Stoichiometry of Gaseous Substances, Mixtures, and Reactions

Learning Outcomes

Use the ideal gas law to compute gas densities and molar masses
Perform stoichiometric calculations involving gaseous substances
State Dalton’s law of partial pressures and use it in calculations involving gaseous mixtures

The study of the chemical behavior 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.”

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.

Gas Density and Molar Mass

The ideal gas law described previously in this chapter relates the properties of pressure P, volume V, temperature T, and molar amount n. This law is universal, relating these properties in identical fashion regardless of the chemical identity of the gas:

[latex]PV=nRT[/latex]

The density d of a gas, on the other hand, is determined by its identity. As described in another chapter of this text, the density of a substance is a characteristic property that may be used to identify the substance.

[latex]d=\dfrac{m}{v}[/latex]

Rearranging the ideal gas equation to isolate V and substituting into the density equation yields

[latex]d=\dfrac{mP}{nRT}=(\dfrac{m}{n})\dfrac{P}{RT}[/latex]

The ratio m/n is the definition of molar mass, ℳ:

[latex]ℳ=\dfrac{m}{n}[/latex]

The density equation can then be written

[latex]d=\dfrac{\mathscr{M}P}{RT}[/latex]

This relation may be used for calculating the densities of gases of known identities at specified values of pressure and temperature as demonstrated in Example 1.

Example 1: Measuring Gas Density

What is the density of molecular nitrogen gas at STP?

Solution

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Check Your Learning

What is the density of molecular hydrogen gas at 17.0 °C and a pressure of 760 torr?

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When the identity of a gas is unknown, measurements of the mass, pressure, volume, and temperature of a sample can be used to calculate the molar mass of the gas (a useful property for identification purposes). Combining the ideal gas equation

[latex]PV=nRT[/latex]

and the definition of molarity

[latex]ℳ=\dfrac{m}{n}[/latex]

yields the following equation:

[latex]ℳ=\dfrac{mRT}{PV}[/latex]

Determining the molar mass of a gas via this approach is demonstrated in Example 2.

Example 2: 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?

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Check Your Learning

Acetylene, a fuel used welding torches, is comprised of 92.3% C and 7.7% H by mass. Find the empirical formula. If 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?

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Example 3: 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 1)

This figure shows four photos each connected by a right-facing arrow. The first photo shows a glass flask with aluminum foil covering the top sitting on a scale. The scale reads 89.516. The second photo shows a syringe being inserted into the flask through the aluminum foil covering. The third photo shows the glass flask being inserted into a beaker of water. The water appears to be heated at 100. The fourth photo shows the glass flask being weighed again. This time the scale reads 89.512.

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?

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Check Your Learning

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 vapor?

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The Pressure of a Mixture of Gases: Dalton’s Law

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 present alone in the container (Figure 2). 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:

[latex]{P}_{Total}={P}_{A}+{P}_{B}+{P}_{C}+\ldots ={\Sigma}_{\text{i}}{P}_{\text{i}}[/latex]

In the equation P_Total is the total pressure of a mixture of gases, P_A is the partial pressure of gas A; P_B is the partial pressure of gas B; P_C is the partial pressure of gas C; and so on.

This figure includes images of four gas-filled cylinders or tanks. Each has a valve at the top. The interior of the first cylinder is shaded blue. This region contains 5 small blue circles that are evenly distributed. The label “300 k P a” is on the cylinder. The second cylinder is shaded lavender. This region contains 8 small purple circles that are evenly distributed. The label “600 k P a” is on the cylinder. To the right of these cylinders is a third cylinder. Its interior is shaded pale yellow. This region contains 12 small yellow circles that are evenly distributed. The label “450 k P a” is on this region of the cylinder. An arrow labeled “Total pressure combined” appears to the right of these three cylinders. This arrow points to a fourth cylinder. The interior of this cylinder is shaded a pale green. It contains evenly distributed small circles in the following quantities and colors; 5 blue, 8 purple, and 12 yellow. This cylinder is labeled “1350 k P a.”

The partial pressure of gas A is related to the total pressure of the gas mixture via its mole fraction, 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):

[latex]{P}_{A}={X}_{A}\times {P}_{Total}\qquad\text{ where }\qquad{X}_{A}=\dfrac{{n}_{A}}{{n}_{Total}}[/latex]

where P_A, X_A, and n_A are the partial pressure, mole fraction, and number of moles of gas A, respectively, and n_Total is the number of moles of all components in the mixture.

Example 4: 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?

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Check Your Learning

A 5.73-L flask at 25 °C contains 0.0388 mol of N₂, 0.147 mol of CO, and 0.0803 mol of H₂. What is the total pressure in the flask in atmospheres?

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Here is another example of this concept, but dealing with mole fraction calculations.

Example 5: The Pressure of a Mixture of Gases

A gas mixture used for anesthesia contains 2.83 mol oxygen, O₂, and 8.41 mol nitrous oxide, N₂O. The total pressure of the mixture is 192 kPa.

What are the mole fractions of O₂ and N₂O?
What are the partial pressures of O₂ and N₂O?

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Check Your Learning

What is the pressure of a mixture of 0.200 g of H₂, 1.00 g of N₂, and 0.820 g of Ar in a container with a volume of 2.00 L at 20 °C?

