TITLE: EXPERIMENT 4 (STOICHIOMETRY AND THEORETICAL YIELD)

PURPOSE
The aim of this experiment is to identify the limiting reactant, excess reactant and to determine the
percent yield.

INTRODUCTION
Stoichiometry is the study of the mass ratio of reactants and products in chemical reactions, as
well as how to quantify those proportions. With adequate information, stoichiometry may be used to
compute masses, moles, and percentages inside a chemical equation. A chemical equation is a graphic
representation of a chemical reaction. To comprehend the structure of a reaction, the link between the
products and reactants in a balanced chemical equation is vital. This relationship tells us which materials
and how much of them are required for a reaction to take place. A reactant in a chemical reaction, in
the same way, can limit the number of products produced by the reaction. When this occurs, the reactant
is referred to as the limiting reactant. The excess reactant is the reactant in a chemical reaction that is
present in larger quantities than are required to completely react with the limiting reactant. The reactants
are the materials that remain after a chemical reaction has achieved equilibrium.
The quantities of substances produced are referred to as yields. The theoretical yield is the
highest quantity of product that may be formed from the given amounts of reactants. The amount of
product produced when the reaction is carried out in the laboratory is referred to as the actual yield. In
fact, the actual yield usually always less than the theoretical yield. The actual yield is typically
represented as a percentage yield, indicating what percent of the predicted yield was achieved.
Typically, percent yields are less than 100% due to spills and other experimental mistakes, and there
are usually losses due to an incomplete reaction or unwanted side reactions. Following the end of
laboratory reaction, you will calculate the amount of product, then compare the theoretical yield
obtained with actual yield to evaluate the percent yield:
Actual yield
Percent yield = × 100%
Theoretical yield
Steps to calculate theoretical yield:
1. Balance the reaction equation and determine the ratios of reactants to products.
2. Calculate the number of moles of each reactant used.
3. Determine which reactant is limiting.
4. Calculate the moles of product expected if the yield were 100% based on limiting reactant.
5. Calculate the mass of product corresponding to the number of moles expected (theoretical
yield).

CHEMICALS AND APPARATUS

0.5 M CaCl2
1.5 M Na2 CO3
Burette
Erlenmeyer / Conical flask
Watch glass
Oven
Analytical balance
Gravity filter set:
- Filter paper
- Funnel
Suction filtration set:
- Unfolded filter paper
- Buchner funnel
- Adapter
- Vacuum filter flask
- Rubber tube

METHOD
1. 0.5 M CaCl2 and 1.5 M Na2 CO3 had been dispensed into a clean conical flask. It would be ideal
to reach as close to the recommended volume as possible. (All volumes should be recorded to
two decimal places since burette had been used).
Table 4.1

Reaction 0.5 M 𝐂𝐚𝐂𝐥𝟐 1.5 M 𝐍𝐚𝟐 𝐂𝐎𝟑

1 20mL 10mL

2 20mL 5mL

2. You might not even notice a precipitate at first. The flask had been swirled and leaved for five
minutes to enable the precipitate to completely be formed.
3. Buchner funnel (Figure 4.1) had been used to suction filter the product or gravity filter the white
solid. This is handled by folding a piece of filter paper into quarters and making a funnel. The
filter paper had been placed inside the glass funnel. The solution had been poured into the center
of the filter paper. Ensure that it does not rise over the level of the filter paper.
4. The sides of conical flask had been washed with a small amount of distilled water (use wash
bottle) and it had been added to the filter paper. There will still be some white solid within the
flask, but the time and effort required to recover it are not worthwhile (One area of inefficiency).
5. The filter paper had been removed carefully (It will still be quite wet) and it had been placed
on pre-weighed watch glass. The product had been dried in the oven for half an hour.
Note: This is an effective way to begin the method for the following reactions.
6. The product (CaCO3) had been scrapped from the filter paper onto the watch glass. It had been
removed without tearing or scrapped off some of the filter paper. The filter paper had been
discarded and the watch glass had been returned to the oven for a further ten minutes. The mass
of the product had been determined. The mass of product had been reheated and determined at
least three times to complete the dryness or until the two weighing’s within 0.02g of one
another.
7. The amount of CaCO3 obtained had been compared with the expected amount. The limiting
reactant and excess reactant and percent yield had been identified and calculated.
 DATA SHEET EXPERIMENT 4
STOICHIOMETRY AND THEORETICAL YIELD

NAME: NOOR AFRINA BINTI MOHD FAZRUL DATE: 18 NOVEMBER 2021

STUDENT ID: 2021458818 GROUP: AS251 1M1

Reactions

1 2
Mass of product, g
0.888 0.542
(1st heat)
Mass of product, g
- -
(After reheating)
Mass of product, g
- -
(After reheating)
Mass of product, g 0.888 0.542
RESULT AND CALCULATION
1) For each of the two reactions:
2) Was the yield you obtain satisfactory? Justify your answer.

