Chemistry Lab Reports: Aspirin, Recrystallization, and Titration

This document presents a series of chemistry lab reports covering several fundamental experiments. The first report details the synthesis of aspirin (acetylsalicylic acid), including the balanced equation, mole ratios, limiting reagent determination, and percentage yield calculation, comparing theoretical and experimental values. The second report focuses on the purification of benzoic acid through recrystallization, discussing the separation of impurities and the determination of melting points. The third report describes the determination of the percentage of sodium carbonate in a sample of commercial washing soda via titration with hydrochloric acid, including balanced equations and calculations of moles and percentage composition. The report also includes the preparation of standard solutions, reactions of acids and bases, and experimental techniques for volumetric analysis, including dilution effects. The experiments provide valuable insights into chemical reactions, purification methods, and analytical techniques.

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Lab reports
Title: Synthesis of Aspirin (Acetylsalicylic acid).
Introduction
This experiment involves production of aspirin. The major reactants which were used in the
experiment include salicylic acid HC7H5O3, and C4H6O3. The reactants reacted in order to produce
acetylsalicylic acid or simply aspirin HC9H7O4 and ethanol acid CH3CHOOH. Asprin has several
uses such as an analgesic or painkiller, as an anti-pyretic or fever reducer and as thinning of the
blood.
Balanced equation HC7H5O3 + C4H6O3 = HC9H7O4 + CH3CHOOH
Moles ratio 1 1 1 1
=
Results and Calculation
Available results:
Salicylic acid mass= 3.2888 g
Experiment yield of Aspirin: =3.625g
Acetic Acid Anhydride density: 1.1 g/mL
Acetic acid anhydride volume: 10mL
Moles of salicylic acid:

HC7H5O3 Molar mass = 138g/mol as given
number of moles = HC7H5O3 grams/ HC7H5O3 molar mass
Moles =3.2888g/138g/mol= 0.0238319 moles
Acetic acid anhydride moles:
C4H6O3 Molar mass =102g/mol
Acetic acid density=1.1g/mL
Mass = density times volume
acetic acid anhydride mass=1.1g/mL × 10mL=11
number of moles =grams/molar mass
Acetic acid anhydride Moles =11g/102g/mol=0.1078431moles
Limiting reagent
In this case, the limiting reagent was found using the moles ratio indicated as the ration of
Salicylic acid moles to acetic acid moles
From the equation, the mole ration is 1:1
Therefore, theoretically, 1 mole of acetic acid anhydride = 1 mole of salicylic acid. This means
that the moles for salicylic acid will be 0.0238319 theoretically.

∴in conclusion, comparing the theoretical and calculated mole ration for
acetic acid anhydride, we find that, the theoretical 0.0238319 mol of acetic acid
anhydride is less than the actual amount of 0.1078431mol of acetic acid anhydride. This means
that salicylic acid will be the limiting reagent while acetic acid anhydride is the excess reagent.
Percentage yield
In this part, we use the limiting reagent and its mole ratio to find the moles of asprin from the
experiment.
The ration of moles of HC7H5O3 to HC9H7O4 moles is 1:1
∴ This means that the moles for the two reagents will be the same. That is 0.0238319 moles for
HC9H7O4 and the same moles for HC7H5O3.
The theoretical yield of ASA (in g)
Finding the molar mass of HC9H7O4:
The molar mass of HC9H7O4 will be 180g/moles as indicated
Calculating theoretical mass of HC9H7O4
HC9H7O4 grams = moles × molar mass
0.0238319 mol ×180g/mol=4.289742g
As from the experiment, the actual yield of HC9H7O4 was 3.625g

Percentage yield will therefore be:
% yield= {(actual yield)/(theoretical yield)}×100
% yield= {(3.625g/4.289742g)}100
% yield=84.503917%
∴ the % yield of aspirin is therefore 84.50% representing 3.625g of a maximum of 4.2897g
theoretical value.
In conclusion, it was clear that the theoretical and experiment values of produced aspirin are
different. The experiment in this case is used to determine the yield and melting point of the
products which are determined through confirmatory test. The limiting reagents in the
experiment were determined in order to help to calculate the theoretical yield.
Title: purification of an organic compound by recrystallisation
Introduction
This experiment involves the purification of benzoic acid through recrystallisation process. The
insoluble impurity used is carbon and water soluble impurity is citric acid. Recrystallisation
method was used in the determination of the melting point of the impure compound and
separation of the solids. Materials can be contained in other and be insoluble when cold but
soluble when hot and vice versa. Comparison of the melting point ranges was also carried out in
this experiment.
Calculation

