1 L The use of this conversion factor is illustrated in Figure 3.6, and the quantities of
3. This leaves only the O atoms to balance. A coefficient of 2 in front of O 2
makes the atoms of O equal on both sides of the equation:
CH4(g) 1 __2 O2(g) S CO2(g) 1 __2 H2O(g)
+ +
element reactant Side product Side Balanced?
C 1 1 ✔
H 4 2 3 2 5 4 ✔
O 2 3 2 5 4 2 1 (2 3 1) 5 4 ✔
Balancing first C, then H, then O is a useful strategy when writing the equations for many combustion reactions involving hydrocarbons.
combustion reaction a heat-producing reaction between oxygen and another element or compound.
FIGURE 3.19 The combustion of methane in a gas stove produces carbon dioxide and water.
3.4 Combustion Reactions 103
SAmpLe eXerCiSe 3.11 Writing a Balanced chemical equation for a combustion reaction
Lo2
Methane (CH4) is the principal ingredient in natural gas, but significant concentrations of the hydrocarbon gases ethane (C2H6) and propane (C3H8) are also present in most natural gas samples. Write a balanced chemical equation describing the complete combustion of C2H6.
Collect, Organize, and Analyze We need to write a balanced chemical equation for the combustion of C2H6. The products of complete combustion of hydrocarbons are carbon dioxide and water.
Solve The preliminary expression with single particles of all known reactants and products describing the combustion of ethane is
C2H6(g) 1 O2(g) S CO2(g) 1 H2O(g)
element reactant Side product Side Balanced?
C 2 1 ✗
H 6 2 ✗
O 2 2 1 1 5 3 ✗
Because this reaction is between oxygen and a hydrocarbon, we balance C first, then H, then O. Balance the carbon atoms first by giving CO2 in the product a coefficient of 2:
__ C2H6(g) 1 __ O2(g) S __2 CO2(g) 1 __ H2O(g) element reactant Side product Side Balanced?
C 2 2 3 1 5 2 ✔
H 6 2 ✗
O 2 (2 3 2) 1 1 5 5 ✗
Then balance the hydrogen atoms by giving H2O a coefficient of 3:
__ C2H6(g) 1 __ O2(g) S __2 CO2(g) 1 __3 H2O(g) element reactant Side product Side Balanced?
C 2 2 3 1 5 2 ✔
H 6 3 3 2 5 6 ✔
O 2 (2 3 2) 1 (3 3 1) 5 7 ✗
At this stage, the oxygen atoms cannot be balanced with a simple whole-number coefficient for O2; we need 72 O2. However, if we give O2 a coefficient of 7 and double the coefficients for ethane, carbon dioxide, and water, we can write
__2 C2H6(g) 1 __7 O2(g) S __4 CO2(g) 1 __6 H2O(g) element reactant Side product Side Balanced?
C 2 3 2 5 4 4 3 1 5 4 ✔
H 2 3 6 5 12 6 3 2 5 12 ✔
O 7 3 2 5 14 (4 3 2) 1 (6 3 1) 5 14 ✔ This equation is balanced.
3.5 Stoichiometric Calculations and the Carbon Cycle
Earth’s atmosphere underwent a major change beginning about 2.5 billion years ago with the evolution of bacteria capable of photosynthesis. Photosynthesis is driven by the energy in sunlight and involves several steps, but the overall reaction is
6 CO2(g) 1 6 H2O(,) S C6H12O6(aq) 1 6 O2(g) Glucose
Although photosynthetic bacteria are believed to have been the initial source of oxygen in our atmosphere, today O2 comes mostly from green plants. The reverse reaction, called respiration, is the major source of energy for all living things on Earth:
C6H12O6(aq) 1 6 O2(g) S 6 CO2(g) 1 6 H2O(,) Glucose
Photosynthesis and respiration are key reactions in the carbon cycle (Figure 3.20).
The two processes are nearly, but not exactly, in balance in Earth’s biosphere. If they were exactly in balance, no net change would have taken place in the concen- trations of atmospheric carbon dioxide or oxygen in the past 2.5 billion years.
However, about 0.01% of the decaying mass of plants and animals (called detritus) is incorporated into sediments and soil when organisms die. Shielded in this way from exposure to oxygen, the carbon in this mass is not converted back into CO2. Although 0.01% may not seem like much, over hundreds of millions of years it has added up to the removal of about 1020 kg of carbon dioxide from the atmo- sphere. About 1015 kg of this buried carbon is in the form of fossil fuels: coal, petroleum, and natural gas.
As a result of human activity and the combustion of fossil fuels, the natural balance that limited the concentration of CO2 in the atmosphere is being altered.
