissociation constant, and is written as Ka. The larger the value of Ka, the greater the extent of ionization and the and the higher the resulting concentration of the hydronium ion. Because the concentration of the hydronium ion is directly correlated with acidity, acids with a large value of Ka are termed strong acids. We will introduce “weak acids” in Chapter 10, but for now the important thing to remember is that strong acids are virtually 100% ionized in solution. That doesn’t mean that the back-reaction does not occur, is simply means that much more favorable and that 99.9999999999% of the acid is present in its ionized form. Because this exceeds the number of significant figures that we typically work with, strong acids are generally described as 100% ionized in solution. Table 8.1 lists the common strong acids that we will study in this text.
Table 8.1. Common Strong Acids
Table 8.1 Common Strong Acids
HCl hydrochloric acid
HNO3 nitric acid
H2SO4 sulfuric acid
HBr hydrobromic acid
HI hydroiodic acid
HClO4 perchloric acid | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/08%3A_Acids_Bases_and_pH/8.2%3A_Ionization_of_Acids_in_Solution.txt |
Acid dissociation reactions are often described in terms of the concepts of conjugate acids and their corresponding conjugate bases. The description of “acids and bases” that we will deal with in this text will be limited to simple dissociation reactions, like those shown above, where a hydronium ion is produced. This description is referred to as the Brønsted-Lowery Acid-Base Theory, and in the Brønsted theory, the conjugate acid is defined as the species that donates a hydrogen in the forward reaction, and the conjugate base is the species that accepts a hydrogen the reverse reaction. Thus for the ionization of HCl, HCl is the conjugate acid and Cl is the conjugate base.
HCl (aq) + H2O ⇄ H3O+ (aq) + Cl (aq)
In the discussion of Brønsted acid-base behavior, the hydrogen atom that is transferred is generally referred to as a proton, because it is transferred as a hydrogen atom without its electron. Thus for the ionization of HCl, HCl (the conjugate acid) is a proton donor and Cl (the conjugate base) is a proton acceptor. In General Chemistry you will learn that acid-base behavior can also be described in terms of electron donors and electron acceptors (the Lewis Acid-Base Theory in which an acid is an electron acceptor and a base is an electron donor), but here we will limit our discussion to simple, strong, Brønsted acids and bases.
Exercise \(1\)
For each of the reactions given below, identify the conjugate acid and the conjugate base. For example (d), also identify the conjugate acid and the conjugate base in the reverse reaction.
1. HClO4 (aq) + H2O ⇄ H3O+ (aq) + ClO4 (aq)
2. H2SO4 (aq) + H2O ⇄ H3O+ (aq) + HSO4 (aq)
3. HSO4 (aq) + H2O ⇄ H3O+ (aq) + SO22- (aq)
4. HNO3 (aq) + NH3 ⇄ NH4+ (aq) + NO3 (aq)
5. H2PO4- + H3O+ ⇄ H2O + H3PO4
6. NH3(g) + H2O ⇄ NH4+(aq) + OH-(aq)
7. H2O(l) + HNO2(aq) ⇄ H3O+(aq) + NO2-(aq)
8.4: Acids-Bases Reactions: Neutralization
In Chapter 5, we examined a special case of a double replacement reaction in which an acid reacted with a base to give water and a pair of ions in solution. In the context of the Brønsted Theory, a base can be thought of as an ionic compound that produces the hydroxide anion in solution. Thus, sodium hydroxide, $\ce{NaOH}$, ionizes to form the sodium cation and the hydroxide anion ($\ce{OH^{-}}$).
$\ce{NaOH (aq) -> Na^{+} (aq) + OH^{-} (aq)}\nonumber$
We have written this equation using a single arrow because sodium hydroxide is a strong base and is essentially 100% ionized in solution. The hydroxide anion ($\ce{OH^{–}}$) reacts with the hydronium ion ($\ce{H3O^{+}}$) to form two moles of water, as shown in the equation given below.
$\ce{H3O+ + OH^{-} -> 2 H2O}\nonumber$
Thus, if you have aqueous solutions of $\ce{HCl}$ and $\ce{NaOH}$, the following process occur:
• $\ce{HCl}$ ionizes to form the hydronium ion: $\ce{HCl (aq) + H2O -> H3O^{+} (aq) + Cl^{-} (aq)}\nonumber$
• $\ce{NaOH}$ ionizes to form the hydroxide anion: $\ce{NaOH (aq) -> Na^{+} (aq) + OH^{-} (aq)}\nonumber$
• $\ce{OH^{-}}$ reacts with $\ce{H3O^{+}}$ to form two moles of water: $\ce{H3O^{+} + OH^{-} -> 2 H2O}\nonumber$
If we add up the set of three equations, we see that hydronium and hydroxide ions appear on both sides of the arrow and cancel, leaving:
$\ce{HCl (aq) + NaOH (aq) -> H2O + NaCl (aq)}\label{1}$
In Equation \ref{1}, we have not shown the additional water from the hydronium ion and we have grouped the sodium and chloride ions as NaCl (aq), with the understanding that it will be fully ionized in aqueous solution. This is an example of a neutralization reaction; an acid and a base have reacted to form water. When we write neutralization equations we generally do not show hydronium or hydroxide ions and we generally show ionic species as distinct compounds. Neutralization equations therefore look very much like the other double replacements that we studied in Chapter 5.
Exercise $1$
For each of the following, write a balanced neutralization equation:
1. The reaction of calcium hydroxide with hydrochloric acid.
2. The reaction of sodium hydroxide with sulfuric acid (both ionizations).
3. The reaction of barium hydroxide with nitric acid.
Exercise $2$
Write a balanced neutralization equation for the reaction of calcium hydroxide with sulfuric acid | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/08%3A_Acids_Bases_and_pH/8.3%3A_Conjugate_Acid-Base_Pairs.txt |
In the previous section, we described the reaction of an acid and a base to form water. When all of the acid and base have been consumed, we are left with water and an aqueous solution containing an ionic compound. Another way to think of this would be to say that, the acid we began with had a high concentration of hydronium cations (H3O+), the base had a high concentration of hydroxide anions (HO) and the neutral solution contains only water. This, however, is a little too simplistic. Just like water can promote the ionization of acids, water can also promote the ionization of itself. Picture two water molecules sharing a hydrogen bond. Just like for HF , the partially bonded hydrogen can transfer along the hydrogen bond to form a hydronium cation and a hydroxide anion. This process occurs very rapidly in pure water, thus, any sample of pure water will always contain a small concentration of hydronium and hydroxide ions. How small is “small”? Very small! In pure water at 25 oC, the concentration of hydronium ions ([H3O+]) and hydroxide ions ([HO]) will both be equal to exactly 1 × 10-7 M. Based on this, we can expand upon our definitions of acidic, basic and neutral solutions:
• A solution is acidic if [H3O+] > 1 × 10-7 M.
• A solution is basic if [H3O+] < 1 × 10-7 M.
• A solution is neutral if [H3O+] = 1 × 10-7 M.
Working with these definitions, if you have a solution with [H3O+] = 4.5 × 10-4 M, it will be acidic (4.5 × 10-4 > 1 × 10-7). If you have a solution with [H3O+] = 1 × 10-4 M, it will be basic (1 × 10-4 < 1 × 10-7). Finally, a neutral solution is one in which [H3O+] and [HO] are both 1 × 10-7 M.
Recalling our discussion of acid dissociation constants from Section 8.2, we can write the ionization equilibrium for water
$\ce{2 H2O(l) <=> H3O^{+}(aq) + OH^{–} (aq)} \nonumber$
and the expression for the dissociation constant, Ka, as shown below:
\begin{align*} K_{a} &=\dfrac{a_{H_3O^+}·a_{OH^-}}{a_{H_2O}^2} \[4pt] &\approx \dfrac{[H_{3}O^{+}][HO^{-}]}{(1)^{2}} \[4pt] &=[H_{3}O^{+}][HO^{-}] \end{align*} \nonumber
where $a$ is the activity of a species. Because water is the solvent, and the solution is assumed to be dilute, the activity of the water is approximated by the activity of pure water, which is defined as having a value of 1. The activity of each of the solutes is approximated by the molarity of the solute. In this reaction, one water molecule acts as an acid and one water molecule acts as a base. Thus, this reaction actually can be designated as the $K_a$ of water and the $K_b$ of water. It is most common, however, to designate this reaction and the associated law of mass action as the $K_w$ of water:
$K_{w}=[H_{3}O^{+}][HO^{-}] \nonumber$
At neutrality and 25 oC, [H3O+] and [HO] are both 1 × 10-7 M, therefore:
$K_{w}=[1\times 10^{-7}][1\times 10^{-7}]=1\times 10^{-14} \nonumber$
This simple relationship is actually quite powerful. Because KW is a constant, if we know either a hydronium ion or a hydroxide ion concentration, we can directly calculate the concentration of the other species. For example, if you are given that [H3O+] is 1 × 10-4 M, [HO] can be calculated as:
\begin{align*} K_{w} &=[H_{3}O^{+}][HO^{-}] \[4pt] &=1\times 10^{-14} \[4pt] &=[1\times 10^{-4}][HO^{-}] \end{align*} \nonumber
$[HO^{-}]=\frac{1\times 10^{-14}}{1\times 10^{-4}}=1\times 10^{-10}M \nonumber$
Exercise $1$: Calculating hydronium and hydroxide concentrations
A solution at 25 oC, is known to have a hydronium ion concentration of 4.5 × 10-5 M; what is the concentration of hydroxide ion in this solution?
Exercise $2$: Calculating hydronium and hydroxide concentrations
A solution at 25 oC, is known to have a hydroxide ion concentration of 7.5 ×10-2 M; what is the concentration of hydronium ion in this solution?
Exercise $3$: Calculating hydronium and hydroxide concentrations
A solution is known to have a hydronium ion concentration of 9.5 × 10-8 M; what is the concentration of hydroxide ion in this solution? | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/08%3A_Acids_Bases_and_pH/8.5%3A_The_Meaning_of_Neutrality_-_The_Autoprotolysis_of_Water.txt |
One thing that you should notice about the numbers in the previous examples is that they are very small. In general, chemists find that working with large negative exponents like these (very small numbers) is cumbersome. To simplify the process, calculations involving hydronium ion concentrations are generally done using logarithms. Recall that a logarithm is simply the exponent that some base number needs to be raised to in order to generate a given number. In these calculations, we will use a base of 10. A number such as 10,000 can be written as 104, so by the definition, the logarithm of 104 is simply 4. For a small number such as 10-7, the logarithm is again simply the exponent, or -7. Before calculators became readily available, taking the logarithm of a number that was not an integral power of 10 meant a trip to “log tables” (or even worse, using a slide rule). Now, pushing the LOG button on an a scientific calculator makes the process trivial. For example, the logarithm of 14,283 (with the push of a button) is 4.15482. If you are paying attention, you should have noticed that the logarithm contains six digits, while the original number (14,283) only contains five significant figures. This is because a logarithm consists of two sets of numbers; the digits to the left of the decimal point (called the characteristic) simply reflect the integral power of 10, and are not included when you count significant figures. The numbers after the decimal (the mantissa) should have the same significance as your experimental number, thus a logarithm of 4.15482 actually represents five significant figures.
There is one other convention that chemists apply when they are dealing with logarithms of hydronium ion concentrations, that is, the logarithm is multiplied by (-1) to change its sign. Why would we do this? In most aqueous solutions, [H3O+] will vary between 10-1 and 10-13 M, giving logarithms of –1 to –13. To make these numbers easier to work with, we take the negative of the logarithm (-log[H3O+]) and call it a pH value. The use of the lower-case “p” reminds us that we have taken the negative of the logarithm, and the upper-case “H” tells us that we are referring to the hydronium ion concentration. Converting a hydronium ion concentration to a pH value is simple. Suppose you have a solution where [H3O+] = 3.46 × 10-4 M and you want to know the corresponding pH value. You would enter 3.46 × 10-4 into your calculator and press the LOG button. The display should read “-3.460923901”. First, we multiply this by (-1) and get 3.460923901. Next, we examine the number of significant figures. Our experimental number, 3.46 × 10-4 has three significant figures, so our mantissa must have three digits. We round our answer and express our result as, pH = 3.461.
The reverse process is equally simple. If you are given a pH value of 7.04 and are asked to calculate a hydronium ion concentration, you would first multiply the pH value by (-1) to give –7.04. Enter this in your calculator and then press the key (or key combination) to calculate “10x”; your display should read “9.120108 × 10–8”. There are only two digits in our original mantissa (7.04) so we must round this to two significant figures, or [H3O+] = 9.1 × 10-8.
Exercise $1$
Calculating [H3O+] and pH Values
1. A solution is known to have a hydronium ion concentration of 4.5 ×10-5 M; what is the pH this solution?
2. A solution is known to have a pH of 9.553; what is the concentration of hydronium ion in this solution?
3. A solution is known to have a hydronium ion concentration of 9.5 ×10-8 M; what is the pH this solution?
4. A solution is known to have a pH of 4.57; what is the hydronium ion concentration of this solution?
There is another useful calculation that we can do by combining what we know about pH and expression
$K_{W}=[H_{3}O^{+}][HO^{-}] \nonumber$
We know that KW = 10-14 and we know that (-log [H3O+]) is pH. If we define (-log [HO]) as pOH, we can take our expression for KW and take the (-log) of both sides (remember, algebraically you can perform the same operation on both sides of an equation) we get:
$K_{W}=10^{-14}=[H_{3}O^{+}][HO^{-}] \nonumber$
$-\log (10^{-14})=(-\log [H_{3}O^{+}])+(-\log [HO^{-}]) \nonumber$
$14=pH+pOH \nonumber$
Which tells us that the values of pH and pOH must always add up to give 14! Thus, if the pH is 3.5, the pOH must be 14 – 3.5 = 11.5. This relationship is quite useful as it allows you to quickly convert between pH and pOH, and therefore between [H3O+] and [HO].
We can now re-address neutrality in terms of the pH scale:
• A solution is acidic if pH < 7.
• A solution is basic if pH > 7.
• A solution is neutral if pH = 7.
The simplest way to determine the pH of a solution is to use an electronic pH meter. A pH meter is actually a sensitive millivolt meter that measures the potential across a thin, sensitive glass electrode that is immersed in the solution. The voltage that develops is a direct function of the pH of the solution and the circuitry is calibrated so that the voltage is directly converted into the equivalent of a pH value. You will most likely use a simple pH meter in the laboratory. The thing to remember is that the sensing electrode has a very thin, fragile, glass membrane and is somewhat expensive to replace. Be careful!
A simple way to estimate the pH of a solution is by using an indicator. A pH indicator is a compound that undergoes a change in color at a certain pH value. For example, phenolphthalein is a commonly used indicator that is colorless at pH values below 9, but is pink at pH 10 and above (at very high pH it becomes colorless again). In the laboratory, a small amount of phenolphthalein is added to a solution at low pH and then a base is slowly added to achieve neutrality. When the phenolphthalein changes from colorless to pink, you know that enough base has been added to neutralize all of the acid that is present. In reality, the transition occurs at pH 9.2, not pH 7, so the resulting solution is actually slightly alkaline, but the additional hydroxide ion concentration at pH 9 (10-5 M) is generally insignificant relative to the concentrations of the solutions being tested.
A convenient way to estimate the pH of a solution is to use pH paper. This is simply a strip of paper that has a mixture of indicators embedded in it. The indicators are chosen so that the paper takes on a slightly different color over a range of pH values. The simplest pH paper is litmus paper that changes from pink to blue as a solution goes from acid to base. Other pH papers are more exotic. In the laboratory, you will use both indicators, like phenolphthalein, and pH papers in neutralization experiments called titrations as described in section 8.7. | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/08%3A_Acids_Bases_and_pH/8.6%3A_pH_Calculations.txt |
One of the standard laboratory exercises in General Chemistry is an acid-base titration. In order to perform an acid-base titration, you must have a solution of acid or base with a known concentration. You then slowly add a known volume of this solution, using a volumetric burette, to an acid or base solution with an unknown concentration until neutrality has been achieved. At that point, you know the volume and concentration of the reactant you have added, which means that you can calculate the number of moles that you added. Based on the stoichiometry of your neutralization reaction, you then know how many moles of acid or base were in the unknown sample. How do you know when you have reached neutrality? Generally an indicator or a pH meter is used (as described in Section 8.5). For example, if we had a solution of NaOH that was exactly 0.100 M and we had a beaker containing an unknown concentration of HCl. To perform the titration we would add a few drops of a stock phenolphthalein solution to our HCl, and then slowly add a measured amount of the NaOH solution until all of the acid had been consumed and the indicator changed from colorless to pink.
HCl (aq) + NaOH (aq) → NaCl (aq) + H2O
Working with the above example, if the volume of base that we added (as measured on the buret) was 12.6 mL, we could calculate the number of moles present in the unknown acid solution. This is equal to the known concentration of our NaOH (0.100 M) multiplied by the volume required for neutrality (0.0126 L), or:
$(0.100moles/L)(0.0126L)=1.26\times 10^{-3}moles \nonumber$
If we had used exactly 100.0 mL of our unknown acid in our titration, the concentration of our acid would be:
$\frac{(1.26\times 10^{-3}moles)}{(0.1000L)}=1.26\times 10^{-2}M \nonumber$
Exercise $1$
1. You are given a solution containing an unknown concentration of HCl. You carefully measure 50.0 mL of this solution into a flask and then add a few drops of phenolphthalein solution. You prepare a buret containing 0.055 M NaOH and note that the initial level of the solution in the buret is 12.6 mL. You slowly add the NaOH solution to the acid until the color change just occurs (as evidence of the color change becomes visible, you carefully stir the solution after each drop has been added). When the acid solution turns (and remains) pink, you note that the volume in the buret is now 28.9 mL. What is the concentration of the unknown acid solution?
2. If 25.00 mL of HCl solution with a concentration of 0.1234 M is neutralized by 23.45 mL of NaOH, what is the concentration of the base?
8.S: Acids Bases and pH (Summary)
• A bond that is formed from a hydrogen atom, which is part of a polar covalent bond (such as the O—H bond) to another, more electronegative atom (that has at least one unshared pair of electrons in its valence shell) is called a hydrogen bond. Hydrogen bonds are weak, partially covalent bonds. The bond dissociation energy of the O—H covalent bond 464 kJ/mole; the bond dissociation energy an O—H••••O hydrogen bond is about 21 kJ/mole.
• Even though hydrogen bonds are relatively weak, the vast network of hydrogen bonds in water makes the energy significant, and hydrogen bonding is generally used to explain the high boiling point of water (100 ˚C), relative to molecules of similar mass that cannot hydrogen bond. The extra energy represents the energy required to break down the hydrogen bonding network.
• Polar molecules, such as acids, strongly hydrogen bond to water. This hydrogen bonding not only stabilizes the molecular dipoles, but also weakens the H—A covalent bond (A represents the acid molecule). As a result of this weakening, the H—A bond in these acids stretches (the bond length increases) and then fully breaks. The hydrogen that was hydrogen-bonded to the water molecule now becomes fully bonded to the oxygen, forming the species H3O+ (the hydronium ion) and the acid now exists as an anion (A); this is the process of acid dissociation.
• The hydronium ion and the acid anion that are formed in an acid dissociation can react to re-form the original acid. This represents a set of forward- and back-reactions that occur together on a very fast time-scale; this type of a set of reactions is called an equilibrium and a double arrow is used in the chemical reaction to show this. This type of reaction is referred to as an acid dissociation equilibrium. HA (aq) + H2O ⇄ H3O+ (aq) + A (aq)
• For any equilibrium, an equilibrium constant can be written that describes whether the products or the reactants will be the predominant species in solution. For the dissociation of the simple acid, HA, the equilibrium constant, Ka, is simply given by the ratio of the concentrations of the products and the reactants, remembering that the molarity of the solutes have been used to approximate their activity, and that solvents, such as water, have an activity of 1. Thus for the ionization of HA; $K_{a}=\frac{[H_{3}O^{+}]{[A^{-}]}}{[HA]} \nonumber$
• According to the Brønsted Acid-Base Theory, any substance that ionizes in water to form hydronium ions (a proton donor) is called an acid; any substance that accepts a proton from a hydronium ion is a base. In an acid-base equilibrium, the conjugate acid is defined as the species that donates a hydrogen (a proton) in the forward reaction, and the conjugate base is the species that accepts a hydrogen (a proton) the reverse reaction. Thus for the ionization of HCl, HCl is the conjugate acid and Cl is the conjugate base. HCl (aq) + H2O ⇄ H3O+ (aq) + Cl (aq)
• Metal hydroxides, such as NaOH, dissolve in water to form metal cations and hydroxide anion. Hydroxide anion is a strong Brønsted base and, therefore, hydroxide anion accepts a proton from the hydronium ion to form two moles of water. The reaction of a Brønsted acid with a Brønsted base to form water is the process of neutralization.
• Just like water can promote the ionization of acids, water can also promote the ionization of itself. This equilibrium process occurs very rapidly in pure water and any sample of pure water will always contain a small concentration of hydronium and hydroxide ions. In pure water, at 25 oC, the concentration of hydronium ions ([H3O+]) and hydroxide ions ([HO]) will both be equal to exactly 1 × 10-7 M. This is referred to as the autoprotolysis of water.
• The equilibrium for the autoprotolysis of water is defined as Kw, according to the equation shown below:
$K_{W}=[H_{3}O^{+}]{[HO^{-}]} \nonumber$
and at neutrality, [H3O+] and [HO] are both 1 × 10-7 M, making the value of Kw
$K_{W}=[1\times 10^{-7}]{[1\times 10^{-7}]}=1\times 10^{-14} \nonumber$
• Based on the autoprotolysis equilibrium, acidic, basic and neutral solutions can be defined as:
• A solution is acidic if [H3O+] > 1 × 10-7 M.
• A solution is basic if [H3O+] < 1 × 10-7 M.
• A solution is neutral if [H3O+] = 1 × 10-7 M.
• A pH value is simply the negative of the logarithm of the hydronium ion concentration (-log[H3O+]).
• Remember that a logarithm consists of two sets of numbers; the digits to the left of the decimal point (the characteristic) reflect the integral power of 10, and are not included when you count significant figures. The numbers after the decimal (the mantissa) have the same significance as your experimental number, thus a logarithm of 4.15482 represents five significant figures.
• In an acid-base titration a solution of acid or base with a known concentration is slowly add to an acid or base solution with an unknown concentration, using a volumetric burette, to until neutrality has been achieved. Typically an indicator or a pH meter is used to signify neutrality.
• At neutrality, the volume and concentration of the reactant you have added is known, which means that you can calculate the number of moles that you added (remember, concentration × volume = moles). Based on the stoichiometry of your neutralization reaction, you then know how many moles of acid or base were in the unknown sample. | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/08%3A_Acids_Bases_and_pH/8.7%3A_Titrations_-_Neutralization_and_Stoichiometry.txt |
Some of the first real breakthroughs in the study of chemistry happened in the study of the gaseous state. In gases, the volume of the actual gas particles is but a tiny fraction of the total volume that the gas occupies. This allowed early chemists to relate parameters such as volume and the number of gas particles, leading to the development of the mole concept. As we have seen in previous chapters, the notion of a chemical mole allows us to do quantitative chemistry and lead us to the point where we can routinely address reaction stoichiometry, etc. In this chapter, we will visit some of the early observations that lead to our current understand of gasses and how they behave. We will see how the relationships between pressure and volume; volume and temperature and volume and moles lead to the ideal gas laws and how these simple rules can allow us to do quantitative calculations in the gas phase.
09: The Gaseous State
In Chapter 2, we learned about the three principle states of matter; solids, liquids and gasses. We explained the properties of the states of matter using the kinetic molecular theory (KMT). Substances in the gaseous state, according to the KMT, have enough kinetic energy to break all of the attractive forces between the individual gas particles and are therefore free to separate and rapidly move throughout the entire volume of their container. Because there is so much space between the particles in a gas, a gas is highly compressible. High compressibility and the ability of gases to take on the shape and volume of its container are two of the important physical properties of gasses.
The gas that we are all most familiar with is the mixture of elements and compounds that we call the “atmosphere”. The air that we breath is mostly nitrogen and oxygen, with much smaller amounts of water vapor, carbon dioxide, noble gasses and the organic compound, methane (Table 9.1).
Table 9.1. Approximate Composition of the Atmosphere
Table 9.1 Approximate Composition of the Atmosphere
Gas Concentration, Parts per Billion Percentage
N2 7.8 × 108 78%
O2 2.0 × 108 20%
H2O About 106 – 107 < 1%
Ar 9.3 × 106 < 1%
CO2 3.5 × 105 < 0.05%
Ne 1.8 × 104 trace
He 5.2 × 103 trace
CH4 1.6 × 103 trace
A gas that is enclosed in a container exerts a pressure on the inner walls of that container. This pressure is the result of the countless collisions of the gas particles with the container wall. As each collision occurs, a small amount of energy is transferred, generating a net pressure. Although we are generally unaware of it, the gasses in the atmosphere generate a tremendous pressure on all of us. At sea level, atmospheric pressure is equal to 14.7 pounds per square inch. Putting this in perspective, for a person of average height and build, the total pressure from the atmosphere pressing on their body is about 45,000 pounds! Why aren’t we squashed? Remember, we also have air inside our bodies and the pressure from the inside balances the pressure outside, keeping us nice and firm, not squishy!
The proper SI unit for pressure is the Pascal (Pa), where 1 Pa = 1 kg m-1 s-2. In chemistry, however, it is more common to measure pressure in terms of atmospheres (atm) where 1 atm is atmospheric pressure at sea level, or 1 atm = 14.7 pounds per square inch (1 atm = 101,325 Pa). Atmospheric pressure is typically measured using a device called a barometer. A simple mercury barometer (also called a Torricelli barometer, after its inventor) consists of a glass column, about 30 inches high, closed at one end and filled with mercury. The column is inverted and placed in an open, mercury-filled reservoir. The weight of the mercury in the tube causes the column to drop to the point that the mass of the mercury column matches the atmospheric pressure exerted on the mercury in the reservoir. The atmospheric pressure is then read as the height of the mercury column. Again, working at sea level, 1 atmosphere is exactly equal to a column height of 760 mm of mercury. The units for the conversion are 1 atm = 760 mm Hg, and this is an exact relationship with regard to significant figures. The unit torr (after Torricelli) is sometimes used in place of mm Hg. | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/09%3A_The_Gaseous_State/9.1%3A_Gasses_and_Atmospheric_Pressure.txt |
The three gas laws that we covered in Section 9.2, 9.3, 9.4 and 9.5 describe the effect of pressure, temperature and the number of moles of a gas on volume. The three independent gas laws are:
• Boyle’s law: ${$/extract_itex]displaystyle V\propto {}^{1}\!\!\diagup \!\!{}_{P}\;} • Charles’s law: ${\$displaystyle V\propto T{\text{ }}} • Avogadro’s law: ${\$displaystyle V\propto n{\text{ }}} If volume (V) is proportion to each of these variables, it must also be proportional to their product: \[V\propto \frac{nT}{P} \nonumber$
If we replace the proportionality symbol with a constant (let’s just choose R to represent our constant), we can re-write the equation as:
$V=R\left ( \frac{nT}{P}\right ) \nonumber$
or
$PV=nRT \nonumber$
Example $1$:
The value of the proportionality constant R, can be calculated from the fact that exactly one mole of a gas at exactly 1 atm and at 0 ˚C (273 K) has a volume of 22.414 L.
Solution
Substituting in the equation:
$PV=nRT\; \; or\; \; R=\frac{PV}{nT} \nonumber$
$R=\frac{(1\; atm)(22.414\; L)}{(1\; mole)(273\; K)}=0.082057\; L\; atm\; mol^{-1}\; K^{-1} \nonumber$
The uncomfortable and somewhat obnoxious constant is called the universal gas constant, and you will need to know it (or look it up) whenever you solve problems using the combined ideal gas law.
Exercise $1$
What volume will 17.5 grams of N2 occupy at a pressure of 876 mm Hg and at 123 ˚C?
Many of the problems that you will encounter when dealing with the gas laws can be solved by simply using the “two-state” approach. Because R is a constant, we can equate an initial and a final state as:
$R=\frac{P_{1}V_{1}}{n_{1}T_{1}}\; \; for\; the\; initial\; state \nonumber$
$R=\frac{P_{2}V_{2}}{n_{2}T_{2}}\; \; for\; the\; final\; state \nonumber$
$\frac{P_{1}V_{1}}{n_{1}T_{1}}=\frac{P_{2}V_{2}}{n_{2}T_{2}} \nonumber$
Using this equation, you can solve for multiple variables within a single problem.
Exercise $1$
A sample of oxygen occupies 17.5 L at 0.75 atm and 298 K. The temperature is raised to 303 K and the pressure is increased to 0.987 atm. What is the final volume of the sample?
If you noticed, we calculated the value of the proportionality constant R based on the fact that exactly one mole of a gas at exactly 1 atm and at 0 ˚C (273 K) has a volume of 22.414 L. This is one of the “magic numbers in chemistry; exactly one mole of any gas under these conditions will occupy a volume of 22.414 L. The conditions, 1 atm and 0 ˚C, are called standard temperature and pressure, or STP. The fact that all gases occupy this same molar volume can be rationalized by realizing that 99.999% of a gas is empty space, so it really doesn’t matter what’s in there, it all occupies the same volume. This realization is attributed to Amedeo Avogadro and Avogadro’s hypothesis, published in 1811, suggested that equal volumes of all gases at the same temperature and pressure contained the same number of molecules. This is the observation that led to the measurement of Avogadro’s number (6.0221415 × 1023), the number of things in a mole. The importance of the “magic number” of 22.414 L per mole (at STP) is that, when combined with the ideal gas laws, any volume of a gas can be easily converted into the number of moles of that gas.
Exercise $1$
1. A sample of methane has a volume of 17.5 L at 100.0 ˚C and 1.72 atm. How many moles of methane are in the sample?
2. A 0.0500 L sample of a gas has a pressure of 745 mm Hg at 26.4˚ C. The temperatureis now raised to 404.4 K and the volume is allowed to expand until a final pressure of 1.06 atm is reached. What is the final volume of the gas?
3. When 128.9 grams of cyclopropane (C3H6) are placed into an 8.00 L cylinder at 298 K, the pressure is observed to be 1.24 atm. A piston in the cylinder is now adjusted so that the volume is now 12.00 L and the pressure is 0.88 atm. What is the final temperature of the gas? | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/09%3A_The_Gaseous_State/9.5%3A_The_Ideal_Gas_Law.txt |
With an understanding of the ideal gas laws, it is now possible to apply these principles to chemical stoichiometry problems. For example, zinc metal and hydrochloric acid (hydrogen chloride dissolved in water) react to form zinc (II) chloride and hydrogen gas according to the equation shown below:
2 HCl (aq) + Zn (s) → ZnCl2 (aq) + H2 (g)
Example $1$:
A sample of pure zinc with a mass of 5.98 g is reacted with excess hydrochloric acid and the (dry) hydrogen gas is collected at 25.0 ˚C and 742 mm Hg. What volume of hydrogen gas would be produced?
Solution
This is a “single state” problem, so we can solve it using the ideal gas law, PV = nRT. In order to find the volume of hydrogen gas (V), we need to know the number of moles of hydrogen that will be produced by the reaction. Our stoichiometry is simply one mole of hydrogen per mole of zinc, so we need to know the number of moles of zinc that are present in 5.98 grams of zinc metal. The temperature is given in centigrade, so we need to convert into Kelvin, and we also need to convert mm Hg into atm.
Conversions:
$25.0\; C+273=298\; K \nonumber$
$(742\; mm\; Hg)\times \left ( \frac{1\; atm}{760\; mm\; Hg} \right )=0.976\; atm \nonumber$
$(5.98\; g\; Zn)\times \left ( \frac{1.00\; mol}{65.39\; g\; Zn} \right )=0.0915\; mol \nonumber$
Substituting:
$PV=nRT \nonumber$
$(0.976\; atm)\times V=(0.0915\; mol)(0.0821\; L\; atm\; mol^{-1}K^{-1})(298\; K) \nonumber$
$V=\frac{(0.0915\; mol)(0.0821\; L\; atm\; mol^{-1}K^{-1})(298\; K)}{(0.976\; atm)}=2.29\; L \nonumber$
We can also use the fact that one mole of a gas occupies 22.414 L at STP in order to calculate the number of moles of a gas that is produced in a reaction. For example, the organic molecule ethane (CH3CH3) reacts with oxygen to give carbon dioxide and water according to the equation shown below:
2 CH3CH3 (g) + 7 O2 (g) → 4 CO2 (g) + 6 H2O (g)
Example $1$:
An unknown mass of ethane is allowed to react with excess oxygen and the carbon dioxide produced is separated and collected. The carbon dioxide collected is found to occupy 11.23 L at STP; what mass of ethane was in the original sample?
Solution
Because the volume of carbon dioxide is measured at STP, the observed value can be converted directly into moles of carbon dioxide by dividing by 22.414 L mol–1. Once moles of carbon dioxide are known, the stoichiometry of the problem can be used to directly give moles of ethane (molar mass 30.07 g mol-1), which leads directly to the mass of ethane in the sample.
$(11.23\; L\; CO_{2})\times \left ( \frac{1\; mol}{22.414\; L} \right )=0.501\; mol\; CO_{2} \nonumber$
Reaction stoichiometry:
$(0.501\; mol\; CO_{2})\times \left ( \frac{2\; mol\; CH_{3}CH_{3}}{4\; mol\; CO_{2}} \right )=0.250\; mol\; CH_{3}CH_{3} \nonumber$
The ideal gas laws allow a quantitative analysis of whole spectrum of chemical reactions. When you are approaching these problems, remember to first decide on the class of the problem:
• If it is a “single state” problem (a gas is produced at a single, given, set of conditions), then you want to use PV = nRT.
• If it is a “two state” problem (a gas is changed from one set of conditions to another) you want to use $\frac{P_{1}V_{1}}{n_{1}T_{1}}=\frac{P_{2}V_{2}}{n_{2}T_{2}} \nonumber$
• If the volume of gas is quoted at STP, you can quickly convert this volume into moles with by dividing by 22.414 L mol-1.
Once you have isolated your approach ideal gas law problems are no more complex that the stoichiometry problems we have addressed in earlier chapters.
Exercise $1$
1. An automobile air bag requires about 62 L of nitrogen gas in order to inflate. The nitrogen gas is produced by the decomposition of sodium azide, according to the equation shown below
2 NaN3 (s) → 2 Na (s) + 3 N2 (g)
What mass of sodium azide is necessary to produce the required volume of nitrogen at 25 ˚C and 1 atm?
1. When Fe2O3 is heated in the presence of carbon, CO2 gas is produced, according to the equation shown below. A sample of 96.9 grams of Fe2O3 is heated in the presence of excess carbon and the CO2 produced is collected and measured at 1 atm and 453 K. What volume of CO2 will be observed?
2 Fe2O3(s) + 3 C (s) → 4 Fe (s) + 3 CO2 (g)
1. The reaction of zinc and hydrochloric acid generates hydrogen gas, according to the equation shown below. An unknown quantity of zinc in a sample is observed
to produce 7.50 L of hydrogen gas at a temperature of 404 K and a pressure of 1.75 atm. How many moles of zinc were in the sample?
Zn (s) + 2 HCl (aq) → ZnCl2 (aq) + H2 (g) | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/09%3A_The_Gaseous_State/9.6%3A_Combining_Stoichiometry_and_the_Ideal_Gas_Laws.txt |
• Gasses are compressible because there is so much space between individual gas particles. Energy transferred from the collision of gas particles with their container exerts a gas pressure. In chemistry, we typically measure gas pressure using units of atmospheres (atm) or in mm Hg (also referred to as torr). One atmosphere of pressure equals exactly 760 mm Hg.
• The volume of a gas varies inversely with the applied pressure; the greater the pressure, the smaller the volume. This is relationship is referred to as Boyles’s Law. For a two-state system where the number of moles of gas and the temperature remain constant, Boyle’s Law can be expressed as $P_{1}V_{1}=P_{2}V_{2} \nonumber$
• The volume of a gas varies directly with the absolute temperature; the higher the temperature, the larger the volume. This is relationship is referred to as Charles’s Law. For a two-state system where the number of moles of gas and the pressure remain constant, Charles’s Law can be expressed as: $\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}} \nonumber$
• In this equation, the absolute temperature in Kelvin (K) must be used. Kelvin is defined as (degrees centigrade + 273.15). Zero degrees Kelvin is referred to as “absolute zero” and it is the temperature at which (theoretically) all molecular motion would cease.
• The volume of a gas varies directly with the number of moles of the gas that are present; the greater the number of moles, the larger the volume. This is relationship is referred to as Avogadro’s Law. For a two-state system where the temperature and the pressure of a gas remain constant, Avogadro’s Law can be expressed as: $\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}} \nonumber$
• Because the volume of a gas varies directly with the number of moles of the gas that are present and with the absolute temperature (in Kelvin), and inversely with the pressure, the gas laws can be combined into a single proportionality; $V\propto \left ( \frac{nT}{P} \right ) \nonumber$
• This proportionality can be converted to an equality by inserting the proportionality constant R (the universal gas constant), where R = 0.082057 L atm mol-1 K-1, and can be re-written as: $V=R\left ( \frac{nT}{P} \right )\; \; or\; \; PV=nRT \nonumber$
• This is referred to as the Ideal Gas Law and is valid for most gasses at low concentrations. For a two-state system where the identity of the gas does not change, the Ideal Gas Law can be expressed as: $\frac{P_{1}V_{1}}{n_{1}T_{1}}=\frac{P_{2}V_{2}}{n_{2}T_{2}} \nonumber$
• The gas constant R, is calculated based on the experimental observation that exactly one mole of any gas at exactly 1 atm and at 0 ˚C (273 K) has a volume of 22.414 L. The conditions, 1 atm and 0 ˚C, are called standard temperature and pressure, or STP.
• The ideal gas laws allow a quantitative analysis of whole spectrum of chemical reactions involving gasses. When you are approaching these problems, remember to first decide on the class of the problem:
• If it is a “single state” problem (a gas is produced at a single, given, set of conditions), then you want to use PV = nRT.
• If it is a “two state” problem (a gas is changed from one set of conditions to another) you want to use $\frac{P_{1}V_{1}}{n_{1}T_{1}}=\frac{P_{2}V_{2}}{n_{2}T_{2}} \nonumber$
• If the volume of gas is quoted at STP, you can quickly convert this volume into moles with by dividing by 22.414 L mol-1. | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/09%3A_The_Gaseous_State/9.S%3A_The_Gaseous_State_%28Summary%29.txt |
The kinetic molecular theory is useful when we are trying to understand the properties and behaviors of gases. The KMT (and related theories) tell us that:
• There is a tremendous amount of distance between individual particles in the gas phase.
• Gas particles move randomly at various speeds and in every possible direction.
• Attractive forces between individual gas particles are negligible.
• Collisions between gas particles are fully elastic.
• The average kinetic energy of particles in the gas phase is proportional to the temperature of the gas.
In reality, these predictions only apply to “ideal gases”. An ideal gas has perfectly elastic collisions and has no interactions with its neighbors or with the container. Real gases deviate from these predictions, but at common temperatures and pressures, the deviations are generally small and in this text we will treat all gases as if they were “ideal”.
Because there is so much empty space between gas molecules, it is easy to see why a gas is so compressible. If you have a container filled with a gas, you can squeeze it down to a smaller volume by applying pressure. The harder you squeeze (the more pressure you apply) the smaller the resulting volume will be. Imagine a bicycle pump compressing air into a tire. As pressure is applied to the pump, the same number of gas molecules are squeezed into a smaller volume.
The dependence of volume on pressure is not linear. In 1661, Robert Boyle systematically studied the compressibility of gasses in response to increasing pressure. Boyle found that the dependence of volume on pressure was non-linear but that a linear plot could be obtained if the volume was plotted against the reciprocal of the pressure, 1/P. This is stated as Boyle’s law.
Boyle’s law
The volume (V) of an ideal gas varies inversely with the applied pressure (P) when the temperature (T) and the number of moles (n) of the gas are constant.
Mathematically, Boyle’s law can be stated as:
$V\propto \frac{1}{P}\; \; at\; constant\; T\; and\; n \nonumber$
$V=constant\left ( \frac{1}{P} \right )\; \; or\; \; PV=constant \nonumber$
We can use Boyle’s law to predict what will happen to the volume of a sample of gas as we change the pressure. Because PV is a constant for any given sample of gas (at constant T), we can imagine two states; an initial state with a certain pressure and volume (P1V1), and a final state with different values for pressure and volume (P2V2). Because PV is always a constant, we can equate the two states and write:
$P_{1}V_{1}=P_{2}V_{2} \nonumber$
Example $1$:
Now imagine that we have a container with a piston that we can use to compress the gas inside. You are told that, initially, the pressure in the container is 765 mm Hg and the volume is 1.00 L. The piston is then adjusted so that the volume is now 0.500 L; what is the final pressure?
Solution
We substitute into our Boyle’s law equation:
$P_{1}V_{1}=P_{2}V_{2} \nonumber$
$(765\; mm\; Hg)(1.00\; L)=P_{2}(0.500\; L) \nonumber$
$P_{2}=\left ( \frac{(765\; mm\; Hg)(1.00\; L)}{(0.500\; L)} \right )=1530\; mm\; Hg \nonumber$
Exercise $1$
1. A container with a piston contains a sample of gas. Initially, the pressure in the container is exactly 1 atm, but the volume is unknown. The piston is adjusted so that the volume is 0.155 L and the pressure is 956 mm Hg; what was the initial volume?
2. The pressure of 12.5 L of a gas is 0.82 atm. If the pressure changes to 1.32 atm, what will the final volume be? A sample of helium gas has a pressure of 860.0 mm Hg. This gas is transferred to a different container having a volume of 25.0 L; in this new container, the pressure is determined to be 770.0 mm Hg. What was the initial volume of the gas? | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/09%3A_The_Gaseous_State/92%3A_The_Pressure-Volume_Relationship%3A_Boyles_Law.txt |
A fun laboratory demonstration in which a fully inflated balloon is placed in liquid nitrogen (at a temperature of –196 ˚C) and it shrinks to about 1/1000th of its former size. If the balloon is carefully removed and allowed to warm to room temperature, it will again be fully inflated.
This is a simple demonstration of the effect of temperature on the volume of a gas. In 1787, Jacques Charles performed a systematic study of the effect of temperature on gases. Charles took samples of gases at various temperatures, but at the same pressure, and measured their volumes.
The first thing to note is that the plot is linear. When the pressure is constant, volume is a direct linear function of temperature. This is stated as Charles’s law.
Charles’s law
The volume (V) of an ideal gas varies directly with the temperature of the gas (T) when the pressure (P) and the number of moles (n) of the gas are constant.
We can express this mathematically as:
$V\propto T\; \; at\; constant\; P\; and\; n \nonumber$
$V=constant(T)\; or\; \frac{V}{T}=constant \nonumber${$/extract_itex]displaystyle V\propto T{\text{ }}\left({\text{at constant }}P{\text{ and }}n\right)} The data for three different samples of the same gas is as follows: 0.25 moles, 0.50 moles and 0.75 moles. All of these samples behave as predicted by Charles’s law (the plots are all linear), but , if you extrapolate each of the lines back to the y-axis (the temperature axis), all three lines intersect at the same point! This point, with a temperature of –273.15 ˚C, is the theoretical point where the samples would have “zero volume”. This temperature, -273.15 ˚C, is called absolute zero. An even more intriguing thing is that the value of absolute zero is independent of the nature of the gas that is used. Hydrogen, oxygen, helium, argon, (or whatever), all gases show the same behavior and all intersect at the same point. The temperature of this intersection point is taken as “zero” on the Kelvin temperature scale. The abbreviation used in the Kelvin scale is K (no degree sign) and there are never negative values in degrees Kelvin. The size of the degree increment in Kelvin is identical to that in Centigrade and Kelvin and centigrade scales are related by the simple conversion: \[Kelvin=Centigrade+273.15 \nonumber$
Note
Please note that whenever you work gas law problems where temperature is one variable, you MUST use the Kelvin scale.
Just like we did for pressure-volume problems, we can use Charles’s law to predict what will happen to the volume of a sample of gas as we change the temperature. Because ${\displaystyle {}^{V}\!\!\diagup \!\!{}_{T}\;}$ is a constant for any given sample of gas (at constant P), we can again imagine two states; an initial state with a certain temperature and volume ( ${\displaystyle {}^{V_{1}}\!\!\diagup \!\!{}_{T_{1}}\;}$ ), and a final state with different values for pressure and volume ([ ${\displaystyle {}^{V_{2}}\!\!\diagup \!\!{}_{T_{2}}\;}$). Because ${\displaystyle {}^{V}\!\!\diagup \!\!{}_{T}\;}$ is always a constant, we can equate the two states and write:
$\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}} \nonumber$
Example $1$:
We have a container with a piston that we can use to adjust the pressure on the gas inside. You are told that, initially, the temperature of the gas in the container is 175 K and the volume is 1.50 L. The temperature is changed to 76 K and the piston is then adjusted so that the pressure is identical to the pressure in the initial state; what is the final volume?