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Collection of Gases over Water

This figure shows a diagram of equipment used for collecting a gas over water. To the left is an Erlenmeyer flask. It is approximately two thirds full of a lavender colored liquid. Bubbles are evident in the liquid. The label “Reaction Producing Gas” appears below the flask. A line segment connects this label to the liquid in the flask. The flask has a stopper in it through which a single glass tube extends from the open region above the liquid in the flask up, through the stopper, to the right, then angles down into a pan that is nearly full of light blue water. This tube again extends right once it is well beneath the water’s surface. It then bends up into an inverted flask which is labeled “Collection Flask.” This collection flask is positioned with its mouth beneath the surface of the light blue water and appears approximately half full. Bubbles are evident in the water in the inverted flask. The open space above the water in the inverted flask is labeled “collected gas.”

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 3), the pressure of the gas is equal to the atmospheric pressure, which can be measured with a barometer.

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 vapor) above a sample of liquid water. As a gas is collected over water, it becomes saturated with water vapor and the total pressure of the mixture equals the partial pressure of the gas plus the partial pressure of the water vapor. The pressure of the pure gas is therefore equal to the total pressure minus the pressure of the water vapor—this is referred to as the “dry” gas pressure, that is, the pressure of the gas only, without water vapor. The vapor pressure of water, which is the pressure exerted by water vapor in equilibrium with liquid water in a closed container, depends on the temperature (Figure 4); more detailed information on the temperature dependence of water vapor can be found in Table 1, and vapor pressure will be discussed in more detail in the next chapter on liquids.

alt="A graph is shown. The horizontal axis is labeled “Temperature ( degrees C )” with markings and labels provided for multiples of 20 beginning at 0 and ending at 100. The vertical axis is labeled “Vapor pressure ( torr )” with marking and labels provided for multiples of 200, beginning at 0 and ending at 800. A smooth solid black curve extends from the origin up and to the right across the graph. The graph shows a positive trend with an increasing rate of change. On the vertical axis is ( 7 60) and an arrow pointing to it. The arrow is labeled, “Vapor pressure at ( 100 degrees C ).”"

Figure 4. This graph shows the vapor pressure of water at sea level as a function of temperature.

Table 1. Vapor Pressure of Ice and Water in Various Temperatures at Sea Level
Temperature (°C)	Pressure (torr)	Temperature (°C)	Pressure (torr)	Temperature (°C)	Pressure (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 6: 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?

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Check Your Learning

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?

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Chemical Stoichiometry and Gases

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 behavior contain the same number of molecules in the same volume (at the same temperature and pressure). Thus, the ratios of volumes of gases involved in a chemical reaction are given by the coefficients in the equation for the reaction, provided that the gas volumes are measured at the same temperature and pressure.

We can extend Avogadro’s law (that the volume of a gas is directly proportional to the number of moles of the gas) to chemical reactions with gases: Gases combine, or react, in definite and simple proportions by volume, provided that all gas volumes are measured at the same temperature and pressure. For example, since nitrogen and hydrogen gases react to produce ammonia gas according to [latex]{\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)\rightarrow{2{\text{NH}}}_{3}\left(g\right)[/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 5. 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.

This diagram provided models of the chemical reaction written with formulas across the bottom of the figure. The reaction is written; N subscript 2 plus 3H subscript 2 followed by an arrow pointing right to NH subscript 3. Just above the formulas, space-filling models are provided. Above NH subscript 2, two blue spheres are bonded. Above 3H subscript 2, three pairs of two slightly smaller white spheres are bonded. Above NH subscript 3, two molecules are shown composed each of a central blue sphere to which three slightly smaller white spheres are bonded. Across the top of the diagram, the reaction is illustrated with balloons. To the left is a light blue balloon, which is labeled “N subscript 2”. This balloon contains a single space-filling model composed of two bonded blue spheres. This balloon is followed by a plus sign, then three grey balloons which are each labeled “H subscript 2.” Each of these balloons similarly contain a single space-filling model composed of two bonded white spheres. These white spheres are slightly smaller than the blue spheres. An arrow follows that points right to two light-green balloons, which are each labeled “2 NH subscript 3.” Each light-green balloon contains a space-filling model composed of a single central blue sphere to which three slightly smaller white spheres are bonded. This diagram provided models of the chemical reaction written with formulas across the bottom of the figure. The reaction is written; N subscript 2 plus 3H subscript 2 followed by an arrow pointing right to NH subscript 3. Just above the formulas, space-filling models are provided. Above NH subscript 2, two blue spheres are bonded. Above 3H subscript 2, three pairs of two slightly smaller white spheres are bonded. Above NH subscript 3, two molecules are shown composed each of a central blue sphere to which three slightly smaller white spheres are bonded. Across the top of the diagram, the reaction is illustrated with balloons. To the left is a light blue balloon, which is labeled “N subscript 2”. This balloon contains a single space-filling model composed of two bonded blue spheres. This balloon is followed by a plus sign, then three grey balloons which are each labeled “H subscript 2.” Each of these balloons similarly contain a single space-filling model composed of two bonded white spheres. These white spheres are slightly smaller than the blue spheres. An arrow follows that points right to two light-green balloons, which are each labeled “2 NH subscript 3.” Each light-green balloon contains a space-filling model composed of a single central blue sphere to which three slightly smaller white spheres are bonded.

Figure 5. One volume of N₂ combines with three volumes of H₂ to form two volumes of NH₃.

Example 7: 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.

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Check Your Learning

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?

