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) *(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*(1molH_(2)O )/(18.02g H_(2)O ) =0.136molH_(2)O`

Moles of C =
`0.341 mol CO_(2)*(1 molC)/(1 molCO_(2) )  =0.341molC`

Moles of H =
`0.136 mol H_(2)O * (2 mol H)/(1 mol H_(2)O ) =0.272 mol H`

Mass of C in the sample =
`0.341 mol C*(12.01g C)/(1 mol C) =  4.095 g C`

Mass of H =
`0.272 mol H *(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 * (1 molO)/(16 g O) =0.0686 mol O`

Simplest mole ratios of the elements in the compound:

`Cx_{(0.341mol)/(0.0686mol)} H_{(0.272mol)/(0.0686mol) }O_{(0.0686mol)/(0.0686mol) }`

Therefore the empirical formula of the compound is
`C_(5)H_(4)O`