The document describes several titration experiments: 1) Titration of ethanoic acid in vinegar with sodium hydroxide to determine its molarity. The vinegar was diluted before titration. 2) Titration of hydrated sodium carbonate with hydrochloric acid to determine its molarity and empirical formula. 3) Iodometric titration of vitamin C with potassium iodate to determine the amount of vitamin C. 4) Titration of iron in an iron tablet with potassium manganate to determine the percentage of iron. 5) Titration of chlorine in bleach with sodium thiosulfate after reaction with potassium iodide to determine its molarity.

1. Equilibrium established when ethanoic acid and ethanol react together
in strong acid, using propanone as solvent. Eqn given.
CH3COOH + C2H5OH ↔ CH3COOC2H5 + H2O
Density ethanoic acid is 1.05 g cm–3.
i. Find amt, mol, of acid present
ii. Conc acid is 1.748 mol dm–3. Find % uncertainty of conc.
Titration performed on acid using a base. Result shown below
Find absolute uncertainty of titre for Titration 1 (27.60 cm3).
Liquid Vol/cm3
Ethanoic acid 5.00 ± 0.05
Ethanol 5.00 ± 0.05
Hydrochloric acid 1.00 ± 0.02
Propanone 39.0 ± 0.5
gmass
voldenstiymass
25.500.505.1 

vol
mass
Density 
molMol
RMM
mass
Mol
0874.0
60
25.5


RMM acid = 60
% uncertainty acid conc = % uncertainty in vol acid + % uncertainty in total vol
vol
mol
acidConc .
(0.05/5.00) x 100 %
= 1 %
(0.62/50) x 100%
= 1.24 %
Total % uncertainty = (1 + 1.24) %
= 2.24%
Uncertainty final – initial vol
(28.80 ±0.05 – 1.20 ±0.05 )
= (27.60 ± 0.1)
Add absolute uncertainty together

2. Method
10 ml of H2O2 measured using 100 ml measuring cylinder and placed in conical flask. A bung and tubing was
attached and the other end connected to a 100 ml gas syringe. 0.2 g of MnO2 catalyst was weighed out.
Catalyst was added to H2O2 . Total vol of O2 produced was measured.
Reaction: 2H2O2 → 2 H2O + O2
Ave vol gas produced = 13.7 ml
Assuming stp, no. of moles in 13.7 mlof gas = 13.7 ml ÷24,000 ml = 5.69 x 10-4 mol
2 mol H2O2 produce 1 mol O2
5.69 x 10-4 mol H2O2 in 10 ml= 1.139 x 10-3 mol
Conc of H2O2 sol = 1.139 x 10-3 mol ÷ 0.010 dm3 = 0.1139M
Accuracy of equipment used is shown below.
Uncertainty listed below, cal total percentage error.
100 ml measuring cylinder ± 0.5ml
100 ml gas syringe ± 0.5 ml
100 ml conical flask ± 5 ml
Top pan balance ± 0.005 g
Error from measuring cylinder = ± 0.5 / 10 x 100% = 5%
Error from gas syringe = ± 0.5 / 13 x 100% = 3.85%
Total error = 5% + 3.85% = 8.85%
Suggest how % error could been reduced
using same equipment
↓
↓
By using a larger vol of H2O2
↓
Larger vol of gas would be produced
↓
Both of above errors would be reduced
(Denominator in both cases is larger)
Comment on use of sig fig in his analysis.
Give final conc of H2O2
Vol gas - 2 sig fig.
Final conc to 2 sig fig - 0.11 M
↓

3. i. Find total uncertainty, in vol of rxn mixture
Mixture contained:
5.0 ± 0.1 cm3 of 2M H2O2
5.0 ± 0.1 cm3 of 1 % starch
20.0 ± 0.1 cm3 of 1M H2SO4
20.0 ± 0.1 cm3 of 0.01 M Na2S2O3
50.0 ± 0.1 cm3 of water with 0.0200 ± 0.0001 g KI
i. Add all vol together: Add all absolute uncertainty together.
(5.0 ± 0.1)+ (5.0 ± 0.1) + (20.0 ± 0.1)+ (20.0 ± 0.1 ) + (50.0 ± 0.1)
= (100 ± 0.5) cm 3
ii. Conc KI =Mass/ vol
% uncertainty conc KI = % uncertainty mass + % uncertainty vol KI
% ∆ mass = (0.0001/0.02) x 100% = 0.5 %
% ∆ vol = (0.1/50) x 100% = 0.2 %
% conc KI = (0.5 + 0.2)% = 0.7 %
iii. Final Conc KI = Conc KI in total mixture
ii. Find % uncertainty for KI conc in final rxn sol.
iii. Find % uncertainty for KI conc in overall rxn mixture
% ∆ final conc KI = % ∆ conc KI + % ∆ total vol KI
% ∆ conc KI
= 0.7 %
% ∆ total vol
= (0.5/100) x 100 %
= 0.5%
% conc KI
= (0.5 + 0.7)
= 1.2 %
Mixture contained:
5.0 ± 0.1 cm3 of 2M H2O2
5.0 ± 0.1 cm3 of 1 % starch
20.0 ± 0.1 cm3 of 1M H2SO4
20.0 ± 0.1 cm3 of 0.01 M Na2S2O3
50.0 ± 0.1 cm3 of water with 0.0200 ± 0.0001 g KI
total vol
Only vol/mass/conc KI
Two rxn kinetic investigated using iodine clock rxn.
Reaction A: H2O2 + 2I− + 2H+ → I2 + 2H2 O
Reaction B: I2 + 2S2O3
2− → 2I− + S4O6
2-

