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Stoichiometry

How do you do stoichiometry calculations?

IntermediateReactions & StoichiometryLast reviewed 3 October 2026
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What is it?

Stoichiometry (stoy-kee-OM-uh-tree) is the part of chemistry that uses a balanced chemical equation to calculate how much of each substance reacts or is produced.

The key is that the coefficients in a balanced equation give the mole ratio between the substances. In

2 HX2+OX2→2 HX2O\ce{2H2 + O2 -> 2H2O}

2 mol of hydrogen react with 1 mol of oxygen to make 2 mol of water.

Key idea

The numbers in front of the formulas compare moles, not grams. To use them, always convert to moles first.

Why does it matter?

  • In the lab: how much of each starting material should you weigh out, and how much product should you expect?
  • In industry: manufacturers of fertilisers, medicines and fuels must know exactly how much raw material they need, because waste costs money.
  • In everyday life: the amount of oxygen a fuel needs to burn completely, or the amount of carbon dioxide it produces, is a stoichiometry calculation.

How does it work?

The three-step method

  1. Convert what you know into moles. For a mass, divide by the molar mass.
  2. Use the mole ratio from the balanced equation to find moles of the substance you want: nB=nA×ban_\text{B} = n_\text{A} \times \frac{b}{a} where aa and bb are the coefficients of A (what you have) and B (what you want).
  3. Convert moles of B into what is asked for. For a mass, multiply by the molar mass of B.

Before you start, check that the equation is balanced. An unbalanced equation gives the wrong mole ratio, and therefore the wrong answer.

Think of it like this

A sandwich recipe says: 2 slices of bread + 1 slice of cheese → 1 sandwich. With 10 slices of bread you need 5 slices of cheese and can make 5 sandwiches. You compared numbers of slices, not their weights. A balanced equation is a recipe that counts particles, and moles are how chemists count particles.

More precisely

These calculations give the theoretical amount of product: the most that could form if the reaction went to completion and nothing was lost. Real reactions often give less. Comparing the two (percent yield), and working out which reactant runs out first (the limiting reactant), are covered in the next lesson.

Visualise it

Stoichiometry road map. Grams of A: divide by the molar mass of A to get moles of A. Multiply by the mole ratio, B over A, from the balanced equation, to get moles of B. Multiply by the molar mass of B to get grams of B. The equation connects moles to moles only; never go straight from grams to grams.
The stoichiometry road map: grams → moles → moles → grams.

Worked example

Worked example: Mass of product from mass of reactant

Question: What mass of water forms when 5.00 g of hydrogen gas reacts completely with oxygen? (H = 1.008, O = 16.00)

2 HX2+OX2→2 HX2O\ce{2H2 + O2 -> 2H2O}
  1. Moles of HX2\ce{H2}: M=2(1.008)=2.016M = 2(1.008) = 2.016 g/mol, so n=5.002.016=2.480n = \dfrac{5.00}{2.016} = 2.480 mol
  2. Mole ratio: 2 mol HX2O2 mol HX2\dfrac{2\ \text{mol}\ \ce{H2O}}{2\ \text{mol}\ \ce{H2}}, so moles of HX2O\ce{H2O} = 2.480 mol
  3. Mass of HX2O\ce{H2O}: 2.480 mol×18.02 g/mol=44.72.480\ \text{mol} \times 18.02\ \text{g/mol} = 44.7 g

Answer: 44.7 g of water.

Check: the reaction also uses 39.7 g of oxygen, and 5.00 g + 39.7 g = 44.7 g. Mass is conserved.

Worked example: Burning propane

Question: Propane burns in oxygen: CX3HX8+5 OX2→3 COX2+4 HX2O\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}. What mass of oxygen is needed to burn 22.0 g of propane, and what mass of carbon dioxide forms? (C = 12.01, H = 1.008, O = 16.00)

  1. Moles of propane: M=3(12.01)+8(1.008)=44.09M = 3(12.01) + 8(1.008) = 44.09 g/mol, so n=22.044.09=0.4990n = \dfrac{22.0}{44.09} = 0.4990 mol
  2. Oxygen: 0.4990×51=2.4950.4990 \times \dfrac{5}{1} = 2.495 mol, and 2.495×32.00=79.82.495 \times 32.00 = 79.8 g
  3. Carbon dioxide: 0.4990×31=1.4970.4990 \times \dfrac{3}{1} = 1.497 mol, and 1.497×44.01=65.91.497 \times 44.01 = 65.9 g

Answer: 79.8 g of oxygen is needed, and 65.9 g of carbon dioxide forms.

Common mistake

Common mistake: Using the coefficients as a mass ratio

In 2 HX2+OX2→2 HX2O\ce{2H2 + O2 -> 2H2O}, the coefficients do not mean that 2 g of hydrogen reacts with 1 g of oxygen. They compare moles. Always convert grams to moles before using the ratio.

Common mistake: Turning the mole ratio upside down

Write the ratio as what you want ÷ what you have. To find oxygen from propane, multiply moles of propane by 5/1, not 1/5. Writing the units on every number makes the mistake easy to spot.

Common mistake: Starting from an unbalanced equation

The ratio is only correct if the equation is balanced. Check that every element has the same number of atoms on both sides before you calculate.

Notation note

  • The numbers in front of formulas are called coefficients or stoichiometric coefficients.
  • The fraction made from two coefficients is called the mole ratio, or stoichiometric factor.
  • “Reacts completely” or “in excess” in a question tells you there is enough of the other reactant, so only the one you’re given limits the amount of product.

Remember this

Remember this

  • Balanced equation first; its coefficients give mole ratios.
  • Road map: grams → moles → (mole ratio) → moles → grams.
  • Mole ratio = want ÷ have.
  • Check your answer: total mass of reactants used = total mass of products.