[latex]2{\text{C}}_{2}{\text{H}}_{2}+5{\text{O}}_{2}\rightarrow 4{\text{CO}}_{2}+2{\text{H}}_{2}\text{O}[/latex]

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Example 8: Volumes of Reacting Gases

Ammonia is an important fertilizer and industrial chemical. Suppose that a volume of 683 billion cubic feet of gaseous ammonia, measured at 25 °C and 1 atm, was manufactured. What volume of H₂(g), measured under the same conditions, was required to prepare this amount of ammonia by reaction with N₂?

[latex]{\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)\rightarrow 2{\text{NH}}_{3}\left(g\right)[/latex]

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Check Your Learning

What volume of O₂(g) measured at 25 °C and 760 torr is required to react with 17.0 L of ethylene, C₂H₄(g), measured under the same conditions of temperature and pressure? The products are CO₂ and water vapor.

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Example 9: Volume of Gaseous Product

What volume of hydrogen at 27 °C and 723 torr may be prepared by the reaction of 8.88 g of gallium with an excess of hydrochloric acid?

[latex]2\text{Ga}\left(s\right)+6\text{HCl}\left(aq\right)\rightarrow 2{\text{GaCl}}_{3}\left(aq\right)+3{\text{H}}_{2}\left(g\right)[/latex]

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Check Your Learning

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?

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Greenhouse Gases and Climate Change

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 favorable 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 6).

alt="This diagram shows half of a two dimensional view of the earth in blue and green at the left of the image. A slight distance outside the hemisphere is a grey arc. A line segment connects the label “Atmosphere” to the region between the hemisphere and the grey arc. In this region, near the surface of the earth the chemical formulas C O subscript 2, C H subscript 3, and N subscript 2 O appear. Five red arrows formed from wavy lines extend from green regions on the earth out into and just beyond the region labeled “Atmosphere.” The label “Infrared radiation” points to one of these red arrows. At a fair distance outside of the grey arc appears a yellow circle with a jagged boundary. This circle is labeled “Sun.” From it extend yellow arrows with wavy lines which extend toward the earth. Three of the arrows extend to the green region on the earth. One of the arrows appears to be reflected off the grey arc, causing its path to turn away from the earth."

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 7).

This figure has the heading “Carbon Dioxide in the Atmosphere.” The first graph has a horizontal axis label “Year ( B C )” and a vertical axis label “Carbon dioxide concentration ( p p m ).” The horizontal axis labels begin at 700,000 on the left and increases by multiples of 100,000 up to 0 on the right. The vertical axis begins at 0 and increases by multiples of 50 extending up to 400. A jagged, cyclical pattern is shown that begins before 600,000 B C at under 200 p p m. Up to 0 B C values appear to vary cyclically up to a high of about 300 p p m. Extending beyond 0 B C to the right, the carbon dioxide concentration appears to be on a steady increase, having reached nearly 400 p p m in recent years. The second graph is shown to magnify the portion of the graph that is most recent. This graph begins just before the year 1960 and includes markings for multiples of 10 up to the year 2010. The vertical axis begins just below 320 p p m and includes markings for all multiples of 20 up to 400 p p m. A smooth black line is shown extending through a jagged red data pattern. The trend is a steady, nearly linear increase from the lower left to the upper right on the graph.

A photograph is shown of Susan Solomon sitting next to a globe.

Figure 8. Susan Solomon’s research focuses on climate change and has been instrumental in determining the cause of the ozone hole over Antarctica. (credit: National Oceanic and Atmospheric Administration)

Portrait of a Chemist: Susan Solomon

Atmospheric and climate scientist Susan Solomon (Figure 8) 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 honors 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.

Key Concepts and Summary

The ideal gas law can be used to derive a number of convenient equations relating directly measured quantities to properties of interest for gaseous substances and mixtures. Appropriate rearrangement of the ideal gas equation may be made to permit the calculation of gas densities and molar masses. Dalton’s law of partial pressures may be used to relate measured gas pressures for gaseous mixtures to their compositions. Avogadro’s law may be used in stoichiometric computations for chemical reactions involving gaseous reactants or products.

Key Equations

[latex]{P}_{Total}={P}_{A}+{P}_{B}+{P}_{C}+\ldots ={\Sigma}_{\text{i}}{P}_{\text{i}}[/latex]
[latex]{P}_{A}={X}_{A}\cdot {P}_{Total}[/latex]
[latex]{X}_{A}=\dfrac{{n}_{A}}{{n}_{Total}}[/latex]

