Wednesday, 18 February 2026

VOLUMETRIC ANALYSISINTRODUCTION note

Volumetric analysis is an aspect of quantitative analysis which involves the measurement of the volume of reacting solutions in order to find the masses of substances dissolved in them. It involves two solutions where one of them is a standard solution ( a solution with a know concentration) while you are required, to standardize the other one by titration.

The other aspect of quantitative analysis is gravimetric analysis,this aspect involves weighing and determining the masses of reactants and products of a chemical reaction The main technique used in Volumetric analysis is the acid – base titration and redox or precipitation reaction.

APPARATUS USED IN VOLUMETRIC ANALYSIS ARE:

The burette, pipette, beaker, flasks, funnel, wash bottle, chemical balance, dropping pipette and retort stand.

Basic volumetric analysis Terms

TITRATION: this is the method used for carrying out volumetric analysis.

CONCENTRATION: - this is the amount of solute in a given volume of the solution.

STANDARD SOLUTION: This is a solution with an accuratly known concentration.

that is, a known amount of solute in a known volume of solution. An example of a standard solution is containing 20grammes of a NaOH in 250cm³ of solution

Iv. MOLAR SOLUTION: It is a solution which contains one mole of solute or the molar mass in 1dm³ of solution.

An example is a solution containing 40grammes (molar mass) of NaOH in 1dm³ of solution.

END POINT: This is the point at which the chemical reaction is complete during titration. The end point is detected with the help of an indicator.

INDICATORS: These are weak organic acids or bases whose colors changes according to the pH of the solution.

Indicators are widely used to monitor titrations involving colorless solutions of acids and bases.

INDICATORS AND THEIR COLOURS IN DIFFERENT MEDIA

Indicator Colour in acid Colour in alkaline Colour in neutral solutions

Methyl Orange Pinkish red yellow Orange

Phenolphthalein Colorless red colorless

Litmus Red blue Pink/purple

Methyl red Pink yellow orange

Screened methyl orange Purple/violet green grey

Bromothymol blue Yellow blue green

UNIVERSAL INDICATOR: This is a mixture of indicators used either as solution on a test paper to test the pH value of a solution. The pH scale is shown below:

Red pink, yellow green blue indigo (deep blue) violet (bluish purpl

1 2 3 4 5 6 7 8 9 10 11 12 13 14

Acidity increases↼↼↼↼↼↼↼↼↼ ⇀⇀⇀⇀⇀⇀⇀⇀⇀⇀⇀⇀Alkalinity increases

EFFECT OF WRONG USE OF INDICATOR

The success of a titration exercises depend on the use of the correct indicator. Wrong use of indicator will definitely give wrong result. For instance, let’s consider a case of the titration of a solution of a strong acid say HCl with that of a weak base say Na₂CO₃, methyl orange is the suitable indicator, but if phenolphthalein indicator is used instead, the end point will appear when only half of the weak base has been used up. This can then be represented with the following equation.

HCl + Na2CO3 Phenolphthalein NaHCO3 + NaCl

Indicator

This happened because the phenolphthalein is sensitive to a weak acid such as Na2CO3.

TITRATION EXAMPLE PH RANGE SUITABLE INDICATOR

1.Strongacid vs. strong base(3-1) O4(aq) and KOH(aq) 3.5 – 9.5 Any indicator is suitable.

2.Weak acid vs. strong base(7-11) H2C2O4and NaOH 7.0-9.5 phenolphthalein

3.strong acid vs. Weak base(3-7) HCl(aq)andNH3(aq) or K2CO3(aq) or Na2CO3(aq) or Ca(OH)2(aq) 3.5-7.0 Methyl orange or screened Methyl orange

4 Weak acid vs. Weak base CH3COOH(aq) and NH3(aq) No sharp change No suitable indicator. Or phenol red.

IMPORTANCE OF VOLUMETRIC ANALYSIS

I. Standardize unknown solution

II. Calculate molar mass, water of crystallization and solubility.

III. Determine the purity of substances.

IV. Determine the masses of substances dissolved.

V. Faster and more convenient

TITRATION PRECUTIONS

The burette must be clamped vertically or not tilted.

Wash the burette and pipette with water andrinse with distiled water.

Rinse the burette with acid,the pipette with base (alkali)before putting these solutions into them.

Ensure no air bubbles in the burette or pipette.

Remove the funnel after putting the acid into the burette(if a funnel is used).

The content of the pipette should be allowed to run into the conical flask without blowing air into it.

Use a drop or two (small amount)of indicators.

Read the lower meniscus.

Ensure the tap of the burette is not leaking.

EXAMPLE 1

A is a solution of tetraoxosulphate {vi} acid.

B is a solution containing 0.0500 mole of anhydrous Na2CO3 per dm3.

(a)Put A in the burette and titrate 20.00 or 25.00 cm3 portions of B using methyl orange as the indicator. Record the size of your pipette. Tabulate the burette readings, and calculate the average volume of the acid used. (b) From your result and data provided, calculate the

Amount of Na2CO3 IN 25.00 CM3 OF B used

Concentration of A in moldm-3

Concentration of A in gdm-3

Number of hydrogen ions in 1.00dm3 of A. {Avogadro number = 6.02x1023 mol1}

The equation of reaction is: H2SO4{aq} +Na2CO3{aq] Na2SO4 {aq} + H2O +CO2

{H =1,O=16, S=32}

SOLUTION

A Volume of pipette: 25cm-3

Titration results {Hypothetical data}

Burette reading 1cm3 II cm3 IIIcm3

Final 24.75 49.15 25.70

Initial 0.00 24.75 1.35

Volume of acid used 24.75 24.40 24.35

Average volume of acid used from titrations II and III:

(24.40+24.35 )/2 = 48.75/2 = 24.38cm3

NOTES: Only the titre values from titrations I and II can be used in averaging, since they are within ± 0.20cm3 of each other. …Rough of first titre can also be used in averaging, if it is within ± 0.20cm3 of any other titre value, and is not crossed.-----Do not round up 24.38cm3 to 24.40cm3

(b )To calculate the amount of Na2CO3IN 25.00CM3 Given: concof B = 0.050moldm3 :Volume = 25/1000dm3

Amount = Conc {moldm3 x Volume {dm3} = 0.050x25/1000 = 0.00125mol

To calculate the concentration of A in moldm3: The various titration variables are:

CA = xmoldm3; VA = 24.38CM3; Na = 1, CB = 0.050moldm3 VB = 25CM3: nB =1

Method 1: proportion method {from the first principle}

From the balanced equation of reaction:

1mol of Na2CO3 = 1 mol of H2SO4

. ∴ 0.00125mol Na2CO = 0.00125molH2so4

i.e. 24.38cm3 of A contained 0.00125 mol H2SO4

∴1000cm3 of A contained {0.00125 x 1000} /24.38 mol = 0.0513mol.Hence, concentration of A is 0.0513mol moldm3.

Method 2:

Mathematic formula method

From the data above, it is safe to use the mole ratio expression, in order to calculate the concentration CA of A, which the required variable.

Using CA VA /CB VB = nA/nB

Substituting; CA X 24.38/0.050 X25 =1/1

Making CA the subject of the formula

∴ CA =1 X0.050X25 /24.38 X1 = 0.0513 moldm-3

To calculate the concentration of A in gdm-3:

Using conc. {gdm-3} x Molar mass {gmol-1}

Concentration of H2SO4, moldm-3 =0.0513moldm-

Molar mass H2SO4, = 2 {1.0} + 32.0 + 4{16.0

= 2.0 +32.0 +64.0 = 98.0gmol-1

Substituting; Mass conc = 0.0513 x98 = 5.0274gdm-3.

=5.03gdm-3 {3 sig fig.}

(Iv).Number of hydrogen ions in 1.00dm3 of A

1 dm3 of A contained 0.0513mol of H2SO4.

H2SO4 ionizes in water completely thus:

H2SO4 (a q) 2H+SO42-(a q)

1mol 2mol

From the equation;

1 mole of H2SO4 produces2 x 0.0513 moles of H+ = 0.103 mol 0f H+

But 1 mole of H+ contains 6.02 x 1023 ions;

Therefore, 0.103 x 6.02 x 1023 ions =6.02 x10 23 ions.{3 sig fig.]

STOICHIOMETRY OF REACTION: This is the study of the quantitative relationship implied by

chemical reactions. Stoichiometry is the term used to describe calculations involving mass –

volume relationship between atoms in a compound and between molecules and atoms

participating in a chemical reaction.

The stoichiometry of the reaction can also be defined as the mole ratio in which the reactants

combine and the products are formed.

Experiments are designed to determine the stoichiometry of reactions and these include.

Precipitation reaction

Displacement of hydrogen from acid

Displacement of metallic ions

Synthesis of metallic oxides

Reduction of metallic oxides

The titration experiment is used to study the stoichiometry of the reaction of acids with

bases/alkalis.

Mole ratio and mass relationship

Consider the equation;

Mg(s) + 2HCl(aq) MgCl(aq) + H2(g)

No. Of moles 1 2 1 1

Mole ratio 1 : 2 1 : 1

Molar mass 24g 36.5g 95g 2g

Reacting mass 24g 2 x (36.5) 95g 2g

From the relationship above;

Reacting mass = no of moles x molar mass

:- No. of moles = reacting mass ...................................... 1

Molar mass

Other formulae for moles are

No. of moles = no of elementary particles ......................... 2

Avogadro’s number

No. of moles = molar concentration x Volume (in cm3) ........ 3

1000

By combining formulae 1 and 3, an important formulae is obtained;

No. of mole = reacting mass = molar concentration x volume

Molar mass 1000

:- reacting mass = molar concentration x volume

Molar mass 1000

Hence; Molar concentration = reacting mass x 1000 ............................. 4

Molar mass x volume

Equation 4 is very important when calculating the concentration of a solution with volumes other

than 1000cm3 (1dm3)

Example

How many moles are contained in 50g of Magnesium trioxocarbonate (IV)? (Mg=24, C=12,

O=16)

Solution

Molar mass of MgCO3 = 24 + 12 + (3 x 16) = 84g/mol

Reacting mass = 50g

No. of moles = reacting mass/molar mass

= 50/84 = 0.6 moles

Calculate the number of moles of CaCl2 that can be obtained from 30g of CaCO3 in the presence

of excess HCl. [Ca=40, C=12, O=16, H=1, Cl=35.5]

Solution

CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g)

Mole ratio 1 : 2 1 : 1 : 1

Molar mass of CaCO3 = 100g/mol

Molar mass of CaCl2 = 111g/mol

From the equation;

1 mole of CaCO3 produced 1 mole of CaCl2

100g of CaCO3 produced 111g of CaCl2

:- 30g of CaCO3 will produce 30 x 111 = 33.3g of CaCl2

100

Convert 33.3g of CaCl2 to moles;

No of moles = reacting mass/molar mass

= 33.3/111 = 0.3 moles

What mass of Na2CO3 is needed to prepare 1 dm3 of a 0.3M solution.