4. 4.32 x 10-5 x 176.14
= 7.61 x10-3 g Vit C
KIO3 + 5KI + 6H+ → 3I2 + 6K+ + 3H2O
3C6H8O6 + 3I2 → 3C6H6O6 + 6I- + 6H+
Iodometric titration on Vit C, (C6H8O6).
Vit C titrated with 0.002M KIO3 , using excess KI and starch.
Redox Titration – Vit C quantification
KIO3
M = 0.002M
Vit C
Amt = ?
Mole ratio (1 :3)
1 mol KIO3 : 3 mol I2 : 3 mol C6H8O6
1 mol KIO3 3 mol C6H8O6
Amt = ?
transfer
1g KI excess + starch
titrated
Vit C
5
3
1032.4)..(
3
1
)..(
0072.0002.0
3
1
).(
)(





CVitMole
CVitMole
CVitMV
KIOMV
i. Find mass, of KIO3, required to prepare 0.250 dm3 of 0.002M KIO3
ii Titration results shown in table below
Find % uncertainty in mean vol of KIO3 used.
Mean vol = (7.20 ± 0.10) cm3
Find amt of KIO3 used
Mol = M x V
= 0.002 x 7.20
1000
= 1.44 x 10-5 mol
4
1000.5
250.0002.0
250.0
002.0
.





mol
mol
mol
vol
mol
acidConc
Convert mole KIO3 → Mass/g
X RMM
5.00 x 10-4 x 214.00 = 0.107 g
% ∆ vol = (0.10/7.20) x 100 %
= 1.4 %
Find amt, Vit C in sample Find mass of Vit C
Convert mole VIT C → Mass
RMM
Vit C – 176.14

5. M x 0.0292 = 2.5 x 10-3 acid
M = 2.5 x 10-3
0.0292
M = 0.0856M
Acid/Base Titration– Ethanoic acid in vinegar
CH3COOH
M = ?
V = 29.2ml
NaOH
M = 0.1M
V = 25.0ml
NaOH + CH3COOH → CH3COONa + H2O
M = 0.1M M = ?
V = 25ml V = 29.2ml
V = 250ml
M = ?
Mole ratio (1 : 1)
1 mole NaOH - 1 mole acid
2.5 x 10-3 mole NaOH - 2.5 x 10-3 acid
Mole ratio – 1: 1
Diluted 10x
V = 25 ml
M = ?
25ml of conc vinegar (ethanoic acid) was diluted to total vol 250 ml in a flask. Diluted vinegar was transfer to a
burette. 25ml, 0.1M NaOH is pipette into a flask, with indicator added. End point reached when average 29.2 ml of
diluted vinegar added. Find its molarity.
mole ratio
Moles bef dilution = Moles aft dilution
M1 V1 = M2V2
M1 = Ini molarity M2= Final molarity
V1 = Ini vol V2 = Final vol
Mole NaOH = MV
= (0.1 x 0.025)
= 2.5 x 10-3
0856.0
1
1
0292.0
025.01.0
1
1




a
aa
bb
M
VM
VM
formula
MM
M
VMVM
856.0
2500856.025
1
1
2211




6. Acid/Base Titration - Empirical formula Na2CO3. x H2O
HCI
M = 0.100 M
V = 48.8ml
Na2CO3
M = ? M
V = 25 ml
2HCI + Na2CO3 → 2NaCI + CO2 + H2O
M = 0.1M M = ?
V = 48.8ml V = 25.0ml
V = 1L
M = ?
25 ml
transfer
Mole ratio – 2: 1
Mass Na2CO3 . x H2O = 27.82 g
Mass Na2CO3 = 10.36 g
Mass of water = (27.82 – 10.36) g
= 17.46 g
Diuted to 1L
27.82g
Na2CO3. xH2O
27.82 g of hydrated (Na2CO3 . x H2O) dissolved in water, making up to 1L. 25 ml of sol was
neutralized by 48.8ml of 0.1 M HCI. Cal molarity and mass of Na2CO3 present in 1L of sol. Find x
Convert mol dm-3 → g dm-3
Empirical formula
Na2CO3 H2O
Mass/g 10.36 17.46
RMM 106 18.02
Mole 10.34/106
= 0.09773
17.46/18.02
= 0.9689
Lowest
ratio
0.09773/0.09733
1
0.9689/0.09733
10
Empirical formula
Na2CO3 . 10 H2O
M
M
VM
VM
b
bb
aa
0976.0
1
2
0250.0
0488.01.0
1
2




