Planning and Designing Chemistry Lab – Determining the solubility of salt at different temperatures.

Planning and Designing Chemistry Lab – Determining the solubility of salt at different temperatures. 

Problem:

You are given a salt and asked if the salt will dissolve in water more at high temperatures.

Hypothesis:

The salt will dissolve more at room temperature.

Aim:

To determine the solubility of a salt at different temperatures.

Apparatus:

salt, distilled water, tripod stand and gauze, Bunsen burner, beakers, ice, filter paper, funnel, thermometer, watch glass and stop-watch

Method:

  1. Weigh out four equal amounts of the salt and place in four (4) separate test-tubes.
  2. Measure out four 25cm3 of distilled water and place in four beakers.
  3. Prepare the water to the following temperatures – 0°C, 30°C, 60 °C and 90°C
  4. Prepare a bath-water and heat the water to 60°C and to a boil to 90°C
  5. Place the salt in each container of water and stir for 20 seconds.
  6. Immediately pour the solution through a filter paper.
  7. Allow the residue to dry and reweigh.
  8. The one with the highest mass was most in-soluble.
  9. Tabulate Results.

Variables:

Controlled:

  • Volume of water used
  • Amount of salt used
  • Time given for salt to dissolve

Manipulated: Temperature

Responding: Amount of salt that remains undissolved

Expected Results:

Chemistry diagram 13

Treatment of Results:

 

To calculate the solubility of each salt at the different temperature, you will need to use the formula:

Chemistry diagram 14

Interpretation of Results:

The solubility of a given solute in a given solvent typically depends on temperature. For many solids dissolved in liquid water, the solubility increases with temperatures up to 100 °C. As the temperature of a solution increases, the average kinetic energy of the molecules that make up the solution also increases. This increase in kinetic energy allows the solvent molecules to more effectively break apart the solute molecules that are held together by intermolecular attractions.

The increased vibration (kinetic energy) of the molecules causes them to be less able to hold together, and thus they dissolve more readily. Therefore if the solubililty of the final salt in the 30°C water has the highest solubilty then the hypothesis is accurate. If the solubility is higher as the temperature increases then the hypothesis is not accurate.

 

Planning and Designing Chemistry Lab – To determine whether green limes have a higher level of acidity than yellow limes using titration

Planning and Designing Chemistry Lab – To determine whether green limes have a higher level of acidity than yellow limes using titration  

Problem Statement
Ms. Lawrence wanted to know if her green limes, had a higher level of acidity than the yellow limes.

Hypothesis
Yellow limes have a higher level of acidity than green limes.

Aim 
To determine whether green limes have a higher level of acidity than yellow limes using titration

Apparatus and Materials
Sodium hydroxide solution (0.1mol dm3), burette, burette stand conical flask, beaker, pipette filler, filter funnel, pipette, phenolphthalein indicator, yellow and green limes.

Method

  1.  In a beaker squeeze 10 yellow limes, and in another beaker squeeze the green limes.
  2. Use a pipette to transfer 10cm3 of yellow lime solution to a conical flask
  3. Add 2 drops of phenolphthalein to the flask
  4. Set up burette on stand appropriately.
  5. Use the funnel to fill the burette with sodium hydroxide (NaOH)
  6. Perform the titration and record all readings
  7. Repeat procedure for a total of three readings
  8. Record all readings and tabulate results
  9. Repeat steps 2-8 with the green limes

Variables

  • Controlled: The volume of lime juice and the amount of indicator used.
  • Manipulated: the types of limes
  • Responding: pH reading

Expected Results

Chemistry diagram 12

Treatment of results

If the lime juice from yellow limes gave the lower pH reading, then the lime juice of yellow limes is more acidic and the hypothesis is correct
Interpretation of results:

Lime juice contains citric acid which is a weak acid and thus contains free hydrogen ions when dissolved in water. A higher level of acidity means a higher concentration of hydrogen ions present in the aqueous solution.
pH is related to hydrogen ion concentration via the formula pH = -log[H+]

This means the higher the hydrogen ion concentration, the lower the pH value

Sources of errors / Assumptions / Limitations:
Passing the end point thus using too much of the carbonate, thus altering the results.

