Question

a welding fuel gas contains carbon and hydrogen only.during a small of it burning in oxygen gives 3.38g carbon dioxide .690g of water and no other products.a volume of 10.0L of this welding gas found weigh 11.6g.calculate molecular formula molar mass and emperical formula of the gas

Answer 1

Answer:

(i) percentage of C can be calculated as follows:

CO2 = C

i.e 44 parts of CO2= 12 parts of C

OR

44g of CO2 = 12 g of C

Therefore according to question

3.38 g of CO2 contains C = 12/44 * 3.38 = 0.921 g

18 g of water contains hydrogen = 2g

Therefore 0.690 g of water contains hydrogen = 2/18 * 0.690 = 0.0767g

0.690 g of water will contain hydrogen

= 0.0767 g