Impure compound weight 1.110
Water 35 cm
Weight of filter glass 26.4199
Weight of clock glass is 26.835
Benzoic acid weight 0.416 gm
% Recovery of Pure Benzoic Acid = Weight of dry pure benzoic acid x 100
Weight of impure compound used
= (0.1416/1.11)*100 = 12.76%
In conclusion, through this experiment, it was clear that the substances with similar molecules
are considered pure and have different melting points with impure substances. Cooling filtrate
makes the pure substances to recrystallise out into small quantities. This was the key method
which was used to purify the substances in this experiment. The recovery percentage was
important to show the rate of purification of the substance.
Title: Determination of the percentage Sodium Carbonate in a sample of commercial
Washing Soda.
Introduction
This experiment involved production of sodium carbonate from washing soda. The preparation
of the Sodium Carbonate is done from commercial washing soda through titration process. In
order to neutralize sodium carbonate, two parts of hydrochloric acid are needed. First, the soda
was measured and then transferred to the clockglass. Water was added and the mixture was

titrated with HCL until color change to pink was experienced. Titration process was also
important in this experiment.
Na2CO3 + 2HCl  2NaCl + H2O + CO2
Calculation
Washing soda of 25ml
Weight of clock glass 23.44g
2 drops of indicator gives 25g
6 of HCL = 10 cm3 x 6 = 60 cm3
Moles of HCL use = 60* 0.1 / 1000 =0.006 moles
The balanced equation of the reaction will be:
Equation Na2CO3+2HCl = 2NaCl + H2O + CO2
Moles representation 1 mole 2 moles 2 moles 1 mole 1 mole
Using the ration, Na2CO3 used moles = 0.006 / 2
= 0.003
Na2CO3 Molar mass = 25g/mol
Mass of Na2CO3 available in 25 cm3 = 0.003 x 25 g
= 0.075 g
Mass of Na2CO3 available in 250 cm3 = 7.5 g

Mass of Na2CO3.xH2O available in 250 cm3 = 25 g
water mass available in this amount of Na2CO3.xH2O =25-7.5 = 17.5 g
Percentage of water available in Na2CO3.xH2O = 70.00%
Moles of water present in this amount of Na2CO3.xH2O = 17.5 / 18
= 0.97722222
Moles of Na2CO3 present in this amount of Na2CO3.xH2O = 0.0097
Value of x in Na2CO3.xH2O = 0.9722 / 0.0097
= 10
Formula of hydrated sodium carbonate: = Na2CO3.10H2O
In conclusion, the production of sodium carbonate was done through the use of several reactants.
HCL was used in the titration process. The amount of sodium carbonate produced was depended
on the reactants and their status.
Lesson 9
1.4 preparations of standard solutions
1
The molar mass of oxalic acid dihydrate is 126 g/mol
Weight needed to prepare 1000 cm3 of a 1.0 molar solution 126 g
Weight needed to prepare 250 cm3 of a 1.0 molar solution 31.5 g
Weight needed to prepare 250 cm3 of a 0.1 molar solution 12.6 g
2. To prepare a standard solution of sodium carbonate

Sodium carbonate is available as pure anhydrous sodium carbonate and is also available in an
impure hydrated form (washing soda).
The molar mass of sodium carbonate, Na2CO3 is 106 g/mol
Weight needed to prepare 250 cm3 of a 1.0 M solution 26.5 g
Weight needed to prepare 250 cm3 of a 0.05 M solution 1.325 g
1.5 Reactions of Acids and Bases
Acids and bases may be strong or weak.
They may be concentrated or dilute.
Common strong acids are:
HCl, which is named as ________hydrocloric acid
H2SO4, which is named as sulphiric acid
HNO3, which is named as nitric acid
Common weak acids are:
Ethanoic acid, which has the formula: _CH3COOH
Oxalic acid, which has the formula: _ (COOH)2
Answer the following:
Which is the stronger acid, 0.1 M HCl or 5 M ethanoic acid?: 0.1 M HCl
Which is the more concentrated acid? 5 M ethanoic acid
1.6 Experimental technique for volumetric analysis
It is important to understand when the dilution of a solution will affect a result and when it will
not.
Think carefully about each step in the following sequence and answer the questions.

Q.1) Orange squash concentrate from a bottle is poured into 2 beakers as shown below. The
concentrate contains 2.0 g of vitamin C per 100 cm3.
Which is the more concentrated solution, that in beaker A or B beaker A_
Which beaker contains the most vitamin C, beaker A or B? beaker B
Q.2) A 25 cm3 aliquot (or portion) of squash from the bottle is pipetted into each of two beakers,
Beaker X is wet and Beaker Y is dry.
Which is the more concentrated, the contents of beaker X or Y? beaker Y
Which beaker contains more vitamin C? beaker X
Q.3) A 25 cm3 sample of squash is transferred from beaker A with two different pipettes, one of
which is dry and one of which is wet, and each sample is placed in a conical flask, identified as
Pdry and Pwet respectively.
Which conical flask contains more vitamin C? Pwet
Give a reason for your answer:
Since it wet, the beaker will need more vitamin C to titrate until the color changes