Assumption:

there are no other substances besides citric acid present that would affect the acidity of the lime.

 

 

 

Planning and Designing Chemistry Lab – To determine the partition coefficient of monocarboxylic acid between water and diethyl ether using extraction.

Planning and Designing Chemistry Lab – To determine the partition coefficient of monocarboxylic acid between water and diethyl ether using extraction.

Problem :

As a young chemist, supplied with 15g of a solid monocarboxylic acid, of molar mass 122 and the solvent diethyl ether; plan and design and experiment to determine the partition coefficient of the acid between water and diethyl ether. You are free to choose the bits of apparatus necessary to achieve the objective of the design, and if needed one other chemical apart from water.

Title:

Partition Coefficient

Hypothesis:

The monocarboxylic acid is more soluble in water than in diethyl ether.

Aim:

To determine the partition coefficient of monocarboxylic acid between water and diethyl ether using extraction.

Background:

Polar and ionic substances generally dissolve more readily in water than in organic solvents and vice versa. If a solute therefore, dissolves in two immiscible solvents, then it will partition itself between the two solvents such that the ratio of its concentration in one solution to its concentration in the other solvent is constant when the liquid mixture is in equilibrium (at a fixed temperature).

Determining the partition coefficient of a particular solute leads to an understanding of its solubility as well as it’s the hydrophilic-hydrophobic nature. Knowing the partition coefficients of various solutes allows us to separate chemicals efficiently by extraction.

Variables:

  • Manipulated Variables: Solute and Solvents Chosen
  • Controlled Variable:  Mass of Solute (Carboxylic Acid)
  • Responding Variables: Partition Coefficient and concentration of acid in each solution

Materials and Apparatus:

Electronic balance, Measuring cylinder , beakers, thermometer , separatory funnel and stopper, retort Stand, burette, pipette, conical flask, distilled water, diethyl ether, monocarboxylic acid

Procedure:

Part A

  1. Weigh an empty dry beaker on the balance and record its mass.
  2. Add a small amount of monocarboxylic acid (solute) to the beaker and record the mass of the beaker and solid.
  3. Subtract the mass of the empty beaker from the mass of the beaker and solid in order to calculate the mass of the solid.
  4. Repeat the process until 5.00g of the solute is in the beaker.
  5. Measure 80cm3 of diethyl ether as well as 80cm3 of distilled water (solvents) and pour the solvents into the separatory funnel.
  6. Add the solute and shake carefully (by inversion) for a prolonged period of time to ensure full agitation of the mixture (pause occasionally and open tap for a few seconds).
  7. Secure the separatory funnel in the retort stand.
  8. Allow the liquid mixture to equilibrate.
  9. After the liquid mixture has settled (two distinct layers can be seen).
  10. Place a beaker below the funnel.
  11. Remove the cover and release the tap so that the solution runs into beaker. Stop at the interface layer.

Part B:

  1. Rinse and fill the burette with dilute sodium hydroxide solution of known concentration and record the initial burette reading.
  2. Using pipette filler rinse the pipette with the solution in the beaker and carefully transfer 25cm3 of the solution to a clean conical flask.
  3. Add 2 drops of phenolphthalein indicator.
  4. Run the sodium hydroxide solution from the burette into the flask with swirling until pale pink color is seen.
  5. Record the burette reading.
  6. Repeat the process until consistent results are obtained.

Expected Results:

Chemistry diagram 10

Treatment of Results:

 

  1. To determine the mass of solute being used subtract the mass of the empty beaker from the mass of the beaker and the solute.
  2. To determine the volume of sodium hydroxide used, subtract the initial burette reading from the final burette reading.
  3. Add three consecutive values obtained for volume used and divide the sum by 3 to obtain the mean liter. The mean liter is the average amount of sodium hydroxide used in the titration.
  4. In order to calculate the amount of solute which dissolved in each solvent:
  • Determine the number of moles of NaOH used in the titration.
  • Determine the number of moles of acid which dissolved in the aqueous layer.
  •  Convert the moles of acid obtained above to mass/grams. 

