4.3 Stoichiometry of gaseous substances, mixtures, and reactions

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By the end of this section, you will be able to:
• 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.” “Quotations by Joseph-Louis Lagrange,” last modified February 2006, accessed February 10, 2015, http://www-history.mcs.st-andrews.ac.uk/Quotations/Lagrange.html

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, $\rho =\phantom{\rule{0.2em}{0ex}}\frac{m}{V}.$ 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 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 [link] .

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

1. PV = nRT
2. Rearrange to get (mol/L) : $\frac{n}{v}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{P}{RT}$
3. Multiply each side of the equation by the molar mass, ℳ. When moles are multiplied by ℳ in g/mol, g are obtained:
$\left(\text{ℳ}\right)\left(\frac{n}{V}\right)=\left(\frac{P}{RT}\right)\left(\text{ℳ}\right)$
4. $g\text{/L}=\rho =\phantom{\rule{0.2em}{0ex}}\frac{P\text{ℳ}}{RT}$

A gas was found to have a density of 0.0847 g/L at 17.0 °C and a pressure of 760 torr. What is its molar mass? What is the gas?

$\rho =\phantom{\rule{0.2em}{0ex}}\frac{P\text{ℳ}}{RT}$
$0.0847\phantom{\rule{0.2em}{0ex}}\text{g/L}=760\phantom{\rule{0.2em}{0ex}}\overline{)\text{torr}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{1\phantom{\rule{0.2em}{0ex}}\overline{)\text{atm}}}{760\phantom{\rule{0.2em}{0ex}}\overline{)\text{torr}}}\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\frac{\text{ℳ}}{\text{0.0821 L}\phantom{\rule{0.2em}{0ex}}\overline{)\text{atm}}\text{/mol K}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{290 K}$
ℳ = 2.02 g/mol; therefore, the gas must be hydrogen (H 2 , 2.02 g/mol)

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 liter changes with temperature or pressure. Gas densities are often reported at STP.

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