Solution

Molar mass of Na2CO3 = 106g/mol

Note: 0.3 M means that 0.3 mole of Na2CO3 is contained in 1dm3(1000cm3) solution.

:- No. of moles = reacting mass/molar mass

0.3 = mass/106

Mass = 0.3 x 106 = 31.8g

Calculations involving gas volume

In the industrial preparation of trioxonitrate (V) acid, ammonia gas is burned in oxygen in the

presence of a catalyst according to the following equation;

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

If 250cm3 of NH3 are burned completely, what volume of (a) oxygen is used up? (b) NO is produced

Solution

(a)

Equation of the reaction;

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

From the equation;

4 moles of NH3 ≡ 5 moles of O2

Thus 4 volume of NH3 ≡ 5 volumes of O2

Therefore 250 volumes of NH3 ≡ 250 x 5 = 312.50 cm3 of O2

(b)

From the equation;

4 volumes of NH3 ≡ 4 volumes of NO

:- 250 volumes of NH3 ≡ 250 volumes of NO

= 250cm3

What volume of dry oxygen gas (measured at s.t.p) will be produced from the decomposition of

3.50g potassium trioxochlorate (V)?

Solution

Equation of the reaction;

2KClO3(s) 2KCl(s) + 3O2(g)

From the equation;

2 moles of KClO3 produced 3 moles of O2

2 moles of KClO3 = 2 x 122.5g = 245g

Molar volumes of O2 at s.t.p. = 22.4dm3

:- 3 moles of O2 occupies 3 x 22.4 = 67.2dm3

Hence 245g of KClO3 produced 67.2dm3 of O2

:- 3.50g of KClO3 will produce 3.50 x 67.2 = 0.96dm3 of O2

245

Calculate the volume of carbon (IV) oxide at 350C and 720mmHg pressure which could be

obtained by heating 19g of copper (II) trioxocarbonate (IV) [ CuCO3 = 123.5, CO2 = 44, molar

volume of a gas at s.t.p. = 22.4dm3]

Solution

Equation of the reaction;

CuCO3(s) CuO(s) + CO2(g)

From the equation;

1 mole of CuCO3 ≡ 1 mole of CO2

Amount (in moles) of CuCO3 that reacted = 19/123.5 (i.e reacting mass/molar mass) = 0.15moles

:- 0.15 moles of CuCO3 yields 0.15 moles of CO2

But at s.t.p, 1 mole of CO2 occupies 22.4dm3

:- 0.15 moles of CO2 occupies 0.15 x 22.4 = 3.36dm3

Using the general gas equation;

P1V1 = P2V2

T1 T2

Where; P1 = 760mmHg, V1 = 3.36dm3 (i.e. 3360cm3), T1 = 273K

P2 = 720mmHg, V2 = ? T2 = 350C = 308K

V2 = P1V1T2 = 760 x 3360 x 308

T1P2 273 x 720

V2 = 4001cm3

Calculations involving liquid volumes

Calculate the volume of 0.1M ammonia which could be obtained by heating 2.7g of ammonium

chloride with excess sodium hydroxide and absorbing all the ammonia evolved.

Solution

Equation of the reaction;

NH4(s) + NaOH(aq) NaCl(aq) + NH3(g) + H2O(l)

From the equation

1 mole of NH4Cl produces 1 mole (22.4dm3) of NH3

Amount of NH4Cl that reacted = 2.7/53.5 (i.e. reacting mass/molar mass) = 0.05moles

:- 0.05 moles of NH4Cl produces 0.05 moles of NH3

Using the formula

Number of moles = molar conc. X volume (cm3)

1000

Where; number of moles = 0.05 moles

Molar conc. = 0.1M

Volume = ?

= 0.05 x volume

1000

Volume of NH3 = 0.1 x 1000 = 500cm3

0.5

Calculate the volume of 2M hydrochloric acid which could be obtained by dissolving 560cm3

hydrogen chloride gas (measured at s.t.p.) in water. [molar volume of gas at s.t.p = 22.4dm3]

Solution

22400cm3 hydrogen chloride gas = 1 mole of hydrogen chloride gas at s.t.p

:- 560 cm3 of hydrogen chloride gas = 560 x 1 = 0.025moles

22400

The number of moles of hydrogen chloride gas = 0.025 moles

The concentration of hydrochloric acid formed is = 2M

Volume of the acid formed = ?

Number of moles = molar conc. X volume (cm3)

1000

0.025 = 2 x volume

1000

Volume = 0.025 x 1000 = 12.5cm3

Calculation involving masses

Calculate the mass of carbon(IV) oxide produced by burning 104g of ethyne ( C=12, O=16, H=1)

Solution

Equation of the reaction

2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(g)

From the equation;

moles of ethyne produces 4 moles of CO2

i.e. 2 x 26g of ethyne produces 4 x 44g of CO2

52g of ethyne produces 176g of CO2

:- 104g of ethyne produces 104 x 176 = 352g of CO2

What mass of lead (II) trioxonitrate (V) would be required to yield 12g of lead (II) chloride on the

addition of excess sodium chloride solution? (Pb = 207, N=14, O=16, Na=23, Cl=35.5)

Solution

Equation of the reaction

Pb(NO3)2(aq) + 2NaCl(aq) PbCl(s) + 2NaNO3(aq)

331g 278g

From the equation

278g of PbCl is produced from 331g of Pb(NO3)2

:- 12g of PbCl will be produced 12 x 331 = 14.29g of Pb(NO3)2

278

Assignment

Answer question 6 (theory) on page 145 of Essential Chemistry

Tutorial questions

In a certain reaction, 15.0g of impure magnesium sample reacted with excess hydrochloric acid

liberating 8.6 dm3 of hydrogen gas at s.t.p.

Write a balanced equation for the reaction

Calculate the:

mass of pure magnesium in the sample

Percentage purity of the magnesium sample

Number of chloride ions produced in the reaction.

[ Mg = 24.0, volume at s.t.p. = 22.4dm3, Avogadro’s constant = 6.02 x 1023 mol-1]

What is the volume of 0.25moldm-3 solution of KOH that would yield 6.5g of solid KOH on

evaporation? [ K = 39.0, O = 16.0, H = 1.00]

Consider the reaction represented by the following equation:

Na2CO3(aq) + 2HCl(aq) 2NaCl(aq) + H2O(l) + CO2(g)

What volume of 0.02 moldm-3 Na2CO3(aq) would be required to completely neutralize 40cm3 of

0.10 moldm-3 HCl(aq)?

4a. Find the volume of oxygen produced by 1 mole of potassium trioxochlorate (V) at s.t.p in the

following reaction.

2KClO3(s) 2KCl(s) + 3O2(g)

b. Find the mass of sodium trioxocarbonate (IV) needed to give 22.4dm3 of carbon (IV) oxide at

s.t.p in the following reaction.

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

[ Na = 23, O = 16, C = 12, molar volume of gas at s.t.p = 22.4dm3]

What mass of anhydrous sodium trioxocarbonate (IV) is present in 500cm3 of 0.1moldm-3 of

the solution? [ Na = 23, C = 12, O = 16]

ACID-BASE REACTION

Determination of the composition of substances either qualitatively or quantitatively is very

important in chemistry. The two approaches to the quantitative analysis are volumetric and

gravimetric approaches. Volumetric analysis is commonly used because it is faster and more

convenient, although less accurate than gravimetric analysis.

TITRATION: it is the method employed in volumetric analysis. The method involves adding a

solution to a known volume of another solution until the chemical reaction between them is

completed. The completion is shown by a colour change in the resulting solution or in an added

indicator. Usually, a standard solution must be used to react with a solution of unknown

concentration. The reacting volumes of the solutions are the used to calculate the unknown

concentration of the solution.

Definition of terms in volumetric analysis.

End point: this is the point at which the neutralization reaction is just complete. The change in the

colour of the indicator is used to monitor the end point.

Indicators: They are weak organic acids or bases which produce different colours in solution

according to the hydrogen ion, H+ concentration in that solution.

Mass concentration: It is the amount in grams of solute present in 1 dm3 (1000cm3) of the

solution. It is expressed in g/dm3. It is calculated with the formula;

Mass conc. = molar concentration x molar mass

Molar concentration (Molarity): It is the amount in moles of solute per dm3 (1000cm3) of solution.

It is expressed in mol/dm3. It is calculated with the formula;

Molarity (Molar conc.) = mass conc

Molar mass

Standard solution: This is a solution of known concentration.

Molar solution: A molar solution of a compound is the one which contains one mole or the molar

mass of the compound in 1dm3 (1000cm3) of the solution.

Volumetric analysis usually involves the titration of;

Acid against base or trioxocarbonate (IV)

Oxidizing agent against reducing agent

One substance against another substance giving a precipitate.

Materials used in acid base titration include;

Weighing bottle

Chemical balance

Pipette

Burette

Retort stand

Filter paper

Funnel

White tile

Standard volumetric flask

Conical flask

CALCULATIONS IN VOLUMETERIC ANALYSIS

Volumetric analysis is used to:

Standardize unknown solution i.e. to get the concentration of a solution whose

concentration is not known

Determine the purity of substances

Calculate the molar mass of a compound

Calculate the molar concentration of a compound

Calculate the mass concentration of a compound

Number of ions/particles present in a solution

Water of crystallization in a solution

Solubility of a compound

The procedures for all these determinations are similar to that of acid-base titration. The following

formula is important in volumetric analysis.

