Empirical Formula Without Percentage: Can You Dig It?

Question

In need of chemistry help... again... these problems are complicated. How exactly can you find the empirical formula WITHOUT percentage?

A 3.62g sample of a compound containing C, H, and O only burned in air and made 5.19g of CO2 and 2.83g of H2O. What is the empirical formula of the compound?

---AND---

Potassium persulfate is a chemical used for photodeveloping. By mass, potassium persulfate has 28.9% potassium, 23.7% sulfur and the rest is oxygen. What is the empirical formula of this compound?

Answer 1

In the first problem, the carbon from the CO2 comes from the CHO compound as does the hydrogen from water. You can determine the amount of carbon and hydrogen in CO2 and water.
The remainder of the 3.62 g was oxygen. You can then figure out an empirical formula. You will have to divide the mass of each element by its g-atomic wt.

Since the calculation is similar for both problems,
I'll show you the one for CO2.
The mol wt is 44 and the g-atomic wt of C is 12.
Then mass of C= (12/44) x 5.19 g

In problem #2, just pick a convenient mass, such as 100 grams.

Answer 2

First find out how much C is in the compound:
5.19 g CO2( 1 mole CO2/44.0 g)=0.118 moles CO2
This is the same number of moles as the C in the compound.
0.118 moles C (12.0 g/mole) =1.415 g C

Then find out how much H is in the compound:
2.83 g H2O ( 1 mole H2O/18.0 g) = 0.157 moles H2O
This will be equivalent to 2(.157) = 0.314 moles H in the compound.
0.314 moles H (1.008g/mole) = 0.317 g H in the compound.

To find out how much O is in the compound:
3.62 g sample - 1.415g C -0.317 g H = mass O
1.888 g O
1.888 g
O (1 mole/16.0 g) = 0.118 moles O
The ratio of H to C in the empirical formula is .314moles/0.118 moles = 2.66 or 2 2/3 or 8 to 3
The ratio of C to O is .118/.118=1.00 or 1 to 1
Empirical formula is C3H8O3

You can do the other problem by assuming that
% O = 100% -%K-%S