MARKING SCHEME
for
Analysis of Zinc Tablets : A Stoichiometry Experiment
Maximum = 14
DATA (–2)
Part I : Titration of K4Fe(CN)6 with ZnSO4
1st titration
2nd titration
3rd titration
initial burette reading
__ 0.00 __
__ 0.00 __
________
final burette reading
__18.74 _
__18.76 _
________
volume difference
(final – initial)
__ 18.74 _
__ 18.76_
________
Part II : Titration of a zinc tablet
Mass of zinc in each tablet according to the label on the bottle of tablets
_ 50 mg _
1st titration
2nd titration 3rd titration
mass of tablet .................................
0.5574 g
0.5643 g 0.5387 g
initial burette reading ......................
0.00 mL
0.00 mL
0.00 mL
final burette reading
2.80 mL
2.93 mL
2.65 mL
2.80 mL
2.93 mL
2.65 mL
.......................
volume difference ..........................
(final – initial)
CALCULATIONS AND ANALYSIS OF RESULTS
Part I
1. Calculate the average volume of zinc sulphate solution used in the titrations in Part I. If two of the
volumes are quite close to each other and the remaining volume is substantially different, average the two
close values and discard the “farther away” value.
(1/2)
average volume of ZnSO4 used = 1/2 (18.74 + 18.76) = 18.75 mL
2. Use the volume of 0.0500 M ZnSO4 found in Calculation 1 to calculate the average moles of ZnSO4 used
in the titrations.
(1/2)
moles ZnSO4 = 0.0500 mol/L x 0.01875 L = 9.375 x 10–4 mol
3. Calculate the moles of K4Fe(CN)6 present in 25.00 mL of 0.0250 M K4Fe(CN)6 .
(1/2)
moles K4Fe(CN)6 = 0.0250 mol/L x 0.02500 L = 6.25 x 10–4 mol
4. Use the moles of K4Fe(CN)6 found in Calculation 3 and the moles of ZnSO4 found in Calculation 2 to
moles of ZnSO 4
calculate the value of the following mole ratio:
mole ratio =
moles of K4Fe(CN)6
Note: do not leave your answer in the form of a ratio; use your calculator to carry out the division and give
the answer in decimal form.
(1)
mole ratio =
moles of ZnSO 4
moles of K4Fe(CN)6
=
9.375 x 10
6.25 x 10
–4
–4
mol
mol
= 1.50
Page 2
5. Calculate the mole ratio of K4Fe(CN)6 to ZnSO4 for each of the four equations listed in the “Introduction”.
For example, if the equation was
5 K4Fe(CN)6(aq) + 7 ZnSO4(aq)  products ,
7
moles of ZnSO 4
the mole ratio would be:
mole ratio =
=
= 1.40
moles of K4Fe(CN)6
5
(1)
mole ratio =
moles of ZnSO 4
moles of K4Fe(CN)6
=
3
2
= 1.50
6. Which of the 4 values of the mole ratio found in Calculation 5 is closest to the value of the mole ratio found
experimentally in Calculation 4? Write down the equation whose mole ratio is closest to the value in
Calculation 4. This is the equation we will use in the calculations for Part II.
The closest equation is number 4:
2 K4Fe(CN)6(aq) + 3 ZnSO4(aq)  K2Zn3[Fe(CN)6]2(s) + 3 K2SO4(aq)
(1)
Part II
7. Calculate the moles of K4Fe(CN)6 present in the 25.00 mL of 0.0250 M K4Fe(CN)6 you used in Procedure
step 10.
moles K4Fe(CN)6 = 0.0250 mol/L x 0.02500 L = 6.25 x 10–4 mol
(0)
8. In Procedure step 10, the Zn2+ in the tablet reacts with some of the K4Fe(CN)6 solution. The ZnSO4
added in Procedure step 12 reacts with the remaining K4Fe(CN)6 .
Therefore: (moles Zn2+ in tablet) + (moles ZnSO4 added in titration) = (total moles Zn2+)
and: (moles Zn2+ in tablet) = (total moles Zn2+) – (moles ZnSO4 added in titration)
...
(5)
Use the volume of 0.0500 M ZnSO4 found in Procedure step 12 for the titration of the 1st zinc tablet to
calculate the “moles ZnSO4 added in titration”.
