Answer
a) $M=305.41\ g/mol$
b) $1.8\times10^{-4}\ mol$
c) C: $70.79\%$, H: $8.91\%$, N: $4.59\%$, O: $15.72\%$
d) $39\ mg$
Work Step by Step
Atomic weights (g/mol): $C: 12.011,\ H:1.0079,\ N:\ 14.007,\ O: 15.999$
a) $M=18\times12.011+27\times1.0079+14.007+3\times15.999=305.41\ g/mol$
b) $55\times10^{-3}\ g \div 305.41\ g/mol =1.8\times10^{-4}\ mol$
c)
C: $18\times12.011/305.41\times100\%=70.79\%$
H: $27\times1.0079/305.41\times100\%=8.91\%$
N: $14.007/305.41\times100\%=4.59\%$
O: $3\times15.999/305.41\times100\%=15.72\%$
d) $55\ mg\ caps.\times\frac{70.79}{100}\ mg\ C/\ mg\ caps. = 39\ mg$
