Stoichiometry Step by Step (With Examples)

A balanced equation tells you ratios of moles, not grams. So you can never jump straight from grams of one substance to grams of another. The roadmap is:

1. Grams to moles: divide by the molar mass of A.
2. Moles to moles: multiply by the mole ratio from the balanced equation.
3. Moles to grams: multiply by the molar mass of B.

Molar masses used below (g/mol): H 1.008, C 12.011, O 15.999. So $\mathrm{CH_4} = 16.043$, $\mathrm{H_2} = 2.016$, $\mathrm{O_2} = 31.998$, and $\mathrm{H_2O} = 18.015$.

Problem: How many grams of water form when 16.0 g of methane burns completely?

1. Check the equation is balanced. C: 1 = 1. H: 4 = 4. O: 4 = 2 + 2. Good.
2. Grams to moles:

3. Mole ratio: the equation shows 2 mol $\mathrm{H_2O}$ for every 1 mol $\mathrm{CH_4}$.

4. Moles to grams:

Answer: about 35.9 g of water. Notice how the units cancel at each step. If they do not cancel, a fraction is upside down.

When you are given amounts of two reactants, one usually runs out first. That one is the limiting reactant, and it decides how much product you get. The other is in excess.

Problem: 4.00 g of $\mathrm{H_2}$ reacts with 16.0 g of $\mathrm{O_2}$. How much water can form?

1. Convert both reactants to moles.
   • $\mathrm{H_2}$: $4.00 \div 2.016 = 1.984$ mol
   • $\mathrm{O_2}$: $16.0 \div 31.998 = 0.5000$ mol
2. Find how much product each reactant could make.
   • From $\mathrm{H_2}$: $1.984 \text{ mol} \times \frac{2}{2} = 1.984$ mol $\mathrm{H_2O}$
   • From $\mathrm{O_2}$: $0.5000 \text{ mol} \times \frac{2}{1} = 1.000$ mol $\mathrm{H_2O}$
3. The smaller amount wins. $\mathrm{O_2}$ makes less water, so $\mathrm{O_2}$ is the limiting reactant.
4. Convert to grams: $1.000 \text{ mol} \times 18.015 \text{ g/mol} = 18.0$ g of water.

This 18.0 g is the theoretical yield. Notice that $\mathrm{H_2}$ had the smaller mass but is not limiting. You must compare moles, not grams. About 1.98 g of $\mathrm{H_2}$ is left over.

In a real lab you rarely get the full theoretical yield, because of spills, side reactions, or incomplete reactions. Percent yield compares what you got to what was possible:

Problem: In the reaction above, a student collects 14.4 g of water. What is the percent yield?

1. Theoretical yield (from the limiting reactant): 18.0 g.
2. Actual yield (what was measured): 14.4 g.
3. Divide and multiply by 100:

A percent yield over 100% usually means the product was still wet or contaminated.

• Always balance the equation first. A wrong ratio ruins everything after it.
• Write units on every number and cancel them as you go.
• Keep extra digits during the calculation and round only at the end.
• Round your final answer to the same number of significant figures as the least precise measurement given.
• For limiting reactant problems, calculate product from each reactant. It is the most reliable method.