What is the percent composition by mass of nitrogen in the compound N2H4 (gram-formula mass = 32 g/mol)?

[tex]\text{\% of element} = \dfrac{\text{mass of element}}{\text{mass of compound}} \times 100 \, \% = \dfrac{\text{28}}{\text{32}} \times 100 \, \% =\mathbf{88 \, \%}\\\\\text{The percentage of N in N$_{2}$H$_{4}$ is $\large \boxed{\mathbf{88\, \% }}$}}[/tex]

Eduard22sly Eduard22sly · Answer 2 · 2021-10-11T13:38:49+02:00

The percentage composition of the nitrogen in N₂H₄ is 87.5%

The percentage composition of an element in a compound can be obtained by using the following formula:

Percentage = [tex]\frac{mass of element }{mass of compound }[/tex] × 100

With the above formula, we can obtain the percentage composition by mass of nitrogen in N₂H₄ as illustrated below:

Molar mass of N₂H₄ = 32 g/mol

Mass of N in N₂H₄ = 2N = 2 × 14 = 28 g/mol

Percentage composition of Nitrogen =?

Percentage = [tex]\frac{mass of element }{mass of compound }[/tex] × 100

Percentage of nitrogen = [tex]\frac{28}{32}[/tex] × 100

Percentage of nitrogen = 0.875 × 100

Percentage of nitrogen = 87.5%

Therefore, the percentage composition of the nitrogen in N₂H₄ is 87.5%

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