Mass-to-Mass Stoichiometry Calculations

A mass-to-mass stoichiometry calculation finds the mass of one substance in a reaction from the mass of another using three steps: convert the given mass to moles, use the mole ratio from the balanced equation, and convert the moles of the wanted substance to mass. In brief: in 2Mg + O2 → 2MgO, 10.0 g of magnesium gives 10.0 ÷ 24.305 = 0.411 mol of Mg, which forms 0.411 mol of MgO (the ratio is 1 : 1), and 0.411 × 40.304 = 16.6 g of MgO.

Mass-to-mass problems are the most typical in laboratories and industry, because masses are what balances measure. The method is called the “gram-mole-mole-gram” route. This article presents the method, a template for setting out the work, six worked problems that include reactions in industry and laboratory, a table that compares reactant and product masses, a check using the conservation of mass, how to put all steps in a single line of conversion factors and some history.

What is the three-step method?

  1. Grams of A → moles of A: divide by the molar mass of A.
  2. Moles of A → moles of B: multiply by the mole ratio (coefficient of B ÷ coefficient of A).
  3. Moles of B → grams of B: multiply by the molar mass of B.

The molar masses are used twice, one for each substance, and the mole ratio is the link between them. The mole ratio is the only place where the equation is used.

Flow diagram from grams of A to moles of A to moles of B to grams of B Grams of A 10.0 g Mg Moles of A 0.411 mol Mg Moles of B 0.411 mol MgO Grams of B 16.6 g MgO ÷ M(A) × ratio × M(B) 2Mg + O2 → 2MgO Mole ratio from the coefficients: 2 mol MgO : 2 mol Mg Example with 10.0 g of Mg shown in the lower row
The three-step path of every mass-to-mass calculation: divide by the molar mass of A, multiply by the mole ratio from the balanced equation, then multiply by the molar mass of B. The lower row follows 10.0 g of Mg in 2Mg + O2 → 2MgO: 0.411 mol Mg gives 0.411 mol MgO, which is 16.6 g.

How do you set out the work?

A clear layout reduces mistakes: (1) write the balanced equation, (2) write the given and wanted quantities below the formulas, (3) calculate the molar masses, (4) do each of the three conversions and label the units, (5) check the result. Using this layout, the final answer should be reported with the right number of significant figures.

Worked examples

  1. Burning magnesium. 2Mg + O2 → 2MgO. What mass of MgO forms from 10.0 g of Mg? n(Mg) = 10.0 ÷ 24.305 = 0.4114 mol. n(MgO) = 0.4114 mol. m(MgO) = 0.4114 × 40.304 = 16.6 g. The oxygen that combined has a mass of 16.58 − 10.0 = 6.58 g.
  2. Limestone and acid. CaCO3 + 2HCl → CaCl2 + H2O + CO2. For 25.0 g of CaCO3: n = 25.0 ÷ 100.09 = 0.2498 mol. The mass of HCl needed is 0.2498 × 2 × 36.461 = 18.2 g. The mass of CaCl2 formed is 0.2498 × 110.98 = 27.7 g, and of CO2 is 0.2498 × 44.009 = 11.0 g.
  3. Check by conservation of mass. Reactants: 25.0 g CaCO3 + 18.2 g HCl = 43.2 g. Products: 27.7 g CaCl2 + 11.0 g CO2 + 4.5 g H2O (0.2498 × 18.015) = 43.2 g. ✓
    25.0 18.2 = 43.2 g 27.7 11.0 4.5 = 43.2 g
  4. Iron production. Fe2O3 + 3CO → 2Fe + 3CO2. How much iron can be made from 1,000 kg of pure iron(III) oxide? n(Fe2O3) = 1,000,000 g ÷ 159.69 g/mol = 6,262 mol (6.262 kmol). n(Fe) = 6,262 × 2 = 12,525 mol. m(Fe) = 12,525 × 55.845 = 699,400 g = 699 kg.
  5. Aspirin. C7H6O3 (salicylic acid) + C4H6O3 (acetic anhydride) → C9H8O4 (aspirin) + C2H4O2 (acetic acid). What mass of aspirin forms from 2.00 g of salicylic acid (138.12 g/mol) with excess anhydride? n = 2.00 ÷ 138.12 = 0.01448 mol. The ratio is 1 : 1, so n(aspirin) = 0.01448 mol, and m = 0.01448 × 180.16 = 2.61 g.
  6. Oxygen needed for combustion. What mass of oxygen is needed to burn 100.0 g of methane, CH4 + 2O2 → CO2 + 2H2O? n(CH4) = 100.0 ÷ 16.043 = 6.233 mol. n(O2) = 12.47 mol. m(O2) = 12.47 × 31.998 = 399 g. The carbon dioxide formed is 6.233 × 44.009 = 274 g.

Masses in a balanced reaction

Masses for the reaction CaCO3 + 2HCl → CaCl2 + H2O + CO2 (from 25.0 g of CaCO3)
Substance Coefficient Moles Molar mass (g/mol) Mass (g)
CaCO310.2498100.0925.0
HCl20.499636.46118.2
CaCl210.2498110.9827.7
H2O10.249818.0154.50
CO210.249844.00911.0

How do you write the calculation in a single line?

The three steps can be written as a chain of conversion factors in which the units cancel:

10.0 g Mg × (1 mol Mg ÷ 24.305 g Mg) × (2 mol MgO ÷ 2 mol Mg) × (40.304 g MgO ÷ 1 mol MgO) = 16.6 g MgO.

The grams of Mg cancel, then the moles of Mg, and then the moles of MgO, leaving grams of MgO.

10.0 g Mg /24.305 0.4114 mol Mg 2 : 2 0.4114 mol MgO ×40.304 16.6 g MgO

Where are mass-to-mass calculations used?

How did these calculations develop?

Lavoisier’s insistence on weighing all substances in the 1780s made mass calculations possible. Jeremias Richter in the 1790s calculated the masses of acids and bases that neutralize each other and published tables of equivalents. After Dalton and Berzelius determined atomic weights, chemists such as Justus von Liebig used stoichiometric calculations routinely in the analysis of compounds. Today, calculations of this kind are done in software for the design of chemical plants.

Which mistakes are common?

Mistake 1: Using the molar mass of the given substance in the last step

Use the molar mass of the wanted substance when converting moles to grams.

Mistake 2: Skipping the mole ratio

The coefficients in the equation tell how the moles of A and B are related. If the ratio is not 1 : 1, the number of moles changes.

Mistake 3: Using grams in the mole ratio

The ratio of grams is not the ratio of coefficients. The ratio applies only to moles.

Mistake 4: Not checking the mass balance

Add the masses of all reactants and products. They should be equal, which is a good check on your arithmetic.