The ionic compund sometimes called yellow uranium is used to produce colored glazes for ceramics. It is 7.252% sodium, 75.084% uranium, and 17.664% oxygen

Question

a. What is the empirical formula for this compound?

b. This compound has a molecular mass of 1268.06 g/mol, what is the molecular compound for this formula?

Answer 1

Take 100 g sample which gives you
7.252 g Na
75.084 g U
17.664 g O
------------
convert to mols by mols = grams/atomic mass.

7.252/23 = approx 0.315
75.084/238 = approx 0.315
17.664/16 = approx 1.10
For the empirical formula you want to find the ratio of these elements to each other with the smaller whole number being 1.00. The easy way to do that is to divide the smallest number by itself (thereby assuring 1.00 for that element) then divide the other numbers by the same small number.
If I do that I have
0.315/0.315 = 1.00 for Na
0.315/0.315 = 1.00 for U
1.10/0.315 = 3.5 for O. Since we want whole numbers it is easy to see that multiplying everything by 2 will give whole numbers (we do that INSTEAD of rounding that 3.5 to either 3.0 or 4.0) so the empirical formula is Na2U2O7.
Empirical formula mass is ? you can add it up.
Then empirical formula x n = molar mass
Plug in empirical formula mass, plug in molar mass of 1268, solve for n, round to a whole number, and the molecular formula is (Na2U2O7)n and n will be 1 or 2 or 3 etc.

Answer 2

make a table such as this, assume you have 1000 grams of the stuff.

read om four columns: element/mass/moles/ratio
Na/ 72.52 / 3.12
U/ 750.84/3.02
O / 176.64/11.04

Now, for the last column, pick the lowest number in mole column (here 3.02) and divide it into the other numbers.
Na: 1
U: 1
O : 3.65
Now the simplest whole number ration
Na: 2
U: 2
O: 7
empirical: Na2U2O7
empirical x(2*23+2*238+7*16)=1268
x(634=1268
x=2
mol formula= Na4U4O14