(1/2)
moles ZnSO4 added in the titration = 0.0500 mol/L x 0.00280 L = 1.40 x 10–4 mol
9. Use the equation you wrote in Calculation 6 to find the moles of Zn2+ that react with the moles of
K4Fe(CN)6 you found in question 7. The moles of Zn2+ that react is the “total moles Zn2+”.
(1)
total moles Zn2+ = 6.25 x 10–4 mol K4Fe(CN)6 x
3 mol ZnSO 4
2 mol K 4Fe(CN) 6
= 9.375 x 10–4 mol
10. Use equation 5 in Calculation 8 to find the “moles of Zn2+ in tablet”, using the “moles ZnSO4 added in
titration” found in Calculation 8 and the “total moles Zn2+” found in Calculation 9.
(1)
moles of Zn2+ in the tablet = 9.375 x 10–4 – 1.40 x 10–4 = 7.975 x 10–4 mol
11. Based on your answer to Calculation 10, how many moles of Zn (in the form of Zn2+) are present in tablet
#1? What is the mass of Zn, in milligrams, present in tablet #1?
(0)
(1)
moles of Zn present = 7.975 x 10–4 mol
65.4 g
1 mg
mass of Zn = 7.975 x 10–4 mol x
x
= 52.2 mg
–3
1 mol
10 g
Page 3
12. Repeat Calculations 8 to 11, using the data for the other tablets.
(1/2)
(1/2)
(1/2)
(1/2)
(0)
2nd tablet:
moles ZnSO4 added in the titration = 0.0500 mol/L x 0.002930 L = 1.465 x 10–4 mol
3 mol ZnSO 4
2 mol K4 Fe(CN) 6
total moles Zn2+ = 6.25 x 10–4 mol K4Fe(CN)6 x
= 9.375 x 10–4 mol
moles of Zn2+ in the tablet = 9.375 x 10–4 – 1.465 x 10–4 = 7.910 x 10–4 mol
moles of Zn present = 7.910 x 10–4 mol
1 mg
65 . 4 g
–3
mass of Zn = 7.910 x 10–4 mol x 1 mol x 10 g = 51.7 mg
3rd tablet:
moles ZnSO4 added in the titration = 0.0500 mol/L x 0.002650 L = 1.325 x 10–4 mol
3 mol ZnSO 4
2 mol K4 Fe(CN) 6
total moles Zn2+ = 6.25 x 10–4 mol K4Fe(CN)6 x
= 9.375 x 10–4 mol
2+
–4
–4
moles of Zn in the tablet = 9.375 x 10 – 1.325 x 10 = 8.05 x 10–4 mol
moles of Zn present = 8.05 x 10–4 mol
1 mg
65 . 4 g
mass of Zn = 8.05 x 10–4 mol x 1 mol x 10
–3
g = 52.6 mg
13. Calculate the average mass of Zn in the tablets you used.
(1/2)
average mass of Zn =
1
3
(52.2 + 51.7 + 52.6) = 52.2 mg
14. Calculate the percentage difference between the mass of Zn claimed on the tablet bottle and the average
mass of Zn found in Calculation 13, as follows:
mass of Zn (claimed on bottle ) – average mass of Zn (found experimentally )
mass of Zn (claimed on bottle )
% difference =
x 100%
(1/2)
50 mg –52 . 2 mg
50 mg
% difference =
x 100% = –4.4%
15. If the Zn2+ present in the tablets is present in the form of zinc gluconate, Zn(C6H11O6)2 , what is the mass
of Zn(C6H11O6)2 in an average tablet, based on the average mass of Zn found in Calculation 13?
(2)
423 . 4 g Zn(C 6 H 11 O 6 ) 2
1 mol Zn(C 6 H 11 O 6 ) 2
1 mol Zn
1 mol Zn(C 6 H 11 O 6 ) 2
1 mol Zn
mass of Zn(C6H11O6)2 = 0.0522 g Zn x 65 . 4 g Zn x
x
= 0.338 g
16. Some medicinal tablets contain a “binder” such as corn starch to help hold together powdered chemical
ingredients when they are compressed into tablet form. Calculate the average mass of the tablets you used.
Based on the results of Calculation 15 and the average mass of the tablets you used, do the zinc tablets contain a
binder?
1
3
(1/2)
average mass of tablets = (0.5574 + 0.5643 + 0.5387) = 0.5535g
(1/2)
Since there was only 0.338 g of zinc gluconate in each tablet, the remaining mass must have been due to
a binder.