Solution
We substitute into our Charles’s law equation:
$\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}} \nonumber$
$\frac{1.50\; L}{175\; K}=\frac{V_{2}}{76\; K} \nonumber$
$V_{2}=\left ( \frac{(76\; K)(1.50\; L)}{175\; K} \right )=0.65\; L \nonumber$
Exercise $1$
1. A container with a piston contains a sample of gas. Initially, the pressure in the container is exactly 1 atm, the temperature is 14.0 ˚C and the volume is 997 mL.
The temperature is raised to 100.0 ˚C and the piston is adjusted so that the pressure is again exactly 1 atm What is the final volume?
2. A 50.0 mL sample of gas at 26.4˚ C, is heated at constant pressure until its volume is 62.4 mL . What is the final temperature of the gas? A sample container of carbon monoxide occupies a volume of 435 mL at a temperature of 298 K. What would its temperature be if the pressure remained constant and the volume was changed to 265 mL? (182 K)
94: The Mole-Volume Relationship: Avogadros Law
A plot of the effect of temperature on the volume of a gas at constant pressure shows that the volume of a gas is directly proportional to the number of moles of that gas. This is stated as Avogadro’s law.
Avogadro’s law
The volume ($V$) of an ideal gas varies directly with the number of moles of the gas (n) when the pressure (P) and the number of temperature (T) are constant.
We can express this mathematically as:
$V\propto n\; \; at\; \; constant\; P\; and\; T \nonumber$
$V=constant\times (n)\; \; or\; \; \frac{V}{n}=constant \nonumber$
As before, we can use Avogadro’s law to predict what will happen to the volume of a sample of gas as we change the number of moles. Because $V/n$ is a constant for any given sample of gas (at constant $P$ and $T$), we can again imagine two states; an initial state with a certain number of moles and volume ($V_1/n_1$), and a final state with values for a different number of moles and volume ($V_2/n_2$). Because $V/n$ is always a constant, we can equate the two states and write:
$\dfrac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}} \nonumber$
Example $1$
We have a container with a piston that we can use to adjust the pressure on the gas inside, and we can control the temperature. You are told that, initially, the container contains 0.20 moles of hydrogen gas and 0.10 mole of oxygen in a volume is 2.40 L. The two gases are allowed to react (a spark ignites the mixture) and the piston is then adjusted so that the pressure is identical to the pressure in the initial state and the container is cooled to the initial temperature; what is the final volume of the product of the reaction?
Solution
First, we need to look at the reaction involved. Hydrogen and oxygen react to form water. Two moles of hydrogen react with one mole of oxygen to give two moles of water, as shown below:
$\ce{2H2 (g) + O2 (g) → 2 H2O (g)} \nonumber$
Initially we have three moles of gas and, after reaction, we have two moles. We can now substitute into Avogadro’s law:
$\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}} \nonumber$
$\frac{2.40\; L}{3\; moles}=\frac{V_{2}}{2\; moles} \nonumber$
$V_{2}=\left ( \frac{(2.40\; L)(2\; moles)}{3\; moles} \right )=1.60\; L \nonumber$
Thus we have described the dependence of the volume of a gas on the pressure (Boyle’s law), the temperature (Charles’s law) and the number of moles of the gas (Avogadro’s law). In the following section, we will combine these to generate the Ideal Gas Law, in which all three variables (pressure, temperature and number of moles) can vary independently. | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/09%3A_The_Gaseous_State/93%3A_The_Temperature-Volume_Relationship%3A_Charless_Law.txt |
As we have studied chemical reactions in this course, we have used a “reaction arrow” to indicate the process of reactants being converted into products. The implication here is that the reaction is “irreversible”, proceeding in the direction of the arrow. Many simple reactions that we encounter in chemistry, however, are not irreversible, but proceed in both directions with products readily be converted back into reactants. When a set of reactions, such as this, proceed so that the rate of conversion in one direction equals the rate of conversion in the other, we say the reactions are in equilibrium. An equilibrium system is shown by using a set of double arrows, proceeding in opposite directions. An understanding of equilibrium is essential to an appreciation of the concepts behind acid-base behavior, solubility phenomena, etc.
10: Principles of Chemical Equilibrium
Pure dinitrogen tetroxide ($\ce{N2O4}$) is a colorless gas that is widely used as a rocket fuel. Although $\ce{N2O4}$ is colorless, when a container is filled with pure $\ce{N2O4}$, the gas rapidly begins to turn a dark brown. A chemical reaction is clearly occurring, and indeed, chemical analysis tells us that the gas in the container is no longer pure $\ce{N2O4}$, but has become a mixture of dinitrogen tetroxide and nitrogen dioxide; $\ce{N2O4}$ is undergoing a decomposition reaction to form $\ce{NO2}$. If the gaseous mixture is cooled, it again turns colorless and analysis tells us that it is again, almost pure $\ce{N2O4}$; this means that the $\ce{NO2}$ in the mixture can also undergo a synthesis reaction to re-form $\ce{N2O4}$. Initially, only $\ce{N2O4}$ is present. As the reaction proceeds, the concentration of $\ce{N2O4}$ decreases and the concentration of $\ce{NO2}$ increases. However, if you examine the figure, after some time, the concentrations of $\ce{N2O4}$ and $\ce{NO2}$ have stabilized and, as long as the temperature is not changed, the relative concentrations of the two gasses remain constant.
The reversible reaction of one mole of $\ce{N2O4}$, forming two moles of $\ce{NO2}$, is a classic example of a chemical equilibrium. We encountered the concept of equilibrium in Chapter 9 when we dealt with the autoprotolysis of water to form the hydronium and hydroxide ions, and with the dissociation of weak acids in aqueous solution.
$\ce{2 H2O <=> H3O^{+} + HO^{–}} \nonumber$
When we wrote these chemical equations, we used a double arrow to signify that the reaction proceeded in both directions. Using this convention, the dissociation of dinitrogen tetroxide to form two molecules of nitrogen dioxide can be shown as:
$\ce{N2O4 ⇄ 2 NO2} \nonumber$
If the temperature of our gas mixture is again held constant and the total pressure of the gas in the container is varied, analysis shows that the partial pressure of $\ce{N2O4}$ varies as the square of the partial pressure of $\ce{NO2}$. The Ideal Gas Laws tell us that the partial pressure of a gas, Pgas, is directly proportional to the concentration of that gas in the container). Mathematically, the relationship between the partial pressures of the two gasses can be expressed by the equation below:
$\frac{(P_{NO_{2}})^{2}}{P_{N_{2}O_{4}}}=K \nonumber$
10.2: The Equilibrium Constant
The constant in the equation in section 10.1 is called the equilibrium constant for the reaction. What the equilibrium constant for this reaction tells us is that, regardless of pressures (or concentrations), a mixture of the two gasses will undergo reaction such that the ratios of the partial pressures reach a constant value, given by the equilibrium constant, K. Once this constant ratio has been reached does this mean the reactions stop? Of course not. In the region of the plot where the concentrations of N2O4 and NO2 are constant (the lines are level) N2O4 is still decomposing to form two molecules of NO2 and two molecules of NO2 are still reacting to synthesize a molecule of N2O4, but the lines are level because the rates of the two chemical reactions have become constant; N2O4 is decomposing at the same rate as two molecules of NO2 are reacting to form N2O4.
In theory, all chemical reactions are equilibria. In practice, however, most reactions are so slow in the reverse direction that they are considered “irreversible”. When a reaction evolves a gas, forms a precipitate or proceeds with the generation of a large amount of heat or light (for example, combustion) the reaction is essentially irreversible. Many chemical reactions are, however, readily reversible and for these reactions the mathematical expression for the equilibrium constant can be written using a simple set of rules.
1. Partial pressures (or molar concentrations) of products are written in the numerator of the expression and the partial pressures (or concentrations) of the reactants are written in the denominator.
2. If there is more that one reactant or more that one product, the partial pressures (or concentrations) are multiplied together.
3. The partial pressure (or concentration) of each reactant or product is then raised to the power that numerically equals the stoichiometric coefficient appearing with that term in the balanced chemical equation.
4. Reactants or products that are present as solids or liquids or solvents have a defined activity of 1. Therefore, although their activity does formally appear in the equilibrium expression, they do not affect the value of the equilibrium constant, and so are often not written in the expression.
Thus, for the reaction of nitrogen with hydrogen gas to form ammonia:
N2 (g) + 3 H2 (g) ⇄ 2 NH3 (g)
The expression for the equilibrium constant will have the partial pressure of ammonia in the numerator, and it will be squared, corresponding to the coefficient “2” in the balanced equation; ${\displaystyle \left(P_{NH_{3}}\right)^{2}}$. Because there are two reactants, the partial pressures for nitrogen and hydrogen will be multiplied in the denominator. The partial pressure of nitrogen will be raised to the “first power” (which is not shown) and the partial pressure of hydrogen will be cubed, corresponding to the coefficient “3”; ${\displaystyle \left(P_{N_{2}}\right)\left(P_{H_{2}}\right)^{3}}$. The final expression for the equilibrium constant is given in the equation below:
$\frac{(P_{NH_{3}})^{2}}{P{N_{2}(P_{H_{2}})^{3}}}=K \nonumber$
Exercise $1$
For the chemical reactions shown below, write an expression for the equilibrium constant in terms of the partial pressures of the reactants and products.
1. PCl5 (g) ⇄ PCl3 (g) + Cl2 (g)
2. 2 NOCl (g) ⇄ 2 NO (g) + Cl2 (g)
3. PCl3 (g) + 3 NH3 (g) ⇄ P(NH2)3 (g) + 3 HCl (g)
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The numeric value of the equilibrium constant tells us something about the ratio of the reactants and products in the final equilibrium mixture. Likewise, the magnitude of the equilibrium constant tells us about the actual composition of that mixture.
In the three equilibrium systems, the first depicts a reaction in which the ratio of products to reactants is very small. Because the expression for the equilibrium constant is given by the pressure (or concentration) of products divided by the pressure (or concentration) of reactants, the equilibrium constant, K, for this system is also small. In the second example, the concentrations of reactants and products are shown to be equal, making the ratio (the equilibrium constant) equal to “1”. In the last example, the products are shown to dominate the equilibrium mixture, making the ratio ${$/extract_itex]displaystyle {}^{\left(P_{Products}\right)}\!\!\diagup \!\!{}_{\left(P_{Reactants}\right)}\;} very large. In these examples, the stoichiometric ratios of the reactants and products are one and there is only one reactant and only one product; if multiple reactants or products are involved, the relationship between their concentrations would be more complex, but that ratio is always given by the expression for K. This fact allows us to take data for an equilibrium reaction and, if K is known, calculate concentrations for reactants and products. Likewise, if all of the equilibrium concentrations are known, we can use these to calculate a value for the equilibrium constant. In these types of problems, an ICE table is often useful. This table has entries for Initial concentrations (or pressures), Equilibrium concentrations and any Change between the initial and equilibrium states. For example, consider the reaction between carbon monoxide and chlorine to form phosgene, a deadly compound that was used as a gas warfare agent in World War I. CO (g) + Cl2 (g) ⇄ COCl2 (g) Example $1$: A mixture of CO and Cl2 has initial partial pressures of 0.60 atm for CO and 1.10 atm for Cl2. After the mixture reaches equilibrium, the partial pressure of COCl2 is 0.10 atm. Determine the value of K. Solution The initial pressures for carbon monoxide and chlorine are placed in the first row and the equilibrium pressure for phosgene is placed in the last row. Initially, the pressure of phosgene was zero, so that goes in the first row; the change for phosgene is therefore “+ 0.10 atm”. Solutions to Example 10.3.1 ${\displaystyle P_{CO}}$ ${\displaystyle P_{Cl_{2}}}$ ${\displaystyle P_{COCl_{2}}}$ === Initial === 0.60 atm 1.10 atm 0 atm Change + 0.10 atm Equilibrium 0.10 atm Because one mole of CO is required to make one mole of COCl2 the partial pressure of CO must have dropped by 0.10 atm (the Change) in order to make COCl2 with a partial pressure of 0.10 atm, giving a final (Equilibrium) pressure of 0.50 atm for carbon monoxide. Likewise, one mole of chlorine is required to make one mole of COCl2 making the Change for chlorine 0.10 atm and the Equilibrium partial pressure 1.00 atm. The completed table is shown below: Solutions to Example 10.3.1 ${\displaystyle P_{CO}}$ ${\displaystyle P_{Cl_{2}}}$ ${\displaystyle P_{COCl_{2}}}$ === Initial === 0.60 atm 1.10 atm 0 atm Change -0.10 atm -0.10 atm + 0.10 atm Equilibrium 0.50 atm 1.00 atm 0.10 atm The equilibrium expression for the phosgene-forming reaction is given by the following equation: \[\frac{P_{COCl_{2}}}{P_{CO}P_{Cl_{2}}}=K \nonumber$
Substituting the values from the Table into this equation:
$\frac{P_{COCl_{2}}}{P_{CO}P_{Cl_{2}}}=\frac{(0.10)}{(0.50)(1.00)}=0.20 \nonumber$
Notice that equilibrium constants for gas phase reactions are not typically written with units, although units are sometimes used in equilibrium constants calculated from molar concentrations. Many textbooks differentiate between equilibrium constants calculated from partial pressures and molar concentrations by affixing subscripts; KP and Kc. In this book, we will simply use K and Kc to represent the two; a value for K will always denote a constant calculated from partial pressure data.
Exercise $1$
1. For the reaction shown below, all four gasses are introduced into a vessel, each with an initial partial pressure of 0.500 atm, and allowed to come to equilibrium; at equilibrium, the partial pressure of SO3 is found to be 0.750 atm. Determine the value of K.
SO2 (g) + NO2 (g) ⇄ SO3 (g) + NO (g)
1. For the reaction shown above, the initial partial pressures of SO3 and NO are 0.500 atm under conditions where the equilibrium constant is, K = 9.00. The equilibrium partial pressure for SO2 is found to be 0.125 atm. Calculate the equilibrium partial pressure for SO3. | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/10%3A_Principles_of_Chemical_Equilibrium/10.3%3A_Calculating_Equilibrium_Values.txt |
As we have pointed out several times in the preceding sections, the Ideal Gas Laws (Chapter 10) tell us that the partial pressure of a gas and the molar concentration of that gas are directly proportional. We can show this simply by beginning with the combined gas law:
$P_{gas}V=nRT \nonumber$
If we divide both sides by the volume, V, and state that V must be expressed in liters, the right side of the equation now contains the term ${$/extract_itex]displaystyle \left({}^{n}\!\!\diagup \!\!{}_{V_{liters}}\;\right)} . Realizing that the number of moles of gas (n) divided by the volume in liters is equal to molarity, M, this expression can be re-written as: \[P_{gas}=MRT \nonumber$
Using this expression, molar concentrations can easily be substituted for partial pressures, and visa versa.
Exercise $1$
1. For the reaction shown below, if the molar concentrations of SO3, NO and SO2 are all 0.100 M, what is the equilibrium concentration of NO2?
2. For the reaction between carbon monoxide and chlorine to form phosgene, the equilibrium constant calculated from partial pressures is K = 0.20. How does this value relate to the equilibrium constant, KC, under the same conditions, calculated from molar concentrations?
CO (g) + Cl2 (g) ⇄ COCl2 (g)
10.5: Equilibria involving Acids and Bases
Consider a simple chemical system that is at equilibrium, such as dinitrogen tetroxide: nitrogen dioxide. The Law of Mass Action states that when this system reaches equilibrium, the ratio of the products and reactants (at a given temperature) will be defined by the equilibrium constant, K. Now imagine that, after equilibrium has been reached, more dinitrogen tetroxide is introduced into the container. In order for the ratio to remain constant (as defined by K) some of the $\ce{N2O4}$ that you added must be converted to NO2. The addition of reactants or products to a system at equilibrium is commonly referred to as a “stress”. The response of the system to this stress is dictated by Le Chatelier's Principle.
Le Chatelier's Principle
Le Chatelier's Principle states that, if a "stress" is applied to a chemical reaction at equilibrium, the system will readjust in the direction that best reduces the stress imposed on the system. Again, stress refers to a change in concentration, a change in pressure or a change in temperature, depending on the system being examined. If pressure or temperature are changed, the numeric value K will change; if only concentration changes are involved, K does not change.
We will consider temperature and pressure effects in General Chemistry, but for now, remember; in a reaction at equilibrium, the introduction of more products will shift the mass balance towards more reactants, but the ratio of Products/Reactants (as defined by the equilibrium expression) does not change, hence, K is unchanged.
In Chapter 8, we learned that a “weak acid” was only partially dissociated in solution, while a “strong acid” was fully dissociated. Now that we better understand the concept of equilibrium, these two classes of Brønsted acids can simply be differentiated based on their equilibrium constants. For an acid, $\ce{BH}$, that dissociates in water to form $\ce{B^{–}}$ and hydronium ion:
$\ce{BH(aq) + H2O(l) <=> B^{–}(aq) + H3O^{+}(aq)} \nonumber$
we can write a simple equilibrium expression, as follows:
$K_{C}=\frac{[H_{3}O^{+}][B^{-}]}{[BH]}=''K_{a}'' \nonumber$
You should note two things in this equation. Because the activity of water, as the solvent, is defined to have a value of 1, the activity for water does not affect the value of the equilibrium constant (remember, solids and liquids and solvents all have an activity of 1, and so do not affect the value of K) and the equilibrium constant for KC is written as Ka to denote that this is an acid dissociation equilibrium. Now, as we learned in Chapter 8, a strong acid is “fully dissociated”, which simply means that [BH] is very, very small, thus Ka for a strong acid is very, very large. A weak acid is only “partially dissociated” which means that there are significant concentrations of both BH and B in solution, thus Ka for a weak acid is “small”. For most common weak acids, the values for Ka will be in the range of 10-3 to 10-6.
Example $1$:
Consider acetic acid (the acidic component of vinegar) where Ka = 1.8 × 10-5.
$\ce{CH3COOH(aq) + H2O(l) <=> CH3COO^{–}(aq) + H3O^{+}(aq)} \nonumber$
Solution
$K_{a}=\frac{[H_{3}O^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]}=1.8\times 10^{-5} \nonumber$
Exercise $1$
1. A series of acids have the following Ka values: rank these in descending order from the strongest acid to the weakest acid.
A. 6.6 × 10–4 B. 4.6 × 10–4 C. 9.1 × 10–8 D. 3.0 × 102
2. At 25.0 oC, the concentrations of H3O+ and OH in pure water are both 1.00 × 10-7 M, making Kc = 1.00 × 10-14 (recall that this equilibrium constant is generally referred
to as Kw). At 60.0o C, Kw increases to 1.00 × 10-13. What is the pH of a sample of pure water at 60.0o C? | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/10%3A_Principles_of_Chemical_Equilibrium/10.4%3A_Using_Molarity_in_Equilibrium_Calculations.txt |
For a solution of a strong acid, calculating the [H3O+] concentration is simple; because the acid is 100% dissociated, the concentration of hydronium ions is equal to the molar concentration of the strong acid (this is, of course, only true for a monoprotic acid such as HCl or HNO3; for H2SO4, [H3O+] = 2 × [H2SO4], etc.). For a weak acid, however, the hydronium ion concentration will be much, much less than the molar concentration of the acid and [H3O+] must be calculated using the value of Ka. We can approach this using an ICE table, like we did for previous equilibrium problems. If we prepared a solution of acetic acid that was exactly 0.50 M, then initially [CH3COOH] is 0.50 M and both [CH3COO] and [H3O+] are zero. A small amount of CH3COOH will ionize; let’s call this x, making the change for [CH3COOH] “-x”, increasing both [CH3COO] and [H3O+] by the amount “+x”. Finally, the equilibrium concentration of [CH3COOH] will be (0.50 M – x) and both [CH3COO] and [H3O+] will be x. The completed table is shown below.
[CH3COOH] [CH3COO] [H3O]+
=== Initial === 0.50 M 0 0
Change
- x + x + x
Equilibrium
0.50 M - x x x
The expression for Ka for acetic acid is given in equation in section 10.5. Substituting for our equilibrium values:
$K_{a}=1.8\times 10^{-5}=\frac{[H_{3}O^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]}=\frac{x^{2}}{0.50-x} \nonumber$
$x^{2}+9.0\times 10^{-6}x-1.8\times 10^{-5}=0 \nonumber$
The above equation is a quadratic equation and we could solve it using the standard quadratic formula. This is not necessary, however, because acetic acid is a weak acid and by definition, very little of the dissociated form will exist in solution, making the quantity x very, very small. If x is much, much less than 0.50 M (our initial concentration of acetic acid), then (0.50 M – x) 0.50 M and the equation simplifies to:
$K_{a}=1.8\times 10^{-5}=\frac{[H_{3}O^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]}=\frac{x^{2}}{0.50} \nonumber$
$x=[H_{3}O^{+}]=\sqrt{(1.8\times 10^{-5})\times 0.50}=3.0\times 10^{-3}M \nonumber$
We can test our assumption by substituting for x; (0.50 – 0.0030) = 0.497, which rounds to 0.50 to two significant figures. Because the concentration of hydronium ion is very small for a weak acid, for most typical solutions, the concentration of hydronium ion can be estimated simply as:
$[H_{3}O^{+}]=\sqrt{(K_{a}\times C_{0}} \nonumber$
where C0 is the initial molar concentration of the weak acid.
Exercise $1$
1. Nitrous acid (HNO2) is a weak acid with a Ka of 4.3 × 10-4. Estimate the hydronium ion concentration and the pH for a 0.50 M solution of nitrous acid in distilled water.
2. Acetic acid is a weak acid with Ka = 1.8 × 10-5. For a solution of acetic acid in water, the [H3O+] is found to be 4.2 × 10-3 M. What is the concentration of unionized acetic acid in this solution?
$\ce{CH3COOH(aq) + H2O(l) <=> CH3COO^{–}(aq) + H3O^{+}(aq)} \nonumber$
1. A solution is prepared in which acetic acid is 0.700 M and its conjugate base, acetate anion is 0.600 M. As shown above, the Ka of acetic acid is 1.8 x 10-5; what will the pH of this solution be?
2. What concentration of the weak acid, acetic acid (Ka = 1.8 × 10-5) must you have in pure water in order for the final pH to be 2.38?
10.7: Solubility Equilibria
In Chapter 5 we learned about a class of reactions that involved the formation of a solid that was “insoluble” in water, and precipitated from the solution. In these “precipitation reactions”, one ionic salt was described as “insoluble”, driving the reaction towards the formation of products. Silver chloride is a classic example of this. If you mix silver nitrate (almost all nitrate salts are “soluble” in water) with sodium chloride, a copious white precipitate of silver chloride formed and the silver nitrate was deemed “insoluble”.
Nonetheless, if you took the clear solution from above the silver chloride precipitate and did a chemical analysis, there will be sodium ions, nitrate ions, and traces of chloride ions and silver ions. The concentrations of silver and chloride ions would be about 1.67 × 10-5 M, far below the concentrations we typically work with, hence we say that silver chloride is “insoluble in water”. That, of course, is not true. Solubility is an equilibrium in which ions leave the solid surface and go into solution at the same time that ions are re-deposited on the solid surface. For silver chloride, we could write the equilibrium expression as:
$\ce{AgCl(s) + H2O(l) <=> Ag^{+}(aq) + Cl^{-}(aq)} \nonumber$
In order to write the expression for the equilibrium constant for this solubility reaction, we need to recall the rules stated in Section 10.2 of this chapter; Rule #4 states, “Reactants or products that are present as solids or liquids or the solvent, all have an activity value of 1, and so they do not affect the value of the equilibrium expression.” Because silver chloride is a solid, and water is the solvent, the expression for the equilibrium constant is simply,
$K_{sp}=[Ag^{+}][Cl^{-}] \nonumber$
Note that we have denoted the equilibrium constant as Ksp, where “sp” refers to solubility equilibrium, or “solubility product” (the product of the concentrations of the ions). We can calculate the value of Ksp for silver chloride from the analytical data that we cited above; an aqueous solution above solid silver chloride has a concentration of silver and chloride ions of 1.67 × 10-5 M, at 25˚ C. Because the concentrations of silver and chloride ions are both 1.67 × 10-5 M, the value of Ksp under these conditions must be:
$K_{sp}=[Ag^{+}][Cl^{-}]=(1.67\times 10^{-5})^{2}=2.79\times 10^{-10} \nonumber$
This is very small, considering that Ksp for sodium chloride is about 29!
For a salt such as PbI2 chemical analysis tells us that the lead concentration in a saturated solution (the maximum equilibrium solubility under a specified set of conditions, such as temperature, pressure, etc.) is about 1.30 × 10-3 M. In order to calculate Ksp for lead (II) iodide, you must first write the chemical equation and then the equilibrium expression for Ksp and then simply substitute for the ionic concentrations.
$\ce{PbI2(s) <=> Pb^{2+}(aq) + 2 I^{-}(aq)} \nonumber$
As you do this, remember that there are two iodide ions for every lead ion, therefore the concentrations for lead (II) and iodide are 1.30 × 10-3 M and 2.60 × 10-3 M, respectively.
$K_{sp}=[Pb^{2+}][I^{-}]^{2}=(1.30\times 10^{-3})(2.60\times 10^{-3})^{2}=8.79\times 10^{-9} \nonumber$ | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/10%3A_Principles_of_Chemical_Equilibrium/10.6%3A_The_pH_of_Weak_Acid_Solutions.txt |
• If two opposing chemical reactions proceed simultaneously at the same rate, the processes are said to be in equilibrium. The two opposing reactions are shown linked with a double arrow (⇄). An example of opposing chemical reactions and their equilibrium expressions are:
$\ce{N2(g) + 3 H2(g) -> 2 NH3(g)}$
and
$\ce{2 NH3(g) -> N2(g) + 3 H2(g)}$
The equilibrium would be written as:
$\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)}$
• An equilibrium constant (K) is a numerical value that relates the concentrations of the products and reactants for a chemical reaction that is at equilibrium. The numeric value of an equilibrium constant is independent of the initial concentrations of reactants, but is dependent on the temperature. Equilibrium constants are generally not written with units (although they may be).
• Because equilibrium constants are written as the concentrations (or partial pressures) of products divided by the concentrations (or partial pressures) of reactants, a large value of K means that there are more products in the equilibrium mixture than there are products. Likewise, a small value of K means that, at equilibrium, there are more reactants than products.
• Equilibrium constants that are based on partial pressures are often written as KP, while equilibrium constants based on molar concentrations are written as Kc.
• An expression for an equilibrium constant can be written from a balanced chemical equation for the reaction. The Law of Mass Action states the following regarding equilibrium expressions:
• Partial pressures (or molar concentrations) of products are written in the numerator of the expression and the partial pressures (or concentrations) of the reactants are written in the denominator.
• If there is more that one reactant or more that one product, the partial pressures (or concentrations) are multiplied together.
• The partial pressure (or concentration) of each reactant or product is then raised to the power that numerically equals the stoichiometric coefficient appearing with that term in the balanced chemical equation.
• Reactants or products that are present as solids or liquids do not appear in the equilibrium expression.
• As an example of an equilibrium expression, consider the reaction of nitrogen and hydrogen to form ammonia. The partial pressure of ammonia will be in the numerator, and it will be squared. Because there are two reactants, the partial pressures for nitrogen and hydrogen will be multiplied in the denominator. The partial pressure of nitrogen will be raised to the “first power” (which is not shown) and the partial pressure of hydrogen will be cubed.
$\ce{N2 (g) + 3 H2 (g) <=> 2 NH3 (g)}$
$\frac{(P_{NH_{3}})^{2}}{P_{N_{2}}(P_{H_{2}})^{3}}=K \nonumber$
• If equilibrium values for a given reaction are known, the equilibrium constant can be calculated simply by substituting those values in the equilibrium expression. Quite often, however, initial and equilibrium values are only given for selected reactants and products. In these cases, initial and equilibrium values are arranged in an ICE Table, and the changes between initial and equilibrium states are calculated based on reaction stoichiometry.
• Because the partial pressure of a gas and the molar concentration of that gas are directly proportional, the ideal gas law can be rearranged as follows, to give an expression relating molarity and partial pressure of a gas.
$P_{gas}V=nRT \nonumber$
Dividing by the volume in liters gives the term ${$/extract_itex]displaystyle \left({}^{n}\!\!\diagup \!\!{}_{V_{liters}}\;\right)} which is equivalent to molarity, M. \[P_{gas}=MRT \nonumber$
• Le Chatelier's Principle states that, if a "stress" is applied to a chemical reaction at equilibrium, the system will readjust in the direction that best reduces the stress imposed on the system. In this context, stress refers to a change in concentration, a change in pressure or a change in temperature, although only concentration is considered here. If pressure or temperature are changed, the numeric value K will change; if only concentration changes are involved, K does not change. In a reaction at equilibrium, the introduction of more products will shift the mass balance towards more reactants, and the introduction of more reactants will lead to the formation of more products, but the ratio of Products/Reactants (as defined by the equilibrium expression) does not change, hence, K is unchanged.
• For a weak acid, dissociating in water to form it's conjugate base and hydronium ion, the equilibrium constant is referred to as Ka. Because a weak acid is only "partially dissociated", the concentration of BH in solution is large, thus Ka for a weak acid is "small" (in the range of 10-3 to 10-6). For example, acetic acid (the acidic component of vinegar), has an acid dissociation constant of Ka = 1.8 x 10-5.
CH3COOH(aq) + H2O(l) ⇄ CH3COO-(aq) + H3O+(aq)
• For a solution of a weak acid in water, the concentration of hydronium ion will be very small. If the concentration of the weak acid is fairly large (typically > 0.01 M) the concentration of the undissociated acid will be much larger than [H3O+]. Because of this, the hydronium ion concentration (hence, the pH) can be fairly accurately estimated from the Ka of the weak acid and the initial concentration of the acid (Co), by the equation:
$[H_{3}O^{+}]=\sqrt{K_{a}\times C_{0}} \nonumber$
• The equilibrium constant defining the solubility of an ionic compound with low solubility is defined as Ksp, where “sp” refers to “solubility product”. Because reactants or products that are present as solids or liquids do not appear in equilibrium expressions, for silver chloride, the expression Ksp will be written as:
AgCl(s) ⇄ Ag+(aq) + Cl-(aq)
$K_{sp}=[Ag^{+}][Cl^{-}] \nonumber$
• For silver chloride, the solubility at 25 oC is 1.67 × 10-5 M. That means the concentrations of silver and chloride ions in solution are each 1.67 × 10-5 M, making the value of Ksp under these conditions: $K_{sp}=[Ag^{+}][Cl^{-}]=(1.67\times 10^{-5})^{2}=2.79\times 10^{-10} \nonumber$ | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/10%3A_Principles_of_Chemical_Equilibrium/10.8%3A_Study_Points.txt |
In today’s society, the term radioactivity conjures up a variety of images. Nuclear power plants producing hydrocarbon-free energy, but with potentially deadly by-products that are difficult to store safely. Bombs that use nuclear reactions to produce devastating explosions with horrible side effects on the earth as we know it and on the surviving populations that would inhabit it. Medical technology that utilizes nuclear chemistry to peer inside living things to detect disease and the power to irradiate tissues to potentially cure these diseases. Fusion reactors that hold the promise of limitless energy with few toxic side products. Radioactivity has a colorful history and clearly presents a variety of social and scientific dilemmas. In this chapter we will introduce the basic concepts of radioactivity, nuclear equations and the processes involved in nuclear fission and nuclear fusion.
• 11.1: Radioactivity
Certain elements spontaneously produced a variety of particles. The three basic classes of particles were identified as “alpha”, “beta”, and “gamma” particles. Alpha particles were positive, relatively massive and were showed to be identical to the nucleus of the helium atom. Beta particles had a very small mass and were of higher energy and they carried a negative charge. Gamma particles were much more energetic, appeared to be neutral and were comparable to a high-energy photon of light.
• 11.2: The Nuclear Equation
To show radioactive decay in a chemical equation, you need to use atomic symbols. In the atomic symbol, the atomic number (the number of protons in the nucleus) appears as a subscript preceding the symbol for the element. The mass number appears as a superscript, also preceding the symbol.
• 11.3: Beta Particle Emission
• 11.4: Positron Emission
A positron, also called an antielectron, is an exotic bit of matter, or more correctly, an example of antimatter. A positron is the antimatter equivalent of an electron. It has the mass of an electron, but it has a charge of +1. Positrons are formed when a proton sheds its positive charge and becomes a neutron.
• 11.5: Radioactive Half-Life
• 11.6: Nuclear Fission
• 11.7: Nuclear Fusion
• 11.S: Nuclear Chemistry (Summary)
11: Nuclear Chemistry
The actual discovery of radioactivity is generally attributed to the French scientist, Henri Becquerel in 1896. As with most discoveries, he was working on something else. In this case it was the nature of phosphorescence; the property of some substances to “glow in the dark” after being exposed to light. In the course of his work, he allowed photographic plates to come in contact with uranium salts, only to find out that the uranium had “fogged” the unexposed plates. Further work by Becquerel and others (including Marie Curie) led to the realization that certain elements spontaneously produced a variety of particles, some of which were charged (both positive and negative) and one class that was of higher energy, but appeared to be neutral. The three basic classes of particles were characterized and identified as “alpha”, “beta”, and “gamma” particles. Alpha particles were positive, relatively massive and, subsequent work showed that they were identical to the nucleus of the helium atom, containing two protons and two neutrons. Beta particles had a very small mass. They were of higher energy and they carried a negative charge; equivalent in mass and charge to an electron. Gamma particles (actually referred to as gamma rays) were much more energetic, appeared to be neutral and were comparable to a high-energy photon of light. Although it was not apparent immediately, one of the most surprising observations regarding radioactive elements was that as they emitted particles, the identity of the element slowly changed; uranium, for example, slowly became enriched with lead.
When alpha, beta or gamma particles collides with a target, some of the energy in the particle is transferred to the target, typically resulting in the promotion of an electron to an “excited state”. In many “targets”, especially gasses, this results in ionization, and alpha, beta and gamma radiation is broadly referred to as ionizing radiation. A Geiger counter (or Geiger-Müller counter) takes advantage of this in order to detect these particles. In a Geiger tube, the electron produced by ionization of a captive gas travels to the anode and the change in voltage is detected by the attached circuitry. Most counters of this type are designed to emit an audible “click” in response to the change in voltage, and to also show it on a digital or analog meter.
Today, we recognize that radioactive decay is actually quite complex, but the basic principles and patterns that were established over 100 years ago still stand. The three basic subatomic particles that occur in radioactive decay are the alpha particle, the beta particle and the gamma ray. The gamma ray is of highest energy (and perhaps the greatest ultimate danger), but from a chemistry standpoint, the alpha and beta particles are of the greatest interest. An alpha particle consists of two protons and two neutrons. It has a mass of four amu and a charge of +2. It is identical with the helium nucleus, and when a radioactive element emits an alpha particle, it loses four amu from its nucleus, including two protons. Because the number of protons in a nucleus define the identity of the element, the atomic number of the element decreases by two when it loses an alpha particle; thus uranium ( \(\ce{_{92}^{238}U}\)) loses an alpha particle and becomes an atom of thorium ( \(\ce{_{90}^{234}Th}\)); we will discuss this process further in the following section. In order for a beta particle (an electron) to emerge from the nucleus, it must be formed by the decomposition of a neutron (on a very simple scale, think of a neutron as being composed of a positive proton bound to a negative electron). When a neutron decays and emits a beta particle, it leaves behind the newly formed proton. Again, this changes the identity of the element in question. | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/11%3A_Nuclear_Chemistry/11.1%3A_Radioactivity.txt |
In Chapter 1, we described the meaning of the atomic symbol for an element. In the atomic symbol, the atomic number (the number of protons in the nucleus) appears as a subscript preceding the symbol for the element. The mass number appears as a superscript, also preceding the symbol. Thus for uranium (atomic number 92) with a mass of 238, the symbol is $\ce{_{92}^{238}U}$. To show radioactive decay in a chemical equation, you need to use atomic symbols. Thus, for the loss of an alpha particle from ${\displaystyle {}_{92}^{238}U}$$\ce{ _{92}^{238}U}$, you need to show uranium on the “reactant” side of the equation and thorium and the alpha particle on the “product” side. Just like any other chemical equation, a nuclear equation must balance. The sum of the mass numbers on the reactant side must equal the sum of the mass numbers on the product side. Because we started with uranium-238 and lost four mass units in the alpha particle, the product (or products) of the decay must have a total mass of (238 – 4) = 234. We have also removed two protons from the uranium nucleus, dropping the atomic number by two. The newly formed element is therefore thorium-234.
$_{92}^{238}\textrm{U}\rightarrow _{2}^{4}\textrm{He}+_{90}^{234}\textrm{Th} \nonumber$
In this equation, we have shown the alpha particle using the atomic symbol for helium ($\ce{_2^4 He}$), but this is often shown using the symbol $\ce{_{2}^{4}\alpha}$. Compounds that emit alpha particles are very toxic, in spite of the poor penetrating ability of the particle. This is especially true if the emitting element is inhaled or ingested. The toxic dose of the alpha-emitter $\ce{^{210}Po}$ in a 175-pound person has been estimated to be about one microgram (1 × 10-6 g).
Exercise $1$
Thorium-230 and polonium-210 both undergo loss of an alpha particle to form different elements. For each of these radioactive decay processes, write the appropriate nuclear equation and show the nature of the elements that are formed.
Exercise $2$
Radium-226 and polonium-214 both undergo loss of an alpha particle to form different elements. For each of these radioactive decay processes, write the appropriate nuclear equation and show the nature of the elements that are formed.
11.3: Beta Particle Emission
In an element with an “excess” of neutrons, one of these neutrons can break down to form an electron and a proton. In this process, an antinutrino is also produced, but because it has no mass, it is generally ignored in this process. The nuclear equation for the decomposition of a neutron can be written:
$\ce{_{0}^{1}n -> _{-1}^{0}\beta +_{1}^{1}p} \nonumber$
where the neutron has the symbol, $\ce{^1_{0}n}$, the proton has the symbol, $\ce{^1_{1}p}$, and the electron that is produced is called a beta particle, with the symbol $\ce{^0_{-1}\beta}$. Because the nuclear equation must balance for mass and atomic numbers, the “atomic number” of the beta particle must be –1. Adding the atomic numbers on the right side of the equation shown above gives {(-1) + (+1) = 0}; identical to the “atomic number” in the neutron ($\ce{^1_{0}n}$); (even though a neutron can break down to produce a proton, there are no actual protons in a neutron, hence its atomic number is zero). Likewise, the “mass number” of the beta particle must be zero because the proton (the product) and the neutron (the reactant) each have a mass of one. Therefore, when a nucleus loses a beta particle, the number of neutrons in the nucleus decreases by one, but the mass number does not change; the neutron is converted into a proton, also having a mass number of one. Because the neutron is converted into a proton, the atomic number of the element increases by one unit, changing the identity of the element to the next highest in the periodic table. For example, thorium-234 undergoes loss of a beta particle to form protactinium-234 by the equation shown below:
$_{90}^{234}\textrm{Th}\rightarrow +_{-1}^{0}\beta +_{91}^{234}Pa \nonumber$
Again, with a beta-particle emission, the mass number does not change, but the atomic number increases by one unit.
Exercise $1$: Beta-Particle Emission
Bismuth-210 and lead-214 both undergo loss of a beta particle to form different elements. For each of these radioactive decay processes, write the appropriate nuclear equation and show the nature of the elements that are formed.
Exercise $2$: Beta-Particle Emission
Chlorine-39 and strontium-90 both undergo loss of a beta particle to form different elements. For each of these radioactive decay processes, write the appropriate nuclear equation and show the nature of the elements that are formed. | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/11%3A_Nuclear_Chemistry/11.2%3A_The_Nuclear_Equation.txt |
A positron, also called an antielectron, is an exotic bit of matter, or more correctly, an example of antimatter. A positron is the antimatter equivalent of an electron. It has the mass of an electron, but it has a charge of +1. Positrons are formed when a proton sheds its positive charge and becomes a neutron, as shown below:
$_{1}^{1}\rho \rightarrow +_{+1}^{0}\beta +_{0}^{1}n \nonumber$
Again, in the nuclear equation for positron emission, the sum of protons (atomic numbers) on the right equals the number of protons on the left and the masses all equal one. When an element emits a positron, the identity of the element changes to the one having one fewer protons on the periodic table. An example of a nuclear equation showing positron emission is shown below:
$_{6}^{11}C \rightarrow +_{+1}^{0}\beta +_{5}^{11}B \nonumber$
Boron has one fewer protons in its nucleus than carbon, but the mass is unchanged because the proton has been replaced by a neutron.
$_{9}^{18}F \rightarrow +_{+1}^{0}\beta +_{8}^{18}O \nonumber$
Positron emission from Fluorine-18, as shown above , has become an important medical diagnostic tool; Positron Emission Tomography (a PET scan). The heart of this technique is based on the fact that positrons undergo instant annihilation when they collide with an electron (an example of matter-antimatter annihilation). When this occurs, two high-energy gamma rays are produced and exit the scene of the annihilation in exactly opposite directions. During a PET scan, a patient is given an injection containing fluorodeoxyglucose (FDG), a sugar analog. The glucose analog is absorbed by metabolically active cells, where the FDG accumulates and undergoes positron decay. After a short waiting period, the patient is scanned using a circular array of gamma-radiation detectors. The fact that the gamma rays are emitted in opposite directions allows the attached computer to “draw a line” through the patient, where the line passes through the point of annihilation. Because this occurs through many directions, the exact location of the emission can be accurately calculated and then imaged as a three-dimensional picture showing the intensity of the emission.
11.5: Radioactive Half-Life
Elements such as ${\displaystyle {}_{92}^{238}U}$ that emit radioactive particles do so at rates that are constant and unique for each element. The rate at which an radioactive element decays is measured by its half-life; the time it takes for one half of the radioactive atoms to decay, emitting a particle and forming a new element. Half-lives for elements vary widely, from billions of years to a few microseconds. On a simple, intuitive level, if you begin with 1.00 gram of a radioactive element, after one half-life there will be 0.500 grams remaining; after two half-lives, half of this has decayed, leaving 0.250 grams of the original element; after three half-lives, 0.125 grams would remain, etc. For those that prefer equations, the amount remaining after n half-lives can be calculated as follows:
$R=I\left ( \frac{1}{2} \right )^{n} \nonumber$
where I represents the initial mass of the element and R represents the mass remaining.
Example $1$:
The half-life of Actinium-225 is 10.0 days. If you have a 1.00 gram sample of Actinium-225, how much is remaining after 60.0 days?
Solution
The number of half-lives is 6.00 (that is n) and I = 1.00 gram. Substituting:
$R=(1.00\; gram)\left ( \frac{1}{2} \right )^{6.00}=(1.00\; gram)(0.0156)=0.0156\; gram \nonumber$
Exercise $1$
The half-life of Antimony-124 is 60.20 days. If you have a 5.00 gram sample of Actinium, how much is remaining after 5.0 half-lives?
One of the interesting uses for half-life calculations involves radiocarbon dating, where the content of carbon-14 in organic (formally living matter) is used to calculate the age of a sample. The process begins in the upper atmosphere, where nitrogen is bombarded constantly by high-energy neutrons from the sun. Occasionally, one of these neutrons collides with a nitrogen nucleus and the isotope that is formed undergoes the following nuclear equation:
$_{0}^{1}n+_{7}^{14}N \rightarrow _{1}^{1}\rho +_{6}^{14}C \nonumber$
Plants take up atmospheric carbon dioxide by photosynthesis, and are ingested by animals, so every living thing is constantly exchanging carbon-14 with its environment as long as it lives. Once it dies, however, this exchange stops, and the amount of carbon-14 gradually decreases through radioactive decay with a half-life of about 5,730 years, following the nuclear equation shown below:
$_{6}^{14}C \rightarrow _{-1}^{0}\beta +_{7}^{14}N \nonumber$
Thus, by measuring the carbon-14/carbon-12 ratio in a sample and comparing it to the ratio observed in living things, the number of half-lives that have passed since new carbon-14 was absorbed by the object can be calculated. | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/11%3A_Nuclear_Chemistry/11.4%3A_Positron_Emission.txt |
The process by which nitrogen is converted to carbon-14 is an example of neutron capture, in which particles are absorbed by the nucleus of another atom to form a new element. These types of reactions are actually quite common in nuclear chemistry. A uranium-235 nucleus captures a “slow-moving” neutron, just like nitrogen captures a neutron, leading to the formation of carbon-14. Initially, uranium-236 is formed, but this nucleus has a neutron-to-proton ratio that makes it exceptionally unstable. The unstable nucleus instantaneously breaks apart (undergoes fission) to form lighter elements and to release additional free neutrons. As the nucleus breaks apart, a significant amount of energy is also released. A nuclear equation showing a typical fission of uranium-235 is shown below:
$_{92}^{235}U+_{0}^{1}n \rightarrow _{56}^{141}Ba +_{36}^{92}Kr+3_{0}^{1}n \nonumber$
The three neutrons that are released are now speeding through the mass of uranium. If these are captured by another nucleus, the process happens again and three more neutrons are released. This represents a chain reaction, and in order to sustain a chain reaction like this, the mass or uranium must be large enough so that the probability of every released neutron being captured by another uranium is high. The mass of uranium (or other fissile element) that is required in order to sustain a chain reaction is called the critical mass.
