Chemistry: Atoms First 2e · Gases

Stoichiometry of Gaseous Substances, Mixtures, and Reactions

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In 30 seconds
Why this matters
The college version
Eli explains
Worked example
Key takeaway
Check yourself
Study tools
Sources & references

In 30 seconds

Stoichiometry is the same whether the substances are solids, solutions, or gases: start from a measured quantity, convert to moles, use the balanced equation's mole ratios, and convert to the desired quantity. What changes for gases is the measurement — you often measure a gas by its volume or pressure rather than its mass. Two tools make that possible:

The of an ideal gas: at (273.15 K, 1 atm), one mole occupies 22.4 L.
The ideal gas law, PV = nRT, which converts pressure–volume–temperature readings into moles at any conditions.

This topic also covers , (each gas in a mixture behaves independently), and the mole fraction, which links a gas's share of the total pressure to its share of the total moles. Together these let you calculate how much gas a reaction produces, how dense a gas is, and how mixtures of gases — like air — behave.

Why this matters

Predicting reaction products: industries that make gases (ammonia, hydrogen, CO₂) need to know exactly how much gas a given mass of reactant will produce — volume at operating conditions is the practical unit.
Airbags: the sodium azide decomposition that inflates an airbag is a classic gas-stoichiometry calculation: solid reactant → liters of nitrogen gas in milliseconds.
Breathing and diving: Dalton's law explains oxygen at altitude and decompression concerns in scuba diving; medical gas mixtures are described by partial pressures.
Environmental chemistry: air-pollution measurements report gas concentrations, and combustion products (CO₂, SO₂) are quantified as gas volumes or densities.
Exams: expect (a) molar-volume conversions at STP, (b) ideal-gas-law mole conversions at non-STP conditions, (c) gas density/molar-mass problems, and (d) partial-pressure calculations — all with dimensional analysis.

The college version

Core Concepts

The molar volume shortcut at STP

At standard temperature and pressure (273.15 K, 1 atm), the ideal gas law gives:

$V = nRT P = (1.00 mol)(0.08206 L atm mol -1 K -1)(273.15 K) 1.00 atm = 22.4 L$

So at STP, moles and liters convert with the factor 22.4 L/mol. This is a genuine conversion factor, valid only at STP; at any other conditions you must use the ideal gas law.

Gas density and molar mass

The density of a gas is mass per volume. Substituting n = m/M into the ideal gas law gives a direct formula:

$PV = nRT \Rightarrow P = m M RT V \Rightarrow d = m V = PM RT$

where d is density (g/L), M is molar mass (g/mol), P is pressure, and T is kelvin temperature. Rearranged, the same equation identifies an unknown gas from its density:

$M = dRT P$

Because gases have small densities (about 1–2 g/L at STP for common gases), these problems demand careful unit tracking: pressure in atm, temperature in kelvin, and R = 0.08206 L atm mol^-1K^-1.

Dalton's law of partial pressures

In a mixture of non-reacting gases, each gas exerts its own pressure — its partial pressure — exactly as if it alone occupied the container. The total pressure is the sum of the partial pressures:

$P total = P A + P B + P C + \dots$

The partial pressure of each gas equals its mole fraction times the total pressure:

$P A = X A P total, X A = n A n total$

This is why the atmosphere's total pressure is the sum of nitrogen's, oxygen's, and other gases' contributions, and why a gas collected "over water" must have the water-vapor pressure subtracted.

Gas volumes and reaction stoichiometry

Avogadro's law means that, at the same temperature and pressure, gas volumes are directly proportional to moles. So for gases at the same T and P, the volume ratios equal the mole ratios from the balanced equation:

$V A V B = n A n B$

This allows elegant "volume-to-volume" stoichiometry, plus the general mass → moles → volume pathway when one substance is a solid or solution.