Try It

What is the density of laughing gas, dinitrogen monoxide, N₂O, at a temperature of 325 K and a pressure of 113.0 kPa?
Calculate the density of Freon 12, CF₂Cl₂, at 30.0°C and 0.954 atm.
Which is denser at the same temperature and pressure, dry air or air saturated with water vapor? Explain.
A cylinder of O₂(g) used in breathing by emphysema patients has a volume of 3.00 L at a pressure of 10.0 atm. If the temperature of the cylinder is 28.0 °C, what mass of oxygen is in the cylinder?
What is the molar mass of a gas if 0.0494 g of the gas occupies a volume of 0.100 L at a temperature 26 °C and a pressure of 307 torr?
What is the molar mass of a gas if 0.281 g of the gas occupies a volume of 125 mL at a temperature 126 °C and a pressure of 777 torr?
How could you show experimentally that the molecular formula of propene is C₃H₆, not CH₂?
The density of a certain gaseous fluoride of phosphorus is 3.93 g/L at STP. Calculate the molar mass of this fluoride and determine its molecular formula.
Consider this question: What is the molecular formula of a compound that contains 39% C, 45% N, and 16% H if 0.157 g of the compound occupies l25 mL with a pressure of 99.5 kPa at 22 °C?
1. Outline the steps necessary to answer the question.
2. Answer the question.
A 36.0–L cylinder of a gas used for calibration of blood gas analyzers in medical laboratories contains 350 g CO₂, 805 g O₂, and 4,880 g N₂. What is the pressure in the flask in atmospheres, in torr, and in kilopascals?
A cylinder of a gas mixture used for calibration of blood gas analyzers in medical laboratories contains 5.0% CO₂, 12.0% O₂, and the remainder N₂ at a total pressure of 146 atm. What is the partial pressure of each component of this gas? (The percentages given indicate the percent of the total pressure that is due to each component.)
A sample of gas isolated from unrefined petroleum contains 90.0% CH₄, 8.9% C₂H₆, and 1.1% C₃H₈ at a total pressure of 307.2 kPa. What is the partial pressure of each component of this gas? (The percentages given indicate the percent of the total pressure that is due to each component.)
A mixture of 0.200 g of H₂, 1.00 g of N₂, and 0.820 g of Ar is stored in a closed container at STP. Find the volume of the container, assuming that the gases exhibit ideal behavior.
Most mixtures of hydrogen gas with oxygen gas are explosive. However, a mixture that contains less than 3.0 % O₂ is not. If enough O₂ is added to a cylinder of H₂ at 33.2 atm to bring the total pressure to 34.5 atm, is the mixture explosive?
A commercial mercury vapor analyzer can detect, in air, concentrations of gaseous Hg atoms (which are poisonous) as low as 2 × 10^-6 mg/L of air. At this concentration, what is the partial pressure of gaseous mercury if the atmospheric pressure is 733 torr at 26 °C?
A sample of carbon monoxide was collected over water at a total pressure of 756 torr and a temperature of 18 °C. What is the pressure of the carbon monoxide? (See Table 9.2 for the vapor pressure of water.)
In an experiment in a general chemistry laboratory, a student collected a sample of a gas over water. The volume of the gas was 265 mL at a pressure of 753 torr and a temperature of 27 °C. The mass of the gas was 0.472 g. What was the molar mass of the gas?
Joseph Priestley first prepared pure oxygen by heating mercuric oxide, HgO: [latex]2\text{HgO}\left(s\right)\rightarrow 2\text{Hg}\left(l\right)+{\text{O}}_{2}\left(g\right)[/latex]
1. Outline the steps necessary to answer the following question: What volume of O₂ at 23° C and 0.975 atm is produced by the decomposition of 5.36 g of HgO?
2. Answer the question.
Cavendish prepared hydrogen in 1766 by the novel method of passing steam through a red-hot gun barrel:[latex]4{\text{H}}_{2}\text{O}\left(g\right)+3\text{Fe}\left(s\right)\rightarrow{\text{Fe}}_{3}{\text{O}}_{4}\left(s\right)+4{\text{H}}_{2}\left(g\right)[/latex]
1. Outline the steps necessary to answer the following question: What volume of H₂ at a pressure of 745 torr and a temperature of 20 °C can be prepared from the reaction of 15.O g of H₂O?
2. Answer the question.
The chlorofluorocarbon CCl₂F₂ can be recycled into a different compound by reaction with hydrogen to produce CH₂F₂(g), a compound useful in chemical manufacturing: [latex]{\text{CCl}}_{2}{\text{F}}_{2}\left(g\right)+4{\text{H}}_{2}\left(g\right)\rightarrow{\text{CH}}_{2}{\text{F}}_{2}\left(g\right)+2\text{HCl}\left(g\right).[/latex]
1. Outline the steps necessary to answer the following question: What volume of hydrogen at 225 atm and 35.5 °C would be required to react with 1 ton (1.000 × 10³ kg) of CCl₂F₂?
2. Answer the question.
Automobile air bags are inflated with nitrogen gas, which is formed by the decomposition of solid sodium azide (NaN₃). The other product is sodium metal. Calculate the volume of nitrogen gas at 27 °C and 756 torr formed by the decomposition of 125 g of sodium azide.
Lime, CaO, is produced by heating calcium carbonate, CaCO₃; carbon dioxide is the other product.
1. Outline the steps necessary to answer the following question: What volume of carbon dioxide at 875° and 0.966 atm is produced by the decomposition of 1 ton (1.000 × 10³ kg) of calcium carbonate?
2. Answer the question.
Before small batteries were available, carbide lamps were used for bicycle lights. Acetylene gas, C₂H₂, and solid calcium hydroxide were formed by the reaction of calcium carbide, CaC₂, with water. The ignition of the acetylene gas provided the light. Currently, the same lamps are used by some cavers, and calcium carbide is used to produce acetylene for carbide cannons.
1. Outline the steps necessary to answer the following question: What volume of C₂H₂ at 1.005 atm and 12.2 °C is formed by the reaction of 15.48 g of CaC₂ with water?
2. Answer the question.
Calculate the volume of oxygen required to burn 12.00 L of ethane gas, C₂H₆, to produce carbon dioxide and water, if the volumes of C₂H₆ and O₂ are measured under the same conditions of temperature and pressure.
What volume of O₂ at STP is required to oxidize 8.0 L of NO at STP to NO₂? What volume of NO₂ is produced at STP?
Consider the following questions:
1. What is the total volume of the CO₂(g) and H₂O(g) at 600 °C and 0.888 atm produced by the combustion of 1.00 L of C₂H₆(g) measured at STP?
2. What is the partial pressure of H₂O in the product gases?
Methanol, CH₃OH, is produced industrially by the following reaction:
[latex]\text{CO}\left(g\right)+2{\text{H}}_{2}\left(g\right)\,\,\,{\xrightarrow{\text{copper catalyst }300^{\circ}\text{ C}, 300\text{ atm}}}\,\,\,{\text{CH}}_{3}\text{OH}\left(g\right)[/latex]
Assuming that the gases behave as ideal gases, find the ratio of the total volume of the reactants to the final volume.
What volume of oxygen at 423.0 K and a pressure of 127.4 kPa is produced by the decomposition of 129.7 g of BaO₂ to BaO and O₂?
A 2.50-L sample of a colorless gas at STP decomposed to give 2.50 L of N₂ and 1.25 L of O₂ at STP. What is the colorless gas?
Ethanol, C₂H₅OH, is produced industrially from ethylene, C₂H₄, by the following sequence of reactions:
[latex]3{\text{C}}_{2}{\text{H}}_{4}+2{\text{H}}_{2}{\text{SO}}_{4}\rightarrow{\text{C}}_{2}{\text{H}}_{5}{\text{HSO}}_{4}+{\left({\text{C}}_{2}{\text{H}}_{5}\right)}_{2}{\text{SO}}_{4}[/latex] [latex]{\text{C}}_{2}{\text{H}}_{5}{\text{HSO}}_{4}+{\left({\text{C}}_{2}{\text{H}}_{5}\right)}_{2}{\text{SO}}_{4}+3{\text{H}}_{2}\text{O}\rightarrow 3{\text{C}}_{2}{\text{H}}_{5}\text{OH}+2{\text{H}}_{2}{\text{SO}}_{4}[/latex]
What volume of ethylene at STP is required to produce 1.000 metric ton (1000 kg) of ethanol if the overall yield of ethanol is 90.1%?
One molecule of hemoglobin will combine with four molecules of oxygen. If 1.0 g of hemoglobin combines with 1.53 mL of oxygen at body temperature (37 °C) and a pressure of 743 torr, what is the molar mass of hemoglobin?
A sample of a compound of xenon and fluorine was confined in a bulb with a pressure of 18 torr. Hydrogen was added to the bulb until the pressure was 72 torr. Passage of an electric spark through the mixture produced Xe and HF. After the HF was removed by reaction with solid KOH, the final pressure of xenon and unreacted hydrogen in the bulb was 36 torr. What is the empirical formula of the xenon fluoride in the original sample? (Note: Xenon fluorides contain only one xenon atom per molecule.)
One method of analyzing amino acids is the van Slyke method. The characteristic amino groups (-NH₂) in protein material are allowed to react with nitrous acid, HNO₂, to form N₂ gas. From the volume of the gas, the amount of amino acid can be determined. A 0.0604-g sample of a biological sample containing glycine, CH₂(NH₂)COOH, was analyzed by the van Slyke method and yielded 3.70 mL of N₂ collected over water at a pressure of 735 torr and 29 °C. What was the percentage of glycine in the sample?
[latex]{\text{CH}}_{2}\left({\text{NH}}_{2}\right){\text{CO}}_{2}\text{H}+{\text{HNO}}_{2}\rightarrow{\text{CH}}_{2}\left(\text{OH}\right){\text{CO}}_{2}\text{H}+{\text{H}}_{2}\text{O}+{\text{N}}_{2}[/latex]

