Stoichiometry is the point in gen chem where many students stop following along. The word itself sounds intimidating, and the problems pile up conversions until you lose track of which number goes where. The good news is that learning how to do stoichiometry mostly means learning one pattern, because almost every problem follows it. So if you’ve been staring at a balanced equation with no idea what to do next, start here. How do you do stoichiometry without getting lost in the numbers?
To do stoichiometry, you follow four steps: balance the equation, convert what you’re given into moles, use the mole ratio from the coefficients to switch to moles of the substance you want, and then convert those moles into the units the question asks for. The mole ratio is the bridge in the middle, and it only works in moles, never in grams. Limiting reactant and percent yield problems use the same four steps with one extra check added on. If you can do a grams-to-grams problem cleanly, you already have most of what you need. The rest of this article walks through the method and three fully worked examples.
What Is Stoichiometry, in Plain Words?
Stoichiometry is the math of how much stuff reacts and how much stuff you make. A balanced chemical equation is basically a recipe, and stoichiometry is how you scale that recipe up or down.
Think about baking cookies. If a recipe calls for 2 eggs per batch and you have 6 eggs, you can make 3 batches. You didn’t need to weigh the eggs. You counted them. Chemists do the same thing, except atoms and molecules are far too small to count one at a time, so we count them in groups called moles.
A mole is just a number, like a dozen. A dozen is 12 of anything. A mole is about 6.022 × 1023 of anything. (OpenStax lists the exact value of Avogadro’s number and the mole concept if you want the full story.) The reason chemists love the mole is that the mass of one mole of a substance, its molar mass, comes straight off the periodic table in grams.
Here’s the key idea: equations tell you how many particles react, not how many grams. That’s why every stoichiometry problem passes through moles.
How to Do Stoichiometry: The Four Steps
This is the whole method. Write it at the top of your scratch paper until you don’t need to anymore.
- Balance the equation. The coefficients are only trustworthy if the equation is balanced.
- Convert the given amount to moles. Grams divided by molar mass gives moles. Liters of solution times molarity also gives moles.
- Use the mole ratio. Multiply by (coefficient of what you want) over (coefficient of what you have).
- Convert moles to the requested unit. Usually that means multiplying by molar mass to get grams.
Each step is a conversion factor, so the whole problem can be written as one chain where units cancel. That chain is really dimensional analysis, the same unit-canceling skill you used for converting inches to centimeters. If your units cancel down to the unit the question asked for, your setup is almost certainly right.
What Is the Mole Ratio?
The mole ratio is the ratio of coefficients between two substances in a balanced equation. It tells you how many moles of one substance go with how many moles of another.
Take the combustion of propane:
C3H8 + 5 O2 → 3 CO2 + 4 H2O
From this one line you can write several mole ratios. For every 1 mole of propane, 5 moles of oxygen react. For every 1 mole of propane, 3 moles of carbon dioxide form. For every 5 moles of oxygen, 4 moles of water form. You pick the ratio that connects what you have to what you want, and you put the thing you want on top.
One warning that saves a lot of points: the coefficients are mole ratios, not mass ratios. You can’t say 1 gram of propane makes 3 grams of CO2. It is one of the most common stoichiometry mistakes in gen chem, and it usually comes from skipping step 2.
Worked Example 1: Grams to Grams
Question: How many grams of carbon dioxide form when 10.0 g of propane burns completely in oxygen?
Step 1, balance. The equation above is already balanced: C3H8 + 5 O2 → 3 CO2 + 4 H2O.
Step 2, grams to moles. The molar mass of C3H8 is 3(12.01) + 8(1.008) = 44.09 g/mol.
10.0 g C3H8 ÷ 44.09 g/mol = 0.2268 mol C3H8
Step 3, mole ratio. We want CO2 and we have propane, so the ratio is 3 mol CO2 / 1 mol C3H8.
0.2268 mol × (3 / 1) = 0.6804 mol CO2
Step 4, moles to grams. The molar mass of CO2 is 12.01 + 2(16.00) = 44.01 g/mol.
0.6804 mol × 44.01 g/mol = 29.9 g CO2
The answer has three significant figures because 10.0 g has three. Notice the answer is about three times the starting mass. That makes sense here only because propane and CO2 happen to have nearly the same molar mass, so the 3-to-1 mole ratio shows up almost unchanged in grams. Usually that won’t be true, which is exactly why you go through moles.
How Do You Find the Limiting Reactant?
You find the limiting reactant by figuring out which reactant would run out first. The fastest way is to convert each reactant to moles, then calculate how much product each one could make on its own. Whichever reactant makes less product is the limiting reactant, and that smaller amount is your answer.
