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Percent composition shows how much of each element is present in a compound by mass. Students need this skill to connect chemical formulas with measured masses in the lab. Hydrate formulas extend the same idea by showing how many water molecules are trapped in an ionic crystal.

This cheat sheet gives worked-example steps for percent composition and hydrate analysis.

Key Facts

  • Molar mass is the sum of the atomic masses in a formula, written in units of g/mol\mathrm{g/mol}.
  • Percent composition by mass is % element=mass of element in 1 mol compoundmolar mass of compound×100%\%\text{ element} = \frac{\text{mass of element in }1\text{ mol compound}}{\text{molar mass of compound}} \times 100\%.
  • The percent compositions of all elements in a compound should add to about 100%100\%, allowing small rounding differences.
  • For a hydrate, the mass of water lost during heating is mwater=mhydratemanhydrous saltm_{\text{water}} = m_{\text{hydrate}} - m_{\text{anhydrous salt}}.
  • Moles of anhydrous salt are found with nsalt=msaltMsaltn_{\text{salt}} = \frac{m_{\text{salt}}}{M_{\text{salt}}}.
  • Moles of water are found with nwater=mwater18.02 g/moln_{\text{water}} = \frac{m_{\text{water}}}{18.02\ \mathrm{g/mol}}.
  • The hydrate coefficient xx in saltxH2O\text{salt}\cdot x\mathrm{H_2O} comes from the mole ratio x=nwaternsaltx = \frac{n_{\text{water}}}{n_{\text{salt}}}, rounded to a small whole number.
  • The percent water in a hydrate is %H2O=mwatermhydrate×100%\%\mathrm{H_2O} = \frac{m_{\text{water}}}{m_{\text{hydrate}}} \times 100\%.

Vocabulary

Percent composition
The percentage by mass of each element in a compound.
Molar mass
The mass of one mole of a substance, found by adding the atomic masses in its chemical formula.
Hydrate
An ionic compound that has a fixed number of water molecules attached within its crystal structure.
Anhydrous salt
The salt that remains after all water of crystallization has been removed from a hydrate.
Water of crystallization
Water molecules that are chemically included in the crystal structure of a hydrate.
Mole ratio
A ratio comparing the moles of substances, used to find formula subscripts or hydrate coefficients.

Common Mistakes to Avoid

  • Using only the atomic mass of one atom when a formula contains subscripts is wrong because subscripts multiply that element's mass, such as 22 oxygen atoms in CO2\mathrm{CO_2}.
  • Dividing by the element's mass instead of the compound's total molar mass is wrong because percent composition compares each part to the whole compound.
  • Forgetting to subtract the anhydrous salt mass from the hydrate mass is wrong because mwaterm_{\text{water}} must represent only the water lost during heating.
  • Rounding the hydrate mole ratio too early is wrong because small rounding errors can change xx in saltxH2O\text{salt}\cdot x\mathrm{H_2O}.
  • Treating H2O\mathrm{H_2O} as massless in a hydrate formula is wrong because the water molecules contribute to the hydrate's molar mass and percent water.

Practice Questions

  1. 1 Find the percent composition of oxygen in CO2\mathrm{CO_2} using MC=12.01 g/molM_{\mathrm{C}} = 12.01\ \mathrm{g/mol} and MO=16.00 g/molM_{\mathrm{O}} = 16.00\ \mathrm{g/mol}.
  2. 2 A sample of CuSO4xH2O\mathrm{CuSO_4}\cdot x\mathrm{H_2O} has a mass of 2.50 g2.50\ \mathrm{g} before heating and 1.60 g1.60\ \mathrm{g} after heating. Find the mass of water lost and the percent water.
  3. 3 A hydrate contains 0.0200 mol0.0200\ \mathrm{mol} of anhydrous salt and 0.100 mol0.100\ \mathrm{mol} of water. Determine xx in saltxH2O\text{salt}\cdot x\mathrm{H_2O}.
  4. 4 Explain why a hydrate formula uses a dot, as in MgSO47H2O\mathrm{MgSO_4}\cdot 7\mathrm{H_2O}, instead of writing the water atoms as part of one large covalent molecule.

Understanding Percent Composition and Hydrate Formula Worked Examples

Accurate formula reading comes first. A subscript applies only to the element just before it. A number outside parentheses applies to every atom inside the parentheses.

For example, in calcium nitrate, the two nitrate groups contain two nitrogen atoms and six oxygen atoms. Missing one oxygen changes the calculated mass and every percentage that follows. Use atomic masses from the same periodic table throughout a problem.

Keep extra decimal places during calculations, then round only at the end. This reduces small errors that can become noticeable after several steps.

A useful way to organize percent work is to make a short list of each element, its number of atoms, the atomic mass, and its total contribution to one mole of the substance. The contribution comes from multiplying the atom count by the atomic mass. Add all contributions to find the total molar mass.

Then compare each contribution with that total and convert the result to a percentage. This method works for simple compounds, large ionic formulas, and substances containing parentheses. It also shows why elements with heavier atoms can make up most of a sample even when their subscripts are small.

Hydrate analysis depends on careful heating rather than just arithmetic. A sample is heated in a crucible so that its water leaves as vapor while the dry ionic solid remains. Students weigh the container and sample before heating, heat the sample, cool it, then weigh it again.

Cooling matters because a hot crucible can create unreliable balance readings. The sample is often reheated and reweighed until two measurements are nearly the same. This is called heating to constant mass.

It gives evidence that the removable water has left. The lost mass is treated as water only when the salt itself has not decomposed or escaped during heating.

The final hydrate formula comes from comparing amounts in moles, not comparing masses directly. Water has a much smaller molar mass than most salts, so equal masses do not mean equal numbers of particles. Divide both mole amounts by the smaller amount to get the simplest ratio.

A result close to two, three, five, or another small whole number is usually expected. A value such as two point five may indicate a ratio that should be multiplied by two. Large or awkward ratios often signal a measurement problem.

Common causes include incomplete heating, loss of solid from the crucible, moisture absorbed while cooling, or rounding too early. These ideas matter in laboratory work because many materials, including some crystals, building materials, and drying agents, change mass when they gain or lose water.