Annually about 6.8 trillion (6.8 3 1012) kg of carbon is reintroduced into the atmosphere as CO2 as a result of the combustion of fossil fuels, and deforestation adds another 2 3 1012 kg each year. The effects of these additions on global cli- mate have been the subject of considerable debate, and we examine them in Chapter 8. Here, though, we will use a typical reaction involving CO2 to learn how to use a balanced chemical equation to calculate the mass of products in a chemical reaction.
If combustion of fossil fuels adds 6.8 3 1012 kg of carbon to the atmosphere each year as CO2, what is the mass of the carbon dioxide added? The mass must be more than 6.8 3 1012 kg, because this amount is only the mass due to carbon.
The balanced chemical equation for the combustion of carbon to carbon dioxide in excess O2 is
C(s) 1 O2(g) S CO2(g)
Chemtour Carbon Cycle
Think About It Balancing chemical equations for combustion reactions of hydrocarbons often requires several iterations of the coefficients before all of the elements are balanced.
d Practice Exercise Write a balanced chemical equation describing the complete combustion of propane (C3H8).
(Answers to Practice Exercises are in the back of the book.)
3.5 Stoichiometric Calculations and the Carbon Cycle 105
To use this equation to determine the mass of CO2 released, we first use the molar mass of carbon, 12.01 g/mol, to convert mass into moles:
6.831012 kg C3 103 g
1 kg 3 1 mol C
12.01 g C 55.731014 mol C
We know from the coefficients for CO2 and C in the balanced equation that the mole ratio of CO2 to C is 1:1, so the amount of CO2 produced is also 5.7 3 1014 mol:
5.731014 mol C3 1 mol CO2
1 mol C 55.731014 mol CO2
To convert moles of CO2 to mass, we first calculate the molar mass of CO2: 12.01 g/mol 1 2(16.00 g/mol) 5 44.01 g/mol CO2
Multiplying the moles of CO2 by the molar mass of CO2 and converting grams to kilograms gives us the mass of CO2:
5.731014 mol CO23 44.01 g CO2
1 mol CO2 3 1 kg
103 g52.531013 kg CO2
Terrestrial plants Terrestrial
plants Terrestrial
animals Carbon
as CO2
Combustion Photosynthesis Respiration
Respiration Photosynthesis
Marine phytoplanktonMarine phytoplankton
Marine animals
Carbonate sediments Fossil fuels
Detritus Detritus
3 1 1 3
2
2
4 4
5 5
6
FIGURE 3.20 The carbon cycle. ① Green plants and marine phytoplankton
incorporate CO2 into their biomass.
② Some of this biomass becomes the biomass of animals. ③ As photosynthetic organisms and animals respire, they release CO2 back into the environment.
④ When they die, the decay of their tissues releases most of their carbon content as CO2, but about 0.01% is incorporated into carbonate minerals and deposits of coal, petroleum, and natural gas (fossil fuels,
⑤). ⑥ Mining and the combustion of fossil fuels for human use are shifting the natural equilibrium that has controlled the concentration of CO2 in the atmosphere.
SAmpLe eXerCiSe 3.12 calculating a Product Mass from a reactant Mass
Lo3
In 2014 electric power plants in the United States consumed about 1.57 3 1011 kg of natural gas. Natural gas is mostly methane, CH4, so we can approximate the combustion reaction generating the energy by the equation:
CH4(g) 1 2 O2(g) S CO2(g) 1 2 H2O(g)
How many kilograms of CO2 were released into the atmosphere from these power plants in 2014?
Collect, Organize, and Analyze We are asked to calculate the mass of CO2 produced from combustion of 1.57 3 1011 kg of CH4. The balanced equation tells us that 1 mole of carbon dioxide is produced for every 1 mole of methane consumed. We can change the mass of CH4 in kilograms to grams and then convert into moles of CH4 by multiplying by the molar mass of CH4. In this reaction the moles of CH4 consumed is equal to the number of moles of CO2 produced, so the mole ratio is 1:1. Finally we convert moles of CO2 into kilograms of CO2. We can combine all of these steps into a single calculation:
kg CH4 103 g
1 kg g CH4 mol CH4
1 mol CO2
1 mol CH4 mol CO2
molar mass1 molar mass g CO2
kg CH4
103 g
1 kg g CH4 mol CH4
1 mol CO2
1 mol CH4 mol CO2
101 kg3 g
molar mass1 molar mass g CO2 kg CO2
Solve The chemical equation is balanced as written (you should always check to be sure). To convert between grams and moles of CH4 and CO2, we need to calculate the molar masses of these compounds. The molar mass of CH4 is
12.01 g/mol 1 4(1.008 g/mol) 5 16.04 g/mol CH4
and the molar mass of CO2 is
12.01 g/mol 1 2(16.00 g/mol) 5 44.01 g/mol CO2
Convert the mass of CH4 into moles:
1.5731011 kg CH43103 g
1 kg 3 1 mol CH4
16.04 g CH4
59.78831012 mol CH4
Convert moles of CH4 into moles of CO2 by using the mole ratio from the balanced chemical equation:
9.78831012 mol CH431 mol CO2
1 mol CH4
59.78831012 mol CO2
Notice that the answer in each of these steps is the starting point for the next step. Therefore we can combine the three separate calculations into a single calculation:
6.831012 kg C3 103 g
1 kg 3 1 mol C
12.01 g C3 1 mol CO2
1 mol C 344.01 g CO2
1 mol CO2 3 1 kg 103 g
52.531013 kg CO2
This procedure can be applied to determining the mass of any substance (reac- tant or product) involved in any chemical reaction if we know (1) the mass of another substance in the reaction and (2) the stoichiometric relation between the two substances, that is, their mole ratio in the balanced chemical equation.