- Both reactions are acceptable since their percent yield is less than 100% due to losses, side
reactions, and incomplete reactions.
DISCUSSION

In this experiment, a reaction takes place between CaCl2 and Na2 CO3 , producing CaCO3 and
NaCl for both reactions 1 and 2. The balanced chemical equation for both reactions 1 and 2 is:

CaCl2 + Na2 CO3 → CaCO3 + 2NaCl

In reaction 1, we used the recommended volume 20 mL of 0.5 M CaCl2 react with 10 mL of

1.5 M Na2 CO3. When the two aqueous solutions react, a white insoluble solid, CaCO3 (product)
precipitates out of solution whereas the colourless solution is NaCl, a byproduct. To get actual yield of
the product in gram, we need to put CaCO3 in the oven for half an hour. After half an hour, we scrape
the product and heat it for another ten minutes. We weigh the CaCO3, then reheat and determine the
amount of product at least three times to ensure complete dryness and we get 0.888g for actual yield.
Limiting reactant is a reactant that will be totally consumed. Once that reactant is consumed, the reaction
cannot continue. As a result, it limits the reaction from proceeding. While the other reactant is excess
reactant that have left over. The theoretical yield is the maximum possible mass of the product that will
result from the reaction. The percent yield is the ratio of the actual yield to the theoretical yield.

From the balanced chemical equation, we can determine the limiting reactant, excess reactant,
and percent yield from the reaction. But first, we need to find the mole of both reactants using the
formula:

MV
n=
1000

From the calculation based on the formula given, the mole of the CaCl2 is 0.01 mol whereas the
mole of Na2 CO3 is 0.015 mol. We can determine the mole of product using stoichiometry by relate the
product and reactant based on the mole obtained. To identify the limiting reactant, we need to calculate
the mole ratio of each reactant present and compare it to the mole ratio of the reactants in the balanced
chemical equation, hence the reactant with the least mole ratio is the limiting reactant. Thus, the other
reactant is the excess reactant. From reaction 1, we could deduce that CaCl2 is limiting reactant while
Na2 CO3 is excess reactant that may continue to react if the other was not consumed. The theoretical
yield is the maximum possible mass of the product that will result from the reaction. Therefore, we can
compute the mass of theoretical yield which is CaCO3 using the formula:

mass of product
n=
molar mass of product

From the formula above, we can calculate that mass of product is 1g. Since we already get the
theoretical yield, we can determine the percent yield by using the formula below:

Actual yield (gram)

× 100% = Percent yield
Theoretical yield (gram)

The amount of product obtained experimentally from a chemical reaction is known as the actual
yield. From the formula, the percent yield calculated is 88.8%.

In reaction 2, we used the recommended volume 20 mL of 0.5 M CaCl2 react with 5 mL of 1.5
M Na2 CO3. We can observe the presence of white precipitate and the colourless solution. A white
insoluble solid, CaCO3 (product) precipitates out of solution whereas the colourless solution is NaCl, a
byproduct. To get actual yield of the product in gram, we need to put CaCO3 in the oven for half an
hour. After half an hour, we scrape the product and heat it for another ten minutes. We weigh the CaCO3,
then reheat and determine the amount of product at least three times to ensure complete dryness and we
get 0.542g for actual yield. Limiting reactant is a reactant that will be totally consumed. Once that
reactant is consumed, the reaction cannot continue. As a result, it limits the reaction from proceeding.
While the other reactant is excess reactant that have left over. The theoretical yield is the maximum
possible mass of the product that will result from the reaction. The percent yield is the ratio of the actual
yield to the theoretical yield.

From the balanced chemical equation, we can determine the limiting reactant, excess reactant,
and percent yield from the reaction. But first, we need to find the mole of both reactants using the
formula:

MV
n=
1000

From the calculation based on the formula given, the mole of the CaCl2 is 0.01 mol whereas the
mole of Na2 CO3 is 0.0075 mol. We can determine the mole of product using stoichiometry by relate
the product and reactant based on the mole obtained. To identify the limiting reactant, we need to
calculate the mole ratio of each reactant present and compare it to the mole ratio of the reactants in the
balanced chemical equation, hence the reactant with the least mole ratio is the limiting reactant. Thus,
the other reactant is the excess reactant. From reaction 1, we could deduce that Na2 CO3 is limiting
reactant while CaCl2 is excess reactant that may continue to react if the other was not consumed. The
theoretical yield is the maximum possible mass of the product that will result from the reaction.
Therefore, we can compute the mass of theoretical yield which is CaCO3 using the formula:

mass of product
n=
molar mass of product

From the formula above, we can calculate that mass of product is 0.75g. Since we already get
the theoretical yield, we can determine the percent yield by using the formula below:

Actual yield (gram)

× 100% = Percent yield
Theoretical yield (gram)

The amount of product obtained experimentally from a chemical reaction is known as the actual
yield. From the formula, the percent yield calculated is 72.27%.