Selected Answers

2. [latex]\rho =\frac{P\mathcal{M}}{RT}=\frac{0.954\cancel{\text{atm}}\left[12.011+2\left(18.9954\right)+2\left(35.453\right)\right]\text{g}\cancel{{\text{mol}}^{-\text{1}}}}{\text{0.08206 L}\cancel{\text{atm}}\cancel{{\text{mol}}^{-\text{1}}}\cancel{{\text{K}}^{-\text{1}}}\times 303.15\cancel{\text{K}}}=\text{4.64 g}{\text{L}}^{-\text{1}}[/latex]

4. [latex]\text{mass}{\text{O}}_{2}=\frac{\left(\text{31.9988 g}\cancel{{\text{mol}}^{-\text{1}}}\right)\left(10.0\cancel{\text{atm}}\right)\left(3.00\cancel{\text{L}}\right)}{\left(0.08206\cancel{\text{L}}\cancel{\text{atm}}\cancel{{\text{mol}}^{-\text{1}}}\cancel{{\text{K}}^{-\text{1}}}\right)\left(301.15\cancel{\text{K}}\right)}=\text{38.8 g}[/latex]

6. From the ideal gas law, PV = nRT, set [latex]n=\frac{\text{mass}}{\text{molar mass}}[/latex] and solve the molar mass. [latex]\text{molar mass}=\frac{\left(\text{0.281 g}\right)\left(0.08206\cancel{\text{L}}\cancel{\text{atm}}{\text{mol}}^{-\text{1}}\cancel{{\text{K}}^{-\text{1}}}\right)\left(399.15\cancel{\text{K}}\right)}{\left(\frac{777\cancel{\text{torr}}}{760\cancel{\text{torr}}\cancel{{\text{atm}}^{-\text{1}}}}\right)\left(0.125\cancel{\text{L}}\right)}=\text{72.0 g}{\text{mol}}^{-\text{1}}[/latex]

