A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate
(i) empirical formula,
(ii) molar mass of the gas, and
(iii) molecular formula.
(i) percentage of C can be calculated as follows:
CO2 = C
i.e 44 parts of CO2= 12 parts of C
OR
44g of CO2 = 12 g of C
Therefore according to question
3.38 g of CO2 contains C = 12/44 * 3.38 = 0.921 g
18 g of water contains hydrogen = 2g
Therefore 0.690 g of water contains hydrogen = 2/18 * 0.690 = 0.0767g
0.690 g of water will contain hydrogen
= 0.0767 g
Since carbon and hydrogen are the only constituents of the compound, the total mass of the compound is:
= 0.9217 g + 0.0767 g = 0.9984 g
Now percentage of carbon = weight of carbon/weight of compound * 100
=0.921 /0.998 * 100= 92.32 %
Also percentage of hydrogen = weight of hydrogen/weight of compound *100
=0.0766/0.998 * 100 = 7.68 %
(ii) Given,
Weight of 10.0L of the gas (at S.T.P) = 11.6 g
Weight of 22.4 L of gas at STP
= 25.984 g
≈ 26 g
Hence, the molar mass of the gas is 26 g.
(iii) empirical formula
Element | Percentage | Atomic mass | Atomic ratio | Simplest ratio | Simplest whole no ratio |
C | 92.32 | 12 | 92.32/12 = 7.69 | 7.69/7.65 = 1.00 | 1 |
H | 7.65 | 1 | 7.65/1 = 7.65 | 7.65/7.65 = 1 | 1 |
Empirical formula of the compound = CH
Now molecular formula calculation
Empirical formula mass = 12 + 1 = 13 amu
Also molecular mass = 26 g (calculated in previous step)
Therefore n = molecular mass/empirical formula mass = 26/13 = 2
Now molecular formula = n x empirical formula = 2 x CH = C2H2
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From where "Weight of 22.4 L of gas" came