The process of nuclear fission is best known within the context of fission bombs and as the process that operates within nuclear power plants. Designing a workable fission bomb presents many technical challenges. A mass of fissile material that exceeds the critical mass is unstable, so you must begin with a smaller, non-critical mass and somehow create one within a few microseconds. In the original design, this was accomplished by taking two non-critical pieces and forcing them together (very rapidly). This is typically referred to as a “gun assembly”, in which one piece of fissile uranium is fired at a fissile uranium target at the end of the weapon, similar to firing a bullet down a gun barrel.
Each of the uranium fragments are less than a critical mass, but when they collide, they form a mass capable of sustaining the nuclear chain reaction. The assembly stays together for a few microseconds before the energy released from the fission blows it to pieces. The trick is designing nuclear devices like this is to keep them together long enough so that enough energy is released. Neutron reflectors and “boosters” are generally used to accomplish this, nonetheless, this basic type of weapon is inefficient, although easy to design and incredibly deadly. A critical mass of uranium-235 is a sphere that is slightly less than 7 inches in diameter.
A much more efficient fission bomb is based on achieving a critical mass of fissile material, not by combining smaller fragments, but by increasing the density of a sub-critical mass to the point that the rate of neutron capture sustains the chain reaction. This design is called the “implosion” bomb and it basically consists of a sphere of fissile material surrounded by shaped explosives that must be detonated simultaneously. The resulting shock wave compresses the fissile material, allowing the chain reaction to occur. This type of design requires much less fissile material, but is technically challenging. Modern devices have neutron reflectors, “neutron initiators”, etc., and sophisticated bombs can be efficient, have a high yield and a relatively small physical size.
In a nuclear reactor designed to heat water, produce steam and electrical power, the chemistry is the same, but control rods are introduced between the pieces of fissile material to absorb some of the neutrons that are produced so that a critical mass is never achieved and the chain reaction can be controlled. In this set-up, as the control rods are withdrawn, the chain reaction speeds up, and as they are inserted, the reaction slows down. Even under “meltdown” conditions, where control rods fail, the critical mass of fissile material would be formed slowly. The resulting explosion would be a bad thing, but would not compare with the energy released from a well-designed fission weapon. | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/11%3A_Nuclear_Chemistry/11.6%3A_Nuclear_Fission.txt |
As we saw in the preceding section, when the nuclei of heavy atoms split, energy is released. For light atoms, the opposite is true; when these nuclei combine (fuse together), energy is released. This is the process of nuclear fusion. Fusion of light elements, mostly hydrogen, is the force that powers energy release in the sun and in sun-like stars. Imagine the sun as a huge sphere of hydrogen. Because a star is so massive, the gravitation pull on the hydrogen atoms is sufficient to overcome the repulsion between the two nuclei to force them together to form an unstable ${\displaystyle {}_{2}^{2}{\text{He}}}$ nucleus. This immediately ejects a positron, leaving deuterium, ${\displaystyle {}_{1}^{2}{\text{H}}}$, and releasing a significant amount of energy. In the cascade of reactions deuterium fuses with another hydrogen to give ${\displaystyle {}_{2}^{3}{\text{He}}}$, and two of these combine to form helium, ejecting two high-energy protons in the process.
In stars that are larger and heavier than our sun, the “triple alpha process” is the dominant nuclear reaction. In this, helium nuclei fuse to eventually form carbon, releasing significant energy in the process.
One of the great challenges in physics and engineering today is to replicate fusion of this sort under controlled conditions, harvesting the energy released and converting it, indirectly, into electrical power. The extremely high temperatures and pressures that are required to initiate and sustain fusion reactions thwarted, thus far, attempts to build a fusion reactor that is “break even” in terms of the energy released relative to the energy required to produce the fusion events. Uncontrolled fusion is certainly possible, and fusion bombs exist, but these typically use an advanced fission bomb to create the temperatures and pressures necessary to promote the fusion of the lighter elements. Clearly, this approach does not work in the laboratory! Work on fusion reactors continues at a fast pace and includes novel approaches such as aneutronic fusion reactions that utilize proton-boron fusion to produce charged particles rather than a barrage of neutrons. The advantage here is that few neutrons are produced, reducing the need for shielding, and the charged particles formed can potentially be captured directly as electricity.
11.S: Nuclear Chemistry (Summary)
• In most atoms, a nucleus containing an “excess” of neutrons (more neutrons than protons) is unstable and the nucleus will decompose by radioactive decay, in which particles are emitted until a stable nucleus is achieved. Common particles emitted during radioactive decay include:
• Alpha particles, consist of two protons and two neutrons. This is equivalent to a helium nucleus and an alpha particle has a charge of 2+. Because it is positive, it will be attracted towards a negative charge in an electric field. The atomic symbol for an alpha particle is ${\displaystyle {}_{2}^{4}He}$, or sometimes ${\displaystyle {}_{2}^{4}\alpha }$. Alpha particles are slow-moving and are easily absorbed by air or a thin sheet of paper. When an element ejects an alpha particle, the identity of the element changes to the element with an atomic number that is two less than the original element. The mass number of the element decreases by four units.
• Beta particles are electrons, are considered to have negligible mass and have a single negative charge. They will be attracted towards a positive charge in an electric field. The atomic symbol for a beta particle is ${\displaystyle {}_{-1}^{0}\beta }$, or sometimes ${\displaystyle {}_{-1}^{0}e}$. Beta particles have “intermediate” energy and typically require thin sheets of metal for shielding. A beta particle is formed in the nucleus when a neutron “ejects” its negative charge (the beta particle) leaving a proton behind. When an element ejects a beta particle, the identity of the element changes to the next higher atomic number, but the mass number does not change.
• Gamma particles (gamma rays) are high-energy photons. They have no mass and can be quite energetic, requiring thick shielding.
• Positrons are anti-electrons, are considered to have negligible mass and have a single positive charge. They will be attracted towards a negative charge in an electric field. The atomic symbol for a positron is symbol ${\displaystyle {}_{+1}^{0}\beta }$. Positrons have “intermediate” energy and typically require thin sheets of metal for shielding. A positron is formed in the nucleus when a proton “ejects” its positive charge (the positron) leaving a neutron behind. When an element ejects a positron, the identity of the element changes to the next lower atomic number, but the mass number does not change.
• In a nuclear equation, elements and sub-atomic particles are shown linked by a reaction arrow. When you balance a nuclear equation, the sums of the mass numbers and the atomic numbers on each side must be the same.
• Radioactive elements decay at rates that are constant and unique for each element. The rate at which an radioactive element decays is measured by its half-life; the time it takes for one half of the radioactive atoms to decay, emitting a particle and forming a new element. The amount of an original element remaining after n half-lives can be calculated using the equation: $R=I\left ( \frac{1}{2} \right )^{n} \nonumber$ where I represents the initial mass of the element and R represents the mass remaining.
• In nuclear fission, a nucleus captures a neutron to form an unstable intermediate nucleus, which then splits (undergoes fission) to give nuclei corresponding to lighter elements. Typically, neutrons are also ejected in the process. For heavy isotopes, the process of fission also releases a significant amount of energy. A nuclear equation for a classical fission reaction is shown below: $_{92}^{235}U+_{0}^{1}n\rightarrow _{56}^{141}Ba+_{36}^{92}Kr+3_{0}^{1}n \nonumber$
• In nuclear fusion, nuclei combine to form a new element. For light isotopes, the process of fusion also releases a significant amount of energy. A nuclear equation for the fusion cascade that typically occurs in stars the size of our sun is shown below:
$2_{1}^{1}\rho \rightarrow _{+1}^{0}\beta +_{2}^{3}He \nonumber$
$2_{2}^{3}He \rightarrow 2_{1}^{1}\rho +_{2}^{4}He \nonumber$ | textbooks/chem/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/11%3A_Nuclear_Chemistry/11.7%3A_Nuclear_Fusion.txt |
• 1.1: Measurements Matter
Measurements in chemistry are reported with a variety of units. The standard units for various measurements are discussed along with the modifying prefixes for these units.
• 1.2: Significant Figures
Significant figures are important in reporting values because the numbers used in chemistry are based on measurements. The precision of a measurement should not be under- or over-reported through the use of the wrong number of significant figures.
• 1.3: Scientific Dimensional Analysis
Dimensional analysis (also called factor label method or unit analysis) is used to convert from one set of units to another. This method uses relationships or conversion factors between different sets of units. While the terms are frequently used interchangeably, conversion factors and relationships are different.
• 1.4: Percentages
Percentages are used in a variety of ways in healthcare including concentrations of solutions, medication dosages, and reporting changes to measured values for patients over time.
• 1.5: Measurements and Problem-Solving (Exercises)
These are homework exercises to accompany Chapter 1 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Solutions are available below the questions.
01: Measurements and Problem-Solving
Learning Outcomes
• Demonstrate use of scientific notation.
• Identify base and derived units for measurements.
• Memorize the metric prefixes, abbreviations, and values (Tables 1.1.1 and 1.1.3).
• Convert between temperatures in Celsius and Fahrenheit.
Many aspects of chemistry use quantitative measurements to describe behavior. In this section, we will look at how we deal with very large or very small numbers, units of measure and prefixes used with these units, and how numbers are reported.
Scientific Notation
Scientific notation is a way to express numbers as the product of two numbers: a coefficient and the number 10 raised to a power. It is a very useful tool for working with numbers that are either very large or very small. As an example, the distance from Earth to the Sun is about 150,000,000,000 meters - a very large distance indeed. In scientific notation, the distance is written as $1.5 \times 10^{11} \: \text{m}$. The coefficient is the 1.5 and must be a number greater than or equal to 1 and less than 10. The power of 10, or exponent, is 11 because you would have to multiply 1.5 by $10^{11}$ to get the correct number. Scientific notation is sometimes referred to as exponential notation.
When working with small numbers, we use a negative exponent. So 0.1 meters is $1 \times 10^{-1}$ meters, 0.01 is $1 \times 10^{-2}$ and so forth. A red blood cell has a diameter of 0.000008 meters which is neither convenient to write nor say. It is much easier to report a diameter of $8 \times 10^{-6} \: \text{m}$. Using scientific notation is one way to make writing very large or very small numbers more convenient. Note the use of a leading zero (the zero to the left of the decimal point) when writing very small numbers. The leading zero is there to help you see the decimal point more clearly. The figure 0.01 is less likely to be misunderstood that .01 where you may not see the decimal point.
Units
SI Base Units
All measurements depend on the use of units that are all 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, five of which are shown in the table below. While some of the base units have a calculation as their standard, the kilogram has a physical standard but there is a movement to change that. Read or listen to more about changes to the standard at http://www.npr.org/templates/story/s...ryId=112003322.
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 $\text{mol}$
These five units are frequently encountered in chemistry and other units of measurement, such as volume, force, and energy, can be derived from the SI base units.
The map in the figure 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.
Derived Units
Some units are combinations of SI base units. A derived unit is a unit that results from a mathematical combination of SI base units. These derived units are the standard for these quantities but they are not always the units used in normal practice. For example, the density of a solid or a liquid is often reported as $\text{g/cm}^3$, or $\text{g/mL}$ rather than $\text{kg/m}^3$ (the derived SI unit). Some common examples of derived units are listed in the table below.
Quantity Symbol Unit Unit Abbreviation Derivation
Table $2$: Derived SI Units
Area $A$ square meter $\text{m}^2$ $\text{length} \times \text{width}$
Volume $V$ cubic meter $\text{m}^3$ $\text{length} \times \text{width} \times \text{height}$
Density $D$ kilograms/cubic meter $\text{kg/m}^3$ $\frac{\text{mass}}{\text{volume}}$
Concentration $c$ moles/liter $\text{mol/L}$ $\frac{\text{amount}}{\text{volume}}$
Speed (velocity) $v$ meters/second $\text{m/s}$ $\frac{\text{length}}{\text{time}}$
Acceleration $a$ meters/second/second $\text{m/s}^2$ $\frac{\text{speed}}{\text{time}}$
Force $F$ newton $\text{N}$ $\text{mass} \times \text{acceleration}$
Energy $E$ joule $\text{J}$ $\text{force} \times \text{length}$
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 $\mu \epsilon \gamma \alpha \varsigma$, meaning "great".
The table below lists the most common metric prefixes and their relationship to the base unit that has no prefix. Length is used as an example to demonstrate the relative size of each prefixed unit. However, these prefixes can be used with any base unit.
Prefix Unit Abbreviation Meaning Example
Table $3$: SI Prefixes
giga $\text{G}$ 1,000,000,000 1 gigameter $\left( \text{Gm} \right) = 10^9 \: \text{m}$
mega $\text{M}$ 1,000,000 1 megameter $\left( \text{Mm} \right) = 10^6 \: \text{m}$
kilo $\text{k}$ 1,000 1 kilometer $\left( \text{km} \right) = 1,000 \: \text{m}$
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 $\left( \text{nm} \right) = 10^{-9} \: \text{m}$
*Micro is often abbreviated "mc" in healthcare to avoid confusion between the $\mu$ and m which often look similar when written by hand.
There are more prefixes, some of them rarely used, that go far beyond what is given here. Have you ever heard of a zeptometer? You can learn more about prefixes at http://www.bipm.org/en/measurement-units/
Since the metric system is not the usual unit system we use, it can be challenging to understand the relative size of each unit. Go to http://learn.genetics.utah.edu/content/cells/scale and use the slider below the image to zoom in and see how the sizes of objects compare.
There are a couple of odd little practices with the use of unit abbreviations. Most abbreviations are lower-case. We use "$\text{m}$" for meter and not "$\text{M}$". However, when it comes to volume, the base unit "liter" is abbreviated as "$\text{L}$" and not "$\text{l}$". So 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 $6500 \: \text{g}$. Both are correct, bu the $\text{kg}$ units in this case make for a small and easily managed number.
Temperature Scales
There are three temperature scales that are commonly used in measurement. Their units are $^\text{o} \text{F}$ (degrees Fahrenheit), $^\text{o} \text{C}$ (degrees Celsius), and $\text{K}$ (Kelvin). The Fahrenheit scale, which is the most commonly used scale in the United States, defines the normal freezing point and boiling point of water as $32^\text{o} \text{F}$ and $212^\text{o} \text{F}$, respectively. The Celsius scale defines the normal freezing point and boiling point of water as $0^\text{o} \text{C}$ and $100^\text{o} \text{C}$, respectively. The Celsius scale is commonly used in most countries across the globe. The Kelvin scale, which is also referred to as the absolute temperature scale, defines absolute zero as the lowest theoretically possible temperature, which means that temperatures expressed in Kelvin cannot be negative numbers.
Converting Temperature Scales
Regardless of the temperature scale used, it is important to be able to convert from one scale to another. Here are the conversions we use.
$^\text{o} \text{F}$ to $^\text{o} \text{C}$
$T_{^\text{o} \text{C}} = \left( T_{^\text{o} \text{F}} - 32 \right) \times \frac{5}{9}$
$^\text{o} \text{C}$ to $^\text{o} \text{F}$
$T_{^\text{o} \text{F}} = \frac{9}{5} \times \left( T_{^\text{o} \text{C}} \right) + 32$
$^\text{o} \text{C}$ to $\text{K}$
$T_\text{K} = T_{^\text{o} \text{C}} + 273.15$
$\text{K}$ to $^\text{o} \text{C}$
$T_{^\text{o} \text{C}} = T_\text{K} - 273.15$
Example $1$
The melting point of mercury is $-38.84^\text{o} \text{C}$. Convert this value to degrees Fahrenheit and Kelvin.
Solution
The formulas above can be used to convert among temperature units. First, the given value can be used to convert from $^\text{o} \text{C}$ to $^\text{o} \text{F}$.
\begin{align*} T_{^\text{o} \text{F}} &= \frac{9}{5} \times \left( -38.84^\text{o} \text{C} \right) + 32 & \ T_{^\text{o} \text{F}} &= -37.91^\text{o} \text{F} & \end{align*}
Then, the same initial temperature can be used to find the temperature in Kelvin.
\begin{align*} T_\text{K} &= -38.84^\text{o} \text{C} + 273.15 & \ T_\text{K} &= 234.75 \: \text{K} & \end{align*}
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/01%3A_Measurements_and_Problem-Solving/1.01%3A_Measurements_Matter.txt |
Learning Outcomes
• Distinguish between accuracy and precision.
• Explain the concept of significant figures.
• Report answers to calculations with the correct number of significant figures.
In chemistry, we are looking at measurements and not just numbers. Therefore, it is necessary to be able to count the number of significant figures and report measurements to the correct level of precision. The reported value should be as precise as possible without adding digits to a value that cannot be measured.
Accuracy and Precision
In everyday speech, the terms accuracy and precision are frequently used interchangeably. However, their scientific meanings are quite different. Accuracy is a measure of how close a measurement is to the correct or accepted value of the quantity being measured. Precision is a measure of how close a series of measurements are to one another. Precise measurements are highly reproducible, even if the measurements are not near the correct value. Darts thrown at a dartboard are helpful in illustrating accuracy and precision (see figure below).
Assume that three darts are thrown at the dartboard, with the bulls-eye representing the true, or accepted, value of what is being measured. A dart that hits the bulls-eye is highly accurate, whereas a dart that lands far away from the bulls-eye displays poor accuracy. Pictured above are the four possible outcomes.
(A) The darts have landed far from each other and far from the bulls-eye. This grouping demonstrates measurements that are neither accurate nor precise.
(B) The darts are close to one another, but far from the bulls-eye. This grouping demonstrates measurements that are precise, but not accurate. In a laboratory situation, high precision with low accuracy often results from a systematic error. Either the measurer makes the same mistake repeatedly or the measuring tool is somehow flawed. A poorly calibrated balance may give the same mass reading every time, but it will be far from the true mass of the object.
(C) The darts are not grouped very near to each other, but they are generally centered around the bulls-eye. This demonstrates poor precision but fairly high accuracy. This situation is not desirable because in a lab situation, we do not know where the "bulls-eye" actually is. Continuing with this analogy, measurements are taken in order to find the bulls-eye. If we could only see the locations of the darts and not the bulls-eye, the large spread would make it difficult to be confident about where the exact center was, even if we knew that the darts were thrown accurately (which would correspond to having equipment that is calibrated and operated correctly).
(D) The darts are grouped together and have hit the bulls-eye. This demonstrates high precision and high accuracy. Scientists always strive to maximize both in their measurements. Turning back to our laboratory situation, where we can see the darts but not the bulls-eye, we have a much narrower range of possibilities for the exact center than in the less precise situation depicted in part C.
Significant Figures in Measurements
Some error or uncertainty always exists in any measurement. The amount of uncertainty depends both upon the skill of the measurer and upon the quality of the measuring tool. While some balances are capable of measuring masses only to the nearest $0.1 \: \text{g}$, other highly sensitive balances are capable of measuring to the nearest $0.001 \: \text{g}$ or even better. Many measuring tools such as rulers and graduated cylinders have small lines which need to be carefully read in order to make a measurement. Pictured below is an object (indicated by the blue arrow) whose length is being measured by two different rulers.
With either ruler, it is clear that the length of the object is between 2 and 3 centimeters. The bottom ruler contains no millimeter markings, so the tenths digit can only be estimated, and the length may be reported by one observer as $2.5 \: \text{cm}$. However, another person may judge that the measurement is $2.4 \: \text{cm}$ or perhaps $2.6 \: \text{cm}$. While the 2 is known for certain, the value of the tenths digit is uncertain.
The top ruler contains marks for tenths of a centimeter (millimeters). Now, the same object may be measured as $2.55 \: \text{cm}$. The measurer is capable of estimating the hundredths digit because he can be certain that the tenths digit is a 5.
Again, another measurer may report the length to be $2.54 \: \text{cm}$ or $2.66 \: \text{cm}$. In this case, there are two certain digits (the 2 and the 5), with the hundredths digit being uncertain. Clearly, the top ruler is a superior ruler for measuring lengths as precisely as possible.
The significant figures in a measurement consist of all the certain digits in that measurement plus one uncertain (estimated) digit. In a correctly reported measurement, the final digit is significant but not certain. Insignificant digits are not reported.
Rules for Determination of Significant Figures
1. All nonzero digits are significant.
• 237 has three significant figures
• 1.897 has four significant figures
2. Zeroes between significant figures are significant.
• 39004 has five significant figures
• 2.03 has three significant figures
3. Zeroes that appear to the left of all of the nonzero digits are called placeholder zeroes and are not significant.
• 0.008 has one significant figure
• 0.0000416 has three significant figures
4. Zeroes that appear after the last nonzero digit may be significant.
a. If the number is greater than 1 and the zero(es) are not followed by a decimal point, then the number of significant figures is ambiguous.
• 140 is ambiguous
b. If the zero is followed by a decimal point, then the zero is significant.
• 140. has three significant figures (note the decimal point after the zero)
c. If the zero is after a decimal point, then the zero is significant.
• 141.0 has four 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.
Calculations
Adding and Subtracting Significant Figures
The sum or difference is determined by the smallest number of digits to the right of the decimal point in any of the original numbers.
Example $1$
What is the result of $89.332 + 1.1$ when answered to the correct number of significant figures?
Solution
The mathematical result is $89.332 + 1.1 = 90.432$. However, we need to round the answer to the correct number of significant figures. For addition and subtraction, the answer should have the same number of decimal places as the starting value with the least number of decimal places. Since 1.1 only has one digit after the decimal place, the answer will be rounded to one decimal place. The first digit to be dropped is less than five so we round down.
The correct answer is 90.4.
Multiplying and Dividing Significant Figures
The number of significant figures in the final product or quotient is equal to the number of significant figures in the starting value that has the fewest significant figures.
Example $2$
What is the product of 2.8 and 4.5039 reported to the correct number of significant figures?
Solution
First, we find the mathematical answer to the calculation which is $2.8 \times 4.5039 = 12.61092$. Next, we need to round the answer to the correct number of significant figures. For multiplication and division, the answer should be rounded to that it has the same number of significant figures as the starting value with the least number of significant figures. The value 2.8 has two significant figures while 4.5039 has five. Therefore, the answer should have two significant figures so we should round the answer to 13 because the first digit to be dropped is greater than 5, so we need to round up.
Example $3$
A bag of chocolate candies has a mass of 8.25 ounces. The candies were to be divided among 4 people. How many ounces should each person get? Report your answer to the correct number of significant figures.
Solution
First, we'll find the mathematical answer to this calculation: $8.25 \div 4 = 2.0625$. Now, we need to round to the correct number of significant figures so we look back at the starting values. There are three significant figures in 8.25 but 4 is an exact number so it is not considered when determining the number of significant figures in the answer. Therefore, the answer should be rounded to three significant figures or 2.06 ounces.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/01%3A_Measurements_and_Problem-Solving/1.02%3A_Significant_Figures.txt |
Learning Outcomes
• Convert values among units.
• Use dimensional analysis to solve problems.
Dimensional analysis (also called factor label method or unit analysis) is used to convert from one set of units to another. This method is used for both simple (feet to inches) and complex ($\text{g/cm}^3$ to $\text{kg}$/gallon) conversions and uses relationships or conversion factors between different sets of units. While the terms are frequently used interchangeably, conversion factors and relationships are different. Conversion factors are quantities that are equal to one another, such as $100 \: \text{cm} = 1 \: \text{m}$, in which both values describe a length. Relationships are between two values that are not necessarily a measure of the same quantity. For example, the density of water is $1.00 \: \text{g/mL}$. Grams are a measure of mass while milliliters measure volume so this is considered a relationship rather than a conversion factor. Either way, we depend on units to help set up and solve the calculation. We will see additional examples of relationships as we explore other details about chemical substance.
Importance of using Correct Conversions
In healthcare professions, a calculation error can quite literally have a life or death consequence. Read the short article, Med-Math Errors and the Nursing Student at http://www.alysion.org/dimensional/matherrors.htm, to better understand the importance of units and correct calculations.
Conversion Factors
Many quantities can be expressed in several different ways. The English system measurement of 4 cups is also equal to 2 pints, 1 quart, and $\ce{1/4}$ of a gallon. $4 \: \text{cups} = 2 \: \text{pints} = 1 \: \text{quart} = 0.25 \: \text{gallon}$. Notice that the numerical component of each quantity is different, while the actual amount of material that it represents is the same. That is because the units are different. We can establish the same set of equalities for the metric system: $1 \: \text{meter} = 10 \: \text{decimeters} = 100 \: \text{centimeters} = 1000 \: \text{millimeters}$. The metric system's use of powers of 10 for all conversions makes this quite simple. We can write conversion factors between any pair of equivalent quantities. In each conversion factor, the numerator and denominator represent equal quantities so they are all valid conversion factors. Additionally, these conversion factors can be inverted or used in combination with other conversion factors in a dimensional analysis problem.
$\dfrac{\text{4 cups}}{\text{2 pints}} =\dfrac{\text{1 quart}}{\text{4 cups}} = \dfrac{\text{0.25 gallen}}{\text{1 quarter}}=1 \nonumber$
$\dfrac{\text{1 meter}}{\text{10 decimeters}} =\dfrac{\text{100 centimeters}}{\text{1000 milimeters}} = \dfrac{\text{1000 milimeters}}{\text{1 meter}}=1 \nonumber$
Example $1$
How many centimeters are in $3.4 \: \text{m}$?
Solution
This problem requires the conversion from one unit to another so we can use dimensional analysis to solve the problem. We need to identify the units that are given $\left( \text{m} \right)$, the units for the answer $\left( \text{cm} \right)$, and any relationships that relate the units of the known and unknown values. In this case, we will use the relationship of $1 \: \text{m} = 100 \: \text{cm}$. Start with the known value and its unit.
$3.4 \: \text{m} \times \dfrac{?}{?} \nonumber$
Then, we look at the units of our relationship to see which value goes in the numerator and which value goes in the denominator. Remember, we are trying to find the value in centimeters. Since our known value is in units of meters, we need meters to be in the denominator so that it will cancel. As a result, centimeters will be in the numerator.
$3.4 \: \text{m} \times \dfrac{100 \: \text{cm}}{1 \: \text{m}} \nonumber$
Note that the numbers stay with the appropriate unit (100 with centimeters and 1 with meters). Now, the meters will cancel out and we are left with units of centimeters. Always check that your problem is set up completely and that your units cancel correctly before you do the actual calculation.
\begin{align*}3.4 \: \cancel{\text{m}} \times \dfrac{100 \: \text{cm}}{1 \: \cancel{\text{m}}} &= 340 \: \text{cm} \[5pt] &= 3.4 \times 10^2 \: \text{cm} \end{align*}
We find the answer to be $340 \: \text{cm}$ or $3.4 \times 10^2 \: \text{cm}$.
Derived Units
Using dimensional analysis with derived units requires special care. When units are squared or cubed as with area or volume, the conversion factors themselves must also be squared or cubed. Two convenient volume units are the liter, which is equal to a cubic decimeter, and the milliliters, which is equal to a cubic centimeter. There are thus $1000 \: \text{cm}^3$ in $1 \: \text{dm}^3$, which is the same thing as saying there are $1000 \: \text{mL}$ in $1 \: \text{L}$. The conversion factor of $1 \: \text{cm}^3 = 1 \: \text{mL}$ is a very useful conversion.
There are $1000 \: \text{cm}^3$ in $1 \: \text{dm}^3$. Since a $\text{cm}^3$ is equal to a $\text{mL}$, and a $\text{dm}^3$ is equal to a $\text{L}$, we can say that there are $1000 \: \text{mL}$ in $1 \: \text{L}$.
Example $2$
Convert $3.6 \: \text{mm}^3$ to $\text{mL}$.
Solution
Determine the units of the known value $\left( \text{mm}^3 \right)$ and the units of the unknown value $\left( \text{mL} \right)$. The starting and ending units will help guide the setup of the problem. Next, list any known conversion factors that might be helpful.
• $1 \: \text{m} = 1000 \: \text{mm}$
• $1 \: \text{mL} = 1 \: \text{cm}^3$
• $1 \: \text{m} = 100 \: \text{cm}$
Now, we can set up the problem to find the value in units of $\text{mL}$. Once we know the starting units, we can then use the conversion factors to find the answer.
$3.6 \: \text{mm}^3 \times \left( \dfrac{?}{?} \right) \nonumber$
Continue to use the conversion factors between the units to set up the rest of the problem. Note that all of the units cancel except $\text{mL}$, which are the requested units for the answer. Since the values in these conversion factors are exact numbers, they will not affect the number of significant figures in the answer. Only the original value (3.6) will be considered in determining significant figures.
$3.6 \: \text{mm}^3 \times \left( \dfrac{1 \: \text{m}}{1000 \: \text{mm}} \right)^3 \times \left( \dfrac{?}{?} \right) \nonumber$
Once you have solved the problem, always ask if the answer seems reasonable. Remember, a millimeter is very small and a cubic millimeter is also very small. Therefore, we would expect a small volume which means $0.0036 \: \text{mL}$ is reasonable.
If you find that you forgot to cube numbers as well as units, you can setup the problem in an expanded form which is the equivalent to the previous method to cube the numerical values.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/01%3A_Measurements_and_Problem-Solving/1.03%3A_Scientific_Dimensional_Analysis.txt |
Learning Outcomes
• Calculate percentages and use in calculations.
In chemistry and healthcare, values are reported with a variety of units. Percentages are often used in the concentration of IV solutions and medications. There are multiple ways to report percentages so it is important to understand the basic meaning of a percent, which will be the focus here, as well as the type of percentage, such as mass or volume percent.
Percent
Percents are used to report many different values, such as a grade on an assignment as $85\%$, the shoes are $20\%$ off their original price, or the concentration of the IV solution is $5.5\%$ $\ce{NaCl}$ by mass. While all of these numbers use percent very differently, the values are all found the same way. For now, we will focus on some basic calculations using percents. Percents are all calculated with the same basic formula using the amount of the "part" and the amount of the "whole". Note that the percent will always be less than or equal to $100\%$ because the "part" must be less than or equal to the "whole". The basic percent equation is shown here.
$\% = \frac{\text{part}}{\text{whole}} \times 100$
Example $1$
What percent of 70 is 14?
Solution
Since this is a percent problem, first look at the equation for percent and see which values are given in the problem. In this case, we know the part (14) and the whole (70). With this information and the formula, we can solve for the percent by inserting the known values into the equation and solving for the unknown.
\begin{align*} \% &= \frac{\text{part}}{\text{whole}} \times 100 \[5pt] \% &= \frac{14}{70} \times 100 \[5pt] \% &= 0.20 \times 100 \[5pt] \% &= 20 \end{align*}
Therefore, 14 is $20\%$ of 70.
Example $2$
What is $35\%$ of 80?
Solution
This problem also involves percent so we will use the same equation as the previous example. However, in this example, we know the percentage (35) and the whole (80) and we are trying to find the value of the part. As before, we will insert the known values into the equation and solve for our unknown.
\begin{align*} \% &= \frac{\text{part}}{\text{whole}} \times 100 \[5pt] 35 &= \frac{\text{part}}{80} \times 100 \[5pt] 0.35 &= \frac{\text{part}}{80} \[5pt] \text{part} &= 28 \end{align*}
So 28 is $35\%$ of 80.
If you need to review how to rearrange equations and solve for an unknown, check out the worked examples and practice problems at https://www.mathisfun.com/algebra/in...-multiply.html.
Example $3$
A patient has a fasting blood sugar level of $150 \: \text{mg/dL}$. The doctor has recommended some dietary changes and is hoping to see at least a $20\%$ reduction in the patient's blood sugar level at their next appointment. If the patient meets this goal, what will their blood sugar level be?
Solution
We can start this problem like we do other percent problems. The problem gives use the percent (20) and the whole (150) and we need to calculate the part which will be the amount of decrease in the patient's blood sugar.
\begin{align*} \% &= \frac{\text{part}}{\text{whole}} \times 100 \[5pt] 20 &= \frac{\text{part}}{150} \times 100 \[5pt] 0.20 &= \frac{\text{part}}{150} \[5pt] \text{part} &= 30 \end{align*}
The part that was calculated is the decrease that should be seen in the patient's blood sugar level. However, it's not the actual blood sugar level that the doctor wants to see at the next visit. To find that, we need to subtract from the original value.
$\text{original} - \text{loss} = \text{final}$
$150 \: \frac{\text{mg}}{\text{dL}} - 30 \: \frac{\text{mg}}{\text{dL}} = 120 \: \frac{\text{mg}}{\text{dL}}$
At the next visit, the patient's blood sugar level should be $120 \: \text{mg/dL}$ (or less!) if they are making sufficient progress towards improving their health and reducing their risk of diabetes.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/01%3A_Measurements_and_Problem-Solving/1.04%3A_Percentages.txt |
These are homework exercises to accompany Chapter 1 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Solutions are available below the questions.
Questions
(click here for solutions)
Q1.1.1
Express the following values in scientific notation.
1. 150,000,000
2. 0.000043
3. 332000
4. 0.0293
5. 932
6. 0.1873
7. 78,000
8. 0.0001
9. 4500
10. 0.00290
11. 6281
12. 0.00700
Q1.1.2
Express the following values in decimal notation.
1. 3.8 x 10-3
2. 9.21 x 105
3. 7.91 x 10-2
4. 2.5 x 106
5. 3.42 x 10-8
6. 5.4 x 105
7. 3 x 10-3
8. 7.34 x 102
9. 9.8 x 10-4
10. 6 x 107
11. 4.20 x 10-6
12. 4.20 x 106
Q1.1.3
What SI base unit would be appropriate for each measurement?
1. the length of a room
2. the amount of carbon in a diamond
3. the mass of NaCl in a bottle
Q1.1.4
List the meaning of each abbreviation of the base units.
1. m
2. K
3. kg
4. s
5. mol
Q1.1.5
What is the the derived unit from the SI base units for the relationship of each pair of quantities?
1. mass and volume
2. distance and time
3. amount of substance and volume
4. area
Q1.1.6
Give the meaning and name of each metrix prefix abbreviation.
1. M
2. m
3. n
4. d
Q1.1.7
Give the abbreviation and meaning of each metrix prefix.
1. kilo
2. centi
3. micro
4. giga
Q1.1.8
Name the prefix with the following numerical meaning.
1. 1/10
2. 1,000,000
3. 1/1,000,000
4. 1/100
5. 1
Q1.1.9
Convert each temperature to the missing one between Celsius and Fahrenheit.
1. 77°F
2. 212°F
3. 37°C
4. 22°C
5. 95°F
6. 15°C
7. 0°F
8. 0°C
9. –10°C
10. –10°F
(click here for solutions)
Q1.2.1
Explain the similarities and differences between accuracy and precision.
Q1.2.2
The density of a copper sample was determined by three different students (shown below). Each performed the measurement three times and is reported below (all values in units of g/cm3). The accepted value for the density of copper is 8.92 g/cm3.
1. Determine if each student's data is accurate, precise, neither or both.
2. What is the average density based on Justin's data?
3. What is the average density based on Jane's data?
• Jane: 8.94, 8.89, 8.91
• Justin: 8.32, 8.31, 8.34
• Julia: 8.64, 9.71, and 9.10
Q1.2.3
Determine the number of significant figures in each of the following values.
1. 406
2. 3.00
3. 3.20
4. 0.25
5. 0.0689
6. 0.002910
7. 3941
8. 46.250
9. 30.21
10. 0.10300
Q1.2.4
Write each value with three significant figures, use scientific notation if necessary.
1. 34500
2. 24
3. 0.0345
4. 0.012
5. 612.8
6. 98.22
7. 0.14928
8. 300
Q1.2.5
Give three examples of exact numbers.
Q1.2.6
Find the result of each of the following calculations and report the value with the correct number of significant figures.
1. 0.23 + 12.2 =
2. 13 - 1.03 =
3. 0.839 + 0.28925 =
4. 28 + 34.4 =
5. 0.8 + 2.3 =
6. 34.9 - 0.583 =
7. 21 - 0.132 =
8. 0.840 + 0.9334
Q1.2.7
Find the result of each of the following calculations and report the value with the correct number of significant figures.
1. 34 x 0.12 =
2. 68.2 / 0.78 =
3. 3.29 x 104 x 16.2 =
4. 0.8449 x 29.7 =
5. 5.92 x 103 / 0.628 =
6. 3.00 x 2.6 =
7. 2.50 x 9.331 =
8. 3.20 / 12.75 =
(click here for solutions)
Q1.3.1
What is a conversion factor?
Q1.3.2
What is the conversion factor between each pair of units?
1. feet and inches
2. mL and cm3
3. kg and g
4. cm and m
5. mm and cm
6. inches and centimeters
7. grams and pounds
8. g and µg (mcg)
Q1.3.3
Complete each of the following conversions.
1. 34 cm to m
2. 3.7 ft to in
3. 345 mg to Mg
4. 5.3 km to mm
5. 4.0 L to mL
6. 3.45 x 103 mm to km
7. 78 cm3 to mL
8. 0.85 kg to dg
9. 13 pints to gallon
10. 0.35 L to cm3
Q1.3.4
Complete each of the following conversions.
1. 342 cm3 to dm3
2. 2.70 g/cm3 to kg/L
3. 34 mi/hr to km/min
4. 0.00722 km2 to m2
5. 4.9 x 105 mcm3 to mm3
6. 80. km/hr to mi/hr
(click here for solutions)
Q1.4.1
Solve each of the following.
1. What percent of 35 is 8.2?
2. What percent of 56 is 12?
3. What percent of 923 is 38?
4. What percent of 342 is 118?
Q1.4.2
Solve each of the following?
1. What is 42% of 94?
2. What is 83% of 239?
3. What is 16% of 45?
4. What is 38% of 872?
Q1.4.3
Solve each of the following?
1. 42 is 34% of what number?
2. 73 is 82% of what number?
3. 13 is 57% of what number?
4. 75 is 25% of what number?
5. 25 is 15% of what number?
6. 98 is 76% of what number?
Q1.4.4
A patient originally weighs 182 pounds and loses 15.0% of their body weight. What is their final weight?
Q1.4.5
A patient's original weight was 135 pounds and they lose 12 pounds. What percent of their body weight did they lose?
Q1.4.6
A patient needs to increase their calcium supplement by 25% a week. If they are currently taking a 300. mg supplement, how much more will they need to take?
Q1.4.7
An infant's birth weight is 7 pounds, 1 ounce. Her discharge weight is 6 pounds, 13 ounces. What percent of her birth weight did she lose?
Q1.4.8
A patient needs a 20.% decrease in their medication dosage from 125 mg. What will his dosage be after the decrease?
Answers
1.1: Measurements Matter
Q1.1.1
1. 1.5 × 108
2. 4.3 × 10–5
3. 3.32 × 105
4. 2.93 × 10–2
5. 9.32 × 102
6. 1.873 × 10–1
7. 7.8 × 104
8. 1 × 10–4
9. 4.5 × 103
10. 2.9 × 10–3
11. 6.281 × 103
12. 7 × 10–3
Q1.1.2
1. 0.0038
2. 921000
3. 0.0791
4. 2500000
5. 0.0000000342
6. 540000
7. 0.003
8. 734
9. 0.00098
10. 60000000
11. 0.00000420
12. 4200000
Q1.1.3
1. meter
2. mole
3. kilogram
Q1.1.4
1. meter
2. Kelvin
3. kilogram
4. second
5. mole
Q1.1.5
1. kg/m3
2. m/s
3. mol/m3 is based on SI base units, but mol/L is also acceptable
4. m2
Q1.1.6
1. Mega, 106
2. milli, 10–3
3. nano, 10–9
4. deci, 10–1
Q1.1.7
1. k, 103
2. c, 10–2
3. µ (or mc), 10–6
4. G, 109
Q1.1.8
1. deci
2. mega
3. micro
4. centi
5. none (base unit)
Q1.1.9
1. 77°F = 25°C
2. 212°F = 100°C
3. 37°C = 98.6°F
4. 22°C = 72°F
5. 95°F = 35°C
6. 15°C = 59°F
7. 0°F = –18°C
8. 0°C = 32°F
9. –10°C = 14°F
10. –10°F = –23°C
1.2 Significant Figures
Q1.2.1
Accuracy is a measure of how close the values are close to the correct value while precision is a measure of how close values are to each other.
Q1.2.2
1.
• Jane: 8.94, 8.89, 8.91 - accurate and precise
• Justin: 8.32, 8.31, 8.34 - precise
• Julia: 8.64, 9.71, and 9.10 - neither accurate nor precise
1. 8.32 g/cm3
2. 8.91 g/cm3
Q1.2.3
1. 3
2. 3
3. 3
4. 2
5. 3
6. 4
7. 4
8. 5
9. 4
10. 5
Q1.2.4
1. 3.45 × 104
2. 2.40 × 101
3. 3.45 × 10–2
4. 1.20 × 10–2
5. 613 or 6.13 × 102
6. 9.82 × 101
7. 0.149 or 1.49 × 10–1
8. 300. or 3.00 × 102
Q1.2.5
Answers will vary. 12 eggs, 100 cm = 1 m, 1 inch = 2.54 cm, 4 people
Q1.2.6
1. 0.23 + 12.2 = 12.43 = 12.4
2. 13 - 1.03 = 11.97 = 12
3. 0.839 + 0.28925 = 1.12825 = 1.128
4. 28 + 34.4 = 62.4 = 62
5. 0.8 + 2.3 = 3.1
6. 34.9 - 0.583 = 34.317 = 34.3
7. 21 - 0.132 = 20.868 = 21
8. 0.840 + 0.9334 = 1.7734 = 1.773
Q1.2.7
1. 34 x 0.12 = 4.08 = 4.1
2. 68.2 / 0.78 = 87.4358974 = 87
3. 3.29 x 104 x 16.2 = 5.32980 × 105 = 5.33 × 105
4. 0.8449 x 29.7 = 25.09353 = 25.1
5. 5.92 x 103 / 0.628 = 9.4267515 × 103 = 9.43 × 103
6. 3.00 x 2.6 = 7.8
7. 2.50 x 9.331 = 23.3275 = 23.3
8. 3.20 / 12.75 = 0.25098 = 0.251
1.3 Scientific Dimensional Analysis
Q1.3.1
A conversion factor is a relationship between two units. The value in the numerator has some equivalence to the value in the denominator.
