Stoichiometry Calculator

Stoichiometry is the arithmetic of a balanced equation. Every problem is the same three-part journey: convert what you were given into moles, cross to the other substance using the mole ratio from the balanced equation, then convert those moles into whatever the question asks for.

$\text{given} \;\longrightarrow\; n_{\text{A}} \;\xrightarrow{\;\text{mole ratio}\;}\; n_{\text{B}} \;\longrightarrow\; \text{asked}$

The three doors into and out of moles are:

$n = \frac{m}{\mathcal{M}} \qquad n = c\,V \qquad n = \frac{N}{N_A}$

• $m$ — mass in g, $\mathcal{M}$ — molar mass in g/mol.
• $c$ — molarity in mol/L, $V$ — solution volume in litres.
• $N$ — number of particles, $N_A = 6.02214 \times 10^{23}\ \mathrm{mol^{-1}}$.

The mole ratio itself comes straight from the coefficients. For $\mathrm{C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O}$, one mole of propane gives three of carbon dioxide.

What this assumes. The equation must be balanced, the reaction must go to completion in the stated stoichiometry with no competing side reactions, and any yield stated as a percentage is applied at the very end. Mass ratios are never used in place of mole ratios.

Mass to mass

1. Balance the equation — the coefficients are the whole calculation.
2. Convert the given mass to moles with its molar mass.
3. Multiply by the mole ratio, written so the given substance cancels: $n_B = n_A \times \dfrac{\text{coeff. B}}{\text{coeff. A}}$.
4. Convert to the requested unit, usually by multiplying by the molar mass of B.

Solutions

When a reactant is a solution, step 2 becomes $n = cV$ instead. This is the whole of titration arithmetic: moles of titrant, mole ratio, moles of analyte.

Limiting reactant

With two amounts given, one runs out first and caps the product.

1. Convert both to moles.
2. Divide each by its coefficient in the balanced equation.
3. The smallest quotient identifies the limiting reactant; base every product amount on it.
4. The excess left over is the other reactant's initial moles minus what the limiting reactant consumed.

Significant figures

Coefficients from a balanced equation are exact counts and never limit precision. Neither does $N_A$. The measured mass, volume or concentration does — usually the one with the fewest significant figures.

• Using an unbalanced equation. Every mole ratio is read off the coefficients, so an unbalanced equation makes every later step wrong.
• Applying the ratio to grams. The coefficients count particles, not mass. Convert to moles first, always.
• Inverting the ratio. Write it as a fraction that cancels the unit you have: mol $\mathrm{CO_2}$ per mol $\mathrm{C_3H_8}$ when converting from propane.
• Assuming the reactant with the smaller mass is limiting. Compare moles divided by coefficients, not masses — a light molecule can supply far more moles per gram.
• Forgetting to convert millilitres to litres before multiplying by molarity.
• Confusing theoretical and actual yield. Stoichiometry gives the theoretical maximum; multiply by the percent yield only at the end, and never build it into the mole ratio.