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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

Respuesta :

Dejanras
Answer is: c. Fe₃O₄.
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.
Kalahira
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.

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