Q1.3.2
1. 1 foot = 12 inches
2. 1 mL = 1 cm3
3. 1 kg = 1000 g or 1 × 10–3 kg = 1 g
4. 100 cm = 1 m or 1 cm = 1 × 10–2 m
5. 10 mm = 1 cm
6. 1 inch = 2.54 cm
7. 454 grams = 1 pound
8. 1 g = 1 × 106 µg (mcg) or 1 × 10–6 g = 1 µg (mcg)
Q1.3.3
1. $34 \; cm \times \frac{1 \; m}{100\;cm} = 0.34\;m$
2. $3.7 \; ft \times \frac{12 \; in}{1\;ft}=44.4\;in=44\;in$
3. $345\;mg \times \frac{1\;g}{1000\;mg} \times \frac{1\;Mg}{1\times {10}^6\;g}=3.45 \times {10}^{-7}\;Mg$
4. $5.3\;km\times\frac{1000\;m}{1\;km}\times\frac{1000\;mm}{1\;m}=5.3\times{10}^6\;mm$
5. $4.0\;L\times\frac{1000\;mL}{1\;L}=4.0\times{10}^3\;mL$
6. $3.45\times{10}^3\;mm\times\frac{1\;m}{1000\;mm}\times\frac{1\;km}{1000\;m}=3.45\times{10}^{-3}\;km$
7. $78\;{cm}^3\times\frac{1\;mL}{{cm}^3}=78\;mL$
8. $0.85\;kg\times\frac{1000\;g}{1\;kg}\times\frac{10\;dg}{1\;g}=8.5\times{10}^3\;dg$
9. $13\;pints\times\frac{1\;quart}{2\;pints}\times\frac{1\;gallon}{4\;quarts}=1.6\;gallons$
10. $0.35\;L\times\frac{1000\;mL}{1\;L}\times\frac{1\;mL}{1\;{cm}^3}=3.5\times{10}^2\;{cm}^3$
Q1.3.4
1. $342\;{cm}^3\times\frac{1\;dm}{10\;cm}\times\frac{1\;dm}{10\;cm}\times\frac{1\;dm}{10\;cm}=0.342\;{dm}^3$ or $342\;{cm}^3\times {\left( \frac{1\;dm}{10\;cm} \right)}^3=342\;{cm}^3\times \frac{{1}^3\;{dm}^3}{{10}^3\;{cm}^3}=0.342\;{dm}^3$
2. $\frac{2.70\;g}{{cm}^3}\times\frac{1\;kg}{1000\;g}\times\frac{1\;{cm}^3}{1\;mL}\times\frac{1000\;mL}{1\;L}=\frac{2.70\;kg}{L}$
3. $\frac{34\;mi}{hr} \times \frac{5280\;ft}{1\;mi} \times \frac{12\;in}{1\;ft}\times \frac{2.54\;cm}{1\;in} \times \frac{1\;m}{100\;cm} \times \frac{1\;km}{1000\;m} \times \frac{1\;hr}{60\;min} = \frac{0.91\;km}{min}$
4. $0.00722\;k{m^2} \times \frac{1000\;m}{1\;km} \times \frac{1000\;m}{1\;km} = 7.22 \times {10^3}\;{m^2}$
5. $4.95 \times {10^5}\;mc{m^3} \times \frac{1\;mm}{1000\;mcm} \times \frac{1\;mm}{1000\;mcm} \times \frac{1\;mm}{1000\;mcm} = 4.95 \times {10^ - }^4\;m{m^3}$
6. $\frac{80.\;km}{hr}\times\frac{1000\;m}{1\;km}\times\frac{100\;cm}{1\;m}\times\frac{1\;in}{2.54\;cm}\times\frac{1\;ft}{12\;in}\times\frac{1\;mi}{5280\;ft}=\frac{50.\;mi}{hr}$
1.4 Percentages
Q1.4.1
1. $\%= \frac{part}{whole} \times 100= \frac{8.2}{35} \times 100= 23\%$
2. $\%= \frac{part}{whole} \times 100= \frac{12}{56} \times 100= 21\%$
3. $\%= \frac{part}{whole} \times 100= \frac{38}{923} \times 100= 4.1\%$
4. $\%= \frac{part}{whole} \times 100= \frac{118}{342} \times 100= 34.5\%$
Q1.4.2
1. $\begin{array}{c} \% = \frac{part}{whole} \times 100\ 42\% = \frac{part}{94} \times 100\ part = 39 \end{array}$
2. $\begin{array}{c} \quad \ \% = \frac{part}{whole} \times 100\ 83\% = \frac{part}{239} \times 100\ part = 198=2.0\times{10}^2 \end{array}$
3. $\begin{array}{c} \quad \ \% = \frac{part}{whole} \times 100\ 16\% = \frac{part}{45} \times 100\ part = 7.2\ \end{array}$
4. $\begin{array}{c} \quad \ \% = \frac{part}{whole} \times 100\ 38\% = \frac{part}{872} \times 100\ part = 3.3\times{10}^2 \end{array}$
Q1.4.3
1. $\begin{array}{c} \quad \ \% = \frac{part}{whole} \times 100\ 34\% = \frac{42}{whole} \times 100\ whole = 1.2\times{10}^2 \end{array}$
2. $\begin{array}{c} \quad \ \% = \frac{part}{whole} \times 100\ 82\% = \frac{73}{whole} \times 100\ whole = 89 \end{array}$
3. $\begin{array}{c} \quad \ \% = \frac{part}{whole} \times 100\ 57\% = \frac{13}{whole} \times 100\ whole = 23 \end{array}$
4. $\begin{array}{c} \quad \ \% = \frac{part}{whole} \times 100\ 25\% = \frac{75}{whole} \times 100\ whole = 3.0\times{10}^2 \end{array}$
5. $\begin{array}{c} \quad \ \% = \frac{part}{whole} \times 100\ 15\% = \frac{25}{whole} \times 100\ whole = 1.7\times{10}^2 \end{array}$
6. $\begin{array}{c} \quad \ \% = \frac{part}{whole} \times 100\ 76\% = \frac{98}{whole} \times 100\ whole = 129 \end{array}$
Q1.4.4
$\begin{array}{l} \% = \frac{part}{whole} \times 100\ 15.0\% = \frac{part}{182\;pounds} \times 100\ part = 27.3\;pounds\;lost\ \ 182\;pounds - 27.3\;pounds = 154.7\;pounds = 155\;pounds \end{array}$
Q1.4.5
$\begin{array}{l} \% = \frac{part}{whole} \times 100\ \% = \frac{12\;pounds}{135\;pounds} \times 100\ \% = 8.9\%\;lost\ \end{array}$
Q1.4.6
$\begin{array}{l} \% = \frac{part}{whole} \times 100\ 25\% = \frac{part}{300.\;mg} \times 100\ part = 75\;mg\;more\ \end{array}$
Q1.4.7
Convert both weights to ounces, find the ounces lost, and then find the percent lost.
Birth weight: $\left( 7\;pounds\times 16 \right) + 1\;ounce=113\;ounces$
Discharge weight: $\left( 6\;pounds\times 16 \right) + 13\;ounces=109\;ounces$
Weight lost: $113\;ounces-109\;ounces=4\;ounces$
Percent lost from original brith weight.
$\begin{array}{l} \% = \frac{part}{whole} \times 100\ \% = \frac{4\;ounces}{113\;ounces} \times 100\ \% = 3.5\% =4\% \end{array}$
Q1.4.8
$\begin{array}{l} \% = \frac{part}{whole} \times 100\ 20\% = \frac{part}{125\;mg} \times 100\ part = 25\;mg\;lost\ \ 125\;mg - 25\;mg = 100.\;mg \end{array}$ | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/01%3A_Measurements_and_Problem-Solving/1.05%3A_Measurements_and_Problem-Solving_%28Exercises%29.txt |
Chemistry is the study of matter and the changes it undergoes. Atoms are the building blocks of matter and are the smallest unit of an element.
• 2.1: Isotopes and Atomic Mass
Atoms are the fundamental building blocks of all matter and are composed of protons, neutrons, and electrons. Isotopes are atoms of the same element with a different mass due to a different number of neutrons.
• 2.2: Matter
Living things are made of matter. In fact, matter is the "stuff" of which all things are made (see figure below. Anything that occupies space and has mass is known as matter. Matter, in turn, consists of chemical substances. Chemistry is the study of matter and the changes it undergoes.
• 2.3: Mole and Molar Mass
A mole is used as a measure of the amount of substance present. Atoms are very small so using more common counting numbers, such as a dozen, are impractical.
• 2.4: Electron Arrangements
The number and arrangement of electrons in an atom control the types and number of bonds that an atom can form so understanding their arrangement helps explain the behavior of an atom in a compound or molecule.
• 2.5: Ion Formation
Ions form from the gain or loss of electrons. The electron arrangements allows for the prediction of how many electrons will be gained or lost.
• 2.6: Ionic Compounds
Ionic compounds form from at least one negatively-charged anion and one positively charged cation. The charges on the ions allow the prediction of the number of each that are combined to form a single formula unit of an ionic compound.
• 2.7: Elements and Ions (Exercises)
These are homework exercises to accompany Chapter 2 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Solutions are available below the questions.
02: Elements and Ions
Learning Outcomes
• Define atomic and mass numbers.
• Determine the number of protons, neutrons, and electrons in an atom.
• Identify the charge and relative mass of subatomic particles.
• Label the location of subatomic particles in the atom.
• Define isotope.
• Write the isotopic symbol of an atom.
• Explain the concept of average atomic mass.
Atoms are the fundamental building blocks of all matter and are composed of protons, neutrons, and electrons. Because atoms are electrically neutral, the number of positively charged protons must be equal to the number of negatively charged electrons. Since neutrons do not affect the charge, the number of neutrons is not dependent on the number of protons and will vary even among atoms of the same element.
Atomic Number
The atomic number (Z) of an element is the number of protons in the nucleus of each atom of that element. An atom can be classified as a particular element based solely on its atomic number. For example, any atom with an atomic number of 8 (its nucleus contains 8 protons) is an oxygen atom, and any atom with a different number of protons would be a different element. The periodic table (see figure below) displays all of the known elements and is arranged in order of increasing atomic number. In this table, an element's atomic number is indicated above the elemental symbol. Hydrogen, at the upper left of the table, has an atomic number of 1. Every hydrogen atom has one proton in its nucleus. Next on the table is helium, whose atoms have two protons in the nucleus. Lithium atoms have three protons, beryllium atoms have four, and so on.
Since atoms are neutral, the number of electrons in an atom is equal to the number of protons. Hydrogen atoms all have one electron occupying the space outside of the nucleus. Helium, with two protons, will have two electrons.
Mass Number
Experimental data showed that the vast majority of the mass of an atom is concentrated in its nucleus, which is composed of protons and neutrons. The mass number is defined as the total number of protons and neutrons in an atom. Consider the table below, which shows data from the first six elements of the periodic table.
Table $1$: Atoms of the First Six Elements
Name Symbol Atomic Number Protons Neutrons Electrons Mass Number
hydrogen $\ce{H}$ 1 1 0 1 1
helium $\ce{He}$ 2 2 2 2 4
lithium $\ce{Li}$ 3 3 4 3 7
beryllium $\ce{Be}$ 4 4 5 4 9
boron $\ce{B}$ 5 5 6 5 11
carbon $\ce{C}$ 6 6 6 6 12
View animations showing the atomic structure of the first 11 elements on the periodic table at http://web.visionlearning.com/custom...imations.shtml
Consider the element helium. Its atomic number is 2, so it has two protons in its nucleus. Its nucleus also contains two neutrons. Since $2 + 2 = 4$, we know that the mass number of the helium atom is 4. Finally, the helium atom also contains two electrons, since the number of electrons must equal the number of protons. This example may lead you to believe that atoms have the same number of protons and neutrons, but a further examination of the table above will show that this is not the case. Lithium, for example, has three protons and four neutrons, giving it a mass number of 7.
Knowing the mass number and the atomic number of an atom allows you to determine the number of neutrons present in that atom by subtraction.
$\text{Number of neutrons} = \text{mass number} - \text{atomic number}$
Atoms of the element chromium $\left( \ce{Cr} \right)$ have an atomic number of 24 and a mass number of 52. How many neutrons are in the nucleus of a chromium atom? To determine this, you would subtract as shown:
$52 - 24 = 28 \: \text{neutrons in a chromium atom}$
The composition of any atom can be illustrated with a shorthand notation using the atomic number and the mass number. Both are written before the chemical symbol, with the mass number written as a superscript and the atomic number written as a subscript. The chromium atom discussed above would be written as:
$\ce{^{52}_{24}Cr}$
Another way to refer to a specific atom is to write the mass number of the atom after the name, separated by a hyphen. The above atom would be written as chromium-52, with the mass number written after the name. The atomic number does not have to be included because all atoms of chromium have the same number of protons but can vary in the atomic mass.
Isotopes
As stated earlier, not all atoms of a given element are identical. Specifically, the number of neutrons in the nucleus can vary for many elements. As an example, naturally occurring carbon exists in three forms, which are illustrated in the figure below.
Each carbon atom has the same number of protons (6), which is equal to its atomic number. Each carbon atom also contains six electrons, allowing the atom to remain electrically neutral. However, the number of neutrons varies from six to eight. Isotopes are atoms that have the same atomic number but different mass numbers due to a change in the number of neutrons. The three isotopes of carbon can be referred to as carbon-12 $\left( \ce{^{12}_6C} \right)$, carbon-13 $\left( \ce{^{13}_6C} \right)$, and carbon-14 $\left( \ce{^{14}_6C} \right)$. Naturally occurring samples of most elements are mixtures of isotopes. Carbon has only three natural isotopes, but some heavier elements have many more. Tin has ten stable isotopes, which is the most of any known element. The nucleus of a given carbon atom will be one of the three possible isotopes discussed above.
While the presence of isotopes affects the mass of an atom, it does not affect its chemical reactivity. Chemical behavior is governed by the number of electrons and the number of protons. Carbon-13 behaves chemically in exactly the same way as the more plentiful carbon-12.
Size of Atoms
The graphite in your pencil is composed of the element carbon, a nonmetal. Imagine taking a small piece of carbon and grinding it until it is a fine dust. Each speck of carbon would still have all of the physical and chemical properties of carbon. Now imagine that you could somehow keep dividing the speck of carbon into smaller and smaller pieces. Eventually, you would reach a point where your carbon sample is as small as it could possibly be. This final particle is called an atom.
Atoms, as you probably know, are extremely small. In fact, the graphite in an ordinary pencil contains about $5 \times 10^{20}$ atoms of carbon. This is an almost incomprehensibly large number. The population of the entire Earth is about $7 \times 10^9$ people, meaning that there are about $7 \times 10^{10}$ times as many carbon atoms in your pencil as there are people on Earth! For this to be true, atoms must be extremely small. We can only see atoms with a modern instrument called a scanning tunneling microscope. (www.nobelprize.org/educationa...opes/scanning/)
Atomic Mass
The masses of individual atoms are very, very small. However, using a modern device called a mass spectrometer, it is possible to measure such miniscule masses. An atom of oxygen-16, for example, has a mass of $2.66 \times 10^{-23} \: \text{g}$. While comparisons of masses measured in grams would have some usefulness, it is far more practical to have a system that will allow us to more easily compare relative atomic masses. Scientists decided on using the carbon-12 nuclide as the reference standard by which all other masses would be compared. By definition, one atom of carbon-12 is assigned a mass of exactly 12 atomic mass units $\left( \text{amu} \right)$. An atomic mass unit is defined as a mass equal to one twelfth of an atom of carbon-12. The mass of any isotope of any element is expressed in relation to the carbon-12 standard. For example, one atom of helium-4 has a mass of $4.0026 \: \text{amu}$. An atom of sulfur-32 has a mass of $31.972 \: \text{amu}$.
The carbon-12 atom has six protons and six neutrons in its nucleus for a mass number of 12. Since the nucleus accounts for nearly all of the mass of the atom, a single proton or single neutron has a mass of approximately $1 \: \text{amu}$. However, as seen by the helium and sulfur examples, the masses of individual atoms are not quite whole numbers. This is because an atom's mass is affected very slightly by the interactions of the various particles within the nucleus and also includes the small mass added by each electron.
As stated in the section on isotopes, most elements occur naturally as a mixture of two or more isotopes. Listed below (see table below) are the naturally occurring isotopes of several elements along with the percent natural abundance of each.
Table $2$: Atomic Masses and Percent Abundances of Some Natural Isotopes
Element Isotope (Symbol) Percent Natural Abundance Atomic mass $\left( \text{amu} \right)$ Average atomic mass $\left( \text{amu} \right)$
Hydrogen $\ce{^1_1H}$ 99.985 1.0078 1.0079
$\ce{^2_1H}$ 0.015 2.0141
$\ce{^3_1H}$ negligible 3.0160
Carbon $\ce{^{12}_6C}$ 98.89 12.000 12.011
$\ce{^{13}_6C}$ 1.11 13.003
$\ce{^{14}_6C}$ trace 14.003
Oxygen $\ce{^{16}_8O}$ 99.759 15.995 15.999
$\ce{^{17}_8O}$ 0.037 16.995
$\ce{^{18}_8O}$ 0.204 17.999
Chlorine $\ce{^{35}_{17}Cl}$ 75.77 34.969 35.453
$\ce{^{37}_{17}Cl}$ 24.23 36.966
Copper $\ce{^{63}_{29}Cu}$ 69.17 62.930 63.546
$\ce{^{65}_{29}Cu}$ 30.83 64.928
For some elements, one particular isotope is much more abundant than any other isotopes. For example, naturally occurring hydrogen is nearly all hydrogen-1, and naturally occurring oxygen is nearly all oxygen-16. For many other elements, however, more than one isotope may exist in substantial quantities. Chlorine (atomic number 17) is yellowish-green toxic gas. About three quarters of all chlorine atoms have 18 neutrons, giving those atoms a mass number of 35. About one quarter of all chlorine atoms have 20 neutrons, giving those atoms a mass number of 37. Were you to simply calculate the arithmetic average of the precise atomic masses, you would get approximately 36.
$\frac{34.969 + 36.966}{2} = 35.968$
As you can see, the average atomic mass given in the last column of the table above is significantly lower. Why? The reason is that we need to take into account the natural abundance percentages of each isotope in order to calculate what is called the weighted average. The atomic mass of an element is the weighted average of the atomic masses of the naturally occurring isotopes of that element. The average atomic masses are the values we see on the periodic table.
$0.7577 \left( 34.969 \right) + 0.2423 \left( 36.966 \right) = 35.453$
The weighted average is determined by multiplying the percent of natural abundance by the actual mass of the isotope. This is repeated until there is a term for each isotope. For chlorine, there are only two naturally occurring isotopes so there are only two terms.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/02%3A_Elements_and_Ions/2.01%3A_Isotopes_and_Atomic_Mass.txt |
Learning Outcomes
• Define matter.
• Classify pure substances as elements or compounds.
• Classify mixtures as homogeneous or heterogeneous.
• Know the names and symbols of elements indicated in "Things to Memorize" on Canvas.
• Distinguish among metals, nonmetals, and metalloids.
Living things are made of matter. In fact, matter is the "stuff" of which all things are made (see figure below. Anything that occupies space and has mass is known as matter. Matter, in turn, consists of chemical substances. Chemistry is the study of matter and the changes it undergoes.
Pure Substances
A pure substance is a material that has a definite chemical composition. It is also homogenous, so the same chemical composition is found uniformly throughout the substance. A pure substance may be an element or a chemical compound.
Elements
An element is a pure substance that cannot be broken down into different types of substances. Examples of elements include carbon, oxygen, hydrogen, and iron. Each element is made up of just one type of atom. An atom is the smallest particle of an element that still characterizes the element. As shown in the figure below, at the center of an atom is a nucleus. The nucleus contains positively charged particles called protons and electrically neutral particles called neutrons. Surrounding the nucleus is a much larger electron cloud consisting of negatively charged electrons. An atom is electrically neutral if it has the same number of protons as electrons. Each element has atoms with a characteristic number of protons. For example, all carbon atoms have six protons, and all oxygen atoms have eight protons.
There are almost 120 known elements (see figure below) and each is represented in the periodic table by a one or two letter symbol. The majority of known elements are classified as metals. Metals are elements that are lustrous, or shiny. They are also good conductors of electricity and heat. Examples of metals include iron, gold, and copper. Fewer than 20 elements are classified as nonmetals. Nonmetals lack the properties of metals. Examples of nonmetals include oxygen, hydrogen, and sulfur. Certain other elements have properties of both metals and nonmetals. They are known as metalloids. Examples of metalloids include silicon and boron.
The New Periodic Table Song can be heard at https://youtu.be/zUDDiWtFtEM
Tom Lehrer performed the original The Element Song in 1967. You can hear it at https://youtu.be/DYW50F42ss8.
Chemical Compounds
A chemical compound is a pure substance that forms when atoms of two or more elements react with one another. A compound always has a unique and fixed chemical composition and the atoms of a compound are held together by chemical bonds. There are different types of chemical bonds, and they vary in how strongly they hold together the atoms of a compound. Two types of bonds are covalent and ionic bonds. Covalent bonds form when atoms share electrons and occur between two or more nonmetals. Ionic bonds form when electrons are transferred from one atom to another and usually form between a metal and a nonmetal.
An example of a chemical compound is water. A water molecule forms when oxygen $\left( \ce{O} \right)$ and hydrogen $\left( \ce{H} \right)$ atoms react and are held together by covalent bonds. Like other compounds, water always has the same chemical composition: a 2:1 ratio of hydrogen atoms to oxygen atoms. This is expressed in the chemical formula for water, $\ce{H_2O}$. The ratio of elements in a compound is given by the chemical formula. For example, $\ce{NaCl}$ has a 1:1 ratio between sodium and chlorine atoms. The absence of a subscripted number within the formula indicates that there is one of that element.
Mixtures
Like a chemical compound, a mixture consists of more than one chemical substance. Unlike a compound, a mixture does not have a fixed chemical composition. The substances in a mixture can be combined in any proportions. One characteristic of mixtures is that they can be separated into their components by physical methods. Since each part of the mixture has not reacted with another part of the mixture, the identities of the different substances remain unchanged.
The following examples illustrate these differences between mixtures and compounds. Both examples involve the same two elements: the metal iron $\left( \ce{Fe} \right)$ and the nonmetal sulfur $\left( \ce{S} \right)$.
• When iron fillings and sulfur powder are mixed together in any ratio, they form a mixture (see figure below). No chemical reaction occurs, and both elements retain their individual properties. A magnet can be used to physically separate the two elements by attracting the iron fillings out of the mixture and leaving the sulfur behind.
• When iron and sulfur are mixed together in a certain ratio and heated, a chemical reaction occurs. This results in the formation of a unique new compound, called iron (II) sulfide $\left( \ce{FeS} \right)$ (see figure below). A magnet cannot be used to mechanically separate the iron from the iron (II) sulfide because neither iron nor sulfur exist in the compound. Instead, another chemical reaction is required to separate the iron and sulfur.
Homogeneous Mixtures
A homogenous mixture is a mixture in which the composition is uniform throughout the mixture. A mixture of salt and water is homogenous because the dissolved salt is evenly distributed throughout the entire salt water sample. Often it is easy to confuse a homogeneous mixture with a pure substance because they are both uniform. The difference is that the composition of the substance is always the same, while the composition of a mixture may vary. The amount of salt in the salt water can vary from one sample to another, while water, for example, always has the same composition.
Heterogeneous Mixtures
A heterogeneous mixture is a mixture in which the composition is not uniform throughout the mixture. One example of a heterogeneous mixture is vegetable soup. Any given spoonful of soup will contain varying amounts of the different vegetables and other components of the soup. Another example of a heterogeneous mixture is soil which is composed of a variety of substances and is often of different composition depending on the sample taken. One shovel may come up with dirt and grass while the next shovel could contain an earthworm.
See more examples of heterogeneous mixtures at www.buzzle.com/articles/homog...-examples.html
Supplemental Resources
Learn More
• David Bodanis, E = mc$^2$: A Biography of the World's Most Famous Question. Walker and Co., 2005
• John Emsley, Nature's Building Blocks: An A-Z Guide to the Elements. Oxford University Press, 2003.
• Nevin Katz, Elements, Compounds, and Mixtures: Middle and High School (Mr. Birdley Teaches Science). Incentive Publications, 2007.
• The Science of Macaroni Salad: What's in a Mixture? http://youtu.be/Vt7IN4QPU0k
• Heterogeneous mixtures antoine.frostburg.edu/chem/se...ogeneous.shtml
• Element Flash Cards http://education.jlab.org/elementflashcards/
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/02%3A_Elements_and_Ions/2.02%3A_Matter.txt |
Learning Outcomes
• Define mole.
• Determine the ratio of elements in a compound from a formula.
• Determine molar mass of an element and compound.
• Convert among mass, moles, and number of particles of a substance.
When you go to the bakery and order a dozen doughnuts, both you and the baker know that means 12. Dozen is a counting number that is defined as 12. There are other counting numbers such as pair (2), ream (500), gross (144), and score (20). The last one was famously used by Abraham Lincoln in the Gettysburg Address when he said "Four score and seven years ago". All of these words define a specific number of things regardless of what it is we are counting. A mole is also a counting number.
Avogadro's Number
It certainly is easy to count bananas or to count elephants (as long as you stay out of their way). However, you would be counting grains of sugar from your sugar canister for a long. long time. Atoms and molecules are extremely small - far, far smaller than grains of sugar. Counting atoms and molecules is not only unwise, it is absolutely impossible. One drop of water contains about $10^{22}$ molecules of water. If you counted 10 molecules every second for 50 years without stopping you would have counted only $1.6 \times 10^{10}$ molecules. Put another way, at that counting rate, it would take you over 30 trillion years to count the water molecules in one tiny drop.
Chemists needed a name that can stand for a very large number of items. Amedeo Avogadro (1776 - 1856), an Italian scientist, provided just such a number. He is responsible for the counting unit of measure called the mole. A mole (abbreviated as $\text{mol}$) is the amount of a substance that contains $6.02 \times 10^{23}$ representative particles of that substance. The mole is the SI unit for amount of a substance. Just like dozen or gross, it is a name that stands for a number. There are $6.02 \times 10^{23}$ atoms in a mole of carbon. There are $6.02 \times 10^{23}$ water molecules in a mole of water molecules. There also would be $6.02 \times 10^{23}$ bananas in a mole of bananas, if such a huge number of bananas ever existed.
The number $6.02 \times 10^{23}$ is called Avogadro's number, the number of representative particles in a mole. It is an experimentally determined number. A representative particle is the smallest unit in which a substance naturally exists. For the majority of elements, the representative particle is the atom. Iron, carbon, and helium consist of iron atoms, carbon atoms, and helium atoms, respectively. Seven elements exist in nature as diatomic molecules and the are $\ce{H_2}$, $\ce{N_2}$, $\ce{O_2}$, $\ce{F_2}$, $\ce{Cl_2}$, $\ce{Br_2}$, and $\ce{I_2}$. The representative particle for these elements is the molecule. Likewise, all molecular compounds such as $\ce{H_2O}$ and $\ce{CO_2}$ exist as molecules and so the molecule is their representative particle. For ionic compounds such as $\ce{NaCl}$ and $\ce{Ca(NO_3)_2}$, the representative particle is the formula unit. A mole of any substance contains Avogadro's number $\left( 6.02 \times 10^{23} \right)$ of representative particles.
Conversions Between Moles and Number of Particles
Using our unit conversion techniques, we can use the mole label to convert back and forth between the number of particles and moles.
Example $1$
The element carbon exists in two primary forms: graphite and diamond. How many moles of carbon atoms is $4.72 \times 10^{24}$ atoms of carbon?
Solution
In this problem, we are given the number of atoms and asked to provide the number of moles. We need to determine what factor we will use to convert from atoms to moles. Avogadro's number gives us the relationship we need, $1 \: \text{mol} = 6.02 \times 10^{23}$ atoms.
$4.72 \times 10^{24} \: \text{atoms} \: \ce{C} \times \frac{1 \: \text{mol} \: \ce{C}}{6.02 \times 10^{23} \: \text{atoms} \: \ce{C}} = 7.84 \: \text{mol} \: \text{C}$
Notice that atoms is on the bottom of the conversion factor because we need atoms to cancel so we are left with moles of carbon. Our final check is to make sure our answer is reasonable. Notice that we have $10^{24}$ atoms of carbon so it is reasonable that the value is greater than 1 mole.
Suppose that you wanted to know how many hydrogen atoms were in a mole of water molecules. First, you would need to know the chemical formula for water, which is $\ce{H_2O}$. Based on the subscript for hydrogen, there are two atoms of hydrogen in each molecule of water. How many atoms of hydrogen would there be in two water molecules? (see figure below) There would be $2 \times 2 = 4$ hydrogen atoms. How about in a dozen? In that case a dozen is 12 so $12 \times 2 = 24$ hydrogen atoms in a dozen water molecules. To get the answers (4 and 24), you had to multiply the given number of molecules by two atoms of hydrogen per molecule. So to find the number of hydrogen atoms in a mole of water molecules, the problem could be solved in a similar manner.
$1 \: \text{mol} \: \ce{H_2O} \times \frac{6.02 \times 10^{23} \: \text{molecules} \: \ce{H_2O}}{1 \: \text{mol} \: \ce{H_2O}} \times \frac{2 \: \text{atoms} \: \ce{H}}{1 \: \text{molecule} \: \ce{H_2O}} = 1.20 \times 10^{24} \: \text{atoms} \: \ce{H}$
The first conversion factor converts from moles of particles to the number of particles. The second conversion factor reflects the number of atoms contained within each molecule.
Two water molecules contain 4 hydrogen atoms and 2 oxygen atoms. A mole of water molecules contains 2 moles of hydrogen atoms and 1 mole of oxygen atoms.
Example $2$
Sulfuric acid has the chemical formula $\ce{H_2SO_4}$. A certain quantity of sulfuric acid contains $4.89 \times 10^{25}$ atoms of oxygen. How many moles of sulfuric acid is in the sample?
Solution
In this problem, the number of atoms of oxygen is given. However, the number of molecules of $\ce{H_2SO_4}$ is not equal to the number of atoms. Therefore, an additional step will be need to solve the problem.
We can write two relationships that will help solve this problem.
• $1 \: \text{mol} = 6.02 \times 10^{23} \: \text{molecules}$
• $1 \: \text{molecule} \: \ce{H_2SO_4} = 4 \: \ce{O} \: \ce{atoms}$
While 1 mole will always equal $6.02 \times 10^{23}$, the second relationship will vary depending on the identity of the compound and the element. For $\ce{H_2SO_4}$, we could also write that 1 molecule of $\ce{H_2SO_4} =$ 2 atoms $\ce{H}$ or 1 molecule $\ce{H_2SO_4} =$ 1 atom $\ce{S}$. Both of these are true, but they just aren't helpful for this problem.
$4.89 \times 10^{25} \: \text{atoms} \: \ce{O} \times \frac{1 \: \text{molecule} \: \ce{H_2SO_4}}{4 \: \text{atoms} \: \ce{O}} \times \frac{1 \: \text{mol} \: \ce{H_2SO_4}}{6.02 \times 10^{23} \: \text{molecules} \: \ce{H_2SO_4}} = 20.3 \: \text{mol} \: \ce{H_2SO_4}$
Note that atoms $\ce{O}$ and molecules $\ce{H_2SO_4}$ both cancel and we are left with units of $\text{mol} \: \ce{H_2SO_4}$.
An alternative way to set up this problem is to use the relationship $1 \: \text{mol} \: \ce{H_2SO_4} = 4 \: \text{mol} \: \ce{O} \: \text{atoms}$. Since a mole is just a counting number, we are simply multiplying both values (1 molecule and 1 atoms) by Avogadro's number to get a relationship in terms of moles. In this example, we convert from atoms to moles of $\ce{O}$, then convert from moles of $\ce{O}$ to moles of $\ce{H_2SO_4}$.
$4.89 \times 10^{25} \: \text{atoms} \: \ce{O} \times \frac{1 \: \text{mol} \: \ce{O}}{6.02 \times 10^{23} \: \text{atoms} \: \ce{O}} \times \frac{1 \: \text{mol} \: \ce{H_2SO_4}}{4 \: \text{mol} \: \ce{O}} = 20.3 \: \text{mol} \: \ce{H_2SO_4}$
Both methods are correct and will give you the same answer. Use the method that makes the most sense to you.
Molar Mass
Molar mass is defined as the mass (in grams) of one mole of representative particles of a substance. By looking at a periodic table, we can conclude that the molar mass of lithium is $6.94 \: \text{g}$, the molar mass of zinc is $65.38 \: \text{g}$, and the molar mass of gold is $196.97 \: \text{g}$. Each of these quantities contains $6.02 \times 10^{23}$ atoms of that particular element. The units for molar mass are grams per mole or $\text{g/mol}$. Notice that these are the same numbers as the atomic mass but with different units. For example 1 atom of lithium has a mass of $6.94 \: \text{amu}$ while 1 mole $\left( 6.02 \times 10^{23} \: \text{atoms} \right)$ has a mass of $6.94 \: \text{g}$. We can use this relationship to determine the moles of an element from its mass or vice versa.
Example $3$
How many moles of carbon are in a $29.3 \: \text{g}$ sample?
Solution
We are given the mass of the sample and look up the molar mass of carbon in the periodic table which is $12.01 \: \text{g/mol}$. Now, we can set up the calculation to solve for moles.
$29.3 \: \text{g} \: \ce{C} \times \frac{1 \: \text{mol} \: \ce{C}}{12.01 \: \text{g} \: \ce{C}} = 2.44 \: \text{mol} \: \ce{C}$
Note that grams will cancel and we will be left with units of moles of carbon for our answer which is what is being asked for in this problem. Since the mass of the carbon sample is greater than the molar mass, it is reasonable that we have more than one mole of the carbon.
Example $4$
What is the mass of aluminum in a $1.95 \: \text{mol}$ sample?
Solution
As in the previous example, we will need the molar mass to solve the problem. For aluminum, the molar mass is $27.0 \: \text{g/mol}$. Now, we can set up the calculation to solve for mass.
$1.95 \: \text{mol} \: \ce{Al} \times \frac{27.0 \: \text{g} \: \ce{Al}}{1 \: \text{mol} \: \ce{Al}} = 52.7 \: \text{g} \: \ce{Al}$
In this example, we put gras on top and moles on the bottom so that the units would cancel correctly. Note that $\text{mol} \: \ce{Al}$ will cancel and we are left with units of grams of aluminum which is what is being asked for in the problem. The answer looks reasonable because we have almost two moles of aluminum and the mass is almost twice the molar mass.
Molar Masses of Compounds
The molecular formula of the compound carbon dioxide is $\ce{CO_2}$. One molecule of carbon dioxide consists of 1 atom of carbon and 2 atoms of oxygen. We can calculate the mass of one molecule of carbon dioxide by adding together the masses of 1 atom of carbon and 2 atoms of oxygen.
$12.01 \: \text{amu} + 2 \left( 16.00 \: \text{amu} \right) = 44.01 \: \text{amu}$
The molecular mass of a compound is the mass of one molecule of that compound. The molecular mass of carbon dioxide is $44.01 \: \text{amu}$.
The molar mass of any compound is the mass in grams of one mole of that compound. One mole of carbon dioxide molecules has a mass of $44.01 \: \text{g}$. The molar mass is $44.01 \: \text{g/mol}$ for $\ce{CO_2}$. For water, the molar mass is $18.02 \: \text{g/mol}$. In both cases, it is the mass of $6.02 \times 10^{23}$ molecules.
Example $5$
Calcium nitrate, $\ce{Ca(NO_3)_2}$, is used as a component in fertilizer. Determine the molar mass of calcium nitrate.
Solution
The molar mass of a compound is found from the molar masses of the elements in the compound.
• $\ce{Ca} = 40.08 \: \text{g/mol}$
• $\ce{N} = 14.01 \: \text{g/mol}$
• $\ce{O} = 16.00 \: \text{g/mol}$
First we need to analyze the formula. Since the $\ce{Ca}$ lacks a subscript, there is one $\ce{Ca}$ atom per formula unit. The 2 outside the parentheses means that there are two nitrate ions per formula unit and each nitrate ion consist of one nitrogen atom and three oxygen atoms. Therefore, there are a total of $1 \times 2 = 2$ nitrogen atoms and $3 \times 2 = 6$ oxygen atoms per formula unit. Thus, $1 \: \text{mol}$ of calcium nitrate contains $1 \: \text{mol}$ of $\ce{Ca}$ atoms, $2 \: \text{mol}$ of $\ce{N}$ atoms, and $6 \: \text{mol}$ of $\ce{O}$ atoms.
Use the molar masses of each atom together with the number of atoms in the formula and add together.
\begin{align} &1 \: \text{mol} \: \ce{Ca} \times \frac{40.08 \: \text{g} \: \ce{Ca}}{1 \: \text{mol} \: \ce{Ca}} = 40.08 \: \text{g} \: \ce{Ca} \ &2 \: \text{mol} \: \ce{N} \times \frac{14.01 \: \text{g} \: \ce{N}}{1 \: \text{mol} \: \ce{N}} = 28.02 \: \text{g} \: \ce{N} \ &6 \: \text{mol} \: \ce{O} \times \frac{16.00 \: \text{g} \: \ce{O}}{1 \: \text{mol} \: \ce{O}} = 96.00 \: \text{g} \: \ce{O} \ &\text{molar mass of} \ce{Ca(NO_3)_2} = 40.08 \: \text{g} + 28.02 \: \text{g} + 96.00 \: \text{g} = 164.10 \: \text{g/mol} \end{align}
Conversions Between Moles and Mass
Like we converted between moles and mass of an element, we can also convert between moles and mass of a compound using the molar mass of the compound. This relationship is frequently used in the laboratory. Suppose that for a certain experiment you need 3.00 moles of calcium chloride $\left( \ce{CaCl_2} \right)$. Since calcium chloride is a solid, it would be convenient to use a balance to measure the mass that is needed. The molar mass of $\ce{CaCl_2}$ is $110.98 \: \text{g/mol}$. The relationship that can be used is then based on the equality that $1 \: \text{mol} = 110.98 \: \text{g} \: \ce{CaCl_2}$. Dimensional analysis will allow you to calculate the mass of $\ce{CaCl_2}$ that you should measure.
$3.00 \: \text{mol} \: \ce{CaCl_2} \times \frac{110.98 \: \text{g} \: \ce{CaCl_2}}{1 \: \text{mol} \: \ce{CaCl_2}} = 333 \: \text{g} \: \ce{CaCl_2}$
When you measure the mass of $333 \: \text{g}$ of $\ce{CaCl_2}$, you are measuring 3.00 moles of $\ce{CaCl_2}$.
Calcium chloride is used as a drying agent and as a road deicer.
Example $6$
A certain reaction produces $2.81 \: \text{g}$ of copper (II) hydroxide, $\ce{Cu(OH)_2}$. Determine the number of moles produced in the reaction.
Solution
The mass of the substance is known and we can determine the moles using the molar mass of the compound, $\ce{Cu(OH)_2}$. First we have to find the molar mass of $\ce{Cu(OH)_2}$ using the molar masses of the elements.
$\text{molar mass} \: \ce{Cu(OH)_2} = 63.55 \: \text{g/mol} + 2 \left( 16.00 \: \text{g/mol} \right) + 2 \left( 1.008 \: \text{g/mol} \right) = 97.57 \: \text{g/mol}$
Now, we can use the molar mass to convert from grams to moles.
$2.81 \: \text{g} \: \ce{Cu(OH)_2} \times \frac{1 \: \text{mol} \: \ce{Cu(OH)_2}}{97.57 \: \text{g} \: \ce{Cu(OH)_2}} = 0.0288 \: \text{mol} \: \ce{Cu(OH)_2}$
The units of grams will cancel and we are left with units of moles which is what is being requested in the question.
Example $7$
What is the mass of water in a $3.50 \: \text{mol}$ sample?
Solution
In this example, the moles of water is known and we need to find the mass. Again, we will use the molar mass which was previously given as $18.02 \: \text{g/mol}$.
$3.50 \: \text{mol} \: \ce{H_2O} \times \frac{18.02 \: \text{g} \: \ce{H_2O}}{1 \: \text{mol} \: \ce{H_2O}} = 63.1 \: \text{g} \: \ce{H_2O}$
The units of $\text{mol} \: \ce{H_2O}$ will cancel and we are left with units of grams which is what is being requested in the question.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/02%3A_Elements_and_Ions/2.03%3A_Mole_and_Molar_Mass.txt |
Learning Outcomes
• Determine the energy levels of electrons for the first 20 elements.
• Explain the relevance of valence electrons in chemical processes.
• Identify the number of valence electrons in an element.
• Describe the stability of an atom as a result of following the octet rule.
The structure of the atom was discussed in the previous unit and now we will focus on the role that electrons play in the formation of compounds. Regardless of the type of compound or the number of atoms or electrons involved, it is the electrons of those atoms that interact to form a compound.
Electron Arrangement
Electrons are not randomly arranged in an atom and their position within the atom can be described using electron arrangements, which are a simplified version of electron configurations. For each element of interest, we look at the number of electrons in a single atom and then determine how those electrons are arranged based on the atomic model. The main idea behind electron arrangements is that electrons can only exist at certain energy levels. By understanding the energy levels of electrons in an atom, we can predict properties and understand behavior of the atom.
As shown in the figure below, there are multiple energy levels where electrons can be found. As the energy level increases, the energy difference between them decreases. A maximum of two electrons can be found in the $n=1$ level; eight electrons can be in the $n=2$ level. Although the $n=3$ and $n=4$ levels show only eight electrons in this diagram, those energy levels can hold more but not until we start looking at the transition metals. We will only be concerned with the electron arrangements of elements through calcium $\left( Z = 20 \right)$ so we will put a maximum of eight electrons in the $n=3$ level and two in the $n=4$ level.
Example $1$
What is the electron arrangement of oxygen?
Solution
Oxygen has eight electrons. The first two electrons will go in the $n=1$ level. Two is the maximum number of electrons for the level so the other electrons will have to go in a higher energy level. The $n=2$ level can hold up to eight electrons so the remaining six electrons will go in the $n=2$ level. The electron arrangement of oxygen is (2, 6).
Example $2$
What is the electron arrangement of chlorine?
Solution
Chlorine has 17 electrons. The first two electrons will go in the $n=1$ level. Two is the maximum number of electrons for the level so the other electrons will have to go in higher energy levels. The $n=2$ level can hold up to eight electrons so the next 8 electrons will go in the $n=2$ level. The remaining 7 electrons can go in the $n=3$ level since it holds a maximum of 8 electrons. The electron arrangement of chlorine is (2, 8, 7).
The electron arrangement also provides information about the number of valence electrons. The valence electrons are the electrons in the highest energy level and the ones involved in ion and bond formation. Knowing the number of valence electrons will allow us to predict how a particular element will interact with other elements. Electrons in lower energy levels are called the core electrons.
Let's look at the figure below which shows the electron diagram for magnesium and its 12 electrons. The first two electrons are found in the $n=1$ energy level, the next eight electrons are found in the $n=2$ level, and the remaining two electrons are found in the $n=3$ level. The electrons always fill the lowest energy levels available until that level is filled, then electrons fill the next energy level until it is filled. This continues for all of the electrons in an atom. We can show the electron arrangement as (2, 8, 2) representing the electrons in the $n=1$, $n=2$, and $n=3$ levels, respectively.
The electron arrangement also shows the number of valence electrons which is two for magnesium because there are two electrons in the $n=3$ energy level which is the highest occupied energy level for magnesium. This corresponds to the $2+$ charge formed when magnesium forms an ion. It is willing to lose 2 electrons so that it has the same electron arrangements as the nearest noble gas, which is neon (2, 8). Atoms will gain or lose electrons to look like the nearest noble gas because the noble gases are unreactive due to the stability of having eight electrons in the highest energy level. This desire of atoms to have eight electrons in their outermost shell is known as the octet rule.
Example $3$
What is the electron arrangement of aluminum? How many valence electrons does it have?
Solution
Aluminum has 13 electrons so it will have the electron arrangement (2, 8, 3) which represents two electrons in the $n=1$ energy level, eight electrons in the $n=2$ level, and three electrons in the $n=3$ level. Aluminum has three valence electrons (indicated by the three electrons in the $n=3$ level).
Example $4$
How many valence electrons does chlorine have? How many electrons will chlorine gain or lose to form an ion?
Solution
Chlorine has 7 electrons in its valence shell. To meet the octet rule, it must either gain one electron or lose seven electrons. Gaining one is easier than losing seven so it will gain one electron to have a total of eight electrons when it forms an ion (i.e. charged particle).
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/02%3A_Elements_and_Ions/2.04%3A_Electron_Arrangements.txt |
Learning Outcomes
• Define ion, cation, and anion.
• Recognize characteristics of monatomic and polyatomic ions.
• Explain how and why cations and anions are formed.
• Determine the number of subatomic particles in an ion.
• Predict the charge on ions formed from representative elements.
In many parts of the country, including parts of Kentucky, the water contains high concentrations of minerals that stain clothes, build up deposits on bathtubs and water heaters, and create problems with soap foaming properly. This problem is caused by what is called "hard water". The water contains excessive amounts of cations such as iron and calcium.
Ions
Ions are charged substances that have formed through the gain or loss of electrons. Cations form from the loss of electrons and have a positive charge while anions form through the gain of electrons and have a negative charge.
Cation Formation
Cations are the positive ions formed by the loss of one or more electrons. The most commonly formed cations of the representative elements are those that involve the loss of all of the valence electrons. Consider the alkali metal sodium $\left( \ce{Na} \right)$. It has one valence electron in the $n=3$ energy level. Upon losing that electron, the sodiu ion now has an octet of electrons from the second energy level and a charge of $1+$.
The electron arrangement of the sodium ion is now the same as that of the noble gas neon. Consider a similar process with magnesium and aluminum. In this case, the magnesium atom loses its two valence electrons in order to achieve the same arrangement as the noble gas neon and a charge of $2+$. The aluminum atom loses its three valence electrons to have the same electron arrangement as neon and a charge of $3+$. For representative elements under typical conditions, three electrons is usually the maximum number that will be los. Representative elements will not lose electrons beyond their valence because they would have to "break" the octet of the previous energy level which provides stability to the ion.
Anions
Anions are the negative ions formed from the gain of one or more electrons. When nonmetal atoms gain elections, they often do so until their outermost principal energy level achieves an octet. For fluorine, which has an electron arrangement of (2, 7), it only needs to gain one electron to have the same electron arrangement as neon. Forming an octet (eight electrons in the outer shell) provides stability to the atom. Fluorine will gain one electron and have a charge of $1-$. The electron arrangement of the fluoride ion (2, 8) will also change to reflect the gain of an electron.
Oxygen has an electron arrangement of (2, 6) and needs to gain two electrons to fill the $n=2$ energy level and achieve an octet of electrons in the outermost shell. The oxide ion will have a charge of $2-$ as a result of gaining two electrons. Under typical conditions, three electrons is the maximum that will be gained in the formation of anions.
Subatomic Particles in an Ion
Since ions form from the gain or loss of electrons, we can also look at the number of subatomic particles (protons, neutrons, and electrons) found in an ion. Remember that the number of protons determines the identity of the element and will not change in a chemical process.
Example $1$
How many protons, neutrons, and electrons in a single oxide $\left( \ce{O^{2-}} \right)$ ion?
Solution
Oxygen has the atomic number 8 so both the atom and the ion will have 8 protons. The average atomic mass of oxygen is 16. Therefore, there will be 8 neutrons $\left( \text{atomic mass} - \text{atomic number} = \text{neutrons} \right)$. A neutral oxygen atom would have 8 electrons. However, the anion has gained two electrons so $\ce{O^{2-}}$ has 10 electrons.
