State Dalton’s law of partial pressures and use it in calculations involving gaseous mixtures
The study of the chemical behaviour of gases was part of the basis of perhaps the most fundamental chemical revolution in history. French nobleman Antoine Lavoisier, widely regarded as the “father of modern chemistry,” changed chemistry from a qualitative to a quantitative science through his work with gases. He discovered the law of conservation of matter, discovered the role of oxygen in combustion reactions, determined the composition of air, explained respiration in terms of chemical reactions, and more. He was a casualty of the French Revolution, guillotined in 1794. Of his death, mathematician and astronomer Joseph-Louis Lagrange said, “It took the mob only a moment to remove his head; a century will not suffice to reproduce it.”[1]
As described in an earlier chapter of this text, we can turn to chemical stoichiometry for answers to many of the questions that ask “How much?” We can answer the question with masses of substances or volumes of solutions. However, we can also answer this question another way: with volumes of gases. We can use the ideal gas equation to relate the pressure, volume, temperature, and number of moles of a gas. Here we will combine the ideal gas equation with other equations to find gas density and molar mass. We will deal with mixtures of different gases, and calculate amounts of substances in reactions involving gases. This section will not introduce any new material or ideas, but will provide examples of applications and ways to integrate concepts we have already discussed.
Density of a Gas
Recall that the density of a gas is its mass to volume ratio, [latex]\rho = \frac{m}{V}[/latex]. Therefore, if we can determine the mass of some volume of a gas, we will get its density. The density of an unknown gas can be used to determine its molar mass and thereby assist in its identification. The ideal gas law, PV = nRT, provides us with a means of deriving such a mathematical formula to relate the density of a gas to its volume in the proof shown in Example 1.
Example 12.4a
Derivation of a Density Formula from the Ideal Gas Law
Use PV = nRT to derive a formula for the density of gas in g/L
Solution
PV = nRT
Rearrange to get (mol/L): [latex]\frac{n}{v} = \frac{P}{RT}[/latex]
Multiply each side of the equation by the molar mass, [latex]\mathcal{M}[/latex]. When moles are multiplied by [latex]\mathcal{M}[/latex] in g/mol, g are obtained:
[latex](\mathcal{M})(\frac{n}{V}) = (\frac{P}{RT})(\mathcal{M})[/latex]
We must specify both the temperature and the pressure of a gas when calculating its density because the number of moles of a gas (and thus the mass of the gas) in a litre changes with temperature or pressure. Gas densities are often reported at STP.
Example 12.4b
Empirical/Molecular Formula Problems Using the Ideal Gas Law and Density of a Gas
Cyclopropane, a gas once used with oxygen as a general anesthetic, is composed of 85.7% carbon and 14.3% hydrogen by mass. Find the empirical formula. If 1.56 g of cyclopropane occupies a volume of 1.00 L at 0.984 atm and 50 °C, what is the molecular formula for cyclopropane?
Solution
Strategy: First solve the empirical formula problem using methods discussed earlier. Assume 100 g and convert the percentage of each element into grams. Determine the number of moles of carbon and hydrogen in the 100-g sample of cyclopropane. Divide by the smallest number of moles to relate the number of moles of carbon to the number of moles of hydrogen. In the last step, realize that the smallest whole number ratio is the empirical formula:
Acetylene, a fuel used welding torches, is comprised of 92.3% C and 7.7% H by mass. Find the empirical formula. If 1.10 g of acetylene occupies of volume of 1.00 L at 1.15 atm and 59.5 °C, what is the molecular formula for acetylene?
Another useful application of the ideal gas law involves the determination of molar mass. By definition, the molar mass of a substance is the ratio of its mass in grams, m, to its amount in moles, n:
[latex]\mathcal{M} = \frac{\text{grams of substance}}{\text{moles of substance}} = \frac{m}{n}[/latex]
The ideal gas equation can be rearranged to isolate n:
[latex]n = \frac{PV}{RT}[/latex]
and then combined with the molar mass equation to yield:
[latex]\mathcal{M} = \frac{mRT}{PV}[/latex]
This equation can be used to derive the molar mass of a gas from measurements of its pressure, volume, temperature, and mass.
Example 12.4c
Determining the Molar Mass of a Volatile Liquid
The approximate molar mass of a volatile liquid can be determined by:
Heating a sample of the liquid in a flask with a tiny hole at the top, which converts the liquid into gas that may escape through the hole
Removing the flask from heat at the instant when the last bit of liquid becomes gas, at which time the flask will be filled with only gaseous sample at ambient pressure
Sealing the flask and permitting the gaseous sample to condense to liquid, and then weighing the flask to determine the sample’s mass (see Figure 12.4a)
Using this procedure, a sample of chloroform gas weighing 0.494 g is collected in a flask with a volume of 129 cm3 at 99.6 °C when the atmospheric pressure is 742.1 mm Hg. What is the approximate molar mass of chloroform?
