# How to solve this chemistry stoichiometry sum? A compund has O=61.32%,S=11.15%,H=4.88% and Zn=22.65%.The relative molecular mass of the compound is 287 a.m.u.Find the molecular formula of the compound,assuming that all the hydrogen is present as water of crystallisation.

Lets assume the compound to be `Zn_xS_yO_z*nH_2O` ` `

Let us say we have 100g of the compound.

Then;

Mass of S = 11.15g

Mass of O = 61.32g

Mass of H = 4.88g

Mass of Zn = 22.65g

Molic weights;

S = 32g/mol

O = 16g/mol

H = 1g/mol

Zn = 65g/mol

It is given that all the hydrogen in the compound is retain as water.

H moles present `= 4.88/1 = 4.88`

one water crystal is formed by two H moles.

Amount of water crystals moles present `= 4.88/2 = 2.44`

Mole ratio of O:H in water `= 1:2`

Amount of O moles used for water crystals `= 2.44`

Total O moles present in 100g of compound `= 61.32/16 = 3.83`

O moles in compound not as water `= 3.83-2.44 = 1.39`

Number of Zn moles in compound `= 22.65/65 = 0.348 mol`

Number of S moles in compound `=11.15/32 = 0.348 mol`

Mole ratio:

`x:y:z = 0.348:0.348:1.39`

Usually we keep the ratios in whole numbers.

`x:y:z = 1:1:3.99`

Approximately we can say;

`x:y:Z = 1:1:4`

So the compound will be;

`ZnSO_4*nH_2O`

It is given that the molar mass of the compound is 287 a.m.u.

`65+32+16*4+n(2+16) = 287`

` n = 7`

So the formula of the compound is;

`ZnSO_4*7H_2O`

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