0.0976 x 106 = 10.36g/dm3
X RMM

7. Redox Titration - % Fe in iron tablet
Iron tablet contain FeSO4.7H2O. One tablet weigh 1.863 g crushed, dissolved in water
and sol made up to 250 ml. 10ml of this sol added to 20 ml of H2SO4 and titrated with
0.002M KMnO4. Ave 24.5ml need to reach end point. Cal % Fe in iron tablet.
1.863 g
250ml
KMnO4
M = 0.002M
V = 24.5 ml
Fe2+
M = ?
V = 30ml
MnO4
- + 5Fe2+ + 8H+ → Mn2+ + 5Fe2+ + 4H2O
M = 0.002M M = ?
V = 24.5ml
Mole ratio – 1: 5
Mass (expt yield) = 1.703g
Mass (Actual ) = 1.863g
% Fe = 1.703 x 100%
1.863
= 91.4%
6.125 x 10-3 x 278.05 = 1.703 g FeSO4
10ml sol contain - 2.45 x 10-4 Fe2+
250ml sol contain - 250 x 2.45 x 10-4 Fe2+
10
= 6.125 x 10-3 mole Fe2+
42
2
1045.2.
5
1
.
0245.0002.0
5
1






FeMole
FeMole
VM
VM
bb
aa
Convert mole → Mass
X RMM

8. Mole bef dil = Mole aft dil
M1 V1 = M2V2
M1 x 10 = 1.78 x 10-2 x 250
M1 = 1.78 x 10-2 x 250
10
M1 = 0.445M
2CIO- + 2I- + 2H+ → I2 + 2CI- + H2O
I2 + 2S2O3
2- → S4O6
2- + 2I-
10ml bleach (CIO-) diluted to a vol of 250 ml. 20ml is added to 1g of KI (excess) and iodine
produced is titrated with 0.0206 M Na2S2O3.Using starch indicator, end point was 17.3ml.
Cal molarity of CIO in bleach.
Redox Titration – CIO- in Bleach
Na2S2O3
M = 0.0206M
V = 17.3ml
I2
M = ?
Mole ratio ( 1 : 1)
2 mole CIO- : 1 mole I2 : 2 mole S2O3
2-
2 mole CIO- 2 mole S2O3
2-
10.0ml CIO-
transfer
V = 250ml
M = 1.78 x 10-2 M
20ml transfer
1g KI excess
added
M x V = Mol CIO-
M x V = 3.56 x 10-4
M x 0.02 = 3.56 x 10-4
M = 3.56 x 10-4
002
M = 1.78 x 10-2 M diluted 25x
Diuted 25x
V = 10
M = ?
titrated
Water added
till 250ml
4
32
1056.3)..(
2
2
0173.00206.0
).(
2
2
)(
)(





CIOMole
CIOMole
OSMV
CIOMV

9. KIO3 + 5KI + 6H+ → 3I2 + 6K+ + 3H2O
3C6H8O6 + 3I2 → 3C6H6O6 + 6I- + 6H+
Iodometric titration on Vit C, (C6H8O6). 25 ml Vit C titrated with 0.002M KIO3 from burette,
using excess KI and starch. Ave vol KIO3 is 25.5ml. Cal molarity of Vit C.
Redox Titration – Vit C quantification
KIO3
M = 0.002M
V = 25.5ml
Vit C
M = ?
V = 25ml
Mole ratio (1 :3)
1 mol KIO3 : 3 mol I2 : 3 mol C6H8O6
1 mol KIO3 3 mol C6H8O6
V = 25ml
M = ?
25ml transfer
1g KI excess + starch
titrated
Vit C
4
3
1053.1)..(
3
1
)..(
0255.0002.0
3
1
).(
)(





CVitMole
CVitMole
CVitMV
KIOMV M x V = Mol Vit C
M x V = 1.53 x 10-4
M x 0.025 = 3.56 x 10-4
M = 3.56 x 10-4
0025
M = 6.12 x 10-3 M

10. 2.82 x 10-3 x 63.5 = 0.179 g Cu in 25ml
1.79 g Cu in 250ml
% Cu = mass Cu x 100%
mass brass
= 1.79 x 100%
2.5
= 71.8%
2Cu2+ + 4I- → I2 + 2CuI
I2 + 2S2O3
2- → S4O6
2- + 2I-
2.5g brass react with 10ml HNO3 producing Cu2+ ion. Sol made up to 250ml using water. Pipette 25ml of sol to
flask. Na2CO3 add to neutralize excess acid. 1g KI (excess) and starch added. Titrate with 0.1M S2O3
2- and end
point, is 28.2 ml. Find molarity Cu 2+ and % Cu found in brass.
Redox Titration - % Cu in Brass
Na2S2O3
M = 0.1M
V = 28.2ml
I2
M = ?
Mole ratio (1 : 1)
2 mol Cu2+ : 1 mol I2 : 2 mol S2O3
2-
2 mol Cu2+ 2 mol S2O3
2-
Pour into
Volumetric flask
V = 250ml
M = ?
25ml transfer
1g KI excess/ starch
10 ml HNO3
titrated
Water added 250ml
2.5g brass
32
2
2
32
2
1082.2).(
2
2
0282.01.0
).(
2
2
)(
)(








