Question

A sample of a compound was found to contain 70% iron and 30% oxygen. Determine the empirical formula. (Fe=56; O=16)

Answer 1

Answer :-

Empirical formula of the compund is Fe₂O₃ .

Explanation :-

For Iron :-

• Percentage = 70 %

• Atomic mass = 56

Relative number of moles :-

= 70/56

= 1.25

For Oxygen :-

• Percentage = 30 %

• Atomic mass = 16

Relative number of moles :-

= 30/16

= 1.875

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Now, we will divide 1.25 and 1.875 with 1.25 [as it is smallest] to get a simple ratio of the moles.

→ Simple ratio (Iron) = 1.25/1.25 = 1

→ Simple ratio (Oxygen) = 1.875/1.25 = 1.5

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For calculation of empirical formula, we require simplest whole ratio . So, we would multiply 1 and 1.5 with 2 to obtain it.

→ Simplest ratio (Iron) = 2 × 1 = 2

→ Simplest ratio (Oxygen) = 2 × 1.5 = 3

Thus, empirical formula of the compound is Fe₂O₃ .

Answer 2

Answer:

Given :-

• A sample of a compound was found to contain 70% iron and 30% oxygen.

To Find :-

• What is the empirical formula.

Solution :-

First, we have to find the relative number of moles of iron and oxygen :

${\normalsize{\bold{\purple{\underline{\bigstar\: In\: case\: of\: iron\: :-}}}}}$

Given :-

• Percentage of iron = 70%
• Atomic mass of iron (Fe) = 56

Then,

$\leadsto \sf\bold{\pink{Relative\: number\: of\: Moles =\: \dfrac{Percentage}{Atomic\: mass}}}$

According to the question by using the formula we get,

$\implies \sf Relative\: number\: of\: Moles =\: \dfrac{70\%}{56}$

$\implies \sf\bold{\green{Relative\: number\: of\: Moles =\: 1.25\%}}$

${\normalsize{\bold{\purple{\underline{\bigstar\: In\: case\: of\: oxygen\: :-}}}}}$

Given :

• Percentage of oxygen = 30%
• Atomic mass of oxygen (O) = 16

According to the question by using the formula we get,

$\implies \sf Relative\: number\: of\: Moles =\: \dfrac{30\%}{16}$

$\implies \sf\bold{\green{Relative\: number \: of\: Moles =\: 1.875\%}}$

Now, we have to divide relative number of moles of iron and oxygen with the smallest number to find simple ratio :

${\normalsize{\bold{\purple{\underline{\bigstar\: In\: case\: of\: iron\: :-}}}}}$

$\implies \sf Simple\: ratio\: of\: iron =\: \dfrac{\cancel{1.25}}{\cancel{1.25}}$

$\implies \sf Simple\: ratio\: of\: iron =\: \dfrac{1}{1}$

$\implies \sf\bold{\green{Simple \: ratio\: of\: iron =\: 1}}$

${\normalsize{\bold{\purple{\underline{\bigstar\: In\: case\: of\: oxygen\: :-}}}}}$

$\implies \sf Simple\: ratio\: of\: oxygen =\: \dfrac{1.875}{1.25}$

$\implies \sf Simple\: ratio\: of\: oxygen =\: \dfrac{\dfrac{1875}{1000}}{\dfrac{125}{100}}$

$\implies \sf Simple\: ratio\: of\: oxygen =\: \dfrac{1875}{1000} \times \dfrac{100}{125}$

$\implies \sf Simple\: ratio\: of\: oxygen =\: \dfrac{1875\cancel{00}}{1250\cancel{00}}$

$\implies \sf Simple\: ratio\: of\: oxygen =\: \dfrac{\cancel{1875}}{\cancel{1250}}$

$\implies \sf \bold{\green{Simple\: ratio\: of\: oxygen =\: 1.5}}$

Now, we have to find the simple ratio of iron to oxygen :

$\implies \sf 1 : 1.5$

$\implies \sf \bold{\red{2 : 3}}$

$\therefore$ The empirical formula of the iron oxide is fe₂O₃.