A compound of C, H and Ncontains 66.70%carbon,7.41%hydrogen and 25.90%

1.	A compound of C, H and Ncontains 66.70%carbon,7.41%hydrogen and 25.90%nitrogen.The molecular mass of the compound was found to be 108Find the empirical and molecular formula of the compound
Answer» Answer: Molecular formula: $\\\tt C_{6}H_{<klux>8</klux>}N_{2}$ Empirical formula: $\\\tt C_{3}H_{4}N_{1}$ Logic: FIND mass of each element $\\\tt \longrightarrow$ Find number of atoms of each element in the COMPOUND Calculation: Given mass of molecule = 108amu. Mass of C = 66.70% of Mass of compound = 72.036 amu ≈ 72amu Mass of H= 7.41% of Mass of compound = 8.0028 amu ≈ 8 amu Mass of N= 25.90% of Mass of compound = 27.972 amu ≈ 28 amu No of atoms of C in compound = 72/12 = 6 No of atoms of H in compound = 8/1= 8 No of atoms of N in compound = 28 /14 = 2 Therefore the molecular formula must be, $\\\tt \longdownarrow$ $\\\tt C_{6}H_{8}N_{2}$ Therefore empirical formula $\\\tt C_{3}H_{4}N_{1}$

A compound of C, H and Ncontains 66.70%carbon,7.41%hydrogen and 25.90%nitrogen.The molecular mass of the compound was found to be 108Find the empirical and molecular formula of the compound

Answer»

Molecular formula: $\\\tt C_{6}H_{<klux>8</klux>}N_{2}$

Empirical formula: $\\\tt C_{3}H_{4}N_{1}$

FIND mass of each element $\\\tt \longrightarrow$ Find number of atoms of each element in the COMPOUND

Calculation:

Given mass of molecule = 108amu.

Mass of C = 66.70% of Mass of compound = 72.036 amu ≈ 72amu

Mass of H= 7.41% of Mass of compound = 8.0028 amu ≈ 8 amu

Mass of N= 25.90% of Mass of compound = 27.972 amu ≈ 28 amu

No of atoms of C in compound = 72/12 = 6

No of atoms of H in compound = 8/1= 8

No of atoms of N in compound = 28 /14 = 2

Therefore the molecular formula must be,

$\\\tt \longdownarrow$ $\\\tt C_{6}H_{8}N_{2}$

Therefore empirical formula $\\\tt C_{3}H_{4}N_{1}$