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



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


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:


    ∴ 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

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.

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


I hope it helps you! Cheers! 
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