**Step 1:** There is **1 mole of C** in **1 mole of CO _{2}**. Finding the moles of C:

MM of CO_{2} is 44.01

$\mathrm{mole}\mathrm{of}\mathrm{C}=33.01\overline{)\mathrm{g}{\mathrm{CO}}_{2}}\times \frac{1\overline{)\mathrm{mol}{\mathrm{CO}}_{2}}}{44.01\overline{)\mathrm{g}{\mathrm{CO}}_{2}}}\times \frac{1\mathrm{mol}\mathrm{C}}{1\overline{)\mathrm{mol}{\mathrm{CO}}_{2}}}$

moles of C = 0.75 mol

There are **2 moles of H** in **1 mole of H _{2}O**. Finding the moles of H:

MM of H_{2}O is 18.02

$\mathrm{moles}\mathrm{of}\mathrm{H}=13.52\overline{)\mathrm{g}{\mathrm{H}}_{2}\mathrm{O}}\times \frac{1\overline{)\mathrm{mol}{\mathrm{H}}_{2}\mathrm{O}}}{18.02\overline{)\mathrm{g}{\mathrm{H}}_{2}\mathrm{O}}}\times \frac{2\mathrm{mol}\mathrm{H}}{1\overline{)\mathrm{mol}{\mathrm{H}}_{2}\mathrm{O}}}$

moles of H = 1.5 mol

Combustion analysis of a hydrocarbon produced 33.01 g CO_{2} and 13.52 g H_{2}O.

Calculate the empirical formula of the hydrocarbon.

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