91Ó°ÊÓ

Since the scale for percentages is 100, it is most convenient to calculate the mass of elements present in a sample weighing 100 g. Considering this definition, the mass percentages provided may be more conveniently expressed as fractions:

\(\begin{align}40.0\% {\rm{ of C = }}\frac{{40{\rm{ g of C}}}}{{100{\rm{ g of compound}}}}\\6.71\% {\rm{ of H = }}\frac{{{\rm{6}}{\rm{.71 g of H}}}}{{100{\rm{ g of compound}}}}\\53.28\% {\rm{ of O = }}\frac{{{\rm{53}}{\rm{.28 g of O}}}}{{100{\rm{ g of compound}}}}\end{align}\)

The molar amounts of carbon, hydrogen, and oxygen in a 100-g sample are calculated by dividing each element’s mass by its molar mass:

\(\begin{align}40.0g{\rm{ of C}}\left( {\frac{{{\rm{mol of C}}}}{{12.01g}}} \right){\rm{ = 3}}{\rm{.30 mol of C}}\\6.71g{\rm{ of H}}\left( {\frac{{{\rm{mol of H}}}}{{1.008g}}} \right){\rm{ = 6}}{\rm{.65 mol of H}}\\53.28g{\rm{ of O }}\left( {\frac{{{\rm{mol of H}}}}{{16.00g}}} \right){\rm{ = 3}}{\rm{.33 mol of O}}\end{align}\)

Coefficients for the tentative empirical formula are derived by dividing each molar amount by the lesser of the three:

\(\begin{align}\frac{{3.30{\rm{ mol C}}}}{{3.30}} = 1\\\frac{{6.65{\rm{ mol H}}}}{{3.30}} = 2\\\frac{{3.33{\rm{ mol O}}}}{{3.30}} = 1\end{align}\)

Hence, the empirical formula will be CH₂O