## Chemistry 12th Edition

(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 $0.469g\space C_9H_{10}O\times (\frac{1mol \space C_9H_{10}O}{134.1g\space C_9H_{10}O})\times )(6.023\times 10^{23}\frac{molecules}{mol})=2.11\times 10^{21}molecules$