Question

• A compound is 21.20%. Nitrogen, 6.06 percent of hydrogen,
24.30% sulpur and 48.45 percent oxygen. Write the
Empirical formula for the compound.​

Answer 1

Solution :-

• % N = 21.02

• %H = 6.06

• %S = 24.30

• %O = 24.30

Relative no of atoms

Divide the % composition of elements with their respective atomic masses

• Atomic mass of N = 14

• Atomic mass of H = 1

• Atomic mass of S = 32

• Atomic mass of O = 16

⇒ N = 21.20 / 14 = 1.5

⇒ H = 6.06 / 1 = 6.06

⇒ S = 24.3 / 32 = 0.75

⇒ O = 48.45 /16 = 3.02

Simple Ratio

Now divide all the values obtained above with the least no in order to get the simple ratio

⇒ N = 1.5 /0.75= 2

⇒ H = 6.06 /0.75= 8

⇒ S  = 0.75/ 0.75= 1

⇒ O  = 3.02/ 0.75= 4

Empirical formula = N2 H8 S O4

The empirical formula of the compound is  N2 H8 S O4