**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.

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Combustion Analysis concept. You can view video lessons to learn Combustion Analysis. Or if you need more Combustion Analysis practice, you can also practice Combustion Analysis practice problems.