Question

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

Answer 1

                      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! 
Qba

Answer 2

CxHy + O2 ==> CO2 + H2O

moles of C:  3.38 g CO2 x 1 mole CO2/44 g x 1 mole C/mole CO2 = 0.0768 moles C

moles of H:  0.690 g H2O x 1 mole H2O/18 g x 2 moles H/mole H2O = 0.0767 moles H

Since the mole ratio of C:H is 1:1, the empirical formula is CH

10.0 L x 1 mole/22.4 L = 0.446 moles

11.6 g/0.446 moles = 26.0 g/mol = molar mass of the gas

Molar mass of empirical formula = 12 + 1 = 13

26.0 g/13 g = 2

Molecular formula = C2H2 (acetylene)