Question

Determine the molecular formula of an oxide of iron, in which the mass percent of iron $\&$ oxygen are $69.9\&30.1$ respectively.

Answer 1

Molar mass of iron oxide:

Description

Given: 𝑚𝑎𝑠𝑠 $\%$ 𝑜𝑓 𝐹𝑒 𝑖𝑛 𝑖𝑟𝑜𝑛 𝑜𝑥𝑖𝑑𝑒$=69.1\%$

𝑚𝑎𝑠𝑠$\%$ 𝑜𝑓 𝑜𝑥𝑦𝑔𝑒𝑛 𝑖𝑛 𝑖𝑟𝑜𝑛 𝑜𝑥𝑖𝑑𝑒$=30.1\%$

Let the 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑖𝑟𝑜𝑛 𝑜𝑥𝑖𝑑𝑒$=100g$

𝑚𝑎𝑠𝑠 $\%$ 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡$=\frac{\text{massofelement}}{\text{massofcompound}}\times 100$

$\therefore \text{massof}Fe=\frac{\text{massofcompound (ironoxide}\times \text{mass \%}}{100}$

$\text{massof}Fe=\frac{100\times 69.1}{100}=69.1g$

Similarly 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑜𝑥𝑦𝑔𝑒𝑛$=\frac{30.1\times 100}{100}=30.1g$

$\therefore$ we know, 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 $Fe=55.845gmo{l}^{–1}$

𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑜𝑥𝑦𝑔𝑒𝑛$=16gmo{l}^{–1}$

$\&$ Number of 𝑚𝑜𝑙𝑒𝑠$=\frac{\text{Massofsubstance}}{\text{Molarmassof substance}}$

$\therefore$Number of 𝑚𝑜𝑙𝑒 𝑜𝑓 $Fe=\frac{69.9}{58.845}=1.25$

Number of 𝑚𝑜𝑙𝑒 𝑜𝑓 𝑜𝑥𝑦𝑔𝑒𝑛$=\frac{30.1}{16}=1.88$

Now,
Elements $Fe$ $O$

Number of Moles $1.25$ $1.88$ $\left(\text{for simplifying divide both by 1.25}\right)$

Ratio of moles $1$ $1.5$ $\left(\text{It is not whole number so multiply both by 2}\right)$

Simple ratio $2$ $3$

Hence, Molecular formula of Iron oxide $\Rightarrow F{e}_2{O}_3$

Final answer: molecular formula of Iron oxide is $F{e}_2{O}_3$