Aim / Objective:

To produce soap using a base-catalyzed saponification of triglycerides.


Soap molecules are the conjugate bases of fatty acids.  Vegetable oils and animal fats are the main materials that are “saponified”. These fats are in fact tri-esters of a glycerol molecule. In the traditional one-step process, the triglyceride is treated with a strong base (e.g., lye), which accelerates cleavage of the ester bond and releases the fatty acid in its conjugate base form, and glycerol.


A General Reaction is as follows:

Saponification using trigylcerides

Saponification using triglycerides

Different alkyl (R) groups are found in different fats and oils.  Depending on which triglyceride (tri-ester) you choose, your soap will have different properties.  For example, some oils make soft or liquid soaps, and some fats make hard soap.


Tripod stand, wire gauze, spatula, breaker, 2 measuring cylinder, glass stirring rod, filter funnel, filter paper, Bunsen burner, 2 test tubes with bungs, test tube rack, evaporating dish, castor oil, concentrated sodium hydroxide (NaOH), distilled water, perfume, dye, saturated solution of sodium Chloride (NaCl)

Method / Procedure:

  1. Half fill a beaker with tap water and set to boil.
  2. Place 2cm3 castor oil into evaporating dish. Use a measuring cylinder to pour 10cm3 of concentrated NaOH into the castor oil
  3. Place the evaporating dish atop the beaker of boiling water
  4. Stir the mixture of oil and alkali with a glass stirring rod for 10-15 minutes
  5. Add 10cm3 of the saturated salt solution to the basin and stir the mixture
  6. Turn the Bunsen burner off and leave to cool for 2-3 hours
  7. Use a spatula to scrape the crust of soap which is formed on the side of the evaporating dish
  8. Put this material in a beaker
  9. Add water to the material in the beaker and heat the beaker.
  10. Add a few drops of dye and perfume to the beaker


1. What is the name given to this process?

>>> Saponification

2. Write the word equation for this reaction

>>> Fat/Oil + NaOH        =         Glycerol + Soap (Sodium Salt of Acid)

3. Why is the product of saponification called a salt?

>>> This experiment is the hydrolysis of a fatty acid, usually from lye and fats. The process produces a carboxylate which is a sodium salt.

4. Why was ethanol added to the mixture of fat and base?

>>> Ethanol (ethyl alcohol) is added to the mixture to make the soap transparent. Transparent soap is also known as glycerin.

5. How does soap emulsify fats and oils?

>>> Grease and oil are nonpolar and insoluble in water. When soap is mixed with oils and fats  the nonpolar hydrocarbon portion of the micelles of the soap break up the nonpolar oil molecules. A different type of micelle then forms, with nonpolar oils and fats molecules in the center. Therefore, grease and oil and the ‘dirt’ attached to them are caught inside the micelle and can be rinsed away.

6. Explain the difference in “hard water” and “soft water”

Hard water is any water containing a great quantity of dissolved minerals while soft water is treated water in which the only cation (positively charged ion) is sodium.

7.  Explain which water is better to use with soap

>>>Soap is less effective in hard water as it will react with the ions in the water to form the calcium or magnesium salt of the organic acid of the soap. These salts are insoluble and form grayish soap scum, but no cleansing lather.


Significant Figures and Rounding Off

Significant Figures and Rounding Off

Definition: What exactly are significant figures?

The number of significant figures in a result is the number of figures that are known with some degree of reliability. Significant figures are very important when reporting scientific data as the will give the reader an idea of how well you measured your data.

RULES When Assigning Significant Figures:

  1. All non-zero (1, 2, 3, 4, 5, 6, 7, 8, 9) digits are ALWAYS significant

Here’s what we mean:

22 has two significant figures

55.3 has three significant figures

14.96 has four significant figures

2. Zeros placed between other digits are always significant

Let’s look at that now:

3005 has four significant figures

208 has three significant figures

3.09 has three significant figures

3. Leading zeros to the left of the first non-zero digits are NOT significant.significant figures pic 1

They simply indicate the position of the decimal point

For Example:

0.0596 has three significant figures

0.0002 has one significant figure

4. Trailing zeros that are to THE RIGHT OF A DECIMAL POINT in a number ARE SIGNIFICANT

Look at these:

0.0150 has three significant figuressignificant figures pic 2

0.200 has three significant figures

0.80 has two significant figures

5. When a number ends in zeros that are not to the right of a decimal point, the zero are not necessarily significant.

To avoid confusions and problems:  the use of scientific notation or exponential is used.

