Cliffs AP Chemistry, 3rd Edition

Analysis:

1. Calculate the theoretical yield of ASA to be obtained in this synthesis.
   .
   ..

g salicylic acid

g salicylic acid

mol salicylic acid

1 mol salicylic acid

200

138 1

1

# =0 0145

..

.

mL acetic anhydride

mL

g

g acetic anhydride

mol acetic anhydride

1

500 108

102 1

1

## 1

= 0.053 mol acetic anhydride

Since 1 mol of salicylic acid is required for each mole of acetic anhydride, salicylic acid is

the limiting reagent.

. mol salicylic acid.
mol salicylic acid

mol ASA

mol ASA

g ASA

1

0 0145

1

1

1

180 17

##= 2.61 g ASA

2. What was the actual percentage yield of the impure product?

%%..

.

yield g %

g

261 100 72 8 yield

190

= # =

3. To analyze the purity of the ASA he produced, the student measured out 0.400 g of pure,
   analytical reagent grade ASA and then treated it with NaOH to create sodium salicylate.
   He then added a FeCl 3 −KCl−HCl solution to create a purple salicylate complex. He then
   diluted this solution to 250. mL with distilled water. Determine the molarity of the stock
   solution.

.

.

L..

g ASA

g ASA

mol ASA M

0 250

0 400

180 17

##^1 =888 10-^3

4. 5.00, 4.00, 3.00, 2.00 and 1.00 mL samples of the stock solution were then diluted to
   100. mL with distilled water. Determine the molarity of each aliquot.

'

'

100. mL standard sol n

5.00 mL stock sol n 8.88 10 M

4.44 10 M

3

:# = # 4

• ^ h _ i -

'

'

100. mL standard sol n

4.00 mL stock sol n 8.88 10 M

3.55 10 M

3

:# = # 4

• ^ h _ i -

'

'

100. mL standard sol n

3.00 mL stock sol n 8.88 10 M

2.66 10 M

3

:# = # 4

• ^ h _ i -

'

'

100. mL standard sol n

2.00 mL stock sol n 8.88 10 M

1.78 10 M

3

:# = # 4

• ^ h _ i -

'

'

100. mL standard sol n

(1.00 mL stock sol n) 8.88 10 M

8.88 10 M

3

:# = # 5

• _i-

5. Samples from these aliquots were then colormetrically analyzed with the spectrophotome-
   ter and a Beer’s plot drawn (see Figure 2).

Laboratory Experiments