Column – I | Column - II | ||
(a) | A gaseous organic compound containing C = 52.17%, H = 13.04% & O = 34.78% (by weight) having molar mass 46 g/mol. | (p) | One mole of compound contains 4NA atoms of Hydrogen. |
(b) | 0.3 g of an organic compound containing C, H and O on combustion yields 0.44 g of CO2 and 0.18 g of H2O, with two O atoms per molecule. | (q) | The empirical formula of the compound is same as its molecule formula. |
(c) | A hydrocarbon containing C = 42.857% and H = 57.143% (by mole) containing 3C atoms per molecule. | (r) | Combustion products of one mole of compound contains larger number of moles of CO2 than that of H2O. |
(d) | A hydrocarbon containing 10.5 g carbon per gram of hydrogen having vapour density 46. | (s) | CO2 gas produced by the combustion of 0.25 mole of compound occupies a volume of 11.2 L at NTP. |
Text Solution
Verified by ExpertsA
(A - q,s); (B - p, s); (C - p, q, r); (D - q, r)
C : H : O = 
: 
: 
= 4 : 12 : 2 or 2 : 6 : 1
Empirical formula = C 2 H 6 O & molar mass = 46 g/mol
Mol formula = C 2 H 6 O
C 2 H 6 O + 3O 2 
2CO 2 + 3H 2 O
1 mole 44.8 L at STP
0.25 mole (11.2 L at STP)
Mass of C in organic compound = mass of C in CO 2 = 
× 12 = 0.12 g
Mass of H in organic compound = Mass of H in H 2 O = 
× 2 = 0.02 g
∴ Mass of O in organic compound = 0.3 – (0.12 + 0.02) = 0.16 g
∴ C : H : O = 
: 
: 
= 0.01 : 0.02 : 0.01 = 1 : 2 : 1
∴ Empirical formula = CH 2 O, but it contains 2 O atom per molecule
∴ Molecular formula = C 2 H 4 O 2
1 mole of C 2 H 4 O 2 contains 4 N A hydrogen atoms.
C 2 H 4 O 2 + 2O 2 
2CO 2 + 2H 2 O
1 mole 44.8 L
0.25 mole 11.2 L
C : H = 42.857 : 57.143
= 3 : x (given)
On solving, x = 4 ∴ molecular formula = C 3 H 4
1 mole of C 3 H 4 contains 4N A hydrogen atoms.
Empirical formula is same as molecular formula
C 3 H 4 + 4O 2 
3CO 2 + 2H 2 O
> 
C: H = 
: 
= 
: 1 = 7 : 8 Empirical formula = C 7 H 8
Mol wt. = 2 × VD = 2 × 46 = 92
Mol formula = Empirical formula = C 7 H 8
C 7 H 8 + 9O 2 
7CO 2 + 4H 2 O
> 
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