If we observe at both reactions, we can see that the actual yield is less than the theoretical yield.
This is because when we handle experiments, the reaction may be incomplete, resulting in a partial
conversion of the reactant, or some products may be lost to the surrounding environment. So far, the
actual yield does not same as the theoretical yield, but it is still satisfactory because most of the reactants
have interacted with one another to produce product.

CONCLUSION
In conclusion, the aim of this experiment is to identify the limiting reactant, excess reactant,
and percent yield using the experiment and stoichiometric calculations from the reaction of:

CaCl2 + Na2 CO3 → CaCO3 + 2NaCl

We were able to determine the limiting reactant in reaction 1 which is CaCl2 and Na2 CO3 in
reaction 2. The excess reactant for reaction 1 and reaction 2 is Na2 CO3 and CaCl2 respectively. Based
on the results of the experiment in reaction 1, the calculated percent yield is 88.8% while in reaction 2,
the calculated percent yield is 72.27%.
REFERENCES

• Goldsby, K., & Chang, R. (2012). Chemistry. McGraw-Hill Education.

• 12.9: Theoretical yield and percent yield. (2021, June 30). Chemistry
LibreTexts. https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book%3A_Intro
ductory_Chemistry_(CK-
12)/12%3A_Stoichiometry/12.09%3A_Theoretical_Yield_and_Percent_Yield

• Lab report of experiment 4 (Stoichiometry and theoretical yield). (n.d.).

StuDocu. https://www.studocu.com/my/document/universiti-teknologi-mara/general-
chemistry/lab-report-of-experiment-4-stoichiometry-and-theoretical-yield/13696467
JOTTER: EXPERIMENT 4 (STOICHIOMETRY AND THEORETICAL YIELD)
TITLE: EXPERIMENT 6 (ACID AND BASES)

PURPOSE
The objective of this experiment is to investigate the characteristics of acidic and basic substances using
indicators and a pH meter.

INTRODUCTION
When acids and bases are dissolved in water, they undergo complete or incomplete ionisation
and are described to as strong acids, strong bases, weak acids, or weak bases. The experimental
determination of the pH of a solution, which is often carried using indicators or a pH meter. Ka or Kb,
the acid or base dissociation (ionisation equilibrium) constant, may be measured experimentally. A
weak acid (HA) sample is dissolved in water and broken down into two equal-volume parts. When one
portion of the mixture is titrated with a sodium hydroxide solution, all the HA molecules present are
transformed into A- ions:
OH − + HA → H2 O + A−

K a = [H + ][A] / [HA] = [H + ] = 10−ph

The amount of A− ion generated is equal to the amount of HA in the initial half-portion, as well as the
amount of HA in the remaining amount of weak acid. The pH of a half-neutralized sample of acid can
be used to calculate the value of Ka.
While doing an acid-base titration, the end point of the titration may be discovered when the indicator
changes colour. A pH meter may be used to measure the change in pH.

CHEMICAL AND APPARATUS

Sample solution A
Sample solution B
Unknown acid solution
Indicators (thymol blue, bromophenol blue, methyl orange and phenolphthalein)
Test tubes
Conical flask
Beaker
Pipette (25mL)
Burette
pH meter
METHOD
A) pH using indicator
1. 1 – 2 drops of one of the indicators given in table 6.1 had been added into the small test
tubes that contained 1 – 2 mL of the sample solution.
2. The colour of the solution had been recorded in table 6.1 after adding the indicator.
3. Steps 1 – 2 had been repeated for other indicators.
4. Steps 1 – 3 had been repeated for other sample solutions.

B) Determination of 𝐊 𝐚 for a weak acid

1. The burette had been filled with 0.2 M of NaOH solution.
2. 25 mL of an unknown weak acid solution (0.2 M) had been pipetted into a conical flask. 3
– 4 drops of phenolphthalein indicator had been added and the acid solution had been
titrated until a faint persistent pink colour appeared.
3. 25 mL of unknown acid solution had been added into the neutralized solution (step 2). The
solution had been stirred well and the pH of the solution had been recorded.

