A 0.4527 g sample of pewter, containing tin, lead, copper, and zinc, was dissolv
ID: 576815 • Letter: A
Question
A 0.4527 g sample of pewter, containing tin, lead, copper, and zinc, was dissolved in acid. Tin was precipitated as SnO2·4H2O and removed by filtration. The resulting filtrate and washings were diluted to a total volume of 200.0 mL. A 20.00 mL aliquot of this solution was buffered, and titration of the lead, copper, and zinc in solution required 34.88 mL of 0.001501 M EDTA. Thiosulfate was used to mask the copper in a second 25.00 mL aliquot. Titration of the lead and zinc in this aliquot required 34.49 mL of the EDTA solution. Finally, cyanide was used to mask the copper and the zinc in a third 30.00 mL aliquot. Titration of the lead in this aliquot required 24.20 mL of the EDTA solution.
A 0.4527 g sample of pewter, containing tin, lead, copper, and zinc, was dissolved in acid. Tin was precipitated as SnO2, 4H20 and removed by filtration. The resulting filtrate and washings were diluted to a total volume of 200.0 mL. A 20.00 mL aliquot of this solution was buffered, and titration of the lead, copper, and zinc in solution required 34.88 mL of 0.001501 M EDTA. Thiosulfate was used to mask the copper in a second 25.00 mL aliquot. Titration of the lead and zinc in this aliquot required 34.49 mL of the EDTA solution. Finally, cyanide was used to mask the copper and the zinc in a third 30.00 mL aliquot. Titration of the lead in this aliquot required 24.20 mL of the EDTA solution. Determine the percent composition by mass of each metal in the pewter sample. Number Number 1.38 3.49 %Zn Number Number 11.07 % Pb 84.06 % SnExplanation / Answer
0.4257 g powder contains all elements
let amount of tine be x moles
initially tin is removed so we are left with ead, copper, and zinc
Volume = 200 ml
We take 20 ml of this 200 ml sol
moles of EDTA used= 34.88 mL * 0.001501 M
=0.052 millimoles
by stoichiometry
moles of lead = a
moles of Cu= b
moles of Zn =c
moles of EDTA = 2 * (moles of lead, copper, and zinc)
0.104 millimloes = a+b+c
but this is for 20 ml sol
for actual sol
a+b+c = 10*.104 = 1.04 milliomoles
in expt 2
we took 25 ml sol
moles of EDTA used= 34.49 mL * 0.001501 M
=0.0517 millimoles
(a+c)/2= .0517
a+c= 0.1034
actually
a+c =(.1034)*(200/25)
= 0.827 millimoles
for trial 3
we took 30 ml sol
moles of EDTA used= 24.20 mL * 0.001501 M
=0.036
a/2= .036
=.072
actually
a= .072*(250/30)
= 0.6
We have 3 eqn
a=0.6 millimoles
a+c= 0.827 millimoles
c= .027 millimoles
a+b+c=1.04
b= 0.213
a= moles of Pb
=0.6 millimoles
= 0.6*207 mg
=124.2 mg
b= moles of Cu
= .213 millimoles
= .213* 63.5
= 13.52 mg
c= moles of Zn
= .027 millimoles
=.027*65
=1.755 mg
Mass of Cu + Pb+Zn= 124.2+1.755+13.52 mg
= 139.475 mg
Total wt= 452.7 mg
Sn weight = 452.7-139.475
=313.225 mg
Mass%
a. Pb= (124.2/452.7)*100
=27.43%
b Cu = (13.52/452.57)*100
=2.98%
c. Zn =(1.755/452.57)*100
=0.387%
d. Sn = (313.225/452.57)*100
=69.21%
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