Magnetite is a binary compound containing only iron and oxygen. the percent, by weight, of iron is 72.360%. what is the empirical formula of magnetite?

a. feo

b. fe2o3

c. fe3o4

d. feo2

Answer is: c. Fe ₃ O ₄.

ω (Fe) = 72,360%.

ω (O) = 100% - 72,36% = 27,64%.

For example, if we the mass of compound is 100 g:

m (Fe) = 72,36 g.

n (Fe) = m (Fe) : M (Fe).

n (Fe) = 72,36 g : 55,85 g/mol.

n (Fe) = 1,296 mol.

n (O) = 27,64 g : 16 g/mol.

n (O) = 1,727 mol.

n (Fe) : n (O) = 1,296 mol : 1,727 mol.

n (Fe) : n (O) = 1 : 1,33 or 3 : 4.

Option C. The percentages can be taken for mass. So mass of Iron = 72.360% = 72.36g So mass of Oxygen = 100 - 72.360 = 27.694% = 27.65g Number of moles of Iron = mass / molar mass = 72.36 / 55.854 = 1.29 Number of moles of Oxygen = 27.65/16 = 1.728 We shall divide by the smallest So we have 1.29/1.29 = 1 while 1.728/1.29 = 1.33 But 1.33 = 4/3. So we multiply by the lowest common number which happens to be a 3. Hence the empirical formula is Fe3O4.