CHEMICAL FORMULAE AND EQUATIONS IN CHEMISTRY

FIRST TERM E-LEARNING NOTES

SUBJECT: CHEMISTRY CLASS: SS1

SCHEME OF WORK

WEEKLY LESSON NOTE 

WEEK: 5

FORMULAE AND EQUATIONS

CONTENTS:

  1. Chemical formulae
  2. Chemical equations
  3. Empirical and Molecular formulae
  4. Relative Molecular Mass and Percentage composition by mass of an element.

PERIOD 1: CHEMICAL FORMULAE

Chemical formula can be defined as a collection of two or more symbols to represent one molecule of the compound. For example, the formula of tetraoxosulphate(VI) acid is H2SO4. This formula shows that in a molecule of tetraoxosulphate(VI) acid, there are two atoms of hydrogen, one atom of sulphur and four atoms of oxygen.

The table below contain examples of compounds with their formulae

Compounds Formulae
Oxygen molecule O2
Hydrogen molecule H2
Hydrogen chloride HCl
Potassium chloride KCl
Magnesium tetraoxosulphate(VI) MgSO4
Hydrogen sulphide H2S
Bromine Molecule Br2
Ozone O3

EVALUATION

Write the chemical formulae of the following:

(1) Sodium oxide, Calcium oxide, aluminium oxide,

(2) Carbon(IV) oxide, iron(II) oxide, copper(I) oxide

(3) Sodium tetraoxosulphate(VI), Calcium trioxonitrate(V), magnesium trioxocarbonate(IV).

 

PERIOD 2: CHEMICAL EQUATIONS

Chemical reactions are represented in form of equations which show the reactants and products in any given chemical reaction. For example, the reaction of aqueous hydrogen chloride and aqueous sodium hydroxide is represented by the equation:

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

Balancing Chemical Equations

All chemical equations must be balanced in order to comply with the law of conservation of mass. For example, to balance the equation for the reaction involving the combustion of ammonia gas in air, the following steps should be followed:

Step 1: The reactants (LHS) are NH3 and O2, while the products (RHS) are NO and H2O

NH3(g) + O2(g) NO(g) + H2O(g)

Step 2: Place a 4 in front of NH3 and a 6 in front of H2O. There are now 12 hydrogen atoms on both sides of the equation.

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

Step 3: Place a 4 in front of NO to balance with the 4 nitrogen atoms of the LHS of the equation.

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

Step 4: Place a 5 in front of O2. An “atom count” shows that the equation is balanced.

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

An equation must be balanced. A balanced equation contains the same number of atoms of the elements of the compounds on both sides of the equation.

The equation must also show the physical states of the reactants and the products i.e. whether in aqueous solution or gaseous or solid state.

Evaluation

Balance the equations below:

  1. CaCO3(s) +HCl(aq) CaCl2(s) + H2O(l) + CO2(s)
  2. CO2(g) + NaOH(aq) Na2CO3(aq) + H2O(l)

PERIOD 3: EMPIRICAL AND MOLECULAR FORMULAE

Empirical formula is the simplest formula which gives the ratio of the number of different atoms present in a compound. It does not give the exact number of each atom whereas the molecular formula gives the exact number of the atoms present in a molecule of a compound.

The molecular formula of a compound is a whole number multiple of its empirical formula.[mediator_tech]

Examples:

  1. Find the empirical formula of a compound which contains 80% carbon and 20% hydrogen by mass.

CHEMICAL FORMULAE AND EQUATIONS IN CHEMISTRY

Hence, the empirical formula is CH3

  1. Find the empirical formula of a compound which on analysis yields the following as the reacting masses carbon= 2.0, hydrogen=0.34g, Oxygen = 2.67g. From your result find the molecular formula of the compound. If it’s relative molecular mass is 60. (C=12, H=1, O=16)

Solution: CHEMICAL FORMULAE AND EQUATIONS IN CHEMISTRY

(CH2O)X= 60

(12 + (2×1) + 16)x =60

(12+2+16)x=60

(30)x= 60

X = 2

Molecular formula is therefore (CH2O)2

=C2H4O2 or CH3COOH

EVALUATION:

Calculate the empirical formula of 15.8% Al, 28.1% S, 56.1%O

PERIOD 4: RELATIVE MOLECULAR MASS, MOLAR MASS AND PERCENTAGE COMPOSITION

If the formulae of a substance and the relative atomic Masses of each of the elements are known, then it is possible to determine the relative molecular mass of that substance.

The relative molecular mass refers to the number of times a mole is heavier than one- twelfth the mass of one atom of carbon -12. It has no unit.

The relative molecular mass of a compound is the sum of the masses of all the atoms present in one molecule of the compound .e.g.

For NaCl, the relative molecular mass= (23 +35.5) = 58.5

For ethanol = C2H5OH (carbon=12, H=1, O =16)

The relative molecular mass of ethanol

= C2H5OH

(12×2) + (1×5) + (16) + (1)

24 + 5 + 16 + 1 = 46

THE MOLAR MASS

This is the relative molecular mass expressed in grams. E.g. the molar mass of ethanol is 46gmol-1

In 12g of carbon-12, there are 6 × 1023 atoms of carbon. This is one mole of carbon -12.

A mole of any substance is the amount of that substance which contains 6 × 1023 particles of that substance e.g. One mole of ethanol has a mass of 46g and contains 6 × 1023 ethanol molecules.

NOTE: The relative molecular mass has no units but the molar mass of any substance is expressed in grams.

PERCENTAGE COMPOSITION OF A COMPOUND

To calculate the percentage position of ethanol whose molecular formula is C2H5OH, given that the relative atomic masses of carbon, hydrogen and oxygen are 12, 1, and 16 respectively?

First calculate the molar mass of C2H5OH =(12×2) +(1×5) + (16)+ (1) = 46gmol-1

Then determine the masses of C H and O present;

Mass of carbon= 12×2=24g

Mass of hydrogen= 6×1 = 6g

Mass of oxygen = 16× 1 = 16g

Molar mass of C2H5OH= 46g

Therefore, percentage of C= × 100 = 52.17%

Percentage of hydrogen = × 100 = 13.04%

Percentage of oxygen = × 100 = 13.04%

  1. Calculation of the chemical formula from percentage composition by mass.

We can determine the simplest chemical formula of a compound, given its percentage composition e.g. If the formula for anhydrous disodium trioxocarbonate (iv) is not known, if its percentage composition by mass is known then it chemical formula could be calculated.

For example, the percentage composition of the compound was found to be Na=43.40%,C= 11.32% and O = 45.28%. This would mean that in every 100g of the compound, the masses of Na, C and O were 43.40g, 11.32g and 45.28g respectively.

The amount in moles of Na, C and O would be.

, and respectively.

Therefore, the amount in moles of Na = = 1.89.

Amount in mole of C= = 0.94.

Amount in mole of O= =2.83.

Molar ratio of H: C: O is 1.89: 0.94: 2.83.

2: 1 : 3

Na: C: O= Na2C O3

The simplest formula is therefore, Na2CO3

GENERAL EVALUATION

ESSAY QUESTION

  1. Diffentiate between valency and oxidation number.
  2. Determine the empirical formula of an oxide of nitrogen containing 70% oxygen, if the relative molecular mass of the oxide is 92, deduce its molecular formula.
  3. Balance this chemical equation

NaOH + H2SO4 Na2SO4 + H2O

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