Common Confusions

Do Not Confuse	With	Difference
22.4 L/mol at STP	22.4 L/mol at any conditions	The molar volume shortcut is valid only at 273.15 K and 1 atm; otherwise use PV = nRT
Density of a gas (g/L)	Density of a liquid (g/mL)	Gas densities are ~1–2 g/L at STP; liquid densities are ~1000× larger. Units always matter
Partial pressure	Total pressure	Partial pressure is one component's share; total is the sum of all partial pressures
Mole fraction X	Mole percent	Mole fraction is a fraction (0 to 1); multiply by 100 for percent. Never plug percent into P_i = X_i P_total without dividing by 100
Volume ratio in gas reactions	Mole ratio in solution reactions	For gases at equal T and P, volume ratios equal mole ratios; for solutions, use molarity × volume
Vapor pressure of water	Total collected-gas pressure	Gas collected over water is a mixture; subtract P_H₂O to get the dry gas pressure
Molar mass from density	Density from molar mass	Same equation, different unknown: M = dRT/P vs d = PM/RT

Eli explains

The same idea, in plain words

Explain it like I’m 10

Think of gas moles like identical balloons: one mole of any gas at STP fills one 22.4-liter balloon, no matter what gas it is. If a recipe (balanced equation) says 1 unit of baking soda makes 3 units of gas, you can count the balloons the gas fills. And when several gases share a container, each one pushes with its own share of the pressure — like kids on a seesaw, each contributing their own weight to the total.

Worked example

Example 1: Gas volume from a solid reactant at STP

The airbag reaction: sodium azide decomposes to sodium metal and nitrogen gas.

$2 NaN 3 (s) \to 2 Na (s) + 3 N 2 (g)$

What volume of N₂ gas (at STP) is produced by the decomposition of 65.0 g of NaN₃? (Molar mass of NaN₃ = 65.01 g/mol.)

Step 1 — Convert grams to moles:

$n NaN 3 = 65.0 g 65.01 g mol -1 = 1.00 mol$

Step 2 — Mole ratio from the balanced equation (3 mol N₂ : 2 mol NaN₃):

$n N 2 = 1.00 mol NaN 3 \times 3 mol N 2 2 mol NaN 3 = 1.50 mol N 2$

Step 3 — Convert moles to volume at STP using the molar volume:

$V N 2 = 1.50 mol \times 22.4 L 1 mol = 33.6 L$

Dimensional check: g → mol → mol N₂ → L. A 65 g airbag charge produces ~34 L of nitrogen, which is why the reaction can fill a bag so quickly.

Example 2: Gas volume at non-STP conditions

Using the ideal gas law, find the volume of the 1.50 mol of N₂ from Example 1 if the airbag inflates at 25.0 °C and 1.20 atm instead of STP.

Step 1 — Convert temperature to kelvin:

$T = 25.0 + 273.15 = 298.2 K$

Step 2 — Write the ideal gas law solved for V:

$V = nRT P$

Step 3 — Substitute:

$V = (1.50 mol)(0.08206 L atm mol -1 K -1)(298.2 K) 1.20 atm = 30.6 L$

The warmer, higher-pressure conditions give 30.6 L rather than 33.6 L — showing why you must use PV = nRT when conditions differ from STP.

Example 3: Density and molar mass of a gas

A sample of an unknown gas has a density of 1.96 g/L at STP. Identify the gas by calculating its molar mass.

Step 1 — Write the molar-mass formula:

$M = dRT P$

Step 2 — Substitute STP values (d = 1.96 g/L, T = 273.15 K, P = 1.00 atm):

$M = (1.96 g L -1)(0.08206 L atm mol -1 K -1)(273.15 K) 1.00 atm = 43.9 g mol -1$

A molar mass of ~44 g/mol identifies the gas as CO₂ (44.01 g/mol). Dimensional check: L and atm and K all cancel, leaving g/mol. Alternatively, at STP the molar volume shortcut gives M = d × 22.4 L/mol = 1.96 × 22.4 = 43.9 g/mol — the same answer faster.

