Stoichiometry: A Step-by-Step Guide with Examples

Stoichiometry is the part of chemistry that measures the amounts of reactants and products in a chemical reaction. It uses a balanced equation to predict how much of each substance you need or make. Every calculation follows the same path, and this guide walks through it one step at a time with fully worked examples so you can check each number yourself.

What is stoichiometry?

Stoichiometry is the study of the quantitative relationships in a chemical reaction. It answers questions like how much product forms and how much reactant you must supply. The coefficients in a balanced chemical equation are the key. They tell you the exact ratio in which substances react and form.

Think of a balanced equation as a recipe. If a recipe makes two cakes from three cups of flour, you can scale it up or down. Chemistry works the same way. In the reaction N₂ + 3H₂ → 2NH₃, one molecule of nitrogen reacts with three molecules of hydrogen. Those numbers hold true for moles as well. So one mole of N₂ reacts with three moles of H₂ to make two moles of NH₃.

The mole is the unit that makes this work. A mole is a fixed count of particles, about 6.022 × 10²³. Because coefficients describe ratios of particles, they also describe ratios of moles. This link between coefficients and moles is the heart of every stoichiometry problem.

Why does the mole ratio matter?

The mole ratio is the bridge between one substance and another in a reaction. It comes straight from the coefficients in the balanced equation. Without it, you cannot connect the amount of a reactant to the amount of a product. Every conversion between two chemicals passes through this ratio.

Here is a simple mole-to-mole example. Use the ammonia synthesis equation again:

N₂ + 3H₂ → 2NH₃

Suppose you have 4.0 moles of H₂ and want to know how many moles of NH₃ can form. The mole ratio of NH₃ to H₂ is 2 to 3. Multiply the moles of hydrogen by that ratio:

• Moles NH₃ = 4.0 mol H₂ × (2 mol NH₃ / 3 mol H₂)
• Moles NH₃ = 2.67 mol NH₃

The ratio always goes with the wanted substance on top and the known substance on the bottom. That way the known unit cancels and the wanted unit remains. Write the ratio carefully and the rest of the math follows.

The step-by-step method

Nearly every stoichiometry problem follows the same three-part path. First convert what you know into moles. Then use the mole ratio to switch substances. Finally convert moles into the unit the question asks for. Keeping this order in mind prevents most mistakes.

Here is the standard workflow written as a checklist:

• Step 1. Balance the equation. Make the atom counts equal on both sides. You cannot get correct ratios from an unbalanced equation.
• Step 2. Convert to moles. If you are given grams, divide by the molar mass to get moles.
• Step 3. Apply the mole ratio. Multiply by the ratio from the coefficients to switch from the known substance to the wanted substance.
• Step 4. Convert to the final unit. If the answer must be in grams, multiply the moles by the molar mass.

This grams to moles to moles to grams pattern is called a mass-to-mass calculation. It is the most common type on exams. The next section works one all the way through.

How do you solve a mass-to-mass problem?

A mass-to-mass problem gives you the mass of one substance and asks for the mass of another. You solve it by moving through moles in the middle. Grams cannot convert to grams directly. You must pass through the mole ratio, so both ends of the problem use molar mass.

Work through the combustion of methane, the main gas in natural gas. The balanced equation is:

CH₄ + 2O₂ → CO₂ + 2H₂O

Question: How many grams of carbon dioxide form when 25.0 g of methane burns completely?

First list the molar masses you need:

Step 1. Convert grams of CH₄ to moles.

• Moles CH₄ = 25.0 g / 16.04 g/mol = 1.559 mol

Step 2. Use the mole ratio. The equation shows 1 CH₄ makes 1 CO₂, a 1 to 1 ratio.

• Moles CO₂ = 1.559 mol CH₄ × (1 mol CO₂ / 1 mol CH₄) = 1.559 mol

Step 3. Convert moles of CO₂ to grams.

• Mass CO₂ = 1.559 mol × 44.01 g/mol = 68.6 g

So 25.0 g of methane produces about 68.6 g of carbon dioxide. Notice the answer is larger than the starting mass. That makes sense because each carbon atom picks up two heavier oxygen atoms during combustion.

Finding the limiting reactant

The limiting reactant is the substance that runs out first. It sets the maximum amount of product a reaction can make. The other reactant is left over and is called the excess reactant. To find the limiting one, compare how much product each reactant could make on its own. The smaller amount wins.

Use ammonia synthesis with real masses:

N₂ + 3H₂ → 2NH₃

Question: You react 14.0 g of N₂ with 6.0 g of H₂. Which reactant limits the reaction, and how many grams of NH₃ form?

Molar masses: N₂ is 28.02 g/mol, H₂ is 2.016 g/mol, and NH₃ is 17.03 g/mol.

Step 1. Convert each reactant to moles.

• Moles N₂ = 14.0 g / 28.02 g/mol = 0.500 mol
• Moles H₂ = 6.0 g / 2.016 g/mol = 2.98 mol

Step 2. Find how much H₂ the nitrogen needs. The ratio of H₂ to N₂ is 3 to 1.

• H₂ required = 0.500 mol N₂ × (3 mol H₂ / 1 mol N₂) = 1.50 mol

You have 2.98 mol of H₂ but only need 1.50 mol. Hydrogen is in excess, so nitrogen is the limiting reactant.

Step 3. Calculate the NH₃ from the limiting reactant. The ratio of NH₃ to N₂ is 2 to 1.

• Moles NH₃ = 0.500 mol N₂ × (2 mol NH₃ / 1 mol N₂) = 1.00 mol
• Mass NH₃ = 1.00 mol × 17.03 g/mol = 17.03 g

The reaction makes about 17.0 g of ammonia. Always base the product amount on the limiting reactant, never the excess one.

How do you calculate percent yield?

Percent yield compares the amount of product you actually collect to the amount the math predicts. The predicted amount is the theoretical yield. The amount you really measure in the lab is the actual yield. Real reactions rarely reach 100 percent because of side reactions, spills, and incomplete conversion.

The formula is short:

Percent yield = (actual yield / theoretical yield) × 100

Continue the ammonia example. The theoretical yield was 17.03 g of NH₃. Suppose the lab only collected 14.5 g. Plug the numbers in:

• Percent yield = (14.5 g / 17.03 g) × 100
• Percent yield = 0.851 × 100 = 85.1%

So the reaction reached about 85 percent of its maximum. A high percent yield means the process was efficient. A low value tells a chemist that product was lost or that the reaction did not finish. You can read more about how chemists report this figure on the yield page.

Frequently asked questions

What is the first step in every stoichiometry problem?

Always balance the chemical equation first. The coefficients in a balanced equation give you the mole ratios you need. If the equation is not balanced, the ratios are wrong and every later step fails. Count the atoms of each element on both sides before you start any math.

Why do you have to convert grams to moles?

Coefficients describe ratios of particles, not ratios of mass. Two substances with equal mass can hold very different numbers of moles. Moles let you count particles fairly. So you convert grams to moles using molar mass, apply the mole ratio, then convert back to grams if the question needs mass.

Can the limiting reactant change if you add more of the other reactant?

Yes. Adding more of the excess reactant does not change which one limits the reaction. But if you add enough of the original limiting reactant, the other one can become limiting instead. The limiting reactant is simply whichever substance would produce the least product with the amounts you have.

Why is percent yield usually less than 100 percent?

Real reactions lose product in many small ways. Some reactant forms side products, some material sticks to glassware, and some reactions stop before finishing. Measurement errors add to the gap. A percent yield above 100 percent usually means the product was still wet or contained impurities that added extra mass.

By the Chemistry Dictionary Editorial Team · Published July 27, 2026