A chromium-containing compound has the formula Fe(x)Cr(y)O4 and is 28.59% oxygen by mass. Find x and y

Please show all work and give explanations. Thank you.

We know that there is 4 oxygens in the empirical formula, and we know that the oxygen makes up 28.59% of the compound. Therefore, we know the relative mass of oxygen in the compound. Using this, we can find the molecular weight of the compound.

4 O = 4 * 16 g/mol = 64 g/mol

If we divide by the molecular weight, M, we would get 28.59%, so:

64 / M = 0.2859

M = 223.85 g/mol

The molecular weight of the compound is 223.85 g/mol.

Oxygen makes up 64 g of this, so Fe and Cr must make up:

223.85 - 64 = 159.6 g/mol

So we'll have the formula:

xFe + yCr = 159.6 g/mol

Plug in the atomic masses for Fe and Cr.

55.85x + 52.00y = 159.6

With two unknowns, we do not know how much of each we need. But we can determine using other information. First, we know that we have at least 1 Fe and 1 Cr, because it is specified in the empirical formula. So let's find out what the max Fe and Cr we can have:

159.6 / 55.85 = 2.85 Fe = 2 Fe
159.6 / 52.00 = 3.06 Cr = 3 Cr

We round down because we can't overstep 159.6. It must be 159.6 and below. So the maximum Fe we can have is 2, and the maximum Cr we can have is 3. However, we MUST have 1 of each at least.

1 Fe and 1 Cr so far

Let's see how much mass we used:

159.6 - 55.85 - 52.00 = 51.75 = 52.0

We have JUST enough to squeeze in 1 more Cr, if we round, so we would have:

1 Fe and 2 Cr

So the empirical formula is:

FeCr2O4