CAVA = na

CBVB nb

Where:

CA = Molarity ( molar concentration) of acid

VA = Volume of acid

na = number of moles of acid

CB = Molarity (molar concentration) of base

VB = Volume of base

nb = number of moles of base

PRACTICAL CLASS

Experiment to determine the concentration of hydrochloric acid using standard sodium hydroxide

solution

Assignment.

What are the precautions observe during acid-base titration?

Calculations on molar solution

Calculate (a) the mass of anhydrous sodium trioxocarbonate (IV) present in 300cm3 of 0.1M

solution (b) the number of Na2CO3 particles present in the solution. (Na = 23, C = 12, O = 16)

Solution

(a)

Molar mass of Na2CO3 = 106g/mol

Molar conc. = 0.1 M

Mass conc. = ?

But mass conc = molar conc. X molar mass

:- mass conc = 0.1 x 106 =10.6g

This means that 10.6g of Na2CO3 is contained in 1000cm3 (1dm3) of 0.1 M Na2CO3 solution

:- in 300cm3 of the solution we have

300 x 10.6

1000

= 3.18g

Alternative method (formula method)

Molar mass of Na2CO3 = 106g/mol

Molar conc. = 0.1 M

Volume = 300cm3

Reacting mass = ?

Molar conc. = reacting mass x 1000

Molar mass x volume

= reacting mass x 1000

106 x 300

Mass = 106 x 300 x 0.1 =3.18g

1000

(b)

Number of moles (n) = number of elementary particles

Avogadro’s number

Number of particles = number of moles x Avogadro’s number

Since reacting mass = 3.18g

Number of moles = reacting mass = 3.18 = 0.03 moles

Molar mass 106

:- number of Na2CO3 particles = 0.03 x 6.02 x 1023 = 0.181 x 1023

Calculate the volume of 0.25M solution of H2SO4 that will contain a mass of 4.5g of the acid.

[H=1, S=32, O=16]

Solution

Molar conc. = 0.25M

Molar mass of H2SO4 = 98g/mol

Mass conc. = molar conc. X molar mass

= 0.25 x 98 = 24.5g/dm3

i.e. 24.5g of H2SO4 is contained in 1000cm3 of the solution

:- 4.5g of H2SO4 will be contained in

4.5 x 1000 = 184cm3

24.5

Alternative method (formula method)

Molar mass ofH2SO4 = 98g/mol

Molar conc. = 0.25 M

Volume = ?

Reacting mass = 4.5g

Molar conc. = mass x 1000

Molar mass x volume

0.25 = 4.5 x 1000

98 x vol

Volume = 4.5 x 1000 = 184 cm3

0.25 x 98

Calculate the volume of hydrogen chloride gas at s.t.p. that would yield 1.2dm3 of 0.15M

aqueous hydrogen chloride solution. (molar volume of all gases at s.t.p. = 22.4dm3)

Solution

Molar conc. Of HCl solution is 0.15M i.e. dm3 of the HCl solution contains 0.15 moles of HCl.

:- 1.2dm3 of the solution will contain

1.2 x 0.15 = 0.18moles of HCl

But 1 mole of HCl gas at s.t.p occupies 22.4dm3

:- 0.18 moles at s.t.p will occupy 0.18 x 22.4 = 4.03dm3

4.03 dm3 of HCl gas at s.t.p would be required to yield 1.2dm3 of 0.15M aqueous solution.

Alternative method (formula method)

Molar conc = 0.15M

Molar mass of HCl = 36.5g/mol

Volume = 1.2dm3 = 1200cm3

Reacting mass = ?

Molar conc. = mass x 1000

Molar mass x volume

0.15 = mass x 1000

36.5 x 1200

Mass = 0.15 x 36.5 x 1200 = 6.57g

1000

At s.t.p 36.5g of HCl occupies 22.4dm3

:- 6.57g of HCl will occupy 6.57 x 22.4 = 4.03dm3

36.5

Calculations on standardization

A is 0.05 mol/dm3 of H2SO4. B is sodium trioxocarbonate (IV) solution. If 37.5cm3 of the acid was

required to neutralize 25.00cm3 of the trioxocarbonate, calculate;

conc. Of B in mol/dm3

conc of B in g/dm3

volume of CO2 librated at s.t.p. during the titration

(Na = 23, C = 12, O = 16, S = 32)

Equation of the reaction;

H2SO4(aq) + Na2CO3(aq) Na2SO4(aq) + H2O(l) + CO2(g)

Solution

(a)

CA = 0.05mol/dm3 VA = 37.5cm3 na = 1

CB = ? VB = 25cm3 nb = 1

CAVA = na

CBVB nb

0.05 x 37.5 = 1

CB x 25 1

CB = 0.05 x 37.5 x 1 = 0.075mol/dm3

25 x 1

(b)

Mass conc. = molar conc. X molar mass

= 0.075 x 106

= 7.95g/dm3

Note: the molar mass of Na2CO3 is 106g/mol

(c)

From the equation;

mole of Na2CO3 liberated 1 mole of CO2

:- 0.075 moles of Na2CO3 would liberate 0.075 x 1 = 0.075 moles of CO2

At s.t.p, 1 mole of CO2 occupies 22.4dm3

:- 0.075 mole will occupy 0.075 x 22.4 = 1.68dm3

Hence, 1.68dm3 of CO2 was liberated during the reaction.

Ycm3 of hydrogen chloride gas at s.t.p. were passed into 60cm3 of 0.1moldm3 sodium

trioxocarbonate (IV) solution. The excess trioxocarbonate (IV) was neutralized by 20cm3 of 0.05

mol/dm3 H2SO4. Calculate (a) the mass of excess Na2CO3 in g/dm3 (b) the value of Y

[molar volume of gas at s.t.p = 22.4dm3, Na2CO3 = 106]

Solution

(a)

Equation of the reaction;

H2SO4(aq) + Na2CO3(aq) Na2SO4(aq) + H2O(l) + CO2(g)

From the equation;

Molar (i.e. 1000cm3 of 1mol/dm3) solution of H2SO4 ≡ 1 Molar solu on of Na2CO3

Thus; 1000cm3 1M H2SO4 ≡ 106g in 1000cm3 Na2CO3

1000cm3 0.05M H2SO4 ≡ 106 x 0.05g in 1000cm3 Na2CO3

:- 20cm3 0.05M H2SO4 ≡ 106 x 0.05 x 20g in 1000cm3 Na2CO3

1000

= 0.106g/dm3 Na2CO3

The mass concentration of the excess Na2CO3 which reacted with H2SO4 = 0.106g/dm3

1000cm3 1M Na2CO3 contains 106g Na2CO3

:- 1000cm3 0.1M Na2CO3 contains 10.6g Na2CO3

Hence 60cm3 0.1M Na2CO3 contains 10.6 x 60 g Na2CO3

1000

= 0.636g/dm3

The mass concentration of the original Na2CO3 solution = 0.636g/dm3

:- the mass concentration of Na2CO3 neutralized by Ycm3 HCl gas = 0.636 – 0.106 = 0.53g/dm3

(b)

Equation of reaction

Na2CO3(aq) + 2HCl(aq) 2NaCl(aq) + H2O(l) + CO2(g)

Form the equation

106g of Na2CO3 ≡ 2 x 22.4 dm3 of HCl gas at s.t.p

:- 0.53g of Na2CO3 ≡ 2 x 22400 x 0.53 cm3

106

= 224cm3

A is a dilute tetraoxosulphate (VI) acid. B contains 1.5g of sodium hydroxide per 250cm3 of

solution. 25cm3 of B requires 15.5cm3 of A for complete neutralization. Calculate

Concentration of B in mol/dm3

Concentration of A in mol/dm3

The number of hydrogen ion ions in 1.0dm3 of solution A

Solution

Equation of reaction

H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l)

(a)

250cm3 of B contains 1.5g of NaOH

:- 1000cm3 of B will contain 1000 x 1.5g of NaOH

250

= 6.00g/dm3

Molar conc. Of B = mass conc/molar mass

The molar mass of NaOH = 40g/mol

:-Molar Conc. (Molarity) of B = 6/40

= 0.15mol/dm3

(b)

CAVA = na

CBVB nb

CA = ? VA = 15.5cm3 na = 1

CB = 0.15mol/dm3 VB = 25cm3 nb = 2

CA x 15.5 = 1

0.15 x 25 2

CA = 0.15 x 25 x 1 = 0.121mol/dm3

x 15.5

(c)

H2SO4 2H+ + SO4

mole of H2SO4 yield 2 moles of H+

Since the Molarity of H2SO4 = 0.121mol/dm3

The concentration of H+ in the acid = 2 x 0.121 = 0.242mol/dm3

Number of H+ present = number of moles x Avogadro’s constant

= 0.242 x 6.02 x1023

= 1.46 x1023 hydrogen ions.

