Question

Combustion analysis of a hydrocarbon produced 33.01g of CO2

and 5.41 g of H2O

Answer 1

Moles of $CO_2=\frac{33.01} {44.01 }=0.75$

Moles of $H_2O=\frac{5.41} {18.01528}=0.3$

Raito $\frac{0.75}{0.3}=2.5$ $\frac{0.3}{0.3}=1$

$C_2H_5$