Goals and prerequisites
You will convert between mass and amount, use coefficients as mole ratios, find a limiting reactant, and compare actual and theoretical yield. These calculations assume a correctly balanced equation and compatible units.
Prerequisites: Complete Chemical reactions and review unit conversion in Matter and measurement. Use supplied atomic masses and molar masses as given.
Counting particles with the mole
A mole is a counting unit. One mole contains exactly 6.02214076 × 1023 specified entities. The Avogadro constant is NA = 6.02214076 × 1023 mol−1. Name the entity: atoms, molecules, ions, or formula units. One mole of NaCl is one mole of formula units, not a collection of NaCl molecules.
Worked example: moles to particles
How many molecules are in 0.250 mol CO2? Multiply 0.250 mol × (6.02214076 × 1023 molecules mol−1) = 1.51 × 1023 molecules, to three significant figures.
Molar mass connects mass and amount
Molar mass is the mass of one mole, in g mol−1. Add atomic masses from the formula: H2O is 2(1.008) + 16.00 = 18.016 g mol−1, usually reported as 18.02 g mol−1. Use the supplied values in a problem; do not impose a blanket rounding rule on every periodic-table value.
The number of grams and the number of moles are different quantities. The conversion factor has both units, so the units show which direction you are converting.
Worked example: mass to particles
How many molecules are in 9.00 g H2O? Keep guard digits: 9.00 g × (1 mol / 18.016 g) = 0.49956 mol. Then 0.49956 mol × 6.02214076 × 1023 molecules mol−1 = 3.008 × 1023, or 3.01 × 1023 molecules. Three significant figures are supported by 9.00 g.
Ratios and dimensional analysis
Coefficients in a balanced equation compare moles, never grams directly. For 2 H2 + O2 → 2 H2O, the ratio is 2 mol H2 : 1 mol O2 : 2 mol H2O. A reliable setup lets units cancel in a chain:
Worked example: mass to mass
What mass of water forms from 4.00 g H2 with excess O2?
- 4.00 g H2 × (1 mol / 2.016 g) = 1.98413 mol H2.
- 1.98413 mol H2 × (2 mol H2O / 2 mol H2) = 1.98413 mol H2O.
- 1.98413 mol H2O × (18.02 g / 1 mol) = 35.75 g H2O, which rounds to 35.8 g.
The equation's ratio is used only after converting to moles; guard digits are retained until the final result.
Limiting reactants
When both reactants have specified amounts, the limiting reactant runs out first and sets the maximum product. The other reactant is in excess. Compare the product amount predicted from each reactant, or compare each available amount to its coefficient ratio.
Worked example: identify the limiter
For N2 + 3 H2 → 2 NH3, suppose 2.00 mol N2 and 4.00 mol H2 are mixed. Nitrogen could form 4.00 mol NH3; hydrogen could form 4.00 × (2/3) = 2.67 mol NH3. Hydrogen predicts less product, so H2 limits and the theoretical amount is 2.67 mol NH3. Nitrogen remains in excess.
In real reactions, incomplete mixing, competing reactions, measurement uncertainty, and product loss can make actual yield lower than theoretical yield.
Yield and sensible rounding
Theoretical yield is the calculated maximum from the limiting reactant. Actual yield is measured. Percent yield = (actual yield ÷ theoretical yield) × 100%. A result over 100% usually signals wet or impure product, a measurement issue, or inconsistent assumptions. Keep guard digits through the calculation and round the final answer to the least precise input.
Worked example: percent yield
If the theoretical yield is 12.5 g and the collected mass is 10.8 g, percent yield = (10.8 g ÷ 12.5 g) × 100% = 86.4%, and the displayed inputs support three significant figures. The grams cancel, leaving a percentage.
Practice with explained answers
- Find the molar mass of CO2.
Answer and reasoning
12.01 + 2(16.00) = 44.01 g mol−1.
- How many moles are in 5.00 g NaCl (58.44 g mol−1)?
Answer and reasoning
5.00 g ÷ 58.44 g mol−1 = 0.0856 mol NaCl.
- For 2 KClO3 → 2 KCl + 3 O2, how many moles O2 form from 0.400 mol KClO3?
Answer and reasoning
0.400 × 3/2 = 0.600 mol O2. The coefficient ratio is dimensionless and moles remain.
- For 2 H2 + O2 → 2 H2O, which limits when 3.0 mol H2 and 2.0 mol O2 react?
Answer and reasoning
3.0 mol H2 needs 1.5 mol O2, so hydrogen is limiting. It makes 3.0 mol water; 0.5 mol O2 remains.
- An actual yield is 7.2 g and theoretical yield 8.0 g. Find percent yield.
Answer and reasoning
(7.2 ÷ 8.0) × 100% = 90%. Both values have two significant figures.
Common misconceptions
A coefficient is a mole ratio only after the equation is balanced. The limiting reactant is not automatically the reactant with the smaller mass or smaller number of moles. Do not multiply grams by a coefficient without converting through molar mass first.