Determine the molecular formula of an oxide of iron in which the mass percent of iron and oxygen are 69.9 and 30.1 respectively. Given that the molar mass of the oxide is 159.69g mol- 1

The correct option/answer is Fe₂O₃.

To determine the molecular formula of the oxide of iron, we first find its empirical formula using the mass percentages of the constituent elements, iron (Fe) and oxygen (O).

Step 1: Calculate the relative number of moles of each element
Let us assume we have a $100\hspace{0.17em}g$ sample of the iron oxide.
This means:
Mass of Iron ($\mathrm{Fe}$) = $69.9\hspace{0.17em}g$
Mass of Oxygen ($O$) = $30.1\hspace{0.17em}g$

Using the atomic masses of iron ($\mathrm{Fe}\approx 55.85\hspace{0.17em}g/\mathrm{mol}$) and oxygen ($O\approx 16.00\hspace{0.17em}g/\mathrm{mol}$), we calculate the number of moles:

Moles of $\mathrm{Fe}$ = $\frac{69.9}{55.85}\approx 1.25\hspace{0.17em}\mathrm{mol}$

Moles of $O$ = $\frac{30.1}{16.00}\approx 1.88\hspace{0.17em}\mathrm{mol}$

Step 2: Determine the simplest molar ratio
To find the simplest ratio, we divide the number of moles of each element by the smallest number of moles obtained (which is $1.25$):

Ratio for $\mathrm{Fe}$ = $\frac{1.25}{1.25}=1$

Ratio for $O$ = $\frac{1.88}{1.25}\approx 1.5$

Since we cannot have fractional atoms in a chemical formula, we multiply both ratios by $2$ to obtain the simplest whole-number ratio:
For $\mathrm{Fe}$: $1\times 2=2$
For $O$: $1.5\times 2=3$

Thus, the empirical formula is Fe₂O₃.

Step 3: Determine the molecular formula
Next, we calculate the empirical formula mass of Fe₂O₃:

Empirical formula mass = $\left(2\times 55.85\right)+\left(3\times 16.00\right)=111.7+48.0=159.7\hspace{0.17em}g/\mathrm{mol}$

Now, we find the multiplier $n$:

$n=\frac{\text{Molar Mass}}{\text{Empirical Formula Mass}}=\frac{159.69}{159.7}\approx 1$

Since $n=1$, the molecular formula is identical to the empirical formula.
Therefore, the molecular formula of the iron oxide is Fe₂O₃.