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Sagot :
Answer:
3. 49.48% C, 5.19% H, 28.85% N, and 16.48% O
Explanation:
What is given?
Molar mass of C = 12 g/mol,
Molar mass of H = 1.008 g/mol,
Molar mass of N = 14 g/mol,
Molar mass of O = 16 g/mol.
Molar mass of C8H10N4O2 = (12 x 8 + 1 x 10 + 14 x 4 + 16 x 2) g/mol = 194 g/mol.
Step-by-step solution:
If we want to find the percent composition, we have to multiply the molar mass of each element by the number of atoms that we have in the molecule and divide this by the molar mass of the molecule.
- To find the % of C, we multiply 12 g/mol by the number of C, which is 8, and divide it by 194 g/mol, and multiply all this by 100 because it is a percentage:
[tex]\%\text{ C=}\frac{8\cdot12\frac{g}{mol}}{194\frac{g}{mol}}\cdot100\%=49.48\text{ }\%\text{ C.}[/tex]- To find the % of H, we do the same as we did before, but in this case, we have 10 hydrogens and its molar mass is 1.008 g/mol:
[tex]\%\text{ H=}\frac{10\cdot1.008\frac{g}{mol}}{194\frac{g}{mol}}=5.19\text{ }\%\text{ H.}[/tex]- To find the % of N we just have to do the same as we did in the last steps. We have 4 nitrogens and its molar mass is 14 g/mol:
[tex]\%\text{ N=}\frac{4\cdot14\frac{g}{mol}}{194\frac{g}{mol}}\cdot100\%=28.85\%\text{ N.}[/tex]- And finally, we do the same for % of O if we have 2 oxygens and the molar mass is 16 g/mol:
[tex]\%\text{ O=}\frac{2\cdot16\frac{g}{mol}}{194\text{ }\frac{g}{mol}}\cdot100\%=16.48\%\text{ O.}[/tex]The answer would be 3. 49.48% C, 5.19% H, 28.85% N, and 16.48% O
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