A welding fuel gas contain carbon dioxide and hydrogen only. Burning a small sample of it oxygen gives 3.38g CO2, 0.690 g of H2O and no other products. A volume of 10.0 L of tgis weilding gas to weight 11.6g. Calculate :
(i) Empirical Formula
(ii) Molar mass of gas
(iii) Molecular Formula
Answers
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)
Answer:
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)