A welding fuel gas contains carbon and hydrogen only

Asked by admin @ 11/08/2021 in Chemistry viewed by 287 People

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.

Ncert solutions for Class 11th Chemistry Part - 1 Chapter 1 Exercise 33

Answered by admin @ 11/08/2021

   Hi, here goes the answer and the exlplanation, enjoy ;)

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I. 1 mole (44g) of $CO_2$ contains 12g of carbon.
  ∴ 3.38g of $CO_2$ will contain $\frac{12g}{18g}*3.38g=0.9217g$ of carbon.

18g of water contains 2g of hydrogen.
  ∴  0.690g of water will contain $\frac{2g}{18g}*0.690g=0.0767g$ of hydrogen.
Since carbon and hydrogen are the only constituents of the compound, the total mass of the compound is:

$0.9217g+0.0767g=0.9984g$

  ∴ Percentage of $C$ in the compound $=\frac{0.9217g}{0.9984g}*100=92.32%$

  Percentage of $H$ in the compound $=\frac{0.0767g}{0.9984g}*100=7.68%$

  Moles of carbon in the compound $=\frac{92.32}{12.00}=7.69$

  Moles of hydrogen in the compound $=\frac{7.68}{1}=7.68$

  ∴ Ratio of carbon to hydrogene in the compound $=7.69:7.68=1:1$
Hence, the empirical formula of the gas is $CH$
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II. Given,

          Weight of 10.0L of the gas (at S.T.P) = 11.6g

  ∴ Weight of 22.4L of gas at STP $=\frac{11.6g}{10.0L}*22.4L=25.984g\approx26g$
Hence, the molar mass of the gas is 26g.
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III. Empirical formula mass of $CH=12g+1g=13g$

  ∴ $n=\frac{Molar \ mass \ of \ gas}{Empirical \ formula \ mass \ of \ gas}=\frac{26g}{13g}\\n=2$

   Molecular formula of gas $=(CH)_n=C_2H_2$

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I hope it helps you! Cheers!
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A welding fuel gas contains carbon and hydrogen only

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