 

  1. Subtract the mass of acid obtained in the previous calculation from the total mass of acid added to the solvents to obtain the amount of solute which dissolved in layer 2.

 

  1. Calculate the partition coefficient:

Chemistry diagram 11

Interpretation of Results:

 

A partition coefficient value of less than one shows that the monocarboxylic acid is more soluble in water than in diethyl ether. However, a partition coefficient value of more than one shows that the monocarboxylic acid is more soluble in diethyl ether than in water. If any mass of carboxylic acid is placed in any volume of a liquid mixture of diethyl ether and water the will partition or distribute themselves according to the partition coefficient.

 

Sources of Error/Limitation

 

  • Some of the interface layer may mix with the aqueous layer during release.
  • Environmental factors may cause the temperature in the room to change. Therefore each successive trial may occur at different temperature affecting the results.
  • Cover of the separating funnel was removed before the tap was opened to prevent air from flowing in and causing the layers to mix. Similarly, during agitation of the liquid mixture, the funnel’s stopper was removed intermittently to relieve the buildup of pressure.

 

Planning and Designing Chemistry Lab- Determining if three different brands of vinegar contain the same concentration of ethanoic acid

Planning and Designing Chemistry Lab- Determining if three different brands of vinegar contain the same concentration of ethanoic acid

Problem:

Do Different brands of vinegar contain the same concentration of ethanoic acid

Hypothesis:

Different brands of vinegar contain varying concentration of ethanoic acid due to the presence of different ingredients in each.

Aim:

To determine whether three (3) different brands of vinegar contain the same concentration of ethanoic acid using titration.

Apparatus / Method:

three different brands of vinegar containing ethanoic acid, 0.1moldm3 Sodium carbonate (Na2CO3), phenolphthalein indicator, pipette, burette, retort stand, conical flask.

Method:

  1. Rinse burette several times with water, then condition it using the sodium carbonate.
  2. Set up titration apparatus and fill the burette with the sodium carbonate solution. Ensure no air bubbles are present in the burette.
  3. Record the volume of sodium carbonate in the burette to two decimal places. This is the initial volume used in the titration.
  4. Pipette 25ml of one brand of vinegar in a clean dry conical flask.
  5. Add 3 drops of phenolphthalein to the flask.
  6. Begin the titration; ensure that the flask is continuously swirled until the end point is reach. (Color change from colorless to pink).
  7. Record the final burette reading.
  8. Repeat the process until two accurate volumes (no more than 0.1 from each other) is obtained.
  9. Repeat Steps 1-8 for the other two brands of vinegar and tabulate results.

Variables:

  • Controlled: the concentration of carbonate used and the amount of indicator used.
  • Manipulated: the brands of vinegar
  • Responding the amount of carbonate used to neutralize the different brand of vinegar

Expected Results:

Chemistry diagram 9

Interpretation of Results:

 

The vinegar solution that contains the least amount of ingredients will take less volume for a complete neutralization of occur as there is a low concentration of ethanoic acid present.

 

The vinegar solution that has more constituents in the ethanoic acid will thus have a higher concentration and will required more carbonate to neutralize the acid. If the data shows this then the hypothesis is accurate.

 

Limitations:

  • Incorrect reading of the volume in the burette
  • Passing the end point thus using too much of the carbonate, thus altering the results.

 

Assumption:

All brands of vinegar contain a different amount of ethanoic acid due to the varying number of other ingredients present in each.

 

Planning and Designing Chemistry Lab – Investigating the alcohol content of store-bought alcohol and homemade alcoholic beverages

Planning and Designing Chemistry Lab – Investigating the alcohol content of store-bought alcohol and homemade alcoholic beverages 

Problem:

Miss Mavis feels that her homemade alcoholic beverage is of a higher alcoholic content than that of the commercial brands in the supermarket but her friends think otherwise.

Hypothesis:

The other commercial brands contain a higher alcoholic content than Miss Mavis’s beverage

Aim:

To investigate if the other brands of alcoholic beverages contain a higher alcoholic content than Miss. Mavis’s beverage.