Assignment

A solution of trioxonitrate (V) acid contained 0.67g in 100cm3. 31.0cm3 of this solution

neutralized 25cm3 of a sodium trioxocarbonate (IV) solution. Calculate the concentration of the

trioxocarbonate (IV) solution. (HNO3 = 63, Na2CO3 = 106)

Tutorial questions

A solution of trioxonitrate (V) acid contained 0.67g in 100cm3. 31.0cm3 of this solution

neutralized 25cm3 of a sodium trioxocarbonate (IV) solution. Calculate the concentration of the

trioxocarbonate (IV) solution. (HNO3 = 63, Na2CO3 = 106)

2.Calculate the mass of pure sodium chloride that will yield enough hydrogen chloride gas to

neutralize 25cm3 of 0.5M potassium trioxocarbonate (IV) solution. ( NaCl = 58.5, HCl = 35.5,

K2CO3 = 138)

Calculations of percentage purity

A piece of limestone, CaCO3 was added to 1dm3 of 0.1mol/dm3 hydrochloric acid. After

effervescence had stopped, 31.25cm3 of the resulting solution required 25cm3 of 0.05mol/dm3

sodium hydroxide for complete neutralization. Calculate the mass of limestone added (CaCO3 =

100, HCl=36.5, NaOH = 40)

Solution

Equation of the reaction

CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g)

100g 2 x 36.5g

NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)

Using CAVA = na

CBVB nb

Where CA = ? VA = 31.25cm3 na = 1

CB = 0.05moldm-3 VB = 25cm3 nb = 1

CA x 31.25 = 1

0.05 x 25 1

CA = 0.05 x 25 x 1 = 0.040mol/dm3

31.25 x 1

The mass concentration of the acid that react with NaOH therefore = 0.04 x 36.5 = 1.46g/dm3

But, the original concentration of the acid is 0.1mol/dm3 = 0.1 x 36.5 = 3.65g/dm3

:- the concentration of the acid that reacted with CaCO3 = 3.65 – 1.46 = 2.19g/dm3

From the equation;

CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g)

100g of CaCO3 ≡ 73gof HCl

:- Yg of CaCO3 ≡ 2.19g of HCl

Y = 100 x 2.19 = 3g

The mass of CaCO3 added to 0.1mol/dm3 HCl is 3g

D. Calculations on molar mass, water of crystallization and solubility of substances

28.0cm3 hydrochloric acid of concentration 4.1g/dm3 neutralized 25.0cm3 of an unknown alkali

YOH, whose concentration was 7.0g/dm3. Calculate (a) the molar concentration of YOH (b) the

relative atomic mass of the element Y. Name the element Y if possible. (HCl = 36.5)

Solution

(a)

Equation of the reaction;

HCl(aq) + YOH(aq) YCl(aq) + H2O(l)

Molar conc. Of HCl = Mass conc/molar mass

= 4.1/36.5 = 0.112mol/dm3

CAVA = na

CBVB nb

CB = 0.112 x 28 x 1 = 0.125mol/dm3

25 x 1

:- molar conc. Of YOH = 0.125mol/dm3

(b)

Molar conc. = mass conc/molar mass

:- 0.125 = 7.0

Y + 16 + 1

Y + 17 = 7.0/0.125

Y = 56 – 17 = 39

The relative atomic mass of Y = 39

The element is Potassium and YOH is KOH

Some crystals of washing soda were exposed to the atmosphere for efflorescence to take place.

6.02g of this partly effloresced washing soda, Na2CO3.yH2O were then dissolved in 500cm3 of

water. 25cm3 of this trioxocarbonate (IV) solution required 32.10cm3 of 0.097mol/dm3

hydrochloric acid for complete neutralization. Calculate y. Hence, write the formula of the

effloresced washing soda. (Na = 23, H = 1, C = 12, Cl = 35.5, O = 16)

Solution

Equation of then reaction

Na2CO3.yH2O(s) + 2HCl(aq) 2NaCl(aq) + H2O(l) + CO2(g) + yH2O(l)

CAVA = na

CBVB nb

0.097 x 32.1 = 2

CB x 25 1

CB = 0.097 32.1 = 0.0623mol/dm3

25 x 2

6.02g of Na2CO3.yH2O is contained in 500cm3

:- in 1000cm3 we have 6.02 x 1000 = 12.04g

500

:- the mass concentration of Na2CO3.yH2O is 12.04g/dm3

Molar mass of Na2CO3.yH2O = (106+ 18y)gmol-1

Molar mass = mass conc./molar conc.

106 + 18y = 12.04/0.0623

106 + 18y = 193.26

18y = 193.26 – 106

18y = 87.3

Y = 4.9

Therefore the formula of the effloresced washing soda is Na2CO3.5H2O

A saturated solution of lead (II) trioxonitrate (V) Pb(NO3)2, was prepared at 220C. 27cm3 of this

solution required 46cm3 of sodium chloride, NaCl solution containing 96g/dm3 for complete

precipitation. Find the solubility of lead (II) trioxonitrate (V) at 220C in (a) mol/dm3 (b) g/dm3

(Na=23, Cl=35.5, Pb=207, N=14, O=16)

Solution

Equation of the reaction

2NaCl(aq) + Pb(NO3)2(aq) PbCl2(s) + 2NaNO3(aq)

(a)

Molarity of the NaCl solution = 96/58.5 = 1.641mol/dm3

CAVA = na

CBVB nb

Where A represent NaCl and B represents Pb(NO3)2

CA = 1.641mol/dm3 VA = 46cm3 na = 2

CB = ? VB = 27cm3 nb = 1

x 46 = 2

CB x 27 1

CB = 1.641 x 46 x 1 = 1.398mol/dm3

27 x 2

The solubility of Pb(NO3)2 at 220C = 1.398mol/dm3

(b)

Mass conc = molar conc x molar mass

= 1.398 x 331

= 462.7g/dm3

The solubility of Pb(NO3)2 at 220C = 462.7g/dm3

E. Calculations involving dilution of a solution

What volume of 2 mol/dm3 NaOH solution is required to prepare 100cm3 of a solution of NaOH

with a concentration of 0.2mol/dm3?

Solution

M1V1 = M2V2

Where; M1 = initial concentration = 2mol/dm3

V1 = initial volume = ?

M2 = final concentration = 100cm3

V2 = final Volume =0.2

V1 = M2V2 = 100 x 0.2

M1 2

V1 = 10cm3

Thus if 10cm3 of a 2mol/dm3 NaOH solution is diluted with water to 100cm3, it will produce a

0.2mol/dm3 solution of NaOH.

Assignment

Water is added to 100cm3 of a 0.25 mol/dm3 NaCl solution to make it up to 1.5dm3. Calculate

the concentration of the solution after dilution.

Tutorial questions

2g of a mixture of NaOH and NaCl (as impurity) were dissolved in 500cm3 of water. If 25cm3 of

this solution was neutralized by 21cm3 0.1mol/dm3 hydrochloric acid, calculate the percentage

of the NaCl impurity. (NaOH = 40, HCl = 36.5, NaCl = 58.5)

An excess of a divalent metal M was dissolved in a limited volume of hydrochloric acid. If

576cm3 of hydrogen were liberated at s.t.p, what was the mass of the metal that produced this

volume of hydrogen? (M = 24, H=1, molar volume of gas at s.t.p = 22.4dm3)

Thursday, 5 February 2026

BENZENE notes for students

BENZENE AND ITS DERIVATIVES

1. Introduction

Benzene is the simplest aromatic hydrocarbon with the molecular formula C₆H₆. It is a colourless, flammable liquid with a pleasant smell and is an important raw material in the chemical industry.

Although benzene contains three double bonds, it is much more stable than ordinary unsaturated compounds. This unusual stability is due to its aromatic nature.

2. Structure of Benzene

(a) Kekulé Structure

Friedrich Kekulé proposed that benzene consists of six carbon atoms arranged in a ring with alternating single and double bonds.

However, this structure could not explain:

The equal length of all C–C bonds in benzene
The exceptional stability of benzene
Why benzene undergoes substitution instead of addition reactions

(b) Resonance Structure

Benzene actually exists as a resonance hybrid of two Kekulé structures. The π-electrons are delocalized around the ring.

This delocalization:

Makes all C–C bonds equal in length
Gives benzene extra stability (resonance energy)

Benzene is often represented as a hexagon with a circle inside to show delocalized electrons.

3. Aromaticity

For a compound to be aromatic, it must:

Be cyclic
Be planar
Have continuous overlap of p-orbitals
Obey Hückel’s rule: contain (4n + 2) π-electrons

Benzene has 6 π-electrons (n = 1), so it is aromatic.

4. Physical Properties of Benzene

i. Molecular formula: C₆H₆

ii. Molecular mass: 78 g mol⁻¹

iii. Colourless liquid

iv. Sweet smell

v. Melting point: 5.5 °C

vi. Boiling point: 80.1 °C

vii. Insoluble in water

viii. Soluble in organic solvents

x. Highly flammable

5. Chemical Properties of Benzene

Benzene mainly undergoes electrophilic substitution reactions to preserve its aromatic ring.

(a) Nitration

Reaction with concentrated nitric acid in the presence of concentrated sulfuric acid:

C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O
(Product: Nitrobenzene)

(b) Halogenation

Reaction with chlorine or bromine in the presence of FeCl₃ or FeBr₃:

C₆H₆ + Cl₂ → C₆H₅Cl + HCl
(Product: Chlorobenzene)

(c) Sulfonation

Reaction with fuming sulfuric acid:

C₆H₆ + H₂SO₄ → C₆H₅SO₃H + H₂O
(Product: Benzenesulfonic acid)

(d) Friedel–Crafts Alkylation

Reaction with an alkyl halide in the presence of AlCl₃:

C₆H₆ + CH₃Cl → C₆H₅CH₃ + HCl
(Product: Toluene)

(e) Friedel–Crafts Acylation

Reaction with an acyl chloride:

C₆H₆ + CH₃COCl → C₆H₅COCH₃ + HCl
(Product: Acetophenone)

6. Benzene Derivatives

Benzene derivatives are compounds formed when one or more hydrogen atoms in benzene are replaced by other atoms or groups (functional groups).

(a) Toluene (Methylbenzene)

Formula: C₆H₅CH₃

Toluene is a derivative of benzene for when one of the hydrogens is substituted by an alkyls group

Toluene (Methylbenzene, C₇H₈) – Physical and Chemical Properties

Toluene is an aromatic hydrocarbon obtained mainly from petroleum and coal tar. It consists of a benzene ring attached to a methyl group (–CH₃), which slightly increases its reactivity compared to benzene.

