.
CLASS B, STUDY GROUP 3
SOLUTIONS TO QUESTIONS 1, 3, 5 AND 7

Question 1
10.0g of acetic acid CH
3
COOH were reacted with 3.0g of ethanol ( C
2
H
5
OH ) to give ethyl acetate ( CH
3
COO C
2
H
5
) and water
(a) What is the limiting reagent?
(b) Calculate the theoretical yield of ethyl acetate.
(c) If the actual yield of ethyl acetate obtained was 3.5g, calculate percentage yield
(d) Why are percentage yield rarely 100%
Solution
C H
3
COOH + C
2
H
5
OH → C H
3
COO C
2
H
5
Mole ratio: 1 1 1
Molar mass of acetic acid =60g
Molar mass of ethanol = 46g
Molar mass of ethyl acetate=88g
(a) No of moles of C H
3
COOH =
10
⁄
60
= 0.1667moles
No of moles of C
2
H
5
OH = 3 46 =0.065217
Since the number of moles of C
2
H
5
OH is less than that of C H
3
COOH and their mole
Ratio is the same, the limiting reagent is therefore C
2
H
5
OH
(b) theoretical yield of C H
3
COO C
2
H
5
No of moles of C H
3
COO C
2
H
5
=0.0652
Mass of C H
3
COO C
2
H
5
produced = 0.065217 × molar mass of C H
3
COO C
2
H
5
Theoretical yield of C H
3
COO C
2
H
5
=0.065217
× 88 = 5.739g 𝑎𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
(c) % yield =
𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
× 100
3.5
=
5.739
= 0.60985 = 60.985
%
(d) percentage yield is rarely 100 % because:
I.
Errors in measurement (weight, concentration etc)
II.
Side reactions may occur

III.
There may be impurities in the system
IV.
No system is 100 % efficient

Question 3
Potassium hydrogen phthalate (KHP), KHC
8
H
4
O
4
, was titrated with a given solution of sodium hydroxide. 23,48mL of NaOH was needed to completely neutralize 0.5468g of KHP.
What is the concentration of the NaOH solution, in i.
mol/L ii.
g/L
Solution
The balanced chemical equation of the reaction
𝐾𝐻𝐶
8
𝐻
4
𝑂
4
+ 𝑁𝑎𝑂𝐻 → 𝑁𝑎𝐾𝐶
8
𝐻
4
𝑂
4
+ 𝐻
2
𝑂
1 1 → 1 1
Mole ratio for a reaction between KHP and NaOH is
Molar mass of KHP = 204g/mole
1:1
Mass of KHP = 0.5468g
Number of moles of KHP=
0.5468
204
= 2.6904 × 10 −3 moles
Molar mass of NaOH = 40g/mole
Volume of NaOH = 23.48mL
(i)
Since the mole ratio for the reaction is 1:1, then the number of moles of NaOH required is
2.6904x 10 -3 moles
Since concentration in mol/L (M) can be expressed as 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 ( 𝑚𝑜𝑙
𝐿
) = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 (𝑁𝑎𝑂𝐻) 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝐿𝑖𝑡𝑟𝑒𝑠 (𝐿)
=
2.6904 × 10 −3
𝑉
=
2.6904 × 10 −3 𝑚𝑜𝑙𝑒𝑠
=
(
23.48
1000 ) 𝐿
0.11458
𝑚𝑜𝑙
𝐿

(ii)
Concentration in g/L 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 ( 𝑔
) = 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 (
𝐿 𝑚𝑜𝑙
𝐿
) × 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 ( 𝑔
) = 0.11458
𝐿 𝑚𝑜𝑙
𝐿
× 40 = 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 ( 𝑔
𝐿
) = 4.5833 ( 𝑔
𝐿
)

Question 5
Calculate the % of P in ferric phosphate
Solution
Molecular structure of ferric phosphate= 𝐹𝑒𝑃𝑂
4
Molar mass of iron Fe = 56 g/mol
Molar mass of phosphate P = 31 g/mol
Molar mass of oxygen O = 16 g/mol
Molar mass of =56 + 31+ 4(16) = 151g/mol
31 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑜𝑓 𝑃 =
151
× 100 = 20.5298%

Question 7
How many atoms of chlorine are there in 7.58mg of carbon tetrachloride?
Solution
Mass of CCl
4
= 7.58mg
=0.00758 gram
Molar mass of CCl
4
= 154 gram/mol
Number of moles of CCl
4
= 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐶𝐶𝑙
4
(𝑔𝑟𝑎𝑚)
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐶𝐶𝑙
4
= 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐶𝐶𝑙
4
(𝑔𝑟𝑎𝑚/𝑚𝑜𝑙)
= 4.99221 × 10 −5 𝑚𝑜𝑙𝑒𝑠
=
0.00758
154
Since there 4 atoms of Cl in 1 molecule of CCl
4
, then the number of moles of chlorine atoms is
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐶𝑙 𝑎𝑡𝑜𝑚𝑠 = 4 × 4.99221 × 10 −5 𝑚𝑜𝑙𝑒𝑠 = 1.96883 × 10 −4 𝑚𝑜𝑙𝑒𝑠
Then, the number of Cl atoms can be determined 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝐶𝑙 𝑎𝑡𝑜𝑚𝑠 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐶𝑙 𝑎𝑡𝑜𝑚𝑠 × 𝐴𝑣𝑜𝑔𝑎𝑑𝑟𝑜 ′ 𝑠 𝑛𝑢𝑚𝑏𝑒𝑟
= 1.96883 × 10 −4 𝑚𝑜𝑙𝑒𝑠 × 6.022 × 10 23
= 1.18566 × 10 20 𝑎𝑡𝑜𝑚𝑠 𝑜𝑓 𝐶𝑙