We can also use information about the subatomic particles to determine the identity of an ion.
Example $2$
An ion with a $2+$ charge has 18 electrons. Determine the identity of the ion.
Solution
If an ion has a $2+$ charge then it must have lost electrons to form the cation. If the ion has 18 electrons and the atom lost 2 to form the ion, then the neutral atom contained 20 electrons. Since it was neutral, it must also have had 20 protons. Therefore the element is calcium.
Polyatomic Ions
A polyatomic ion is an ion composed of two or more atoms that have a charge as a group (poly = many). The ammonium ion (see figure below) consists of one nitrogen atom and four hydrogen atoms. Together, they comprise a single ion with a $1+$ charge and a formula of $\ce{NH_4^+}$. The hydroxide ion (see figure below) contains one hydrogen atom and one oxygen atom with an overall charge of $1-$. The carbonate ion (see figure below) consists of one carbon atom and three oxygen atoms and carries an overall charge of $2-$. The formula of the carbonate ion is $\ce{CO_3^{2-}}$. The atoms of a polyatomic ion are tightly bonded together and so the entire ion behaves as a single unit. The figures below show several examples.
The table below lists a number of polyatomic ions by name and by structure. The heading for each column indicates the charge on the polyatomic ions in that group. Note that the vast majority of the ions listed are anions - there are very few polyatomic cations.
$1-$ $2-$ $3-$ $1+$
Table $1$: Common Polyatomic Ions
acetate, $\ce{CH_3COO^-}$ carbonate, $\ce{CO_3^{2-}}$ arsenate, $\ce{AsO_3^{3-}}$ ammonium, $\ce{NH_4^+}$
bromate, $\ce{BrO_3^-}$ chromate, $\ce{CrO_4^{2-}}$ phosphite, $\ce{PO_3^{3-}}$
chlorate, $\ce{ClO_3^-}$ dichromate, $\ce{Cr_2O_7^{2-}}$ phosphate, $\ce{PO_4^{3-}}$
chlorite, $\ce{ClO_2^-}$ hydrogen phosphate, $\ce{HPO_4^{2-}}$
cyanide, $\ce{CN^-}$ oxalate, $\ce{C_2O_4^{2-}}$
dihydrogen phosphate, $\ce{H_2PO_4^-}$ peroxide, $\ce{O_2^{2-}}$
hydrogen carbonate, $\ce{HCO_3^-}$ silicate, $\ce{SiO_3^{2-}}$
hydrogen sulfate, $\ce{HSO_4^-}$ sulfate, $\ce{SO_4^{2-}}$
hydrogen sulfide, $\ce{HS^-}$ sulfite, $\ce{SO_3^{2-}}$
hydroxide, $\ce{OH^-}$
hypochlorite, $\ce{ClO^-}$
nitrate, $\ce{NO_3^-}$
nitrite, $\ce{NO_2^-}$
perchlorate, $\ce{ClO_4^-}$
permanganate, $\ce{MnO_4^-}$
The vast majority of polyatomic ions are anions, many of which end in -ate or -ite. Notice that in some cases such as nitrate $\left( \ce{NO_3^-} \right)$ and nitrite $\left( \ce{NO_2^-} \right)$, there are multiple anions that consist of the same two elements. In these cases, the difference between the ions is the number of oxygen atoms present, while the overall charge is the same. As a class, these are called oxyanions. When there are two oxyanions for a particular element, the one with the greater number of oxygen atoms gets the -ate suffix, while the one with the fewer number of oxygen atoms gets the -ite suffix. The four oxyanions of chlorine are shown below, which also includes the use of the prefixes hypo- and per-.
• $\ce{ClO^-}$, hypochlorite
• $\ce{ClO_2^-}$, chlorite
• $\ce{ClO_3^-}$, chlorate
• $\ce{ClO_4^-}$, perchlorate
Not your usual ion
"Drink you milk. It's good for your bones." We're told this from early childhood, and with good reason. Milk contains a good supply of calcium, part of the structure of bone. However, there are two other ionic components of hydroxyapatite, the mineral component. Phosphate ion and hydroxide ion make up the remainder of the inorganic material in bone.
News You Can Use
• Bone is a very complex structure. It is composed of protein (mainly collagen), hydroxyapatite (a calcium-phosphate-hydroxide mixture), some other minerals, and contains $10$-$20\%$ water. The calcium/phosphate ratios are not stoichiometric, but vary somewhat from one portion of bone to the next.
• Bones are very strong but will break under enough stress. Regular exercise and proper nutrition help to increase bone strength. Watch a video about bone structure at http://www.youtube.com/watch?v=d9owEvYdouk
• Nitrate is an anion with a complex bonding structure. Major sources for this ion in drinking water are runoff from fertilizer, septic tank leakage, sewage, and natural deposits. High concentrations of nitrates represent a significant health hazard, especially to infants. The nitrate in the body is converted to nitrite, which then binds to hemoglobin. This binding decreases the ability of hemoglobin to transport oxygen, thus depriving the cells of the $\ce{O_2}$ needed for proper functioning.
• Cyanide production is widespread throughout nature. Forest fires will produce significant amounts of cyanide. Many plants contain cyanide, and it is produced by a number of bacteria, algae, and fungi. Cyanide is used industrially in metal finishing, iron and steel mills, and in organic synthesis processes. This material is also an important component for the refining of precious metals. Formation of a complex between cyanide and gold allows extraction of this metal from a mixture.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/02%3A_Elements_and_Ions/2.05%3A_Ion_Formation.txt |
Learning Outcomes
• Describe the types of elements that form an ionic bond.
• Explain how an ionic bond is formed.
• Predict the formula of an ionic compound.
Chemistry can be broadly divided into tow main classes based on the identity of the elements present in the chemical compounds. Organic chemistry is the branch of chemistry that deals with compounds containing carbon. In this section and the following one, we will be discussing inorganic chemistry, which is the branch of chemistry dealing with compounds that do not contain carbon. Ionic compounds are examples of inorganic compounds.
Types of Formulas
Recall that a molecules includes two or more atoms that have been chemically combined. A chemical formula that indicates how many of each type of atom are present in a single molecule is referred to as a molecular formula. For example, a molecule of ammonia contains one nitrogen atom and three hydrogen atoms, so it has the following molecular formula.
Another type of chemical formula, the empirical formula, shows the elements in a compound in their lowest whole-number ratio. Glucose is an important simple sugar that cells use as their primary source of energy. Its molecular formula is $\ce{C_6H_{12}O_6}$. Since each of the subscripts is divisible by 6, the empirical formula for glucose is $\ce{CH_2O}$. When chemists analyze an unknown compound, often the first step is to determine its empirical formula. There are a great many compounds whose molecular and empirical formulas are the same. If the molecular formula cannot be simplified into a smaller whole-number ratio, as in the case of $\ce{H_2O}$ or $\ce{P_2O_5}$, then the empirical formula is also the molecular formula.
Unlike molecular compounds, ionic compounds are quite different. Water and other molecules exist as collections of individual molecules (see figure below) while ionic compounds do not exist as discrete molecular units. Instead,an ionic compound consists of a large three-dimensional array of alternating cations and anions. For example, sodium chloride $\left( \ce{NaCl} \right)$ is composed of many $\ce{Na^+}$ and $\ce{Cl^-}$ ions arranged into a structure like the one pictured (see figure below). The formula for ionic compounds is always an empirical formula because it shows the smallest, whole-number ratio between the cations and anions. The actual number of ions in a sample of an ionic compound will be very large and will vary from sample to sample.
The most straightforward way to describe this structure with a chemical formula is to give the lowest whole-number ratio between the two ions. In the case of $\ce{NaCl}$, there are equal numbers of sodium ions and chloride ions in the salt crystal. In contrast, a crystal of magnesium chloride has twice as many chloride ions as magnesium ions, so it has a formula of $\ce{MgCl_2}$.
Writing Formulas for Binary Ionic Compounds
If you know the elements that form a binary ionic compound, you can write its formula. Start by writing the metal ion and its charge, followed by the nonmetal ion with its charge. Because the overall compound must be electrically neutral, decide how many ions of each type are needed in order for the positive and negative charges to cancel each other out. Consider the compound aluminum nitride. The charges on each of these ions can be determined by looking at the groups in which aluminum and nitrogen are found on the periodic table. The ions are:
$\ce{Al^{3+}} \: \: \: \: \: \: \: \: \: \: \ce{N^{3-}}$
Since the ions have charges that are equal in magnitude (3, but different signs), 1:1 is the lowest ratio of ions that will produce a neutral compound. Since the charge on aluminum is 3+ and the charge on nitrogen is $3-$, the sum of their charges is zero ($+3\;+\; -3=0$). As a result, the formula of aluminum nitride is $\ce{AlN}$. Another compound, lithium oxide, contains the following ions:
$\ce{Li^+} \: \: \: \: \: \: \: \: \: \: \ce{O^{2-}}$
In this case, two lithium ions, each with a 1+ charge, are required to balance out the charge of each oxide ion, which has a $2-$ charge. The formula of lithium oxide is $\ce{Li_2O}$ because the compound must be neutral. Therefore, $\left( 2\times+1\right) \;+\left( 1\times -2\right)=0$.
For compounds in which the ratio of ions is not as obvious, an alternative way to determine the correct formula is to use the "crisscross" method. In this method, the numerical value of each charge crosses over to become the subscript of the opposite ion. The signs of the charges are dropped. The crisscross method is demonstrated below for aluminum oxide.
The red arrows indicate that the 3 from the $3+$ charge will cross over to become the subscript for $\ce{O}$, while the 2 from the $2-$ charge will cross over to become the subscript for $\ce{Al}$. The formula for aluminum oxide is $\ce{Al_2O_3}$.
For aluminum oxide, the crisscross method directly produces the correct formula, but in some cases, another step is required. Because ionic compounds are always described by their empirical formulas, they must be written as the lowest whole-number ratio of the ions. In the case of aluminum nitride, the crisscross method would yield a formula of $\ce{Al_3N_3}$, which is not correct. A second step must be performed in which the subscripts are reduced but the ratio is kept the same. $\ce{Al_3N_3}$ can be reduced to $\ce{AlN}$, because both formulas describe a 1:1 ratio of aluminum ions to nitride ions. Following the crisscross method to write the formula for lead (IV) oxide would involve the following steps:
Some transition metals can have more than one possible charge. When this happens, the charge on the transition metal cation is included in parentheses in the name. For example, lead(IV) oxide has $\ce{Pb^{4+}}$ as its metal cation.
The crisscross method first yields $\ce{Pb_2O_4}$ for the formula, but that must be reduced to $\ce{PbO_2}$, which is the correct formula.
Writing Formulas for Ionic Compounds Containing Polyatomic Ions
Writing a formula for an ionic compound with polyatomic ions involves the same steps as for a binary (two element) ionic compound. Write the symbol and charge of the cation followed by the symbol and charge of the anion. Use the crisscross method to ensure that the final formula is neutral. Calcium nitrate is composed of calcium cations and nitrate anions.
The charge is balanced by the presence of two nitrate ions and one calcium ion. Parentheses are used around the nitrate ion because more than one polyatomic ion is needed. If only one polyatomic ion is present in a formula, parentheses are not used. For example, the formula for calcium carbonate is $\ce{CaCO_3}$. The carbonate ion carries a $2-$ charge, so it exactly balances the $2+$ charge of the calcium ion and parentheses are not needed around the polyatomic ion.
Example $1$
Write the formula for zinc phosphate which is composed of $Zn^{2+}$ and $PO_4^{3-}$ ions.
Solution
Write the metal cation followed by the nonmetal anion. Crisscross the ion charges in order to make the ionic compound neutral. Use parentheses around the polyatomic ion if more than one is present in the final formula. Reduce to the lowest ratio if necessary.
The formula for zinc phosphate is $\ce{Zn_3(PO_4)_2}$. Three zinc cations with $2+$ charges balance out two phosphate anions with $3-$ charges.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/02%3A_Elements_and_Ions/2.06%3A_Ionic_Compounds.txt |
These are homework exercises to accompany Chapter 2 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Solutions are available below the questions.
Questions
(click here for solutions)
Q2.1.1
Give the names and symbols of three elements.
Q2.1.2
Describe where protons, neutrons, and electrons are located in an atom.
Q2.1.3
Why are all atoms electrically neutral?
Q2.1.4
How many protons are in the nucleus of each of the following atoms?
1. neon (Ne)
2. gold (Au)
3. strontium (Sr)
4. uranium (U)
Q2.1.5
A certain atom contains 28 protons, 28 electrons, and 31 neutrons. Provide the following:
1. atomic number
2. mass number
3. name of element
Q2.1.6
How many protons, neutrons, and electrons are in an atom of cesium-133?
Q2.1.7
How many protons, neutrons, and electrons are there in the atom $\ce{^{19}_9F}$?
Q2.1.8
How many protons, neutrons, and electrons are there in an atom of lead-207?
Q2.1.9
A certain atom has an atomic number of 36 and a mass number of 84. Write out the designation for this isotope in both nuclide symbol form and in hyphenated form.
Q2.1.10
An atom has a mass number of 59 and contains 32 neutrons in its nucleus. What element is it?
Q2.1.11
Complete the Table below:
Isotope Isotope Symbol Atomic Number Mass Number
sodium-23
$\ce{^{75}_{33}As}$
silver-108
Q2.1.12
Which one is an isotope of $\ce{^{40}_{18}Ar}$? Explain.
1. $\ce{^{40}_{20}Ca}$
2. $\ce{^{39}_{18}Ar}$
3. $\ce{^{40}_{18}Ar}$
Q2.1.13
Fill in Table below:
Isotope Number of Protons Number of Electrons Number of Neutrons Isotope Symbol
hydrogen-1
hydrogen-2
beryllium-9
aluminum-27
Q2.1.14
Fill in Table below:
Element Symbol Atomic Number Mass Number # of Protons # of Electrons # of Neutrons Isotope Symbol
nitrogen 14
B 11
30 35
77 116
$\ce{^{56}_{26}Fe}$
(click here for solutions)
Q2.2.1
Define matter and give three examples of matter.
Q2.2.2
Explain the differences between compounds and mixtures.
Q2.2.3
Explain the differences between pure substances and mixtures.
Q2.2.4
Identify each of the following as a pure substance or a mixture.
1. salt water
2. NaCl
3. brewed coffee
4. air
Q2.2.5
Label each of the following as an element, compound, homogeneous mixture, or heterogeneous mixture.
1. silicon
2. sulfuric acid
3. air
4. soda
5. sugar
6. muddy water
7. chicken noodle soup
8. scoop of sand from the beach
Q2.2.6
Identify each of the following elements as a metal, nonmetal, or metalloid.
1. iron (Fe)
2. gallium (Ga)
3. phosphorus (P)
4. potassium (K)
5. neon (Ne)
6. germanium (Ge)
Q2.2.7
Identify each of the following elements as a metal, nonmetal, or metalloid.
1. chlorine (Cl)
2. hydrogen (H)
3. antimony (Sb)
4. titanium (Ti)
5. nitrogen (N)
6. selenium (Se)
Q2.2.8
Describe how you identify molecular and ionic compounds.
Q2.2.9
Label each as an ionic or molecular compound.
1. H2O2
2. N2O5
3. PF3
4. MgCl2
5. KBr
6. AlCl3
Q2.2.10
Label each as an ionic or molecular compound.
1. CaO
2. Na2S
3. NH3
4. CO2
5. N2H4
6. N2O
(click here for solutions)
Q2.3.1
Define "counting number".
Q2.3.2
What is the value of Avogadro's number?
Q2.3.3
What seven elements exist as diatomic molecules in nature?
Q2.3.4
How many atoms of helium are present in each of the following samples?
1. 1 mole
2. 2 moles
3. 2.5 moles
4. 0.5 moles
5. 0.35 moles
Q2.3.5
How many molecules of water are present in each of the following samples?
1. 1 mole
2. 2 moles
3. 2.5 moles
4. 0.5 moles
5. 0.35 moles
Q2.3.6
Compare the answers for each part of questions 4 and 5 to one another. How do they compare? Explain why.
Q2.3.7
How many moles of silicon is 6.73 x 1025 atoms of silicon?
Q2.3.8
How many moles of sodium is 4.29 x 1022 atoms of sodium?
Q2.3.9
How many atoms of each element are in one unit of each compound?
1. H2O2
2. N2O5
3. PF3
4. MgCl2
5. KBr
6. AlCl3
7. CaO
8. Na2S
9. NH3
10. CO2
11. N2H4
12. N2O
Q2.3.10
How many moles of each element are in one mole of each compound?
1. H2O2
2. N2O5
3. PF3
4. MgCl2
5. KBr
6. AlCl3
7. CaO
8. Na2S
9. NH3
10. CO2
11. N2H4
12. N2O
Q2.3.11
How do the answers in questions 9 and 10 compare to one another? Explain the similarities or differences.
Q2.3.12
How many moles of carbon are in 0.75 moles of CCl4? How many moles of chlorine?
Q2.3.13
How many atoms of carbon are in 0.75 moles of CCl4? How many atoms of chlorine?
Q2.3.14
How many moles of hydrogen are in 2.5 moles of H2O? How many moles of oxygen?
Q2.3.15
How many atoms of hydrogen are in 2.5 moles of H2O? How many atoms of oxygen?
Q2.3.16
A sample of CaNO3 contains 3.87 x 1025 atoms of oxygen. How many molecules of CaNO3 are in the sample?
Q2.3.17
A sample of propane gas (C3H8) contains 5.39 x 1024 atoms of carbon. How many atoms of hydrogen are in the sample?
Q2.3.18
What is the molar mass of each of the following elements (in atomic form)?
1. carbon
2. nitrogen
3. sodium
4. hydrogen
5. potassium
6. phosphorus
Q2.3.19
How many moles of each element listed in the previous question are present in a 25.0 g sample of the element?
Q2.3.20
For question 19, all of the samples have the same mass. Are the moles the same? Why or why not?
Q2.3.21
What is the mass of each of the following samples?
1. 0.35 moles sodium
2. 0.75 moles carbon
3. 1.34 moles potassium
4. 1.21 moles silicon
5. 0.95 moles calcium
6. 2.85 moles helium
Q2.3.22
Determine the molar mass of each of the following compounds?
1. CO2
2. N2H4
3. CaF2
4. C6H12O6
5. CH4
6. C6H6
7. Na2SO4
8. K3PO4
9. Al(NO3)3
10. Mg3(PO4)2
Q2.3.23
Calculate the moles of each of the following samples.
1. 25.0 g CO2
2. 10.0 g N2H4
3. 85.0 g CaF2
4. 15.5 g C6H12O6
5. 20.0 g CH4
6. 100.0 g C6H6
7. 30.0 g Na2SO4
8. 75.0 g K3PO4
9. 50.0 g Al(NO3)3
10. 47.2 g Mg3(PO4)2
Q2.3.24
Calculate the mass of each of the following samples.
1. 3.5 mol CO2
2. 0.45 mol N2H4
3. 2.25 mol CaF2
4. 1.75 mol C6H12O6
5. 4.9 mol CH4
6. 8.75 mol C6H6
7. 2.35 mol Na2SO4
8. 0.672 mol K3PO4
9. 0.95 mol Al(NO3)3
10. 1.15 mol Mg3(PO4)2
(click here for solutions)
Q2.4.1
What is the electron arrangement for each of the elements?
a. Na
b. Ne
c. Be
d. N
e. S
f. Cl
Q2.4.2
How many valence electrons are in each element?
a. K
b. P
c. F
d. S
e. Li
f. B
Q2.4.3
What is the octet rule?
(click here for solutions)
Q2.5.1
Define ion.
Q2.5.2
How are anions and cation the same? Different?
Q2.5.3
What is the most common ion formed from each element?
a. Li
b. Na
c. Ca
d. B
e. P
f. S
g. Cl
h. Br
Q2.5.4
How many protons, neutrons, and electrons are present in the ions indicated in the previous question?
Q2.5.5
Identify the following elements.
a. An ion with a 3+ charge and two electrons.
b. An ion with a 1$-$ charge and 18 electrons.
c. An ion with a 1+ charge and 18 electrons.
d. An ion with a 3$-$ charge and 10 electrons.
Q2.5.6
Describe a polyatomic ion.
Q2.5.7
Which are polyatomic ions?
1. NO3
2. O2–
3. NH4+
4. Mg2+
5. Na+
6. O22–
(click here for solutions)
Q2.6.1
What element is present in all organic compounds?
Q2.6.2
Give three examples of metallic elements.
Q2.6.3
Give three examples of nonmetallic elements.
Q2.6.4
What types of elements form an ionic compound?
Q2.6.5
How do the electrons behave in the formation of an ionic bond?
Q2.6.6
What is the overall charge of an ionic compound?
Q2.6.7
What is the formula for the ionic compound formed from each of the following pairs?
a. potassium and sulfur
b. silver and chlorine (silver has a 1+ charge)
c. calcium and oxygen
d. aluminum and iodine
e. barium and nitrogen
f. sodium and phosphorus
g. lithium and fluorine
h. magnesium and nitrogen
i. calcium and sulfur
j. beryllium and bromine
k. zinc and nitrogen (zinc has a 2+ charge)
l. tin and iodine (tin has a 4+ charge)
Q2.6.8
Write the formula for the compound formed between sodium and each of these polyatomic ions. You can look up the formula and charge for each polyatomic ion.
a. carbonate
b. chlorate
c. chlorite
d. phosphate
e. nitrate
f. sulfate
g. chromate
h. dichromate
Q2.6.9
Write the formula for the compound formed between magnesium and each of the polyatomic ions listed in the previous question.
Q2.6.10
Explain when parentheses should and should not be used in the formulas of ionic compounds.
Answers
2.1: Isotopes and Atomic Mass
Q2.1.1
Answers will vary.
Q2.1.2
Protons and neutrons are in the nucleus and electrons are located outside the nucleus.
Q2.1.3
The sum of the charges on ions in an ionic compound must equal zero.
Q2.1.4
1. 10
2. 79
3. 38
4. 92
Q2.1.5
A certain atom contains 28 protons, 28 electrons, and 31 neutrons. Provide the following:
1. 28
2. 59
3. nickel
Q2.1.6
55 protons, 78 neutrons, 55 electrons
Q2.1.7
9 protons, 10 neutrons, 9 electrons
Q2.1.8
82 protons, 125 neutrons, 82 electrons
Q2.1.9
$\ce{^{84}_{36}Kr}$, krypton-84
Q2.1.10
cobalt
Q2.1.11
Isotope Isotope Symbol Atomic Number Mass Number
sodium-23 $\ce{^{23}_{11}Na}$ 11 23
aresenic-75 $\ce{^{75}_{33}As}$ 33 75
silver-108 $\ce{^{108}_{47}Ag}$ 47 108
Q2.1.12
1. $\ce{^{40}_{20}Ca}$ - not an isotope because it is a different element
2. $\ce{^{39}_{18}Ar}$ - isotope because it has the same atomic number but a different atomic mass
3. $\ce{^{40}_{18}Ar}$ - not an isotope because it has the same atomic number and the same atomic mass
Q2.1.13
Isotope Number of Protons Number of Electrons Number of Neutrons Isotope Symbol
hydrogen-1 1 1 0 $\ce{^{1}_{1}H}$
hydrogen-2 1 1 1 $\ce{^{2}_{1}H}$
beryllium-9 4 4 5 $\ce{^{9}_{4}Be}$
aluminum-27 13 13 14 $\ce{^{27}_{13}Al}$
Q2.1.14
Element Symbol Atomic Number Mass Number # of Protons # of Electrons # of Neutrons Isotope Symbol
nitrogen N 7 14 7 7 7 $\ce{^{14}_{7}N}$
boron B 5 11 5 5 6 $\ce{^{11}_{5}B}$
zinc Zn 30 65 30 30 35 $\ce{^{65}_{30}Zn}$
iridium Ir 77 193 77 77 116 $\ce{^{193}_{77}Ir}$
iron Fe 26 56 26 26 30 $\ce{^{56}_{26}Fe}$
2.2: Matter
Q2.2.1
Matter is anything that has mass and occupies space. Examples of matter will vary and can be any object from an atom to a macroscopic object.
Q2.2.2
A compound is a combination of elements with a fixed composition. The elements in the compound do not retain their individual identity by have the properties of the compound. A mixture does not have a fixed composition and each component of the mixture retains its identity and properties.
Q2.2.3
A pure substance contains only one component, either an element or compound, while a mixture contains multiple pure substances.
Q2.2.4
1. mixture
2. pure substance
3. mixture
4. mixture
Q2.2.5
1. element
2. compound
3. homogeneous mixture
4. heterogeneous mixture
5. compound
6. heterogeneous mixture
7. heterogeneous mixture
8. heterogeneous mixture
Q2.2.6
1. metal
2. metal
3. nonmetal
4. metal
5. nonmetal
6. metalloid
Q2.2.7
1. nonmetal
2. nonmetal
3. metalloid
4. metal
5. nonmetal
6. nonmetal
Q2.2.8
Ionic compounds are generally formed between a metal and nonmetal or between a polyatomic ion and another ion. Molecular compounds are composed of two ore more nonmetals.
Q2.2.9
1. molecular
2. molecular
3. molecular
4. ionic
5. ionic
6. ionic
Q2.2.10
1. ionic
2. ionic
3. molecular
4. molecular
5. molecular
6. molecular
2.3: Mole and Molar Mass
Q2.3.1
A counting number is a word that is associated with a specific number.
Q2.3.2
$6.022\times10^{23}$
Q2.3.3
H2, N2, O2, F2, Cl2, Br2, I2
Q2.3.4
1. $6.022\times10^{23}$ atoms
2. $1.204\times10^{24}$ atoms
3. $1.506\times10^{24}$ atoms
4. $3.011\times10^{23}$ atoms
5. $2.108\times10^{23}$ atoms
Q2.3.5
1. $6.022\times10^{23}$ molecules
2. $1.204\times10^{24}$ molecules
3. $1.506\times10^{24}$ molecules
4. $3.011\times10^{23}$ molecules
5. $2.108\times10^{23}$ molecules
Q2.3.6
The numbers are the same for the same number of moles because moles are a counting number. Regardless of what is being counted, a mole will have the same number of items.
Q2.3.7
$6.73 \times {10^{25}}\;atoms\;{\rm{Si}}\left( {\frac{1\;mole}{6.022 \times {10}^{23}\;atoms}} \right) = 112\;moles\;{\rm{Si}}$
Q2.3.8
$4.29 \times {10^{22}}\;atoms\;{\rm{Na}}\left( {\frac{1\;mole}{6.022 \times {10}^{23}\;atoms}} \right) = 0.0712\;moles\;{\rm{Na}}$
Q2.3.9
1. 2 hydrogen atoms, 2 oxygen atoms
2. 2 nitrogen atoms, 5 oxygen atoms
3. 1 phosphorus atom, 3 fluorine atoms
4. 1 magnesium atom, 2 chlorine atoms
5. 1 potassium atom, 1 bromine atom
6. 1 aluminum atom, 3 chlorine atoms
7. 1 calcium atom, 1 oxygen atom
8. 2 sodium atoms, 1 oxygen atom
9. 1 nitrogen atom, 3 hydrogen atoms
10. 1 carbon atom, 2 oxygen atoms
11. 2 nitrogen atoms, 4 hydrogen atoms
12. 2 nitrogen atoms, 1 oxygen atom
Q2.3.10
1. 2 moles hydrogen, 2 moles oxygen
2. 2 moles nitrogen, 5 moles oxygen
3. 1 mole phosphorus, 3 moles fluorine
4. 1 mole magnesium, 2 moles chlorine
5. 1 mole potassium, 1 mole bromine
6. 1 mole aluminum, 3 moles chlorine
7. 1 mole calcium, 1 mole oxygen
8. 2 moles sodium, 1 mole oxygen
9. 1 mole nitrogen, 3 moles hydrogen
10. 1 mole carbon, 2 moles oxygen
11. 2 moles nitrogen, 4 moles hydrogen
12. 2 moles nitrogen, 1 mole oxygen
Q2.3.11
The numbers are the same because the ratios are the same between atoms and moles. Moles are a counting number so they are a multiple of the number of atoms.
Q2.3.12
$0.75\;mol\;{\rm{CCl}_4}\left ( \frac{1\;mol\;{\rm{C}}}{1\;mol\;{\rm{CCl}_4}} \right ) =0.75\;mol\;\rm{C}$
$0.75\;mol\;{\rm{CCl}_4}\left ( \frac{4\;mol\;{\rm{Cl}}}{1\;mol\;{\rm{CCl}_4}} \right ) =3.0\;mol\;\rm{Cl}$
Q2.3.13
$0.75\;mol\;{\rm{CCl}_4}\left ( \frac{1\;mol\;{\rm{C}}}{1\;mol\;{\rm{CCl}_4}} \right )\left ( \frac{6.02\times 10^{23} \;atoms}{1\;mol\;{\rm{C}} }\right ) =4.52\times10^{23}\;atoms\;\rm{C}$
$0.75\;mol\;{\rm{CCl}_4}\left ( \frac{4\;mol\;{\rm{Cl}}}{1\;mol\;{\rm{CCl}_4}} \right )\left ( \frac{6.02\times 10^{23} \;atoms}{1\;mol\;{\rm{Cl}}} \right ) =1.81\times10^{24}\;atoms\;\rm{Cl}$
Q2.3.14
$2.5\;mol\;{\rm{H}_2\rm{O}}\left ( \frac{2\;mol\;{\rm{H}}}{1\;{mol}\;\rm{H}_2\rm{O}} \right ) =5.0\;mol\;\rm{H}$
$2.5\;mol\;{\rm{H}_2\rm{O}}\left ( \frac{1\;mol\;{\rm{O}}}{1\;{mol}\;\rm{H}_2\rm{O}} \right ) =2.5\;mol\;\rm{O}$
Q2.3.15
$2.5\;mol\;{\rm{H}_2\rm{O}}\left ( \frac{2\;mol\;{\rm{H}}}{1\;{mol}\;\rm{H}_2\rm{O}} \right ) \left ( \frac{6.02\times 10^{23} \;atoms}{1\;mol\;{\rm{H}} }\right )=3.01\times10^{24}\;atoms\;\rm{H}$
$2.5\;mol\;{\rm{H}_2\rm{O}}\left ( \frac{1\;mol\;{\rm{O}}}{1\;{mol}\;\rm{H}_2\rm{O}} \right ) \left ( \frac{6.02\times 10^{23} \;atoms}{1\;mol\;{\rm{O}} }\right )=1.51\times10^{24}\;atoms\;\rm{O}$
Q2.3.16
$3.87\times10^{25}\;atoms\;{\rm{O}}\left ( \frac{1\;molec\;{\rm{CaNO}}_3}{3\;atoms\;\rm{O}} \right )=1.29\times10^{25}\;molec\;\rm{CaNO}_3$
Q2.3.17
$5.39\times10^{24}\;atoms\;{\rm{C}}\left ( \frac{8\;atoms\;{\rm{H}}}{3\;atoms\;\rm{C}} \right )=1.44\times10^{25}\;atoms\;\rm{H}$
Q2.3.18
1. 12.01 g/mol
2. 14.01 g/mol
3. 22.99 g/mol
4. 1.008 g/mol
5. 39.10 g/mol
6. 30.97 g/mol
Q2.3.19
1. $25.0\;g\;{\rm{C}}\left ( \frac{1\;mol\;{\rm{C}}}{12.01\;\frac{g}{mol}} \right )=2.08\;mol\;{\rm{C}}$
2. $25.0\;g\;{\rm{N}}\left ( \frac{1\;mol\;{\rm{N}}}{14.01\;\frac{g}{mol}} \right )=1.78\;mol\;{\rm{N}}$
3. $25.0\;g\;{\rm{Na}}\left ( \frac{1\;mol\;{\rm{Na}}}{22.99\;\frac{g}{mol}} \right )=1.09\;mol\;{\rm{Na}}$
4. $25.0\;g\;{\rm{H}}\left ( \frac{1\;mol\;{\rm{H}}}{1.008\;\frac{g}{mol}} \right )=24.8\;mol\;{\rm{H}}$
5. $25.0\;g\;{\rm{K}}\left ( \frac{1\;mol\;{\rm{K}}}{39.10\;\frac{g}{mol}} \right )=0.639\;mol\;{\rm{K}}$
6. $25.0\;g\;{\rm{P}}\left ( \frac{1\;mol\;{\rm{P}}}{30.97\;\frac{g}{mol}} \right )=0.807\;mol\;{\rm{P}}$
Q2.3.20
The moles are different because a mole of atoms of each element has a different mass. Although they have the same mass, the number of atoms of each sample is different. Compare a ton of feathers to a ton of books. The same total mass but a different quantity of each.
Q2.3.21
What is the mass of each of the following samples?
1. $0.35\;mol\;{\rm{Na}}\left ( \frac{22.99\;g}{mol\;{\rm{Na}}} \right )=8.0\;g\;\rm{Na}$
2. $0.75\;mol\;{\rm{C}}\left ( \frac{12.01\;g}{mol\;{\rm{C}}} \right )=9.0\;g\;\rm{C}$
3. $1.34\;mol\;{\rm{K}}\left ( \frac{39.10\;g}{mol\;{\rm{K}}} \right )=52.4\;g\;\rm{K}$
4. $1.21\;mol\;{\rm{Si}}\left ( \frac{28.09\;g}{mol\;{\rm{Si}}} \right )=34.0\;g\;\rm{Si}$
5. $0.95\;mol\;{\rm{Ca}}\left ( \frac{40.08\;g}{mol\;{\rm{Ca}}} \right )=38\;g\;\rm{Ca}$
6. $2.85\;mol\;{\rm{He}}\left ( \frac{4.003\;g}{mol\;{\rm{He}}} \right )=11.4\;g\;\rm{He}$
Q2.3.22
1. CO2 has 1 mole of carbon and 2 moles of oxygen; $\left ( 12.01\frac{g}{mol}\times1 \right )+\left ( 16.00\frac{g}{mol}\times2 \right )=44.01\frac{g}{mol}$
2. N2H4 has 2 moles of nitrogen and 4 moles of hydrogen; $\left ( 14.01\frac{g}{mol}\times2 \right )+\left ( 1.008\frac{g}{mol}\times4 \right )=32.05\frac{g}{mol}$
3. CaF2 has 1 mole of calcium and 2 moles of fluorine; $\left ( 40.08\frac{g}{mol}\times1 \right )+\left ( 19.00\frac{g}{mol}\times2 \right )=78.08\frac{g}{mol}$
4. C6H12O6 has 6 moles of carbon, 12 moles of hydrogen, and 6 moles of oxygen; $\left ( 12.01\frac{g}{mol}\times6 \right )+\left ( 1.008\frac{g}{mol}\times12 \right )+\left ( 16.00\frac{g}{mol}\times6 \right )=180.16\frac{g}{mol}$
5. CH4 has 1 mole of carbon and 4 moles of hydrogen; $\left ( 12.01\frac{g}{mol}\times1 \right )+\left ( 1.008\frac{g}{mol}\times4 \right )=16.04\frac{g}{mol}$
6. C6H6 has 6 moles of carbon and 6 moles of hydrogen; $\left ( 12.01\frac{g}{mol}\times6 \right )+\left ( 1.008\frac{g}{mol}\times6 \right )=78.11\frac{g}{mol}$
7. Na2SO4 has 2 moles of sodium, 1 mole of sulfur, and 4 moles of oxygen; $\left ( 22.99\frac{g}{mol}\times2 \right )+\left ( 32.06\frac{g}{mol}\times1 \right )+\left ( 16.00\frac{g}{mol}\times4 \right )=142.04\frac{g}{mol}$
8. K3PO4 has 3 moles of potassium, 1 mole of phosphorus, and 4 moles of oxygen; $\left ( 39.10\frac{g}{mol}\times3 \right )+\left ( 30.97\frac{g}{mol}\times1 \right )+\left ( 16.00\frac{g}{mol}\times4 \right )=212.27\frac{g}{mol}$
9. Al(NO3)3 has 1 mole of aluminum, 3 moles of nitrogen, and 9 moles of oxygen; $\left ( 26.98\frac{g}{mol}\times1 \right )+\left ( 14.01\frac{g}{mol}\times3 \right )+\left ( 16.00\frac{g}{mol}\times9 \right )=213.01\frac{g}{mol}$
10. Mg3(PO4)2 has 3 moles of magnesium, 2 moles of phosphorus, and 8 moles of oxygen; $\left ( 24.31\frac{g}{mol}\times3 \right )+\left ( 30.97\frac{g}{mol}\times2 \right )+\left ( 16.00\frac{g}{mol}\times8 \right )=262.87\frac{g}{mol}$
Q2.3.23
1. $25.0\;g\;{\rm{CO_2}}\left ( \frac{1\;mol\;{\rm{CO_2}}}{44.01\;g} \right )=0.568\;mol\;\rm{CO_2}$
2. $10.0\;g\;{\rm{N_2H_4}}\left ( \frac{1\;mol\;{\rm{N_2H_4}}}{32.05\;g} \right )=0.312\;mol\;\rm{N_2H_4}$
3. $85.0\;g\;{\rm{CaF_2}}\left ( \frac{1\;mol\;{\rm{CaF_2}}}{78.08\;g} \right )=1.09\;mol\;\rm{CaF_2}$
4. $15.5\;g\;{\rm{C_6H_{12}O_6}}\left ( \frac{1\;mol\;{\rm{C_6H_{12}O_6}}}{180.16\;g} \right )=0.0860\;mol\;\rm{C_6H_{12}O_6}$
5. $20.0\;g\;{\rm{CH_4}}\left ( \frac{1\;mol\;{\rm{CH_4}}}{16.04\;g} \right )=1.25\;mol\;\rm{CH_4}$
6. $100.0\;g\;{\rm{C_6H_6}}\left ( \frac{1\;mol\;{\rm{C_6H_6}}}{78.11\;g} \right )=1.28\;mol\;\rm{C_6H_6}$
7. $30.0\;g\;{\rm{Na_2SO_4}}\left ( \frac{1\;mol\;{\rm{Na_2SO_4}}}{142.04\;g} \right )=0.211\;mol\;\rm{Na_2SO_4}$
8. $75.0\;g\;{\rm{K_3PO_4}}\left ( \frac{1\;mol\;{\rm{K_3PO_4}}}{212.27\;g} \right )=0.353\;mol\;\rm{K_3PO_4}$
9. $50.0\;g\;{\rm{Al(NO_3)_3}}\left ( \frac{1\;mol\;{\rm{Al(NO_3)_3}}}{213.01\;g} \right )=0.235\;mol\;\rm{Al(NO_3)_3}$
10. $47.2\;g\;{\rm{Mg_3(SO_4)_2}}\left ( \frac{1\;mol\;{\rm{Mg_3(SO_4)_2}}}{262.87\;g} \right )=0.180\;mol\;\rm{Mg_3(SO_4)_2}$
Q2.3.24
1. $3.5\;mol\;{\rm{CO_2}}\left ( \frac{44.01\;g}{mol\;{\rm{CO_2}}} \right )=1.50\times10^2\;g\;\rm{CO_2}$
2. $0.45\;mol\;{\rm{N_2H_4}}\left ( \frac{32.05\;g}{mol\;{\rm{N_2H_4}}} \right )=14\;g\;\rm{N_2H_4}$
3. $2.25\;mol\;{\rm{CaF_2}}\left ( \frac{78.08\;g}{mol\;{\rm{CaF_2}}} \right )=176\;g\;\rm{CaF_2}$
4. $1.75\;mol\;{\rm{C_6H_{12}O_6}}\left ( \frac{180.16\;g}{mol\;{\rm{C_6H_{12}O_6}}} \right )=315\;g\;\rm{C_6H_{12}O_6}$
5. $4.9\;mol\;{\rm{CH_4}}\left ( \frac{16.04\;g}{mol\;{\rm{CH_4}}} \right )=79\;g\;\rm{CH_4}$
6. $8.75\;mol\;{\rm{C_6H_6}}\left ( \frac{78.11\;g}{mol\;{\rm{C_6H_6}}} \right )=683\;g\;\rm{C_6H_6}$
7. $2.35\;mol\;{\rm{Na_2SO_4}}\left ( \frac{142.04\;g}{mol\;{\rm{Na_2SO_4}}} \right )=334\;g\;\rm{Na_2SO_4}$
8. $0.672\;mol\;{\rm{K_3PO_4}}\left ( \frac{212.27\;g}{mol\;{\rm{K_3PO_4}}} \right )=143\;g\;\rm{K_3PO_4}$
9. $0.95\;mol\;{\rm{Al(NO_3)_3}}\left ( \frac{213.01\;g}{mol\;{\rm{Al(NO_3)_3}}} \right )=2.0\times10^2\;g\;\rm{Al(NO_3)_3}$
10. $1.15\;mol\;{\rm{Mg_3(PO_4)_2}}\left ( \frac{262.87\;g}{mol\;{\rm{Mg_3(PO_4)_2}}} \right )=302\;g\;\rm{Mg_3(PO_4)_2}$
2.4: Electron Arrangements
Q2.4.1
a. 2, 8, 1
b. 2, 8
c. 2, 2
d. 2, 5
e. 2, 8, 6
f. 2, 8, 7
Q2.4.2
a. 1
b. 5
c. 7
d. 6
e. 1
f. 3
Q2.4.3
The octet rule predicts the stability of an atom based on having eight electrons in its electron shell.
2.5: Ion Formation
Q2.5.1
An ion is a charged species which results from the gain or loss of one ore more electrons.
Q2.5.2
Anions and cations are both charged species which results from the change in the number of electrons. Anions have a negative charge while cations have a positive charge.
Q2.5.3
1. $\rm{Li}^+$
2. $\rm{Na}^+$
3. $\rm{Ca}^{2+}$
4. $\rm{B}^{3+}$
5. $\rm{P}^{3-}$
6. $\rm{S}^{2-}$
7. $\rm{Cl}^{-}$
8. $\rm{Br}^-$
Q2.5.4
1. 3 protons, 4 neutrons, 2 electrons
2. 11 protons, 12 neutrons, 10 electrons
3. 20 protons, 20 neutrons, 18 electrons
4. 5 protons, 6 neutrons, 2 electrons
5. 15 protons, 16 neutrons, 18 electrons
6. 16 protons, 16 neutrons, 18 electrons
7. 17 protons, 18 neutrons, 18 electrons
8. 35 protons, 45 neutrons, 36 electrons
Q2.5.5 ( $\therefore$ = therefore)
1. ion with 3+ charge and 2 electrons $\therefore$ neutral atom had 5 electrons $\therefore$ atom has 5 protons $\therefore$ boron (B)
2. ion with 1$-$ charge and 18 electrons $\therefore$ neutral atom had 17 electrons $\therefore$ atom has 17 protons $\therefore$ chlorine (Cl)
3. ion with 1+ charge and 18 electrons $\therefore$ neutral atom had 19 electrons $\therefore$ atom has 19 protons $\therefore$ potassium (K)
4. ion with a 3$-$ charge and 10 electrons $\therefore$ neutral atom had 7 electrons $\therefore$ atom has 7 protons $\therefore$ nitrogen (N)
Q2.5.6
A polyatomic ion contains multiple atoms working together as a group and has an overall charge.
Q2.5.7
1. polyatomic
2. monatomic
3. polyatomic
4. monatomic
5. monatomic
6. polyatomic
2.6: Ionic Compounds
Q2.6.1
carbon
Q2.6.2
Answers will vary. Most metals are in the first two columns of the periodic table or in the transition metal block.
Q2.6.3
Answers will vary. Nonmetallic elements are located in the upper right corner of the periodic table (examples include nitrogen, oxygen, phosphorus, chlorine, bromine, etc)
Q2.6.4
Ionic compounds are composed of ions of metals and nonmetals. Ionic compounds can also include a polyatomic ion.
Q2.6.5
To form an ionic bond, electrons are transferred from the metal to the nonmetal.
Q2.6.6
Zero
Q2.6.7
What is the formula for the ionic compound formed from each of the following pairs?
1. K2S
2. AgCl
3. CaO
4. AlI3
5. Ba3N2
6. Na3P
7. LiF
8. Mg3N2
9. CaS
10. BeBr2
11. Zn3N2
12. SnI4
Q2.6.8
1. Na2CO3
2. NaClO3
3. NaClO2
4. Na3PO4
5. NaNO3
6. Na2SO4
7. Na2CrO4
8. Na2Cr2O7
Q2.6.9
1. MgCO3
2. Mg(ClO3)2
3. Mg(ClO2)2
4. Mg3(PO4)2
5. Mg(NO3)2
6. MgSO4
7. MgCrO4
8. MgCr2O7
Q2.6.10
Parentheses are used when there is more than one of a polyatomic ion in the formula of an ionic compound. | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/02%3A_Elements_and_Ions/2.07%3A_Elements_and_Ions_%28Exercises%29.txt |
Molecular compounds are chemical compounds that take the form of discrete molecules like water $\left( \ce{H_2O} \right)$ and carbon dioxide $\left( \ce{CO_2} \right)$. These compounds are very different from ionic compounds like sodium chloride $\left( \ce{NaCl} \right)$, which are formed when metal atoms lose one or more of their electrons to nonmetal atoms and the resulting ions are mutually attracted to each other.
