A welding fuel gas combines carbon and Hydrogen only. On burning it gives 3.38g of CO2 and 0.690g of water and no other products. A volume of 10L (STP) of this gas found to weigh 11.6g. Calculate empirical formula.

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Solution

(i) Empirical formula

1 mole of CO2 contains 12 g of carbon

Therefore, 3.38 g of CO2 will contain carbon

= 12g44g×3.38g

= 0.9217 g

18 g of water contains 2 g of hydrogen

Therefore, 0.690 g of water will contain hydrogen

= 2g18g×0.690

= 0.0767 g

As hydrogen and carbon are the only elements of the compound. Now, the total mass is:

= 0.9217 g + 0.0767 g

= 0.9984 g

Therefore, % of C in the compound

= 0.9217g0.9984g×100

= 92.32 %

% of H in the compound

= 0.0767g0.9984g×100

= 7.68 %

Moles of C in the compound,

= 92.3212.00

= 7.69

Moles of H in the compound,

= 7.681

= 7.68

Therefore, the ratio of carbon to hydrogen is,

7.69: 7.68

1: 1

Therefore, the empirical formula is CH.

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Similar questions

Q. 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.

A welding fuel gas contains 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