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Chemistry Unit 3: Moles and stoichiometry

The mole lets chemists count atoms by weighing them. Balanced equations give the ratios for how much product forms.

Covers: Molar mass, mole conversions, balancing equations, mole ratios, limiting reactant, percent yield

Study guide

Stoichiometry is the math of chemical quantities. The molar mass of a compound is the sum of its atoms' atomic masses, in g/mol (H₂O = 2(1.008) + 16.00 = 18.02 g/mol). It converts between grams and moles: moles = grams ÷ molar mass. Avogadro's number converts moles to particles. Percent composition is each element's mass share of the compound, and empirical formulas come from converting percent composition to mole ratios.

A balanced chemical equation obeys conservation of mass: the same number of each type of atom appears on both sides. Balance by changing coefficients (the numbers in front of formulas), never subscripts, which would change the substances.

The coefficients give mole ratios. To find how much product forms, convert the given mass to moles, multiply by the mole ratio from the balanced equation, then convert moles of product to mass. This grams → moles → moles → grams path solves most stoichiometry problems.

When reactants aren't in the exact ratio, the limiting reactant runs out first and determines how much product forms; the other reactant is in excess. To find it, calculate how much product each reactant could make, and the smaller amount wins.

The theoretical yield is the maximum product predicted by stoichiometry. The actual yield measured in lab is usually less, and percent yield = actual ÷ theoretical × 100%.

In combustion analysis, all the carbon in a sample ends up in CO₂ and all the hydrogen in H₂O. Convert those product masses to moles of C and H, find oxygen by subtracting the masses of C and H from the sample mass, and then divide by the smallest mole value to get the empirical formula. Every nonzero digit is significant, and so are zeros between significant digits and trailing zeros after a decimal point (20.40 has four). Leading zeros are never significant. Round products and quotients to the fewest significant figures in the data, and round sums and differences to the fewest decimal places.

The molecular formula is a whole-number multiple of the empirical formula. Divide the molar mass by the empirical formula mass to find that multiple. For example, CH₂O (30 g/mol) and a molar mass of 180 g/mol give C₆H₁₂O₆.

Key ideas

• Molar mass (g/mol) = sum of atomic masses in the formula.
• grams ÷ molar mass = moles; moles × 6.022 × 10²³ = particles.
• Balance equations by changing coefficients, never subscripts.
• Use mole ratios from coefficients to convert between substances.
• The limiting reactant runs out first and sets the maximum product.

Formulas and rules

Common mistakes

• Changing subscripts to balance an equation.
• Using grams directly in mole ratios instead of converting to moles first.

Words to know

Molar mass

The mass of one mole of a substance, in g/mol (H₂O = 18.02).

Moles from grams

grams ÷ molar mass

Limiting reactant

The reactant that runs out first.

Percent yield

Actual ÷ theoretical × 100%.

Coefficient

The number in front of a formula; the only thing you change to balance.

Stoichiometry

The math of quantities in chemical reactions.

Mole ratio

The ratio of coefficients in a balanced equation.

Percent composition

Each element's mass share of a compound.

Empirical formula

The simplest whole-number ratio of atoms.

Molecular formula

The actual number of atoms in a molecule.

Theoretical yield

The maximum product predicted by stoichiometry.

Excess reactant

The reactant left over after the limiting reactant runs out.

Conservation of mass

Atoms are neither created nor destroyed in a reaction.

Sample practice questions

53 more questions like these, from level 1 to 5, are in the practice set.