We’re being asked to **determine the empirical formula** of a compound composed of C, H, and O, given **C 75.69%, H 8.80%, O 15.51%.**

Recall that ** mass percent** is given by:

$\overline{){\mathbf{\%}}{\mathbf{}}{\mathbf{mass}}{\mathbf{=}}\frac{\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{X}}{\mathbf{total}\mathbf{}\mathbf{mass}}{\mathbf{\times}}{\mathbf{100}}}$

Assuming we have **100 g** of the compound, this means we have **75.69 g C, 8.80 g H,** and **15.51 g O**.

Now, we need to get the moles of each element in the compound.

The atomic masses are **12.01 g/mol C, 1.01 g/mol H,** and **16 g/mol O**.

**$\mathbf{75}\mathbf{.}\mathbf{69}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{C}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}\mathbf{C}}{\mathbf{12}\mathbf{.}\mathbf{01}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{C}}}$ = 6.30 mol C**

**$\mathbf{8}\mathbf{.}\mathbf{80}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{H}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}\mathbf{H}}{\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{H}}}$ = 8.71 mol H**

**$\mathbf{15}\mathbf{.}\mathbf{51}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{O}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}\mathbf{O}}{\mathbf{16}\mathbf{.}\mathbf{00}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{O}}}$ = 0.97 mol O**

Ibuprofen has the following mass percent composition:

C 75.69%, H 8.80%, O 15.51%.

What is the empirical formula of ibuprofen?

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