Gravimetric Analysis Worked Examples Cheat Sheet

Gravimetric analysis uses careful mass measurements to find the amount of a substance in a sample. Students need this cheat sheet because the calculations combine balanced equations, mole ratios, molar mass, and percent composition. Worked examples help organize each problem from measured precipitate mass to the final quantity requested.

This topic is especially important for lab-based chemistry and quantitative analysis.

The core idea is to convert the measured mass into moles, use the balanced chemical equation, then convert to the desired mass or percent. Important formulas include $n = \frac{m}{M}$ , $\%\text{ by mass} = \frac{\text{part mass}}{\text{total mass}} \times 100\%$ , and $\%\text{ yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%$ . In hydrate and empirical formula problems, mole ratios reveal the formula of the compound.

Accuracy depends on complete precipitation, correct drying, and careful unit tracking.

Key Facts

Convert a measured solid mass to moles using $n = \frac{m}{M}$ , where $m$ is mass in grams and $M$ is molar mass in $\text{g mol}^{-1}$ .
Use the balanced equation to set the mole ratio, such as $\frac{n_{\text{AgCl}}}{n_{\text{Cl}^-}} = \frac{1}{1}$ for $\text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl}(s)$ .
Mass percent is calculated with $\%\text{ by mass} = \frac{\text{mass of analyte}}{\text{mass of sample}} \times 100\%$ .
Percent yield is calculated with $\%\text{ yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%$ .
For a chloride example, $0.287\text{ g AgCl} \times \frac{1\text{ mol AgCl}}{143.32\text{ g AgCl}} \times \frac{1\text{ mol Cl}^-}{1\text{ mol AgCl}}$ gives the moles of $\text{Cl}^-$ .
In hydrate analysis, moles of water are found with $n_{\text{H}_2\text{O}} = \frac{m_{\text{lost}}}{18.02\text{ g mol}^{-1}}$ .
An empirical formula is found by dividing each element amount by the smallest mole value, then multiplying if needed to get whole-number subscripts.
A precipitate must be filtered, washed, dried, and weighed to constant mass before its mass is used in calculations.

Vocabulary

Gravimetric analysis: A quantitative method that determines the amount of an analyte by measuring the mass of a related pure solid.
Analyte: The substance in a sample that is being measured or calculated.
Precipitate: An insoluble solid that forms when ions in solution react.
Molar mass: The mass of one mole of a substance, usually measured in $\text{g mol}^{-1}$ .
Stoichiometric ratio: A mole ratio from a balanced chemical equation used to convert between reactants and products.
Constant mass: A condition reached when repeated heating and weighing give nearly the same mass, showing the sample is dry.

Common Mistakes to Avoid

Using the precipitate mass as the analyte mass, which is wrong because the precipitate often contains other atoms besides the analyte.
Skipping the balanced equation, which is wrong because mole ratios such as $1:1$ or $1:2$ control the conversion between substances.
Using grams directly in mole ratios, which is wrong because stoichiometric coefficients compare moles, not masses.
Forgetting to subtract the container or filter paper mass, which is wrong because only the precipitate mass should enter $n = \frac{m}{M}$ .
Stopping before the final percent calculation, which is wrong when the question asks for $\%\text{ by mass}$ or $\%\text{ yield}$ rather than moles.

Practice Questions

1 A $0.500\text{ g}$ sample containing chloride ions forms $0.287\text{ g}$ of $\text{AgCl}$ . Using $M_{\text{AgCl}} = 143.32\text{ g mol}^{-1}$ and $M_{\text{Cl}} = 35.45\text{ g mol}^{-1}$ , calculate the mass percent of chloride in the sample.
2 A hydrate sample has mass $2.50\text{ g}$ before heating and $1.60\text{ g}$ after heating. If the anhydrous salt has molar mass $160.0\text{ g mol}^{-1}$ , find the value of $x$ in $\text{salt}\cdot x\text{H}_2\text{O}$ .
3 A reaction is expected to produce $1.25\text{ g}$ of precipitate, but the dried precipitate actually has mass $1.10\text{ g}$ . Calculate the percent yield using $\%\text{ yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%$ .
4 Explain why a gravimetric analysis result can be too high if the precipitate is not dried to constant mass.

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Gravimetric Analysis Worked Examples Cheat Sheet

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