How to Calculate Partial Pressures

Why partial pressures actually matter

Every breath you take is a Dalton’s Law problem. Atmospheric air at sea level sits at 760 mmHg total, but your alveoli only care about the ~160 mmHg of oxygen — the partial pressure, not the total. Drop the partial pressure of O₂ below ~60 mmHg and you pass out, regardless of how much “air” is technically present. That’s why pilots use supplemental oxygen above 12,500 ft even though the cabin is full of gas.

In the lab, partial pressures show up the moment you collect a gas over water, run a reaction in a sealed flask, or try to predict equilibrium in a gas-phase system. The math is simple — each gas in a non-reacting mixture acts like it owns the container — but the bookkeeping around mole fractions and water-vapor corrections is where lab reports lose points.

The one equation you need

For any non-reacting mixture of ideal gases:

P_i = X_i × P_total

where X_i is the mole fraction of gas i (its moles divided by total moles). The partial pressures add up to the total:

P_total = P₁ + P₂ + P₃ + …

That’s the entire law. Every problem reduces to finding mole fractions and multiplying.

Walk through a clean problem first

A 3.00 L flask at constant temperature contains 2.00 mol N₂, 0.50 mol O₂, and 0.50 mol CO₂ at 3.00 atm total. Find each partial pressure.

Sum your moles: 2.00 + 0.50 + 0.50 = 3.00 mol total.

Mole fractions:

• X(N₂) = 2.00 / 3.00 = 0.667
• X(O₂) = 0.50 / 3.00 = 0.167
• X(CO₂) = 0.50 / 3.00 = 0.167

(They sum to 1.001 — that’s rounding, not an error.)

Multiply by 3.00 atm:

• P(N₂) = 2.00 atm
• P(O₂) = 0.500 atm
• P(CO₂) = 0.500 atm

Sanity check: 2.00 + 0.500 + 0.500 = 3.00 atm. Done.

Gas collected over water — the lab trap

When you collect H₂ from a Zn/HCl reaction by displacing water in an inverted graduated cylinder, the gas you trap is not pure hydrogen. The headspace is saturated with water vapor at the bath temperature, and that vapor contributes its own partial pressure. Forget to subtract it and your moles of H₂ come out too high.

The fix: look up the vapor pressure of water at your temperature and subtract.

You collect O₂ over water at 25 °C. The leveled cylinder reads a total pressure of 758 mmHg. Vapor pressure of water at 25 °C is 23.8 mmHg.

P(O₂) = 758 − 23.8 = 734.2 mmHg

Use 734.2 mmHg, not 758, in any subsequent PV = nRT calculation. This is the single most-missed step in every general chemistry gas lab.

Going the other direction with PV = nRT

If you know moles, volume, and temperature, you can compute each partial pressure directly without ever finding total pressure first:

P_i = n_i RT / V

A 10.0 L vessel at 300 K holds 0.40 mol He and 0.60 mol Ar.

• P(He) = (0.40)(0.08206)(300) / 10.0 = 0.985 atm
• P(Ar) = (0.60)(0.08206)(300) / 10.0 = 1.477 atm
• P_total = 2.46 atm

Same answer either way — pick whichever route has the data you already have.

Working from masses instead of moles

If a problem hands you grams (the usual textbook curveball), convert to moles before doing anything else. Mass fraction is not mole fraction unless every gas happens to share a molar mass.

A 5.0 L vessel at 25 °C holds 4.0 g He (M = 4.00) and 32.0 g O₂ (M = 32.00).

Moles: He = 1.00 mol, O₂ = 1.00 mol. Total = 2.00 mol.

By moles they’re equal — so X(He) = X(O₂) = 0.500 — even though by mass O₂ is 8× heavier. If you’d used mass fractions you’d have called He’s contribution 11% instead of 50%.

P_total = nRT/V = (2.00)(0.08206)(298) / 5.0 = 9.78 atm P(He) = P(O₂) = 4.89 atm each

Traps to watch for

Mass fractions ≠ mole fractions. Always convert to moles first. This is the single most common error on Dalton’s Law problems.

Water vapor never disappears on its own. Any gas collected over water needs the vapor-pressure subtraction. Look up the value for your bath temperature — it’s not 23.8 mmHg unless you’re at exactly 25 °C.

Volume isn’t shared the way you might think. Each gas occupies the full container volume, not some fraction proportional to its amount. That’s the whole point of “partial” pressure — partial in pressure contribution, not partial in space.

Reacting mixtures break Dalton’s Law. If your gases react (NO + ½O₂ → NO₂), mole counts change as the reaction proceeds and you need an ICE table, not a static partial-pressure calculation.

Try it

Run these through the Dalton’s Law Calculator to check your work:

1. Air is 78% N₂ and 21% O₂ by mole. At 1.00 atm, find the partial pressure of each.
2. A flask holds 1.0 mol H₂, 2.0 mol He, and 3.0 mol Ne at 600 kPa total. Find each P_i.
3. H₂ collected over water at 30 °C; total pressure 745 mmHg; P(H₂O) = 31.8 mmHg. Find P(H₂).
4. 4.0 g He plus 32.0 g O₂ in a 5.0 L vessel at 25 °C — total and partial pressures.
5. In a mixture P(N₂) = 0.80 atm, P(CO₂) = 0.30 atm. Find both mole fractions.