Chemistry questions empirical formula

Give the empirical formula
13.5% C and 86.4%F
57.2$S and 42.8%O

Metals react with oxygen in the air to form metal oxides. This process is accelerated if the metal is heated. If 1.50 grams of sodium is heated, a white powder results whose mass is 2.03 grams. Assuming the sodium only reacted with oxygen, give the empirical formula of the compound.

A chemist has a compound that consists of sodium, phosphorus, and oxygen. She heats 5.00 grams of the compound, leaving only 3.38 grams sodium phosphide.(the oxygen was released by the heating.) Of the remaining sodium phosphide, 1.05 grams is phosphorus. Give the empirical formula and name of the original compound.

Thanks very much.

For the first two, assume you have 100 g of the compound. That means, for the first one, that mass of the compound contains 13.5 g C and 86.5 g F

Now, convert each to moles:
moles C = 13.5 / 12 = 1.125
moles F = 86.5 / 19 g/mol = 4.55

Now, divide each of these by the smaller one:

C = 1.125 / 1.125 = 1
F = 4.55 / 1.125 = 4

The empirical formula is CF4.

Do the sulfur and oxygen one the same way:
mol S = 57.2 / 32.1 = 1.78
mol O = 42.8 / 16 = 2.68

Divide both by 1.78
S = 1.78/1.78 = 1
O = 2.68/1.78 = 1.5
Since O comes out to be 1.5, multiply both by 2 to get:
S = 2
O = 3
S2O3 is the formula of this compound.
_____

After the reaction you have 2.03 - 1.50 = 0.53 grams O

mol Na = 1.5 / 23 = 0.065
mol O = 0.53/16 = 0.033

Dividing by the smaller gives:
Na = 2
O = 1
The formula is Na2O

You should be able to do this one ultimately the same way. You know that 5.00 - 3.38 = 1.62 g = mass of Oxygen
3.38 - 1.05 = 2.33g = mass of Na and
1.05g = mass of P

Again convert each to moles, divide by the smallest to get the ratio of atoms.