Apparatus / Materials:

three (3) different brands of alcohol, beakers, boiling flask, water,  bunsen burner, fractionating column, glass beads, Miss. Mavis’s beverage.

300px-Fractional_distillation_lab_apparatus.svg

Method:

  1. Pour 30ml of one of the brands of alcohol in the boiling flask.
  2. Set up apparatus as shown in the diagram above
  3. Apply heat and note your observations
  4. Use a beaker to collect the distillate and measure the volume that was collected.
  5. Repeat the procedure for the other brands of alcohol and Miss. Mavis’s beverage.
  6. Tabulate results

Variables

  • Controlled – amount of each beverage used.
  • Manipulated- different brands of alcohol used
  • Responding- volume of distillate

Expected Results:

Chemistry diagram 8

Interpretation of Results

If after distilling the four (4) beverages, the store bought brands each had more alcohol than Miss Mavis’s beverage, then the hypothesis is true.

Sources of Error / Limitations

The beaker contained impurities which may have affected the results

Assumption:

All alcoholic beverages contain the same amount of alcohol

 

 

 

Planning and Designing Chemistry Lab- To investigate the rate of reaction when calcium carbonate powder and lumps reacts with an acid

Planning and Designing Chemistry Lab- To investigate the rate of reaction when calcium carbonate powder and lumps reacts with an acid

Problem:

You are provided with calcium carbonate in the form of lumps and as a powdered substance, both dilute hydrochloric acid and sulphuric acid are available to you. Plan and design an investigation that you would carry out to compare the rate of reactions of both forms of calcium carbonate with an acid

Hypothesis: Powdered calcium carbonate will react faster with an acid than calcium carbonate lumps

Aim:

To investigate the rate of reaction when calcium carbonate powder and lumps reacts with an acid.

Apparatus/Materials:

thistle funnel, tap, gas syringe, flask, hydrochloric acid, sulphuric acid, calcium carbonate (lumps and powder) stop-watch,

Method:

  1. Set up apparatus as shown in diagram below
  2. Measure 10g of CaCO3 lumps and add it to the flask. Add 50cm3 of HCl to the flask. Record the volume of gas produced every 30sec
  3. Repeat this step for the powdered CaCO3

Please Note: HCl was used in preference to the H2SO4 as, CaCO3 when mixed with H2SO4 forms a slightly soluble salt and thus the reaction will end quickly.

Chemistry diagram 6

Variables:

  • Controlled – volume of acid, concentration of acid, and mass of CaCO3 lumps and powder
  • Manipulated: sample of CaCO3 lumps and CaCO3 powder
  • Responding – volume of gas produced

 Expected Results:Chemistry diagram

Chemistry diagram 7

Interpretation of Results:

The graph shows how much CO2 gas evolved from the experiment. The difference in shapes is because the powder caused the reaction to go at a faster rate than the lumps. The reactions finished at the same time because the masses were the same.

If at the end of the experiment the CaCO3 powder will react at a faster rate than the CaCO3 lumps then the hypothesis would be acceptable.

Sources of Error / Limitations:

Some of the gas escaped whilst being measured this skewing the results.

Assumption: Calcium carbonate reacts faster in the powdered form.

 

 

Planning and Design Chemistry Lab – Plan and Design an experiment that can be used to determine the contents of four bottles found in your home

Planning and Design Chemistry Lab – Plan and Design an experiment that can be used to determine the contents of four bottles found in your home

Problem:

You venture in a cellar one day and found four stoppered bottles. The labels were found on the floor next to the bottles and were marked: dilute sulphuric acid, sodium carbonate, calcium hydroxide and distilled water. If you were handed a pack of blue litmus paper and a rack of test tube, plan and design an experiment that can be used to determine the contents of each bottle. Label the bottles A, B, C and D before you begin your investigation.

Hypothesis:

Bottle A contains Sodium carbonate (Na2CO3), Bottle B contains (H2O), Bottle C contains Sulphuric acid (H2SO4) and Bottle D contains Ca(OH)2.

