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.

Expert Answers

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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`

Approved by eNotes Editorial Team

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