Answer :
To find the percentage composition of each element in dinitrogen monoxide ([tex]\(\text{N}_2\text{O}\)[/tex]), follow these steps:
1. Determine the molar masses of each element:
- The molar mass of Nitrogen (N) is [tex]\(14.01 \, \text{g/mol}\)[/tex].
- The molar mass of Oxygen (O) is [tex]\(16.00 \, \text{g/mol}\)[/tex].
2. Calculate the molar mass of dinitrogen monoxide ([tex]\(\text{N}_2\text{O}\)[/tex]):
- The formula [tex]\(\text{N}_2\text{O}\)[/tex] indicates that there are 2 Nitrogen atoms and 1 Oxygen atom.
- Calculate the total molar mass of [tex]\(\text{N}_2\text{O}\)[/tex]:
[tex]\[ \text{Molar mass of } \text{N}_2\text{O} = (2 \times 14.01 \, \text{g/mol}) + (1 \times 16.00 \, \text{g/mol}) \][/tex]
[tex]\[ \text{Molar mass of } \text{N}_2\text{O} = 28.02 \, \text{g/mol} + 16.00 \, \text{g/mol} \][/tex]
[tex]\[ \text{Molar mass of } \text{N}_2\text{O} = 44.02 \, \text{g/mol} \][/tex]
3. Calculate the percentage composition of Nitrogen (N):
- The contribution of nitrogen to the molar mass of [tex]\(\text{N}_2\text{O}\)[/tex] is from 2 Nitrogen atoms, so:
[tex]\[ \text{Mass of Nitrogen in } \text{N}_2\text{O} = 2 \times 14.01 \, \text{g} \][/tex]
[tex]\[ \text{Percentage of Nitrogen} = \left( \frac{2 \times 14.01}{44.02} \right) \times 100 \][/tex]
[tex]\[ \text{Percentage of Nitrogen} \approx 63.65\% \][/tex]
4. Calculate the percentage composition of Oxygen (O):
- The contribution of oxygen to the molar mass of [tex]\(\text{N}_2\text{O}\)[/tex] is from 1 Oxygen atom, so:
[tex]\[ \text{Mass of Oxygen in } \text{N}_2\text{O} = 16.00 \, \text{g} \][/tex]
[tex]\[ \text{Percentage of Oxygen} = \left( \frac{16.00}{44.02} \right) \times 100 \][/tex]
[tex]\[ \text{Percentage of Oxygen} \approx 36.35\% \][/tex]
Therefore, the percentage composition of dinitrogen monoxide [tex]\((\text{N}_2\text{O})\)[/tex] is approximately:
[tex]\[ \text{63.65}\% \text{ Nitrogen (N)} ; \text{36.35}\% \text{ Oxygen (O)} \][/tex]
Matching this to the given choices, we select:
[tex]\[ \boxed{63.64\% \, \text{N} ; 36.36\% \, \text{O}} \][/tex]
1. Determine the molar masses of each element:
- The molar mass of Nitrogen (N) is [tex]\(14.01 \, \text{g/mol}\)[/tex].
- The molar mass of Oxygen (O) is [tex]\(16.00 \, \text{g/mol}\)[/tex].
2. Calculate the molar mass of dinitrogen monoxide ([tex]\(\text{N}_2\text{O}\)[/tex]):
- The formula [tex]\(\text{N}_2\text{O}\)[/tex] indicates that there are 2 Nitrogen atoms and 1 Oxygen atom.
- Calculate the total molar mass of [tex]\(\text{N}_2\text{O}\)[/tex]:
[tex]\[ \text{Molar mass of } \text{N}_2\text{O} = (2 \times 14.01 \, \text{g/mol}) + (1 \times 16.00 \, \text{g/mol}) \][/tex]
[tex]\[ \text{Molar mass of } \text{N}_2\text{O} = 28.02 \, \text{g/mol} + 16.00 \, \text{g/mol} \][/tex]
[tex]\[ \text{Molar mass of } \text{N}_2\text{O} = 44.02 \, \text{g/mol} \][/tex]
3. Calculate the percentage composition of Nitrogen (N):
- The contribution of nitrogen to the molar mass of [tex]\(\text{N}_2\text{O}\)[/tex] is from 2 Nitrogen atoms, so:
[tex]\[ \text{Mass of Nitrogen in } \text{N}_2\text{O} = 2 \times 14.01 \, \text{g} \][/tex]
[tex]\[ \text{Percentage of Nitrogen} = \left( \frac{2 \times 14.01}{44.02} \right) \times 100 \][/tex]
[tex]\[ \text{Percentage of Nitrogen} \approx 63.65\% \][/tex]
4. Calculate the percentage composition of Oxygen (O):
- The contribution of oxygen to the molar mass of [tex]\(\text{N}_2\text{O}\)[/tex] is from 1 Oxygen atom, so:
[tex]\[ \text{Mass of Oxygen in } \text{N}_2\text{O} = 16.00 \, \text{g} \][/tex]
[tex]\[ \text{Percentage of Oxygen} = \left( \frac{16.00}{44.02} \right) \times 100 \][/tex]
[tex]\[ \text{Percentage of Oxygen} \approx 36.35\% \][/tex]
Therefore, the percentage composition of dinitrogen monoxide [tex]\((\text{N}_2\text{O})\)[/tex] is approximately:
[tex]\[ \text{63.65}\% \text{ Nitrogen (N)} ; \text{36.35}\% \text{ Oxygen (O)} \][/tex]
Matching this to the given choices, we select:
[tex]\[ \boxed{63.64\% \, \text{N} ; 36.36\% \, \text{O}} \][/tex]
