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 litre (Measured at STP) of this welding gas is found weigh `11.6 g`. Calculate
(i) empirical formula,
(ii) molar mass of the gas, and
(iii) molecular formula.

A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gas 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. Calcuate

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 S.T.P.) of this welding gas is found to weigh 11.6 g. Calculate (i) empirical fomula (ii) molar mass of the gas, and (iii) molecular formula.

A gaseous hydrocarbon when burns in oxygen gives 3.38g CO_(2) , 0.690g water and no other product.At STP,volume of 10L of this hydrocarbon is found to weigh 11.6g .What will be the molecular mass of the hydrocarbon.

A gaseous hydrocarbon on complete combustion gives 3.38 g of CO_(2) and 0.690 g of H_(2)O and no other product. The empirical formula of the hydrocarbon is

(i) Butyric acid contains only C, H and O.A4.24 mg sample of butyric acid is completely burned. It gives 8.45 mg of carbon dioxide (CO_(2)) and 3.46 mg of water. What is the mass percentage of each element in butyric acid? (ii) If the elemental composition of butyric acid is found to be 54.2% C, 9.2% H and 36.6%O , determine the empirical formula. (iii) The molecular mass of butyric acid was determined of experiment to be 88 . What is the moleculare formula ?

0.22 g of a hydrogen (i.e., a compound conatining carbon and hydrogen only) on complete combustion with oxygen gave 0.9 g water and 0.44 g carbon dioxide. Show that these results are in accordance with the law of conservation of mass (atomic mass of C = 12, H = 1, O = 16) .