✅ Physical Properties of Toluene

1. Appearance

Colourless, clear liquid

2. Odour

Sweet, pleasant, benzene-like smell

3. Molecular formula

C₇H₈

4. Molar mass

92 g mol⁻¹

5. Boiling point

110–111°C

6. Melting point

−95°C

7. Density It has a density0.87 g cm⁻³ (lighter than water; floats on water)

8. Solubility: It is Insoluble in water but Soluble in organic solvents (ether, benzene, alcohol, chloroform)

9. Volatility: It is Volatile and evaporates easily

10. Flammability: It is ighly flammable; burns with a smoky (sooty) flame

Uses:

Solvent
Manufacture of TNT
Paint thinner

(b) Phenol (Hydroxybenzene)

Formula: C₆H₅OH

Uses:

Antiseptics
Plastics
Dyes

(c) Aniline (Aminobenzene)

Formula: C₆H₅NH₂

Uses:

Dyes
Drugs
Rubber chemicals

(d) Nitrobenzene

Formula: C₆H₅NO₂

Uses:

Manufacture of aniline
Dyes
Perfumes

(e) Chlorobenzene

Formula: C₆H₅Cl

Uses:

Solvent
Manufacture of pesticides
Dyes

7. Orientation in Substituted Benzene

Substituents on a benzene ring affect the position of further substitutions.

(a) Ortho–Para Directors

Electron-donating groups direct substitution to the ortho (1,2) and para (1,4) positions.

Examples:

–OH
–NH₂
–CH₃

(b) Meta Directors

Electron-withdrawing groups direct substitution to the meta (1,3) position.

Examples:

–NO₂
–COOH
–SO₃H

8. Uses of Benzene

Manufacture of plastics
Synthetic fibres
Detergents
Dyes
Pharmaceuticals

⚠ Safety Note:
Benzene is toxic and carcinogenic. Prolonged exposure can cause serious health problems.

9. Summary

Benzene is an aromatic hydrocarbon with formula C₆H₆.
Its structure is best described by resonance.
It undergoes electrophilic substitution reactions.
Benzene derivatives are formed by replacing hydrogen atoms with functional groups.
Substituents influence the position of further substitutions.

Tuesday, 20 January 2026

FATS AND OILS – COMPLETE STUDENT NOTE

Introduction

Fats and oils are lipids, a class of organic compounds that are insoluble in water but soluble in organic solvents like alcohol or ether.

They are important macronutrients in food, providing energy, insulation, and cell structure.

Composition of Fats and Oils

Fats and oils are composed of:

Glycerol (C₃H₈O₃) – a 3-carbon alcohol
Fatty acids – long chains of carboxylic acids (R-COOH)

General formula of a triglyceride (fat/oil):

Glycerol+ 3 Fatty acids → Triglyceride + 3H₂O

This is formed through a condensation reaction (esterification) where water is released.

Hence Fats and Oils are triglycerides (esters) of long-chain carboxylic acids (fatty acids and propan-1,2,3-triol (glycerol)

Types of Fats and Oils

1. Saturated Fats

No double bonds between carbon atoms in fatty acid chains
Usually solid at room temperature
Found in: Butter, lard, coconut oil
Example: Stearic acid (C₁₇H₃₅COOH)

2. Unsaturated Fats

Have one or more double bonds in the fatty acid chains
Usually liquid at room temperature (oils)
Found in: Vegetable oil, olive oil, fish oil
Example: Oleic acid (C₁₇H₃₃COOH)

Subtypes:

Monounsaturated – 1 double bond
Polyunsaturated – 2 or more double bonds

Structural Formulae

1. Triglyceride (generalized)

       H   H   H
       |   |   |
HO–CH₂–CH–CH₂–OH (Glycerol backbone)
       |   |   |
       R1  R2  R3 (Fatty acids)

2. Example – Triolein (unsaturated oil)

CH₂–O–CO–C₁₇H₃₃
CH–O–CO–C₁₇H₃₃
CH₂–O–CO–C₁₇H₃₃

3. Example – Tristearin (saturated fat)

CH₂–O–CO–C₁₇H_{35
CH–O–CO–C₁₇H₃₅
CH₂–O–CO–C₁₇H₃₅}

Properties of Fats and Oils

1. they are insoluble in water

2. they are soluble in organic solvents like ether and chloroform

3. they have High calorific value (~9 kcal/g)

4. they are Non-polar molecules

Chemical Properties

1. they reaction with alkali to form soap (saponification reaction)

2. Functions of Fats and Oils

Energy source – They provide twice as much energy as carbohydrates per gram
Insulation – they maintain body temperature
Protection – they cushion internal organs
Cell structure – phospholipids in cell membranes
Vitamin absorption – they help in absorption of fat-soluble vitamins (A, D, E, K)

Examples of Fats and Oils

Type	Example	Source
Saturated	Stearic acid	Animal fat, butter
Saturated	Palmitic acid	Palm oil, meat fat
Unsaturated	Oleic acid	Olive oil, groundnut oil
Unsaturated	Linoleic acid	Sunflower oil, corn oil

Tests for Fats and Oils

Solubility test:
- Insoluble in water, soluble in alcohol
Emulsion test:
- Shake sample with ethanol, then add water → milky emulsion indicates fat
Saponification test:
- Heat fat with NaOH → soap + glycerol

Summary

Fats and oils = triglycerides (glycerol + 3 fatty acids)
Saturated fats → solid, no double bonds
Unsaturated fats → liquid, one or more double bonds
Functions: energy, insulation, protection, vitamin absorption
Can be tested using solubility, emulsion, and saponification tests

Objective Questions

1. Fats and oils are classified as:
A. Carbohydrates
B. Proteins
C. Lipids
D. Nucleic acids

2. The monomer unit of fats and oils is:
A. Glucose
B. Glycerol + fatty acids
C. Amino acid
D. Nucleotide

3. The reaction by which glycerol reacts with fatty acids to form a triglyceride is called:
A. Hydrolysis
B. Condensation / Esterification
C. Oxidation
D. Saponification

4. Which of the following fats is saturated?
A. Olive oil
B. Fish oil
C. Butter
D. Sunflower oil

5. Which of the following is an unsaturated fatty acid?
A. Stearic acid
B. Palmitic acid
C. Oleic acid
D. Butyric acid

6. Fats and oils are:
A. Soluble in water
B. Insoluble in water but soluble in organic solvents
C. Ionic compounds
D. Proteins

7. Which test is used to detect fats and oils?
A. Benedict’s test
B. Biuret test
C. Emulsion test
D. Iodine test

8. A fat that is solid at room temperature is usually:
A. Unsaturated
B. Saturated
C. Polyunsaturated
D. Monounsaturated

9. Which products are formed when fats undergo saponification?
A. Soap + Glycerol
B. Water + Glycerol
C. Soap + Fatty acids
D. Alcohol + Soap

10. The function of fats and oils in the body includes:
A. Energy storage
B. Insulation and protection
C. Vitamin absorption
D. All of the above

Theory Questions

Short Answer

Define fats and oils.
State the difference between saturated and unsaturated fats.
Write the general formula for a triglyceride.
Name two sources of saturated fats and two sources of unsaturated fats.
Explain why fats and oils are insoluble in water.

Structured / Calculation-Oriented

Draw the general structure of a triglyceride.
Describe the formation of a triglyceride from glycerol and three fatty acids.
Explain the emulsion test for fats and oils.
State the products of saponification of fats using NaOH.
Compare the physical state at room temperature of saturated and unsaturated fats.

Higher-Level / Application

Explain the importance of fats and oils in the human diet.
A student observes that a sample of fat is solid at room temperature. Identify the type of fat and give one example.
Why are unsaturated fats generally considered healthier than saturated fats?
Describe how triglycerides can be hydrolyzed in the body to release energy.
Compare the chemical structure of stearic acid and oleic acid, indicating the difference in saturation.

PROTEINS – COMPLETE NOTE

Introduction

Proteins are essential biological molecules made of amino acids linked together by peptide bonds.
They are one of the macronutrients needed for growth, repair, and maintenance of the body.

Proteins play important roles in living organisms, such as

Building and repairing tissues
Acting as enzymes and hormones
Supporting the immune system
Providing energy when necessary

Structure of Proteins

Proteins have four levels of structure:

1. Primary Structure

The sequence of amino acids in a polypeptide chain.
Determines the unique characteristics of the protein.

2. Secondary Structure

Folding of the polypeptide chain into patterns like:
- Alpha-helix (α-helix)
- Beta-pleated sheet (β-sheet)
Stabilized by hydrogen bonds.

3. Tertiary Structure

Three-dimensional shape of the protein.
Determined by interactions among R-groups (side chains) of amino acids.

4. Quaternary Structure

Present in proteins with more than one polypeptide chain.
Example: Hemoglobin (made of 4 polypeptide chains).

Composition of Proteins

Proteins are made of carbon (C), hydrogen (H), oxygen (O), and nitrogen (N).
Some proteins also contain sulfur (S), phosphorus (P), or iron (Fe).

Types of Proteins

Simple Proteins
- Made entirely of amino acids.
- Example: Albumin, Globulin
Conjugated Proteins
- Contain a non-protein group (prosthetic group) in addition to amino acids.
- Example: Glycoprotein (protein + carbohydrate), Hemoglobin (protein + heme)

Functions of Proteins

Function	Example/Explanation
Structural	Collagen in connective tissues, keratin in hair
Enzymatic	Amylase, lipase (catalyze reactions)
Hormonal	Insulin, glucagon (regulate metabolism)
Transport	Hemoglobin (carries oxygen)
Immune	Antibodies (fight infection)
Storage	Ferritin (stores iron)
Energy	Proteins can be broken down for energy if carbs/fats are low

Classification Based on Shape

Fibrous Proteins: Long, strand-like, structural proteins.
- Example: Keratin, Collagen
Globular Proteins: Spherical, functional proteins.
- Example: Enzymes, Hemoglobin, Antibodies

AMINO ACIDS

Amino acids are the building blocks of proteins. Each amino acid has the general formula:

H₂N-CH(R)-COOH

Where R = side chain unique to each amino acid

Examples of Amino Acids

Essential amino acids (cannot be made by the body):
- Lysine, Methionine, Leucine, Valine, etc.
Non-essential amino acids (can be made by the body):
- Glycine, Alanine, Serine, etc.

1. Essential Amino Acids

These cannot be made by the body and must be obtained from the diet.

