Question

Determine the empirical formula of a compound that is 39.99% carbon, 53.28% Oxygen and 6.73% Hydrogen

Answer 1

Answer:
`COH_2`

Step-by-step explanation:

39.99% C

53.28% O

6.73% H

First step:

Change % to g

39.99g C

53.28g O

6.73g H

Second step:

Calculate the number of g/mol using atomic mass.

`39.99gC((1mol)/(12g) )=3.332mol`

`53.28gO((1mol)/(16g))=3.330mol`

`6.73gH((1mol)/(1g) )=6.73mol`

Third step:

Determine the smallest number of moles and divide each compound by it. When you get the result, simply put the whole number, not decimals.

`C=(3.332mol)/(3.330mol)=1`

`O=(3.330mol)/(3.330mol) =1`

`H=(6.73mol)/(3.330mol)=2`

Fourth step:

Rewrite those number under the compounds.

This empirical formula should be:
`COH_2`

Answer 2

CH2O

Step-by-step explanation:

Pretend you have a 100 gram sample.

39.99 g C * (1 mol/12.01 g) = 3.33 moles C

53.28 g O * (1 mol/16 g) = 3.33 moles O

6.73 g H * (1 mol/ 1.008 g) = 6.676 moles H

You can see from this data that there are double the number of moles of H compared to the moles of C and O.

To find the empirical formula, you divide all the moles by the least number of moles. In this case, the least number of moles is 3.33. So for carbon and oxygen, divide 3.33 by 3.33... this equals 1. For H, divide 6.676 by 3.33... this is about equal to 2. Therefore, the empirical formula is CH2O. :)

Hope this was helpful!