The easy way to solve this is to write two equations, solve them simultaneously, AND use chemical factors to write the equations. Schools don't teach chemical factors now (at least I'm not aware of any that do) but here is what you do.
Let X = weight ethanol
Let Y = weight ether
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X + Y = 1.005 is the first equation.
Now convert X to grams CO2 and Y to grams CO2 since you know the total CO2 is 1.953 grams.
X(2*molar mass CO2/molar mass EtOH) + Y(4*molar mass CO2/molar mass ether) = 1.963
Solve the simultaneously for Y, then
%ether = (Yin grams/mass sample which is 1.005)*100 = ?
The manufacture of ethanol (CH3CH2OH) yields diethyl ether ((C2H5)2O) as a by product. the complete combustion of a 1.005 g sample of the manufactured product yielded 1.963 g of carbon dioxide. what are the weight percents of ethanol and diethyl ether in the sample?
1 answer
1 answer