Aim:

To determine if the contents in Bottle A is Na2CO3, Bottle B is H2O, Bottle C is H2SO4 and Bottle D is Ca(OH)2 using litmus paper

Apparatus / Materials:

blue litmus paper, test-tube, test-tube rack and solutions in bottle

Method:

  1. Label the test-tube A-D
  2. Pour out 5cm of each solution in its corresponding test-tube.
  3. Place a blue litmus paper in each test-tube and observe for a color change to red
  4. The solution that changed the litmus paper to red would be the H2SO4, which is the acid.
  5. Label the corresponding solution as H2SO4.
  6. The test-tubes that were not identified will now be tested.
  7. Use the H2SO4 to change 3 litmus papers to red.
  8. Place the now red litmus papers in the remaining solutions. Observe for a color change to blue. Two of your solution will change to blue. The one that does not change is the water
  9. Label the corresponding bottle as H2O, and set it aside.
  10. The remaining two (2) solutions are the calcium hydroxide and the sodium carbonate.
  11. Add 2 cm of the H2SO4 to the 2 unknown test-tube and note your observations.
  12. The test-tube that showed signs of bubbling/effervescence would be the carbonate, Na2CO3.
  13. Label the corresponding bottle as Na2CO3 and the final bottle as hydroxide.

Expected Results:

Chemistry diagram 3

This test showed that one of the solutions turned the blue litmus paper red. This was identified as the sulphuric acid (H2SO4).

Chemistry diagram 4

 

This test would show two solutions changing the litmus paper to blue, the one that was not changed would be the water. Use the acid to test the remaining two unknowns to differentiate between them.

Chemistry diagram 5

Variables

 

  • Manipulated- Test carried out on the solutions
  • Controlled- Amount of solutions used
  • Responding- reaction with the litmus paper and the acid

 

Treatment of results

 

Label the bottles corresponding to the observations made when each was tested with litmus paper and the sulphuric acid.

 

Interpretation of results

 

If the contents in Bottle A reacted with sulphuric acid then it should be the sodium carbonate, if the contents of Bottle B did not change the color of the red litmus paper then it was the water. If the contents of Bottle C changed the blue litmus paper to red then it would be the acid and if the contents of Bottle D did not react with the acid then it would be the hydroxide.

 

The data collected would then support the hypothesis. If the contents of each of the bottles reacted differently with the blue litmus paper then the data collected does not support your hypothesis.

 

Planning and Designing Chemistry Lab- Different brands of permanent markers contain the same dyes in their black inks; however the markers are insoluble in water

Planning and Designing Chemistry Lab- Different brands of permanent markers contain the same dyes in their black inks; however the markers are insoluble in water

Problem:

Different brands of permanent markers contain the same dyes in their black inks; however the markers are insoluble in water.

Hypothesis:

Different brands of markers contain the same dyes in their black ink

Aim:

To investigate if four (4) different brands of permanent markers contains the same dyes in their black inks using paper chromatography.

Apparatus:

4 different brands of markers, alcohol, filter paper, scissors, ruler, beakers

Method

1. Fill a beaker with 30ml alcohol

2. Cut a rectangular strip of filter paper

3. Using one of the makers place a dot 1cm from the end of the filter paper.

4. Partially immerse the tip of the filter paper closest to the ink spot in the beaker with the alcohol.

5. Record your observations.

6. Allow the filter paper to dry.

7. Repeat the procedure for the remaining three (3) markers and ensure that the dot is of the same size for each marker.

8. Tabulate results and calculate the retention factor (Rf) of each of the markers

Variables

  • Manipulated- different brands of marker
  • Controlled- the volume of alcohol and the size of the dots
  • Dependent/Responding – the retention factor

Chemistry diagram 2

All four brands of marker will contain the same dyes therefore the hypothesis is true

Assumption:

The four brands of markers contain the same types of dye.