Solution
Since [latex]\mathcal{M} = \frac{m}{n}[/latex] and [latex]n = \frac{PV}{RT}[/latex], substituting and rearranging gives [latex]\mathcal{M} = \frac{mRT}{PV}[/latex],
A sample of phosphorus that weighs 3.243 × 10−2 g exerts a pressure of 31.89 kPa in a 56.0-mL bulb at 550 °C. What are the molar mass and molecular formula of phosphorus vapour?
Unless they chemically react with each other, the individual gases in a mixture of gases do not affect each other’s pressure. Each individual gas in a mixture exerts the same pressure that it would exert if it were present alone in the container (Figure 12.4b). The pressure exerted by each individual gas in a mixture is called its partial pressure. This observation is summarized by Dalton’s law of partial pressures: The total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases:
In the equation PTotal is the total pressure of a mixture of gases, PA is the partial pressure of gas A; PB is the partial pressure of gas B; PC is the partial pressure of gas C; and so on.
Figure 12.4b If equal-volume cylinders containing gas A at a pressure of 300 kPa, gas B at a pressure of 600 kPa, and gas C at a pressure of 450 kPa are all combined in the same-size cylinder, the total pressure of the mixture is 1350 kPa (credit: Chemistry (OpenStax),CC BY 4.0).
The partial pressure of gas A is related to the total pressure of the gas mixture via its mole fraction (X), a unit of concentration defined as the number of moles of a component of a solution divided by the total number of moles of all components:
where PA, XA, and nA are the partial pressure, mole fraction, and number of moles of gas A, respectively, and nTotal is the number of moles of all components in the mixture.
Example 12.4d
The Pressure of a Mixture of Gases
A 10.0-L vessel contains 2.50 × 10−3 mol of H2, 1.00 × 10−3 mol of He, and 3.00 × 10−4 mol of Ne at 35 °C.
What are the partial pressures of each of the gases?
What is the total pressure in atmospheres?
Solution
The gases behave independently, so the partial pressure of each gas can be determined from the ideal gas equation, using [latex]P = \frac{nRT}{V}[/latex]:
A simple way to collect gases that do not react with water is to capture them in a bottle that has been filled with water and inverted into a dish filled with water. The pressure of the gas inside the bottle can be made equal to the air pressure outside by raising or lowering the bottle. When the water level is the same both inside and outside the bottle (Figure 12.4c), the pressure of the gas is equal to the atmospheric pressure, which can be measured with a barometer.
Figure 12.4c When a reaction produces a gas that is collected above water, the trapped gas is a mixture of the gas produced by the reaction and water vapour. If the collection flask is appropriately positioned to equalize the water levels both within and outside the flask, the pressure of the trapped gas mixture will equal the atmospheric pressure outside the flask (see the earlier discussion of manometers) (credit: Chemistry (OpenStax),CC BY 4.0).
However, there is another factor we must consider when we measure the pressure of the gas by this method. Water evaporates and there is always gaseous water (water vapour) above a sample of liquid water. As a gas is collected over water, it becomes saturated with water vapour and the total pressure of the mixture equals the partial pressure of the gas plus the partial pressure of the water vapour. The pressure of the pure gas is therefore equal to the total pressure minus the pressure of the water vapour—this is referred to as the “dry” gas pressure, that is, the pressure of the gas only, without water vapour. The vapour pressure of water, which is the pressure exerted by water vapour in equilibrium with liquid water in a closed container, depends on the temperature (Figure 12.4d); more detailed information on the temperature dependence of water vapour can be found in Table 12.4a, and vapour pressure will be discussed in more detail in the next chapter on liquids.
Figure 12.4d This graph shows the vapour pressure of water at sea level as a function of temperature (credit: Chemistry (OpenStax),CC BY 4.0).