For Example:

170 may be two or three significant figures

86,000 may be two or three significant figures

90,200 may be three or five significant figures

So it is recommended that you change the number to exponential and then use that figure to determine how many significant figures are there.

Using or examples above:

1.7 x 10-2 has two significant figures

8.6 x 10-4 has two significant figures or 8.60 x10-4 has three significant figures.

9.02 x 10-4 has three significant figures

Significant figures in Multiplication, Division, Trig. Functions

In a calculation involving multiplication, division, trigonometric function etc., the number of significant digits in the answer should be equal to the least number of significant digits in any one of the numbers being multiplied, divided etc.


2.13 (3 sig fig)  x 4.236(4 sig fig) = 9.02 (the answer has 3 sig figs.)

483 / 68.60 = 7.04

Significant figures in Addition and Subtraction

When performing addition and subtraction calculations, the number of decimal places in the answer should be the same as the least number of decimal places in any of the numbers in the calculation.


1.96      + 8.7 = 14.6

48.1 – 32.05 = 16.1

Rounding Off

(1)    If the number to be dropped is greater than 5, the last digit is increased by one.

For example, 11.6 is rounded to 12.

(2)    If the number to be dropped is less than 5, the last digit is left as it is.

For example, 10.2 is rounded to 10.

(3)    If the number to be dropped is 5, and if any digit following it is not zero, the last remaining digit is increased by one.

For example, 9.52 is rounded to 10.

(4)    If the digit to be dropped is 5 and is followed only by zeroes, the last remaining digit is increased by one if it is odd, but left as it is if even.

For example,

13.50 is rounded to 14,
12.5 is rounded to 12.

Now that we’ve examine all the rules and shown examples then using the rules of significant figures and rounding off try these:

  1. 47.76 + 3.907 + 227.2 =
  2. 378.26 – 42.970 =
  3. 104.230 + 25.07795 + 0.70 =
  4. 122 – 0.25 + 4.109 =
  5. 2.02 x 3.4 =
  6. 800.0 / 5.2405 =
  7. 0.1174 x 418 =
  8. 48 x 5.00 =
  9. 5.3=
  10. 0.555 x (85 + 12.3) =


Answers to the Questions above:

  1. 47.76 + 3.907 + 227.2 = 278.9
  2. 378.26 – 42.970 = 335.29
  3. 104.230 + 25.07795 + 0.70 = 130.01
  4. 122 – 0.25 + 4.109 = 118
  5. 2.02 x 3.4 = 6.9
  6. 800.0 / 5.2405 = 152.7
  7. 0.1174 x 418 = 49.0
  8. 48 x 5.00 =2.4 x102
  9. 5.34 = 7.9 x 10
  10. 0.555 x (85 + 12.3) = 54

The only way to master this is by practicing. Just practice and practice and practice some more until you master it. You know what they say : Practice makes perfect!!

Let us know how it goes.


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


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


  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

Electrical conductivity of Iodine|Copper|Zinc|Aluminum Foil|Plastic|Plastic Ruler|Rubber|Graphite|Hydrochloric Acid|Sodium Chloride|Sodium Chloride Crystals|Acetone and Ethanol

Electrical conductivity of  Iodine|Copper|Zinc|Aluminum Foil|Plastic|Plastic Ruler|Rubber|Graphite|Hydrochloric Acid|Sodium Chloride|Sodium Chloride Crystals|Acetone and Ethanol

Aim / Objective:
To investigate the electrical conductivity of various substances


•Electrical Conductivity is defined as the measure of a substance/materials ability to transport of an electric charge or current. It is represented by the Greek letters- σ (sigma), κ (kappa), or γ (gamma).
•When a difference of electrical potential is placed across the conductor, an electric current appears. Conductors such as metals have high conductivity while insulators like that of glass have low conductivity.
•Electrical resistivity measures how strongly a material/substance opposes the flow of electric current. It is represented by the Greek letter ρ (rho), the lower the resistivity the more readily the substance will allow the flow of electric charge.
The following table shows the electrical conductivity and electrical resistivity of various materials
electrical conductivity

Materials/ Apparatus:

electrical circuit, electrolytic cell, samples of metals (copper (Cu), Zinc (Zn), aluminum oil (Al), Samples of non-metals (iodine (I), plastic, rubber, plastic ruler, graphite ), sodium chloride crystals (NaCl), aqueous sodium chloride (NaCl), dilute hydrochloric acid (HCl), organic solvents (ethanol & acetone)

Method / Procedure:

  1. Draw a table with three columns, with the names of the test samples in the first column.
  2. Label the second column “Predictions”, and indicate whether the test samples are conductors or insulators based on your previous knowledge.
  3. Label the third column as “Results”.
  4. Connect the circuit as shown in the diagram to determine which of the test samples are electrical conductors.
    For the solid test samples, use circuit 1. For liquid samples, use circuit 2.
  5. Record test results in the third column of the table.
  6. Compare test results with your predictions.