Amino Acid	3-Letter Code	Source / Notes
Lysine	Lys	Meat, eggs, beans
Methionine	Met	Fish, eggs, nuts
Leucine	Leu	Meat, soy, dairy
Isoleucine	Ile	Meat, eggs, cheese
Valine	Val	Meat, dairy, legumes
Threonine	Thr	Eggs, meat, milk
Phenylalanine	Phe	Dairy, soy, meat
Tryptophan	Trp	Milk, chocolate, turkey
Histidine	His	Meat, poultry, fish

2. Non-Essential Amino Acids

These can be synthesized by the body.

Amino Acid	3-Letter Code	Source / Notes
Glycine	Gly	Gelatin, collagen
Alanine	Ala	Meat, poultry
Serine	Ser	Soy, dairy, eggs
Cysteine	Cys	Poultry, eggs, garlic
Tyrosine	Tyr	Meat, eggs, dairy
Aspartic acid	Asp	Meat, eggs
Glutamic acid	Glu	Meat, cheese, soy
Proline	Pro	Collagen-rich foods
Asparagine	Asn	Dairy, asparagus
Glutamine	Gln	Meat, eggs

3. Special Amino Acids

Selenocysteine (Sec, U) – sometimes called the 21st amino acid; contains selenium.
Pyrrolysine (Pyl, O) – found in some microorganisms.

4. Key Points

Amino acids are linked together by peptide bonds to form proteins.
Each amino acid has:
- Amino group (–NH₂)
- Carboxyl group (–COOH)
- Hydrogen atom (H)
- Side chain (R group)
Essential amino acids are important in diet; non-essential can be made in the body.

Tests for Proteins

Biuret Test
- Add Biuret reagent (NaOH + CuSO₄)
- Purple color indicates presence of protein
Xanthoproteic Test
- Add concentrated nitric acid
- Yellow color indicates aromatic amino acids (like tyrosine, tryptophan)

Summary

Proteins are polymers of amino acids linked by peptide bonds.
They have primary, secondary, tertiary, and quaternary structures.
Proteins serve structural, functional, enzymatic, and regulatory roles.
Proper protein intake is essential for growth, repair, immunity, and metabolism.

Keratin = fibrous, structural protein
Hemoglobin = globular, functional, conjugated protein
✅ Tips for Students:
Always link structure → function when answering protein questions.
Remember fibrous = structural, globular = functional.
Include examples and chemical/structural details for high marks.

Objective Questions

1. Proteins are polymers of:
A. Sugars
B. Amino acids
C. Fatty acids
D. Nucleotides

2. The bond linking amino acids in a protein is called:
A. Glycosidic bond
B. Peptide bond
C. Hydrogen bond
D. Ionic bond

3. Which of the following is an essential amino acid?
A. Glycine
B. Alanine
C. Lysine
D. Serine

4. Which protein is primarily structural in hair and nails?
A. Hemoglobin
B. Albumin
C. Keratin
D. Globulin

5. A protein that contains a non-protein prosthetic group is called:
A. Simple protein
B. Conjugated protein
C. Fibrous protein
D. Globular protein

6. Which of the following is non-essential?
A. Methionine
B. Leucine
C. Glycine
D. Isoleucine

7. The secondary structure of proteins can be:
A. Alpha-helix or beta-pleated sheet
B. Single chain of nucleotides
C. Fatty acid tail
D. Glycosidic linkage

8. Which test is used to detect proteins?
A. Benedict’s test
B. Biuret test
C. Iodine test
D. Molisch test

9. Hemoglobin is an example of:
A. Fibrous protein
B. Conjugated protein
C. Simple protein
D. Enzyme

10. Which element is found in all amino acids but not in carbohydrates or fats?
A. Carbon
B. Nitrogen
C. Hydrogen
D. Oxygen

Theory Questions

Short Answer

Define proteins.
Define amino acids.
Distinguish between essential and non-essential amino acids.
Name four functions of proteins in living organisms.
Define peptide bond and explain how it is formed.

Structured Questions

Draw and label the general structure of an amino acid.
Explain the four levels of protein structure (primary, secondary, tertiary, quaternary).
List three examples of simple proteins and three examples of conjugated proteins.
State the Biuret test procedure for detecting proteins.
Explain the difference between fibrous and globular proteins.

Higher-Level / Application Questions

Explain why amino acids are called the building blocks of proteins.
Hemoglobin contains iron as part of its structure. What type of protein is hemoglobin and why?
A student carries out a Biuret test and gets a violet color. What does this indicate?
Explain the importance of essential amino acids in the human diet.
Compare the structural features of keratin and hemoglobin in terms of protein classification.

STOICHIOMETRY

Stoichiometry is the branch of chemistry that deals with the quantitative relationship between reactants and products in a chemical reaction.

That is, the relationship between the number of moles of reactants and products in a chemical reaction

In simple terms, stoichiometry helps us to calculate:

How much reactant is needed
How much product will be formed during a chemical reaction

Stoichiometry is based on the law of conservation of mass, which states that matter can neither be created nor destroyed in a chemical reaction.

Importance of Stoichiometry

Stoichiometry is used to:

Calculate masses of reactants and products
Determine the amount of substances in reactions
Predict product yield
Find limiting and excess reactants
Design industrial chemical processes

Basic Terms in Stoichiometry

1. Mole

A mole is the amount of substance that contains
6.02 × 10²³ particles (Avogadro’s number).

The mole is also the unit of measurement in chemistry.

2. Molar Mass

The molar mass is the mass of one mole of a substance in grams (g/mol).

Example:
Molar mass of H₂O = 2(1) + 16 = 18 g/mol

3. Chemical Equation

A chemical equation shows the relationship between reactants and products.

Example:

2H₂ + O₂ → 2H₂O

2mols 1mol 2mol

or 2g (2x16)g (2x18)g

or 2g 32g 22.4dm3

This means:
2 moles of hydrogen react with 1 mole of oxygen to form 2 moles of water.

Types of Stoichiometric Calculations

1. Mole-to-Mole Calculations

This involves using the ratio of moles in a balanced equation. That is, mole-mole relationship

Example 1:

How many moles of oxygen are needed to react with 4 moles of hydrogen?

2H₂ + O₂ → 2H₂O

From equation:

2 moles H₂ react with 1 mole O₂
So, 4 moles H₂ will need:

1/2 x 4 = 2 moles of O₂

2. Mass-to-Mole Calculations

Example 2:

What is the number of moles in 44 g of CO₂?

Molar mass of CO₂ = 12 + 2(16) = 44 g/mol

Moles = Mass =

Molar mass

44 = 1mole
44

3. Mass-to-Mass Calculations

Example 3:

What mass of CO₂ is produced when 10 g of CaCO₃ decomposes?

Equation:

CaCO₃ →CaO + CO₂

Step 1: Molar masses
CaCO₃ = 100 g/mol
CO₂ = 44 g/mol

From equation:
100 g CaCO₃ → 44 g CO₂

So,
10 g CaCO₃ →?

10 = 4.4 g of CO₂
100

4. Volume-to-Volume (Gaseous Reactions)

At the same temperature and pressure, equal volumes of gases contain equal number of molecules.

Example 4:

What volume of oxygen is needed to react with 40 cm³ of hydrogen?

2H₂ + O₂ →2H₂O

2 volumes H₂ react with 1 volume O₂
So,
40 cm³ H₂ will need:

1/2x 40 = 20 cm³ of O₂

5. Limiting Reactant Calculations

The limiting reactant is the reactant that is completely used up first and stops the reaction.

Example 5:

If 2 g of hydrogen reacts with 16 g of oxygen, which is limiting?

2H₂ + O₂ →2H₂O

Moles:
H₂ = 2 ÷ 2 = 1 mole
O₂ = 16 ÷ 32 = 0.5 mole

Required ratio:
2H₂ : 1O₂
Actual ratio:
1H₂ : 0.5 O₂ → correct ratio

So, no reactant is in excess — both are completely used up.

Percentage Yield

Not all reactions give maximum product.

Formula:

Percentage Yield = Actual Yield x 100
Theoretical Yield

Example 6:

If theoretical yield = 10 g and actual yield = 8 g

8 = 100 = 80%
10

Summary

Stoichiometry helps chemists:

Predict quantities in reactions
Save materials
Improve industrial efficiency
Avoid wastage

OBJECTIVE QUESTIONS (WAEC/NECO)

1. Stoichiometry deals with the
A. speed of reactions
B. colour of substances
C. quantitative relationship between reactants and products
D. energy changes in reactions

2. The number of particles in one mole of a substance is
A. 3.01 × 10²³
B. 6.02 × 10²³
C. 1.00 × 10²³
D. 12.00 × 10²³

3. The molar mass of CO₂ is
A. 12 g/mol
B. 16 g/mol
C. 28 g/mol
D. 44 g/mol

4. How many moles are present in 18 g of water?
A. 0.5
B. 1
C. 2
D. 18

5. In the equation
2H₂ + O₂ → 2H₂O
the mole ratio of H₂ to O₂ is
A. 1:1
B. 1:2
C. 2:1
D. 2:2

6. What mass of NaCl contains 1 mole of NaCl?
A. 23 g
B. 35.5 g
C. 58.5 g
D. 46 g

7. At the same temperature and pressure, equal volumes of gases contain
A. equal masses
B. equal densities
C. equal number of molecules
D. equal pressures

8. Which of the following is the limiting reactant?
A. The reactant in excess
B. The reactant completely used up
C. The product formed
D. The catalyst

9. The formula for calculating percentage yield is
A. Actual × Theoretical
B. Actual ÷ Theoretical × 100
C. Theoretical ÷ Actual × 100
D. Actual − Theoretical

10. How many moles are in 44 g of CO₂?
A. 0.5
B. 1
C. 2
D. 44

11. What volume of oxygen is required to react with 40 cm³ of hydrogen?
(2H₂ + O₂ → 2H₂O)
A. 10 cm³
B. 20 cm³
C. 30 cm³
D. 40 cm³

12. Which law is the basis of stoichiometry?
A. Law of definite proportion
B. Law of multiple proportions
C. Law of conservation of mass
D. Law of gaseous volumes

13. The molar mass of CaCO₃ is
A. 40
B. 56
C. 84
D. 100

14. How many moles are present in 32 g of O₂?
A. 0.5
B. 1
C. 2
D. 16

15. Which of the following is NOT used in stoichiometric calculations?
A. Balanced equation
B. Molar mass
C. Temperature only
D. Mole ratio

THEORY QUESTIONS (WAEC/NECO)

Short Answer Questions

1. Define stoichiometry.
2. What is a mole?
3. State Avogadro’s number.
4. Define molar mass.
5. What is a limiting reactant?