CuMole
CuMole
OSMV
CuMV
M x V = Mol Cu 2+
M x V = 2.82 x 10-3
M x 0.025 = 2.82 x 10-3
M = 2.82 x 10-3
0025
M = 1.13 x 10-3 M
Convert mole Cu → Mass Cu
X RMM
X 10

11. 2Cu2+ + 4I- → I2 + 2CuI
I2 + 2S2O3
2- → S4O6
2- + 2I-
0.456 g brass react with 25ml HNO3 producing Cu2+ ions. Sol was titrate with 0.1M S2O3
2-
and end point, reached when 28.5ml added. Cal mol, mass, molarity of Cu 2+ and % Cu in brass.
Redox Titration - % Cu in Brass
Na2S2O3
M = 0.1M
V = 28.5ml
I2
M = ?
Mole ratio (1 : 1)
2 mol Cu2+ : 1 mol I2 : 2 mol S2O3
2-
2 mol Cu2+ 2 mol S2O3
2-
transfer
1g KI excess starch
25ml HNO3
titrated
0.456g
brass
32
2
2
32
2
1085.2).(
2
2
0285.01.0
).(
2
2
)(
)(








CuMole
CuMole
OSMV
CuMV
M x V = Mol Cu 2+
M x V = 2.85 x 10-3
M x 0.025 = 2.85 x 10-3
M = 2.85 x 10-3
0025
M = 1.14 x 10-3 M
Convert mole Cu → Mass Cu
2.85 x 10-3 x 63.5 = 0.18 g Cu
X RMM
% Cu = mass Cu x 100%
mass brass
= 0.18 x 100%
0.456
= 39.7 %

12. % Calcium carbonate in egg shell - Back Titration
250ml,
2M HNO3
Amt of HNO3 added
Amt of base (egg)
Amt of
HNO3 left
Titrate NaOH
M = 1.0
V = 17.0ml
Amt HNO3 react = Amt HNO3 – Amt HNO3
add left
HNO3 left
Transfer
to flask
Left overnight in acid
added
25 g of egg shell (CaCO3) dissolved in 250 ml, 2 M HNO3. Sol titrated with NaOH.
17 ml, 1M NaOH needed to neutralize excess acid. Cal % CaCO3 by mass in egg shell.
NaOH + HNO3 → NaNO3 + H2O
M = 1.00M mol = ?
V = 17 ml
Amt HNO3 add = M x V
= 2.0 x 0.250
= 0.50 mol
Amt HNO3 react = Amt HNO3 add – Amt HNO3 left
= 0.50 – 1.7 x 10-2
= 0.483 mol
2HNO3 + CaCO3 → (CaNO3)2 + H2O + 2CO2
Mole Mole
0.483 ?
Mole ratio (2 : 1)
2 mol HNO3 - 1 mol CaCO3
0.483 mol HNO3 - o.242 mol CaCO3
2
107.1..
1
1
).(
017.000.1
1
1





acidMole
acidMole
VM
VM
aa
bb
25 g impure
CaCO3 in egg shell
Convert mole CaCO3 → Mass /g
X RMM
0.242 x 100 = 24.2 g CaCO3
% CaCO3 = mass CaCO3 x 100%
mass egg
= 24.2 x 100%
25.0
= 96.8 %

13. % Calcium carbonate in egg shell - Back Titration
Amt of HCI added
Amt of base (egg)
Amt of
HCI left
Titrate NaOH
M = 0.10
V = 23.8 ml
Amt HCI react = Amt HCI – Amt HCI
add left
HCI left
Transfer
to flask
Left overnight in acid
added
NaOH + HCI → NaCI + H2O
M = 0.1 M mol = ?
V = 23.8 ml
Amt HCI add = M x V
= 0.2 x 0.272
= 0.0544 mol
Amt HCI react = Amt HCI add – Amt HCI left
= 0.0544 – 2.38 x 10-3
= 0.00306 mol
2HCI + CaCO3 → CaCI3 + H2O + CO2
Mole Mole
0.00306 ?
Mole ratio (2 : 1)
2 mol HCI - 1 mol CaCO3
0.00306 mol HCI - o.00153 mol CaCO3
3
1038.2..
1
1
).(
238.01.0
1
1





acidMole
acidMole
VM
VM
aa
bb
Convert mole CaCO3 → Mass /g
X RMM
0.00153 x 100 = 0.153 g CaCO3
% CaCO3 = mass CaCO3 x 100%
mass egg
= 0.153 x 100%
0.188
= 81.4 %
0.188g of egg shell (CaCO3) dissolved in 27.2 ml, 0.2 M HCI.
Sol was titrated with NaOH. 23.8ml, 0.10M NaOH need to neutralize excess acid.
Cal mol, mass and % of CaCO3 by mass in egg shell.
0.188g impure
CaCO3 in egg shell
27.20ml, 0.2M
HCI

