Mole-to-Mole and Mass-to-Mole Calculations
A mole-to-mole calculation uses the mole ratio from a balanced equation to find the moles of one substance from the moles of another, and a mass-to-mole calculation adds a first step, converting the given mass to moles with the molar mass. In brief: to find the moles of water formed from 5.00 g of H2 in 2H2 + O2 → 2H2O, convert 5.00 g to 5.00 ÷ 2.016 = 2.48 mol of H2, then apply the 1 : 1 ratio to get 2.48 mol of H2O.
These are the first two types of stoichiometry problems, and they are the building blocks of all the others. This article presents the method for each type, a flow chart in a table, a set of worked examples with different reactions, a short section on how to check the answers, how significant figures work in a multi-step calculation and a short history of the calculations.
How do you solve a mole-to-mole problem?
- Write the balanced equation.
- Underline the given substance and the wanted substance.
- Write the mole ratio (wanted ÷ given) using the coefficients.
- Multiply the moles of the given substance by the ratio.
Example. 2Na + 2H2O → 2NaOH + H2. How many moles of hydrogen gas form from 0.800 mol of sodium? 0.800 mol Na × (1 mol H2 ÷ 2 mol Na) = 0.400 mol H2.
How do you solve a mass-to-mole problem?
- Write the balanced equation.
- Convert the mass of the given substance to moles: n = m ÷ M.
- Use the mole ratio to find the moles of the wanted substance.
Example. How many moles of CO2 are produced when 50.0 g of CaCO3 decomposes, CaCO3 → CaO + CO2? n(CaCO3) = 50.0 ÷ 100.09 = 0.4996 mol. The ratio is 1 : 1, so n(CO2) = 0.500 mol.
Types of conversions
| Type | Given | Wanted | Steps |
|---|---|---|---|
| Mole to mole | Moles of A | Moles of B | Mole ratio |
| Mass to mole | Mass of A | Moles of B | Molar mass of A, then mole ratio |
| Mole to mass | Moles of A | Mass of B | Mole ratio, then molar mass of B |
| Mass to mass | Mass of A | Mass of B | Molar mass of A, mole ratio, molar mass of B |
Worked examples
- Mole to mole. How many moles of oxygen are needed to react with 0.600 mol of aluminum, 4Al + 3O2 → 2Al2O3? 0.600 × (3 ÷ 4) = 0.450 mol O2.
- Mole to mole with a product. How many moles of Al2O3 form from the same 0.600 mol of Al? 0.600 × (2 ÷ 4) = 0.300 mol.
- Mass to mole. How many moles of water form from 5.00 g of hydrogen, 2H2 + O2 → 2H2O? n(H2) = 5.00 ÷ 2.016 = 2.480 mol. n(H2O) = 2.480 × (2 ÷ 2) = 2.48 mol.
- Mass to mole with a ratio. How many moles of water are produced when 10.0 g of oxygen reacts with excess hydrogen? n(O2) = 10.0 ÷ 31.998 = 0.3125 mol. n(H2O) = 0.3125 × (2 ÷ 1) = 0.625 mol.
- Moles of reactant needed. How many moles of HCl react with 2.00 g of magnesium, Mg + 2HCl → MgCl2 + H2? n(Mg) = 2.00 ÷ 24.305 = 0.08229 mol. n(HCl) = 0.08229 × 2 = 0.165 mol.
- Decomposition. How many moles of oxygen are produced by 10.0 g of KClO3 in 2KClO3 → 2KCl + 3O2? n(KClO3) = 10.0 ÷ 122.55 = 0.0816 mol. n(O2) = 0.0816 × (3 ÷ 2) = 0.122 mol.
- Combustion. How many moles of CO2 form when 11.0 g of propane burns, C3H8 + 5O2 → 3CO2 + 4H2O? M(C3H8) = 44.097 g/mol. n = 11.0 ÷ 44.097 = 0.2494 mol. n(CO2) = 0.2494 × 3 = 0.748 mol.
What about mole-to-mass problems?
In a mole-to-mass problem, the mole ratio comes first, and the molar mass of the wanted substance is applied last. For example, what mass of NaOH forms from 0.800 mol of Na in the reaction above? n(NaOH) = 0.800 × (2 ÷ 2) = 0.800 mol. m = 0.800 × 39.997 = 32.0 g. The next article covers the mass-to-mass type in detail.
How do you check an answer?
- Check the units: the given unit must cancel, and the answer must have the unit you want.
- Compare the ratio: if the coefficient of the wanted substance is larger than that of the given one, the number of moles of the wanted substance should be larger.
- Check that the amount is reasonable: for example, a reaction cannot form more moles of a product than the stoichiometry allows.
- For mass problems, the total mass of reactants should equal the total mass of the products.
What about significant figures?
Do not round intermediate values. Keep at least two extra digits and round the final answer to the number of significant figures of the least precise given quantity. The coefficients in the equation are exact numbers and do not limit the precision. In example 3, the given mass has three significant figures (5.00 g), so the final answer is 2.48 mol.
How were these methods used in early chemistry?
Before the mole concept, chemists compared “equivalent weights”, the masses of substances that react with a fixed mass of another. William Hyde Wollaston introduced a scale of equivalents in 1814, and Dalton and Berzelius used atomic weights to calculate quantities in reactions. With the development of reliable formulas and atomic masses after 1860, stoichiometric calculations became general, and the mole, introduced by Ostwald, made them simpler.
Which mistakes are common?
Mistake 1: Skipping the conversion to moles
You cannot apply a mole ratio to grams. Convert first.
Mistake 2: Using the molar mass of the wrong substance
Use the molar mass of the substance whose mass is given in the first step.
Mistake 3: Forgetting that the ratio uses coefficients
In Al2O3 there are two Al atoms per formula unit, but the mole ratio of Al to Al2O3 in the equation above is 4 : 2, from the coefficients.
Mistake 4: Rounding in the middle of the calculation
Round only at the end.