Question

When 5.467 grams of compound z are burned in excess oxygen, 15.02 grams of co2 and 2.458 grams of h2o are produced. determine its empirical formula?

Answer 1

Given the mass of CO2 =15.02 g

Moles of $CO_{2}$=$15.02 gCO_{2} *\frac{1 molCO_{2} }{44.01 gCO_{2} } =0.341 mol CO_{2}$

Mass of H2O = 2.458 g

Moles of $H_{2}O$=$2.458 g H_{2}O*\frac{1molH_{2}O }{18.02g H_{2}O } =0.136molH_{2}O$

Moles of C = $0.341 mol CO_{2}*\frac{1 molC}{1 molCO_{2} } =0.341molC$

Moles of H = $0.136 mol H_{2}O * \frac{2 mol H}{1 mol H_{2}O } =0.272 mol H$

Mass of C in the sample = $0.341 mol C*\frac{12.01g C}{1 mol C} = 4.095 g C$

Mass of H = $0.272 mol H *\frac{1.01 g H}{1 molH}$=0.275 gH

Mass of O in the sample = 5.467 g - (4.095 g +0.275 g) = 1.097 g O

Moles of O = $1.097 g O * \frac{1 molO}{16 g O} =0.0686 mol O$

Simplest mole ratios of the elements in the compound:

$Cx_{\frac{0.341mol}{0.0686mol}} H_{\frac{0.272mol}{0.0686mol} }O_{\frac{0.0686mol}{0.0686mol} }$

Therefore the empirical formula of the compound is $C_{5}H_{4}O$