Question

A sample of a compound contains 0.78g of K, 1.10 g of Mn, and 1.28g O. Find out its empirical formula. ( Ar: K= 39, Mn= 55, 0= 16)

Answer 1

Given :

▪ A sample of a compound contains

• 0.78g of K
• 1.1g of Mn
• 1.28g of O

To Find :

▪ Empirical formula of compound.

SoluTionN :

✴ No. of moles :

$\dashrightarrow\sf\:K=\dfrac{0.78}{39}=\bf{0.02}\\ \\ \dashrightarrow\sf\:Mn=\dfrac{1.10}{55}=\bf{0.02}\\ \\ \dashrightarrow\sf\:O=\dfrac{1.28}{16}=\bf{0.08}$

✴ Mole ratio :

$\implies\sf\:K=\dfrac{0.02}{0.02}=\bf{1}\\ \\ \implies\sf\:Mn=\dfrac{0.02}{0.02}=\bf{1}\\ \\ \implies\sf\:O=\dfrac{0.08}{0.02}=\bf{4}$

✴ Simple ratio :

$\twoheadrightarrow\sf\:K:Mn:O=1:1:4\\ \\ \therefore\:\underline{\boxed{\bf{\orange{Empirical\:formula=KMnO_4}}}}\:\gray{\bigstar}$

Additional Information :

• Empirical formula depicting constituent atom in their simplest ratio.
• Molecular formula depicting actual number of atoms in one molecule of the compound.

Answer 2

Given:

• Mass of pottasium ( K ) = 0.78 g
• Mass of Manganese ( Mn ) = 1.10 g
• Mass of oxygen ( O ) = 1.28 g

To find:

• Empirical formula of compound

Solution:

Firstly finding number of moles

Using

Number of moles = Given mass/Actual mass

→ No. of moles (K) = 0.78/39 = 0.02

→ No. of moles (Mn) = 1.10/55 = 0.02

→ No. of moles (O) = 1.28/16 = 0.08

Now , seek the lowest whole number = 0.02

→ 0.02/0.02 = 1

→ 0.02/0.02 = 1

→ 0.08/0.02 = 4

Now , ratio of atoms

K : Mn : O = 1 : 1 : 4

KMnO4

Hence , empirical formula = KMnO4