Answer
a) $C_{10}H_{15}NO$
$M=165.23\ g/mol$
b) $72.69\%$
c) $7.57\times10^{-4}\ mol$
d) $4.56\times10^{20}\ molecules$
$4.56\times10^{21}\ atoms\ of\ C$
Work Step by Step
Atomic weights (g/mol): $C: 12.011,\ H: 1.0079,\ O: 15.999,\ N: 14.007$
a) $C_{10}H_{15}ON$
$M=10\times12.011+15\times1.0079+15.999+14.007 = 165.23\ g/mol$
b) $10\times12.011/165.23\times100\%=72.69\%$
c) $0.125\ g \div 165.23\ g/mol = 7.57\times10^{-4}\ mol$
d) 0.125g: $7.57\times10^{-4}\ mol\times6.022\times10^{23}\ molecules/mol=4.56\times10^{20}\ molecules$
$4.56\times10^{20}\ molecules\times10\ atoms\ of\ C/molecule=4.56\times10^{21}\ atoms\ of\ C$
