Find the molecular formula of each compound CCl , 189.83 g/mol C3H2N , 156.23 g/mol

Explanation:

The given formula is empirical formula

Let the molecular formula of first be

molecular weight = n x ( 12 + 35.5 )

= 47.5 n

Given molecular weight = 189.83 so

47.5 n = 189.83

n = 3.99 or 4 approx

Molecular formula =

= C₄ Cl₄

Let the molecular formula of second compound  be

molecular weight = n x ( 3 x 12 +2+14 )

= 52 n

Given molecular weight = 156.23  so

52 n = 156.23

n = 3.0044 or 3 approx

Molecular formula =

= C₉H₆ N₃

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