3.5 Stoichiometric Calculations and the Carbon Cycle 107
SAmpLe eXerCiSe 3.13 calculating the Masses of reactants Lo3 Copper was among the first metals to be refined from minerals collected by ancient metalworkers. The production of copper was already an industry by 3500 bce. Cuprite is a copper mineral commonly found near Earth’s surface, making it a likely resource for Bronze Age coppersmiths. Cuprite has the formula Cu2O and can be converted to copper metal by reacting it with charcoal:
2 Cu2O(s) 1 C(s) S 4 Cu(s) 1 CO2(g)
How much cuprite and how much carbon are needed to prepare a 256 g copper bracelet such as the one shown in Figure 3.21?
Collect and Organize We are given a balanced chemical equation and the mass of product. We are asked to find the masses of the reactants. Because the chemical equation relates moles, not masses, of reactants and products, we need to find the molar mass of each substance in the reaction.
Analyze The balanced chemical equation tells us that for every 4 moles of copper we produce, 2 moles of Cu2O and 1 mole of C must react. To use these mole ratios, we first need to convert the mass of copper to the equivalent number of moles of Cu by using the molar mass of Cu. We then work a separate conversion for each reactant: the first conversion uses the mole ratio of Cu2O to Cu (2:4) from the balanced equation, and the second uses the mole ratio of C to Cu (1:4). Finally, we use the molar masses of Cu2O and C to find the mass of each required for the reaction.
g Cu mol Cu
2 mol Cu2O
4 mol Cu mol Cu2O g Cu2O
g Cu
1 mol C 4 mol Cu
mol C g C
molar mass1
mol Cu molar mass1
molar mass molar mass
g Cu mol Cu
2 mol Cu2O
4 mol Cu mol Cu2O g Cu2O
g Cu
1 mol C
4 mol Cu mol C g C
molar mass1
mol Cu molar mass1
molar mass
molar mass
Solve Carrying out the steps described in the analysis of the problem:
256 g Cu3 1 mol Cu
63.55 g Cu32 mol Cu2O
4 mol Cu 3143.09 g Cu2O
1 mol Cu2O 5288 g Cu2O 256 g Cu3 1 mol Cu
63.55 g Cu3 1 mol C
4 mol Cu312.01 g C
1 mol C 512.1 g C Convert moles of CO2 into mass of CO2:
9.78831012 mol CO2344.01 g CO2
1 mol CO2
3 1 kg
103 g54.3131011 kg CO2
We can combine the three separate calculations into a single calculation, and in subsequent problems we may not show the individual steps:
1.5731011 kg CH43103 g
1 kg 3 1 mol CH4
16.04 g CH4
31 mol CO2
1 mol CH4
344.01 g CO2
1 mol CO2
3 1 kg 103 g
54.3131011 kg CO2
Think About It The answer 4.31 3 1011 kg of CO2 is about 1.7% of the mass of CO2 we calculated for total annual CO2 production by combustion of fossil fuels.
d Practice Exercise Disposable lighters burn butane (C4H10) and produce CO2 and H2O. Balance the chemical equation for this combustion reaction, and determine how many grams of CO2 are produced by burning 1.00 g of C4H10. (Answers to Practice Exercises are in the back of the book.)
FIGURE 3.21 Copper bracelet.
3.6 Determining Empirical Formulas from Percent Composition
Natural sources of methane such as swamps, bogs, rice paddies, and animals introduce 0.570 teragrams (1 teragram 5 1 Tg 5 1 million metric tons 5 1012 g) or 5.70 3 1011 g of CH4 into the atmosphere every day. Hydrocarbons are also emitted into the atmosphere by other sources, such as conifers like the longleaf pine shown in Figure 3.22. Conifers emit a variety of hydrocarbons into the air, including pinene, which has the chemical formula C10H16. One way to distinguish between two hydrocarbons such as CH4 and C10H16 is in terms of percent composition: the composition of a compound expressed in terms of the percentages of the masses of the constituent elements with respect to the total mass of the compound.
