SSS 1: PARTICULATE NATURE OF MATTER (II)
Relative Atomic Mass
The Relative Atomic Mass (RAM) of an element is the average mass of its atoms compared with one-twelfth (¹/₁₂) of the mass of one atom of carbon-12.
The carbon-12 isotope (¹²C) is used as the standard for measuring atomic masses.
Relative atomic mass of an element
= average mass of one atom of element
¹/₁₂ × the mass one carbon -12 atom
Relative Moleculer Mass
Relative Molecular Mass (RMM), also called Molecular Weight, is the sum of the relative atomic masses (RAM) of all the atoms present in a molecule. The RMM has no unit (it’s a ratio).
It shows how heavy a molecule is compared to ¹/₁₂ of the mass of one atom of carbon-12 (¹²C).
Examples:
1. Water (H=1, O=16)
H₂O = (2 × H) + (1 × O)
= (2 × 1) + (16)
= 18
RMM of H₂O = 18
2. Carbon dioxide (C= 12, O=16)
CO₂ = (1 × C) + (2 × O)
= (12) + (2 × 16)
= 12 + 32
= 44
RMM of CO₂ = 44
3. Ammonia (N=14, H=1)
NH₃ = (1 × N) + (3 × H)
= (14) + (3 × 1)
= 14 + 3
= 17
RMM of NH₃ = 17
4. Glucose (C=12, H=1, O=16)
C₆H₁₂O₆ = (6 × C) + (12 × H) + (6 × O)
= (6 × 12) + (12 × 1) + (6 × 16)
= 72 + 12 + 96
= 180
RMM of C₆H₁₂O₆ = 180
5. Copper(II) tetraoxosulphate VI pentahydrate (Cu=63.5, S = 32, H=1, O=16)
CuSO₄·5H₂O = (1 × 63.5) + (1 × 32) + (4 ×
16) + 5 × [(2 × 1) + 16]
= 63.5 + 32 + 64 + 90
= 249.5
RMM of CuSO₄·5H₂O = 249.5
Molar Mass
Molar mass is the mass of one mole of a substance (an element or compound).
It tells us the mass (in grams) of
6.022 × 10²³ particles (Avogadro’s number) of that substance.
The S.I Unit of molar Mass is gmol⁻¹
Example:
1. Water (H=1, O=16)
H₂O = (2 × H) + (1 × O)
= (2 × 1) + (16)
= 18 gmol⁻¹
2. Glucose (C=12, H=1, O=16)
C₆H₁₂O₆ = (6 × C) + (12 × H) + (6 × O)
= (6 × 12) + (12 × 1) + (6 × 16)
= 72 + 12 + 96
= 180 gmol⁻¹
Percentage Compositions
Percentage composition is the percentage by mass of each element present in a compound.
It tells us how much (by mass) of each element makes up a compound.
