The plan for this calculation is similar to others used in stoichiometric calculations, the central step being the connection between the moles of BaSO₄ and CaSO₄ through their stoichiometric factor. Once the mass of CaSO₄ is computed, it may be used along with the mass of the sample mixture to calculate the requested percentage concentration.

The mass of CaSO₄ that would yield the provided precipitate mass is

[latex]0.6168\cancel{{\text{g BaSO}}_{4}}\times \frac{1\cancel{{\text{mol BaSO}}_{4}}}{233.43\cancel{{\text{g BaSO}}_{4}}}\times \frac{1\cancel{{\text{mol CaSO}}_{4}}}{1\cancel{{\text{mol BaSO}}_{4}}}\times \frac{136.14{\text{g CaSO}}_{4}}{1\cancel{{\text{mol CaSO}}_{4}}}=0.3597{\text{g CaSO}}_{4}[/latex]

The concentration of CaSO₄ in the sample mixture is then calculated to be

[latex]\begin{array}{l}\\ {\text{percent CaSO}}_{4}=\frac{{\text{mass CaSO}}_{4}}{\text{mass sample}}\times 100\%\\ \frac{\text{0.3597 g}}{0.4550 g}\times 100\%=79.05\%\end{array}[/latex]

The primary assumption in this exercise is that all the carbon in the sample combusted is converted to carbon dioxide, and all the hydrogen in the sample is converted to water:

[latex]{\text{C}}_{\text{x}}{\text{H}}_{\text{y}}\text{(}s\text{)}+{\text{excess O}}_{2}\text{(}g\text{)}\rightarrow x{\text{CO}}_{2}\text{(}g\text{)}+y{\text{H}}_{2}\text{O}\text{(}g\text{)}[/latex]

Note that a balanced equation is not necessary for the task at hand. To derive the empirical formula of the compound, only the subscripts x and y are needed.

First, calculate the molar amounts of carbon and hydrogen in the sample, using the provided masses of the carbon dioxide and water, respectively. With these molar amounts, the empirical formula for the compound may be written as described in the previous chapter of this text. An outline of this approach is given in the following flow chart:

[latex]\begin{array}{l}\\ \text{mol C}=0.00394{\text{g CO}}_{2}\times \frac{1{\text{mol CO}}_{2}}{\text{44.01 g/mol}}\times \frac{1\text{mol C}}{1{\text{mol CO}}_{2}}=8.95\times {10}^{-5}\text{mol C}\\ \text{mol H}=0.00161{\text{g H}}_{2}\text{O}\times \frac{1{\text{mol H}}_{2}\text{O}}{18.02\text{g/mol}}\times \frac{2\text{mol H}}{1{\text{mol H}}_{2}\text{O}}=1.79\times {10}^{-4}\text{mol H}\end{array}[/latex]

The empirical formula for the compound is then derived by identifying the smallest whole-number multiples for these molar amounts. The H-to-C molar ratio is

[latex]\frac{\text{mol H}}{\text{mol C}}=\frac{1.79\times {10}^{-4}\text{mol H}}{8.95\times {10}^{-5}\text{mol C}}=\frac{2\text{mol H}}{1\text{mol C}}[/latex]

and the empirical formula for polyethylene is CH₂.