CHEMISTRY: 💧Atoms, Molecules & Stoichiometry💧

EDUCATION THROUGH COMMUNICATION

CHEMISTRY: Atoms, Molecules & Stoichiometry

Introduction: Lesson 1

Definitions

• An atomic mass unit (amu) is defined as 1/12 the mass of one atom of carbon-12. It has a value of 1.6606х10^-24g.

• Atom: the simplest particle of an element that cannot be broken down into simpler particles.

• Molecule: a substance formed when combining non-metal elements (to form covalent bonds) for example, Cl2, O2, H2O.
• Stoichiometry: the study of how substances, elements react

Avogadro constant and the Mole Concept

• Avogadro constant: the number of particles in one mole of a substance and the number is equal to 6.02х10^23. 
• Unit = mol^-1

• Mole: the amount of a substance that contains 6.02х10^23 particles (the avogadro number of particles).
• Particles could be in the form of atoms, ions, molecules, electrons, etc.

Calculations

Formulas

LET;

N be the number of particles

n  be the number of moles

L  be the avogadro constant

With these three components we can derive the equation; N = n х L

LET;

M be the molar mass

m be the mass of the substance

n  be the number of moles 

With these three components, we can derive the equation; m = n х M

Worked Examples

1. How many oxide ions are present in 0.5g of Al2O3?

• So n = m/M 
• Therefore n(Al2O3) = 0.5/(27х2)+(16х3) = 0.5/54+48 = 0.5/102   So n(Al2O3) = 0.0049

• Al2O3 has the ions: Al^3+ and O^2-;        the ratio of the molecule is:  2:3. Therefore 2Al^3+ + 3O^2-  →  {so the formula unit is:}  (Al^3+)2 (O^2-)3

• Mole ratio;  2:3:1
• Use the ratio 3:1, therefore; 1→ 0.0049        3 → 0.0049х3 = 0.0147
• Number of O^2- ions = 0.0049 х 0.0147 х 6.02х10^23(number of particles 1mol of a substance) = 4.338х10^19mol^-1

2. Calculate the mass of phosphate ions PO4^3- in 1.505х10^19 of Ca3(PO4)2? 

• (Ca^2+) + (PO4^3-)
• = Ca3(PO4)2       mole ratio = 3:2
• Ca3(PO4)2 : Ca^2+ : PO4^3-         therefore the ratio is: 1:3:2
• n(Ca3(PO4)2) = N/L                       derived from: N = nL
• 1.505х10^19/6.02х10^23 = 0.000025
• n(PO4^3-) = 0.000025 х 2 = 0.00005mol
• PO4  →          P                O4

                    31.0            16х4

                    31.0     +     64     = 95gmol^-1

• m = n х M   =  0.00005mol х 95gmol^-1  =  0.00475g

Try The Following

1. How many Nitrogen atoms are there in 10.0g of nitrogen gas molecules?
2. Calculate the mass of 24х10^23 molecules of ammonia gas (NH3)

Lesson 3

Definitions

• Molecular formula: The formula that shows the actual number of each of the different atoms present in a molecule (eg. for Butanoic Acid: C4H8O2).

• Empirical formula: The formula of a compound which has the simplest whole number ratio of the elements present in one molecule or formula unit of the compound (eg. for Butanoic Acid: C2H4O).  

For example: BUTANOIC ACID (C4H8O2)

Molecular formula	C4H8O2
Empirical formula	C2H4O

The factor used to deduce the empirical formula of C4H8O2, was 2

Calculations

Worked Examples

Calculate the percentage of H (hydrogen) in Ethanol, C2H6O:

Use the Mr and the number of molecules of each element in C2H6O

[Mr: C=12; H=1; O=16]

H => (6x1) = 6                                          6/46 x 100 = 13.04%

C2H6O => [(12x2)+(6x1)+(16x1)] = 46   

 Try The Following

Calculate the percentage of the following in Ethanol, C2H6O:

1. C (carbon)
2. O (oxygen)

Worked Example

(the following question comes from certain important sources)

Calculate the empirical formula of the following compounds from their following compositions by mass:

[Cu 37.1%; Cl 41.7%; H2O 21.2% (total: 100%)]

Remember Mr of H2O = (2x1)+(16) = 18

	Cu	Cl	H2O
% by mass	37.1	41.7	21.2
mass in 100g	37.1	41.7	21.2
Divide by Ar	37.1/63.5 =0.5843	41.7/35.5 =1.1746	21.2/18 =1.1778
Divide by smallest	0.5843/0.5843 =1.000	1.1746/0.5843 =2.010	1.1778/0.5843 =2.015
The ratio	1	2	2

Therefore the empirical formula = CuCl2(H2O)2

(CuCl2(H2O)2 also presented as Cl2CuH4O2, is known as Copper (II) chloride dihydrate)