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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=mMn = \frac{m}{M}, % by mass=part masstotal mass×100%\%\text{ by mass} = \frac{\text{part mass}}{\text{total mass}} \times 100\%, and % yield=actual yieldtheoretical yield×100%\%\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=mMn = \frac{m}{M}, where mm is mass in grams and MM is molar mass in g mol1\text{g mol}^{-1}.
  • Use the balanced equation to set the mole ratio, such as nAgClnCl=11\frac{n_{\text{AgCl}}}{n_{\text{Cl}^-}} = \frac{1}{1} for Ag++ClAgCl(s)\text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl}(s).
  • Mass percent is calculated with % by mass=mass of analytemass of sample×100%\%\text{ by mass} = \frac{\text{mass of analyte}}{\text{mass of sample}} \times 100\%.
  • Percent yield is calculated with % yield=actual yieldtheoretical yield×100%\%\text{ yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%.
  • For a chloride example, 0.287 g AgCl×1 mol AgCl143.32 g AgCl×1 mol Cl1 mol AgCl0.287\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 Cl\text{Cl}^-.
  • In hydrate analysis, moles of water are found with nH2O=mlost18.02 g mol1n_{\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 g mol1\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 1:11:1 or 1:21:2 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=mMn = \frac{m}{M}.
  • Stopping before the final percent calculation, which is wrong when the question asks for % by mass\%\text{ by mass} or % yield\%\text{ yield} rather than moles.

Practice Questions

  1. 1 A 0.500 g0.500\text{ g} sample containing chloride ions forms 0.287 g0.287\text{ g} of AgCl\text{AgCl}. Using MAgCl=143.32 g mol1M_{\text{AgCl}} = 143.32\text{ g mol}^{-1} and MCl=35.45 g mol1M_{\text{Cl}} = 35.45\text{ g mol}^{-1}, calculate the mass percent of chloride in the sample.
  2. 2 A hydrate sample has mass 2.50 g2.50\text{ g} before heating and 1.60 g1.60\text{ g} after heating. If the anhydrous salt has molar mass 160.0 g mol1160.0\text{ g mol}^{-1}, find the value of xx in saltxH2O\text{salt}\cdot x\text{H}_2\text{O}.
  3. 3 A reaction is expected to produce 1.25 g1.25\text{ g} of precipitate, but the dried precipitate actually has mass 1.10 g1.10\text{ g}. Calculate the percent yield using % yield=actual yieldtheoretical yield×100%\%\text{ yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%.
  4. 4 Explain why a gravimetric analysis result can be too high if the precipitate is not dried to constant mass.

Understanding Gravimetric Analysis Worked Examples

A gravimetric result is only as good as the solid that is collected. In precipitation work, the chosen reagent should form a compound with very low solubility. This makes most of the target ion leave the solution and enter the solid.

Chemists usually add the precipitating reagent slowly while stirring. Slow addition avoids tiny particles that can pass through filter paper. Warming the mixture for a short time can help small crystals dissolve and reform as larger crystals.

This process is called digestion. Larger crystals are easier to filter and are less likely to trap dissolved impurities inside them.

Several sources of error can change the measured mass. If some precipitate remains on the beaker wall or passes through the filter, the final mass is too low. The calculated amount of the original substance then becomes too low.

If the solid contains leftover solution, the mass is too high. Washing removes soluble ions that stuck to the crystal surfaces. The wash liquid must be used carefully because too much washing can dissolve a small amount of the precipitate.

Drying is equally important. A sample is weighed, heated, cooled, then weighed again.

When two successive masses are nearly the same, the solid has reached constant mass. Cooling in a dry container prevents the solid from picking up water from air before weighing.

Hydrate problems use mass change as evidence about the water held in a crystal. Heating removes water, but the remaining salt must not decompose. A student should compare the mass before heating with the mass after repeated heating.

The lost mass is treated as water only when the heating conditions support that assumption. The amount of water is then compared with the amount of dry salt. For example, a ratio near two moles of water for every one mole of salt gives a hydrate formula containing two water molecules.

Empirical formula work follows the same ratio idea. A ratio such as one, one point five, and two is not a final formula ratio. Multiplying all values by two gives whole numbers of two, three, and four.

Percent yield has a different purpose from composition calculations. It compares what was actually recovered with what should have formed if every step were perfect. A low yield may result from incomplete reaction, loss during transfer, dissolved product, or poor filtration.

A value above one hundred percent is a warning sign rather than a success. It often means water or impurities added extra mass. In school problems, make a short path before calculating.

Write the known mass, identify the substance represented by that mass, find its moles, follow the equation ratio, then convert only at the end to the requested unit. Keep extra digits during the work and round only in the final answer. This method is useful in water testing, mineral analysis, food chemistry, and any lab where a mass measurement is used to make a claim about composition.