Question

Determine the empirical formula of an oxide of iron, which has $69.9\%$ iron and $30.1\%$ dioxygen by mass.

Answer 1

Step 1: Convert the mass percent into gram:

Given: 𝑚𝑎𝑠𝑠 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑜𝑓 𝑖𝑟𝑜𝑛 $=69.9\%$

𝑀𝑎𝑠𝑠 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑜𝑓 𝑜𝑥𝑦𝑔𝑒𝑛 $=30.1\%$

Let the mass of compound is $100g$

We know, mass percent of compound $=\frac{\text{massofthatelementinthecompound}}{\text{molarmassofthecompound}}\times 100$

So, mass of iron $=69.9g$ and mass of oxygen $=30.1g$

Step 2 : Calculation of number of moles

Description:

We know, 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒$=\frac{\text{Weightof element}}{\text{Molar mass ofelement}}$

$(Mole{)}_{iron}=\frac{69.9}{55.8}=1.25$

$(Mole{)}_{oxygen}=\frac{30.1}{16}=1.88$

Step 3: Calculate the simple molar ratio

Description:

To calculate a simple molar ratio, we divided by least value of mole.

$({\text{simplest molar ratio)}}_{iron}=\frac{1.25}{1.25}=1and$

$(\text{simplest molar ratio}{)}_{oxygen}=\frac{1.88}{1.25}=1.5$

Step 4: Calculate the simplest whole number ratio

Description

Convert the simplest molar ratio to the nearest whole number by multiplying the simplest atomic ratio by a suitable integer.

$Fe$ $O$

$1\times 2$ $1.5\times 2$

$2$ $3$

Step 5 : Calculate the simplest whole number ratio

Description

Insert numerical value of the simplest whole number.

According to the following steps, for iron and oxygen is solved in table:

$\begin{array}{ccccccc}Element& Symbol& Element& \text{Atomic mass of element}& \text{Mole of the element}& \text{Simple molar ratio}& \text{Simple whole ratio}\\ Iron& Fe& 69.9& 69.9& \frac{69.9}{55.8}=1.25& \frac{1.25}{1.25}=1& 69.9\\ Oxygen& O& 30.1& 30.1& \frac{30.1}{16}=1.88& \frac{1.88}{1.25}=1.5& 30.1\end{array}$

Hence, the empirical formula is $F{e}_2{O}_3$

Final answer: The empirical formula of an oxide of iron is $F{e}_2{O}_3$