Question

A certain alcohol contains only three elements, carbon, hydrogen, and oxygen. Combustion of a 60.00 gram sample of the alcohol produced 114.6 grams of CO2 and 70.44 grams of H2O. What is the empirical formula of the alcohol

Answer 1

Answer:

$C_2H_6O$

Explanation:

The first step is the calculation of the moles of $H_2O$ and $CO_2$, so:

$114.6~g~CO_2\frac{1~mol~CO_2}{44~g~CO_2}=2.6~mol~of~CO_2$

$70.44~g~H_2O\frac{1~mol~H_2O}{18~g~H_2O}=~3.9~mol~H_2O$

Now, in 1 mol of CO2 we have 1  mol of C and in 1 mol of $H_2O$ we have 1 mol of H. Additionally, if we want to calculate the moles of oxygen we need to calculate the grams of C and O and then do the substraction form the initial amount, so:

$2.6~mol~CO_2\frac{1~mol~C}{1~mol~CO_2}\frac{12~g~C}{1~mol~C}=31.25~g~of~C$

$3.9~mol~H_2O\frac{2~mol~H}{1~mol~H_2O}\frac{1~g~H}{1~mol~H}=7.82~g~of~H$

$Total~grams=~31.25~+~7.82=39.08~g$

$grams~of~O=60.00~g-~39.08~g=20.92~g~of~O$

Now we can convert the grams of O to moles, so:

$20.92~g~of~O\frac{1~mol~O}{16~g~O}=1.30~mol~O$

The next step is to divide all the mol values by the smallest one:

$O=\frac{1.30~mol~O}{1.30~mol~O}=~1$

$C=\frac{2.6~mol~C}{1.30~mol~O}=~2$

$H=\frac{7.82~mol~H}{1.30~mol~O}=6$

Therefore the formula is $C_2H_6O$