C) Strong acid-base titration

1. The burette had been filled with 0.2 M HCl.
2. 25 mL of 0.2 M NaOH had been transferred into the 100 mL beaker by using a pipette and
3-5 drops of phenolphthalein had been added.
3. A pH meter had been calibrated at pH 7 and pH 4.
4. The pH of NaOH had been measured and recorded in table 6.2. For the entire experiment,
the pH meter's tip should be immersed.
5. 10 mL of HCl (burette) had been carefully added into the NaOH solution. The solution had
been mixed gently and the pH of the solution had been measured.
6. The acid had been added continuously and the pH had been measured as indicated in the
table 6.2 of the data sheet of the experiment.
 DATA SHEET EXPERIMENT 6
ACID AND BASES

NAME: NOOR AFRINA BINTI MOHD FAZRUL DATE: 13 JANUARY 2022

STUDENT ID: 2021458818 GROUP: AS2511M1

A) pH using indicator
Table 6.1: pH using indicator

Indicator Colour change pH of colour Colour change Colour change

change in sample 1 in sample 2

Thymol blue Red - Yellow 1.2 – 2.8 Pink Blue

Bromophenol Yellow - Blue 3 – 4.7 Yellow Purple

blue

Methyl orange Red - Yellow 3.2 – 4.4 Red Yellow

B) Determination of 𝐊 𝒂 for a weak acid

Determination of K 𝑎 for a weak acid = 4.159 x 10−10

pH of half neutralized solution of unknown weak acid = 5.04

C) Strong acid-base titration

Table 6.2: Strong acid-base titration

a. HCl 0 10 15 20 23 25 26 27 29 31 33 35
added
(mL)

b. Measured 12.0 12.14 12.12 11.99 11.63 7.25 3.00 2.64 2.31 2.24 2.15 2.10
pH
RESULT AND CALCULATION

QUESTIONS:
1. Estimate the pH of Sample 1 and Sample 2.
- Sample 1 = pH 3.00 – 3.20
- Sample 2 = pH 4.00

2. Which indicators bracketed the pH colour change of Sample 1?

- Bromophenol Blue

3. Which indicators bracketed the pH colour change of Sample 2?

- Methyl Orange

4. From the observed pH of the unknown weak acid (experiment B), calculate

a) [H⁺] in the solution and Ka.

b) Percent ionization.
5. Construct a titration curve by plotting measured pH versus volume HCl (mL) added (refer to
Table 6.2).

(a) What is the pH range for the colour change of phenolphthalein as shown in the plotted
graph?
- pH 3.00 – pH 11.63

(b) What is the pH of the equivalence point in this titration?

(c) Explain why phenolphthalein was used in this experiment?

- In this experiment, phenolphthalein was chosen since its colour varies depending on the pH
of the substances. The equivalence point for titrating a strong acid with a strong base is pH
7. The quantity of base required to create a pH of 7 is about the same as the amount required
to make phenolphthalein pink. As a conclusion, while conducting an acid-base titration, the
change in pH may be seen visually by analysing the colour of the solution.
DISCUSSION
In step A, the pH value was observed by three different indicators which is thymol blue,
bromophenol blue, and methyl orange. Thymol blue is a brownish-green or reddish-brown crystalline
powder indication. It is soluble in alcohol but insoluble in water, and it may also be used to dilute
alkaline solutions. Thymol blue displays colour changes from red to yellow at pH 1.2 - 2.8 and from
yellow to blue at pH 8.0 - 9.6, where in alkaline solution is blue colour, in neutral solution is yellow,
and in acidic solution is red. Both sample’s colours have been altered using thymol blue, with sample
1 turning pink and sample 2 becoming blue. This implies that sample 1 is acidic solution whereas sample
2 is alkaline. At pH 3.0 - 4.7, bromophenol blue turns from yellow to blue, while alkaline solution is
blue, in neutral solutions is green, and yellow in acidic solutions. The colour both samples has affected
by bromophenol blue, which turns yellowish colour in sample 1 and purplish colour in sample 2. Data
reveals that sample 1 is acidic whereas sample 2 is alkaline solution. Following that, methyl orange is
a common indicator because it generates a clear and visible colour variation at different pH levels. The
colour of methyl orange transforms from red to yellow at pH 3.2 - 4.4, where it is yellow in alkaline
and neutral solution, while red in acidic solution. Methyl orange has influenced the colour between both
samples, causing sample 1 to turn red and sample 2 to turn yellow. This may deduce that sample 1 is
acidic and sample 2 is alkaline.
Step B yielded a value of 4. 159 x 10−10 for the measurement of Ka for a weak acid, with 5.04
being the pH of a half-neutralized solution of an unknown weak acid. In step C, a strong acid-base
titration was performed. In 12 experiments, different quantities of mL of HCl were added to 25 mL of
0.2 M NaOH solution. A pH meter was then used to ascertain the pH value. I can infer from the When
0 mL to 15 mL of HCl was applied, the pH value raised significantly, and then when 20 mL to 35 mL
of HCl was administered, the concentration began to fall.

