Q.1.00 g of a hydrated salt contains 0.2014 g of iron, 0.1153 g of sulfur, 0.2301 g of oxygen and 0.4532 g of water of crystallisation. Find the empirical formula. (At. wt. : Fe = 56; S = 32; O = 16) (Ans.: FeSO4)

Step 1. Moles: Fe $= 0.2014/56 = 0.003596$; S $= 0.1153/32 = 0.003603$; O $= 0.2301/16 = 0.014381$; H2O $= 0.4532/18 = 0.025178$.

Step 2. Divide every value by the smallest (Fe, 0.003596): Fe $= 1.00$; S $= 1.002 \approx 1$; O $= 4.00$; H2O $= 7.00$.

Step 3. Ratio Fe : S : O : H2O $= 1 : 1 : 4 : 7$, giving the full empirical formula FeSO4.7H2O (this is the same green-vitriol hydrate as in problem F). The book's printed short answer, 'FeSO4', names only the anhydrous salt component and appears to omit the water of crystallisation the data itself supports — this is …

Convert each measured mass (Fe, S, the 4 'oxide' oxygens, and the separately-tracked water of crystallisation) to mol …

• Only computing the Fe:S:O ratio and forgetting to also include the separately-given water-of-crystallisation mass in the empirical formula. …