Question

3. A compound of C, H and N contains 66.70% carbon, 7.41% hydrogen and
25.90% Nitrogen. The Molecular mass of the compound was found
to be 108.
(i) Find the empirical formula of the compound. [Ans: CHAN]
(ii) Find the Molecular formula of the compound. (Ans: CH3N2]
Determin solve its with step by step .​

Answer 1

Answer:

for carbon

12x/12x+y+14z=66.70 where x y z are number of atoms

similarly

x/12x+......=7.41

14z/12x+.....=25.9

also given that 12x+y+14z=108

so just replace the denominator by 108

so solve these equations u will gt ur answer

Answer 2

Given :-

• %C = 66.7
• %H = 7.41

• %N = 25.9

Relative no of atoms

Simply divide the percentage composition of elements with their atomic mass

• Atomic mass of C = 12 g

• Atomic mass of H = 1 g

• Atomic mass of N = 14 g

Carbon

C = 66.7 / 12 = 5.55

Hydrogen

H = 7.41 / 1 = 7.41

Nitrogen

N = 25.9 / 14 = 1.85

Simple ratio

Now let's divide the above values obtained with the least value ( i.e. 1.85) in this case

Carbon

⇛ C  = 5.55 / 1.85 = 3

Hydrogen

⇛ H  = 7.41 / 1.85 = 4

Nitrogen

⇛ N = 1.85 / 1.85 = 1

• Empirical formula = C3 H4 N

• Empirical mass = 3(C) + 4(H) + N =  3(12) + 4(1) + 14 = 36 + 4 + 14 = 54 g

• Molecular mass = 108 g

we know that ,

⇛ Molecular mass = n x Empirical mass

⇛ 108 = n x 54

⇛ n = 108 / 54

⇛ n = 2

Similarly ,

⇛ Molecular formula = n x Empirical formula

⇛ Molecular formula = 2 ( C3 H4 N )

⇛ Molecular formula = C6 H 8 N2