Solution:
Step 1: Calculate the molar mass of C₆H₅NH₂.
molar mass = (12.01 g/mol × 6) + (1.008 g/mol × 7) + (14.01 g/mol × 1)
molar mass = 93.126 g/mol
Step 2: Calculate the percent composition of C₆H₅NH₂.
General Formula:
[tex]\text{percent composition} = \frac{\text{atomic mass of an element × number of atoms in a compound}}{\text{molar mass of compound}} × 100[/tex]
For %C
[tex]\text{\%C} = \frac{\text{12.01 g/mol × 6}}{\text{93.126 g/mol}} × 100[/tex]
[tex]\boxed{\text{\%C = 77.38\%}}[/tex]
For %H
[tex]\text{\%H} = \frac{\text{1.008 g/mol × 7}}{\text{93.126 g/mol}} × 100[/tex]
[tex]\boxed{\text{\%H = 7.58\%}}[/tex]
For %N
[tex]\text{\%N} = \frac{\text{14.01 g/mol × 1}}{\text{93.126 g/mol}} × 100[/tex]
[tex]\boxed{\text{\%N = 15.04\%}}[/tex]
#CarryOnLearning
