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

`C_2H_6O`

Step-by-step explanation:

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

`114.6~g~CO_2(1~mol~CO_2)/(44~g~CO_2)=2.6~mol~of~CO_2`

`70.44~g~H_2O(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(1~mol~C)/(1~mol~CO_2)(12~g~C)/(1~mol~C)=31.25~g~of~C`

`3.9~mol~H_2O(2~mol~H)/(1~mol~H_2O)(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(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=(1.30~mol~O)/(1.30~mol~O)=~1`

`C=(2.6~mol~C)/(1.30~mol~O)=~2`

`H=(7.82~mol~H)/(1.30~mol~O)=6`

Therefore the formula is
`C_2H_6O`