Table 12.4a Vapour Pressure of Ice and Water in Various Temperatures at Sea Level
Temperatures -10 to 16 (°C)
Pressures at -10 to 16°C (torr)
Temperatures 18 to 29 (°C)
Pressures at 18 to 29°C (torr)
Temperatures 30 to 101.0 (°C)
Pressures at 30 to 101.0 °C (torr)
–10
1.95
18
15.5
30
31.8
–5
3.0
19
16.5
35
42.2
–2
3.9
20
17.5
40
55.3
0
4.6
21
18.7
50
92.5
2
5.3
22
19.8
60
149.4
4
6.1
23
21.1
70
233.7
6
7.0
24
22.4
80
355.1
8
8.0
25
23.8
90
525.8
10
9.2
26
25.2
95
633.9
12
10.5
27
26.7
99
733.2
14
12.0
28
28.3
100.0
760.0
16
13.6
29
30.0
101.0
787.6
Example 12.4f
Pressure of a Gas Collected Over Water
If 0.200 L of argon is collected over water at a temperature of 26 °C and a pressure of 750 torr in a system like that shown in Figure 3, what is the partial pressure of argon?
Solution
According to Dalton’s law, the total pressure in the bottle (750 torr) is the sum of the partial pressure of argon and the partial pressure of gaseous water:
A sample of oxygen collected over water at a temperature of 29.0 °C and a pressure of 764 torr has a volume of 0.560 L. What volume would the dry oxygen have under the same conditions of temperature and pressure?
Chemical stoichiometry describes the quantitative relationships between reactants and products in chemical reactions.
We have previously measured quantities of reactants and products using masses for solids and volumes in conjunction with the molarity for solutions; now we can also use gas volumes to indicate quantities. If we know the volume, pressure, and temperature of a gas, we can use the ideal gas equation to calculate how many moles of the gas are present. If we know how many moles of a gas are involved, we can calculate the volume of a gas at any temperature and pressure.
Avogadro’s Law Revisited
Sometimes we can take advantage of a simplifying feature of the stoichiometry of gases that solids and solutions do not exhibit: All gases that show ideal behaviour contain the same number of molecules in the same volume (at the same temperature and pressure). Thus, the ratios of volumes of gases involved in a chemical reaction are given by the coefficients in the equation for the reaction, provided that the gas volumes are measured at the same temperature and pressure.
We can extend Avogadro’s law (that the volume of a gas is directly proportional to the number of moles of the gas) to chemical reactions with gases: Gases combine, or react, in definite and simple proportions by volume, provided that all gas volumes are measured at the same temperature and pressure. For example, since nitrogen and hydrogen gases react to produce ammonia gas according to [latex]\text{N}_2(g) + 3\text{H}_2(g) \longrightarrow 2\text{NH}_3(g)[/latex], a given volume of nitrogen gas reacts with three times that volume of hydrogen gas to produce two times that volume of ammonia gas, if pressure and temperature remain constant.
The explanation for this is illustrated in Figure 12.4e. According to Avogadro’s law, equal volumes of gaseous N2, H2, and NH3, at the same temperature and pressure, contain the same number of molecules. Because one molecule of N2 reacts with three molecules of H2 to produce two molecules of NH3, the volume of H2 required is three times the volume of N2, and the volume of NH3 produced is two times the volume of N2.
Figure 12.4e One volume of N2 combines with three volumes of H2 to form two volumes of NH3 (credit: Chemistry (OpenStax),CC BY 4.0).
Example 12.4g
Reaction of Gases
Propane, C3H8(g), is used in gas grills to provide the heat for cooking. What volume of O2(g) measured at 25 °C and 760 torr is required to react with 2.7 L of propane measured under the same conditions of temperature and pressure? Assume that the propane undergoes complete combustion.
Solution
The ratio of the volumes of C3H8 and O2 will be equal to the ratio of their coefficients in the balanced equation for the reaction:
A volume of 13.5 L of O2 will be required to react with 2.7 L of C3H8.
Exercise 12.4g
An acetylene tank for an oxyacetylene welding torch provides 9340 L of acetylene gas, C2H2, at 0 °C and 1 atm. How many tanks of oxygen, each providing 7.00 × 103 L of O2 at 0 °C and 1 atm, will be required to burn the acetylene?
Ammonia is an important fertilizer and industrial chemical. Suppose that a volume of 683 billion cubic feet of gaseous ammonia, measured at 25 °C and 1 atm, was manufactured. What volume of H2(g), measured under the same conditions, was required to prepare this amount of ammonia by reaction with N2?
Because equal volumes of H2 and NH3 contain equal numbers of molecules and each three molecules of H2 that react produce two molecules of NH3, the ratio of the volumes of H2 and NH3 will be equal to 3:2. Two volumes of NH3, in this case in units of billion ft3, will be formed from three volumes of H2:
The manufacture of 683 billion ft3 of NH3 required 1020 billion ft3 of H2. (At 25 °C and 1 atm, this is the volume of a cube with an edge length of approximately 1.9 miles.)
