1.

A crystalline compound when heated become anhydrous by losing 51.2 % of the mass. On analysis, the anhydrous compound gave the following percentage composition : `Mg = 20.0 % , S = 26.66 % and O = 53.33 %`. Calculate the molecular formula of the anhydrous compound and crystalline compound. The molecular mass of anhydride compound is 120 u.

Answer» Step I. Determination of empirical formula of the anhydrous compound
`{:("Element","Percentage","Atomic mass","Gram atoms (Moles)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("Mg",20.0,24,(24.0)/(24)=0.83,(0.83)/(0.83)=1,1),("S",26.66,32,(26.66)/(32)=0.83,(0.83)/(0.83)=1,1),("O",53.33,16,(53.33)/(16)=3.33,(3.33)/(0.83)=4.01,4):}`
`:.` The empirical formula of anhydrous compound is `MgSO_(4)`
Step II. Determination of the molecular formula of the anhydrous compound
Empirical formula mass `= 24+32 +4 xx 16 = 24 + 32 + 64 = 120` u
Molecular mass = 120 u (Given) , `n=("Molecular mass")/("Empirical formula mass")=(120)/(120)=1`
`:.` Molecular formula `=nxx` Empirical formula `=1 xx MgSO_(4)=MgSO_(4)`
`:.` Molecular formula of anhydrous compound `= MgSO_(4)`
Step III. Determination of molecular formula of the crystalline compound
Let us first calculate the number of molecules of water of crystallisation in the compound. For that,
Let the weight of hydrated compound = 100 u
Weight of water lost = 51.2 u
`:.` Weight of anhydrous compound `= 100-51.2 = 48.9` u
Now, if the weight of anhydrous compound is 48.8 u, then that of water 51.2 u
If the weight of anhydrous compound is 120 u, then that of water `= (51.2)/(48.8)xx120 = 125.9` u
Molecular mass of water `(H_(2)O)=2xx1+16 = 18` u
`:.` No. of `H_(2)O` molecules `= ("Weight of water")/("Molecular mass of water")=(125.9)/(18)=7` (approx)
Thus, the formula of crystalline compound `= MgSO_(4).7H_(2)O`.


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