The composition of iron (Fe) and manganese (Mn) in some metal alloys can be determined by traditional gravimetric analysis using 8-hydroxyquinoline (C9H7NO) as the precipitation reagent. In a typical analysis, a 250 mg sample of alloy containing iron (Fe), manganese (Mn) and other metals was dissolved in acid and treated with 8-hydroxyquinoline (C9H7NO). The iron and manganese was precipitated and isolated as Fe(C9H6NO)3 and Mn(C9H6NO)2, yielding a total mass of 1704.5 mg. The total amount of 8-hydroxylquinoline required for two precipitates was determined to be 10.363 mmol.
(a) Let x and y be the mass of Fe and Mn in the original alloy sample respectively. Using the given information, express the total mass of precipitate isolated and amount of 8-hydroxylquinoline in two precipitates in terms of x and y.
Chemistry?
2016-10-10 3:16 pm
回答 (1)
2016-10-11 5:37 pm
molar mass of Fe = 55.8 g/mol = 55.8 mg/mmol
molar mass of Mn = 54.9 mg/mmol
molar mass of Fe(C9H6NO)3 = 55.8 + (12*9 + 1*6 + 14 + 16)*3 = 55.8 + 144*3 = 487.8 mg/mmol
molar mass of Mn(C9H6NO)2 = 54.9 + 144*2 = 342.9 mg/mmol
mass of Fe in precipitate = x mg
no. of mmol of Fe = x/55.8 (mmol)
mass of Mn in precipitate = y mg
no. of mmol of Mn = y/54.9 mmol
mass of 8-hydroxyquinoline = 1704.5 - x - y (mg)
since mole ratio of Fe & Mn to their respective 8-hydroxyquinoline complex = 1:1 ,
no. of mmol of the Fe-complex = x/55.8
mass of the Fe-complex = (x/55.8) * 487.8 = 487.8x/55.8 mg
similarly, mass of the Mn-complex = 342.9y/54.9 mg
total mass = 1704.5 mg
so, 487.8x/55.8 + 342.9y/54.9 = 1704.5
Given that no. of mmol of 8-hydroxyquinoline has been given (10.363 mmol), and mass of it has been calculated as above, the second equation involving x & y can be set up.
Solve for x & y with this set of simultaneous equations.
molar mass of Mn = 54.9 mg/mmol
molar mass of Fe(C9H6NO)3 = 55.8 + (12*9 + 1*6 + 14 + 16)*3 = 55.8 + 144*3 = 487.8 mg/mmol
molar mass of Mn(C9H6NO)2 = 54.9 + 144*2 = 342.9 mg/mmol
mass of Fe in precipitate = x mg
no. of mmol of Fe = x/55.8 (mmol)
mass of Mn in precipitate = y mg
no. of mmol of Mn = y/54.9 mmol
mass of 8-hydroxyquinoline = 1704.5 - x - y (mg)
since mole ratio of Fe & Mn to their respective 8-hydroxyquinoline complex = 1:1 ,
no. of mmol of the Fe-complex = x/55.8
mass of the Fe-complex = (x/55.8) * 487.8 = 487.8x/55.8 mg
similarly, mass of the Mn-complex = 342.9y/54.9 mg
total mass = 1704.5 mg
so, 487.8x/55.8 + 342.9y/54.9 = 1704.5
Given that no. of mmol of 8-hydroxyquinoline has been given (10.363 mmol), and mass of it has been calculated as above, the second equation involving x & y can be set up.
Solve for x & y with this set of simultaneous equations.
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