• 3.1: Molecular Compounds
Molecular compounds are chemical compounds that take the form of discrete molecules. These compounds are very different from ionic compounds like sodium chloride (NaCl) . Ionic compounds are formed when metal atoms lose one or more of their electrons to nonmetal atoms. Rather than forming ions, the atoms of a molecule share their valence electrons in such a way that a bond forms between pairs of atoms.
• 3.2: Straight-Chain Alkanes
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.
• 3.3: Compounds (Exercises)
These are homework exercises to accompany Chapter 3 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Solutions are available below the questions.
03: Compounds
Learning Outcomes
• Describe types of elements that form covalent bonds.
• Explain how a covalent bond is formed.
• Distinguish between ionic and molecular compounds.
• Know the meaning of prefixes used in naming molecular compounds.
• Distinguish between molecules and compounds.
Water and diamonds - two very different materials. Water can be found almost everywhere. It is in lakes, creeks, rivers, and oceans. We get water from the sky when it rains. Diamonds, on the other hand, are very rare. They are only found in a few locations on the earth and must be mined to become available. Major diamond mines are located in various African countries, Australia, and Russia. The United States has several underground sources of diamonds in Alaska, Colorado, Minnesota, Montana, and Wyoming, but the only "active" U.S. mine is the Crater of Diamonds mine in a state park near Murfreesboro, Arkansas. For a small fee, visitors can dig for diamonds. You won't get rich by visiting, though - only a few hundred carats of low-grade diamonds are found each year.
The two materials do have at least one thing in common. The atoms in the materials are held together by covalent bonds. These bonds consist of electrons shared between two or more atoms. Unlike ionic bonds, where electrons are either lost or gained by an atom to form charged ions, electrons in covalent compounds are shared between the two atoms, giving rise to properties that are quite different from those seen in ionic materials.
Molecular Compounds
Molecular compounds are chemical compounds that take the form of discrete molecules. Examples include such familiar substances as water $\left( \ce{H_2O} \right)$ and carbon dioxide $\left( \ce{CO_2} \right)$ (Figure $1$). These compounds are very different from ionic compounds like sodium chloride $\left( \ce{NaCl} \right)$. Ionic compounds are formed when metal atoms lose one or more of their electrons to nonmetal atoms. The resulting cations and anions are electrostatically attracted to each other.
So what holds the atoms of a molecule together? Rather than forming ions, the atoms of a molecule share their valence electrons in such a way that a bond forms between pairs of atoms. In a carbon dioxide molecule, there are two of these bonds, each occurring between the carbon atom and one of the two oxygen atoms.
Binary Molecular Compound Names
Recall that a molecular formula shows the number of atoms of each element that a molecule contains. A molecule of water contains two hydrogen atoms and one oxygen atom, so its formula is $\ce{H_2O}$. A molecule of octane, which is a component of gasoline, contains eight atoms of carbon and eighteen atoms of hydrogen. The molecular formula of octane is $\ce{C_8H_{18}}$.
A binary molecular compound is a molecular compound that is composed of two elements. In general, the elements that combine to form binary molecular compounds are both nonmetals. This contrasts with ionic compounds, which involve bonds between metal cations and nonmetal anions. One difference between ionic and molecular compounds is that two nonmetal atoms will frequently combine with one another in a variety of ratios. Consider the elements nitrogen and oxygen. They combine to make several binary compounds, including $\ce{NO}$, $\ce{NO_2}$, (see figure below), and $\ce{N_2O}$. Obviously they can't all be called nitrogen oxide! How would someone know which one you were talking about? Each of the three compounds has very different properties and reactivity. A system to distinguish between compounds such as these is necessary.
Prefixes are used in the names of binary molecular compounds to identify the number of atoms of each element. Although we will not cover nomenclature rules for molecular compounds you should be familiar with the prefixes used in their names. Listed below (see table below) are the prefixes up to ten.
Number of Atoms Prefix
Table $1$: Numerical Prefixes
1 mono-
2 di-
3 tri-
4 tetra-
5 penta-
6 hexa-
7 hepta-
8 octa-
9 nona-
10 deca-
Some examples of molecular compounds are listed below (see table below). Note the use of prefixes in their names.
Formula Name
Table $2$: Examples of Molecular Compounds
$\ce{NO}$ nitrogen monoxide
$\ce{N_2O}$ dinitrogen monoxide
$\ce{S_2Cl_2}$ disulfur dichloride
$\ce{Cl_2O_7}$ dichlorine heptoxide
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/03%3A_Compounds/3.01%3A_Molecular_Compounds.txt |
Learning Outcomes
• Distinguish between organic and inorganic compounds.
• Describe features of alkane structure.
• Pair each prefix with its numerical definition.
• Distinguish between an alkane and a cycloalkane.
Organic Compounds
All living things on earth are formed mostly of carbon compounds. The prevalence of carbon compounds in living things has led to the epithet “carbon-based” life. The truth is we know of no other kind of life. Early chemists regarded substances isolated from organisms (plants and animals) as a different type of matter that could not be synthesized artificially, and these substances were thus known as organic compounds. The widespread belief called vitalism held that organic compounds were formed by a vital force present only in living organisms. The German chemist Friedrich Wohler was one of the early chemists to refute this aspect of vitalism, when, in 1828, he reported the synthesis of urea, a component of many body fluids, from nonliving materials. Since then, it has been recognized that organic molecules obey the same natural laws as inorganic substances, and the category of organic compounds has evolved to include both natural and synthetic compounds that contain carbon. Some carbon-containing compounds are not classified as organic, for example, carbonates and cyanides, and simple oxides, such as CO and CO2. Although a single, precise definition has yet to be identified by the chemistry community, most agree that a defining trait of organic molecules is the presence of carbon as the principal element, bonded to hydrogen and other carbon atoms. Organic compounds often contain other elements such as nitrogen, oxygen, sulfur, or phosphorus. The number of carbon and hydrogen atoms in an organic compound is usually much greater than the number of other atoms.
Today, organic compounds are key components of plastics, soaps, perfumes, sweeteners, fabrics, pharmaceuticals, and many other substances that we use every day. The value to us of organic compounds ensures that organic chemistry is an important discipline within the general field of chemistry. The element carbon gives rise to a vast number and variety of compounds that are found in biological and industrial settings. These compounds typically contain functional groups which include atoms other than carbon and hydrogen. A particular arrangement of atoms within a compound is known as a functional group. Types of functional groups will be discussed in the next chapter. For now, we will identify organic and inorganic (which means not organic) compounds and look at structures for the simplest of organic compounds, hydrocarbons.
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 have single, double, or triple bonds between the carbon atoms. Alkanes will be discussed here and other hydrocarbons will be covered later.
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 $\left( \ce{C_4H_{10}} \right)$ 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 $\ce{n}$ is the number of carbons in the chain. The table below lists the first ten members of the alkane series.
Table $1$: First Ten Members of the Alkane Series
Name Molecular Formula Condensed Structural Formula
Methane $\ce{CH_4}$ $\ce{CH_4}$
Ethane $\ce{C_2H_6}$ $\ce{CH_3CH_3}$
Propane $\ce{C_3H_8}$ $\ce{CH_3CH_2CH_3}$
Butane $\ce{C_4H_{10}}$ $\ce{CH_3CH_2CH_2CH_3}$
Pentane $\ce{C_5H_{12}}$ $\ce{CH_3CH_2CH_2CH_2CH_3}$
Hexane $\ce{C_6H_{14}}$ $\ce{CH_3CH_2CH_2CH_2CH_2CH_3}$
Heptane $\ce{C_7H_{16}}$ $\ce{CH_3CH_2CH_2CH_2CH_2CH_2CH_3}$
Octane $\ce{C_8H_{18}}$ $\ce{CH_3CH_2CH_2CH_2CH_2CH_2CH_2CH_3}$
Nonane $\ce{C_9H_{20}}$ $\ce{CH_3CH_2CH_2CH_2CH_2CH_2CH_2CH_2CH_3}$
Decane $\ce{C_{10}H_{22}}$ $\ce{CH_3CH_2CH_2CH_2CH_2CH_2CH_2CH_2CH_2CH_3}$
Note that the table shows a variation of a structural formula called a condensed structural formula, which we will discuss later. 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.
Cycloalkanes
Alkanes can also exist in a ring structure which is known as a cycloalkane. For example, pentane is a chain of five carbon atoms while cyclopentane is a ring of five carbons as shown in the figure below.
3.03: Compounds (Exercises)
These are homework exercises to accompany Chapter 3 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Solutions are available below the questions.
(click here for solutions)
Q3.1.1
What types of elements typically form binary molecular compounds?
Q3.1.2
Describe the similarities and differences between ionic and molecular compounds.
Q3.1.3
Give the prefix that would be used in the name a molecular compound for each of the following quantities of atoms.
a. 6
b. 3
c. 9
d. 5
e. 8
Q3.1.4
Write the formula for each compound.
1. diphosphorus pentoxide
2. dinitrogen monoxide
3. trisilicon tetranitride
4. dinitrogen pentoxide
5. tetraphosphorus decasulfide
6. disulfur hexafluoride
7. triboron dicarbide
8. tetraselenium tetranitride
(click here for solutions)
Q3.2.1
What elements are found in a hydrocarbon?
Q3.2.2
Describe an alkane.
Q3.2.3
How many carbon atoms are found in octane? propane? nonane?
Q3.2.4
What is the name for the alkane with six carbons? four carbons? two carbons?
Q3.2.5
What is the difference between an alkane and a cycloalkane?
3.1: Molecular Compounds
Q3.1.1
Binary molecular compounds are composed of two nonmetallic elements.
Q3.1.2
They both form as a result of bonding between atoms. Ionic compounds result from the transfer of electrons from one element to another while molecular compounds form bonds through the sharing of electrons.
Q3.1.3
a. hexa
b. tri
c. nona
d. penta
e. octa
Q3.1.4
1. P2O5
2. N2O
3. Si3N4
4. N2O5
5. P4S10
6. S2F6
7. B3C2
8. Se4N4
3.2: Straight-Chain Alkanes
Q3.2.1
carbon and hydrogen
Q3.2.2
An alkane contains only carbon and hydrogen atoms with the carbons connected by single bonds.
Q3.2.3
octane, 8; propane, 3; nonane, 9
Q3.2.4
6, hexane; 4, butane; 2, ethane
Q3.2.5
An alkane contains a chain of carbon atoms while a cycloalkane contains carbons in a ring structure. | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/03%3A_Compounds/3.02%3A_Straight-Chain_Alkanes.txt |
The three dimensional shape or configuration of a molecule is an important characteristic. This shape is dependent on the preferred spatial orientation of covalent bonds to atoms having two or more bonding partners.
• 4.1: Lewis Electron Dot Structures
Ionic substances are completely held together by ionic bonds. The full charges of the ions (for example, Na+ and Cl− in sodium chloride) cause electrostatic interactions that result in a stable crystal lattice. Ionic compounds exist as extended, orderly arrangements of ions. This is quite different from the structure of molecular substances, which take the form of collections of individual molecules.
• 4.2: Representing Structures
Lewis structures are great for small molecules but can be complicated when dealing with large molecules. There are a variety of ways that we can represent molecules that provide use with enough information without creating a complicated figure. T
• 4.3: Electron Group Geometry
Electron group geometry is the three-dimensional arrangement of atoms in a molecule. The molecular geometry, or shape, of a molecule is an important factor that affects the physical and chemical properties of a compound. Those properties include melting and boiling points, solubility, density, and the types of chemical reactions that a compound undergoes. In this section, you will learn a technique to predict molecular shape based on a molecule's Lewis electron dot structure.
• 4.4: Functional Groups
With over twenty million known organic compounds in existence, it would be very challenging to memorize chemical reactions for each one. Fortunately, molecules with similar functional groups tend to undergo similar reactions. A functional group is defined as an atom or group of atoms within a molecule that has similar chemical properties whenever it appears in various compounds. Even if other parts of the molecule are quite different, certain functional groups tend to react in certain ways.
• 4.5: Structure and Function (Exercises)
These are homework exercises to accompany Chapter 4 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Solutions are available below the questions.
04: Structure and Function
Learning Outcomes
• Determine Lewis structure of a compound.
Ionic substances are completely held together by ionic bonds. The full charges of the ions (for example, $\ce{Na^+}$ and $\ce{Cl^-}$ in sodium chloride) cause electrostatic interactions that result in a stable crystal lattice. Ionic compounds exist as extended, orderly arrangements of ions. This is quite different from the structure of molecular substances, which take the form of collections of individual molecules.
The electrons that form a covalent bond are not fully possessed by a single atom (as the electrons in an ion would be) but are shared between the two atoms involved in the bond. The concept of the covalent bond was first proposed in 1916 by the American chemist G. N. Lewis (1875 - 1946), who suggested that sharing electrons was one way that atoms could attain a complete octet of valence electrons. This idea was expanded upon by Linus Pauling (1901 - 1994), who eventually won the Nobel Prize in Chemistry in 1954 for his work on chemical bonding.
Single Covalent Bonds
The simplest covalent bond is formed between two hydrogen atoms. Each hydrogen atom has a single electron, and each needs two electrons for a full outer shell. The hydrogen molecule, $\ce{H_2}$, consists of two hydrogen atoms sharing their two valence electrons. Hydrogen can also form covalent bonds with other atoms. For example, hydrogen and chlorine each need one more electron to achieve a noble gas configuration. By sharing valence electrons (each atom donates one), the stable $\ce{HCl}$ molecule is formed.
We will use a simplified representation of covalent bonds known as Lewis structures. These drawings are also know by various other names, including Lewis dot structures or electron dot structures. Each dot in the structure represents one valence electron in the compound. For example, $\ce{H_2}$ could be drawn as $\ce{H} : \ce{H}$. Each dot represents one valence electron, and the fact that they are placed between the two atoms means that they are being shared bas a covalent bond. For larger molecules, it can become cumbersome to draw out all of the valence electrons, so a bonding pair of electrons can also be drawn as a straight line. Thus, $\ce{H_2}$ can also be represented as $\ce{H-H}$.
If we wanted to show the Lewis structure of $\ce{HCl}$, we would draw the following:
We can see that the covalent bond consists of two electrons between the $\ce{H}$ and the $\ce{Cl}$. The $\ce{H}$ has a full outer shell of two electrons and the chlorine has a full outer shell of eight electrons. Covalent bonds with other halogens can be written the same way.
Similar types of Lewis structures can be written for other molecules that form covalent bonds. Many compounds that contain $\ce{O}$, $\ce{N}$, $\ce{C}$, $\ce{S}$, and $\ce{P}$ are held together by covalent bonds. The number of covalent bonds an atom will form can generally be predicted by the number of electrons an atom requires to fill its valence shell. For example, oxygen has 6 electrons in its outer shell and needs two more to fill this shell, so it will only form two covalent bonds with other atoms. If we look at the water molecule $\left( \ce{H_2O} \right)$ (see figure below), we see that the oxygen atom makes two total bonds (one with each hydrogen atom).
As you can see, there are two pairs of electrons not involved in covalent bonding. These unbonded pairs of electrons are known as lone pairs and contribute to the overall shape of the molecule. Similarly, nitrogen needs three electrons to complete its valence shell, so it tends to make three covalent bonds, with one lone pair of non-bonding electrons left over.
Again, each of the lines stands for a pair of bonding electrons (a single bond), and the lone pair of nitrogen is drawn as two dots.
Double and Triple Bonds
So far we have considered only single bonds, formed by the sharing of one electron from each atom. Many molecules contain double bonds, in which each atom shares two electrons, or triple bonds, in which each atom shares three electrons. These are represented by drawing two or three lines between the bonded atoms. For example, a carbon-carbon double bond can be written as $\ce{C} :: \ce{C}$ or $\ce{C=C}$. A carbon-carbon triple bond is shown as $\ce{C ::: C}$ or with three lines between the two carbon atoms, as seen in the structure of an organic molecule called acetylene (shown in the figure below). Just as $\ce{N}$ wants to form 3 bonds, other elements tend to form the same number of bonds in different compounds.
We will build Lewis structures that satisfy the octet rule to determine how atoms are attached to one another using the components shown in the table below to create structures that follow the octet rule.
Table $1$: Components to Build Lewis Structures
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky)
4.02: Representing Structures
Learning Outcomes
• Use condensed structural formulas to represent molecules.
• Use skeletal structures to represent molecules.
Lewis structures are great for small molecules but can be complicated when dealing with large molecules. There are a variety of ways that we can represent molecules that provide use with enough information without creating a complicated figure. The structures in the figure below all represent the molecules pentane, which is a straight-chain alkane with five carbon atoms.
Lewis Structures
The Lewis structure shows bonds to represent the sharing of electrons and include all atoms in the molecule so we have information about connectivity of atoms but not the molecular shape or geometry.
Space-Filling Model
A 3-D space-filling model shows connectivity of atoms and the shape of the molecule. These work well for smaller molecules, but can be a challenge to interpret since atoms are so close together. This is the most accurate representation of what a molecule looks like.
Skeletal Structure
These are the simplest way to represent the structure of a molecule. Hydrogen atoms are omitted (except when attached to \(\ce{O}\) or \(\ce{S}\)) and carbon atoms are at the intersection of each pair of lines and at the terminal end of each line. Each of the circles in the figure below indicate where a carbon is located.
Condensed Structures
These provide more information than a chemical formula (like \(\ce{C_5H_{12}}\) for pentane) but are condensed as given by their name, so they can be written out rather than in a drawing. The best way to write a condensed structure is to "read" a Lewis structure and see how atoms are connected.
Example \(1\)
Draw the skeletal and condensed structure of 2-butanol based on the given Lewis structure.
Skeletal Structure
Count the number of carbon atoms and draw the carbon chain with the correct number of carbon atoms. This represents a chain of four carbon atoms.
Now look for any atoms other than hydrogen that are attached to the carbon chain. The second carbon has an \(\ce{-OH}\) group. Add that to the second carbon in the chain.
The skeletal structure is complete and all of the remaining hydrogen atoms are implied because every carbon must have four bonds.
Condensed Structure
For the condensed structure, we "read" the Lewis structure from left to right.
The first carbon has three hydrogen atoms attached so we start with
\[\ce{CH_3}\]
The second carbon has one hydrogen attached and one \(\ce{-OH}\) group. Since the \(\ce{-OH}\) group is to the side, we put it in parentheses
\[\ce{CH_3CH(OH)}\]
The third carbon has two hydrogen atoms attached so we add \(\ce{CH_2}\)
\[\ce{CH_3CH(OH)CH_2}\]
The fourth, and last, carbon has three hydrogen atoms attached, so we add \(\ce{CH_3}\)
\[\ce{CH_3CH(OH)CH_2CH_3}\]
The result is the condensed structure of 2-butanol.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/04%3A_Structure_and_Function/4.01%3A_Lewis_Electron_Dot_Structures.txt |
Learning Outcomes
• Explain the basis of VSEPR theory.
• Predict the shapes of molecules using VSEPR theory.
• Predict bond angles in a molecule.
• Account for variations in bond angles on the relative repulsive forces exerted by lone pairs and bonding pairs of electrons.
Electron group geometry is the three-dimensional arrangement of atoms in a molecule. The geometry of a molecule is an important factor that affects the physical and chemical properties of a compound. Those properties include melting and boiling points, solubility, density, and the types of chemical reactions that a compound undergoes. In this section, you will learn a technique to predict the geometry of a molecule based on a its Lewis electron dot structure.
VSEPR Theory
The valence shell is the outermost occupied shell of electrons in an atom. This shell holds the valence electrons, which are the electrons that are involved in bonding and shown in a Lewis structure. Valence-shell electron pair repulsion theory, or VSEPR theory, states that a molecule will adjust its shape so that the valence electron pairs stay as far apart from each other as possible. This makes sense, based on the fact that negatively charged electrons repel one another. We will systematically classify molecules according to the number of bonding pairs of electrons and the number of nonbonding or lone pairs around the central atom. For the purposes of the VSEPR model, a double or triple bond is no different in terms of repulsion than a single bond. We will begin by examining molecules in which the central atom does not have any lone pairs.
Central Atom with No Lone Pairs
In order to easily understand the types of molecules possible, we will use a simple system to identify the parts of any molecule.
$\ce{A} =$ central atom in a molecule
$\ce{B} =$ atoms surrounding the central atom
Subscripts after the $\ce{B}$ will denote the number of $\ce{B}$ atoms that are bonded to the central $\ce{A}$ atom. For example, $\ce{AB_4}$ is a molecule with a central atom surrounded by four covalently bonded atoms. Again, it does not matter if those bonds are single, double, or triple bonds.
$\ce{AB_2}$
Beryllium hydride $\left( \ce{BeH_2} \right)$ consists of a central beryllium atom with two single bonds to hydrogen atoms. Note that it violates the octet rule, because the central atom has only 4 valence electrons. This is acceptable because beryllium only has two valence electrons to begin with, so it is not possible for it to create more than two covalent bonds with hydrogen atoms.
According to the requirement that electron pairs maximize their distance from one another, the two bonding pairs in the $\ce{BeH_2}$ molecules will arrange themselves on directly opposite sides of the central $\ce{Be}$ atom. The resulting geometry is a linear molecule, shown in a "ball-and-stick" model in the figure below.
The $\ce{H-Be-H}$ bond angle is $180^\text{o}$ because of its linear geometry.
Carbon dioxide is another example of a molecule which falls under the $\ce{AB_2}$ category. Its Lewis structure consists of double bonds between the central carbon atom and each oxygen atom.
The repulsion between the two double bonds on either side of the carbon atom is no different than the repulsion between the two single bonds on either side of the beryllium in the previous example. Therefore carbon dioxide is also linear, as this achieves the maximum distance between the electron pair bonds.
$\ce{AB_3}$
Boron trifluoride $\left( \ce{BF_3} \right)$ consists of a central boron atom with three single bonds to fluorine atoms. The boron atom is an exception to the octet rule, and generally only needs 6 electrons to be stable in a bonded molecule.
The geometry of the $\ce{BF_3}$ molecule is called trigonal planar. The fluorine atoms are positioned at the vertices of an equilateral triangle. The $\ce{F-B-F}$ angle is $120^\text{o}$, and all four atoms lie in the same plane.
$\ce{AB_4}$
Methane $\left( \ce{CH_4} \right)$ is an organic compound that is the primary component of natural gas. Its structure consists of a central carbon atom with four single bonds to hydrogen atoms.
In order to maximize their distance from one another, the four groups of bonding electrons do not lie in the same plane. Instead, each of the hydrogen atoms lies at the corners f a geometrical shape called a tetrahedron. The carbon atom is at the center of the tetrahedron. Each face of a tetrahedron is an equilateral triangle.
The electron group geometry of the methane molecule is referred to as tetrahedral. The $\ce{H-C-H}$ bond angles are $109.5^\text{o}$, which is larger than the $90^\text{o}$ that they would be if the molecule was planar. This way, the bonds are as far apart as possible to minimize electron repulsion. When drawing a structural formula for a molecule such as methane, it is advantageous to be able to indicate the three-dimensional character of its shape. The structural formula in the figure below is called a perspective drawing. The dotted line bond should be visualized as going back into the page, while the solid triangle bond should be visualized as coming out of the page.
There are structures with five and six bonds that we will not explore because they are not seen in biological molecules.
Table $1$: Geometries of Molecules in Which the Central Atom Has No Lone Pairs
Atoms Around Central Atom Electron Group Geometry Example
2 $\ce{AB_2}$ Linear $\ce{BeCl_2}$
3 $\ce{AB_3}$ Trigonal Planar $\ce{BF_3}$
4 $\ce{AB_4}$ Tetrahedral $\ce{CH_4}$
Central Atom with One or More Lone Pairs
The number of bonds to the central atom plus the number of lone pairs on the central atom gives us what is called the electron group geometry. Electron group geometries refer to the five geometries: linear, trigonal planar, tetrahedral, trigonal bipyramidal, or octahedral. If one or more of the bonding pairs of electrons is replaced with a lone pair, the electron geometry does not change but the the shape of the molecule is altered. Understanding the impact of lone pairs of electrons will be important when looking at properties of compounds that are affected by polarity of a molecule. We will focus on the tetrahedral electron group geometry since it is most relevant to organic and biological molecules. $\ce{E}$ represents a non-bonding pair of electrons.
$\ce{AB_3E}$
The ammonia molecule contains three single bonds and one lone pair on the central nitrogen atom.
The electron group geometry for a molecule with four electron pairs is tetrahedral, as was seen with $\ce{CH_4}$. In the ammonia molecule, one of the electron pairs is a lone pair rather than a bonding pair. Although the lone pair is not visible, it will affects the location and bond angles among other atoms in the molecule.
Recall that the bond angles in the tetrahedral $\ce{CH_4}$ molecule are all equal to $109.5^\text{o}$. One might expect the $\ce{H-N-H}$ bond angles in ammonia to be $109.5^\text{o}$ as well, but slight adjustments need to be made for the presence of lone pairs. Within the context of the VSEPR model, lone pairs of electrons are considered to be slightly more repulsive than bonding pairs of electrons, due to their closer proximity to the central atom. In other words, lone pairs "take up more space". Therefore the $\ce{H-N-H}$ angle is slightly less than $109.5^\text{o}$. Its actual value is approximately $107^\text{o}$.
$\ce{AB_2E_2}$
A water molecule consists of two bonding pairs and two lone pairs of electrons.
The water molecule, like the ammonia and methane molecules, has a tetrahedral electron group geometry. In the water molecule, two of the electron pairs are lone pairs rather than bonding pairs. The $\ce{H-O-H}$ bond angle is $104.5^\text{o}$, which is smaller than the bond angle in $\ce{NH_3}$.
Table $2$: Geometries of Molecules in Which the Central Atom Has One or More Lone Pairs
Atoms Plus Lone Pairs Around Central Atom Number of Surrounding Atoms Number of Lone Pairs Electron Group Geometry Example
3 $\ce{AB_2E}$ 2 1 Trigonal Planar $\ce{O_3}$
4 $\ce{AB_3E}$ 3 1 Tetrahedral $\ce{NH_3}$
4 $\ce{AB_2E_2}$ 2 2 Tetrahedral $\ce{H_2O}$
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/04%3A_Structure_and_Function/4.03%3A_Electron_Group_Geometry.txt |
Learning Outcomes
• Describe the importance and purpose of functional groups in organic reactions.
• Identify and name functional groups in compounds.
With over twenty million known organic compounds in existence, it would be very challenging to memorize chemical reactions for each one. Fortunately, molecules with similar functional groups tend to undergo similar reactions. A functional group is defined as an atom or group of atoms within a molecule that has similar chemical properties whenever it appears in various compounds. Even if other parts of the molecule are quite different, certain functional groups tend to react in certain ways.
We've already looked at alkanes, but they are generally unreactive. We primarily use alkanes as a source of energy when they are combusted. While the majority of functional groups involve atoms other than carbon and hydrogen, we will also look at some that include only carbon and hydrogen. Some of the most common functional groups are presented in the following sections.
Organic molecules vary greatly in size and when focusing on functional groups, we want to direct our attention to the atoms involved in the functional group. As a result, the abbreviation R is used in some examples. The letter R is used in molecular structures to represent the “Rest of the molecule”. It consists of a group of carbon and hydrogen atoms of any size. It is used as an abbreviation since a group of carbon and hydrogen atoms does not affect the functionality of the compound. In some molecules, you will see R, R’, or R’’ which indicates that the R groups in the molecule can be different from one another. For example, R might be –CH2CH3 while R’ is –CH2CH2CH2CH3.
Alkenes and Alkynes
Alkenes are hydrocarbons that contain one or more double bonds between neighboring carbon atoms. Alkynes contain one or more triple bonds between neighboring carbon atoms. The presence of a double or triple bond results in reactivity not present in alkane and alkynes (see figure below) undergo characteristic reactions which will be covered later.
Aromatics
The additional functional group that contains only carbon and hydrogen is an aromatic ring which is a six-carbon ring with alternative double bonds. The aromatic ring is also shown as a ring with a circle in the middle representing the double bonds. Aromatic rings (see figure below) are found in many compounds including steroids and medications.
Alcohols
The alcohol functional group involves an oxygen atom that is bonded to one hydrogen atom and one carbon atom. The carbon atom will be part of a larger organic structure. One way to indicate a generic alcohol would be with the formula $\ce{R-OH}$. $\ce{R}$ represents any organic fragment in which a carbon atom is directly bonded to the explicitly indicated functional group (in this case, $\ce{OH}$). The $\ce{R}$ group is typically a chain of carbon atoms.
Alcohols can be classified as primary, secondary, or tertiary based on the characteristics of the carbon to which it is attached. In a primary alcohol, the carbon bonded directly to the oxygen atom is also bonded to exactly one carbon atom, with the other bonds generally going to hydrogen atoms. In a secondary alcohol, the carbon is attached to two other carbon atoms, and in a tertiary alcohol, the carbon is bonded to three other carbon atoms. The type of alcohol being used will determine the product of certain reactions. Note the naming of alcohols as illustrated in the figure above. The location of the $\ce{-OH}$ group is indicated with the number of the carbon to which it is attached.
We are already familiar with several common alcohols. For example, ethanol $\left( \ce{CH_3CH_2OH} \right)$ is the alcohol present in alcoholic beverages. It is also widely used in the industrial manufacture of other chemicals. Methanol $\left( \ce{CH_3OH} \right)$ is used as a gasoline additive or alternative. Additionally, methanol can be used to manufacture formaldehyde, which is employed in the production of plastics, paints, and other useful substances. Isopropanol is commonly known as rubbing alcohol. In addition to its industrial uses, isopropanol is used to clean various surfaces, including computer monitors, whiteboards, and even skin (e.g., before getting blood drawn).
Ethers
The ether functional group consists of an oxygen atom that forms single bonds with two carbon atoms.
Ethers are good solvents for other organic compounds because of their low reactivity. They readily dissolve nonpolar molecules. Diethyl ether is perhaps the best known ether. It is widely used as a solvent and has been used as an inhalable anesthetic.
Although ethers themselves are relatively unreactive, they can be converted to peroxides after prolonged exposure to oxygen. Peroxides are very reactive and are often explosive at elevated temperatures. Many commercially available ethers come with a small amount of a peroxide scavenger dissolved in them to help prevent this type of safety hazard.
Thiol
The thiol functional group contains a sulfur atom bonded to a hydrogen atom. It is very similar to an alcohol functional group with the sulfur replacing the O.
Thiols are also called mercaptans which is derived from the Latin phrase for "capturing mercury" because of the strong bonds it forms with mercury-containing compounds. Some thiol compounds have a distinctive smell similar to rotten eggs. They are often added to natural gas, which itself has no odor, as a way to detect leaks since its odor can be detected by humans in very small amounts. A thiol group is also present in the amino acid cysteine which will be discussed later.
Amines
An amine consists of a nitrogen atom bonded to some combination of carbons and hydrogens.
Like alcohols, amines can be classified as primary, secondary, or tertiary. However, the rules for assigning these categories are slightly different. In an alcohol, the oxygen atom is always bonded to exactly one carbon atom, so we look at the branching on the adjacent carbon, not the oxygen atom itself. In a neutral amine, the nitrogen can be bonded to one, two, or three carbon atoms, and this is how we decide whether it is called a primary, secondary, or tertiary amine.
Neutral amines are weak bases, because the lone pair on nitrogen can act as a proton acceptor. Many smaller amines have very strong and offensive odors. For example, the aptly-named compounds cadaverine and putrescine are foul-smelling amines, formed as a part of the decay process after death.
Amines serve a wide variety of uses. Diphenylamine acts as a stabilizer for certain types of explosives. Amines are found as components in some lubricating materials, in developers, and are a part of waterproofing textiles. Some amines, such as novocaine, are used as anesthetics. Many pharmaceutical compounds contain amines, including 8 of the 10 most prescribed medications in 2012.
Aldehydes
A very common structural component of organic structures is the carbonyl, which is simply a carbon atom and an oxygen atom connected by a double bond. The reactivity of carbonyls is primarily dictated by the polarization of the $\ce{C=O}$ bond, but the surrounding atoms also play a role in its specific reaction pathways. While carbonyl is a component of many functional groups, it is not itself a functional group.
An aldehyde is a carbonyl in which the carbon atom is bonded to at least one hydrogen atom. The other group attached to the carbonyl may be an $\ce{R}$-group or a hydrogen atom. Because the hydrogen atom is so small, the partial positive charge on the carbonyl carbon is very easy for other molecules to approach, making aldehydes a particularly reactive type of carbonyl. Aldehydes are versatile reactants for a wide variety of organic syntheses. Many aldehydes also have distinctive flavors and aromas. For example, the flavor of cinnamon is primarily due to the molecule cinnamaldehyde, and vanillin is the aldehyde most responsible for the smell and taste of vanilla extract.
A special aldehyde is the molecule in which the carbonyl is bonded to two hydrogen atoms. This molecule, called formaldehyde, has a wide variety of uses. By itself, it can be used as a tissue preservative or as a very harsh disinfectant. It is also used as a precursor to various materials, including plastics, resins, and other polymers.
Ketones
A ketone involves a carbonyl in which the carbon atom makes single bonds with two $\ce{R}$-groups. Ketones undergo most of the same reactions as aldehydes, but they tend to be slightly less reactive. The simplest ketone is acetone, in which the carbonyl carbon is bonded to two $\ce{CH_3}$ groups. This ketone is commonly used to remove fingernail polish and serves as an industrial solvent. Methyl ethyl ketone is used as a paint stripper and a solvent. Ketones are also used in the production of various polymers, either as a building block or as a solvent. The $\ce{R}$-group in a ketone can be the same or different as seen in the example.
Carboxylic Acids
Carboxylic acids are another carbonyl-containing functional group, in which the carbon atom is bonded to an $\ce{OH}$ group on one side and either a carbon or hydrogen atom on the other.
As the name implies, carboxylic acids are weak acids. An $\ce{OH}$ group that is directly connected to a carbonyl will ionize to a small extent when dissolved in water. The reason for this is the relative stability of the resulting anion. A carboxylate ion (see figure below), in which the negative charge is spread over two different oxygen atoms through resonance structures, is more stable than an isolated oxygen-centered anion. The carboxylic acid and carboxylate ion are interchangeable. Carboxylate ions are often present in amino acids.
Carboxylic acids are used in a variety of environments. Formic acid acts as a protective chemical for many stinging insects and plants. Acetic acid gives vinegar its characteristic smell and flavor and is a fundamental biological and industrial building block. Carboxylic acids with longer carbon chains (fatty acids) are used by animals as a way of storing energy and are widely used in the manufacture of soaps. Some compounds contain multiple carboxylic acids within a single molecule. For example, citric acid (three carboxyl groups) is especially abundant in citrus fruits and it used as a flavoring and preservative in many foods and beverages.
Esters
An ester is similar to a carboxylic acid, in that it contains a carbonyl where the carbon is bonded to one additional oxygen atom and one carbon or hydrogen atom. However, the second oxygen atom is bonded to another carbon instead of to an acidic hydrogen atom. Structurally, carboxylic acids and esters are related to one another in the same way as alcohols and ethers.
Esters can be formed by heating carboxylic acids and alcohols in the presence of an acid catalyst. This process is reversible, and the starting materials can be regenerated by reacting an ester with water in the presence of a weak base.
Some esters have very pleasant odors, so they are used in the manufacture of many perfumes. Propyl acetate contributes to the odor of pears, while isoamyl acetate gives bananas their smell. This ester also serves as an alarm signal for honeybees. Esters are employed in the manufacture of fabrics (polyesters) and Plexiglass. Anesthetics such as procaine and benzocaine also contain esters.
Amides
An amide is a carbonyl in which the carbon is attached to one nitrogen atom and one carbon or hydrogen atom. Alternatively, we could define an amide as an amine in which one of the carbon atoms attached to the nitrogen is part of a carbonyl.
An amide can be formed by combining a carboxylic acid and an amine. Only primary and secondary amines can be sued to form amides, since they have a hydrogen that can be replaced with the carbonyl carbon; tertiary amines will not form amides. The amide shown in the figure above was formed from a carboxylic acid and a primary amine.
Amides are used as coloring agents in crayons, pencils, and ink. They are employed in the paper, plastic, and rubber industries. Polyacrylamide is a very widely used amide; it is involved in t he treatment of drinking water and sewage, and in plastics manufacture. The amide Kevlar is widely employed for the production of body armor, and nylon is another type of amide-based polymer.
Haloalkanes
The haloalkanes, also known as alkyl halides, are a group of chemical compounds comprised of an alkane with one or more hydrogens replaced by a halogen atom (Group 17 atom). There is a fairly large distinction between the structural and physical properties of haloalkanes and the structural and physical properties of alkanes. A
Haloalkanes are found in fire extinguishers, refrigerants, propellants, solvents, and medications. They are also a significant source of pollution and their use has been reduced or eliminated in some products. Chlorofluorocarbons (CFCs) were used as refrigerants in air-conditioners but were found to be a major cause of the depletion of the ozone layer. Research and development of alternatives began in the 1970s. Hydrochlorofluorocarbons (HCFCs) have been used for many years since they cause less damage to the ozone layer, but many countries agreed to eliminate HCFCs by 2020.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky)
4.05: Structure and Function (Exercises)
These are homework exercises to accompany Chapter 4 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Solutions are available below the questions.
Questions
(Click here for solutions)
Q4.1.1
Describe a Lewis structure.
Q4.1.2
How many bonds are usually formed by each element?
1. sulfur
2. hydrogen
3. nitrogen
4. phosphorus
Q4.1.3
How many bonds does oxygen tend to form? What combinations of bonds (single, double, triple) can it have?
Q4.1.4
How many bonds does carbon tend to form? What combinations of bonds (single, double, triple) can it have?
Q4.1.5
Using Table 4.1.1, draw the Lewis structure for each molecule.
a. H2S
b. CCl4
c. PF3
d. C2H4
(click here for solutions)
Q4.2.1
Given one type of structure (Lewis, condensed, or skeletal), draw the other two.
a. CH3CH2CH3
b.
c.
d. CH3CHBrCH2CH2CHClCH3
e. CH2(OH)CH2CH2CH3
f.
g.
(click here for solutions)
Q4.3.1
How does the presence of a lone pair affect the shape of the molecule?
Q4.3.2
What is the "ideal" bond angle in a tetrahedral (AB4) molecule? How does that change in trigonal pyramidal (AB3E)? Bent (AB2E2)?
Q4.3.3
What is the ideal bond angle for trigonal planar (AB3) molecules? What would be the expected bond angle for a bent (AB2E) molecule?
Q4.3.4
Determine the electron geometry for each of the following molecules.
a.
b.
c.
d.
e.
f.
(click here for solutions)
Q4.4.1
Circle and label the functional groups in each molecule. There may be more than one functional group in a molecule.
a.
b.
c.
d.
e.
f.
Q4.4.2
Draw a molecule which contains an ester, an alcohol, and an aldehyde.
Q4.4.3
Label each alcohol as primary, secondary, or tertiary.
a.
b.
c.
d.
Q4.4.4
Label each amine as primary, secondary, or tertiary.
a.
b.
c.
d.
e.
Answers
4.1: Lewis Electron Dot Structures
Q4.1.1
A Lewis structure shows the connections between atoms (single, double, or triple bonds) as well as any non-bonding electrons.
Q4.1.2
1. 2
2. 1
3. 3
4. 3
Q4.1.3
2 bonds total; 2 single bonds or 1 double bond
Q4.1.4
4 bonds total; 4 single or 2 single and 1 double or 2 double or 1 single and 1 triple
Q4.1.5
a.
b.
c.
d.
4.2: Representing Structures
Q4.2.1
Given one type of structure (Lewis, condensed, or skeletal), draw the other two.
Lewis Structure Condensed Structure Skeletal Structure
a. CH3CH2CH3
b. CH3CH(OH)C(CH3)(OH)CH2CH3
c. CH3C(O)CH3
d. CH3CHBrCH2CH2CHClCH3
e. CH2(OH)CH2CH2CH3
f. CH3CH(CH3)CH2CH2CH3
g. CH3CCCH3
4.3: Molecular Shapes
Q4.3.1
A lone pair causes the compression of other bond angles in the molecule.
Q4.3.2
AB4: 109.5°
AB3E: 107° (less than AB4)
AB2E2: 104.5° (less than AB3E)
Q4.3.3
AB3: 120°
AB2E: less than 120°
Q4.3.4
1. trigonal planar
2. linear
3. tetrahedral
4. linear
5. trigonal planar
6. tetrahedral
4.4: Functional Groups
Q4.4.1
a.
b.
c.
d.
e.
f.
Q4.4.2
Answers will vary. One example is shown.
Q4.4.3
1. secondary
2. primary
3. secondary
4. tertiary
Q4.4.4
1. primary
2. primary
3. secondary
4. tertiary
5. secondary | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/04%3A_Structure_and_Function/4.04%3A_Functional_Groups.txt |
• 5.1: 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.
• 5.2: Carbohydrate Structures
Carbohydrates are organic compounds that contain only carbon (C) , hydrogen (H) , and oxygen (O) . They contain a chain of carbons, an aldehyde or a ketone, and hydroxyl groups. Every carbon atom is attached to one oxygen atom. There are thousands of different carbohydrates, but they all consist of one or more smaller units called monosaccharides.
• 5.3: Polarity and Intermolecular Forces
In an ionic bond, one or more electrons are transferred from one atom to another. In a covalent bond, one or more pairs of electrons are shared between atoms. However, bonding between atoms of different elements is rarely purely ionic or purely covalent.
• 5.4: Chromatography
When chemists have a multi-component solution which may contain traces of important chemical species, they are faced with the challenge of detecting whether these chemicals are present in solution. To deal with these difficulties, chemists employ different methods to separate solutions into their components. Two essential techniques are distillation and chromatography.
• 5.5: Properties of Compounds (Exercises)
These are homework exercises to accompany Chapter 5 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Solutions are available below the questions.
05: Properties of Compounds
Learning Outcomes
• Define isomer.
• Determine the isomeric relationship between a pair of molecules.
• Identify the chiral centers in a molecule.
• Describe different types of 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. There are several different types of isomers which will be described and a flowchart (see figure below) can help you determine which type of isomers are present.
Conformational Isomers
Conformational isomers, also known as conformers, differ from one another by their rotation around a single bond. Rotations occur freely around single carbon-carbon bonds. Unlike double and triple bonds, which are "locked" in their orientation, single bonds have no such restrictions.
Structural Isomers
A structural isomer, also known as a constitutional isomer, is one in which two or more organic compounds have the same molecular formulas but different structures. The two molecules below have the same chemical formula, but are different molecules because they differ 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.
Stereoisomers
Stereoisomers have the same connectivity in their atoms but a different arrangement in three-dimensional space. There are different classifications of stereoisomers depending on how the arrangements differ from one another. Notice that in the structural isomers, there was some difference in the connection of atoms. For example, 1-butene has a double bond followed by two single bonds while 2-butene has a single bond, then a double bond, then a single bond. A stereoisomer will have the same connectivity among all atoms in the molecule.
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 bond. Therefore, there are two different ways to construct the 2-butene molecule (see figure below). 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. This occurs with a double bond or a ring. In addition, the two carbon atoms must each have two different groups attached in order for there to be geometric isomers. Propene (see figure below) has no geometric isomers because one of the carbon atoms (the one on the far left) involved in the double bond has two single hydrogens bonded to it.
Physical and chemical properties of geometric isomers are generally different. 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.
Optical Isomers
Stereoisomers that are not geometric isomers are known as optical isomers. Optical isomers differ in the placement of substituted groups around one or more atoms of the molecule. They were given their name because of their interactions with plane-polarized light. Optical isomers are labeled enantiomers or diastereomers.
Enantiomers are non-superimposable mirror images. A common example of a pair of enantiomers is your hands. Your hands are mirror images of one another but no matter how you turn, twist, or rotate your hands, they are not superimposable.
Objects that have non-superimposable mirror images are called chiral. When examining a molecule, carbon atoms with four unique groups attached are considered chiral. Look at the figure below to see an example of a chiral molecule. Note that we have to look beyond the first atom attached to the central carbon atom. The four circles indicate the four unique groups attached to the central carbon atom, which is chiral.
Another type of optical isomer are diastereomers, which are non-mirror image optical isomers. Diastereomers have a different arrangement around one or more atoms while some of the atoms have the same arrangement. As shown in the figure below, note that the orientation of groups on the first and third carbons are different but the second one remains the same so they are not the same molecule. The solid wedge indicates a group coming out of the page/screen towards you and the dashed line indicates that a group is going away from you "behind" the page/screen.
Epimers are a sub-group of diastereomers that differ at only one location. All epimers are diastereomers but not all diastereomers are epimers.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/05%3A_Properties_of_Compounds/5.01%3A_Isomers.txt |
Learning Outcomes
• Describe the structure and function of carbohydrates.
• Identify functional groups of carbohydrates.
• Give general name for a carbohydrate molecule (i.e. aldotetrose, ketopentose, etc)
• Label carbohydrates as either D- or L-enantiomers.
• Draw the mirror image of a carbohydrate molecule.
• Distinguish between monosaccharides, disaccharides, and polysaccharides.
