A compound containing Mg, S, O and H. It is observed that 492 g of it contains 48 g of Mg, 64 g sulphur,
352 g of oxygen and remaining is hydrogen. Given that its molar mass is 246 g/mol. All the hydrogen present
in it is in the form of water of crystallisation. Write its molecular formula showing water of crystallisation.
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Mass of Mg = 48g
mass of S = 64g
mass of O = 352g
and mass of H = 492g - (48 + 64 + 352)g
= 492g - 464g
= 28g
we know,
atomic mass of Mg = 24 g/mol
atomic mass of S = 32g/mol
atomic mass of O = 16g/mol
and atomic mass of H = 1g/mol
so, mole of Mg in compound = mass of Mg/atomic mass of Mg
= 48/24 = 2
similarly, mole of S in compound = 64/32 = 2
mole of O in compound = 352/16 = 22
mole of H in compound = 28/1 = 28
so, ratio of mole of given atoms is 2 : 2 : 22 : 28 or, 1 : 1 : 11 : 14
hence, molecular formula of compound is .
a/c to question, all the hydrogen are present in the form of water crystallization.
so, molecular formula will be