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Academic session 2026-27, Chapter 1 in the current curriculum.
Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter Some Basic Concepts of Chemistry. This page offers a step-by-step solution to the specific question from Exercise 1, Question 34:
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.
. With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.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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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 34: A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3....
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Comments
Nice answer
Good
great question backed with a great answer
From where 22.4 value came
Excellent answer
tysm, this answer was really helpful.
Thanks for making me understand this Question.â˺
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Why so lengthy I can explain better
Yes your answer is very simple and helping and you know the style of answering very well.You explained each and every step.