14. Amt of HCI added
Amt of base
Amt of
HCI left
Titrate NaOH
M = 0.1108
V = 33.64 ml
Amt HCI react = Amt HCI – Amt HCI
add left
HCI left
Transfer
to flask
Left overnight in acid
added
NaOH + HCI → NaCI + H2O
M = 0.1108 M mol = ?
V = 33.64 ml
Amt HCI add = M x V
= 0.250 x 0.05
= 0.0125 mol
Amt HCI react = Amt HCI add – Amt HCI left
= 0.0125 – 3.727 x 10-3
= 0.008773 mol
2HCI + Ca(OH)3 → CaCI3 + H2O
Mole Mole
0.008773 ?
Mole ratio (2 : 1)
2 mol HCI - 1 mol Ca(OH)2
0.008773 mol HCI - o.004386 mol Ca(OH)2
3
10727.3..
1
1
).(
03364.01108.0
1
1





acidMole
acidMole
VM
VM
aa
bb
Convert mole Ca(OH)2 → Mass /g
X RMM
0.004386 x 74.1 = 0.325g Ca(OH)2
% Ca(OH)2 = mass Ca(OH)2 x 100%
mass impure
= 0.325 x 100%
0.5214
= 62.3 %
50 ml, 0.250M
HCI
% Calcium hydroxide in antacid tablet - Back Titration
0.5214 g of impure Ca(OH)2 from antacid was dissolved in 50 ml, 0.250M HCI.
33.64ml, 0.1108 M NaOH need to neutralize excess acid. Cal % Ca(OH)2 in tablet.
0.5214g impure
Ca(OH)2

15. Amt of NaOH added
Amt of acid
Amt of
NaOH left
Titrate HCI
M = 0.5
V = 17.6 ml
Amt NaOH react = Amt NaOH – Amt NaOH
add left
NaOH left
Transfer
to flask
Left overnight in acid
added
HCI + NaOH → NaCI + H2O
M = 0.5 M mol = ?
V = 17.6 ml
Amt NaOH add = M x V
= 2 x 0.02
= 0.04 mol
Amt NaOH react = Amt NaOH add – Amt NaOH left
= 0.04 – 8.8 x 10-3
= 0.0312 mol
2NaOH + H2A → Na3 A+ 2H2O
Mole Mole
0.0312 ?
Mole ratio (2 : 1)
2 mol NaOH - 1 mol acid
0.0312 mol NaOH - 0.0156 mol acid
3
108.8.
1
1
).(
0176.05.0
1
1





baseMole
acidMole
VM
VM
bb
aa
Molar mass of insoluble acid in tablet -Back Titration
2.04 g insoluble acid dissolve in 20 ml, 2M NaOH. Excess NaOH require
17.6 ml, 0.5M HCI to neutralize it. Find molar mass acid
2.04 g impure
acid H2A
20 ml, 2M NaOH
Molar Mass Acid
0.0156 mol acid - 2.04 g
1 mol acid - 2.04
0.0156
= 131

16. Element H B O
Step 1 Percentage/% 4.8% 17.7% 77.5%
RAM/RMM 1 11 16
Step 2 Number
moles/mol
4.8/1
= 4.8
17.7/11
= 1.6
77.5/16
= 4.84
Step 3 Simplest ratio 4.8/1.6
= 3
1.6/1.6
= 1
4.84/1.6
= 3
Boric acid used to preserve food contains 4.8% hydrogen, 17.7% boron and rest is oxygen.
Find EF of boric acid.
Empirical formula - H3B1O3.
Empirical Formula Calculation
Empirical Formula Calculation
Step 1: Write mass/ % of each element
Step 2: Find number of moles of each element (divide with RAM)
Step 3: Obtain the simplest ratio
2.5 g of X combined with 4 g of Y to form compound formula XY2.
RAM of Y is 80, Find RAM of X.
Element X Y
Step 1 Mass/g 2.5 4
RAM/RMM RAM 80
Step 2 Number
moles/mol
2.5/RAM
= ?
4/80
= 0.05
Step 3 Simplest ratio 1 2 Empirical formula given as X1Y2.
RAM = 100
100
2
05.05.2
2
1
05.0
/5.2



