Formulas and Stoichiometry: Quick Revision

A brief note on Formulas and Stoichiometry. Read it before a test. Try each question at the end before you open its answer.

Formulas

• The molecular formula counts the atoms in one molecule. The empirical formula gives their simplest whole-number ratio.
• Mass percent of an element $= \dfrac{\text{its mass in 1 mol}}{M} \times 100$.
• Empirical formula: take $100\ \text{g}$, change each mass to moles, divide by the smallest, and multiply to whole numbers ($.5$ needs $2$, $.33$ needs $3$, $.25$ needs $4$).
• Molecular formula: $n = \dfrac{\text{molar mass}}{\text{empirical formula mass}}$.

Stoichiometry

• The coefficients of a balanced equation give the mole ratio, never a mass ratio.
• Mass to moles, moles by the mole ratio, moles to mass.
• At STP, one mole of a gas fills $22.7\ \text{L}$. Gas volumes follow the coefficients.
• The limiting reagent has the smallest value of moles divided by its coefficient. It is used up, and it fixes the product.

Concentration

• Mass percent $= \dfrac{\text{mass of solute}}{\text{mass of solution}} \times 100$
• Mole fraction: $x_A = \dfrac{n_A}{n_A + n_B}$, and all the fractions add up to $1$.
• Molarity $= \dfrac{\text{mol of solute}}{\text{L of solution}}$, in $\text{mol L}^{-1}$
• Molality $= \dfrac{\text{mol of solute}}{\text{kg of solvent}}$, in $\text{mol kg}^{-1}$. It does not change with temperature.

A worked example

Sulphur burns in oxygen: $\text{S} + \text{O}_2 \rightarrow \text{SO}_2$. $8\ \text{g}$ of sulphur burns in $12\ \text{g}$ of oxygen. How much sulphur dioxide forms, and what is left? Use $\text{S} = 32$ and $\text{O} = 16$.

1. Moles. Sulphur: $\dfrac{8}{32} = 0.25\ \text{mol}$. Oxygen: $\dfrac{12}{32} = 0.375\ \text{mol}$.
2. Limiting reagent. Both coefficients are $1$, so sulphur, with fewer moles, runs out first.
3. Product. $0.25\ \text{mol}$ of $\text{SO}_2$, which is $0.25 \times 64 = 16\ \text{g}$.
4. Excess. $0.375 - 0.25 = 0.125\ \text{mol}$ of oxygen, which is $4\ \text{g}$, is left.
5. Check. $8 + 12 = 20\ \text{g}$ before, and $16 + 4 = 20\ \text{g}$ after.

Practice with new numbers

A program to try

How much blue copper sulphate do you weigh out to make a solution of a given molarity? The water in the crystals counts in the molar mass.

Mistakes to avoid

• Percents used as moles. Divide each mass by its atomic mass first.
• Rounding a ratio of $1.5$. Multiply every number by $2$ instead.
• A mass ratio from the equation. The coefficients are moles.
• The limiting reagent by mass. Compare moles divided by the coefficients.
• Millilitres in a molarity. Change them to litres first.
• The solution in a molality. Use the mass of the solvent only.

Check yourself

1. Ethanoic acid, $\text{C}_2\text{H}_4\text{O}_2$, and glucose share one empirical formula. Which?

2. In $2\text{C}_2\text{H}_6 + 7\text{O}_2 \rightarrow 4\text{CO}_2 + 6\text{H}_2\text{O}$, how many moles of carbon dioxide come from $3$ mol of ethane?

3. Which reactant limits the reaction when $1$ mol of $\text{N}_2$ meets $2$ mol of $\text{H}_2$?

4. Why is molality preferred over molarity for work over a range of temperatures?

5. How many grams of sodium hydroxide ($40\ \text{g mol}^{-1}$) are in $1\ \text{L}$ of a $0.1\ \text{M}$ solution?