CONCLUSION
In this titration solution experiment, thymol blue, bromophenol blue, and methyl orange are
used to demonstrate that the acid-base reaction is complete. When the chemical reacts with Thymol
blue, it develops transparent pink, yellow when it reacts with Bromophenol blue, and red when it reacts
with Methyl orange. Acidic compounds are those that have been defined as such. Once the material is
exposed to Thymol blue, it turns clear blue; when exposed to Bromophenol blue, it turns purple; and
when exposed to Methyl orange, it turns yellow. These substances are known as basic substances.
Acidic substances have a pH ranging from 0 to 7, whereas basic substances have a pH ranging from 7
to 14.
REFERENCES

• Definitions of pH, pKa, Ka, pKb, and kb. (2016, April 14).
ThoughtCo. https://www.thoughtco.com/ph-pka-ka-pkb-and-kb-explained-4027791

• Overview of acids and bases. (2020, August 21). Chemistry LibreTexts.

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Ma
ps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Acid/O
verview_of_Acids_and_Bases

• Acids and bases | Chemistry library | Science | Khan Academy. (n.d.). Khan
Academy. https://www.khanacademy.org/science/chemistry/acids-and-bases-topic
JOTTER: EXPERIMENT 6 (ACID AND BASES)
TITLE: EXPERIMENT 7 (REDOX TITRATION)

PURPOSE
The aim of this experiment is to standardize the potassium permanganate solution and ascertain the
metal composition by titration.

INTRODUCTION
Electrons are transferred from one molecule to another during oxidation-reduction reactions.
Potassium permanganate is a common oxidizing agent that does not require an indicator unless
extremely dilute solutions are utilized. Potassium permanganate oxidizes iron (II) while being reduced
to manganese in acidic solution (II). The following are the process's reactions:

MnO4− + 8H + + 5e− → Mn2+ + 4H2 O

Fe2+ → Fe3+ + e−

MnO4− + 8H + + 5Fe2+ → Mn2+ + 4H2 O + 5Fe3+

CHEMICALS AND APPARATUS

0.1 M potassium permanganate solution
Solid sodium oxalate (Na2 C2 O4)
1 M sulfuric acid
Iron ore
Burette
Conical flask
Hot plate
Volumetric flask
METHOD
1. Standardization of the potassium permanganate solution
- 0.2 – 0.25 g of dried sodium oxalate, Na2 C2 O4 had been weighed accurately into a conical
flask.
- 40 mL of 1 M sulfuric acid solution has been added into the conical flask.
- The solution had been heated gently to 60°C and immediately titrated with permanganate
solution until the first persistent pink colour occurred.
- The temperature had been maintained at 55-57°C.
- The titration had been repeated to obtain 2 acceptable results and the data had been recorded
in table 7.1.

2. Analysis of Ore
- 0.20 - 0.40g of iron ore had been weighed accurately into a conical flask.
- 30 mL of distilled water and 30 mL of 1 M of sulphuric acid solution had been added into
the conical flask.
- A glass funnel had been placed on the conical flask and had been heated gently on a hot
plate for 20 – 30 minutes until ore dissolved.
- The mixture had been filtered into a 100 mL volumetric flask and had been filled up to the
mark with 1.0 M sulfuric acid solution.
- 25 mL of the solution had been pipetted into a conical flask.
- The solution had been heated gently to 60°C and had been titrated with permanganate
solution until the first persistent pink colour.
- The titration had been repeated to obtain 3 acceptable results and the data had been recorded
in table 7.2.
 DATA SHEET EXPERIMENT 7
REDOX TITRATION

NAME: NOOR AFRINA BINTI MOHD FAZRUL DATE: 13 JANUARY 2022

STDUENT ID: 2021458818 GROUP: AS2511M1

1. Standardization of potassium permanganate solution

Table 7.1

Trial 1 Trial 2

Weight of Na2 C2 O4 (g) 0.2004 0.2004

Final burette reading (mL) 43.30 43.20

Initial burette reading (mL) 50.00 50.00

Volume of KMnO4 (mL) 6.70 6.80

2. Analysis of ore

Table 7.2

Trial 1 Trial 2 Trial 3

Weight of iron ore (g) 0.2004 0.2004 0.2004

Final burette reading (mL) 49.80 49.50 49.20

Initial burette reading 50.00 49.80 49.50

(mL)

Volume of KMnO4 (mL) 0.20 0.30 0.30

RESULT AND CALCULATION
QUESTION
1. Calculate the following:

(a) Actual molarity of potassium permanganate solution. Titration reaction:

(b) Concentration of iron(ll) in iron ore solution (initial volume of l00mL).

