Question

what is the empirical formula for a compound with 70% iron (fe) and 30% oxygen (o)?
a. fe2o
b. fe2o3
c. feo2
d. feo

Explanation:

Step1: Assume the mass of the compound is 100g

So, the mass of Fe is 70g and the mass of O is 30g.

Step2: Calculate the number of moles of each element

The molar mass of Fe is 56g/mol, so the number of moles of Fe, \(n_{Fe}=\frac{70g}{56g/mol}=\frac{5}{4}mol\).
The molar mass of O is 16g/mol, so the number of moles of O, \(n_{O}=\frac{30g}{16g/mol}=\frac{15}{8}mol\).

Step3: Find the mole - ratio

Divide each number of moles by the smaller number of moles.
\(n_{Fe}:n_{O}=\frac{5}{4}:\frac{15}{8}\)
Multiply each ratio by 8 to get rid of the fractions: \(n_{Fe}:n_{O} = 10:15\)
Divide by 5: \(n_{Fe}:n_{O}=2:3\)

Answer:

B. \(Fe_{2}O_{3}\)