8. [latex]\mathcal{M}=\frac{mRT}{PV}D=\frac{m}{V}\mathcal{M}=\frac{DRT}{P}[/latex] [latex]\mathcal{M}=\frac{\text{3.93 g}{\text{L}}^{-\text{1}}\left(\text{0.08206 L}\cancel{\text{atm}}{\text{mol}}^{-\text{1}}\cancel{{\text{K}}^{-\text{1}}}\right)\left(273.15\cancel{\text{K}}\right)}{1.00\cancel{\text{atm}}}=\text{88.1 g}{\text{mol}}^{-\text{1}}[/latex] ℳ_phosphorous = 30.97376 g/mol
ℳ_fluorine = 18.998403 g/mol

[latex]\begin{array}{rrl}\text{molecular formula:}&\text{ phosphorous: }&30.97376\\&{}\text{flourine:}&\underline{3\left(18.998403\right)}\\{}&{}&87.968969\end{array}[/latex]

The molecular formula is PF₃.

To find this answer you can either use trial and error, or you can realize that since phosphorus is in group 5, it can fill its valence shell by forming three bonds. Fluorine, being in group 7, needs to form only one bond to fill its shell. Thus it makes sense to start with PF₃ as a probable formula.

10. Calculate the moles of each gas present and from that, calculate the pressure from the ideal gas law. Assume 25°C. The calibration gas contains:

[latex]\frac{350\cancel{\text{g}}{\text{CO}}_{2}}{44.0098\cancel{\text{g}}{\text{mol}}^{-\text{1}}{\text{CO}}_{2}}=\text{7.953 mol}{\text{ CO}}_{2}[/latex]
[latex]\frac{805\cancel{\text{g}}{\text{O}}_{2}}{31.9988\cancel{\text{g}}{\text{mol}}^{-\text{1}}{\text{O}}_{2}}=\text{25.157 mol}{\text{ O}}_{2}[/latex]
[latex]\frac{4880\cancel{\text{g}}{\text{N}}_{2}}{28.01348\cancel{\text{g}}{\text{mol}}^{-\text{1}}{\text{N}}_{2}}=\text{174.202 mol}{\text{ N}}_{2}[/latex]

Total moles = 7.953 + 25.157 + 174.202 = 207.312 mol

[latex]P=\frac{nRT}{V}=\frac{207.312\cancel{\text{mol}}\times 0.08206\cancel{\text{L}}\text{atm}\cancel{{\text{mol}}^{-\text{1}}}\cancel{{\text{K}}^{-\text{1}}}\times 298.15\cancel{\text{K}}}{36.0\cancel{\text{L}}}=\text{141 atm}[/latex]

12. Since these are percentages of the total pressure, the partial pressure can be calculated as follows:

CH⁴: 90% of 307.2 kPa = 0.900 × 307.2 = 276 kPa
C² H⁶: 8.9% of 307.2 kPa = 0.089 × 307.2 = 27 kPa
C³ H⁸: 1.1% of 307.2 kPa = 0.011 × 307.2 = 3.4 kPa

14. The oxygen increases the pressure within the tank to (34.5 atm – 33.2 atm =) 1.3 atm. The percentage O₂ on a mole basis is [latex]\frac{1.3}{34.5}\times 100\%=3.77\%.[/latex] The mixture is explosive. However, the percentage is given as a weight percent. Converting to a mass basis increases the percentage of oxygen even more, so the mixture is still explosive.

16. The vapor pressure of water at 18 °C is 15.5 torr. Subtract the vapor pressure of water from the total pressure to find the pressure of the carbon monoxide:

P_T = P_gas + P_water

Rearrangement gives: P_T – P_{water =}P_gas

756 torr – 15.5 torr = 740 torr

18. The answers are as follows:

Determine the moles of HgO that decompose; using the chemical equation, determine the moles of O₂ produced by decomposition of this amount of HgO; and determine the volume of O₂ from the moles of O₂, temperature, and pressure.
[latex]\begin{array}{l}\\ \\ 5.36\cancel{\text{g HgO}}\times \frac{\text{1 mol HgO}}{\left(200.59+15.9994\right)\cancel{\text{g HgO}}}=\text{0.0247 mol HgO}\\ 0.0247\cancel{\text{mol HgO}}\times \frac{\text{1 mol}{\text{O}}_{2}}{2\cancel{\text{mol HgO}}}=\text{0.01235 mol}{\text{O}}_{2}\end{array}[/latex]

PV = nRT
P = 0.975 atm
T = (23.0 + 273.15) K

[latex]V=\frac{nRT}{P}=\frac{0.01235\cancel{\text{mol}}\left(\text{0.08206 L}\cancel{\text{atm}}\cancel{{\text{mol}}^{-\text{1}}}\cancel{{\text{K}}^{-\text{1}}}\right)\left(296.15\cancel{\text{K}}\right)}{0.975\cancel{\text{atm}}}=0.308\text{ L}[/latex]

