33.87% Cu, 14.93% N And 51.2% Oxygen. What Is The Empirical Formula For A Compound That Contains 33.87% Cu, 14.93% N And 51.2% Oxygen.?

What is the empirical formula for a compound that contains 33.87% Cu, 14.93% N and 51.2% oxygen.? - 33.87% cu, 14.93% n and 51.2% oxygen.

First, you should assume that this percentage is more than 100 grams of the sample, and what that should not be confused with the symbol% will appear only in grams per gram.
So you know the individual masses of all its elements in their enclosure, and the dough can find moles with the molecular weight of these elements.

33.87gx 1 mol Cu Cu / Cu 63.55g = 0.532966 moles Cu
14.93g NN x 1 mol / 14.01g N = 1.065667 mol N
51.20g OO x 1 mol / 16.00g O = 3.20000 mol O

Now we have the moles of all elements of the simplicity of his division from here on out.
Next, take the moles of Cu and report them to themselves, because the smallest number of moles, and then divided by everything else.

0.532966 moles Cu/0.532966 mole Cu = 1
1.065667 mol N/0.532966 mole Cu = 1.9995 = 2
O/0.532966 mol = 3.20000 moles Cu 6.004 = 6

Thus, the empirical formula for the connection is CuN2O6