Answer
Please see the work below.
Work Step by Step
(a) We know that molar mass of $C_9H_{10}O=134.1g$
Now $\%C=\frac{9(12.0)}{134.1}\times 100\%=80.5\%$
$\%H=\frac{10(1.01)}{134.1}\times 100\%=7.5\%$
$\%O=\frac{16.0}{134.1}\%=11.9\%$
(b) We can find the number of molecules of $C_9H_{10}O$as follows
$0.469g\space C_9H_{10}O\times (\frac{1mol \space C_9H_{10}O}{134.1g\space C_9H_{10}O})\times )(6.023\times 10^{23}\frac{molecules}{mol})=2.11\times 10^{21}molecules$