20. The answers are as follows:

Determine the molar mass of CCl₂F₂. From the balanced equation, calculate the moles of H₂ needed for the complete reaction. From the ideal gas law, convert moles of H₂ into volume.
Molar mass of CCl₂ F₂ = 12.011 + 2 × 18.9984 + 2 × 35.4527 = 120.913 g/mol
[latex]\text{mol}{\text{H}}_{2}=1.000\times {10}^{6}\text{g}\times \frac{\text{1 mol}{\text{CCL}}_{2}{\text{F}}_{2}}{\text{120.913 g}}\times \frac{\text{4 mol}{\text{H}}_{2}}{\text{1 mol}{\text{CCl}}_{2}{\text{F}}_{2}}=3.308\times {10}^{4}\text{mol}[/latex]
[latex]V=\frac{nRT}{P}=\frac{\left(3.308\times {10}^{4}\cancel{\text{mol}}\right)\left(\text{0.08206 L}\cancel{\text{atm}}\cancel{{\text{mol}}^{-\text{1}}}{\cancel{\text{K}}}^{-\text{1}}\right)\left(308.65\cancel{\text{K}}\right)}{225\cancel{\text{atm}}}=3.72\times {10}^{3}\text{L}[/latex]

22. The answers are as follows:

Balance the equation. Determine the grams of CO₂ produced and the number of moles. From the ideal gas law, determine the volume of gas.
[latex]{\text{CaCO}}_{3}\left(s\right)\rightarrow\text{CaO}\left(s\right)+{\text{CO}}_{2}\left(g\right)[/latex]
[latex]\text{mass}{\text{CO}}_{2}=1.00\times {10}^{6}\text{g}\times \frac{\text{1 mol}{\text{CaCO}}_{2}}{\text{100.087 g}}\times \frac{\text{44.01 g}{\text{CO}}_{2}}{\text{1 mol}{\text{CO}}_{2}}\times \frac{\text{1 mol}{\text{CO}}_{2}}{\text{1 mol}{\text{CaCO}}_{2}}=4.397\times {10}^{5}\text{g}[/latex]
[latex]\text{mol}{\text{CO}}_{2}=\frac{4.397\times {10}^{5}\text{g}}{\text{44.01 g}{\text{mol}}^{-\text{1}}}=\text{9991 mol}[/latex]
[latex]V=\frac{nRT}{P}=\frac{\left(\text{9991 mol}\right)\left(\text{0.08206 L atm}{\text{mol}}^{-\text{1}}{\text{K}}^{-\text{1}}\right)\left(\text{875 K}\right)}{\text{0.966 atm}}=7.43\times {10}^{5}\text{L}[/latex]

24. [latex]2{\text{C}}_{2}{\text{H}}_{6}\left(\text{g}\right)+7{\text{O}}_{2}\left(\text{g}\right)\rightarrow 4{\text{CO}}_{2}\left(\text{g}\right)+6{\text{H}}_{2}\text{O}\left(\text{g}\right)[/latex]

From the balanced equation, we see that 2 mol of C₂H₆ requires 7 mol of O₂ to burn completely. Gay-Lussac’s law states that gases react in simple proportions by volume. As the number of liters is proportional to the number of moles,

[latex]\frac{\text{12.00 L}}{\text{2 mol}{\text{C}}_{2}{\text{H}}_{6}}=\frac{V\left({\text{O}}_{2}\right)}{\text{7 mol}{\text{O}}_{2}}[/latex]
[latex]V\left({\text{O}}_{2}\right)=\frac{\text{12.00 L}\times 7}{2}=\text{42.00 L}[/latex]

26. The answers are as follows:

(a) The scheme to solve this problem is:

[latex]{\text{volume C}}_{2}{\text{H}}_{6}\left(g\right)\stackrel{\begin{array}{l}\text{ideal gas}\\ \text{equation}\end{array}}{\to }{\text{mol C}}_{2}{\text{H}}_{6}\left(g\right)\stackrel{\begin{array}{l}\text{reaction}\\ \text{stoichiometry}\end{array}}{\to }{\text{mol CO}}_{2}+{\text{H}}_{2}\text{O}\stackrel{\begin{array}{l}\text{ideal gas}\\ \text{equation}\end{array}}{\to }{\text{volume CO}}_{2}+{\text{H}}_{2}\text{O}[/latex]
[latex]\begin{array}{l}{\text{C}}_{2}{\text{H}}_{6}\left(g\right)+3\frac{1}{2}{\text{O}}_{2}\left(g\right)\rightarrow 2{\text{CO}}_{2}\left(g\right)+3{\text{H}}_{2}\text{O}\left(g\right)\\ \text{1}.n\left({\text{C}}_{2}{\text{H}}_{6}\right)=\frac{PV}{RT}=\frac{1.00\cancel{\text{atm}}\times 1.00\cancel{\text{L}}}{0.08206\cancel{\text{L}}\cancel{\text{atm}}{\text{mol}}^{-\text{1}}\cancel{{\text{K}}^{-\text{1}}}\left(273.15\cancel{\text{K}}\right)}=\text{0.0446 mol}\\ \text{2}.0.0446\cancel{\text{mol}{\text{C}}_{2}{\text{H}}_{6}}\times \frac{\text{5 mol products}}{1\cancel{\text{mol}{\text{C}}_{2}{\text{H}}_{6}}}=\text{0.223 mol products}\\ \text{3}.V=nRT=\frac{\left(0.223\cancel{\text{mol}}\right)\left(\text{0.08206 L}\cancel{\text{atm}}\cancel{{\text{mol}}^{-\text{1}}}\cancel{{\text{K}}^{-\text{1}}}\right)\left(873.15\cancel{\text{K}}\right)}{0.888\cancel{\text{atm}}}=\text{18.0 L}\end{array}[/latex]

