Question

What is the percent by mass of carbon in c10h14n2

Answer 1

`M_{C_(10)H_(14)N_(2)}=120+14+28=162(g)/(mol)\\\\ \%C=(120*100\%)/(162)\approx74,1\% \\\\ \%H=(14*100\%)/(162)\approx 8,7\% \\\\ \%N=100\%-74,1\%-8,7\%=17,2\%`

Answer 2

74.07% is the percent by mass of carbon in
`C_(10)H_(14)N_2` .

Step-by-step explanation:

Molar mass of compound
`C_(10)H_(14)N_2` = M

`M=10* 12 g/mol+14* 1g/mol+2* 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}* \text{Atomic mass}}{\text{molar mas of compound}}* 100`

Mass percentage of Carbon :

`=(10* 12g/mol)/(162 g/mol)* 100=74.07\%`

74.07% is the percent by mass of carbon in
`C_(10)H_(14)N_2` .