 (c) Percent purity of iron (ll) in unknown iron sample.

2. Explain why no indicator is needed in this experiment.

- When the titration reached its end point, the solution was fully oxidized. The solution's
colour changes from colourless to a faint pale pink that is permanent. The presence of
kmno4 leads in colour change. There is no longer any solution available to oxidize it. Thus,
KMnO4 performs as a self-indicator.

3. Typically, a solid iron ore is dried in an oven before analysis. How would the percentage of Fe
determined be affected if the ore had not been dried?
- Since of the excess moisture, the percent of iron would be lower because it has a larger
mass.
DISCUSSION

1. Standardization of potassium permanganate solution

MnO4− is reduced to Mn2+ by the equations, while C2 O4− is oxidized to CCO2 . The oxidation
number of carbon in C2 O4− has changed from +3 to +4. In this context, sodium oxalate is used to reduce
the colour of the standard solution. When the standard solution has been entirely oxidized, it will reach
the end point if a faint pale pink colour appears owing to the addition of a little excess of unreacted
potassium permanganate. Aside from that, the temperature of sodium oxalate and H2 SO4 must be kept
at 55-57°C. When there is a high temperature, the reaction occurs.

2. Analysis of Ore

The oxidation number of iron ore is +2 according to the equation. When the standard solution has
been totally oxidized, it will reach the end point if a faint pale pink colour is formed related to the
addition of a little excess of unreacted potassium permanganate with the addition of one extra drop of
KMnO4 beyond what is indicated with Fe2+ .
Sulfuric acid serves to supply the proton H + while preventing from producing a manganese (IV)
oxide precipitate in this experiment. Due to the formation of MnO2 , there are errors and quantitative
problems that do not provide a solid end point. Dilute sulfuric acid exhibits no redox reactions in the
subsequent titration. A potassium permanganate is used as a self-indicator in this titration.
There are some precautions that must be done to achieve an accurate result. To obtain accurate
reading of the burette, the air gaps must be cleared from the burette, and the funnel must be removed
before observing the initial reading of the burette.

CONCLUSION
At the end of this experiment, we can prepare and standardize the potassium permanganate.
The molarity of potassium permanganate calculated is 0.0886 M. The composition of metal has been
determined by titration method which is 12.87%.
REFERENCES

• Redox titration. (2018, September 2). BYJUS. https://byjus.com/chemistry/redox-titration/

• Experiment 16 help!!! (n.d.). https://faculty.uml.edu/james_hall/84124/16.htm

• Redox titrations (video). (n.d.). Khan Academy. https://www.khanacademy.org/science/ap-

chemistry-beta/x2eef969c74e0d802:chemical-reactions/x2eef969c74e0d802:introduction-to-
titration/v/redox-titration
JOTTER: EXPERIMENT 7 (REDOX TITRATION)
TITLE: EXPERIMENT 8 (GAS LAWS)

PURPOSE
1. To evaluate Graham's law by determining the lengths travelled by two different gases of known
molecular mass during the same period.
2. To find the molar mass of a volatile liquid, measure the amount of liquid vapour required to fill
a flask of known volume at a given temperature and pressure.

INTRODUCTION
A. Graham’s Law

Thomas Graham is the one who discovered experimentally that the rate of diffusion of two distinct
gases is inversely related to the square root of the mass of its particles and invented the graham's law.
The equation can be written in the form:

Rate1 M2
= √
Rate2 M1

in which rate 1 indicates the rate of diffusion of the first gas and rate 2 represents the rate of diffusion
of the second gas. The molar mass of gas 1 is denoted by M1 . The molar mass of gas 2 is M2 .
Throughout this experiment, you need to measure the distance that two gases travel at the same
time to analyse the relative rate of diffusion of hydrogen chloride and ammonia gases. A lighter gas
ought to be able to disperse further than a heavier gas during a specified period. During this experiment,
the two gases will distribute toward each other via the tube. As when the gases collide and react, a white
ring of ammonium chloride forms in the tube.

HCl (g) + NH3 (g) → NH4 Cl (s)

If you apply the position of the white ring throughout the length of the tube, you will be able to
recognize which of the two gases has diffused furthest.