 

Percentage of water in a hydrate- Copper(II) sulphate pentahydrate (CuSO4 ·5 H2O)

 Percentage of water in a hydrate- Copper(II) sulphate pentahydrate (CuSO4 ·5 H2O)

Aim / Objective:

To determine the percentage of water in a hydrate

Introduction: 

Many pure substances combine with water in a fixed mole ratio to yield compounds called hydrates. For example, copper sulphate combines with water to form crystalline, CuSO4. 5H2O, which is a stable compound at normal atmospheric conditions. All pure samples of this hydrate show the same percentage of water by analysis. Thus, this hydrated compound obeys the law of constant composition. Upon heating a sample of such a hydrate, it may lose all its water of hydration and revert to the anhydrous salt. Substances which have adsorbed water on the surface do not show constant composition and therefore are not hydrates. An example of this would be common table salt, NaCl, which becomes very sticky on humid, summer days. In these cases, the percentage of water is not constant for all samples of a particular compound, and the water is not chemically bonded as part of the crystal structure.

Other examples of hydrates are Nickel (II) sulfate hexahydrate – (NiSO4 ·6 H2O), lithium perchlorate trihydrate ( LiClO4 . 3H2), aluminum potassium sulfate dodecahydrate – (AlK(SO4)2 ·12 H2O) and magnesium carbonate pentahydrate – (MgCO3 ·5 H2O.)

In this experiment the percentage of water in a hydrate will be determined in an known hydrate. Water is removed from the hydrate by heating an accurately weighed hydrate sample until the residue has reached a constant weight. The percentage of water in the sample is calculated by using the weight of water lost and the initial hydrate sample weight multiplied by 100.

Materials/ Apparatus:

Porcelain crucible and cover, clay triangle, tripod stand, Bunsen burner, flint, tongs,1.000g  Copper(II) sulphate pentahydrate (CuSO4 ·5 H2O(s) , watch glass

 Method / Procedure:

  1. Clean and dry a porcelain crucible and cover.
  2. Place the empty crucible on a covered crucible on a clay triangle supported by a ring on a ring stand.
  3. Heat the crucible and cover in the hottest flame of the Bunsen burner for 5 minutes. Ensure that a dull red glow is observed on the crucible and cover.
  4. Cool the crucible and cover to room temperature for approximately 15 minutes.
  5. Using crucible tongs transfer the crucible and cover to a watch glass and weigh them to the nearest 0.001g.
  6. Add 1g of the 1.000g CuSO4 ·5 H2O(s) to the crucible and weigh the covered crucible to the nearest 0.001g.
  7. Place the covered crucible on the clay triangle with the cover slightly opened.
  8. Heat the crucible gently for a few minutes. Continue to heat for 15 minutes.
  9. Then allow the crucible to cool on the triangle after removing the flame until it reaches room temperature.
  10. Transfer it to the watch glass and weigh the covered crucible to the nearest 0.001g.
  11. Reheat the crucible and contents for about 5minutes. Cool, and then weigh again.
  12. Repeat this heating, cooling and weighing sequence for a total of two readings .
  13. Tabulate results, and complete calculations for the percentage of water in the copper(II) sulphate.

Suggested Results: 

Table showing the results of the heating and cooling of Copper(II) sulphate pentahydrate (CuSO4 ·5 H2O) to find the percentage hydrate in water

Table showing the results of the heating and cooling of Copper(II) sulphate pentahydrate (CuSO4 ·5 H2O) to find the percentage hydrate in water

Use the results from your experiment. Your teacher may request that you use a evaporating dish instead of a crucible and cover which would may the results above slightly different but the concept is the same.

percentage hydrate snippet

Calculations/Answers/Discussion:

  1. Write the formula for the reaction

>>>CuSO4 ·5 H2O(s)  +  HEAT =  CuSO4 (s)  +  5 H2O (g)

2. Calculate the experimental measurement of the percent hydration:

  •       Mass of hydrate before heating =  1.000g
  •       Mass of hydrate after heating = 0.6400g
  •       Difference- mass of water lost = 0.3600g

Experimental Measurement of percent hydrate

  • (0.3600g/1.000g) x 100=36%

3. Calculate the theoretical percentage hydration from the formula.

percentage hydrate snippet2

 

 

 

 

 

 

 

4.Using the theoretical value and the experimental values calculate the percent error

percentage hydrate snippet3

 

 

 

 

 

 

Source of Error/ Limitations/ Assumptions: 

– Allowing the crucible to cool to room temperature before weighting as if not cooled then convection currents will lower the mass and resulting in incorrect results