RAM
RAM
RAM

17. CHO + O2 CO2 + H2O
X contain 85.7% of carbon by weight. 4.2 g of gas X occupy vol of 3.36 dm3 at stp.
Find EF, RMM and MF of X
Element C H
Step 1 Percentage/% 85.7 14.3
RAM/RMM 12 1
Step 2 Number
moles/mol
85.7/12
= 7.14
14.3/1
= 14.3
Step 3 Simplest ratio 7.14/7.14
= 1
14.3/7.14
= 2
a) Empirical Formula = C1H2
b) Vol of 3.36 dm3 at stp – Mass, 4.2 g
Vol of 22.4dm3 at stp – Mass 1 mol (RMM)
3.36 dm3 – 4.2 g
22.4 dm3 - (4.2 x 22.4)/3.36 = 28
c) Assume molecular formula of X - (CH2)n
RMM of X is (12+2)n = 28
n = 2
Molecular formula X = C2H4
X contain carbon, hydrogen and oxygen. 0.50 g of compound on combustion, yield
0.6875 g of carbon dioxide and 0.5625 g of water. Find EF.
Element C H O
Step 1 Mass/g 0.1875 0.0625 0.25
RAM/RMM 12 1 16
Step 2 Number
moles/mol
0.1875/1 2
= 0.01562
0.0625/1
= 0.0625
0.25/16
= 0.01562
Step 3 Simplest ratio 0.01562
0.01562
= 1
0.0625
0.01562
= 4
0.01562
0.01562
= 1
Conservation of mass
Mass C atom before = Mass C atom after
Mass H atom before = Mass C atom after
Mol C atom in CO2
= 0.6875 = 0.0156 mol
44
Mass C = mol x RAM C
atom = 0.015625 x 12
= 0.1875 g
Mol H atom in H2O
= 0.5625 = 0.03125 x 2 = 0.0625 mol
18
Mass H = mol x RAM H
atom = 0.0625 x 1
= 0.0625 g
0.6875g 0.5625g0.50g 0.75g
Mass of O = (Mass CHO – Mass C – Mass H)
= 0.5 – 0.1875 - 0.0625 = 0.25 g
Empirical formula – C1H4O1
Empirical Formula Calculation

18. Element C H O
Step 1 Mass/g 0.731 0.0730 0.195
RAM 12.01 1.01 16.01
Step 2 Number
moles/mol
0.731/12.01
= 0.0609
0.0730/1.01
= 0.0730
0.195/16.01
= 0.0122
Step 3 Simplest
ratio
0.0609
0.0122
= 5
0.0730
0.0122
= 6
0.0122
0.0122
= 1
CHO + O2 CO2 + H2O
2.68 g 0.657 g1.00g
Empirical formula – C5H6O1
X contain elements carbon, hydrogen and oxygen.
Find EF of X, if 1 g X form 2.68 g of carbon dioxide and 0.657 g water when combust with O2.
Find MF of X, if 0.3 mol has mass of 98.5 g.
Empirical Formula = C5H6O1
Mole → Mass
0.3 mol → 98.5 g
1 mol → 98.5/0.3
RMM = 328 gmol-1
Assume MF - (C5H6O1)n = 328
RMM = ( (5 x 12.01) +(6 x 1.01) + ( 1 x 16.01) )n = 328
82.11 x n = 328 = 4
MF = (C5H6O1)4 C20H24O4
Find MF, given 0.3 mol X has mass of 98.5 g.
Empirical Formula Calculation
Conservation of mass
Mass C atom before = Mass C atom after
Mass H atom before = Mass C atom after
Mol C atom in CO2
= 2.68 = 0.0609 mol
44
Mass C = mol x RAM C
= 0.0609 x 12
= 0.731 g
Mol H atom in H2O
= 0.657 x 2 = 0.0729 mol
18
Mass H = mol x RAM H
= 0.0729 x 1
= 0.0736 g
Mass O = (Mass CHO – Mass C – Mass H)
= 1.0 – 0.731 - 0.0736 = 0.195 g

19. Find EF for X with composition by mass. S 23.7 %, O 23.7 %, CI 52.6 %
Given, T- 70 C, P- 98 kNm-2 density - 4.67g/dm3
What molecular formula?
Empirical formula - SO2CI2
Density ρ = m (mass)
V (vol)
Ideal Gas Equation
Element S O CI
Composition 23.7 23.7 52.6
Moles 23.7
32.1
= 0.738
23.7
16.0
= 1.48
52.6
35.5
= 1.48
Mole ratio 0.738
0.738
1
1.48
0.738
2
1.48
0.738
2
P
RT
M
P
RT
V
m
M
RT
M
m
PV
nRTPV




Density = 4.67 gdm-3
= 4.67 x 10-3 gm-3
M = (4.67 x 10-3) x 8.31 x (273 +70)
9.8 x 104
M = 135.8
135.8 = n [ 32 + (2 x 16)+(2 x 35.5) ]
135.8 = n [ 135.8]
n = 1
MF = SO2CI2
P = 98 kN-2
= 9.8 x 104 Nm-2
3.376 g gas occupies 2.368 dm3 at T- 17.6C, P - 96.73 kPa.
Find molar mass
PV = nRT
PV = mass x RT
M
M = mass x R x T
PV
= 3.376 x 8.314 x 290.6
96730 x 2.368 x 10-3
= 35.61
Vol = 2.368 dm3
= 2.368 x 10-3 m3
P – 96.73 kPa → 96730Pa
T – 290.6K
6.32 g gas occupy 2200 cm3, T- 100C , P -101 kPa.
Calculate RMM of gas
PV = nRT
n = PV
RT
n = (101 x 103) (2200 x 10-6)
8.31 x ( 373 )
n = 7.17 x 10-2 mol
Vol = 2200 cm3
= 2200 x 10-6 m3
RMM = mass
n
RMM = 6.32
7.17 x 10-2
= 88.15

