Question

A welding fuel gas contain carbon and hydrogen only. burning a small sample of it in oxygen gives 3.38g carbon dioxide,0.690g water and no other products. A volume of 10.0L (measured at STP of this welding gas is found to weidh 11.6g. caluculate (I) empirical formula (ii)molar mass of the gas and (iii) molecular formula​

Answer 1

Explanation:

3.38 g of CO₂ is obtained.

∴ Mass of C = $\frac{12}{44} * 3.38$ = 0.92 g

Similarly, Mass of H = $\frac{2}{18} *0.690$ = 0.077 g

∴ Percentage of C = $\frac{0.92}{0.92 + 0.077} * 100$ = 92.3 %

and Percentage of H = $\frac{0.077}{0.92 + 0.077}*100$ = 7.7 %

(i) Empirical Formula:-

No. of moles of C = $\frac{92.2}{12}$ = 7.7

No. of moles of H = $\frac{7.7}{1}$ = 7.7

∴  C : H = 7.7 : 7.7 = 1

Hence, Empirical Formula = (CH)₁ = CH (Ans.)

(ii) The weight of 10 L of gas at STP = 11.6 g

   ∴ 22.4 L of gas at STP = $\frac{11.6}{10} * 22.4$ g mol⁻¹ = 26 g mol⁻¹ (Ans.)

(iii)   Let us suppose the molecular formula of the compound (CH)$x$

     ∴ (CH)$x$ = 26

   or, (12 + 1)$x$ = 26

   or, $x$ = $\frac{26}{13}$ = 2

   Hence, Molecular formula = (CH)₂ = C₂H₂ (Ans.)

Answer 2

3.38 g of CO₂ is obtained.

∴ Mass of C =12/44*3.38g  = 0.92 g

Similarly, Mass of H =  1/44*3.38= 0.077 g

∴ Percentage of C = 0.92/0.92+0.077 = 92.3 %

and Percentage of H =  100-93.3= 7.7 %

(i) Empirical Formula:-

No. of moles of C = 92.2/12 = 7.7

No. of moles of H =  7.7/1= 7.7

∴  C : H = 7.7 : 7.7 = 1

Hence, Empirical Formula = (CH)₁ = CH (Ans.)

(ii) The weight of 10 L of gas at STP = 11.6 g

  ∴ 22.4 L of gas at STP = 11.6/10*22.4 g mol⁻¹ = 26 g mol⁻¹ (Ans.)

(iii)   Let us suppose the molecular formula of the compound (CH)

    ∴ (CH)n = 26

  or, (12 + 1) n= 26

  or,  = n = 2

  Hence, Molecular formula = C₂H₂