To calculate the percent composition of both CH4 and C10H16, we must cal- culate the carbon content of both compounds. Suppose we have 1 mole of CH4, which has a molar mass of
4(1.008 g H/mol) 1 12.01 g C/mol 5 16.04 g/mol
Of this 16.04 g, C accounts for 12.01 g. Therefore the C content of CH4 is mass of C
total mass5 12.01 g C
16.04 g CH4 50.7488 or
0.7488 3 100% 5 74.88% C It follows that the hydrogen content is
mass of H
total mass 5 411.008 g H2 16.04 g CH4
50.2512 or 25.12% H
Because CH4 contains only C and H, we could also determine the percent H by subtracting the percent C from 100%:
100.00% 2 74.88% 5 25.12%
Thus the percent composition of CH4 is 74.88% C and 25.12% H.
How does this compare with the percent C in pinene? If we repeat the calcu- lation for C10H16, the molar mass is
16(1.008 g H/mol) 1 10(12.01 g C/mol) 5 136.23 g/mol Chemtour
Percent Composition
We are allowed only three significant figures in our answer, so the final masses of the reactants we need are 288 g of Cu2O and 12.1 g of C.
Think About It To prepare a given mass of copper, we must have sufficient amounts of both copper ore and charcoal. Usually there is a limited supply of ore and more than enough charcoal.
d Practice Exercise The copper mineral chalcocite, Cu2S, can be converted to copper simply by heating in air: Cu2S(s) 1 O2(g) S 2 Cu(s) 1 SO2(g). How much Cu2S is needed to make 256 g of Cu? How many grams of SO2 are produced?
(Answers to Practice Exercises are in the back of the book.)
CH3
CH2 CH2
CH HC C C
C
H H3C
H3C
FIGURE 3.22 The odor of the longleaf pine (Pinus palustris) comes from the volatile hydrocarbon pinene.
3.6 Determining Empirical Formulas from Percent Composition 109
Note in Sample Exercise 3.14 that the three percentages sum to 100.00%.
Percent composition values should always sum to 100%, or very close to 100%, if we have accounted for all the elements that make up the total mass of the com- pound. The total may deviate slightly from 100% because of rounding.
We typically determine the percent composition of a substance in the labora- tory by measuring the amount of each element in a given mass of the substance.
The C content of C10H16 is mass of C
total mass 5 10112.01 g C2
136.23 g C10H16 50.8816 588.16%
and the hydrogen content is
100.00% 2 88.16% 5 11.84%
Pinene contains a higher percent carbon by mass, and a correspondingly lower percent hydrogen, than methane. Actually, methane has the highest percent hydrogen of all hydrocarbons.
SAmpLe eXerCiSe 3.14 calculating Percent composition from a chemical Formula
Lo4
The compound C2H2F4 is a propellant in the inhalers used by asthma sufferers (Figure 3.23). What is the percent composition of C2H2F4?
Collect, Organize, and Analyze We are asked to calculate the percent composition of C2H2F4. We need the molar masses of each of the elements (C, H, and F) to calculate the molar mass of C2H2F4 and the relative contribution of each element to the molar mass. To calculate the percent composition of C2H2F4, we must determine the percentage by mass of each element in one mole of C2H2F4. These percentages can be calculated by dividing the mass of each element in 1 mole of C2H2F4 by the molar mass of C2H2F4.
Solve The molar mass of C2H2F4 is
2(12.01 g/mol) 1 2(1.008 g/mol) 1 4(19.00 g/mol) 5 102.04 g/mol Thus the percent composition of this compound is
%C524.02 g C
102.04 g 3100%523.54% C %H52.016 g H
102.04 g 3100%51.98% H %F576.00 g F
102.04 g 3100%574.48% F
Think About It Although C and F have similar molar masses, the observation that the percentage of F is nearly three times the percentage of C makes sense because there are two moles of F for every 1 mole of C in C2H2F4. It also makes sense that the percentage of the lightest element, hydrogen, is very small.
d Practice Exercise Determine the percent composition of Fluosol-DA, C10F18, the only FDA-approved synthetic blood substitute (Figure 3.24).
(Answers to Practice Exercises are in the back of the book.)
percent composition the composition of a compound expressed in terms of the percentage by mass of each element in the compound.
FIGURE 3.23 Patients suffering from asthma and other respiratory ailments use inhalers to relieve their discomfort. The medication is delivered using compounds such as C2H2F4 as the propellant.
FIGURE 3.24 The fluorocarbon C10F18 is the only FDA-approved synthetic blood substitute.
We can use these data to determine the empirical formula of the substance. There are four steps to deriving an empirical formula from percent composition data obtained in the laboratory: