Asked by Ava
I completed this question like you told me to. Are the answers correct? Is my method for finding the molecular formula correct?
2.4g of a compound of carbon, hydrogen and oxygen gave on combustion, 3.52g of CO2 and 1.44g of H2O. The relative molecular mass of the compound was found to be 60.
a)What are the masses of carbon, hydrogen and oxygen in 2.4g of the compound?
b)What are the emperical and molecular formulae of the compound?
Ans:
a) 3.52g CO2 x (atomic mass C/molar mass CO2) = 0.96g C.
1.44g H2O x (2*atomic mass H/molar mass H2O)= 0.16g H.
g of O = 2.4 - gC - g H
2.4 - 0.96- 0.16= 1.28g O
b) e.f
C H O
0.96/12 0.16/1 1.28/16
mol=0.08 0.16 0.08
e.f= 1 2 1
emperical formula= CH2O
m.f
(CH2O)n= 60 **60/30=2**
molecular formula= C2H4O2
2.4g of a compound of carbon, hydrogen and oxygen gave on combustion, 3.52g of CO2 and 1.44g of H2O. The relative molecular mass of the compound was found to be 60.
a)What are the masses of carbon, hydrogen and oxygen in 2.4g of the compound?
b)What are the emperical and molecular formulae of the compound?
Ans:
a) 3.52g CO2 x (atomic mass C/molar mass CO2) = 0.96g C.
1.44g H2O x (2*atomic mass H/molar mass H2O)= 0.16g H.
g of O = 2.4 - gC - g H
2.4 - 0.96- 0.16= 1.28g O
b) e.f
C H O
0.96/12 0.16/1 1.28/16
mol=0.08 0.16 0.08
e.f= 1 2 1
emperical formula= CH2O
m.f
(CH2O)n= 60 **60/30=2**
molecular formula= C2H4O2
Answers
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