Moles to grams:
one multiplication, one molar mass

The conversion pair

Chemical equations are written in moles, but a balance reads in grams. Between the two sits one number for each substance: its molar mass, the mass in grams of one mole. Convert with that number and an equation becomes a weighing instruction.

If a recipe says react 2 mol of water, the molar mass of 18.015 g/mol says you need 36.03 g. No other information is required beyond the correct chemical formula.

The units cancel neatly: mol × g/mol leaves g, and g ÷ g/mol leaves mol. If your answer's units do not come out right, the formula is upside down.

Worked example: moles of table salt to grams

Say a reaction needs 0.35 mol of sodium chloride. Its molar mass is 58.44 g/mol, from 22.990 for sodium plus 35.45 for chlorine.

Going the other direction, how many moles are in 250 g of calcium carbonate (CaCO₃, 100.087 g/mol)? Divide: 250 ÷ 100.087 = 2.498 mol, or about 2.50 mol.

A quick reference table

Notice that one mole of different substances weighs very different amounts, but each mole contains the same number of particles, about 6.022 × 10²³.

From the ideal to the real flask

The formula assumes a pure substance of the exact formula you used. Real reagents are rarely like that. A bottle labelled 98% pure holds 2% of something else, so to deliver 20.45 g of actual compound you would need 20.45 ÷ 0.98 = 20.87 g of reagent.

• Purity: divide the target mass by the purity fraction.
• Hydrates: include the water of crystallisation in the molar mass, or you will weigh too little compound.
• Moisture: hygroscopic solids such as sodium hydroxide absorb water from the air, so weigh them quickly.
• Significant figures: match the answer to the least precise input.

Using moles in a reaction

The conversion usually serves a bigger question: how much of one reagent do I need for a given amount of another? The balanced equation gives the mole ratio, and molar masses translate each side to grams.

Consider burning methane: CH₄ + 2 O₂ → CO₂ + 2 H₂O. Suppose you start with 16.0 g of methane. Its molar mass is 12.011 + 4 × 1.008 = 16.043 g/mol, so you have 16.0 ÷ 16.043 = 0.997 mol. The equation shows 1 mol of methane yields 1 mol of carbon dioxide, which weighs 44.009 g/mol. So the carbon dioxide produced is 0.997 × 44.009 = 43.9 g.

Written out as a chain, the route is grams of A, then moles of A, then moles of B through the coefficient ratio, then grams of B. Every stoichiometry problem is a version of this four-step chain, and the moles-to-grams step appears at the beginning and at the end.

Limiting reagent and percentage yield

When two reagents are mixed, the one that runs out first sets the amount of product. Convert both to moles, divide each by its coefficient, and the smaller value marks the limiting reagent. Convert the moles of product from that value back to grams to find the theoretical yield.

Real experiments rarely hit that figure because of incomplete reactions, side products and handling losses. The percentage yield compares what you got with the maximum possible: actual mass ÷ theoretical mass × 100. If the methane example gave 41.0 g of carbon dioxide, the yield would be 41.0 ÷ 43.9 = 93.4%.

• Always base the theoretical yield on the limiting reagent, never on the excess one.
• Keep an extra significant figure through the chain and round only at the end.
• A yield above 100% points to impurities, residual solvent or a weighing error rather than creating matter.

Moles in gases and solutions

Mass is not the only handle on a mole. For gases at standard conditions, an ideal gas occupies about 22.4 L per mole at 0 °C and 1 atm, or about 24.5 L at 25 °C. So 2 mol of oxygen would occupy roughly 49 L at room temperature, and weighs 2 × 31.998 = 64.0 g. The mass-based conversion is the same whichever route you use to count moles.

For solutions, moles come from concentration and volume, n = M × V. A 250 mL flask of 0.100 mol/L sodium chloride holds 0.250 × 0.100 = 0.0250 mol, and multiplying by 58.44 gives 1.461 g of salt. That is the mass you weigh out to make the solution: the molarity, volume and molar mass chain together in one line.

Whichever route you take, the molar mass stays the bridge between counting particles and weighing them. It is worth writing its units next to every number so the cancellations are visible.

• g ÷ (g/mol) = mol
• mol × (g/mol) = g
• mol ÷ (mol/L) = L

Units and slips

Watch for milligrams and kilograms: a molar mass in g/mol only returns grams, so convert afterwards if you need another unit. Also be sure the formula matches the species: oxygen gas is O₂ at 31.998 g/mol, while a single oxygen atom is 15.999.