• Describe the structure of complex carbohydrates.
• Recognize how carbohydrates determine blood type.
The brain is a marvelous organ. And it's a hungry one, too. The major fuel for the brain is the carbohydrate glucose. 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. Something to think about.
As a child, you may have been told that sugar is bad for you. Well, that's not exactly true. Essentially, carbohydrates are made of sugar, from a single sugar molecule to thousands of sugar molecules all attached together. Why? One reason is to store energy. But that does not mean you should eat it by the spoonful.
Carbohydrates
Carbohydrates are organic compounds that contain only carbon $\left( \ce{C} \right)$, hydrogen $\left( \ce{H} \right)$, and oxygen $\left( \ce{O} \right)$. They contain a chain of carbons, an aldehyde or a ketone, and hydroxyl groups. Every carbon atom is attached to one oxygen atom. There are thousands of different carbohydrates, but they all consist of one or more smaller units called monosaccharides.
Monosaccharides
The general formula for a monosaccharide is $\left( \ce{CH_2O} \right) _n$, where $n$ can be any number greater than two. For example, if $n$ is 6, then the formula can be written $\ce{C_6H_{12}O_6}$. This is the formula for the monosaccharide glucose. Another monosaccharide, fructose, has the same chemical formula as glucose, but the atoms are arranged differently. Carbohydrates have many isomers because of the arrangement of the $\ce{-OH}$ groups in their structures. Compare the glucose and fructose molecules in the figure below. Can you identify their differences? The only differences are the positions of some of the atoms. These differences affect the properties of the two monosaccharides.
Monosaccharides can be classified by the number of carbon atoms they contain: diose (2), triose (3), tetrose (4), pentose (5), hexose (6), heptose (7), and so on. They can also be classified based on whether or not they contain an aldehyde (aldose) or ketone (ketose). We can also combine these two designations to refer to classes of carbohydrates. For example, an aldohexose is a carbohydrate (indicated by the -ose ending) with six carbons (hex) and an aldehyde group (aldo). A ketopentose is a carbohydrate with a ketone and 5 carbons. Both glucose and fructose are hexoses because they contain six carbons but glucose is an aldohexose while fructose (also known as "fruit sugar") is a ketohexose. Other common monosaccharides include galactose (part of lactose), xylose ("wood sugar"), ribose (in RNA), and deoxyribose (in DNA).
Fischer Projections
There are several ways to draw the structure of carbohydrate molecules. The Fischer projection (straight chain) makes it appear that the molecule is flat but it is a three-dimensional molecule. Although we will not be concerned with the 3D orientation, know that the arrangement in the Fischer projection does provide information about the orientation of atoms around each carbon atom.
These projections simplify the drawing of molecules yet retain important information about the arrangement of atoms within the structure. The figure below shows the Fischer projections for the enantiomers (non-superimposable mirror images) of ephedrine and pseudoephedrine. While it may appear that the molecules are the same, they are not because the Fischer projection does not explicitly show the three-dimensional geometry of the molecule.
Fischer projections provide an easy way to distinguish among the many, similar carbohydrate molecules that exist. For example, there are sixteen aldohexoses (see figure below). Note the different patterns of the $\ce{-OH}$ bonds on the left and right sides of the Fischer projection for each. Changing the orientation of one or more of the $\ce{-OH}$ groups changes the identity of the molecule.
Each carbohydrate molecule also has an enantiomer and the two are designated as the D- and L- versions of the compound. The designation is based on the orientation of the $\ce{-OH}$ group on the chiral carbon farthest from the aldehyde or ketone. The structures of D-glucose and L-glucose are shown in the figure below. The orientation of all $\ce{-OH}$ groups are reversed but only the arrangement of at the carbon indicated by the arrow determines whether the sugar is a D-sugar with the $\ce{-OH}$ group on the right or an L-sugar with the $\ce{-OH}$ group on the left.
Haworth Structures
Like Fischer projections, the Haworth structures provide information about a molecule's three-dimensional structure without explicitly showing it in the drawing. Carbohydrates are present in the body in both the chain and ring forms with the latter being more common. Haworth projections provide a simple way to display the ring structures and may or may not show the hydrogen atoms attached to each carbon. Remember, every carbon has four bonds so hydrogens are implied when the structure does not show all four bonds. When the cyclic monosaccharide forms, there are two versions that can form, called $\alpha$ (alpha\) and $\beta$ (beta) (see figure below). The arrow in the figure indicates the anomeric carbon which it the location where the ring forms and where the orientation of the $\ce{-OH}$ group can change. The orientation of the other $\ce{-OH}$ groups are fixed because they are determined by the orientation of the $\ce{-OH}$ groups in the particular monosaccharide (compare to the orientation of the $\ce{-OH}$ groups on the left and right sides of the Fischer projections). Each monosaccharide can exist in either $\alpha$ or $\beta$ form and the two forms will interconvert as the ring opens and closes. The $\alpha$ form occurs when the $\ce{-OH}$ group on the anomeric carbon is pointing down and the $\beta$ version exists when the $\ce{-OH}$ group on the anomeric carbon is pointing up.
As a result of these different orientations, we can have four forms of each monosaccharide. For example, glucose can exist as $\alpha$-D-glucose, $\alpha$-L-glucose, $\beta$-D-glucose, or $\beta$-L-glucose. While the $\alpha$ and $\beta$ forms can interconvert, the same cannot be said for D and L versions. Naturally occurring monosaccharides are in the D version, called "D sugars". The arrangement within the D or L form is fixed and they cannot interconvert.
Disaccharides
If two monosaccharides bond together, they form a carbohydrate called a disaccharide. Two monosaccharides will bond together through a dehydration reaction, in which a water molecule is lost. A dehydration reaction is a condensation reaction, a chemical reaction in which two molecules combine to form one single molecule, losing a small molecule in the process. In the dehydration reaction, this small molecule is water. The bond between two monosaccharides is known as a glycosidic bond.
An example of a disaccharide is sucrose (table sugar), which consists of the monosaccharides glucose and fructose (see figure below). Other common disaccharides include lactose ("milk sugar") and maltose. Monosaccharides and disaccharides are also called simple sugars. They provide the major source of energy to living cells.
Got milk?
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. 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.
Oligosaccharides
An oligosaccharide is a saccharide polymer containing a small number (typically two to ten) of monosaccharides. Oligosaccharides can have many functions; for example, they are commonly found on the plasma membrane of animal cells where they can play a role in cell-cell recognition. In general, they are found attached to compatible amino acid side-chains in proteins or to lipids.
Oligosaccharides are often found as a component of glycoproteins or glycolipids. They are often used as chemical markers on the outside of cells, often for cell recognition. Oligosaccharides are also responsible for determining blood type.
Blood Type
Carbohydrates attached to red blood cells also determine blood type (see figure below). Of the four blood types, type O has the fewest types of saccharides attached to it while type AB has the most. As a result, type O blood is considered the universal donor because it doesn't have any saccharides present that will appear as foreign when transfused into blood of another type. The reverse is not true. For example, if type A blood is given to a patient with type O blood, it will be rejected by the body because there is an unknown species being introduced to the body. Type A blood cells contain N-acetyl-galactosamine which is not present in type O blood. A person with type O blood would undergo rejection upon receiving type A blood. The Rhesus factor (Rh) in blood also affects donor and acceptor properties but it does not depend on carbohydrates. The Rh factor is determined by the presence (Rh+) or absence (Rh-) of a specific protein on the surface of red blood cells.
Polysaccharides
Polysaccharides are long carbohydrate molecules of repeated monomer units joined together by glycosidic bonds. A polysaccharide may contain anywhere from a few monosaccharides to several thousand monosaccharides. Polysaccharides are also called complex carbohydrates. Polysaccharides have a general formula of $\ce{C_x(H_2O)_y}$, where $x$ is usually a large number between 200 and 2500.
Starches are one of the more common polysaccharides. Starch is made up of a mixture of amylose $\left( 15 \right.$-$\left. 20\% \right)$ and amylopectin $\left( 80 \right.$-$\left. 85\% \right)$. Amylose consists of a linear chain of several hundred glucose molecules and amylopectin is a branched molecules made of several thousand glucose units. Starches can be digested by hydrolysis reactions, catalyzed by enzymes called amylases, which can break the glycosidic bonds. Humans and other animals have amylases, so they can digest starches. Potato, rice, wheat, and maize are major sources of starch in the human diet. The formations of starches are the ways that plants store glucose. Glycogen is sometimes referred to as animal starch. Glycogen is used for long-term energy storage in animal cells. Glycogen is made primarily by the liver and the muscles.
Are we there yet?
As the weather warms up, the runners come out. Not just the casual joggers, but those really serious ones 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. The rest of us will just sit at the sidewalk restaurant eating our spaghetti and enjoying watching other people work hard.
The main functions of polysaccharides are to store energy and form structural tissues. Examples of several other polysaccharides and their roles are listed in the table below. These complex carbohydrates play important roles in living organisms.
Table $1$: Complex Carbohydrates
Complex Carbohydrate Function Organism
Starch Stores energy
Plants
Amylose Stores energy Plants
Glycogen Stores energy
Animals
Cellulose Forms cell walls
Plants
Chitin Forms an exoskeleton
Some animals
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/05%3A_Properties_of_Compounds/5.02%3A_Carbohydrate_Structures.txt |
Learning Outcomes
• Define electronegativity.
• Describe how the electronegativity difference between two atoms in a covalent bond results in the formation of a nonpolar covalent, polar covalent, or ionic bond.
• Predict polarity of a molecule.
• Describe how molecular geometry plays a role in determining whether a molecule is polar or nonpolar.
• Distinguish between the following three types of intermolecular forces: dipole-dipole forces, London dispersion forces, and hydrogen bonds.
• Identify types of intermolecular forces in a molecule.
• Describe how chemical bonding and intermolecular forces influence the properties of various compounds.
In an ionic bond, one or more electrons are transferred from one atom to another. In a covalent bond, one or more pairs of electrons are shared between atoms. However, bonding between atoms of different elements is rarely purely ionic or purely covalent.
Bond Polarity
Bond polarity is determined by the difference in electronegativity and is defined as the relative ability of an atom to attract electrons when present in a compound. The electronegativities of various elements are shown below. Note that elecronegativity values increase from left to right and from bottom to top on the periodic table. The degree to which a given bond is ionic or covalent is determined by calculating the difference in electronegativity between the two atoms involved in the bond.
As an example, consider the bond that occurs between an atom of potassium and an atom of fluorine. Using the table, the difference in electronegativity is $4.0 - 0.8 = 3.2$. Because the difference in electronegativity is relatively large, the bond between the two atoms is primarily ionic. Since the fluorine atom has a much larger attraction for electrons than the potassium atom does, the valence electron from the potassium atom is considered to have completely transferred to the fluorine atom. The figure below shows how the difference in electronegativity relates to the ionic or covalent character of a chemical bond.
According to the figure above, a difference in electronegativity ($\Delta$ EN) greater than 1.7 results in a bond that is mostly ionic in character.
Nonpolar Covalent Bonds
A bond in which the electronegativity difference is less than 1.7 is considered to be mostly covalent in character. However, a distinction is often made between two general types of covalent bonds. A nonpolar covalent bond is a covalent ond in which the onding electrons are shared equally between the two atoms. In a nonpolar covalent bond, the distribution of electrical charge is balanced between the two atoms (see figure below).
The two chlorine atoms share the pair of electrons in the single covalent bond equally, and the electron density surrounding the $\ce{Cl_2}$ molecule is symmetrical. Any diatomic molecule in which the two atoms are the same element must be joined by a nonpolar covalent bond.
There are seven diatomic elements, which are elements whose natural form is of a diatomic molecule. They are hydrogen $\left( \ce{H_2} \right)$, nitrogen $\left( \ce{N_2} \right)$, oxygen $\left( \ce{O_2} \right)$, fluorine $\left( \ce{F_2} \right)$, chorine $\left( \ce{Cl_2} \right)$, bromine $\left( \ce{Br_2} \right)$, and iodine $\left( \ce{I_2} \right)$. By forming a diatomic molecule, both atoms in each of these molecules satisfy the octet rule, resulting in a structure that is much more stable than the isolated atoms
Notice from the figure above that molecules in which the electronegativity difference is very small (<0.4) are also considered nonpolar covalent. An example would be a bond between chlorine and bromine ($\Delta$ EN $= 3.16 - 2.96 = 0.20$).
Polar Covalent Bonds
A bond in which the electronegativity difference between the atoms is between 0.4 and 1.7 is called a polar covalent bond. A polar covalent bond is a covalent bond in which the atoms have an unequal attraction for electrons, so the sharing is unequal. In a polar covalent bond, sometimes simply called a polar bond, the distribution of shared electrons within the molecule is no longer symmetrical (see figure below).
The hydrogen fluoride molecule has an electronegativity difference of 1.9, which places it in the category of being slightly ionic. However, the hydrogen ion $\left( \ce{H^+} \right)$ is so very small that it is not capable of adopting the crystal lattice structure of an ionic compound. Hydrogen fluoride is a highly polar molecule. Because of its greater electronegativity, the electron density around the fluorine atom is much higher than the electron density around the hydrogen atom.
An easy way to illustrate the uneven electron distribution in a polar covalent bond is to use the Greek letter delta $\left( \delta \right)$ along with a positive or negative sign to indicate that an atom has a partial positive or negative charge.
The atom with the greater electronegativity acquires a partial negative charge, while the atom with the lesser electronegativity acquires a partial positive charge. The delta symbol is used to indicate that the quantity of charge is less than one. A crossed arrow can also be used to indicate the direction of greater electron density.
Example $1$
Which type of bond will form between each of the following pairs of atoms?
1. $\ce{C}$ and $\ce{O}$
2. $\ce{Na}$ and $\ce{N}$
3. $\ce{B}$ and $\ce{H}$
Solution
Step 1: List the known quantities and plan the problem.
Known
Using the electronegativity chart:
• $\ce{C} = 2.5, \: \ce{O} = 3.5$
• $\ce{Na} = 0.9, \: \ce{N} = 3.0$
• $\ce{B} = 2.0, \: \ce{H} = 2.1$
Step 2: Solve.
Calculate the difference and use the diagram above to identify the bond type.
$3.5 - 2.5 = 1.0 \rightarrow \ce{C-O} \: \text{bond is polar covalent}$
$3.0 - 0.9 = 2.1 \rightarrow \ce{Na-N} \: \text{bond is ionic}$
$2.1 - 2.0 = 0.1 \rightarrow \ce{B-H} \: \text{bond is nonpolar covalent}$
Step 3: Think about your result.
Bonds between nonmetal atoms are generally covalent in nature (A and C), while bond between a metal atom and a nonmetal atom are generally ionic.
Molecular Polarity
A polar molecule is a molecule in which one end of the molecule is slightly positive, while the other end is slightly negative. A diatomic molecule that consists of a polar covalent bond, such as $\ce{HF}$, is a polar molecule. The two electrically charged regions on either end of the molecule are called poles, similar to a magnet having a north and a south pole. A molecule with two poles is called a dipole. Hydrogen fluoride is a dipole. A simplified way to depict molecules is pictured below (see figure below).
When placed between oppositely charged plates, polar molecules orient themselves so that their positive ends are closer to the negative plate and their negative ends are closer to the positive plate (see figure below).
Experimental techniques involving electric fields can be used to determine if a certain substance is composed of polar molecules and to measure the degree of polarity.
For molecules with more than two atoms, the molecular geometry must also be taken into account when determining if the molecule is polar or nonpolar. Pictured below (see figure below) is a comparison between carbon dioxide and water. Carbon dioxide $\left( \ce{CO_2} \right)$ is a linear molecule. The oxygen atoms are more electronegative than the carbon atom, so there are two individual dipoles pointing outward from the $\ce{C}$ atom to each $\ce{O}$ atom. However, since the dipoles are of equal strength and are oriented in this way, they cancel each other out, and the overall molecular polarity of $\ce{CO_2}$ is zero.
Water is a bent molecule because of the two lone pairs on the central oxygen atom. The individual dipoles point from the $\ce{H}$ atoms toward the $\ce{O}$ atom. Because of the shape the dipoles do not cancel each other out, and the water molecule is polar. In the figure below, the net dipole is shown in blue and points upward.
Some other molecules are shown below (see figure below). Notice that a tetrahedral molecule such as $\ce{CH_4}$ is nonpolar. However, if one of the peripheral $\ce{H}$ atoms is replaced by another atom that has a different electronegativity, the molecule becomes polar. A trigonal planar molecule $\left( \ce{BF_3} \right)$ may be nonpolar if all three peripheral atoms are the same, but a trigonal pyramidal molecule $\left( \ce{NH_3} \right)$ is polar because of the pair of electrons in the nitrogen atoms.
Intermolecular Forces
Covalent and ionic bonds can be called intramolecular forces: forces that act within a molecule or crystal. Molecules also attract other molecules. Intermolecular forces are attractions that occur between molecules. Intermolecular forces are weaker than either ionic or covalent bonds. However, the varying strengths of different types of intermolecular forces are responsible for physical properties of molecular compounds such as melting and boiling points and the amount of energy needed for changes in state.
London Dispersion Forces
Dispersion forces are the weakest of all intermolecular forces. They are often called London forces after Fritz London (1900 - 1954), who first proposed their existence in 1930. London dispersion forces are intermolecular forces that occur between all atoms and molecules due to the random motion of electrons.
For example, the electron cloud of a helium atom contains two electrons, and, when averaged over time, these electrons will distribute themselves evenly around the nucleus. However, at any given moment, the electron distribution may be uneven, resulting in an instantaneous dipole. This weak and temporary dipole can subsequently influence neighboring helium atoms through electrostatic attraction and repulsion. The formation of an induced dipole is illustrated below.
The instantaneous and induced dipoles are weakly attracted to one another. The strength of dispersion forces increases as the total number of electrons in the atoms or nonpolar molecules increases. The halogen group consists of four elements that all take the form of nonpolar diatomic molecules. Listed below is a comparison of the melting and boiling points for each.
Table $1$: Melting and Boiling Points of Halogens
Molecule Total Number of Electrons Melting Point $\left( ^\text{o} \text{C} \right)$ Boiling Point $\left( ^\text{o} \text{C} \right)$ Physical State at Room Temperature
$\ce{F_2}$ 18 -220 -188 gas
$\ce{Cl_2}$ 34 -102 -34 gas
$\ce{Br_2}$ 70 -7 59 liquid
$\ce{I_2}$ 106 114 184 solid
The dispersion forces are strongest for iodine molecules because they have the greatest number of electrons. The relatively stronger forces result in melting and boiling points which are the highest of the halogen group. These forces are strong enough to hold iodine molecules close together in the solid state at room temperature. The dispersion forces are progressively weaker for bromine, chlorine, and fluorine, as illustrated by their steadily lower melting and boiling points. Bromine is a liquid at room temperature, while chlorine and fluorine are gases. Because gaseous molecules are so far apart from one another, intermolecular forces are nearly nonexistent in the gas state, and so the dispersion forces in chlorine and fluorine only become measurable as the temperature decreases and they condense into the liquid state.
Dipole-Dipole Forces
Dipole-dipole forces are the attractive forces that occur between polar molecules (see figure below). A molecule of hydrogen chloride has a partially positive hydrogen atom and a partially negative chlorine atom. A collection of many hydrogen chloride molecules will align themselves so that the oppositely charged regions of neighboring molecules are near each other.
Hydrogen Bonding
The attractive force between water molecules is an unusually strong type of dipole-dipole interaction. Water contains hydrogen atoms that are bound to a highly electronegative oxygen atom, making for very polar bonds. The partially positive hydrogen atom of one molecule is then attracted to the oxygen atom of a nearby water molecule (see figure below).
A hydrogen bond is an intermolecular attractive force in which a hydrogen atom, that is covalently bonded to a small, highly electronegative atom, is attracted to a lone pair of electrons on an atom in a neighboring molecule. Hydrogen bonds are very strong compared to other dipole-dipole interactions, but still much weaker than a covalent bond. A typical hydrogen bond is about $5\%$ as strong as a covalent bond.
Hydrogen bonding occurs only in molecules where hydrogen is covalently bonded to one of three elements: fluorine, oxygen, or nitrogen. These three elements are so electronegative that they withdraw the majority of the electron density from the covalent bond with hydrogen, leaving the $\ce{H}$ atom very electron-deficient. Because the hydrogen atom does not have any electrons other than the ones in the covalent bond, its positively charged nucleus is almost completely exposed, allowing strong attractions to other nearby lone pairs of electrons.
The hydrogen bonding that occurs in water leads to some unusual, but very important properties. Most molecular compounds that have a mass similar to water are gases at room temperature. However, because of the strong hydrogen bonds, water molecules are able to stay condensed in the liquid state. The figure below shows how its bent shape and the presence of two hydrogen atoms per molecule allows each water molecule to hydrogen bond with several other molecules.
In the liquid state, the hydrogen bonds of water can break and reform as the molecules flow from one place to another. When water is cooled, the molecules begin to slow down. Eventually, when water is frozen to ice, the hydrogen bonds become more rigid and form a well-defined network (see figure below).
The bent shape of the molecules leads to gaps in the hydrogen bonding network of ice. Ice has the very unusual property that its solid state is less dense than its liquid state. As a result, ice floats in liquid water. Virtually all other substances are denser in the solid state than in the liquid state. Hydrogen bonds also play a very important biological role in the physical structures of proteins and nucleic acids.
Boiling Points and Bonding Types
In order for a substance to enter the gas phase, its particles must completely overcome the intermolecular forces holding them together. Therefore, a comparison of boiling points is essentially equivalent to comparing the strengths of the attractive intermolecular forces exhibited by the individual molecules. For small molecular compounds, London dispersion forces are the weakest intermolecular forces. Dipole-dipole forces are somewhat stronger, and hydrogen bonding is a particularly strong form of dipole-dipole interaction. However, when the mass of a nonpolar molecule is sufficiently large, its dispersion forces can be stronger than the dipole-dipole forces in a lighter polar molecule. Thus, nonpolar $\ce{Cl_2}$ has a higher boiling point than polar $\ce{HCl}$.
Table $2$: Intermolecular Forces and Boiling Points
Substance Strongest Intermolecular Force Boiling Point $\left( ^\text{o} \text{C} \right)$
$\ce{H_2}$ dispersion -253
$\ce{Ne}$ dispersion -246
$\ce{O_2}$ dispersion -183
$\ce{Cl_2}$ dispersion -34
$\ce{HCl}$ dipole-dipole -85
$\ce{HBr}$ dipole-dipole -66
$\ce{H_2S}$ dipole-dipole -61
$\ce{NH_3}$ hydrogen bonding -33
$\ce{HF}$ hydrogen bonding 20
$\ce{H_2O}$ hydrogen bonding 100
Supplemental Resources
• Electronegativity: www.chemguideco.uk/atoms/bond...elecroneg.html
• Intermolecular Bonding - van der Waals Forces: www.chemguidecouk/atoms/bonding/vdw.html
• Intermolecular Bonding - Hydrogen Bonds: www.chemguide.co.uk/bonding/hbond.html
• Ionic bond formation: www.dlt.ncssm/edu/core/Chapte...icBonding.html
• Polar covalent bond formation: http://www.dlt.ncssm.edu/core/Chapte...arBonding.html
• Nonpolar covalent bond formation: www.dlt.ncssm/edu/core/Chapte...ntBonding.html
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/05%3A_Properties_of_Compounds/5.03%3A_Polarity_and_Intermolecular_Forces.txt |
Learning Outcomes
• Define chromatography.
• Distinguish between the stationary and mobile phases.
• Explain how components of a substance are separated based on attraction to each phase.
When a pure substance is mixed with another pure substance in which it is soluble, the substances become completely interspersed at the molecular level. In thinking about making solutions at the molecular level an analogy to a can of marbles may be useful. In the analogy, a layer of red marbles is placed in the bottom of a can and covered with a second layer of white marbles. After shaking the can for a short time, the marbles are mixed randomly.
Now let us imagine that you want to collect all the red marbles again. If you simply shake the can, it is unlikely that you will ever divide the marbles into two layers, each with only one kind of marble. Similarly, if two miscible liquids are combined, a chemist cannot simply un-mix the liquids into pure components.
Continuing the analogy, what if a few green marbles and blue marbles are placed into the can? Given enough red and white marbles, it may be difficult to determine that the green marbles and blue marbles are actually there. Similarly, when chemists have a multi-component solution which may contain traces of important chemical species, they are faced with the challenge of detecting whether these chemicals are present in solution.
To deal with these difficulties, chemists employ different methods to separate solutions into their components. Two essential techniques are distillation and chromatography.
Another useful set of techniques for separating mixtures is called chromatography. Perhaps the simplest of these techniques to describe is paper chromatography, an example of which is shown in the video below.
Video: Simple paper chromatography. A simple demonstration on paper chromatography using marker ink and water
Three substances are applied to a strip of chromatography-grade paper (the stationary phase of this experiment). As the liquid level rises and meets the spots, the sample partially dissolves in the liquid (the mobile phase because it is moving) and travels up the plate within the solution. Different substances will travel different distances along the plate. The distance that a substance will travel depends on how strongly it adheres to the stationary phase (a process called adsorption) versus how much time it spends dissolved in the mobile phase.
The more a substance adsorbs, the less it dissolves and the less it moves along the plate. The pink and blue spots at the end of the video are examples of substances highly adsorbed to the stationary phase. The less a substance adsorbs, the more it dissolves and the farther it travels, such as the yellow on the far left. The process is continued until a good separation is created. In this manner, a mixture of substances may be separated such as the middle sample, which was originally green but separated into blue and yellow. Notice that while it was not initially obvious that the middle spot contained both substances, this fact is clear after performing paper chromatography.
All forms of chromatography work on the same general principle as paper chromatography. There is always a stationary phase which does not move and a mobile phase which does. The various components in the mixture being chromatographed separate from each other because they are more strongly held by one phase or the other. Those which have the greatest affinity for the mobile phase move along the fastest.
The most important form of chromatography is gas chromatography or vapor-phase chromatography. A long column is packed with a finely divided solid whose surface has been coated with an inert liquid. This liquid forms the stationary phase. The mobile phase is provided by an inert carrier gas, such as He or N2, which passes continuously through the column (seen below), among the solid particles. A liquid sample can be injected into the gas stream at the sample injector and vaporized just before it enters the tube. As this sample is carried through the column by the slow stream of gas, those components which are most soluble in the inert liquid are held up, while the less-soluble components move on more rapidly. The components thus emerge one by one from the end of the tube, into the detector. In this way it is possible to separate and analyze mixtures of liquids which it would be impossible to deal with by distillation or any other technique.
The development of chromatography is one of the major revolutions in technique in the history of chemistry, comparable to that which followed the development of an accurate balance. Separations which were previously considered impossible are now easily achieved, sometimes with quite simple apparatus. This technique is particular essential to the science of biochemistry, in which complex mixtures are almost always encountered. In the field of environmental chemistry, chromatography has helped us separate and detect very low concentrations of contaminants like DDT or PCB (polychlorinated biphenyls). The major drawback to chromatography is that it does not lend itself to large-scale operation. As a result it remains largely a laboratory, rather than an industrial, technique for separating mixtures. | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/05%3A_Properties_of_Compounds/5.04%3A_Chromatography.txt |
These are homework exercises to accompany Chapter 5 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Solutions are available below the questions.
Questions
(click here for solutions)
Q5.1.1
Define isomer.
Q5.1.2
Circle the chiral carbons in each structure.
a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
Q5.1.3
Determine whether each molecule can have a geometric isomer.
Q5.1.4
For each pair, determine if they are isomers. If they are isomers, identify the type.
a.
b.
c.
d.
e.
f.
g.
h.
Q5.1.5
Draw two isomers with the formula C2H6O
(click here for solutions)
Q5.2.1
A monosaccharide has 4 carbon atoms. What is its chemical formula?
Q5.2.2
What are the differences between monosaccharides, disaccharides, oligosaccharides, and polysaccharides?
Q5.2.3
Compare and contrast simple and complex carbohydrates.
Q5.2.4
Draw each of the following structures.
1. aldoheptose
2. ketopentose
3. aldotriose
4. ketotetrose
Q5.2.5
Draw the Fischer projections of two aldohexoses that are different from the four given in Figure 5.2.3.
Q5.2.6
Draw the enantiomers of D-glucose, D-allose, D-mannose, and D-galactose. (See Figure 5.2.3).
Q5.2.7
Determine if each of the following pairs are diastereomers? Epimers? (Look up the structures in the chapter or online.)
1. D-glucose and D-allose
2. D-glucose and D-galactose
3. D-allose and D-mannose
4. D-allose and D-galactose
Q5.2.8
For each of the given Haworth structures, identify the sugar as being in the α or β form.
a.
b.
c.
d.
e.
Q5.2.9
Refer to Figure 5.2.7 when answering these questions.
1. Which blood type is the universal donor? Why?
2. Which blood type is the universal acceptor? Why?
3. Type A blood can be donated to recipients with which types of blood?
4. Type B recipients can accept which types of blood?
Q5.3.1
Define electronegativity.
Q5.3.2
Describe the periodic trends for electronegativity values.
Q5.3.3
Distinguish between nonpolar and polar covalent bonds.
Q5.3.4
Describe the bond between each pair of elements as ionic, polar covalent, or nonpolar covalent. Refer to Table 5.3.1 when answering this question. For the exam, if you need specific values, a table of electronegativity values will be provided. Note that many questions can be answered without the table by knowing the periodic trends.
1. N and O
2. C and P
3. Si and Cl
4. Al and F
5. Al and I
6. P and S
7. C and N
8. B and Cl
9. Be and Br
10. Si and P
Q5.3.5
Place the following bonds in order from least to most polar. Refer to Table 5.3.1 when answering this question.
1. Fe-N
2. H-Cl
3. Ca-O
4. C-S
Q5.3.6
Place the following bonds in order from least to most polar. Use periodic trends to determine the correct order without looking at electronegativity values.
1. PCl
2. SCl
3. PBr
4. CBr
Q5.3.7
Label each of the molecules as nonpolar or polar covalent.
1. CO2
2. CCl4
3. NH3
Q5.3.8
Describe the types of molecules that have the following types of intermolecular forces.
1. London dispersion forces
2. dipole-dipole forces
3. hydrogen bonding
Q5.3.9
Why are the intermolecular forces in H2O and H2S so different from one another?
Q5.3.10
What type(s) of intermolecular forces are present in each of the molecules in the question 5.3.7?
Q5.3.11
What is the relationship between the strength of intermolecular forces in a molecule and its boiling point?
Q5.3.12
Rank the following in order of increasing (smallest to greatest) boiling point.
1. N2
2. CH3OH
3. PH3
Q5.4.1
Define chromatography.
Q5.4.2
List two types of chromatography.
Q5.4.3
Distinguish between the stationary and mobile phases.
5.1: Isomers
Q5.1.1
Isomers are molecules with the same chemical formula but a different structure or arrangement of atoms.
Q5.1.2
Circle the chiral carbons in each structure.
a.
b.
c. No chiral carbons
d. No chiral carbons
e.
f.
g. No chiral carbons.
h.
i.
j.
Q5.1.3
1. No, because of two H atoms on left side of double bond.
2. Yes, because the chlorine atoms are shown trans but could be drawn cis.
3. Yes, because the two groups could be on the same side of the ring or on opposite sides.
4. No, because you need a double bond or a ring to have a geometric isomer.
5. No, because there is only one non-hydrogen group on the ring.
6. Yes, because each carbon in the double bond has two different groups so it could be described as cis or trans.
Q5.1.4
For each pair, determine if they are isomers. If they are isomers, identify the type.
1. structural
2. conformational
3. not isomers
4. structural
5. structural
6. not isomers
7. enantiomers
8. same molecule
Q5.1.5
5.2: Carbohydrate Structures
Q5.2.1
C4H8O4
Q5.2.2
Monosaccharides are a single carbohydrate molecule. Disaccharides have two carbohydrate molecules bonded together. Polysaccharides have three or more carbohydrate molecules bonded together.
Q5.2.3
Simple carbohydrates are monosaccharides or disaccharides and are easily broken down in the body. Complex carbohydrates are polysaccharides such as starches and fiber and take longer to break down in the body.
Q5.2.4
Answers will vary due to orientation of H and OH at each chiral carbon.
a.
b.
c.
d.
Q5.2.5
Any two of the structures shown here in red. The structures in black are from Figure 5.2.3.
Q5.2.6
L-glucose L-allose L-mannose L-galactose
Q5.2.7
1. Epimers because the orientation differs at a single chiral carbon.
2. Diastereomers because the orientation differs at multiple (but not all) chiral carbons.
3. Diastereomers because the orientation differs at multiple (but not all) chiral carbons.
4. Diastereomers because the orientation differs at multiple (but not all) chiral carbons.
Q5.2.8
1. beta
2. alpha
3. beta
4. alpha
5. beta
Q5.2.9
1. O because it has the fewest types of carbohydrates.
2. AB because it has all carbohydrates found on red blood cells.
3. A can give to A or AB.
4. B can receive B or O.
5.3: Polarity and Intermolecular Forces
Q5.3.1
Electronegativity is an atom's attraction to an electron in a bond.
Q5.3.2
Electronegativity increases from left to right and from bottom to top with a maximum at fluorine.
Q5.3.3
Nonpolar covalent bonds have even sharing of electrons between atoms while polar electrons share electrons unevenly.
Q5.3.4
1. polar
2. nonpolar
3. polar
4. ionic
5. polar
6. nonpolar
7. polar
8. polar
9. polar
10. nonpolar
Q5.3.5
C-S < H-Cl < Fe-N < Ca-O
Calculate the difference in electronegativity for each bond. The smaller the difference, the more nonpolar; the greater the difference, the more polar. If a bond has a large enough difference, it is so polar that it is considered ionic.
Q5.3.6
S-Cl < P-Cl < P-Br < C-Br
• S and Cl are closest together so they will have the smallest difference in electronegativity and be the least polar.
• S-Cl and P-Cl can be compared since they share a common element. P is farther left than S, so the electronegativity of P must be less than that of S. Therefore, the difference between P and Cl must be greater than the difference between S and Cl since we are comparing them both to the same element (Cl).
• Compare P-Cl and P-Br. Since Br is further down the periodic table, it has a lower electronegativity value than Cl. Therefore the difference between P and Cl is less than the difference between P and Br. P-Br is more polar than P-Cl.
• Compare P-Br and C-Br. C is further from Br than P is so the difference in electronegativity is greater for C-Br than P-Br making C-Br the more polar bond.
Q5.3.7
1. nonpolar
2. nonpolar
3. polar
Q5.3.8
1. all molecules
2. polar molecules
3. molecules with an H bonded to F, O, or N which is attracted to the F, O, or N on another molecule
Q5.3.9
H2O has hydrogen bonding while H2S does not.
Q5.3.10
1. dispersion
2. dispersion
3. dispersion, dipole-dipole, and hydrogen bonding
Q5.3.11
The stronger the intermolecular forces, the higher the boiling point.
Q5.3.12
N2 < PH3 < CH3OH
All three of these molecules are comparable in size so the differences in dispersion forces are small. N2 is nonpolar and has only dispersion forces. PH3 is polar, so it has dispersion and dipole-dipole forces but no hydrogen bonding. CH3OH has dispersion, dipole-dipole, and hydrogen bonding forces.
5.4: Chromatography
Q5.4.1
Chromatogaphy is used to separate a mixture into its components when they travel at different rates in the mobile phase.
Q5.4.2
Paper, thin-layer, liquid, and gas are the most well-known types of chromatography
Q5.4.3
The mobile phase is a fluid which moves through the stationary phase. The stationary phase holds the sample until the mobile phase moves it along the stationary phase. | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/05%3A_Properties_of_Compounds/5.05%3A_Properties_of_Compounds_%28Exercises%29.txt |
• 6.1: Heat Flow
Chemical reactions are accompanied by transfers of energy. Keeping track of heat flow and energy requirements is important for a full understanding of chemical processes. Energy is the capacity for doing work or supplying heat. When you fill your car with gas, you are providing it with potential energy. Chemical potential energy is the energy stored in the chemical bonds of a substance. The various chemicals in gas contain a much chemical potential energy that is released when gas is burned
• 6.2: Energy and Properties (Exercises)
These are homework exercises to accompany Chapter 6 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Solutions are available below the questions.
06: Energy and Properties
Learning Outcomes
• Describe how chemical potential energy relates to heat and work.
• Use the specific heat equation to perform calculations that relate mass, specific heat, change in temperature, and the amount of heat absorbed or released.
• Convert among joule, calorie, and Calorie.
• Define endothermic and exothermic.
• Explain the difference between heat and specific heat capacity.
Chemical reactions are accompanied by transfers of energy. Keeping track of heat flow and energy requirements is important for a full understanding of chemical processes. Energy is the capacity for doing work or supplying heat. When you fill your car with gasoline, you are providing it with potential energy. Chemical potential energy is the energy stored in the chemical bonds of a substance. The various chemicals in gasoline contain a large amount of chemical potential energy that is released when the gasoline is burned in a controlled way in the engine of the car. The release of that energy does two things. Some of the potential energy is transformed into work, which is used to move the car (see figure below). At the same time, some of the potential energy is converted to heat, making the car's engine very hot. The energy changes of a system occur as either heat or work, or some combination of both.
Heat is energy that is transferred from one object or substance to another because of a difference in temperature between them. Heat always flows from an object at a higher temperature to an object at a lower temperature (see figure below). The flow of heat will continue until the two objects are at the same temperature.
Thermochemistry is the study of energy changes that occur during chemical reactions and during changes of state. When chemical reactions occur, some chemical bonds are broken, while new chemical bonds form. As a result of the rearrangement of atoms, the total chemical potential energy of the system either increases or decreases.
Exothermic and Endothermic Processes
When physical or chemical changes occur, they are generally accompanied by a transfer of energy. The law of conservation of energy states that in any physical or chemical process, energy is neither created nor destroyed. In other words, the entire energy in the universe is conserved. In order to better understand the energy changes taking place during a reaction, we need to define two parts of the universe, called the system and the surroundings. The system is the specific portion of matter in a given space that is being studied during an experiment or an observation. The surroundings is everything in the universe that is not part of the system. In practical terms for a laboratory chemist, the system is generally the reaction being investigated, while the surroundings include the immediate vicinity within the room. During most processes, energy is exchanged between the system and the surroundings. If the system loses a certain amount of energy, that same amount of energy is gained by the surroundings. If the system gains a certain amount of energy, that energy is supplied by the surroundings.
In the study of thermochemical processes, things are viewed from the point of view of the system. A chemical reaction or physical change is endothermic if heat is absorbed by the system from the surroundings. In the course of an endothermic process, the system gains heat from the surroundings, so the temperature of the surroundings decreases. The quantity of heat for a process is represented by the letter $q$. The sign of $q$ for an endothermic process is positive because the system is gaining heat. A chemical reaction or physical change is exothermic if heat is released by the system into the surroundings. Because the surroundings are gaining heat from the system, the temperature of the surroundings increases. The sign of $q$ for an exothermic process is negative because the system is losing heat. The difference between an endothermic reaction and an exothermic reaction is illustrated below (see figure below).
Units of Heat
Heat flow 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.
$1 \: \text{J} = 0.2390 \: \text{cal} \: \text{or} \: 1 \: \text{cal} = 4.184 \: \text{J}$
Example $1$
How many joules of energy are released when a 400. Calorie hamburger is digested?
Solution
Use the relationship between Calories and calories which is 1000 calories $=$ 1 Calorie and the conversion factor for joules to calories to find the value in joules. Note that all of the units cancel except joules which is what is being asked for in the question.
$400 \: \text{Cal} \left( \frac{1000 \: \text{cal}}{1 \: \text{Cal}} \right) \left( \frac{4.184 \: \text{J}}{1 \: \text{cal}} \right) = 1.67 \times 10^6 \: \text{J}$
Heat Capacity and Specific Heat
If a swimming pool and a bucket, both full of water at the same temperature, were subjected to the same input of heat energy, the bucket of water would certainly rise in temperature more quickly than the swimming pool. Heat capacity is the amount of heat required to raise the temperature of an object by $1^\text{o} \text{C}$. The heat capacity of an object depends both on its mass and its chemical composition. Because of its much larger mass, the swimming pool of water has a larger heat capacity than the bucket of water.
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. 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 table below lists the specific heat of some common substances. 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 $\left( \text{J/g} \cdot ^\text{o} \text{C} \right)$, or calories per gram per degree $\left( \text{cal/g} \cdot ^\text{o} \text{C} \right)$. This text will use $\text{J/g} \cdot ^\text{o} \text{C}$ for specific heat. Note that the specific heat of a substance depends not only on its identity but also its state. For example, ice, liquid water, and steam all have different specific heat values.
Table $1$: Specific Heats of Some Common Substances
Substance Specific Heat $\left( \text{J/g} \cdot ^\text{o} \text{C} \right)$
Water $\left( l \right)$ 4.18
Water $\left( s \right)$ 2.06
Water $\left( g \right)$ 1.87
Ammonia $\left( g \right)$ 2.09
Ethanol $\left( l \right)$ 2.44
Aluminum $\left( s \right)$ 0.897
Carbon, graphite $\left( s \right)$ 0.709
Copper $\left( s \right)$ 0.385
Gold $\left( s \right)$ 0.129
Iron $\left( s \right)$ 0.449
Lead $\left( s \right)$ 0.129
Mercury $\left( l \right)$ 0.140
Silver $\left( s \right)$ 0.233
Notice that water has a very high specific heat compared to most other substance. Water is commonly used as a coolant for machinery because it is able to absorb large quantities of heat (see figure below). 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.
Specific Heat Calculations
The specific heat of a substance can be used to calculate the temperature change that a given substance will undergo when it is either heated or cooled. The equation that relates heat $\left( q \right)$ to specific heat $\left( C_p \right)$, mass $\left( m \right)$, and temperature change $\left( \Delta T \right)$ is shown below.
$q = m \times C_p \times \Delta T$
The heat that is either absorbed or released is measured in joules. The mass is measured in grams. The change in temperature is given by $\Delta T = T_f - T_i$, where $T_f$ is the final temperature and $T_i$ is the initial temperature.
Example $2$
A $15.0 \: \text{g}$ piece of cadmium metal absorbs $134 \: \text{J}$ of heat as its temperature is increased from $24.0^\text{o} \text{C}$ to $62.7^\text{o} \text{C}$. Calculate the specific heat of cadmium.
Solution
Step 1: List the known quantities and plan the problem.
Known
• Heat $= q = 134 \: \text{J}$
• Mass $= m = 15.0 \: \text{g}$
• $\Delta T = 62.7^\text{o} \text{C} - 24.0^\text{o} \text{C} = 38.7^\text{o} \text{C}$
Unknown
• $C_p$ of cadmium $= ? \: \text{J/g} \cdot ^\text{o} \text{C}$
The specific heat equation can be rearranged to solve for the specific heat.
Step 2: Solve.
$C_p = \frac{q}{m \times \Delta T} = \frac{134 \: \text{J}}{15.0 \: \text{g} \times 38.7^\text{o} \text{C}} = 0.231 \: \text{J/g} \cdot ^\text{o} \text{C}$
Step 3: Think about your result.
The specific heat of cadmium, a metal, is fairly close to the specific heats of other metals in the table above. The result has three significant figures.
Since most specific heats are known, they can be used to determine the final temperature attained by a substance when it is either heated or cooled. Suppose that a $60.0 \: \text{g}$ sample of water at $23.52^\text{o} \text{C}$ was cooled by the removal fo $813 \: \text{J}$ of heat. The change in temperature can be calculated using the specific heat equation.
$\Delta T = \frac{q}{C_p \times m} = \frac{-813 \: \text{J}}{4.18 \: \text{J/g} \cdot ^\text{o} \text{C} \times 60.0 \: \text{g}} = -3.24^\text{o} \text{C}$
Since the water was being cooled, heat is removed from the system. Therefore, $q$ is negative, and the temperature decreases. The final temperature is:
$T_f = 23.52^\text{o} \text{C} - 3.24^\text{o} \text{C} = 20.28^\text{o} \text{C}$
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/06%3A_Energy_and_Properties/6.01%3A_Heat_Flow.txt |
These are homework exercises to accompany Chapter 6 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Solutions are available below the questions.
Questions
(click here for solutions)
Q6.1.1
Define potential energy and chemical potential energy.
Q6.1.2
What is one potential use for substances that have a large amount of chemical potential energy?
Q6.1.3
Describe what happens when two objects that have different temperatures come into contact with one another.
Q6.1.4
Distinguish between system and surroundings.
Q6.1.5
Distinguish between endothermic and exothermic.
Q6.1.6
Two different reactions are performed in two identical test tubes. In reaction A, the test tube becomes very warm as the reaction occurs. In reaction B, the test tube becomes cold. Which reaction is endothermic and which is exothermic? Explain.
Q6.1.7
What is the sign of q for an endothermic process? For an exothermic process?
Q6.1.8
Classify the following as endothermic or exothermic processes.
1. Boiling water
2. Sweating
3. Burning paper
4. Water freezing
Q6.1.9
Convert each value to the indicated units.
