Consider the compound [tex]NH_4NO_3[/tex]. Calculate the percent by mass of each element in the compound. Type in each answer using one decimal place.

[tex]\ \textless \ br/\ \textgreater \ \begin{array}{l}\ \textless \ br/\ \textgreater \ \% N = \square \% \\\ \textless \ br/\ \textgreater \ \% H = \square \% \\\ \textless \ br/\ \textgreater \ \% O = \square \%\ \textless \ br/\ \textgreater \ \end{array}\ \textless \ br/\ \textgreater \ [/tex]



Answer :

To find the percent by mass of each element in the compound [tex]\(NH_4NO_3\)[/tex], we will follow these steps:

1. Find the molar mass of each element:

- Nitrogen (N) has a molar mass of 14.01 g/mol.
- Hydrogen (H) has a molar mass of 1.01 g/mol.
- Oxygen (O) has a molar mass of 16.00 g/mol.

2. Calculate the molar mass of [tex]\(NH_4NO_3\)[/tex]:

The compound [tex]\(NH_4NO_3\)[/tex] has the formula consisting of:
- 2 Nitrogens ([tex]\(N\)[/tex])
- 4 Hydrogens ([tex]\(H\)[/tex])
- 3 Oxygens ([tex]\(O\)[/tex])

Therefore, its molar mass is calculated as follows:
[tex]\[ \text{Molar mass of } NH_4NO_3 = (2 \times 14.01) + (4 \times 1.01) + (3 \times 16.00) \][/tex]
This simplifies to:
[tex]\[ \text{Molar mass of } NH_4NO_3 = 28.02 + 4.04 + 48.00 = 80.06 \text{ g/mol} \][/tex]

3. Calculate the percent by mass of each element in [tex]\(NH_4NO_3\)[/tex]:

- Percent by mass of Nitrogen ([tex]\(\%\text{N}\)[/tex]):
[tex]\[ \%\text{N} = \left(\frac{2 \times 14.01}{80.06}\right) \times 100 \approx 35 \% \][/tex]

- Percent by mass of Hydrogen ([tex]\(\%\text{H}\)[/tex]):
[tex]\[ \%\text{H} = \left(\frac{4 \times 1.01}{80.06}\right) \times 100 \approx 5.0 \% \][/tex]

- Percent by mass of Oxygen ([tex]\(\%\text{O}\)[/tex]):
[tex]\[ \%\text{O} = \left(\frac{3 \times 16.00}{80.06}\right) \times 100 \approx 60 \% \][/tex]

Therefore, the percent by mass of each element in [tex]\(NH_4NO_3\)[/tex] is:

[tex]\[ \begin{array}{l} \% N = 35.0 \% \\ \% H = 5.0 \% \\ \% O = 60.0 \% \end{array} \][/tex]