Suggested Results: 

Table showing the results of a circuit test done on various substances to determine if they are conductors or insulators

Table showing the results of a circuit test done on various substances to determine if they are conductors or insulators


1.Define  electrical conductors:

>>> Electrical Conductors are substances which allow a current to flow through them eg. Graphite while insulators do not allow the passage of an electric current. Eg. plastic.

2. Identify and differentiate between the two types of conductions?

>>> There are two types of conductions; metallic conduction and electrolytic conduction. Metallic conduction involves the flow of free electrons through a metal; these electrons are present in the electron pool and the metal remains chemically unchanged after an electric current is passed through. Electrolytic conduction involves the movement of free ions through a molten substance or solution.

3. Explain why solid ionic compounds do not conduct electricity?

>>> these compounds do not conduct electricity in their solid state because the ions are rigid an tightly held in a crystal lattice and are not able to move. However in its molten state or in a solution the lattice is broken down and the ions are free to move, an example of this is Sodium Chloride crystal.

4. Explain why insulators do not conduct electricity?

>>> They do not conduct electricity simply because of the absence of ions, for when ions are set free they move to an oppositely charged electrode when a voltage is applied. Insulators usually contain molecules.

Remember to keep reading guys. It’s the only way that the exams won’t be difficult. See our Additional information page for some very interesting articles to read as well.

Reactions of Acids with Carbonates Hydrogen Carbonates and Bases

Reactions of Acids with Carbonates, Hydrogen Carbonates, and Bases

Aim / Objective:
To investigate the reactions of Acids with Carbonates, Hydrogen Carbonates, and Bases

Acids are groups of compounds that have at least one hydrogen atom as a part of its molecules. This hydrogen atom plays a major role in its reactions. The strength of an acid is given by its pH value.
When an acid reacts with a carbonate, a salt is produced with the evolution of carbon dioxide.
This gas can be tested for by bubbling the gas through lime-water (dilute solution of calcium hydroxide). If a white precipitate (turns milky) is produced the CO2 is present. The salt made is dependent on the acid and the carbonate that is used.
Bases react with acids to produce a salt and water only. No gas is produced in the reaction so be sure to look for the bases dissolving in the acid.
See Acids, Bases and Indicators under additional information for more details about Acids and Bases.

Sodium Hydrogen Carbonate (NaHCO3) Copper (II) Oxide (CuO) Sodium Carbonate (Na2CO3) / marble chips, dilute Sulphuric acid (H2S04), Lime Water Ca(OH)2, dilute Nitric Acid (HNO3) dilute Hydrochloric Acid(HCl), 3 test tubes, test tube rack, spatula,

Method / Procedure:
Place the three test tubes in a test tube rack and label 1, 2, and 3.
To test tube 1, half fill with HCL and add a spatula full of NaCO3
Test with lime water and record results
Half full test tube 2 with HNO3 and add a spatula full of NaCO3. Test with lime water and record results.
To test tube 3, half full with H2SO4 and add a spatula full of CuO. Then gently shake mixture, leave to settle then record results.
Tabulate the results.

Suggested Results:

Table showing reactions of acid with carbonates, hydrogen carbonates and bases

Table showing reactions of acid with carbonates, hydrogen carbonates and bases


1. Write the molecular and ionic equations for each reaction.

Reaction 1:

2 HCL (aq) + Na2CO3(s)        =           2NaCl (aq) + CO2 (g) + H2O (l)

2H+ (aq) + CO2-3(aq)                    =         CO2 (g) + H2O (l)


Reaction 2:

2HNO3 (aq) + NaHCO3(s)              =      2NaNO3 (aq) + CO2 (g) + H2O (l)

H+ (aq)     +      HCO3-(aq)                        =      CO2(g) + H2O(l)


Reaction 3:

H2SO4 (aq)    +    CuO(s)               =       CuSO4 (aq) + H2O (l)

2H+ (aq)   +    O2-(aq)                        =           H2O (l)


2.Write the equation to explain the reactions that occurs when excess carbon dioxide is passed through lime water.

  1. Ca(OH)2(l)   +    CO2(g)                    =                CaCO3(s) + H2O(l)
  2. CaCO3(s)   +   H2O(l)    +     CO2(g)                =        Ca(HCO3)2(aq)