B. Molar mass of volatile liquid

According on the ideal gas law formula, PV = nRT, it is directly proportional to the quantity of gas
(mol) in the sample. For a given container or fixed volume at a certain temperature and pressure, the
maximum amount of ideal gas that may be stored in the container is one:
 PV
n=
RT

in which, P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant,
T is the absolute temperature (in Kelvin). The mass of the gas sample may be computed by carefully
evaluating it, since the molar mass, M, denotes the amount of grammes, g, of the volatile chemical per
mol.

g
Number of moles = n = M

A small amount of easily volatilized liquid will be added to the flask of known volume within
that experiment. The flask will be equilibrated with atmospheric pressure since being heated in a boiling
water bath. The number of moles of gas trapped in the flask may be computed using the volume of the
flask, the temperature of the boiling water bath, and the atmospheric pressure. The molar mass of the
liquid may be determined by subtracting the quantity of liquid necessary to fill the flask with vapour
while it is in the boiling-water bath.

CHEMICAL AND APPARATUS

NH3 (conc.) Stopper

HCl (conc.) Medicine dropper

Acetone (to dry the glass tube) Tweezers or crucible tongs

Retort stand Stock watch

Glass tube (50cm length, 10mm internal diameter) Marker pen

Cotton Long ruler (1 meter)

Boiling chips Retort stand

Unknown volatile liquid – acetone (or other compounds 400/500 mL beaker

recommended by lecturer)

Thermometer 100/125 conical flask

Aluminium foil (square shape just enough to cover the 500mL graduated cylinder
mouth of a conical flask)

Bunsen burner

Pin (to make a hole)

METHOD
A. Graham’s Law

1. A 50 cm length of glass tubing (10 mm internal diameter) had been obtained. Ensure that
it is entirely dried. Two cotton swabs had been obtained and prepared. A setup of figure 8.1
and 8.2 had been prepared.
2. Around 10 to 15 drops of concentrated HCl (number of drops should be same for conc. HCl
and conc. NH3 ) had been placed on a cotton swab using dropper pipettes, and 10 to 15
drops of concentrated NH3 solution had been placed on another cotton swab using dropper
pipettes.
3. The soaked ends of the cotton swabs had been inserted instantly and identically into
opposite ends of the tube using tweezers. The time it took for the faint white cloud (ring)
of ammonium chloride to form had been recorded.
4. Within the first few minutes, a white ring appeared within the tube where the gases HCl
and NH3 mixed to form the white chemical NH4 Cl (ammonium chloride). The region on
the tube where the white ring forms had been recorded. Each gas's distance travelled had
been determined. Datasheet 8A used to display the data.
5. The cotton swabs had been removed with tweezers and immersed in a beaker of tap water.
6. Water had been used to rinse the tube. After washing the tubing with acetone, it had been
dried.

B. Molar mass of volatile liquid

1. A 400 mL beaker with 250 mL of water had been used to set up a boiling water up (or
enough water to immerse the flask).
2. A 125 mL conical flask with boiling chips had been obtained. A small piece of aluminium
foil had been securely wrapped around the flask's mouth. On the foil cap, a tiny hole had
been made with a straight pin.
3. The boiling chips and the empty, sealed flask had been weighed together.
4. The foil cap had been taken off. The flask had been filled with a 2 mL sample of the liquid
to be investigated, and the foil had been replaced.
5. A single burette clamp had been used to secure the flask. The flask had been dipped and
heated after being moved to the boiling water bath. Since volatile liquids are combustible,
the flames must not become too hot. If required, hot chips had been added to the water bath.
6. The liquid within the flask had been observed to be refluxing. The flask had been
maintained at a slight angle because the liquid vanishes would have been immediately
detected.
7. The flask had been heated until no liquid could be observed, and no vapour could be
detected escaping through the pinhole. The heating had been extended for an additional 30
seconds.
8. The flask had been taken out. It had been placed on a hot pad or tile. The clamp had been
released, and the flask had been allowed to cool to room temperature.
9. The flask had dried out. The flask, cap and condensed vapor was weighed.
10. The flask's contents had been disposed of in a waste bottle or as directed. The flask had
been filled halfway with tap water (to overflowing). The water had been put into a 500mL
graduated cylinder, the volume had been measured, and the results had been recorded.
11. The contents of the flask had been disposed away in a waste bottle or as recommended. The
flask had been half-filled with tap water (to overflowing). The water had been placed in a
500 mL graduated cylinder, the volume had been measured, and the findings had been
recorded.
 DATA SHEET EXPERIMENT 8
GAS LAWS

Name: NOOR AFRINA BINTI MOHD FAZRUL Date: 20 JANUARY 2022

Student ID: 2021458818 Group: AS2511M1

A. Graham’s Law
Observation of NH4 Cl appearance:
- The NH4 Cl ring emerges closer to the HCl end of the tube. The NH4 Cl ring is a white ring.

