Question

its urgent............​

Answer 1

Answer:

$\bigstar$ Given:

An Iron Oxide has,

⇒ 69.9% iron by mass

⇒ 30.1% oxygen by mass

$\bigstar$ To Find:

⇒ Determine the empirical formula of the oxide of iron

$\bigstar$ Solution:

As the composition is given in percentage,

Therefore,

100 g of iron oxide contains 69.9 g iron and 30.1 g oxygen

⇒ Calculating the number of moles of iron in 100g of iron oxide

Therefore, number of moles of iron present in 100 g of iron oxide are

$\rm \implies \dfrac{69.9}{55.8} = 1.25$

⇒ Calculating the number of moles of oxygen in 100g of iron oxide

Therefore, number of moles of oxygen present in 100 g of iron oxide are

$\rm \implies \dfrac{30.1}{16} = 1.88$

For the mole ratio, divide each value of moles by the smallest number of moles calculated.

→ (dividing the moles by 1.25)

Simplest molar ratio of Fe to O

We get,

1 : 1.5

For a whole ratio

→ (multiply with 2)

We get,

2 : 3

Therefore, Empirical formula of iron oxide is Fe₂O₃