## Chemistry (7th Edition)

(a) 1. Calculate the number of moles of $C_9H_8O_4$: 12.01* 9 + 1.008* 8 + 16.00* 4 = 180.154g/mol $10.0g \times \frac{1 mol}{ 180.154g} = 0.0555mol (C_9H_8O_4)$ - The ratio of $C_9H_8O_4$ to $C_7H_6O_3$ is 1 to 1: $0.0555 mol (C_9H_8O_4) \times \frac{ 1 mol(C_7H_6O_3)}{ 1 mol (C_9H_8O_4)} = 0.0555mol (C_7H_6O_3)$ 2. Calculate the mass of $C_7H_6O_3$: 12.01* 7 + 1.008* 6 + 16* 3 = 138.118g/mol $0.0555 mol \times \frac{ 138.118 g}{ 1 mol} = 7.67g (C_7H_6O_3)$ (b) 1. Calculate the number of moles of $C_9H_8O_4$: - We have already calculated: $0.0555$ mol. - The ratio of $C_9H_8O_4$ to $CH_3CO_2H$ is 1 to 1: $0.0555 mol (C_9H_8O_4) \times \frac{ 1 mol(CH_3CO_2H)}{ 1 mol (C_9H_8O_4)} = 0.0555mol (CH_3CO_2H)$ 2. Calculate the mass of $CH_3CO_2H$: 12.01* 1 + 1.008* 3 + 12.01* 1 + 16* 2 + 1.008* 1 = 60.052g/mol $0.0555 mol \times \frac{ 60.052 g}{ 1 mol} = 3.33g (CH_3CO_2H)$