Question

An oxide ($Fe_2O_n$) of iron contains 70% iron by mass. Calculate value of n.

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (C)

The value of n is 3.

Answer 2

The given oxide of iron has the formula $Fe_2O_n$. It contains 70% iron by mass. We need to find the value of $n$.

The molar mass of Fe is approximately 56 g/mol, and the molar mass of O is approximately 16 g/mol.

The mass percentage of iron in the compound $Fe_2O_n$ is given by: Mass % of Fe = $\frac{\text{Mass of 2 Fe atoms}}{\text{Molar mass of } Fe_2O_n} \times 100\%$ Mass % of Fe = $\frac{2 \times M_{Fe}}{2 \times M_{Fe} + n \times M_O} \times 100\%$

Given that the mass percentage of Fe is 70%, we have: $70 = \frac{2 \times 56}{2 \times 56 + n \times 16} \times 100$ $70 = \frac{112}{112 + 16n} \times 100$

Divide both sides by 100: $0.70 = \frac{112}{112 + 16n}$

Rearrange the equation to solve for $n$: $0.70 \times (112 + 16n) = 112$ $0.70 \times 112 + 0.70 \times 16n = 112$ $78.4 + 11.2n = 112$

Subtract 78.4 from both sides: $11.2n = 112 - 78.4$ $11.2n = 33.6$

Divide by 11.2: $n = \frac{33.6}{11.2}$ $n = \frac{336}{112}$ $n = 3$

Thus, the value of $n$ is 3. The oxide is $Fe_2O_3$.

The mass percentage of iron in $Fe_2O_n$ is given by $(2 \times M_{Fe}) / (2 \times M_{Fe} + n \times M_O)$. Setting this equal to 70% and using atomic masses $M_{Fe}=56$ and $M_O=16$, we get the equation $0.70 = \frac{112}{112 + 16n}$. Solving this equation for $n$ yields $n=3$.