Calculate the percentage Compositions of each element in the following compounds
1. Ammonia
2. Carbon(IV)oxide
3. Magnesium tetraoxosulphate(VI)
4. Sodium trioxocarbonate(IV)
5. Calcium hydroxide
6. Copper(II) tetraoxosulphate (VI) pentahydrate
[Cu=63.5, S = 32, H=1, O=16, N= 14, C= 12, Mg= 24, Na=23, Ca= 40]
1. Ammonia (NH₃)
Molar mass = 14 + (3 × 1)
= 14 + 3
= 17 g mol⁻¹
% of N = (14 /17) × 100
= 82.35%
% of H = [(3 x1)/17] × 100
= [3/17] × 100
= 17.65%
2. Carbon(IV)oxide (CO₂)
Molar mass = 12 + (2 × 16)
= 12 + 32
= 44 g mol⁻¹
% of C = (12 /44) × 100
= 27.27%
% of O =[(2x16) /44] × 100
= [32/44] × 100
= 72.73%
3. Magnesium tetraoxosulphate(VI)
(MgSO₄)
Molar mass = 24 + 32 + (4 × 16)
= 24 + 32 + 64
= 120 g mol⁻¹
% of Mg = (24/120) × 100
= 20%
% of S = (32/120) × 100
= 26.67%
% of O = [(4x16)/120] × 100
= [64/120] × 100
= 53.33%
4. Sodium trioxocarbonate(IV) (Na₂CO₃)
Molar mass = (2 × 23) + 12 + (3 × 16)
= 46 + 12 + 48
= 106 g mol⁻¹
% of Na = (46/106) × 100
= 43.40%
% of C = (12/106) × 100
= 11.32%
% of O = [(3×16)/106] × 100
= [48/106] × 100
= 45.28%
5. Calcium hydroxide [Ca(OH)₂]
Molar mass = 40 + (2 × 16) + (2 × 1)
= 40 + 32 + 2
= 74 g mol⁻¹
% of Ca = (40/74) × 100
= 54.05%
% of O = [(16×2)/74]× 100
= [32/74] × 100
= 43.24%
% of H = [(2×1)/74] × 100
= [2/74] × 100
= 2.70%
6. Calculate the percentage Composition of oxygen in Copper(II) tetraoxosulphate (VI) pentahydrate (CuSO₄·5H₂O)
Molar Mass = (1 × 63.5) + (1 × 32)
+ (4 × 16) + 5 × [(2 × 1) + 16]
= 63.5 + 32 + 64 + 90
= 249.5g mol⁻¹
% of O = [(9×16)/249.5] ×100
= [144/249.5] ×100
= 57.72%
Vapour density and relative Molecular
mass
Vapour Density (V.D.) is defined as the mass of a certain volume of a gas or vapour compared with the mass of an equal volume of hydrogen gas, measured under the same conditions of temperature and pressure.
Mathematically,
Molar mass of a gas = 2 × Vapour density
Example 1:
If the molar mass of carbon dioxide is 44 g mol⁻¹, find its vapour density.
Solution:
Molar mass of a gas = 2 × Vapour density
V.D. = Molar mass/2
= 44/2
= 22
Example 2:
A gas has a vapour density of 16. Find its molar mass.
Solution:
Molar mass of a gas = 2 × Vapour density
Molar mass = 2 × 16
= 32 g mol⁻¹
Empirical Formula
The empirical formula of a compound is the simplest whole-number ratio of the atoms of the different elements present in the compound.
Example 1:
1.60g of a compound contains 0.64g of copper, 0.32g of sulphur and the rest oxygen. Calculate its empirical formula.
[Cu=64, S=32, O= 16]
Solution:
Example 2:
A compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen. Find its empirical formula. [C=12, H=1, O= 16]
Solution:
A compound contains 31.91% potassium, 28.93% chlorine and the rest oxygen. What is the chemical formula of the compound?
[K = 39, Cl = 35.5, O = 16]
Solution:
An inorganic compound contains 16.1% sodium, 4.2% carbon, 16.8% oxygen and 62.9% water of crystallization. What is the empirical formula of the compound?
[Na = 23, C = 12, O = 16, H₂O = 18]
Solution:
The empirical formula of a hydrated salt is CuSO₄·xH₂O and the percentage of water of crystallization is 36%. Find x.
[Cu=64, S=32, O=16, H=1]
Solution:
Relative Molecular Mass of CuSO₄
= 64 + 32 + (4×16)
= 64 + 32 + 64
= 160
Relative Molecular Mass of water (H₂O)
= (2×1) + 16
= 2 + 16
= 18
Percentage of H₂O = 36%
Percentage of CuSO₄ = 100 - 36
= 64%
Molecular Formula
The molecular formula of a compound is the formula which shows the actual number of atoms of each element in a molecule of the compound.