6. Calculate the number of moles in 22 g of CO₂.

7. What mass of CO₂ is produced when 50 g of CaCO₃ decomposes? according to the equation
CaCO₃ → CaO + CO₂

8. How many moles of oxygen are needed to react completely with 6 moles of hydrogen? given the equation below
2H₂ + O₂ → 2H₂O

9. Explain stoichiometry and state three of its applications.

10. Describe how to calculate the mass of a product formed from a given mass of reactant using a balanced chemical equation.

11. In a reaction, 10 g of calcium carbonate was heated and produced 3.5 g of carbon dioxide.
(a) Calculate the theoretical yield
(b) Calculate the percentage yield
CaCO₃ → CaO + CO₂

12. Explain the term “limiting reactant” and show with an example.

Monday, 19 January 2026

REDOX REACTIONS (Oxidation–Reduction Reactions) at a glance

Meaning of Redox Reaction

A redox reaction is a chemical reaction in which oxidation and reduction occur simultaneously.

The word redox comes from:

RED → Reduction
OX → Oxidation

In every redox reaction:

One substance is oxidized
Another substance is reduced

They always happen together — you cannot have oxidation without reduction.

Definitions of Oxidation

Oxidation can be defined in several ways depending on the concept used:

1. Oxidation in terms of Oxygen

Oxidation is the addition of oxygen to a substance.

Example:

2Mg + O₂ →2MgO

Magnesium is oxidized because it gains oxygen.

2. Oxidation in terms of Hydrogen

Oxidation is the removal of hydrogen from a substance.

Example:

H₂S → S + H₂

Hydrogen sulphide is oxidized because hydrogen is removed.

3. Oxidation in terms of Electrons

Oxidation is the loss of electrons by a substance.

Example:

Zn →Zn²⁺ + 2e^-

Zinc is oxidized because it loses electrons.

4. Oxidation in terms of Oxidation Number

Oxidation is an increase in oxidation number.

Example:
Fe²⁺ → Fe³⁺
Iron is oxidized because its oxidation number increases from +2 to +3.

5. Oxidation in terms of Electrochemistry

Oxidation occurs at the anode in an electrochemical cell.

Definitions of Reduction

Reduction is the opposite of oxidation and can also be defined in different ways:

1. Reduction in terms of Oxygen

Reduction is the removal of oxygen from a substance.

Example:

CuO + H₂ →Cu + H₂O

Copper (II) oxide is reduced because oxygen is removed.

2. Reduction in terms of Hydrogen

Reduction is the addition of hydrogen to a substance.

Example:

N₂ + 3H₂ → 2NH₃

Nitrogen is reduced because it gains hydrogen.

3. Reduction in term of Electrons

Reduction is the gain of electrons by a substance.

Example:

Cu²⁺ + 2e^- →Cu

Copper ions are reduced because they gain electrons.

4. Reduction in terms of Oxidation Number

Reduction is a decrease in oxidation number.

Example:
Cl₂⁰ → Cl⁻^-1
0 -1

Chlorine is reduced because its oxidation number decreases.

5. Reduction in terms of Electrochemistry

Reduction occurs at the cathode in an electrochemical cell.

Oxidizing and Reducing Agents

Oxidizing Agent

An oxidizing agent is a substance that:

Causes oxidation of another substance
Itself gets reduced

Example:
KMnO₄, O₂, Cl₂

Reducing Agent

A reducing agent is a substance that:

Causes reduction of another substance
Itself gets oxidized

Example:
Zn, H₂, CO

Example of a Redox Reaction

Zn + CuSO₄ → ZnSO₄ + Cu

Zinc is oxidized (loses electrons)
Copper (II) ions are reduced (gain electrons)
Zinc is the reducing agent
Copper (II) sulphate is the oxidizing agent

Importance and Examples of Redox Reactions

Redox reactions are very important in daily life and industry, such as:

Respiration in living cells
Rusting of iron
Burning of fuels
Electrolysis
Production of metals
Batteries and cells
Photosynthesis

Summary Table

Oxidation	Reduction
Gain of oxygen	Loss of oxygen
Loss of hydrogen	Gain of hydrogen
Loss of electrons	Gain of electrons
Increase in oxidation number	Decrease in oxidation number
Occurs at anode	Occurs at cathode

Test for oxidizing agents

To common test or reactions that are used to test for an oxidizing agent involves the action on iron (II) chloride and hydrogen sulphide.

a) Reaction with FeCl₂

When an oxidizing agent is added to green iron (II chloride; the iron (II) ions become oxidized to yellow or brown Fe³⁺.

Fe²⁺ → Fe³⁺ + e^–

green yellow/brown

b) Reaction with hydrogen sulphide

When hydrogen sulphide is bubbled through a solution of an oxidizing agent, the sulphide ions S^2– becomes oxidized to elemental sulphur; and this is seen or observed as yellow deposits sulphur, i.e. S^2– → S(s) + 2e^–.

Test for reducing agents

Two commonest reagents that are used to test for a reducing agent are

1 Acidified potassium tetraoxomanganate(VI) (KMnO₄) and acidified potassium heptaoxodichromate(I) (K₂Cr₂O₇).

a) Action of potassium hyptaoxodichromate (VI) (K₂Cr₂O₇)

When acidified potassium heptaoxodichromate (VI) (K₂Cr₂O₇) is added to a sample of a reducing agent, its colour changes from orange to green, due to the reduction of the dichromate (VI) ion (Cr⁶⁺) (orange) to chromium (III) (Cr³⁺) ion green

Cr⁶⁺ + 3e^– → Cr³⁺
Orange green

b) Test using acidified potassium tetraoxomangane(VI) (KMnO₄)

When acidified potassium tetraoxomanganate (VII) to a sample of reducing agent, the purple colour changes to colourless: due to the reduction of the manganate ion from (+7) which is purple to (+2) which is colourless and a more stable oxidation state.

MnO₄^- + 8H⁺ + 5e^– →Mn²⁺+ 4H₂O
purple colourless

Mn⁷⁺ + 5e^–→ Mn²
purple colourless

This reaction is reversible as the purple colour is restored when an oxidizing agent is reintroduced into the mixture.

Mn²⁺ + 5e →Mn⁷⁺
colourless purple

OBJECTIVE QUESTIONS

1. A redox reaction is one in which

A. acids react with bases
B. electrons are transferred
C. salts are formed
D. heat is evolved

2. Oxidation is the process involving
A. gain of electrons
B. loss of electrons
C. gain of neutrons
D. loss of protons

3. Reduction is best defined as
A. loss of hydrogen
B. gain of oxygen
C. gain of electrons
D. loss of oxygen and hydrogen

4. In the reaction
Zn + Cu²⁺ → Zn²⁺ + Cu
the species oxidized is
A. Cu²⁺
B. Zn²⁺
C. Zn
D. Cu

5. Which of the following is a reducing agent?
A. A substance that gains electrons
B. A substance that is reduced
C. A substance that loses electrons
D. A substance that gains oxygen

6. Which of the following statements is correct?
A. Oxidation and reduction occur separately
B. Reduction involves loss of electrons
C. Oxidation involves gain of electrons
D. Oxidation and reduction occur simultaneously

7. In terms of oxygen, oxidation is defined as
A. removal of oxygen
B. addition of oxygen
C. removal of hydrogen
D. addition of hydrogen

8. In the reaction
2Mg + O₂ → 2MgO
Magnesium is
A. reduced
B. oxidized
C. neutralized
D. displaced

9. Which of the following is an oxidizing agent?
A. A substance that is oxidized
B. A substance that loses electrons
C. A substance that gains electrons
D. A substance that donates protons

10. The oxidation number of chlorine in Cl₂ is
A. +1
B. −1
C. 0
D. +2

11. Reduction involves
A. increase in oxidation number
B. decrease in oxidation number
C. no change in oxidation number
D. increase in atomic mass

12. Which reaction is a redox reaction?
A. NaOH + HCl → NaCl + H₂O
B. AgNO₃ + NaCl → AgCl + NaNO₃
C. Zn + H₂SO₄ → ZnSO₄ + H₂
D. CaCO₃ → CaO + CO₂

13. The electrode at which oxidation occurs is called the
A. cathode
B. anode
C. electrolyte
D. salt bridge

14. Which of the following processes is an example of oxidation?
A. Rusting of iron
B. Freezing of water
C. Melting of ice
D. Dissolving sugar

15. A substance that causes reduction and is itself oxidized is called
A. oxidizing agent
B. electrolyte
C. reducing agent
D. catalyst

16. In the reaction
Fe²⁺ → Fe³⁺ + e⁻
iron is
A. reduced
B. oxidized
C. neutralized
D. displaced

17. Which of the following is NOT a redox reaction?
A. Combustion
B. Respiration
C. Electrolysis
D. Neutralization

18. In a redox reaction, electrons are transferred from
A. oxidizing agent to reducing agent
B. reducing agent to oxidizing agent
C. acid to base
D. salt to water

19. Which of the following changes represents reduction?
A. Cu → Cu²⁺
B. Fe²⁺ → Fe³⁺
C. Cl₂ → 2Cl⁻
D. Na → Na⁺

20. The main feature of a redox reaction is
A. formation of precipitate
B. evolution of gas
C. transfer of electrons
D. formation of salt

THEORY QUESTIONS

Short Answer Questions

1. Define oxidation and reduction using the electron concept.

2. What is a redox reaction?

3. State two ways by which oxidation can be defined.

4. State two ways by which reduction can be defined.

5. What is meant by an oxidizing agent?

6. What is meant by a reducing agent?

7. Why must oxidation and reduction always occur together?

8. Identify the oxidizing and reducing agents in the reaction:
Zn + CuSO₄ → ZnSO₄ + Cu

Structured / Calculation-Based Questions

9. For the reaction:
2Fe³⁺ + Sn²⁺ → 2Fe²⁺ + Sn⁴⁺
(a) State the species oxidized
(b) State the species reduced
(c) Identify the oxidizing agent
(d) Identify the reducing agent

10. In the reaction:
Mg + 2HCl → MgCl₂ + H₂
(a) Which substance is oxidized?
(b) Which substance is reduced?
(c) State the role of magnesium.

