The mole
A mole is 6.022 × 10^23 particles. The molar mass of a substance (grams per mole) is the sum of the atomic masses in its formula, and it is the bridge between a mass you weigh and an amount you count. For calcium chloride, CaCl2: 40.08 + 2(35.45) = 110.98 g/mol. Always multiply a subscript by everything inside its bracket.
Conversions you need constantly: moles = mass / molar mass, and for a solution, molarity M = moles of solute / litres of solution. When you dilute, the moles of solute stay the same, so M1 V1 = M2 V2.
Balanced equations as recipes
Coefficients in a balanced equation give mole ratios. In 2 H2 + O2 → 2 H2O, two moles of hydrogen react with one mole of oxygen to make two moles of water. Balance by adjusting coefficients (never subscripts), leaving the most complicated formula for first and elements that appear alone for last.
The standard route
- Convert the given mass to moles.
- Use the mole ratio from the balanced equation.
- Convert moles of the wanted substance to grams or litres.
Limiting reactant and yield
When reactants are not in the exact ratio, the one that runs out first is the limiting reactant; it controls the theoretical yield. Test each reactant by asking how much product it could make, and the smaller answer wins. Percent yield = actual yield / theoretical yield × 100.
An empirical formula is the simplest whole-number ratio of atoms: convert percent by mass to moles, divide by the smallest, and scale to whole numbers if needed. Mass percent of an element = (atoms × atomic mass) / molar mass × 100.
Worked example
Iron(III) oxide reacts with carbon monoxide: Fe2O3 + 3 CO → 2 Fe + 3 CO2. A mixture holds 40.0 g Fe2O3 and 30.0 g CO. How much iron can form? Use Fe 55.85, O 16.00, C 12.01.
- Molar mass Fe2O3 = 2(55.85) + 3(16.00) = 159.70 g/mol, so moles = 40.0 / 159.70 = 0.2505 mol. Moles CO = 30.0 / 28.01 = 1.071 mol.
- CO needed for all the oxide: 3 × 0.2505 = 0.7514 mol. We have 1.071 mol, which is more than enough, so Fe2O3 is limiting and CO is in excess.
- Iron formed: 0.2505 mol Fe2O3 × (2 Fe / 1 Fe2O3) = 0.5009 mol Fe.
- Mass: 0.5009 × 55.85 = 27.98 g, reported as 28.0 g to three sig figs.
If the lab collects 24.5 g, the percent yield is 24.5 / 27.98 × 100 = 87.6 percent. A frequent error is to compare grams of reactants directly; always compare moles after applying the mole ratio.