Asked by Alex
When 1.63 g sample of an unknown hydrocarbon is completely burned in oxygen, 4.97 g of CO2 are produced. What is the percentage by mass of hydrogen in this hydrocarbon?
THANK YOUU
Found out the formula:
4CH+5O2--> 4CO2+2H2O
then found out the moles of CH= 0.1251
I don't know where to go from here.
THANK YOUU
Found out the formula:
4CH+5O2--> 4CO2+2H2O
then found out the moles of CH= 0.1251
I don't know where to go from here.
Answers
Answered by
DrBob222
What's 4CH? 4CH doesn't make sense to me. The hydrocarbon must be CxHy.
Convert CO2 to mols.
mols CO2 = g/molar mass = 4.97/44 = about 0.11 but you need to do it more accurately. Now convert mols CO2 to grams C. That is mols C x atomic mass C = 0.11 x 12 = about 1.3
%C = (mass C/mass sample)*100 = about (1.3/1.63)*100 = about 83%
So %H must be 100 - %C = ?
Convert CO2 to mols.
mols CO2 = g/molar mass = 4.97/44 = about 0.11 but you need to do it more accurately. Now convert mols CO2 to grams C. That is mols C x atomic mass C = 0.11 x 12 = about 1.3
%C = (mass C/mass sample)*100 = about (1.3/1.63)*100 = about 83%
So %H must be 100 - %C = ?
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