Answer:
A. 53.3% N, 46.7% O
Step-by-step explanation:
1st) It is necessary to add both quantities (10.48g of nitrogen and 11.96g of oxygen) to obtain the total mass of the sample, which represents 100%:
10.48g + 11.96g = 22.44g
22.44g represents the 100% of the sample.
2nd) Now we can calculate the percentage composition of the gas, using a mathematical rule of three and the amount of nitrogen and oxygen:
• Nitrogen:
![\begin{gathered} 22.44g-100\% \\ 10.48g-x=(10.48g*100\%)/(22.44g) \\ x=46.7\% \end{gathered}](https://img.qammunity.org/2023/formulas/chemistry/college/48y82vzx9mj4oojsda5e823vdqzyrt646x.png)
• Oxygen:
![\begin{gathered} 22.44g-100\% \\ 11.96g-x=(11.96g*100\%)/(22.44g) \\ x=53.3\% \end{gathered}](https://img.qammunity.org/2023/formulas/chemistry/college/ypojw443im5lhn5q23p5fzq1kujiobjsvi.png)
We can also calculate the oxygen percentage by subtracting 100% minus the nitrogen percentage (46.7%):
100% - 4607% = 53.3%
So, the percentage composition is 53.3% N and 46.7% O.