Solution (1)
Step 1: Calculate the molar mass of CH₄.
molar mass = (12.01 g/mol × 1) + (1.008 g/mol × 4)
molar mass = 16.042 g/mol
Step 2: Calculate the percent by mass of CH₄.
For %C
[tex]\text{\%C} = \frac{\text{12.01 g/mol × 1}}{\text{16.042 g/mol}} × 100[/tex]
[tex]\boxed{\text{\%C = 74.87\%}}[/tex]
For %H
[tex]\text{\%H} = \frac{\text{1.008 g/mol × 4}}{\text{16.042 g/mol}} × 100[/tex]
[tex]\boxed{\text{\%H = 25.13\%}}[/tex]
Solution (2)
Step 1: Calculate the molar mass of NaNO₃.
molar mass = (23.0 g/mol × 1) + (14.0 g/mol × 1) + (16.0 g/mol × 3)
molar mass = 85.0 g/mol
Step 2: Calculate the percent by mass of NaNO₃.
For %Na
[tex]\text{\%Na} = \frac{\text{23.0g/mol × 1}}{\text{85.0 g/mol}} × 100[/tex]
[tex]\boxed{\text{\%Na = 27.06\%}}[/tex]
For %N
[tex]\text{\%N} = \frac{\text{14.0 g/mol × 1}}{\text{85.0 g/mol}} × 100[/tex]
[tex]\boxed{\text{\%N = 16.47\%}}[/tex]
For %O
[tex]\text{\%O} = \frac{\text{16.0 g/mol × 3}}{\text{85 g/mol}} × 100[/tex]
[tex]\boxed{\text{\%O = 56.47\%}}[/tex]
#CarryOnLearning
