Question

Dr. Moore has discovered a new vaccine composed of carbon, hydrogen, and oxygen. When the product was purified and sent off for elemental analysis it gave the following mass percents: 68.85% C, 4.95% H. Determine the empirical formula of this vaccine.

Answer 1

`C_7H_6O_2`

Step-by-step explanation:

We have to assume that we have 100 g of the unknow as first step. With this in mind we can calculate the grams of C, H and O in the sample.

`100~g~(68.85)/(100)=68.85~g~of~C`

`100~g~(4.95)/(100)=4.95~g~of~H`

For the calculation of O we have to know the percentage of O therefore we can add the percentages of H and C and then do a substraction from 100 %, so:

`%~of~O=~100-(68.85+4.95)=~26.2~%~of~O`100-(68.85+4.95)=26.2 %

With this value we can calculate the amount of O in grams in the sample:

`100~g~(26.2)/(100)=26.2~g~of~O`

The next step is the calculation of moles for each atom, for this we have to know the atomic mass of each atom ( C: 12 g/mol; H 1 g/mol; O 16 g/mol):

`68.85~g~of~C(1~mol~C)/(12~g~C)=~5.74~mol~C`

`4.95~g~of~H(1~mol~H)/(1~g~H)=4.95~mol~of~H`

`26.2~g~of~O(1~mol~O)/(16~g~O)=1.64~mol~of~O`

Now we have to divide by the smallest value, so:

`(5.74~mol~C)/(1.64)=3.5`

`(4.95~mol~H)/(1.64)=3`

`(1.64~mol~O)/(1.64)=1`

We obtain an decimal number for C, therefore we have to multiply all by "2", so:

`3.5x2= 7~mol~C`

`3x2= 6~mol~H`

`1x2= 2~mol~O`

Therefore the formula would be:
`C_7H_6O_2`

Answer 2

C4H3O

Step-by-step explanation:

Data obtained from the question include:

Carbon (C) = 68.85%

Hydrogen (H) = 4.95%

Since the vaccine contains carbon, hydrogen and oxygen, the remaining percentage will be for oxygen. The percentage of oxygen will be:

Oxygen (O) = 100 - (68.85 + 4.95)

= 26.2%

The empirical formula for the vaccine can be obtained as follow:

Carbon (C) = 68.85%

Hydrogen (H) = 4.95%

Oxygen (O) = 26.2%

Divide the above by their individual molar mass as shown below:

C = 68.85/12 = 5.7375

H = 4.95/1 = 4.95

O = 26.2/16 = 1.6375

Next, divide by the smallest number as shown below:

C = 5.7375/1.6375 = 4

H = 4.95/1.6375 = 3

O = 1.6375/1.6375 = 1

Therefore, the empirical formula for the vaccine is C4H3O