Essay Questions

11. Define oxidation and reduction. Explain each using:
(a) oxygen
(b) hydrogen
(c) electrons
(d) oxidation numbers.

12. With examples, explain what is meant by oxidizing and reducing agents.

13. Describe an experiment to show that oxidation and reduction occur simultaneously.

14. Explain the importance of redox reactions in everyday life and industry.

15. Using suitable examples, distinguish between oxidation and reduction.

Practical-Oriented Questions

16. Explain how rusting of iron is a redox reaction.

17. Describe the redox processes that occur during electrolysis of molten sodium chloride.

18. Explain how respiration in living cells is a redox process.

19. Balance the following redox equation using the ion-electron method:

MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ (in acidic medium)

20. State three industrial applications of redox reactions.

Sunday, 18 January 2026

ALDEHYDES (R–CHO)

Definition

Aldehydes are organic compounds that contain the formyl functional group (–CHO), in which a carbon atom is double-bonded to oxygen and single-bonded to hydrogen.

General formula:
R–CHO

where R is an alkyl or aryl group.

Nomenclature (Naming of Aldehydes)

IUPAC Naming Rules

Identify the longest carbon chain containing the –CHO group.
The carbon of the –CHO group is always numbered carbon 1.
Replace the –e ending of the parent alkane with –al.

Examples:

Molecular Formula	IUPAC Name	Common Name
HCHO	Methanal	Formaldehyde
CH₃CHO	Ethanal	Acetaldehyde
C₂H₅CHO	Propanal	Propionaldehyde
C₆H₅CHO	Benzaldehyde	Benzaldehyde

Occurrence of Aldehydes

Found in natural substances such as vanilla (vanillin) and cinnamon (cinnamaldehyde).
Present in essential oils, perfumes, and flavorings.
Produced during oxidation of alcohols.

Preparation of Aldehydes

1. Oxidation of Primary Alcohols

Aldehydes are prepared by controlled oxidation of primary alcohols.

R–CH₂OH + [O] → R–CHO + H₂O

Oxidizing agents:

Acidified potassium dichromate (VI) (K₂Cr₂O₇/H₂SO₄)
Acidified potassium permanganate (KMnO₄)

NOTE: Excess oxidation converts aldehydes to carboxylic acids.

2. Dehydrogenation of Alcohols

Passing alcohol vapour over heated copper (300°C).

R–CH₂OH ---{Cu}--> R–CHO + H₂

3. Hydrolysis of Geminal Dihalides

R–CHX₂ + ---{H₂O}--->R–CHO

Physical Properties of Aldehydes

Lower aldehydes are colorless liquids with sharp smells.
They are polar compounds.
Lower members are soluble in water due to hydrogen bonding.
Boiling points are higher than alkanes but lower than alcohols.
Formaldehyde exists as a gas at room temperature.

Chemical Properties of Aldehydes

1. Oxidation (Reducing Nature)

Aldehydes are easily oxidized to carboxylic acids.

R–CHO + [O] → R–COOH

a. Tollens’ Test (Silver Mirror Test)

Aldehydes reduce ammoniacal silver nitrate to silver.
Produces a silver mirror.

✔️ Distinguishes aldehydes from ketones.

b. Fehling’s Solution Test

Aldehydes reduce Fehling’s solution to a brick-red precipitate (Cu₂O).

2. Reduction

Aldehydes are reduced to primary alcohols.

R–CHO + [H] → R–CH₂OH

Reducing agents:

NaBH₄
LiAlH₄
Hydrogen/Nickel catalyst

3. Addition Reactions

Due to the polar C=O bond, aldehydes undergo nucleophilic addition.

Example:
R–CHO + HCN → R–CH(OH)CN

4. Reaction with Sodium Hydrogen Sulphite

Forms crystalline addition compounds (used for purification).

Uses of Aldehydes

1. Formaldehyde (Methanal)

Used as formalin for preserving specimens.
Manufacture of plastics, resins, and disinfectants.
Used in textile and leather industries.

2. Ethanal (Acetaldehyde)

Manufacture of acetic acid.
Used in perfumes and dyes.

3. Benzaldehyde

Used in flavorings and perfumes (almond smell).

Differences Between Aldehydes and Ketones

Aldehydes	Ketones
Have –CHO group	Have –CO– group
Easily oxidized	Not easily oxidized
Give Tollens’ test	Do not
Carbonyl carbon attached to H	No H attached

Important tip

Aldehydes are reducing agents.
Prepared from primary alcohols only.
Distinguished using Tollens’ and Fehling’s tests.
General formula: R–CHO

OBJECTIVE QUESTIONS

The functional group present in aldehydes is
A. –COOH
B. –CO–
C. –CHO
D. –OH
The general formula of aliphatic aldehydes is
A. R–CO–R
B. R–COOH
C. R–CHO
D. R–OH
The IUPAC name of CH₃CHO is
A. Methanal
B. Ethanal
C. Propanal
D. Acetone
Which of the following alcohols on oxidation produces an aldehyde?
A. Tertiary alcohol
B. Secondary alcohol
C. Primary alcohol
D. Phenol
Which reagent gives a silver mirror with aldehydes?
A. Fehling’s solution
B. Benedict’s solution
C. Tollens’ reagent
D. Acidified KMnO₄
Aldehydes can be oxidized to form
A. Ketones
B. Alcohols
C. Carboxylic acids
D. Esters
Which aldehyde exists as a gas at room temperature?
A. Ethanal
B. Methanal
C. Propanal
D. Benzaldehyde
Aldehydes act as reducing agents because they
A. accept hydrogen easily
B. are easily oxidized
C. are acidic
D. have low boiling points
Which of the following does NOT react with aldehydes?
A. Tollens’ reagent
B. Fehling’s solution
C. Sodium hydrogen sulphite
D. Schiff’s reagent
Reduction of an aldehyde produces a
A. secondary alcohol
B. tertiary alcohol
C. primary alcohol
D. carboxylic acid
Which of the following aldehydes has an aromatic ring?
A. Methanal
B. Ethanal
C. Propanal
D. Benzaldehyde
The oxidation of aldehydes to acids involves
A. loss of hydrogen
B. gain of hydrogen
C. loss of oxygen
D. no chemical change
Aldehydes undergo nucleophilic addition mainly due to the
A. non-polar nature of C=O
B. polarity of the carbonyl group
C. presence of –OH group
D. acidic nature of hydrogen
Which of the following can be used to purify aldehydes?
A. Dilute acids
B. Sodium hydroxide
C. Sodium hydrogen sulphite
D. Calcium chloride
The common name of methanal is
A. Acetaldehyde
B. Formaldehyde
C. Propionaldehyde
D. Benzaldehyde

easykemistry

Wednesday, 18 February 2026

VOLUMETRIC ANALYSISINTRODUCTION note

Thursday, 5 February 2026

BENZENE notes for students

1. Introduction

2. Structure of Benzene

(a) Kekulé Structure

(b) Resonance Structure

3. Aromaticity

4. Physical Properties of Benzene

i. Molecular formula: C₆H₆

ii. Molecular mass: 78 g mol⁻¹

iii. Colourless liquid

iv. Sweet smell

v. Melting point: 5.5 °C

vi. Boiling point: 80.1 °C

vii. Insoluble in water

viii. Soluble in organic solvents

x. Highly flammable

5. Chemical Properties of Benzene

(a) Nitration

(b) Halogenation

(c) Sulfonation

(d) Friedel–Crafts Alkylation

(e) Friedel–Crafts Acylation

6. Benzene Derivatives

(a) Toluene (Methylbenzene)

Toluene (Methylbenzene, C₇H₈) – Physical and Chemical Properties

✅ Physical Properties of Toluene

1. Appearance

2. Odour

3. Molecular formula

4. Molar mass

5. Boiling point

6. Melting point

7. Density It has a density0.87 g cm⁻³ (lighter than water; floats on water)

8. Solubility: It is Insoluble in water but Soluble in organic solvents (ether, benzene, alcohol, chloroform)

9. Volatility: It is Volatile and evaporates easily

10. Flammability: It is ighly flammable; burns with a smoky (sooty) flame

(b) Phenol (Hydroxybenzene)

(c) Aniline (Aminobenzene)

(d) Nitrobenzene

(e) Chlorobenzene

7. Orientation in Substituted Benzene

(a) Ortho–Para Directors

(b) Meta Directors

8. Uses of Benzene

9. Summary

Tuesday, 20 January 2026

FATS AND OILS – COMPLETE STUDENT NOTE

Composition of Fats and Oils

Hence Fats and Oils are triglycerides (esters) of long-chain carboxylic acids (fatty acids and propan-1,2,3-triol (glycerol)

Types of Fats and Oils

1. Saturated Fats

2. Unsaturated Fats

Subtypes:

Structural Formulae

Properties of Fats and Oils

1. they are insoluble in water

2. they are soluble in organic solvents like ether and chloroform

3. they have High calorific value (~9 kcal/g)

4. they are Non-polar molecules

Chemical Properties

1. they reaction with alkali to form soap (saponification reaction)

2.

Functions of Fats and Oils

Examples of Fats and Oils

Tests for Fats and Oils

Summary

Objective Questions

Theory Questions

Short Answer

Structured / Calculation-Oriented

Higher-Level / Application

PROTEINS – COMPLETE NOTE

Structure of Proteins

1. Primary Structure

2. Secondary Structure

3. Tertiary Structure

H₂N-CH(R)-COOH

Where R = side chain unique to each amino acid