8A

Trial 1 Trial 2

Start time (s) 0.00 0.00

Finish time (first visible smoke) 721.25 724.25

(s)

Distance travelled by NH3 (cm) 62.0 62.2

Distance travelled by HCl (cm) 38.0 37.8

Ammonia diffusion rate 0.03597 0.03591
(cm/sec)

HCl diffusion rate (cm/sec) 0.05269 0.05220

B. Molar mass of volatile liquid

8B

Mass of flask, foil, boiling chips and 73.5370g

condensed vapour

Mass of flask, boiling chips and foil 73.2610g

Mass of condensed vapour (mass of vapour) 0.3260g

Temperature of vapour 65.0°C

Barometric pressure (pressure of vapour) 760mmHg (Torr)

Volume of flask (volume of vapour) 150mL

RESULT AND CALCULATION
QUESTIONS

1. Experiment 1

(a) Calculate the rate of diffusion for each gas by dividing the distance travelled (cm) by the
time required (sec) for the appearance of the white deposit.

(b) Calculate the ratio of the rate of diffusion of NH3 to the rate of diffusion of HCI.
(c) Using the molecular masses of NH3 and HCI, calculate the theoretical ratio of the rates of
diffusion of these gases.

(d) Calculate the % error in your experimentally determined value for the ratio of the rates of
diffusion of NH3 and HCl. Use the theoretical ratio calculated in (c) as the accepted value
for the ratio. [% error = absolute value of (theoretical ratio - experimental theoretical ratio)
/ theoretical ratio)] × 100%.
 2. Experiment 2

Calculate the molecular weight of the unknown liquid. Show your calculations and include
units of the different quantities in your calculations.

DISCUSSION

A. Graham’s Law

The reaction that is involved is:

NH3 (g) + HCl(g) → NH4 Cl(s)

According to this experiment, ammonia reacts with hydrogen chloride to obtain ammonium
chloride as a result. A ring of white smoke appeared in the tube, showing the synthesis of ammonia
chloride. The number of drops of NH3 should be equal to the number of drops of HCl throughout the
experiment, and both cotton swabs with NH3 and HCl should be placed into the tube around the same
period. This is essential for gaining an exact time and measurement.
Based on observations, the white ring grew up to the end of the HCl cotton swab tube. This is
probably the fact that NH3 diffuses rapidly than HCl. As shown in the calculation, the rate of diffusion
of NH3 to HCl is 5:3. As an outcome, HCl has approximately twice the molecular weight of ammonia.
The rate of diffusion was inversely proportional to the square root of the gas's molecular mass.
In this experiment, it is vital that the tube must clean and totally dry. If possible, dry the tube
by putting an acetone-soaked cotton wool pad through it and keeping it for a few minutes. Concentrated
NH3 solution and concentrated HCl, and perhaps even the fumes they emit, are very caustic and
corrosive. Gloves should be worn to prevent these liquids from getting into touch with your skin. This
step must be carried out under a fume hood.

B. Molar mass of volatile liquid

The temperature of the gas, 99°C, was determined by observing the temperature of the water
bath that surrounded the flask. Weighing the condensate left in the flask gets 0.3260 g of vapour mass.
To compute the mass of condensed vapour, subtract the mass of flask, boiling chips, and foil from the
mass of flask, boiling chips, foil, and condensed vapour. The volume of the vapour is 150 mL, that is
the same as the volume of the flask.
To minimize influencing the mass obtained while weighing the flask, the boiled chips must be
entirely dried and free of excess water. The flask must also be weighed multiple times to achieve an
accurate result.
Volatile liquid was selected because it has a low boiling point and vaporises quickly. To aid
any excess vapour to escape the flask until the volume of the flask is fully filled with the gas, a small
hole must be formed in the aluminium foil that seals the mouth of the flask. To limit the rate of effusion,
a straight pin was utilized to generate as small a hole as possible.
The unknown liquid applied may be damaging to the skin. If the liquid is leaked, avoid contact
and clean instantly. To protect your hands from hot glasses, use tongs or a towel.

CONCLUSION
As ammonia does indeed have a smaller molecular weight than hydrogen chloride, it travels a
longer length of the tube and diffuses quicker. By determining the mass of the liquid's vapour, the molar
mass of the volatile liquid is 60.26 g mol−1 .
REFERENCES

• Gas laws | Chemistry [Master]. (n.d.). Lumen Learning – Simple Book

Production. https://courses.lumenlearning.com/trident-boundless-chemistry/chapter/gas-laws/

• Gas laws. (n.d.). Chemistry & Biochemistry - Department of Chemistry &

Biochemistry. https://www.chem.fsu.edu/chemlab/chm1045/gas_laws.html

• Gas laws - Equations and formulas. (2016, July 9). YouTube. https://youtu.be/IE8-izIba40
JOTTER: EXPERIMENT 8 (GAS LAWS)