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a 5.49g sample of iron(iii) chloride hydrate was found to contain 2.20g of water. what is the complete formula of the hydrate?
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so you have 2.2g of water.. 2.2g/18g/mol = 0.1222 moles of H2O and you have 5.49-2.2= 3.29 g iron(III)chloride since you know Fe is in oxidation state of 3+, the formula is FeCl3 ie. Fe +3 and Cl - so you need 3 Cl's to balance the charge.. now to find the moles of FeCl3 = 3.29g/162.19g/mol =0.0203 divide the larger by the smaller 0.12222/0.0203 = 6 so the ratio of FeCl3 : H2O is 1:6 FeCl3(H2O)6
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