Answer:
Mass percentage of NO in nitroglycerin is 39.63%.
Mass percent of NO in isoamyl nitrate is 22.55\%.
Step-by-step explanation:
Molar mas of nitroglycerin ,
![C_3H_5N_3O_9=227.08 g/mol](https://img.qammunity.org/2020/formulas/chemistry/college/4hyewutvvplmhmyamjfkict8z435k57mx0.png)
Molar mas of isoamyl nitrate,
![C_5H_(11)NO_3=117.15 g/mol](https://img.qammunity.org/2020/formulas/chemistry/college/q8mjqks7hcm38fo0x884jb8k76vlpdi581.png)
Given: If each compound releases one molecule of NO per atom of N.
1)In nitroglycerin, there are 3 nitrogen atoms. then number of NO will be :
3 × 1= 3 NO molecules
Mass percent of NO in
![C_3H_5N_3O_9](https://img.qammunity.org/2020/formulas/chemistry/college/c2fcjh0udbo2qz2nty3w3gncvutt5rfmvq.png)
![\% NO=(3* (14 g/mol+16 g/mol))/(227.08 g/mol)=39.63\%](https://img.qammunity.org/2020/formulas/chemistry/college/umujgz03i1ds3tuo1cawzigo2l47gku8s9.png)
2)In isoamyl nitrate. , there are 1 nitrogen atom. then number of NO will be :
1 × 1= 3 NO molecules
Mass percent of NO in
![C_3H_5N_3O_9](https://img.qammunity.org/2020/formulas/chemistry/college/c2fcjh0udbo2qz2nty3w3gncvutt5rfmvq.png)
![\% NO=(1* (14 g/mol+16 g/mol))/(133 g/mol)=22.55\%](https://img.qammunity.org/2020/formulas/chemistry/college/f9qskww2uv5b2utbd0fzm3ugujvfcnwrgi.png)