Best chemistry question ever!!?
2.4 g of a compound of carbon, hydrogen and oxygen on combustion gave 3.52 g of carbon dioxide and 1.44 g of water. The relative molecular mass of the compound was found to be 60.
(i)What are the masses of carbon, hydrogen and oxygen in 2.4 g of the compound?
(ii)What are the empirical formulae of the compound?
(iii)Given that the relative molecular mass of the compound was found to be 60, what is the molecular formula of the compound?
Answer:
3.52 g / 44 g/mol = 0.08 moles CO2
0.08 moles x 12 g/mol = 0.96 g C
1.44 / 18 = 0.08 moles H2O >> moles H = 0.16
0.16 moles x 1 = 0.16 g H
0.16 + 0.96 = 1.12 g
2.4 - 1.12 = 1.12 g O
1.12 / 16 = 0.07 moles O
C(0.08) H(0.16)O(0.07)
we divide for the smallest number
CH2O ( empirical formula : mass = 30 )
60 / 30 = 2
the molecular formula is C2H4O2
i) Atomic mass of carbon-12, hydrogen-1, oxygen-16
Molecular mass of CO2-44, H2O-18
44g of CO2 is obtained from 12g of C.
3.52g of CO2 is obtained from 12/44 X 3.52 = 0.96g of C
18g of H2O is obtained from 2g of H
1.44g of H2O is obtained from 2/18 X 1.44
=0.16g of H
18g of H2O is obtained from 16g of O
1.44g of H2O is obtained from 16/18 X 1.44
=1.28g of O (= 2.4-0.16-0.96)
Therefore 2.4g contains-0.96g of C
0.16g of H
1.28g of O
ii)No. of moles of H in the compound=0.16/1=0.16
No. of moles of C in the compound=0.96/12=0.08
No. Of moles of O in the compound=1.28/16=0.08
Dividing all three no.s by the smallest-(0.08)
H - 0.16/0.08=2
C - 0.08/0.08 = 1
O-0.08/0.08=1
Empirical formula-CH2O
iii)Empirical mass= 12+2+16
=30
n=molecular mass/empirical mass
=60/30
=2
Molecular formula=(empirical formula)n
=(CH2O)2
=C2H4O2
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