Answer:
74.07% is the percent by mass of carbon in \(C_{10}H_{14}N_2\) .
Explanation:
Molar mass of compound\(C_{10}H_{14}N_2\) = M
\(M=10\times 12 g/mol+14\times 1g/mol+2\times 14 g/mol= 162g/mol\)
Number of carbon atom = 10
Atomic mass of carbon = 12 g/mol
Percentage of element in compound :
\(=\frac{\text{Number of atoms}\times \text{Atomic mass}}{\text{molar mas of compound}}\times 100\)
Mass percentage of Carbon :
\(=\frac{10\times 12g/mol}{162 g/mol}\times 100=74.07\%\)
74.07% is the percent by mass of carbon in \(C_{10}H_{14}N_2\) .
