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  • 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 (i)

1.34    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

Answers (1)

best_answer

The amount of carbon in 3.38 g of CO_{2} :

= \frac{12}{44}\times3.38g = 0.9218g

The amount of hydrogen in 0.690g of H_{2}O:

= \frac{2}{18}\times0.690g = 0.0767g

The compound contains only C and H, 

Therefore, the total mass of the compound will be:

= 0.9218+0.0767 = 0.9985g

Now, the percentage of Carbon in the compound:

= (\frac{0.9218}{0.9985})\times100 = 92.32

and the percentage of Hydrogen in the compound:

= (\frac{0.0767}{0.9985})\times100 = 7.68

Now, the empirical formula,

Moles of carbon in the compound:

= \frac{92.32}{12} = 7.69

Moles of hydrogen in the compound:

= \frac{7.68}{1} = 7.68

So, the simplest molar ratio will be = 7.69:7.68 = 1:1

Therefore, the empirical formula is CH.

 

Posted by

Divya Prakash Singh

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