1. 150. kcal to Cal
2. 355 J to cal
3. 200. Cal to J
4. 225 kcal to cal
5. 3450. cal to kcal
6. 450. Cal to kJ
7. 175 kJ to cal
Q6.1.10
Equal amounts of heat are applied to 10.0 g samples of iron and aluminum, both originally at 25°C. Which one will be at the higher temperature?
Q6.1.11
Which sample will require more heat to increase the temperature by 10°C?
1. 25.0 g copper
2. 25.0 g lead
Q6.1.12
How much energy is required to heat 50.0 g of silver from 30°C to 50°C?
Q6.1.13
What is the final temperature when 125 J is applied to 20.0 g of lead, initially at 15°C?
Q6.1.14
How much energy is required to raise the temperature of 13.7 g of aluminum from 25.2°C to 61.9°C?
Q6.1.15
A 274 g sample of air is heated with 2250 J of heat and its temperature rises by 8.11°C. What is the specific heat of air at these conditions?
Q6.1.16
98.3 J of heat is supplied to 12.28 g of a substance, and its temperature rises by 5.42°C. What is the specific heat of the substance?
Q6.1.17
A quantity of ethanol is cooled from 47.9°C to 12.3°C and releases 3.12 kJ of heat. What is the mass of the ethanol sample?
Answers
6.1: Heat Flow
Q6.1.1
Potential energy is usually described as the energy of position. Chemical potential energy is energy stored within the chemical bonds of a substance.
Q6.1.2
Answers will vary. The most common example in every day life is the burning of fossil fuels to generate electricity or to run a vehicle.
Q6.1.3
The temperature of the hot object decreases and the temperature of the cold object increases as heat is transferred from the hot object to the cold object. The change in temperature of each depends on the identity and properties of each substance.
Q6.1.4
The system is the specific portion of matter being observed in an experiment and is designated by the experimenter. The surroundings is everything that is not the system.
Q6.1.5
Endothermic processes result in the gain of heat to the system while exothermic processes are associated with the loss of heat from the system.
Q6.1.6
Reaction A is exothermic because heat is leaving the system making the test tube feel hot. Reaction B is endothermic because heat is being absorbed by the system making the test tube feel cold.
Q6.1.7
q is positive for endothermic processes and q is negative for exothermic processes.
Q6.1.8
Classify the following as endothermic or exothermic processes.
1. Endothermic because heat is being added to the water to get it from the liquid state to the gas state.
2. Endothermic because energy is consumed to evaporate the moisture on your skin which lowers your temperature.
3. Exothermic because burning (also known as combustion) releases heat.
4. Exothermic because energy is exiting the system in order to go from liquid to solid. Another way to look at it is to consider the opposite process of melting. Energy is consumed (endothermic) to melt ice (solid to liquid) so the opposite process (liquid to solid) must be exothermic.
Q6.1.9
Convert each value to the indicated units.
1. $150\; kcal\left ( \frac{1\;Cal}{1\;kcal} \right )=150\;Cal$
2. $355\;J\left ( \frac{1\;cal}{4.184\;J} \right )=84.8\;cal$
3. $200.\;Cal\left ( \frac{1000\;cal}{1\;Cal} \right )\left ( \frac{4.184\;J}{1\;cal} \right )=8.37\times 10^{5}\;J$
4. $225\;kcal\left ( \frac{1000\;cal}{1\;kcal} \right )=2.25\times 10^{5}\;cal$
5. $3450.\;cal\left ( \frac{1 kcal\;kJ}{1000\;cal} \right )=3.450\;kcal$
6. $450.\;Cal\left ( \frac{1000\;cal}{1\;Cal} \right )\left ( \frac{4.184\;J}{1\;cal} \right )\left ( \frac{1\;kJ}{1000\;J} \right )=1.88\times 10^{3}\;kJ$ or $450.\;Cal\left ( \frac{4.184\;kJ}{1\;Cal} \right )=1.88\times 10^{3}\;kJ$
7. $175\;kJ\left ( \frac{1000\;J}{1\;kJ} \right )\left ( \frac{1\;cal}{4.184\;J} \right )=4.18\times 10^{4}\;cal$
Q6.1.10
Iron has a specific heat capacity of 0.449 $\text{J/g} \cdot ^\text{o} \text{C}$ which means it takes 0.449 J of energy to raise 1 gram of iron by 1$^\text{o} \text{C}$. Aluminum has a specific heat capacity of 0.897 $\text{J/g} \cdot ^\text{o} \text{C}$ which means it takes 0.897 J of energy to raise 1 gram of aluminum by 1$^\text{o} \text{C}$. When equal amounts of heat are applied, the temperature of the iron will increase more because it takes less energy (heat) to raise its temperature so iron will be at a higher temperature since they both start at 25$^\text{o} \text{C}$.
Q6.1.11
Both samples are the same mass so a comparison of the specific heat must be compared. Copper has a specific heat of 0.385 $\text{J/g} \cdot ^\text{o} \text{C}$ which means it takes 0.385 J of energy to raise 1 gram of copper by 1^\text{o} \text{C}\). Lead has a specific heat of 0.129 $\text{J/g} \cdot ^\text{o} \text{C}$ which means it takes 0.129 J of energy to raise 1 gram of copper by 1^\text{o} \text{C}\). More energy is needed to raise the temperature of copper so more heat will be needed to increase the temperature of copper by 10 ^\text{o} \text{C}\).
Q6.1.12
$\begin{array}{c} q=m\cdot C_{p}\cdot \Delta T\ q=50.0\;g\cdot 0.233\; \frac{J}{g\cdot^\text{o} \text{C} }\cdot \left (50^\text{o} \text{C}-30^\text{o} \text{C} \right )\ q=50.0\;g\cdot 0.233\;\frac{J}{g\cdot^\text{o} \text{C} }\cdot \left (20^\text{o} \text{C} \right )\ q=233\; J \end{array}$
Q6.1.13
$\begin{array}{c} q=m\cdot C_{p}\cdot \Delta T\ 125\;J=20.0\;g\cdot 0.129\;\frac{J}{g\cdot^\text{o} \text{C} }\cdot \left (T_{f}-15^\text{o} \text{C} \right )\ 48.4^\text{o} \text{C}=T_{f}-15^\text{o} \text{C}\ T_{f}=63^\text{o} \text{C} \end{array}$
Q6.1.14
$\begin{array}{c} q=m\cdot C_{p}\cdot \Delta T\ q=13.7\;g\cdot 0.897\;\frac{J}{g\cdot^\text{o} \text{C} }\cdot \left (61.9^\text{o} \text{C}-25.2^\text{o} \text{C} \right )\ q=13.7\;g\cdot 0.897\;\frac{J}{g\cdot^\text{o} \text{C} }\cdot \left (36.7^\text{o} \text{C} \right )\ q=451\; J \end{array}$
Q6.1.15
$\begin{array}{c} q=m\cdot C_{p}\cdot \Delta T\ 2250\;J=274\;g\cdot C_{p}\ \cdot 8.11^\text{o} \text{C}\ C_{p}=1.01 \frac{J}{g\cdot^\text{o} \text{C} } \end{array}$
Q6.1.16
$\begin{array}{c} q=m\cdot C_{p}\cdot \Delta T\ 98.3\;J=12.28\;g\cdot C_{p}\ \cdot 5.42^\text{o} \text{C}\ C_{p}=1.48 \frac{J}{g\cdot^\text{o} \text{C} } \end{array}$
Q6.1.17
$\begin{array}{c} q=m\cdot C_{p}\cdot \Delta T\ -3.12\;kJ=m\cdot 2.44\frac{J}{g\cdot^\text{o} \text{C} }\left (12.3^\text{o} \text{C}-47.9^\text{o} \text{C} \right )\ -3.12\times10^{3}\;J=m\cdot 2.44\frac{J}{g\cdot^\text{o} \text{C} }\left (-35.6^\text{o} \text{C} \right )\ m=35.9\;g \end{array}$ | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/06%3A_Energy_and_Properties/6.02%3A_Energy_and_Properties_%28Exercises%29.txt |
• 7.1: States of Matter
Another way that we can describe the properties of matter is the state (also called phase). The amount of energy in molecules of matter determines the state of matter. Matter can exist in one of several different states, including a gas, liquid, or solid state.
• 7.2: State Changes and Energy
Energy must be supplied to a solid in order to melt or vaporize it. On a microscopic level melting or vaporization involves separating molecules which are attracted to each other. The amount of energy needed to separate the molecules is proportional to the intermolecular forces between the molecules.
• 7.3: Kinetic-Molecular Theory
The kinetic-molecular theory is a theory that explains the states of matter and is based on the idea that matter is composed of tiny particles that are always in motion. The theory helps explain observable properties and behaviors of solids, liquids, and gases. However, the theory is most easily understood as it applies to gases. The theory applies specifically to a model of a gas called an ideal gas.
• 7.4: The Ideal Gas Equation
Properties of gases such as pressure (P), volume (V), temperature (T), and moles(n) are relatively easy to measure. Unlike with liquids and solids, the particles (molecules or atoms) in a gas phase sample are very far apart from one another. As a result, their behavior is much more predictable because intermolecular forces become insignificant for most samples in the gas phase even over a wide range of conditions. The presence of intermolecular forces makes their behavior harder to predict.
• 7.5: Aqueous Solutions
A solution is a homogenous mixture consisting of a solute dissolved into a solvent. The solute is the substance that is being dissolved, while the solvent is the dissolving medium. Solutions can be formed with many different types and forms of solutes and solvents. In this chapter, we will focus on solution where the solvent is water. An aqueous solution is water that contains one or more dissolved substance. The dissolved substances in an aqueous solution may be solids, gases, or other liquids.
• 7.6: Colloids and Suspensions
A suspension is a heterogeneous mixture in which some of the particles settle out of the mixture upon standing. The particles in a suspension are far larger than those of a solution, so gravity is able to pull them down out of the dispersion medium (e.g., water). A colloid is a heterogeneous mixture in which the dispersed particles are intermediate in size between those of a solution and a suspension. The particles are spread evenly throughout the medium, which can be a solid, liquid, or gas.
• 7.7: Solubility
Previously, we looked at the primary characteristics of a solution and how water is able to dissolve solid solutes. There are many examples of solutions that do not involve water at all, or solutions that involve solutes that are not solids.
• 7.8: Solutions (Exercises)
These are homework exercises to accompany Chapter 7 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Solutions are available below the questions.
07: Solids Liquids and Gases
Learning Outcomes
• List the three states of matter and give examples of each.
• Describe the properties of each state of matter.
• Identify and describe each type of change in state.
• Recognize that boiling and melting points vary with each substance.
• Recognize that a substance's boiling point depends on the pressure.
States of Matter
Another way that we can describe the properties of matter is the state (also called phase). The amount of energy in molecules of matter determines the state of matter. Matter can exist in one of several different states, including a gas, liquid, or solid state. These different states of matter have different properties, which are illustrated in the figure below.
• 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.
What Determines a Substance's State?
Which state a substance is in depends partly on temperature and air pressure. For example, at the air pressure found at sea level, water exists as a liquid at temperatures between $0^\text{o} \text{C}$ and $100^\text{o} \text{C}$. Above $100^\text{o} \text{C}$, water exists as a gas (water vapor). Below $0^\text{o} \text{C}$, water exists as a solid (ice). Different substances have a different range of temperatures at which they exist in each state. For example, oxygen is a gas above $-183^\text{o} \text{C}$, but iron is a gas only above $2861^\text{o} \text{C}$. These differences explain why some substances are always solids at normal Earth temperatures, whereas others are always gases or liquids.
Water can take many forms. At low temperatures (below $0^\text{o} \text{C}$), it is a solid. It is a liquid between $0^\text{o} \text{C}$ and $100^\text{o} \text{C}$. 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.
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
Technically speaking a fourth state of matter called plasma exists, but it does not naturally occur on earth, so we will omit it from our study here.
Changing States
Matter constantly goes through cycles that involve changing states (also called change phases). Water and all the elements important to organisms, including carbon and nitrogen, are constantly recycled on Earth. As matter moves through its cycles, it changes state repeatedly. For example, in the water cycle, water repeatedly changes from a gas to a liquid or solid and back to a gas again. How does this happen?
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.
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, including the strength of its intermolecular forces. As an example, the values for water are given in Table $1$. Note how the boiling point of water varies greatly with pressure.
Table $1$: Boiling point of water as a function of pressure
Altitude $\left( \text{ft} \right)$ Pressure $\left( \text{atm} \right)$ Boiling Point $\left( ^\text{o} \text{C} \right)$
-500 1.05 100.5
0 1.00 100
4000 0.892 96
7000 0.797 93
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/07%3A_Solids_Liquids_and_Gases/7.01%3A_States_of_Matter.txt |
Learning Outcomes
• Identify whether energy is consumed or released during a phase change.
• Calculate the amount of energy consumed or released during a phase change.
• Recognize that the enthalpies of vaporization and condensation are equal in magnitude but opposite in sign.
• Recognize that the enthalpies of fusion and freezing are equal in magnitude but opposite in sign.
Heating Curve
When heat energy is supplied to a solid (like ice) at a steady rate by means of an electrical heating coil, we find that the temperature climbs steadily until the melting point is reached and the first signs of liquid formation become evident. Thereafter, even though we are still supplying heat energy to the system, the temperature remains constant as long as both liquid and solid are present. On the graph below, this is represented by the flat line, where energy is being added to the ice, but no change is occurring in the temperature. All energy added to the system at this stage is used to convert solid ice into liquid water.
Once all of the sample is in the liquid phase, the addition of energy now increases the temperature until the boiling point is reached and the first signs of gas formation are seen. The temperature remains constant even though energy is being added to the system. The energy is being used to convert the liquid to a gas. Once all of the sample is in the gas phase, additional energy can be added to increase the temperature of the gas.
This macroscopic behavior demonstrates quite clearly that energy must be supplied to a solid in order to melt or vaporize it. On a microscopic level melting or vaporization involves separating molecules which are attracted to each other. The amount of energy needed to separate the molecules is proportional to the intermolecular forces between the molecules.
Enthalpy
The heat energy which a solid absorbs when it melts is called the enthalpy of fusion (or heat of fusion and is usually quoted on a molar basis. (The word fusion means the same thing as “melting.”) When 1 mol of ice, for example, is melted, we find from experiment that 6.01 kJ of energy is needed. The molar enthalpy of fusion of ice is thus +6.01 kJ per mol ($\dfrac{6.01\;kJ}{mol}$), and we can write
$\text{H}_{2}\text{O}(s) \rightarrow \text{H}_{2}\text{O}(l)$
$\triangle H_{fus} = \dfrac{6.01\;kJ}{mol}$
If two moles of water are melted, then it would require twice as much energy (see below). The amount of energy needed to melt a substance will depend on the amount of the substance.
$2\;mol\times \dfrac{6.01\;kJ}{mol}=\dfrac{12.0\;kJ}{mol}$
Selected molar enthalpies of fusion are tabulated in Table $1$. Solids like ice which have strong intermolecular forces have much higher values than those like CH4 with weak ones.
When a liquid is boiled (or vaporized), energy is required to move the molecules apart to go from the liquid phase to the gas phase. The energy which a liquid absorbs when it vaporizes is known as the enthalpy of vaporization (. In the case of water, the molar enthalpy of vaporization is ($\dfrac{40.67\;kJ}{mol}$). In other words
$\text{H}_{2}\text{O}(s) \rightarrow \text{H}_{2}\text{O}(l)$
$\triangle H_{vap} = 40.67 \dfrac {\text{kJ}}{\text{mol}}$
Heat energy is absorbed to vaporize a liquid because molecules which are held together by intermolecular forces in the liquid are separated as the gas is formed. Such a separation requires energy. As with melting, the amount of energy needed to vaporize a substance is proportional to the amount of substance present. The more liquid there is, the more energy required to vaporize it.
In general the energy needed differs from one liquid to another depending on the magnitude of the intermolecular forces. We can expect liquids with strong intermolecular forces to have larger enthalpies of vaporization. See the examples in the table below. Note that the enthalpy of vaporization of a substance is always higher than its enthalpy of fusion. When a solid melts, the molecules are not separated from each other to nearly the same extent as when a liquid boils.
Melting and boiling points are given in units of Kelvin ($K = T_{^{\circ}C}\;+\;273.15$)
Table $1$: Molar Enthalpies of Fusion and Vaporization of Selected Substances.
Substance Formula $\triangle H_{fus}$
($\dfrac{kJ}{mol}$)
Melting Point (K) $\triangle H_{vap}$
($\dfrac{kJ}{mol}$)
Boiling Point (K)
neon Ne 0.33 24 1.80 27
oxygen O2 0.44 54 6.82 90.2
methane CH4 0.94 90.7 8.18 112
ethane C2H6 2.85 90.0 14.72 184
chlorine Cl2 6.40 172.2 20.41 239
carbon tetrachloride CCl4 2.67 250.0 30.00 350
water* H2O
6.01
273.1 40.7 373.1
n-nonane C9H20 19.3 353 40.5 491
lead Pb 4.77 601 178 2022
ammonia NH3 5.636 195 23.35 240
ethanol CH3COOH 4.9 159 38.56 351
Example $1$: Heat Energy
Calculate the heat energy required to (a) vaporize 100. g of lead, (b) melt 100. g of lead, (c) vaporize 100. g water, and (d) melt 100. g of water.
Solution
(a)To vaporize 100. g of lead:
$\text{Pb}(l) \rightarrow\text{Pb}(g)\;\;\;\;\;\;\;\;\;\; \triangle H_{vap} = 178 \dfrac {kJ}{mol}$
$100.\;g\times\dfrac {1\;mol\;\text{Pb}}{207.2\;g\;\text{Pb}}\times \dfrac{178\;{kJ}}{mol} = 85.9\; kJ$
(b) To melt 100. g of lead:
$\text{Pb}(s) \rightarrow\text{Pb}(l)\;\;\;\;\;\;\;\;\;\;\triangle H_{fus} = 4.77 \dfrac {kJ}{mol}$
$100.\;g\times\dfrac {1\;mol\;\text{Pb}}{207.2\;g\;\text{Pb}}\times \dfrac{4.77\;{kJ}}{mol} = 2.30\; kJ$
(c) To vaporize 100. g of water:
$\text{H}_2\text{O}(s) \rightarrow\text{H}_2\text{O}(l)\;\;\;\;\;\;\;\;\;\;\triangle H_{vap} =40.657\dfrac {kJ}{mol}$
$100.\;g\times\dfrac {1\;mol\;\text{H}_2\text{O}}{18.0\;g\;\text{H}_2\text{O}}\times \dfrac{40.657\;{kJ}}{mol} = 226\; kJ$
(d) To melt 100. g of water:
$\text{H}_2\text{O}(s) \rightarrow\text{H}_2\text{O}(l)\;\;\;\;\;\;\;\;\;\;\triangle H_{fus} = 6.01 \dfrac {kJ}{mol}$
$100.\;g\times\dfrac {1\;mol\;\text{H}_2\text{O}}{18.0\;g\;\text{H}_2\text{O}}\times \dfrac{6.01\;{kJ}}{mol} = 33.4\; kJ$
It might be surprising that the heat required to melt or vaporize 100 g of lead is so much less than that require to melt or vaporized water. First, the temperature at which the substance melts has nothing to do with the enthalpy of fusion. Remember, we are only looking at the energy required to change the phase, not the energy required to get the substance to the melting or boiling point.
Also note that the enthalpies of fusion and vaporization are given as kJ per mole. Although the water and the lead have the same mass, the moles of each substance is very different (5.5 moles of water vs. 0.48 moles of lead).
Freezing and Condensation
The discussion here has focused on fusion (melting) and vaporization. But what about freezing and condensation? Fusion (solid to liquid) and freezing (liquid to solid) are opposite processes. As a result, the magnitude of energy for each is the same, but the sign is different. With fusion, energy is a positive value because it is endothermic (consuming energy) while the energy for freezing is negative because it is an exothermic process (releasing energy).
Let's compare the enthalpies of fusion and freezing for methane. Both values have the same magnitude of 0.94 but they have the opposite sign.
$\triangle H_{fusion}=0.94\;\dfrac{kJ}{mol}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\triangle H_{freezing}=-0.94\;\dfrac{kJ}{mol}$
The same concept applies to vaporization (liquid to gas) and condensation (gas to liquid). Energy is consumed during vaporization (positive energy) and released during condensation (negative energy). The enthalpies of vaporization and condensation are equal in magnitude but opposite in sign.
$\triangle H_{vaporization}=8.18\;\dfrac{kJ}{mol}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\triangle H_{condensation}=-8.18\;\dfrac{kJ}{mol}$
Energy is consumed to change a substance from solid to liquid to gas. Energy is released to change a substance from gas to liquid to solid. | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/07%3A_Solids_Liquids_and_Gases/7.02%3A_State_Changes_and_Energy.txt |
Learning Outcomes
• Define pressure and describe how gases exert pressure.
• Convert between units of gas pressure ($\text{atm}$, $\text{mm} \ce{Hg}$, $\text{torr}$, and $\text{Pa}$).
• Relate temperature to average kinetic energy.
• Relate mass to molecular speed.
• Describe the behavior of an ideal gas.
The kinetic-molecular theory is a theory that explains the states of matter and is based on the idea that matter is composed of tiny particles that are always in motion. The theory helps explain observable properties and behaviors of solids, liquids, and gases. However, the theory is most easily understood as it applies to gases. The theory applies specifically to a model of a gas called an ideal gas. An ideal gas is an imaginary gas whose behavior perfectly fits all the assumptions of the kinetic-molecular theory. In reality, gases are not ideal, but they are very close to being so under most everyday conditions.
The kinetic-molecular theory, as it applies to gases, has five basic assumptions.
1. Gases consist of very large numbers of tiny spherical particles that are far apart from one another compared to their size. The particles of a gas may be either atoms or molecules. The distance between the particles of a gas is much, much greater than the distances between the particles of a liquid or a solid. Most of the volume of a gas, therefore, is composed of the empty space between the particles. In fact, the volume of the particles themselves is considered to be insignificant compared to the volume of the empty space.
2. Gas particles are in constant rapid motion in random directions. The fast motion of gas particles gives them a relatively large amount of kinetic energy. Recall that kinetic energy is the energy that an object possesses because of its motion. The particles of a gas move in a straight line until they collide with another particle or with one of the walls of their container (see figure below).
3. Collisions between gas particles and between particles and the container walls are elastic collisions. An elastic collision is one in which there is no overall loss of kinetic energy. Kinetic energy may be transferred from one particle to another during an elastic collision, but there is no change in the total energy of the colliding particles.
4. There are no forces of attraction or repulsion between gas particles. Attractive forces are responsible for particles of a real gas condensing together to form a liquid. It is assumed that the particles of an ideal gas have no such attractive forces. The motion of each particle is completely independent of the motion of all other particles.
5. The average kinetic energy of gas particles is dependent upon the temperature of the gas. As the temperature of a gas is increased, its component particles begin to move faster, resulting in an increase in their kinetic energies. Not all particles in a given sample have the same speed, so the sample will contain particles with a range of different kinetic energies. However, the average kinetic energy of the particles in a sample is proportional to its temperature.
Gas Pressure
Pressure is defined as the force per unit area on a surface.
$\text{Pressure} = \frac{\text{force}}{\text{area}}$
When a person stands on the floor, his feet exert pressure on the surface. That pressure is related to both the mass of the person and the surface area of his feet. If the person were holding a heavy object, the pressure would increase because of a greater force. Alternatively, if the person stands on his toes, the pressure also increases because of a decrease in the surface area.
Gas molecules also exert pressure. Earth's atmosphere exerts pressure because gravity acts on the huge number of gas particles contained in the atmosphere, holding it in place. Pressure is also exerted by small samples of gas, such as the outward pressure exerted by the gas inside a balloon. Gas pressure is the pressure that results from collisions of gas particles with an object. Inside the balloon, the gas particles collide with the balloon's inner walls. It is those collisions that keep the balloon inflated. If the gas particles were to suddenly stop moving, the balloon would instantly deflate. The figure below is an illustration of gas particles exerting pressure inside a container.
Measuring Pressure
Atmospheric pressure is the pressure exerted by the gas particles in Earth's atmosphere as those particles collide with objects. A barometer is an instrument used to measure atmospheric pressure. A traditional mercury barometer consists of an evacuated tube immersed in a container of mercury. Air molecules from the atmosphere push down on the outer surface of the mercury, but, because the inside of the tube is a vacuum, there is no corresponding downward push on the mercury in the tube. As a result, the mercury rises inside the tube. The height to which the mercury rises is dependent on the external air pressure.
At sea level, a mercury column will rise a distance of $760 \: \text{mm}$. This atmospheric pressure is reported as $760 \: \text{mm} \: \ce{Hg}$ (millimeters of mercury). At higher altitudes, the atmospheric pressure is lower, so the column of mercury will not rise as high. For example, on the summit of Mt. Everest (at an elevation of $8848 \: \text{m}$), the air pressure is $253 \: \text{mm} \: \ce{Hg}$. Atmospheric pressure is also slightly dependent on weather conditions.
A more convenient barometer, called an aneroid barometer, measures pressure by the expansion and contraction of a small spring within an evacuated metal capsule.
Units of Gas Pressure
A barometers measures gas pressure by the height of the column of mercury. One unit of gas pressure is the millimeter of mercury $\left( \text{mm} \: \ce{Hg} \right)$. An equivalent unit to the $\text{mm} \: \ce{Hg}$ is called the torr, in honor of the inventor of the barometer, Evangelista Torricelli. The pascal $\left( \text{Pa} \right)$ is the standard unit of pressure. A pascal is a very small amount of pressure, so a more useful unit for everyday gas pressures is the kilopascal $\left( \text{kPa} \right)$. A kilopascal is equal to 1000 pascals. Another commonly used unit of pressure is the atmosphere $\left( \text{atm} \right)$. Standard atmospheric pressure is called $1 \: \text{atm}$ of pressure and is equal to $760 \: \text{mm} \: \ce{Hg}$ and $101.3 \: \text{kPa}$. The relationships between the most common units of pressure are shown below.
$1 \: \text{atm} = 760 \: \text{mm} \: \ce{Hg} = 760 \: \text{torr} = 101.3 \: \text{kPa}$
Example $1$
The atmospheric pressure in a mountainous location is measured to be $613 \: \text{mm} \: \ce{Hg}$. What is this pressure in $\text{atm}$ and in $\text{kPa}$?
Solution
Step 1: List the known quantities and plan the problem.
Known
• Given: $613 \: \text{mm} \: \ce{Hg}$
• $1 \: \text{atm} = 760 \: \text{mm} \: \ce{Hg}$
• $101.3 \: \text{kPa} = 760 \: \text{mm} \: \ce{Hg}$
Unknown
• Pressure $= ? \: \text{atm}$
• Pressure $= ? \: \text{kPa}$
Use conversion factors from the equivalent pressure units to convert from $\text{mm} \: \ce{Hg}$ to $\text{atm}$ and from $\text{mm} \: \ce{Hg}$ to $\text{kPa}$.
Step 2: Solve.
$613 \: \text{mm} \: \ce{Hg} \times \frac{1 \: \text{atm}}{760 \: \text{mm} \: \ce{Hg}} = 0.807 \: \text{atm} \nonumber$
$613 \: \text{mm} \: \ce{Hg} \times \frac{101.3 \: \text{kPa}}{760 \: \text{mm} \: \ce{Hg}} = 81.7 \: \text{kPa} \nonumber$
Step 3: Think about your result.
The air pressure is about $80\%$ of the standard atmospheric pressure at sea level. The standard pressure of $760 \: \text{mm} \: \ce{Hg}$ can be considered to have three significant figures.
Kinetic Energy and Temperature
As stated in the kinetic-molecular theory, the temperature of a substance is related to the average kinetic energy of the particles of that substance. When a substance is heated, some of the absorbed energy is stored within the particles, while some of the energy increases the speeds at which the particles are moving. This is observed as an increase in the temperature of the substance.
Average Kinetic Energy
At any given temperature, not all of the particles in a sample of matter have the same kinetic energy. Instead, the particles display a wide range of kinetic energies. Most of the particles have a kinetic energy near the middle of the range. However, some of the particles have kinetic energies a great deal lower or a great deal higher than the average (see figure below).
The blue curve shown (see figure above) is for a sample of matter at a relatively low temperature, while the red curve is for a sample at a relatively high temperature. In both cases, most of the particles have intermediate kinetic energies, close to the average. Notice that as temperature increases, the range of kinetic energies increases and the distribution curve "flattens out".
At a given temperature, the particles of any substance have the same average kinetic energy. At room temperature, the molecules in a sample of liquid water have the same average kinetic energy as the molecules in a sample of oxygen gas or the ions in a sample of sodium chloride.
Absolute Zero
As a sample of matter is continually cooled, the average kinetic energy of its particles decreases. Eventually, one would expect the particles to stop moving completely. Absolute zero is the temperature at which the motion of particles theoretically ceases. Absolute zero has never been attained in the laboratory, but temperatures on the order of $1 \times 10^{-10} \: \text{K}$ have been achieved. The Kelvin temperature scale is based on this theoretical limit, so absolute zero is equal to $0 \: \text{K}$. The Kelvin temperature of a substance is directly proportional to the average kinetic energy of the particles of the substance. For example, the particles in a sample of hydrogen gas at $200 \: \text{K}$ have twice the average kinetic energy as the particles in a hydrogen sample at $100 \: \text{K}$.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/07%3A_Solids_Liquids_and_Gases/7.03%3A_Kinetic-Molecular_Theory.txt |
Learning Outcomes
• State the ideal gas law and identify the variables.
• Use the ideal gas law to solve for an unknown.
• State the combined gas law.
• Simplify the combined gas law for any values that are constant.
• Use the combined gas law to solve for an unknown value.
Individual Gas Laws
Properties of gases such as pressure (P), volume (V), temperature (T), and moles(n) are relatively easy to measure. Unlike with liquids and solids, the particles (molecules or atoms) in a gas phase sample are very far apart from one another. As a result, their behavior is much more predictable because intermolecular forces become insignificant for most samples in the gas phase even over a wide range of conditions. The presence of intermolecular forces in liquid and solid samples makes their behavior harder to predict.
Experiments with gas phase samples over time showed the relationship between pairs of variables (P, V, T, and n) and individual gas laws (equations) show the quantitative relationship between those variables. Avogadro's law tells us that at constant P and T, the volume of a gas is directly proportional to the amount of gas. Boyle's law says that volume is inversely proportional to pressure at constant T and n. Charles' indicates that volume is directly proportional to temperature at constant P and n.
The video below shows a situation where 3 variables, pressure, volume, and amount of substance (moles) are all interrelated: inside our lungs.
As you can see in the video, when the pink balloon on the bottom (the "diaphragm") is pulled down, the balloon inside expands. This expansion causes a decrease in pressure (Boyle's Law). The pressure decrease causes a pressure differential, drawing air in through the straw, an increase in the amount of air (moles). So in your lungs, volume, pressure, and amount of air are all related. But none of the current laws explain the relation between 2 variables. How can this be resolved?
The Ideal Gas Law
These three laws may all be applied at once if we write ($\propto$ means "proportional to"):
$V\propto n\text{ }\times \text{ }\dfrac{\text{1}}{P}\text{ }\times \text{ }T\label{1}$
or, introducing a constant of proportionality R,
$V=R\text{ }\dfrac{nT}{P}\label{2}$
This is known as the ideal gas law which results from the combination of the individual gas laws. Equation $\ref{2}$ applies to all gases at low pressures and high temperatures and is a very good approximation under nearly all conditions. The value of R, the gas constant, is independent of the kind of gas, the temperature, or the pressure and has a value of $\frac{0.08206\;L\cdot atm}{mol\cdot K}$.
Equation $\ref{2}$ is usually rearranged by multiplying both sides by P, so that it reads
$PV = nRT \label{4}$
This is called the ideal gas equation or the ideal gas law. With the ideal gas equation we can convert from volume of a gas to amount of substance (provided that P and T are known). This is very useful since the volume, pressure, and temperature of a gas are easier to measure than mass, and because knowledge of the molar mass is unnecessary.
Note that for any gas law calculations, the temperature must be in units of Kelvin. The relationship between oC and K is K = °C + 273.15.
Example $1$ : MOLES of Gas
Calculate the moles of gas in a 0.100 L sample at a temperature of 300 K and a pressure of 0.987 atm.
Solution
$PV=nRT$
$\left ( 0.987\; atm \right ) \left ( 0.100\; L\right )=n\left (\frac{0.08206\;L\cdot atm}{mol\cdot K} \right )\left (300 \;K\right )$
$n=0.00401\; mol$
Example $2$ : unit considerations
A sample of benzene (C6H6) was heated to 100.°C in an evacuated flask whose volume was 247.2 ml, a sample of benzene vaporized. When the benzene was condensed to a liquid, its mass was found to be 0.616 g. What was the pressure in the flask?
Solution
The problem gives values for temperature, volume, and mass of the sample. Since R has units of $\left (\frac{L\cdot atm}{mol\cdot K} \right )), we need to have the temperature in units of Kelvin, the volume in liters, and the amount of sample in moles. Temperature: K = °C + 273.15 K = 100.°C + 273.15 K = 373 K Volume: \(247.2\;mL\left ( \frac{10^{-3}\; L}{1\;mL} \right )=0.2472\;L$
Moles: $0.616\;g\left ( \frac{1\;mol}{78.11\; g} \right )=7.89 \times10^{-3}\;mol$
Now that all of the values are in the correct units, the value for the unknown pressure can be determined.
$PV=nRT$
$P \left ( 0.2472\; L\right )=\left (7.89\times10^{-3}\;mol \right )\left (\frac{0.08206\;L\cdot atm}{mol\cdot K} \right )\left (373 \;K\right )$
$P=.977 atm$
Combined Gas law
While the ideal gas law is useful in solving for a single unknown when the other values are known, the combined gas law is useful when comparing initial and final situations. The ideal gas law can be rearranged to solve for R, the gas constant.
$R=\frac{PV}{nT}$
Under the initial conditions, $R=\frac{P_iV_i}{n_iT_i}$ and under final conditions, $R=\frac{P_fV_f}{n_fT_f}$. Since both expressions are equal to R, they are equal to each other.
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_f}{n_fT_f}$
This equation is typically used when one or more of the variables is constant. As a result, that variable is canceled from the equation. For example, the equation 2x2 = 2y can be simplified to x2 = y since the 2 is on both sides of the equation.
What happens to the combined gas law equation when the initial and final pressures are equal (Pi = Pf)? Since they are equal, Pi can replace Pf.
$\frac{P_iV_i}{n_iT_i}=\frac{P_iV_f}{n_fT_f}$
which simplifies to
$\frac{V_i}{n_iT_i}=\frac{V_f}{n_fT_f}$
If two variables are constant, the equation can be simplified even more. If temperature and volume are constant, then Ti = Tf and Vi = Vf. Then,
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_i}{n_fT_i}$
simplifies to
$\frac{P_i}{n_i}=\frac{P_f}{n_f}$
Example $3$
Imagine a 1855 L balloon initially at 30°C and 745 mmHg. The balloon rises to an altitude of 23,000 ft and that the pressure and temperature at that altitude were 312 mmHg and −30°C, respectively. To what volume would the balloon have to expand to hold the same amount of hydrogen gas at the higher altitude?
Solution:
Begin by setting up a table of the two sets of conditions (note that some values will need to be converted to different units):
Initial Final
$P_i=745\;\rm mmHg=0.980\;atm$ $P_f=312\;\rm mmHg=0.411\;atm$
$T_i=\rm30\;^\circ C=303\;K$ $T_f=\rm-30\;^\circ C=243\;K$
$V_i=\rm1855\;L$ $V_f=?$
By eliminating the constant property ($n$) of the gas, the combined gas law is simplified to
$\dfrac{P_iV_i}{T_i}=\dfrac{P_fV_f}{T_f}$
By solving the equation for $V_f$, we get:
$\frac{P_iV_i}{T_i}=\frac{P_fV_f}{T_f}$
$\frac{0.980\;atm\cdot1855\;L}{303\;K}=\frac{0.411\;atm \cdot V_f}{243\;K}$
$V_f=3.55\times10^3\;L$ | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/07%3A_Solids_Liquids_and_Gases/7.04%3A_The_Ideal_Gas_Equation.txt |
Learning Outcomes
• Define a solution and describe the parts of a solution.
• Describe how an aqueous solution is formed from both ionic compounds and molecular compounds.
• Recognize that some compounds are insoluble in water.
• Describe the differences among strong electrolytes, weak electrolytes, and nonelectrolytes.
Forming a Solution
When one substance dissolves into another, a solution is formed. A solution is a homogenous mixture consisting of a solute dissolved into a solvent. The solute is the substance that is being dissolved, while the solvent is the dissolving medium. Solutions can be formed with many different types and forms of solutes and solvents. In this chapter, we will focus on solution where the solvent is water. An aqueous solution is water that contains one or more dissolved substance. The dissolved substances in an aqueous solution may be solids, gases, or other liquids.
In order to be a true solution, a mixture must be stable. When sugar is fully dissolved into water, it can stand for an indefinite amount of time, and the sugar will not settle out of the solution. Further, if the sugar-water solution is passed through a filter, it will remain with the water. This is because the dissolved particles in a solution are very small, usually less than $1 \: \text{nm}$ in diameter. Solute particles can be atoms, ions, or molecules, depending on the type of substance that has been dissolved.
The Dissolving Process
Water typically dissolves most ionic compounds and polar molecules. Nonpolar molecules, such as those found in grease or oil, do not dissolve in water. We will first examine the process that occurs when an ionic compound, such as table salt (sodium chloride), dissolves in water.
Water molecules move about continuously due to their kinetic energy. When a crystal of sodium chloride is placed into water, the water's molecules collide with the crystal lattice. Recall that the crystal lattice is composed of alternating positive and negative ions. Water is attracted to the sodium chloride crystal because water is polar; it has both a positive and a negative end. The positively charged sodium ions in the crystal attract the oxygen end of the water molecules because they are partially negative. The negatively charged chloride ions in the crystal attract the hydrogen end of the water molecules because they are partially positive. The action of the polar water molecules takes the crystal lattice apart (see figure below).
After coming apart from the crystal, the individual ions are then surrounded by solvent particles in a process called solvation. Note in the figure above that the individual $\ce{Na^+}$ ions are surrounded by water molecules with the oxygen atom oriented near the positive ion. Likewise, the chloride ions are surrounded by water molecules with the opposite orientation. Hydration is the process of solute particles being surrounded by water molecules arranged in a specific manner. Hydration helps to stabilize aqueous solutions by preventing the positive and negative ions from coming back together and forming a precipitate.
Table sugar is made of the molecular compound sucrose $\left( \ce{C_{12}H_{22}O_{11}} \right)$. Solid sugar consists of individual sugar molecules held together by intermolecular attractive forces. When water dissolves sugar, it separates the individual sugar molecules by disrupting the attractive forces, but it does not break the covalent bonds between the carbon, hydrogen, and oxygen atoms. Dissolved sugar molecules are also hydrated. The hydration shell around a molecule of sucrose is arranged so that its partially negative oxygen atoms are near the partially positive hydrogen atoms in the solvent, and vice versa.
Insoluble Compounds
Not all compounds dissolve well in water. Some ionic compounds, such as calcium carbonate $\left( \ce{CaCO_3} \right)$ and silver chloride $\left( \ce{AgCl} \right)$, are nearly insoluble. This is because the attractions between the ions in the crystal lattice are stronger than the attraction that the water molecules have for the ions. As a result, the crystal remains intact. The solubility of ionic compounds can be predicted using the solubility rules as shown in Table $1$.
Table $1$: Solubility rules for ionic compounds in water.
Nonpolar compounds also do not dissolve in water. The attractive forces that operate between the particles in a nonpolar compound are weak dispersion forces. In order for a nonpolar molecule to dissolve in water, it would need to break up some of the hydrogen bonds between adjacent water molecules. In the case of an ionic substance, these favorable interactions are replaced by other attractive interactions between the ions and the partial charges on water. However, interactions between nonpolar molecules and water are less favorable than the interactions that water makes with itself. When a nonpolar liquid such as oil is mixed with water, two separate layers form, because the liquids will not dissolve into each other (see figure below). When a polar liquid like ethanol is mixed with water, they completely blend and dissolve into one another. Liquids that dissolve in one another in all proportions are said to be miscible. Liquids that do not dissolve in one another are called immiscible. The general rule for deciding if one substance is capable of dissolving another is "like dissolves like", where the property being compared is the overall polarity of the substance. For example, a nonpolar solid such as iodine will dissolve in nonpolar lighter fluid, but it will not dissolve in polar water.
Electrolytes and Nonelectrolytes
An electrolyte is a compound that conducts an electric current when it is dissolved in water or melted. In order to conduct a current, a substance must contain mobile ions that can move from one electrode to the other. All ionic compounds are electrolytes. When ionic compounds dissolve, they break apart into ions, which are then able to conduct a current. Even insoluble ionic compounds, such as $\ce{CaCO_3}$, are considered electrolytes because they can conduct a current in the molten (melted) state.
A nonelectrolyte is a compound that does not conduct an electric current in either aqueous solution or in the molten state. Many molecular compounds, such a sugar or ethanol, are nonelectrolytes. When these compounds dissolve in water, they do not produce ions. Illustrated below is the difference between an electrolyte and a nonelectrolyte.
Dissociation
Earlier, you saw how an ionic crystal lattice breaks apart when it is dissolved in water. Dissociation is the separation of ions that occurs when a solid ionic compound dissolves. Simply undo the crisscross method that you learned when writing chemical formulas for ionic compounds, and you are left with the components of an ionic dissociation equation. The subscripts for the ions in the chemical formulas become the coefficients of the respective ions on the product side of the equations. Shown below are dissociation equations for $\ce{NaCl}$, $\ce{Ca(NO_3)_2}$, and $\ce{(NH_4)_3PO_4}$.
\begin{align} &\ce{NaCl} \left( s \right) \rightarrow \ce{Na^+} \left( aq \right) + \ce{Cl^-} \left( aq \right) \ &\ce{Ca(NO_3)_2} \left( s \right) \rightarrow \ce{Ca^{2+}} \left( aq \right) + 2 \ce{NO_3^-} \left( aq \right) \ &\ce{(NH_4)_3PO_4} \left( s \right) \rightarrow 3 \ce{NH_4^+} \left( aq \right) + \ce{PO_4^{3-}} \left( aq \right) \end{align}
One formula unit of sodium chloride dissociates into one sodium ion and one chloride ion. The calcium nitrate formula unit dissociates into one calcium ion and two nitrate ions, because the $2+$ charge of each calcium ion requires two nitrate ions (each with a charge of $1-$) to form an electrically neutral compound. The ammonium phosphate formula unit dissociates into three ammonium ions and one phosphate ion.
Do not confuse the subscripts of the atoms within the polyatomic ion for the subscripts that result from the crisscrossing of the charges that make the original compound neutral. The 3 subscript of the ntirate ion and the 4 subscript of the ammonium ion are part of the polyatomic ion and remain a part of the ionic formula after the compound dissociates. Notice that the compounds are solids $\left( s \right)$ that become ions when dissolved in water, producing an aqueous solution $\left( aq \right)$.
Nonelectrolytes do not dissociate when forming an aqueous solution. An equation can still be written that simply shows the solid going into solution. For example, the process of dissolving sucrose in water can be written as follows:
$\ce{C_{12}H_{22}O_{11}} \left( s \right) \rightarrow \ce{C_{12}H_{22}O_{11}} \left( aq \right)$
Strong and Weak Electrolytes
Some polar molecular compounds are nonelectrolytes when the are in their pure state but become electrolytes when they are dissolved in water. Hydrogen chloride $\left( \ce{HCl} \right)$ is a gas in its pure molecular state and is a nonelectrolyte. However, when $\ce{HCl}$ is dissolved in water, it conducts a current well because the $\ce{HCl}$ molecule ionizes into hydrogen and chloride ions.
$\ce{HCl} \left( g \right) \rightarrow \ce{H^+} \left( aq \right) + \ce{Cl^-} \left( aq \right)$
When $\ce{HCl}$ is dissolved into water, it is called hydrochloric acid. Ionic compounds and some polar compounds are completely broken apart into ions and thus conduct a current very well. A strong electrolyte is a solution in which almost all of the dissolved solute exists as ions.
Some other polar molecular compounds become electrolytes upon being dissolved into water but do not ionize to a very great extent. For example, nitrous acid $\left( \ce{HNO_2} \right)$ only partially ionizes into hydrogen ions and nitrite ions when dissolved in water. Aqueous nitrous acid is composed of only about $5\%$ ions and $95\%$ intact nitrous acid molecules A weak electrolyte is a solution in which only a small fraction of the dissolved solute exists as ions. The equation showing the ionization of a weak electrolyte utilizes an equilibrium arrow, indicating an equilibrium between the reactants and products.
$\ce{HNO_2} \left( aq \right) \rightleftharpoons \ce{H^+} \left( aq \right) + \ce{NO_2^-} \left( aq \right)$
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/07%3A_Solids_Liquids_and_Gases/7.05%3A_Aqueous_Solutions.txt |
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