Test yourself

Check your understanding before moving on.

Flashcards

9 cards

  1. Question
    What is stoichiometry?
    Answer

    Using a balanced chemical equation to calculate the amounts of substances that react or are produced.

  2. Question
    What do the coefficients in a balanced equation tell you?
    Answer

    The mole ratio between the substances (not the mass ratio).

  3. Question
    What are the three steps of a stoichiometry calculation?
    Answer

    1. Convert what you know to moles. 2. Multiply by the mole ratio from the balanced equation. 3. Convert moles to what is asked for.

  4. Question
    How do you write the mole ratio so it isn't upside down?
    Answer

    Want ÷ have: (coefficient of the substance you want) ÷ (coefficient of the substance you know).

  5. Question
    In 2 HX2+OX2→2 HX2O\ce{2H2 + O2 -> 2H2O}, how many moles of OX2\ce{O2} react with 4 mol of HX2\ce{H2}?
    Answer

    4×12=24 \times \dfrac{1}{2} = 2 mol of OX2\ce{O2}

  6. Question
    In NX2+3 HX2→2 NHX3\ce{N2 + 3H2 -> 2NH3}, how many moles of NHX3\ce{NH3} form from 3.0 mol of HX2\ce{H2}?
    Answer

    3.0×23=2.03.0 \times \dfrac{2}{3} = 2.0 mol of NHX3\ce{NH3}

  7. Question
    Why must the equation be balanced before you calculate?
    Answer

    Only a balanced equation gives the correct mole ratio. An unbalanced one gives the wrong answer.

  8. Question
    Can you convert grams of A directly into grams of B using the coefficients?
    Answer

    No. The equation connects moles to moles. Go grams → moles → moles → grams.

  9. Question
    How can you check a stoichiometry answer?
    Answer

    Mass is conserved: the total mass of reactants used equals the total mass of products formed.

Quiz

7 questions

  1. Question 1EasyIn 2 HX2+OX2→2 HX2O\ce{2H2 + O2 -> 2H2O}, what is the mole ratio of OX2\ce{O2} to HX2O\ce{H2O}?
    Show answer

    Answer: 1 : 2

    The coefficients are 1 for OX2\ce{O2} and 2 for HX2O\ce{H2O}, so 1 mol of oxygen gives 2 mol of water. The answer 16 : 18 mixes up masses with moles.

  2. Question 2EasyIn NX2+3 HX2→2 NHX3\ce{N2 + 3H2 -> 2NH3}, how many moles of NHX3\ce{NH3} form from 3.0 mol of HX2\ce{H2} (with enough NX2\ce{N2})?
    Show answer

    Answer: 2.0 mol

    Mole ratio = want ÷ have = 2/3. So 3.0×23=2.03.0 \times \dfrac{2}{3} = 2.0 mol. The answer 4.5 mol uses the ratio upside down (3/2).

  3. Question 3EasyPropane burns as CX3HX8+5 OX2→3 COX2+4 HX2O\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}. How many moles of OX2\ce{O2} are needed to burn 2.00 mol of propane?
    Show answer

    Answer: 10.0 mol

    2.00×51=10.02.00 \times \dfrac{5}{1} = 10.0 mol of OX2\ce{O2}. The answer 0.400 mol divides by 5 instead of multiplying.

  4. Question 4MediumUsing the same equation, what mass of COX2\ce{CO2} forms when 1.00 mol of propane burns completely? (M(COX2)M(\ce{CO2}) = 44.01 g/mol)
    Show answer

    Answer: 132 g

    1.00 mol of propane gives 3.00 mol of COX2\ce{CO2} (ratio 3/1). Mass = 3.00×44.01=1323.00 \times 44.01 = 132 g.

  5. Question 5MediumWhat mass of ammonia forms from 28.0 g of nitrogen with excess hydrogen? NX2+3 HX2→2 NHX3\ce{N2 + 3H2 -> 2NH3} (N = 14.01, H = 1.008)
    Show answer

    Answer: 34.0 g

    Moles of NX2\ce{N2} = 28.0 ÷ 28.02 = 0.9993 mol. Moles of NHX3\ce{NH3} = 0.9993 × 2 = 1.999 mol. Mass = 1.999 × 17.03 = 34.0 g. The answer 28.0 g wrongly assumes the masses are equal.

  6. Question 6MediumIn 4 Fe+3 OX2→2 FeX2OX3\ce{4Fe + 3O2 -> 2Fe2O3}, how many moles of FeX2OX3\ce{Fe2O3} form from 6.00 mol of iron?
    Show answer

    Answer: 3.00 mol

    Mole ratio = want ÷ have = 2/4. So 6.00×24=3.006.00 \times \dfrac{2}{4} = 3.00 mol of FeX2OX3\ce{Fe2O3}.

  7. Question 7HardA student uses the coefficients in 2 HX2+OX2→2 HX2O\ce{2H2 + O2 -> 2H2O} to say that 2 g of hydrogen reacts with 1 g of oxygen. What is wrong?
    Show answer

    Answer: The coefficients give a mole ratio, not a mass ratio

    Coefficients compare numbers of particles (moles). 2 mol of HX2\ce{H2} (4.03 g) reacts with 1 mol of OX2\ce{O2} (32.00 g), so the mass ratio is very different from 2 : 1.

Notes and downloads

  • Worksheet

    Stoichiometry Worksheet

    8 questions on mole ratios and mass calculations from balanced equations. Answer key included.

    IntermediateFree

References

  1. Brown, T. L.; LeMay, H. E., Jr.; Bursten, B. E.; Murphy, C. J.; Woodward, P. M.; Stoltzfus, M. W. Chemistry: The Central Science, 15th ed.; Pearson, 2022.

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