Stoichiometry & Molar Mass Cheat Sheet

A printable reference covering molar mass, mole conversions, Avogadro's number, percent composition, empirical formulas, and stoichiometric ratios for grades 10-11.

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Stoichiometry connects the amount of one substance in a chemical reaction to the amount of another substance. This cheat sheet helps students organize mole conversions, molar mass calculations, and balanced equation ratios. These skills are needed to predict reactants used, products formed, and quantities measured in the lab. A clear reference makes multi-step problems easier to set up and check.

Key Facts

Molar mass is found by adding atomic masses from the periodic table, so $M = \sum \left(\text{atomic mass} \times \text{subscript}\right)$ .
Convert between mass and moles using $n = \frac{m}{M}$ , where $n$ is moles, $m$ is mass in grams, and $M$ is molar mass in $\text{g/mol}$ .
Convert between particles and moles using $N = nN_A$ , where $N_A = 6.022 \times 10^{23}\ \text{particles/mol}$ .
A balanced chemical equation gives mole ratios from coefficients, such as $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ , where $2\ \text{mol H}_2$ reacts with $1\ \text{mol O}_2$ .
The main stoichiometry path is $\text{grams A} \rightarrow \text{moles A} \rightarrow \text{moles B} \rightarrow \text{grams B}$ .
Percent composition is calculated with $\%\text{ element} = \frac{\text{mass of element in compound}}{\text{molar mass of compound}} \times 100\%$ .
Empirical formulas use the simplest whole-number mole ratio of elements, while molecular formulas use $\text{molecular formula} = \left(\text{empirical formula}\right)_n$ .
The multiplier for a molecular formula is $n = \frac{\text{molar mass of molecular formula}}{\text{molar mass of empirical formula}}$ .

Vocabulary

Mole: A counting unit equal to $6.022 \times 10^{23}$ representative particles of a substance.
Molar Mass: The mass of $1\ \text{mol}$ of a substance, usually measured in $\text{g/mol}$ .
Avogadro's Number: The constant $N_A = 6.022 \times 10^{23}\ \text{particles/mol}$ used to convert between particles and moles.
Stoichiometry: The use of a balanced chemical equation to calculate amounts of reactants and products.
Mole Ratio: A conversion factor made from coefficients in a balanced chemical equation, such as $\frac{2\ \text{mol H}_2\text{O}}{1\ \text{mol O}_2}$ .
Empirical Formula: The simplest whole-number ratio of atoms or moles of elements in a compound.

Common Mistakes to Avoid

Using subscripts as mole ratios, which is wrong because stoichiometric mole ratios come from coefficients in the balanced equation, not from formulas within compounds.
Skipping equation balancing, which makes every mole ratio incorrect because coefficients must represent conservation of atoms.
Using grams directly in a mole ratio, which is wrong because balanced equations compare moles, so grams must first be converted using $n = \frac{m}{M}$ .
Rounding too early in multi-step problems, which can change the final answer noticeably. Keep extra digits until the final step, then round to the correct significant figures.
Forgetting to multiply atomic masses by subscripts, which gives an incorrect molar mass. For example, $\text{H}_2\text{O}$ contains $2$ hydrogen atoms, not $1$ .

Practice Questions

1 Calculate the molar mass of $\text{Ca}(\text{OH})_2$ using $\text{Ca} = 40.08\ \text{g/mol}$ , $\text{O} = 16.00\ \text{g/mol}$ , and $\text{H} = 1.008\ \text{g/mol}$ .
2 How many moles are in $25.0\ \text{g}$ of $\text{CO}_2$ if the molar mass of $\text{CO}_2$ is $44.01\ \text{g/mol}$ ?
3 For $2\text{Al} + 3\text{Cl}_2 \rightarrow 2\text{AlCl}_3$ , how many moles of $\text{AlCl}_3$ form from $4.50\ \text{mol}$ of $\text{Cl}_2$ ?
4 Explain why a balanced chemical equation is required before using mole ratios in a stoichiometry problem.