Example 4: Partial pressures in a mixture

A 10.0 L container at 300 K holds 2.00 mol of N₂ and 1.00 mol of O₂. What is the total pressure, and what is the partial pressure of each gas?

Step 1 — Find total moles and mole fractions:

$n total = 2.00 + 1.00 = 3.00 mol, X N 2 = 2.00 3.00 = 0.667, X O 2 = 1.00 3.00 = 0.333$

Step 2 — Total pressure from the ideal gas law:

$P total = nRT V = (3.00 mol)(0.08206 L atm mol -1 K -1)(300 K) 10.0 L = 7.39 atm$

Step 3 — Partial pressures from mole fractions:

$P N 2 = X N 2 P total = 0.667 \times 7.39 atm = 4.93 atm$

$P O 2 = X O 2 P total = 0.333 \times 7.39 atm = 2.46 atm$

Check: 4.93 + 2.46 = 7.39 atm = P_total. ✓ Each gas contributes pressure in proportion to its mole fraction.

Key takeaways

At STP, 1 mol of any ideal gas = 22.4 L. Use this only at STP; otherwise use PV = nRT.
Gas density: d = PM/RT (units g/L); molar mass from density: M = dRT/P.
Dalton's law: P_total = ∑P_i; each partial pressure P_i = X_i P_total.
Mole fraction: X_i = n_i / n_total, always between 0 and 1; the sum of all mole fractions is 1.
Gases collected over water are saturated with water vapor: P_gas = P_total - P_H₂O (vapor pressure of water at the collection temperature).
Same T and P ⇒ gas volumes are proportional to moles, so balanced-equation volume ratios apply directly.
Always convert °C → K and choose R to match the pressure unit (0.08206 L atm mol⁻¹ K⁻¹ for atm).

Check yourself

6 review questions from the chapter. Try each one, then open the answer.

What volume does 2.50 mol of an ideal gas occupy at STP?
Show answer
2.50 mol × 22.4 L/mol = 56.0 L (at STP).
Write the equation for gas density and the rearranged equation for molar mass. What units must pressure and temperature be in?
Show answer
d = PM/RT and M = dRT/P; pressure in atm, temperature in kelvin, R = 0.08206 L atm mol^-1K^-1.
A gas collected over water at 25.0 °C has a total pressure of 750 torr. If the at 25.0 °C is 23.8 torr, what is the pressure of the dry gas?
Show answer
P_{dry gas} = 750 - 23.8 = 726 torr.
Air is about 21% oxygen by mole. If total pressure is 1.00 atm, what is the partial pressure of O₂?
Show answer
P_O₂ = X_O₂P_total = 0.21 × 1.00 atm = 0.21 atm.
In the reaction 2H₂(g) + O₂(g) → 2H₂O(g), what volume of O₂ reacts with 10.0 L of H₂ at the same temperature and pressure?
Show answer
Volume ratio = mole ratio: 1 mol O₂ : 2 mol H₂, so V_O₂ = 10.0/2 = 5.00 L.
A gas has density 1.25 g/L at STP. Estimate its molar mass, and suggest a likely identity.
Show answer
M = d × 22.4 = 1.25 × 22.4 = 28.0 g/mol — consistent with N₂ (28.02 g/mol).

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

Molar volume: Volume of one mole of gas: 22.4 L at STP
Gas density: Mass of gas per unit volume, d = PM/RT
Partial pressure: The pressure one gas in a mixture would exert alone
Dalton's law of partial pressures: Total pressure = sum of all partial pressures
Mole fraction X: n_i / n_total, the fraction of moles belonging to one gas
Vapor pressure of water: Pressure exerted by water vapor in equilibrium with liquid water
STP: 273.15 K and 1 atm
Volume ratio = mole ratio: At equal T and P, gas volumes follow the balanced equation's coefficients

Sources & references

openstax.org — Chemistry Atoms First 2e

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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