Experiments In Chemistry

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 _____________
Which beaker contains the most vitamin C, beaker A or 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? ___________ Which beaker contains more vitamin C?

Q.3) A 25 cm3
sample of squash is transferred from beaker A with two different zipettes, 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.
 

Production and Purification of Aspirin

This experiment involves production of aspirin. The major reactants which were used in the experiment include salicylic acid HC₇H₅O_3, 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 

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.

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.

Recrystallisation of Benzoic Acid

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.

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.  

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 à    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

Determination of Percentage Sodium Carbonate in Washing Soda

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

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

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? P_wet 

Give a reason for your answer:

Since it wet, the beaker will need more vitamin C to titrate until the color changes