A compound contains 32%carbon 4% hydrogen and rest oxygen vapour density is 75 what is empirical and molecular formula
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molecular mass = 2×VD
= 2×75
= 150.
now carbon is 32%
=> mass of carbon in the compound = 32%
mass of carbon = 32/100 ×150
= 48gm = 4 mole ofC atoms
mass of 4% of Hydrogen = 4/100 × 150
= 6gm = 6 moles of H atoms
mass of rest 64 % of oxygen = 64/100 × 150
= 96gm = 6 moles of O atoms
so we have 4 carbon, 6 Hydrogen and 6 oxygen.
so we can take 2 common out of them.
empirical formula C2H3O3.
= 2×75
= 150.
now carbon is 32%
=> mass of carbon in the compound = 32%
mass of carbon = 32/100 ×150
= 48gm = 4 mole ofC atoms
mass of 4% of Hydrogen = 4/100 × 150
= 6gm = 6 moles of H atoms
mass of rest 64 % of oxygen = 64/100 × 150
= 96gm = 6 moles of O atoms
so we have 4 carbon, 6 Hydrogen and 6 oxygen.
so we can take 2 common out of them.
empirical formula C2H3O3.
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