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.
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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.
Mass of S = 11.15g
Mass of O = 61.32g
Mass of H = 4.88g
Mass of Zn = 22.65g
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`
`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;
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;
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