What is the empirical formula of a compound composed of 3.25% hydrogen (H), 19.36% carbon (C), and 77.39% oxygen (O) by mass?

2 Answers
May 7, 2016

CH_2O_3CH2O3

Explanation:

We assume 100*g100g of unknown compound, and work out the elemental proportions in terms of moles:

H:H: (3.25*g)/(1.00794*g*mol^-1)3.25g1.00794gmol1 == 3.22*mol;3.22mol;

C:C: (19.36*g)/(12.01*g*mol^-1)19.36g12.01gmol1 == 1.61*mol;1.61mol;

O:O: (77.39*g)/(15.99*g*mol^-1)77.39g15.99gmol1 == 4.84*mol;4.84mol;

We divide thru by the SMALLEST molar quantity, that of carbon, to give an empirical formula:

CH_2O_3CH2O3

May 7, 2016

CH_2O_3CH2O3

Explanation:

Assume that there is 100g of the substance, so there would be

3.25g\ H 19.36g\ C and 77.39g\ O

Divide each mass by the molar mass of their respective element.

(3.25g)/1 H (19.36g)/12 C (77.39g) /16 O

=3.25 H =1.61 C =4.84 O

Divide by the smallest number. In this case, 1.61

=2 H =1 C =3O

So the empirical formula is CH_2O_3