Empirical Chemical formula

Question

I just wanted to double-check if this was right before I submitted it.

1. Determine the empirical formula of a compound with the following composition: 54.52% C, 9.17% H and 36.31% O. If the molecular weight of this compound is 88 g/mol, what is the molecular formula for the compound? Show your work.

54.52/100 = .5452 * 88g/mol = 47.9776g C  

9.17/100 = .0971 * 88g/mol = 8.0696g H

36.31/100 = .3631 * 88g/mol = 31.9528g O

Double check: 47.9776g + 8.0696g + 31.9528g = 88g 

47.9776g C + 1mol/12.011g = 3.994mol C

8.0696g * 1mol/1.0078g = 8.0071mol H

31.9528g * 1mol/15.999g = 1.997mol O

3.994mol / 1.997mol = 2 

8.0071mol / 1.997mol = 4

1.997mol / 1.997mol = 1

C2H4O

Answer 1

Yes! That is what I got. Here is my process (pretty much the exact same as yours).

C: 54.52/100 = .5452 * 88g/mol = 47.98g C / 12.01g = 3.99 mol C

H: 9.17/100 = .0917 * 88g/mol = 8.07g H / 1.008g = 8.005 mol H

O: 36.31/100 = .3631 * 88g/mol = 31.95g O / 15.99g = 1.99 mol O

3.99 / 1.99 = 2 mol C

8.005 / 1.99 = 4 mol H

1.99 / 1.99 = 1 mol O

C₂H₄O

Great job!

Answer 2

Here's a simple way to find empirical formula without using info about molecular formula by assuming 100 g of the unknown:

moles C = 54.52 g C x 1 mol C / 12 g = 4.54 mols C

moles H = 9.17 g H x 1 mol H / 1 g = 9.17 mols H

moles O = 36.31 g O x 1 mol O / 16 g = 2.27 mols O

Divide all by 2.27 to get whole numbers

=> 4.54 / 2.27 = 2 mols C

=> 9.17 / 2.27 = 4 mols H

=>2.27 / 2.27 = 1 mol O

Empirical formula = C₂H₄O

And just to complete the answer, since it does ask for the molecular formula in addition to the empirical formula.

The empirical formula works out to be C₂H₄O, and this has a molar mass of ...

C: 2x12 = 24

H: 4x1 = 4

O: 1x16 = 16

Total: 24 + 4 + 16 = 44 g/mol = molar mass of empirical formula

Since the molar mass of the molecular formula is 88 g/mole, it will take 2 of the empirical formulae to make this

Thu, molecular formula = C₄H₈O₂