37./0.262g of a substance gave, on combustion,0.36lg of Co, and 0.147g

1.	37./0.262g of a substance gave, on combustion,0.36lg of Co, and 0.147g of H,0. What isthe empirical formula of the substance(E - 1996)M) CH,O 2) C,H,O 3) C, H, O, 4) C,H,O,
Answer» mass of CO2 :0.361gmass of water 0.147g moles of CO2 = .361 / 44 =0.0082 = moles of carbon in the compound; moles of water = .147 /18 = 0.0082; there are 2 moles of H in each mole of H2O so moles of H = 0.0163moles molar ratio of C : H = 0.0082:0.0163 or after dividing by the smallest ,molar ratio of C : H = 1.000:1.99 empirical formula is CH2o

37./0.262g of a substance gave, on combustion,0.36lg of Co, and 0.147g of H,0. What isthe empirical formula of the substance(E - 1996)M) CH,O 2) C,H,O 3) C, H, O, 4) C,H,O,

Answer»

mass of CO2 :0.361gmass of water 0.147g

moles of CO2 = .361 / 44 =0.0082 = moles of carbon in the compound;

moles of water = .147 /18 = 0.0082; there are 2 moles of H in each mole of H2O so moles of H = 0.0163moles

molar ratio of C : H = 0.0082:0.0163

or after dividing by the smallest ,molar ratio of C : H = 1.000:1.99

empirical formula is CH2o