Page 41 - Modern Analytical Chemistry

1400-CH02 9/8/99 3:48 PM Page 24

24 Modern Analytical Chemistry

Solving for g H 2 C 2 O 4 gives
M Fe 3+ ´V Fe 3 + ´ FW H 2 C O 4 = (.
0 0130 M)(0.03644 L)(90.03 g/mole)
2
2 2
2 –
2 132 ´
= . 10 g H 2 CO 4
2
and the weight percent oxalic acid is
2 –
2 132 ´ 10 g C 2 HO 4
.
2
´ 100 = 0 201% / C 2 H O 4
.
ww
2
g
.
10 62 rhubarb
EXAMPLE 2.8
One quantitative analytical method for tetraethylthiuram disulfide, C 10H 20N 2S 4
(Antabuse), requires oxidizing the sulfur to SO 2 , and bubbling the resulting
SO 2 through H 2 O 2 to produce H 2 SO 4 . The H 2 SO 4 is then reacted with NaOH
according to the reaction
H 2 SO 4 (aq) +2NaOH(aq) ® Na 2 SO 4 (aq) +2H 2 O(l)
Using appropriate conservation principles, derive an equation relating the
moles of C 10 H 20 N 2 S 4 to the moles of NaOH. What is the weight percent
C 10 H 20 N 2 S 4 in a sample of Antabuse if the H 2 SO 4 produced from a 0.4613-g
portion reacts with 34.85 mL of 0.02500 M NaOH?
SOLUTION
The unbalanced reactions converting C 10 H 20 N 2 S 4 to H 2 SO 4 are

C 10 H 20 N 2 S 4 ® SO 2
SO 2 ® H 2 SO 4
Using a conservation of mass we have

4 ´moles C 10 H 20 N 2 S 4 =moles SO 2 =moles H 2 SO 4
A conservation of protons for the reaction of H 2 SO 4 with NaOH gives

2 ´moles H 2 SO 4 =moles of NaOH
Combining the two conservation equations gives the following stoichiometric
equation between C 10H 20N 2S 4 and NaOH

8 ´moles C 10 H 20 N 2 S 4 =moles NaOH
Now we are ready to finish the problem. Making appropriate substitutions for
moles of C 10 H 20 N 2 S 4 and moles of NaOH gives

8 ´g C 10 HN S
2 4
20
V
= M NaOH ´ NaOH
F W C 10 H 20 N S
2 4
Solving for g C 10 H 20 N 2 S 4 gives
1
g C 10 HN S 4 = ´ M NaOH ´V NaOH F ´ W C 10 H N S 4
20
2
20
2
8
1
.
( 0 02500 M)(0.03485 L)(296.54 g/mol) = 0.032295 g C 10 HN S
2 4
20
8