Question

An unknown compound containing only carbon and hydrogen was burned in excess oxygen. When 8.561 grams of the compound was burned, 26.863 grams of carbon dioxide and 10.999 grams of water were produced. If the unknown compound has a formula mass of 126.23 g/mol, what is the molecular formula of the compound?

Answer 1

We need to determine the number of moles of Carbon and Hydrogen. Here is the carbon:
\[26.863gCO_2\frac{1 molCO_2}{44.0095gCO_2}\frac{1molC}{1molCO_2}=0.61039molC\]
And Hydrogen:
\[10.999gH_2O\frac{1molH_2O}{18.0152gH_2O}\frac{2molH}{1molH_2O}=1.2211molH\]
The empirical formula is determined by using the lowest number of moles which in this case is Carbon which is approximately 1:2 or:
\[CH_2\]
We need to determine the molar mass of the empirical formula:
\[2(1.0079)+12.0107=14.0265\frac{g}{mol}\]
Now we divide the unknown compound by the empirical mass to get a multiplier.
\[\frac{126.23\frac{g}{mol}}{14.0265\frac{g}{mol}}\approx9\]
Now that we have the multiplier, we multiplier the number of components in the empirical formula and get:
\[C_9H_{18}\]