324
Stolchlometry IV: Equivalent Weight and Normality
PROBLEM:
A 0.4462 g sample of iron ore is dissolved in H 2 SO 4 , and the iron is reduced to
Fe
2+ with metallic zinc. The solution requires 38.65 ml of 0.1038 N KMnO 4 for
titration. Calculate the percentage of iron as Fe 2 O 3 in the sample.
SOLUTION:
Again, we do not need the chemical equation for the reaction — only the
knowledge that there is 1 equiv/mole for Fe
2+ as it is oxidized to Fe
3+ by
the KMnO 4 and that there are 2 moles of Fe/mole of Fe 2 O 3 . At the endpoint,
equiv of Fe 2 O 3 = equiv of KMnO 4
= ( 0.1038 £2H1X\ (0.03865 liter)
\
liter /
= 0.004012 equiv Fe 2 O 3
From the number of equivalents, we find
wtof Fe 2 O 3 = (0.004012 equiv) ( *
mole W 159.6
\ 2 equiv/ \
I
.
equiv/ \
mole/
= 0.3202 g Fe 2 O 3
Percentage of Fe 2 O 3 = ' tAf ,
8 x 100 = 71.76% Fe 2 O 3
0.4462 g
PROBLEM:
A radiochemist isolates 8.6 /ug of a chloride of neptunium (atomic weight =
237 g/mole), which she proves has the formula NpCl 3 . In trying to find the
possible valence states of Np, she finds that titration of the 8.6 /u.g sample
requires 37.5 ^il of 0.00200 N KMnO 4 solution. To what electrical charge must
the Np have been oxidized?
SOLUTION:
The number of //.moles of NpCl 3 in the sample is
= 0.0250 /imole NpCl
343.5 /amole
The number of/xequiv of KMnO 4 required for titration is
0.00200
1010"
(37 . 5 1} = 0.0750
liter A
/ul / \
equiv /
Therefore,
0.0750 juequiv _ equiv
0.0250 /Limole ~
mole