1. Copper (II) sulphate pentahydrate, CuSO4*5H2O, has 5 moles of water for each
ID: 836279 • Letter: 1
Question
1. Copper (II) sulphate pentahydrate, CuSO4*5H2O, has 5 moles of water for each mole of CuSO4 in the solid crystal. The molar mass for the pentahydrate is 249.68 g/mol. Copper (II) sulphate without water in the crystal structure is called anhydrous, and would have molar mass of 159.61 g/mol. How many grams of CuSO4*5H2O should be dissolved in a volume of 500.0 mL to make 8.00 mM Cu2+?
2. An iron ore is analyzed for iron content by dissolving in acid, converting the iron to Fe2+, and then titrating with standard 0.0150 M K2Cr2O7 solution. If 35.6 mL is required to titrate the iron in a 1.68 g sample, how much Fe2+ is in the sample?
The titration reaction is: 6 Fe2+ + Cr2O72- + 14 H+
Explanation / Answer
1)
CuSO4*5H2O=249.68g/mole
CuSO4=159.61 g/mol
fraction of Cu in CuSO4*5H2O=63.53/249.68=0.254
mmoles of Cu in solution=0.5L*8mg/L=4mgCu2+
Mass of CuSO4*5H2O required=4mg/0.2544=15.723mg=0.0157g
2)
6 Fe2++ Cr2O72- + 14 H+ ----> 6 Fe3++ 2 Cr3+ + 7 H2O.
moles of K2Cr2O7 =0.015moles/L*0.0356L=0.000534moles
since ratio in balance equation is 1:6
so moles of Fe3+=0.000534*6=0.0032moles
Mass of Fe3+=0.0032moles*56g/mole=0.18g Fe3+
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