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3 years ago

8

Which expression represents the percent by mass of nitrogen in NH4NO3?

2 answers:

guajiro [1.7K]3 years ago

8 0

Answer:

This question is incomplete. The completed question is below

Which expression represents the percent by mass of nitrogen in NH4NO3?

A) 14 g N/80 g NH₄ NO₃ ₓ 100%

B) 28 g N/80 g NH₄ NO₃ ₓ 100%

C) 80 g NH₄ NO₃ /14 g N ₓ 100%

D) 80 g NH₄ NO₃ /28 g N ₓ 100%

The correct option is B

Explanation:

To calculate the percentage by mass of nitrogen in the compound, the formula will be

total mass of nitrogen/total mass of compound × 100

Atomic mass of nitrogen (N) is 14 g/mol

There are 2 nitrogen atoms present in the compound as highlighted below

<u><em>Total mass of nitrogen</em></u> (N) present in NH₄ NO₃ = 14 × 2 = 28g

Atomic mass of hydrogen (H) is 1 g/mol. There are 4 hydrogen atoms present in the compound, NH₄NO₃.

Atomic mass of oxygen (O) is 16 g/mol. There are 3 oxygen atoms present in the compound, NH₄NO₃

<u><em>Total mass of the compound</em></u> =

(14 × 2) + (1 ×4) + (16 × 3) = 80g

Hence, the percentage by mass of nitrogen in the compound (as shown in the formula earlier) is

28g of N/80g of NH₄NO₃ × 100%

podryga [215]3 years ago

7 0

W(N)={2M(N)/M(NH₄NO₃)}*100%

w(N)={2*14.0 g/mol/80.0 g/mol}*100% = 35%

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Indeed, the two samples should contain about the same number of gas particles. However, the molar mass of $\rm O_2\; (g)$ is larger than that of $\rm H_2\; (g)$ (by a factor of about $16$ .) Therefore, the mass of the $\rm O_2\; (g)$ sample is significantly larger than that of the $\rm H_2\; (g)$ sample.

Explanation:

The $\rm O_2\; (g)$ and the $\rm H_2\; (g)$ sample here are under the same pressure and temperature, and have the same volume. Indeed, if both gases are ideal, then by Avogadro's Law, the two samples would contain the same number of gas particles ( $\rm O_2\; (g)$ and $\rm H_2\; (g)$ molecules, respectively.) That is:

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In other words,

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The ratio between the mass of the $\rm O_2\; (g)$ and that of the $\rm H_2\; (g)$ sample would be:

$\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})}\end{aligned}$ .

Since $n(\mathrm{O_2}) = n(\mathrm{H}_2)$ by Avogadro's Law:

$\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})} = \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}$ .

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