20. P = 101 kNm-2
= 101 x 103 Nm-2
Calculate RMM of gas
Mass empty flask = 25.385 g
Mass flask fill gas = 26.017 g
Mass flask fill water = 231.985 g
Temp = 32C, P = 101 kPa
Find molar mass gas by direct weighing, T-23C , P- 97.7 kPa
Mass empty flask = 183.257 g
Mass flask + gas = 187.942 g
Mass flask + water = 987.560 g
Mass gas = (187.942 – 183.257) = 4.685 g
Vol gas = Vol water = Mass water = (987.560 – 183.257) = 804.303 cm3
RMM determination
PV = nRT
PV = mass x R x T
M
M = mass x R x T
PV
= 4.685 x 8.314 x 296
97700 x 804.303 x 10-6
= 146.7
Vol gas = 804.303 cm3
= 804.303 x 10-6 m3
P = 97.7 kPa
= 97700 Pa
Density water = 1g/cm3
M = m x RT
PV
= 0.632 x 8.314 x 305
101 x 103 x 206 x 10-6
= 76.8
m gas = (26.017 – 25.385)
= 0.632 g
vol gas = (231.985 – 25.385)
= 206 x 10-6 m3
X contain C, H and O. 0.06234 g of X combusted,
0.1755 g of CO2 and 0.07187 g of H2O produced.
Find EF of X
Element C H O
Step 1 Mass/g 0.0479 0.00805 0.006384
RAM/RMM 12 1 16
Step 2 Number
moles/mol
0.0479/1 2
= 0.00393
0.00805/1
= 0.00797
0.006384/16
= 0.000393
Step 3 Simplest ratio 0.00393
0.000393
= 10
0.00797
0.000393
= 20
0.000393
0.000393
= 1
Conservation of mass
Mass C atom before = Mass C atom after
Mass H atom before = Mass C atom after
CHO + O2 CO2 + H2O
Mol C atom in CO2
= 0.1755 = 0.00393 mol
44
Mass C = mol x RAM C
= 0.00393 x 12
= 0.0479 g
Mol H atom in H2O
= 0.07187 = 0.0039 x 2 = 0.00797 mol
18
Mass H = mol x RAM H
= 0.00797 x 1.01
= 0.00805 g
Mass of O = (Mass CHO – Mass C – Mass H)
= 0.06234 – 0.0479 - 0.00805 = 0.006384 g
0.06234 g 0.1755 g 0.07187 g
Empirical formula – C10H20O1

21. Sodium azide, undergoes decomposition rxn to produce N2 used in air bag
2NaN3(s) → 2Na(s) + 3N2(g)
Temp, mass and pressure was collected in table below
i. State number of sig figures for Temp, Mass, and Pressure
i. Temp – 4 sig fig Mass – 3 sig fig Pressure – 3 sig fig
Temp/C Mass NaN3/kg Pressure/atm
25.00 0.0650 1.08
ii. Find amt, mol of NaN3 present
ii.
iii. Find vol of N2, dm3 produced in these condition
RMM NaN3 – 65.02
molMol
RMM
mass
Mol
00.1
02.60
0.65


P
nRT
V
nRTPV

 n = 1.50 mol
P – 1.08 x 101000 Pa
= 109080 Pa
2NaN3(s) → 2Na(s) + 3N2(g)
T – 25.00 + 273.15
= 298.15K
2 mol – 3 mol N2
1 mol – 1.5 mol N2
33
1.340341.0
109080
15.29831.850.1
dmmV
V
P
nRT
V





22. Sodium azide, undergoes decomposition rxn to produce N2 used in air bag
2NaN3(s) → 2Na(s) + 3N2(g)
Temp, mass and pressure was collected in table below
Temp/C Volume N2/L Pressure/atm
26.0 36 1.15
Find mass of NaN3 needed to produce 36L of N2
RMM NaN3 – 65.02
RT
PV
n
nRTPV


1.1 x 65.02 = 72 g NaN3
P – 1.15 x 101000 Pa
= 116150 Pa
2NaN3(s) → 2Na(s) + 3N2(g)
T – 26.0 + 273.15
= 299.15K
3 mol N2 – 2 mol NaN3
1.7 mol N2 – 1.1 mol NaN3
moln
n
7.1
15.29931.8
1036116150 3





Vol = 36 dm3
= 36 x 10-3 m3
Convert mole NaN3 → Mass /g

23. Copper carbonate, CuCO3, undergo decomposition to produce a gas.
Determine molar mass for gas X
CuCO3(s) → CuO(s) + X (g)
Temp, mass, vol and pressure was collected in table below
Temp/K Vol gas/ cm3 Pressure/kPa Mass gas/g
293 38.1 101.3 0.088
Find Molar mass for gas X
P – 101300 Pa
T – 293 K
Vol = 38.1 cm3
= 38.1 x 10-6 m3
PV
mRT
M
RT
M
m
PV
nRTPV



