Question

8. The combustion analysis of 0.314 g of the hydrocarbon durene (molecular mass, 134.22 g/mol) gives the following results: 1.030 g CO2 and 0.2949 g H2O. What are the empirical and molecular formulas of durene? Empirical formula C$_a$H$_b$ Molecular formula $\frac{\%C_{CO_2} = 12}{44} \times \frac{(1.030 \times 100)}{0.314} = 89.4614\% CO_2$ $\frac{\%H_{H_2O} = 2}{18} \times \frac{(0.2949 \times 100)}{0.314} = 10.4352\% H_2O$ $CO_2 \frac{89.46\%}{12C} = 7.45512...$ $H_2O \frac{10.43\%}{1.008H} = 10.35X$ Molecular mass =$\frac{134.22}{94.15}$

          8. The combustion analysis of 0.314 g of the hydrocarbon durene (molecular mass, 134.22 g/mol) gives the following results: 1.030 g CO2 and 0.2949 g H2O. What are the empirical and molecular formulas of durene?
Empirical formula C$_a$H$_b$
Molecular formula
$\frac{\%C_{CO_2} = 12}{44} \times \frac{(1.030 \times 100)}{0.314} = 89.4614\% CO_2$
$\frac{\%H_{H_2O} = 2}{18} \times \frac{(0.2949 \times 100)}{0.314} = 10.4352\% H_2O$
$CO_2 \frac{89.46\%}{12C} = 7.45512...$
$H_2O \frac{10.43\%}{1.008H} = 10.35X$
Molecular mass =$\frac{134.22}{94.15}$

8. The combustion analysis of 0.314 g of the hydrocarbon durene (molecular mass, 134.22 g/mol) gives the following results: 1.030 g CO2 and 0.2949 g H2O. What are the empirical and molecular formulas of durene?
Empirical formula CaHb
Molecular formula
(%CCO2 = 12)/(44)×((1.030 × 100))/(0.314) = 89.4614% CO2
(%HH2O = 2)/(18)×((0.2949 × 100))/(0.314) = 10.4352% H2O
CO2 (89.46%)/(12C) = 7.45512...
H2O (10.43%)/(1.008H) = 10.35X
Molecular mass =(134.22)/(94.15)

Added by Milagros M.

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