Here’s my favorite analogy. In a lot of grocery stores, hot dogs come in packs of 10 and buns come in packs of 8. Say you buy 3 packs of hot dogs (30 dogs) and 3 packs of buns (24 buns). How many complete hot dogs can you serve? Only 24. The buns run out first, so the buns are the limiting reactant. The 6 leftover hot dogs are the excess reactant. They’re still sitting in the fridge, and they didn’t make any extra sandwiches.
Chemistry works the same way. Reactions don’t care which reactant you added more grams of. They care which one runs out of particles first.
Worked Example 2: Limiting Reactant
Question: 10.0 g of hydrogen gas reacts with 64.0 g of oxygen gas to form water. Which reactant is limiting, and what mass of water forms?
Step 1, balance. 2 H2 + O2 → 2 H2O
Step 2, convert both reactants to moles.
- H2: 10.0 g ÷ 2.016 g/mol = 4.960 mol
- O2: 64.0 g ÷ 32.00 g/mol = 2.000 mol
Step 3, mole ratio for each reactant. How much water could each one make by itself?
- From H2: 4.960 mol × (2 mol H2O / 2 mol H2) = 4.960 mol H2O
- From O2: 2.000 mol × (2 mol H2O / 1 mol O2) = 4.000 mol H2O
Oxygen makes less water, so O2 is the limiting reactant. Look at that for a second. There were more grams of oxygen than hydrogen, yet oxygen still runs out first. Grams can fool you. Moles can’t.
Step 4, moles to grams. The molar mass of H2O is 2(1.008) + 16.00 = 18.02 g/mol.
4.000 mol × 18.02 g/mol = 72.1 g H2O
Want to check the excess? The 2.000 mol of O2 uses 4.000 mol of H2, so 0.960 mol of H2 (about 1.94 g) is left over. Those are the extra hot dogs.
Worked Example 3: Percent Yield
Percent yield compares what you actually got in the lab to what the math says you should get. The math answer is called the theoretical yield. The lab answer is the actual yield.
Percent yield = (actual yield ÷ theoretical yield) × 100%
Question: Heating 25.0 g of calcium carbonate produces calcium oxide and carbon dioxide. A student collects 12.5 g of CaO. What is the percent yield?
Step 1, balance. CaCO3 → CaO + CO2 (already balanced, every coefficient is 1).
Step 2, grams to moles. The molar mass of CaCO3 is 40.08 + 12.01 + 3(16.00) = 100.09 g/mol.
25.0 g ÷ 100.09 g/mol = 0.2498 mol CaCO3
Step 3, mole ratio. 1 mol CaO per 1 mol CaCO3, so you get 0.2498 mol CaO.
Step 4, moles to grams. The molar mass of CaO is 40.08 + 16.00 = 56.08 g/mol.
0.2498 mol × 56.08 g/mol = 14.0 g CaO (theoretical yield)
Now the extra step: 12.5 g ÷ 14.01 g × 100% = 89.2% yield. (Use the unrounded theoretical yield here. Dividing by the rounded 14.0 g gives 89.3%, a small slip caused by rounding too early.)
A percent yield over 100% is a red flag. It usually means the product was still wet or contaminated, or there was a math error somewhere upstream.
Common Stoichiometry Mistakes (and Quick Fixes)
Most lost points in stoichiometry come from a short list of habits. Here are the ones worth watching for.
- Using an unbalanced equation. Always balance first. Every later step depends on the coefficients.
- Putting the mole ratio upside down. Write the units on every number. The substance you want goes on top so the old one cancels.
- Using the coefficient inside the molar mass. The molar mass of O2 is 32.00 g/mol whether the coefficient is 1 or 5. Coefficients only enter in step 3.
- Picking the limiting reactant by grams. Compare product amounts (or moles divided by coefficients), never raw masses.
- Rounding too early. Keep an extra digit or two through the chain and round once at the end.
If the equation itself is what trips you up, that’s a balancing problem, not a stoichiometry problem. It’s worth getting comfortable with balancing before you push hard on this skill.
How to Do Stoichiometry Every Time: Same Four Steps
Every stoichiometry problem runs on the same four steps: balance, convert to moles, use the mole ratio, convert to the answer’s units. Limiting reactant problems run step 2 and 3 for each reactant and keep the smaller result. Percent yield problems add one division at the end. Once you’ve done a dozen of these, you’ll notice you’re really doing the same problem with different numbers.
The best next step is to work problems from your textbook with the four steps written out every time, even the easy ones. Speed comes later. Stoichiometry shows up all over first-semester finals, and our Gen Chem 1 practice exam shows how it’s tested in the style of the ACS final, with a video solution for every question.
Start with one grams-to-grams problem tonight. Write all four steps, label every unit, and watch them cancel.