Question

1. a compound consists of 52.14% carbon, 13.13% of hydrogen, and 34.73% of oxygen by mass. (a) what is the empirical formula? element c h o mass (g) 52.14 13.13 34.73 atomic mass 12.01 1.008 16.00 mole = mass / atomic mass 52.14/12.01≈4.34 13.13/1.008≈13.02 34.73/16.00≈2.17 mole ratio = mole / smallest mole empirical formula.

Explanation:

Step1: Assume 100g of the compound

This allows us to directly use the mass - percentages as masses. So, mass of C = 52.14g, mass of H = 13.13g, mass of O = 34.73g.

Step2: Calculate the moles of each element

Use the formula $n=\frac{m}{M}$, where $n$ is the number of moles, $m$ is the mass, and $M$ is the atomic mass. For C: $n_C=\frac{52.14g}{12.01g/mol}\approx4.34mol$. For H: $n_H=\frac{13.13g}{1.01g/mol}\approx13.0mol$. For O: $n_O=\frac{34.73g}{16.00g/mol}\approx2.17mol$.

Step3: Find the mole - ratio

Divide each number of moles by the smallest number of moles (2.17mol in this case). For C: $\frac{4.34mol}{2.17mol}=2$. For H: $\frac{13.0mol}{2.17mol}\approx6$. For O: $\frac{2.17mol}{2.17mol}=1$.

Answer:

The empirical formula is $C_2H_6O$