Relationship between empirical and molecular formula
Empirical Formula × n =Molecular Formula
(Empirical Formula) n = Relative molecular mass
Where n = number of moles.
n=1,2,3,4,etc
Example 1:
The empirical formula of a compound is CH₂, and its molecular mass is 28 gmol⁻¹. Find its molecular formula. (C=12, H=1)
Solution:
(Empirical Formula) n = Relative molecular mass
[CH₂]n = 28 gmol⁻¹
[12 + (2×1)]n = 28 gmol⁻¹
[12 + 2]n = 28
[14]n = 28
n = 28/14
n = 2
Molecular formula= [CH₂]n
= [CH₂]₂
= C₂H₄
Example 2:
An organic compound contains 64.8% carbon, 13.5% hydrogen and 21.7% oxygen. If the relative molecular mass is 74, calculate its,
(I) empirical formula
(II) molecular formula
[C= 12, H=1, O=16]
Solution
(II) (Empirical Formula) n = Relative molecular mass
(C₄H₁₀O)ₙ = 74
[(4×12) + (1×10) + 16]ₙ = 74
[48 + 10 + 16]ₙ = 74
74ₙ = 74
n = 74/74
n = 1
Molecular formula= (Empirical Formula) n
= (C₄H₁₀O)1
= C₄H₁₀O
Example 3:
A hydrocarbon contains 88.9% carbon. If the vapour density of the compound is 28. Find its:
(I) empirical formula
(II) molecular formula
[H = 1 C = 12]
Solution:
Hydrocarbon contains Carbon (C) and Hydrogen (H)
Percentage of carbon = 88.9%
Percentage of hydrogen= 100 - 88.9
= 11.1%
(II) Relative molecular mass= 2× Vapour
Density
RMM = 2 × V.D
= 2 × 28
= 56
(Empirical Formula) n = Relative molecular mass
(CH₂)ₙ = 56
[12 + (2×1)]ₙ = 56
[12 + 2]ₙ = 56
14ₙ = 56
n = 56/14
n = 4
Empirical Formula × n =Molecular Formula
Molecular Formula = (CH₂)ₙ
= (CH₂)₄
= C₄H₈
Laws of chemical combination
The laws of chemical combination describe how elements combine to form compounds. These laws are:
I. Law of conservation of matter
II. Law of definite (constant) composition
III. Law of multiple proportion
IV. Law of reciprocal proportion
1. Law of Conservation of Matter.
This law states that matter can neither be created nor destroyed during a chemical reaction. It is also called the law of conservation of mass. The total mass of the reactants equals the total mass of the products.
Example:
When hydrogen burns in oxygen to form water:
2H₂ + O₂ → 2H₂O
RMM= 2(2×1) + (16×2) → 2[(2×1)+16)]
2(2) + 32 → 2[2+16]
4 + 32 → 2[18]
36 → 36
2. Law of Definite (Constant) Proportion
This law states that all pure samples of the same chemical compound always contains the same elements in the same fixed proportion by mass, regardless of its source or method of preparation.
Example:
Water (H₂O) always contains hydrogen and oxygen in the ratio 1:8 by mass, whether it is from rain, a river, or prepared in a lab.
3. Law of Multiple Proportions
States that when two elements A and B combine to form more than one compound, the different masses of element A that combine with a fixed mass of B are in simple whole-number ratios.
Example:
Carbon and oxygen form two compounds:
Carbon monoxide (CO): 12 g of C combines with 16 g of O
Carbon dioxide (CO₂): 12 g of C combines with 32 g of O
The ratio of oxygen masses = 16:32 = 1:2
4. Law of Reciprocal Proportions
States that if two elements A and B combine separately with a third element C, then the ratio in which A and B combine with each other will be the same or a simple multiple of the ratio in which they combine with C
Example:
Hydrogen (H) combines with oxygen (O) to form H₂O → 2g H combines with 16g O
Hydrogen combines with sulfur (S) to form H₂S → 2g H combines with 32g S
Then, O and S combine in the ratio 16:32 = 1:2, which is a simple ratio.
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