Difficulty: Intermediate | Prerequisites: Moles, molar mass, balancing equations
This material sits at the intersection of two foundational chemistry skills: using the ideal gas law to find the amount of a gas under known conditions, and then using stoichiometric ratios to convert between substances in a balanced equation. If you can do these two things fluently, you can handle most quantitative gas-phase problems on the AP exam. You should already be comfortable with the concept of moles, molar mass, and reading coefficients from a balanced equation.
The ideal gas law (PV = nRT) lets you calculate moles of a gas when you know pressure, volume, and temperature. Once you have moles, stoichiometric ratios from the balanced equation let you convert to moles (or grams) of any other substance in the reaction.
Ideal gas law (PV = nRT)
The relationship between pressure (P), volume (V), moles (n), the gas constant (R), and temperature (T) for an ideal gas. Think of it as the master equation that ties together everything about a gas sample's physical state.
Gas constant (R)
R = 0.08206 L·atm/(mol·K). This value applies when pressure is in atm, volume in litres, and temperature in kelvin. In simple terms, it is the proportionality constant that makes the ideal gas law work in those units.
Stoichiometric ratio (mole ratio)
The ratio of coefficients in a balanced chemical equation, used to convert moles of one substance to moles of another. Think of it as the recipe: for every 1 mol of CS₂ you need 3 mol of Cl₂, just as a recipe might call for 3 eggs per 1 cup of flour.
Molar mass
The mass, in grams, of one mole of a substance. For CS₂, it is 76.13 g/mol. In simple terms, it is the conversion factor between grams and moles.
Always convert temperature to kelvin before substituting into PV = nRT. Add 273 to the Celsius value (e.g. 120°C becomes 393 K).
Rearrange to solve for the unknown. To find moles: n = PV / RT.
Substitute with consistent units. If R uses L·atm/(mol·K), then P must be in atm and V in litres.
The reaction is:
CS₂(g) + 3 Cl₂(g) → CCl₄(g) + S₂Cl₂(g)
Part (a)(i): Finding moles of Cl₂
Given: P = 0.40 atm, V = 25.0 L, T = 120°C = 393 K
n = PV / RT = (0.40 × 25.0) / (0.08206 × 393) = 10.0 / 32.25 ≈ 0.31 mol Cl₂
Part (a)(ii): Grams of CS₂ needed
From the balanced equation, 1 mol CS₂ reacts with 3 mol Cl₂.
Moles of CS₂ = 0.31 mol Cl₂ × (1 mol CS₂ / 3 mol Cl₂) ≈ 0.103 mol CS₂
Grams of CS₂ = 0.103 mol × 76.13 g/mol ≈ 7.9 g CS₂
The pressure (0.40 atm) has 2 significant figures, so the final answer should reflect 2 significant figures.
On the AP exam, losing a mark for sig figs is common and avoidable.
PV = nRT, rearranged: n = PV / RT
Molar mass of CS₂ = 12.01 + 2(32.06) = 76.13 g/mol
Stoichiometric conversion:
mol Cl₂ × (1 mol CS₂ / 3 mol Cl₂) × (76.13 g CS₂ / 1 mol CS₂) = grams CS₂
The ideal gas law is the basis for calculating how much gas is in a pressurised container, which matters in everything from scuba tank fills to industrial chemical reactors. Stoichiometry tells chemical engineers exactly how much of each reagent to load so nothing is wasted.
Students often forget to convert Celsius to kelvin. Using 120 instead of 393 will give an answer roughly three times too large.
Some students invert the mole ratio, multiplying by 3 instead of dividing by 3. Always check: the coefficient in front of Cl₂ is larger, so fewer moles of CS₂ are needed, not more.
Rounding too early in multi-step calculations introduces error. Carry at least one extra significant figure through intermediate steps.
Using the wrong value of R (e.g. 8.314 J/(mol·K)) when pressure is in atm will produce nonsensical results.
⚠️ The AP exam will not remind you to convert to kelvin. If the temperature is given in Celsius, the conversion is on you.
⚠️ You must show the setup of PV = nRT with values substituted to earn full marks, not just the final number.
⚠️ Sig fig errors are penalised. Match the least precise measurement in the problem.
True or false: You can use T = 120 directly in PV = nRT if pressure is in atm.
Fill in the blank: R = ________ L·atm/(mol·K).
True or false: If the balanced equation shows 3 mol Cl₂ per 1 mol CS₂, then you need more moles of CS₂ than Cl₂.
Fill in the blank: To convert moles to grams, multiply by the substance's ________.
Answers: 1. False (must convert to 393 K). 2. 0.08206. 3. False (you need fewer moles of CS₂). 4. Molar mass.
Q: A rigid 10.0 L container holds N₂ gas at 2.00 atm and 27°C. How many moles of N₂ are present?
A: T = 27 + 273 = 300 K. n = PV/RT = (2.00 × 10.0) / (0.08206 × 300) = 20.0 / 24.62 = 0.812 mol.
Q: Given the reaction 2 H₂(g) + O₂(g) → 2 H₂O(g), if you have 0.50 mol O₂, how many grams of H₂ are needed?
A: 0.50 mol O₂ × (2 mol H₂ / 1 mol O₂) = 1.0 mol H₂. 1.0 mol × 2.016 g/mol = 2.0 g H₂.
Q: Why is it important that the container in the original problem is described as "rigid"?
A: A rigid container means the volume is constant. This matters because if the container could expand, the volume would change as gas is produced, complicating the calculation.
This connects to gas stoichiometry problems involving limiting reagents, where you must determine which reactant runs out first. It also links to thermochemistry: once you know moles reacting, you can calculate enthalpy changes using ΔH values. The ideal gas law reappears in equilibrium and kinetics when working with partial pressures.
PV = nRT, ideal gas law, gas constant R, stoichiometry, mole ratio, molar mass, significant figures, AP Chemistry free response, gas-phase reactions, CS₂, Cl₂, CCl₄, unit conversion, kelvin, moles of gas, dimensional analysis