Exercise 12.4h
What volume of O2(g) measured at 25 °C and 760 torr is required to react with 17.0 L of ethylene, C2H4(g), measured under the same conditions of temperature and pressure? The products are CO2 and water vapour.
Sulfur dioxide is an intermediate in the preparation of sulfuric acid. What volume of SO2 at 343 °C and 1.21 atm is produced by burning l.00 kg of sulfur in oxygen?
The thin skin of our atmosphere keeps the earth from being an ice planet and makes it habitable. In fact, this is due to less than 0.5% of the air molecules. Of the energy from the sun that reaches the earth, almost [latex]\frac{1}{3}[/latex] is reflected back into space, with the rest absorbed by the atmosphere and the surface of the earth. Some of the energy that the earth absorbs is re-emitted as infrared (IR) radiation, a portion of which passes back out through the atmosphere into space. However, most of this IR radiation is absorbed by certain substances in the atmosphere, known as greenhouse gases, which re-emit this energy in all directions, trapping some of the heat. This maintains favourable living conditions—without atmosphere, the average global average temperature of 14 °C (57 °F) would be about –19 °C (–2 °F). The major greenhouse gases (GHGs) are water vapor, carbon dioxide, methane, and ozone. Since the Industrial Revolution, human activity has been increasing the concentrations of GHGs, which have changed the energy balance and are significantly altering the earth’s climate (Figure 12.4f).
Figure 12.4f Greenhouse gases trap enough of the sun’s energy to make the planet habitable—this is known as the greenhouse effect. Human activities are increasing greenhouse gas levels, warming the planet and causing more extreme weather events (credit: Chemistry (OpenStax),CC BY 4.0).
There is strong evidence from multiple sources that higher atmospheric levels of CO2 are caused by human activity, with fossil fuel burning accounting for about [latex]\frac{3}{4}[/latex] of the recent increase in CO2. Reliable data from ice cores reveals that CO2 concentration in the atmosphere is at the highest level in the past 800,000 years; other evidence indicates that it may be at its highest level in 20 million years. In recent years, the CO2 concentration has increased from historical levels of below 300 ppm to almost 400 ppm today.
Figure 12.4g CO2 levels over the past 700,000 years were typically from 200–300 ppm, with a steep, unprecedented increase over the past 50 years (credit: Chemistry (OpenStax),CC BY 4.0).
Video source: U.S. Environmental Protection Agency. (2015, April 3). The Greenhouse Effect [Video]. YouTube.
Scientists in Action: Dr. Susan Solomon
Figure 12.4h Atmospheric and climate scientist Susan Solomon (credit: work by Bengt Nyman, CC BY 2.0)
Atmospheric and climate scientist Dr. Susan Solomon (Figure 12.4h) is the author of one of The New York Times books of the year (The Coldest March, 2001), one of Time magazine’s 100 most influential people in the world (2008), and a working group leader of the Intergovernmental Panel on Climate Change (IPCC), which was the recipient of the 2007 Nobel Peace Prize. She helped determine and explain the cause of the formation of the ozone hole over Antarctica, and has authored many important papers on climate change.
She has been awarded the top scientific honours in the US and France (the National Medal of Science and the Grande Medaille, respectively), and is a member of the National Academy of Sciences, the Royal Society, the French Academy of Sciences, and the European Academy of Sciences. Formerly a professor at the University of Colorado, she is now at MIT, and continues to work at NOAA.
“Quotations by Joseph-Louis Lagrange,” last modified February 2006, accessed February 10, 2015, http://www-history.mcs.st-andrews.ac.uk/Quotations/Lagrange.html ↵
[latex]\rho = \frac{P \mathcal{M}}{RT}[/latex]
[latex]0.0847 \;\text{g/L} = 760 \;\rule[0.5ex]{1.7em}{0.1ex}\hspace{-1.7em}\text{torr} \times \frac{1 \;\rule[0.25ex]{1.2em}{0.1ex}\hspace{-1.2em}\text{atm}}{760 \;\rule[0.25ex]{1.2em}{0.1ex}\hspace{-1.2em}\text{torr}} \times \frac{\mathcal{M}}{0.0821 \;\text{L} \;\rule[0.25ex]{1.2em}{0.1ex}\hspace{-1.2em}\text{atm/mol K}} \times 290 \;\text{K}[/latex]
[latex]\mathcal{M}[/latex] = 2.02 g/mol; therefore, the gas must be hydrogen (H2, 2.02 g/mol)
↵