A crystalline when heated became anhydrous and loses 51.2℅ of its weight .The anhydrous salt on analysis gave the percentage composition as Mg = 20.0℅ S = 26.66℅ and O = 53.33℅ calculate the molecular formula of anhydrous salt and crystalline salt.( molecular mass of anhydrous salt is 120u)
Answers
Answer:
M
Explanation:
Mg = 20.0% ,S = 26.66% ,O = 53.33%.
suppose 100 mg of the compound is present. This changes the percentage to grams..
Mg = 20g , S = 26.66g , O = 53.33g
convert the masses to mole,
Mg = 20 /24 = 0.8333 moles
S = 26.66/ 32 = 0.8331 moles
O = 53.33 /16 = 3.3331 moles.
divide by lowest no .of moles
Mg = 0.8333/ 0.8331 = 1 , S = 0.8331 / 0.8331 = 1 , O = 3.3331 / 0.8331 = 1
Empirical formula = MgSO4
Molecular mass = 120
Molecular formula = MgSo4
As crystalline salt on becoming anhydrous losses 51.2% by mass , this means 48.8g of anhydrous salt contains H2O = 51.2g
Therefore, 120 mg of anhydrous salt contains H2O = (51.2 /48.8)× 120 = 126g
No.of molecules = 126 /18 = 7 molecules.
Hence, molecular formula of crystalline salt = MgSO4.7H2O.
Explanation:
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