We’re being asked to determine the empirical formula for a compound composed only of carbon and hydrogen (Hydrogen is 25.2% by mass.)

This means we need to do the following steps:

*Step 1:* Calculate the mass and moles of C and H in the compound.

*Step 2:* Determine the lowest whole number ratio of C and H to get the empirical formula.

**Step 1:** The compound is composed of C and H and we’re given the mass percent of **H (25.2% C)**. The total mass percentage of a compound is 100%.

This means:

$\overline{){\mathbf{100}}{\mathbf{\%}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{mass}}{\mathbf{\%}}{\mathbf{}}{\mathbf{C}}{\mathbf{}}{\mathbf{+}}{\mathbf{}}{\mathbf{mass}}{\mathbf{\%}}{\mathbf{}}{\mathbf{H}}}\phantom{\rule{0ex}{0ex}}\mathbf{mass}\mathbf{}\mathbf{\%}\mathbf{}\mathbf{C}\mathbf{}\mathbf{=}\mathbf{}\mathbf{100}\mathbf{\%}\mathbf{}\mathbf{-}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{\%}\mathbf{}\mathbf{H}\phantom{\rule{0ex}{0ex}}\mathbf{mass}\mathbf{}\mathbf{\%}\mathbf{}\mathbf{C}\mathbf{}\mathbf{=}\mathbf{}\mathbf{100}\mathbf{\%}\mathbf{}\mathbf{-}\mathbf{}\mathbf{25}\mathbf{.}\mathbf{2}\mathbf{\%}\phantom{\rule{0ex}{0ex}}\mathbf{mass}\mathbf{}\mathbf{\%}\mathbf{}\mathbf{C}\mathbf{=}\mathbf{}\mathbf{74}\mathbf{.}\mathbf{8}\mathbf{\%}\mathbf{}\mathbf{C}\phantom{\rule{0ex}{0ex}}$

Recall that * mass percent *is given by:

$\overline{){\mathbf{\%}}{\mathbf{mass}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{mass}\mathbf{}\mathbf{X}}{\mathbf{Total}\mathbf{}\mathbf{mass}}{\mathbf{x}}{\mathbf{100}}}$

Assuming we have **100 g** of the compound, this means we have **74.8 g C and 25.2g H. **Now, we need to get the moles of each element in the compound.

The atomic masses are **12 g/mol C and 1 g/mol H**

A compound composed of only carbon and hydrogen is 25.2% hydrogen by mass. The empirical formula for this compound is.

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