5.55
101.38101300
29331.8088.0
6



 
M
M
Potassium chlorate, KCIO3, undergo decomposition to produce a O2.
Find amt O2 collected and mass of KCIO3 decomposed
KCIO3
Temp/K Vol gas/ dm3 Pressure/kPa
299 0.250 101.3
2KCIO3(s) → 2KCI(s) + 3O2 (g)
RT
PV
n
nRTPV


2
3
.010.0
29931.8
10250.0101300
Omoln
n





Vol = 0.250 dm3
= 0.250 x 10-3 m3
P – 101300 Pa
Convert mole KCIO3 → Mass
2KCIO3 → 2KCI+ 3O2
2 mol – 3 mol O2
0.0066 mol – 0.01 mol O2
0.0066 x 122.6 = 0.81 g KCIO3
RMM KCIO3 – 122.6

24. Biological Oxygen Demand
2Mn2+ + O2 + 4OH- → 2MnO2 + 2H2O
2MnO2 + 4I- + 4H+ → 4I2 + 2Mn2- + 4H2O
4I2 + 4S2O3
2- → 4I- + 2S4O6
2-
Dissolve O2 reacts with alkaline manganese (Mn2+) to form (Mn4+)
4Mn2+ + 4OH- → 2Mn(OH)2
1 mol 2 mol
2Mn(OH)2 + O2 → 2MnO(OH)2
2MnO(OH)2 + 8H+ + 6I- → 2I3
- + 6H2O
2 mol 2 mol
2I3
- + 4S2O3
2- → 6I- + 2S4O6
2-
2 mol 4 mol
DO bottle
Mn2+ salt
1g KI excess
alkaline/OH-
shake
White ppt Mn(OH)2
Conc
H2SO4
White ppt dissolve in acid
Na2S2O3
M = 0.05M
V = 12.5ml
titrated S2O3
2-
1 O2 + 4 S2O3
2- → products
M = ? M = 0.05M
V = 12.5ml
I- oxidized to I2 by Mn2+
O2
M = ?
V = 500ml
2 mol 4 mol
4 mol 4 mol
1 mol O2 : 4 mol I2 : 4 mol S2O3
2-
1 mol O2 4 mol S2O3
2-
Brown I2 sol form
Starch added
Water sample
added
1 mol O2 : 4 mol S2O3
2-
Iodometric titration
I2/thiosulphate/starch
↓
Mn2+ oxidized by O2 to Mn4+
↓
Mn4+ oxidized I- to I2
I2 react with starch
(blue black colour)
↓
S2O3
2- added to reduce I2
↓
I2 used up – blue black
disappear
1 mol 2 mol
4
1
)(
)(
32
2

OSMV
OMV
Redox titration Winkler Method

25. Biological Oxygen Demand
Dissolve O2 reacts with alkaline manganese (Mn2+) to form (Mn4+)
DO bottle
Mn2+ salt
1g KI excess
alkaline/OH-
shake
White ppt Mn(OH)2
Conc
H2SO4
White ppt dissolve in acid
Na2S2O3
M = 0.05M
V = 12.5ml
titrated S2O3
2-
1 O2 + 4 S2O3
2- → products
M = ? M = 0.05M
V = 12.5ml
I- oxidized to I2 by Mn2+
O2
M = ?
V = 500ml
1 mol O2 : 4 mol I2 : 4 mol S2O3
2-
1 mol O2 4 mol S2O3
2-
Brown I2 sol
form
Starch added
Water sample
added
1 mol O2 : 4 mol S2O3
2-
4
1
)(
)(
32
2

OSMV
OMV
500ml water tested for dissolve oxygen by adding Mn2+ in alkaline sol, followed by addition of KI and acid. I2 produced is reduced
by titrating with 0.05M S2O3
2-. Ave vol S2O3
2- used is 12.50ml. Cal dissolved oxygen in g/dm3.
molOMole
OSMVOMole
OSMV
OMV
4
2
322
32
2
1056.10125.005.0
4
1
.
)(
4
1
.
4
1
)(
)(




Mass O2 = (1.56 x 10-4 x 32.0) g
= 0.005 g in 500ml
= 0.01 g in 1000ml
= 0.01 g/dm3
Convert mole O2 → Mass /g
RMM O3 – 32
Redox titration Winkler Method

26. Student A
Weigh 1.00 g X and add 100ml water – 1.00%
Perform 10 x serial dilution
Transfer 1 ml of 1 % to 9 ml = 0.1%
Transfer 1 ml from 0.1% to 9 ml = 0.01%
Transfer 1 ml from 0.01% to 9ml = 0.001%
How both student overcome this problem.
Using serial dilution technique.
Dilution factor 10x.
Given conc of 1.00 % = 1.00 g in 100 ml water.
Student A told to prepare 0.001% of X = (0.001g X in 100 ml)
Student B told to prepare 0.001% of Y = (0.001g Y in 100 ml)
Electronic balance has only precision of 0.00 ± 0.01.
Student B
Weigh 1.00 g Y and add 100ml – 1.00%
Transfer 1 ml of 1% to 999 ml = 0.001%
1%
0.1% 0.01% 0.001%
1 ml 1 ml 1 ml
1 ml
0.001%1%