(b) First, calculate the mol H₂O produced:

[latex]\text{0.0446 mol}{\text{C}}_{2}{\text{H}}_{6}\times \frac{\text{3 mol products}}{\text{1 mol}{\text{C}}_{2}{\text{H}}_{6}}=\text{0.1338 mol}[/latex]

Second, calculate the pressure of H₂O:

[latex]P=\frac{nRT}{V}=\frac{\left(0.1338\cancel{\text{mol}}\right)\left(0.8206\cancel{\text{L}}\text{atm}\cancel{{\text{mol}}^{-\text{1}}}\cancel{{\text{K}}^{-\text{1}}}\right)\left(873.15\cancel{\text{K}}\right)}{18.0\cancel{\text{L}}}=\text{0.533 atm}[/latex]

28. First, we must write a balanced equation to establish the stoichiometry of the reaction:

[latex]2{\text{BaO}}_{2}\rightarrow 2\text{BaO}+{\text{O}}_{2}[/latex]
We are given the mass of BaO₂ that decomposes, so the scheme for solving this problem will be:

Mass (BaO₂) = 137.33 + 2(15.9994) = 169.33 g/mol

[latex]n\left({\text{O}}_{2}\right)=\text{129.7 g}{\text{BaO}}_{2}\times \frac{\text{1 mol}{\text{BaO}}_{2}}{\text{169.33 g}{\text{BaO}}_{2}}\times \frac{\text{1 mol}{\text{O}}_{2}}{\text{2 mol}{\text{BaO}}_{2}}=\text{0.3830 mol}{\text{O}}_{2}[/latex]
[latex]V\left({\text{O}}_{2}\right)=\frac{nRT}{P}=\frac{\text{0.3830 mol}\left(\text{8.314 L kPa}{\text{mol}}^{-\text{1}}{\text{K}}^{-\text{1}}\right)\left(\text{423.0 K}\right)}{\text{127.4 kPa}}=\text{10.57 L}{\text{O}}_{2}[/latex]

30. At 90.1% conversion, a 1.000 × 10⁶ g final yield would require a [latex]\left(\frac{1.000\times {10}^{6}}{0.901}\right)=1.1099\times {10}^{6}\text{g}[/latex] theoretical yield.
3C₂H₄ produces 3C₂H₅OH, giving a 1:1 ratio:

[latex]\text{mol}\left({\text{C}}_{2}{\text{H}}_{4}\right)=1.1099\times {10}^{6}\cancel{\text{g}{\text{C}}_{2}{\text{H}}_{5}\text{OH}}\times \frac{1\cancel{\text{mol}{\text{C}}_{2}{\text{H}}_{2}\text{OH}}}{46.069\cancel{\text{g}{\text{C}}_{2}{\text{H}}_{5}\text{OH}}}\times \frac{\text{1 mol}{\text{C}}_{2}{\text{H}}_{4}}{1\cancel{\text{mol}{\text{C}}_{2}{\text{H}}_{2}\text{OH}}}=2.409\times {10}^{4}\text{mol}[/latex]

V (C₂ H₄) = 22.4 L/mol × 2.409 × 10⁴ mol = 5.40 × 10⁵ L

32. The reaction is:

[latex]{\text{XeF}}_{x}+\frac{x}{2}{\text{H}}_{2}\rightarrow\text{Xe}+x\text{HF}[/latex]

The pressure of H₂ that reacts is 48 torr – 24 torr = 24 torr

The number of moles of gas is proportional to the partial pressures. The reaction used 24 torr of XeF_x and 24 torr of H₂ so: [latex]\frac{x}{2}=1[/latex] and x = 2

The formula for the xenon compound is XeF₂.

Immediately after the H₂ is added (before the reaction):

[latex]\begin{array}{lll} \hfill {P}_{\text{Total}}& =& {P}_{{\text{XeF}}_{2}}+{P}_{{\text{H}}_{2}}\hfill \\ \hfill {P}_{{\text{H}}_{2}}& =& {P}_{\text{Total}}-{P}_{{\text{XeF}}_{2}}\hfill \\ & =& \text{72 torr}-\text{24 torr}\hfill \\ & =& 48\text{torr}\hfill \end{array}[/latex]

After the reaction:

[latex]{P}_{\text{Xe}}=\text{24 torr}\text{ }\left(\text{1 mol}{\text{XeF}}_{\text{x}}\rightarrow\text{1 mol Xe}\right)[/latex]

And the partial pressure of unreacted H₂ is:

[latex]\begin{array}{ll}\hfill {P}_{{\text{H}}_{2}}& ={P}_{\text{Total}}-{P}_{\text{Xe}}\hfill \\ & =\text{48 torr}-\text{24 torr}\hfill \\ & =\text{24 torr}\hfill \end{array}[/latex]

Glossary

Dalton’s law of partial pressures: total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases.

mole fraction: concentration unit defined as the ratio of the molar amount of a mixture component to the total number of moles of all mixture components

partial pressure: pressure exerted by an individual gas in a mixture

vapor pressure of water: pressure exerted by water vapor in equilibrium with liquid water in a closed container at a specific temperature

Module 9: Gases