Question

6.00 g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 176. g/mol, is burned completely in excess oxygen, and the mass of the products carefully measured: product carbon dioxide mass 9.00 g water 2.46 g Use this information to find the molecular formula of X. x 6 ?

Answer 1

molecular formula of X : C6H8O6

Explanation

mass of CO2 = 9.00 g

moles of CO2 = (mass of CO2) / (molar mass of CO2)

moles of CO2 = (9.00 g) / (44.01 g/mol)

moles of CO2 = 0.2045 mol

moles of carbon = moles of CO2

moles of carbon = 0.2045 mol

mass of carbon = (moles of carbon) * (molar mass carbon)

mass of carbon = (0.2045 mol) * (12.01 g/mol)

mass of carbon = 2.456 g

mass of water = 2.46 g

moles of water = (mass of water) / (molar mass H2O)

moles of water = (2.46 g) / (18.01528 g/mol)

moles of water = 0.13655 mol

moles of H = 2 * (moles of water)

moles of H = 2 * (0.13655 mol)

moles of H = 0.2731 mol

mass of H = (moles of H) * (molar mass H)

mass of H = (0.2731 mol) * (1.008 g/mol)

mass of H = 0.2753 g

mass of O = (total mass) - (mass of H + mass of C)

mass of O = (6.00 g) - (0.2753 g + 2.456 g)

mass of O = 3.2687 g

moles of O = (mass of O) / (molar mass O)

moles of O = (3.2687 g) / (16.0 g/mol)

moles of O = 0.2043 mol

moles C / moles O = (0.2045 mol) / (0.2043 mol)

moles C / moles O = 1

moles H / moles O = (0.2731 mol) / (0.2043 mol)

moles H / moles O = 1.33

empirical formula : CH1.33O

empirical formula : C3H4O3

empirical formula mass = 88 g/mol

n = (molecular formula mass) / (empirical formula mass)

n = (176 g/mol) / (88 g/mol)

n = 2

molecular formula = n * (empirical formula)

molecular formula = 2 * (C3H4O3)

molecular formula = C6H8O6