## Higher Tier Only

**Avogadro’s Constant**

Specification Point 1.50:

## Recall that one mole of particles of a substance is defined as:

a) the Avogadro constant number of particles (6.02 × 10^{23}atoms, molecules, formulae or ions) of that substance

b) a mass of ‘relative particle mass’ g

**Avogadro´s Constant**

- This is the mass of substance containing the same number of fundamental units as there are atoms in exactly 12.000 g of
^{12}C - The mole is the unit representing the amount of atoms, ions, or molecules.
- One mole is the amount of a substance that contains
**6.02 x 10**particles (Atoms, Molecules or Formulae) of a substance (6.02 x 10^{23}^{23 }is known as the**Avogadro Number).** - For example:
- 1 mole of Sodium (Na) contains 6.02 x 10
^{23}**Atoms**of Sodium - 1 mole of Hydrogen (H
_{2}) contains 6.02 x 10^{23}**Molecules**of Hydrogen - 1 mole of Sodium Chloride (NaC
*l*) contains 6.02 x 10^{23 }**Formula units**of Sodium Chloride

**Relative Mass**

- One mole of any element is equal to the
**relative atomic mass**of that element in grams. - For example, one mole of carbon, that is if you had 6.02 x 10
^{23}atoms of carbon placed on a balance then it would have a mass of 12g. - So one mole of helium atoms would have a mass of 4g, lithium 7g etc.
- For a compound we
**add**up the relative atomic masses. - So one mole of water would have a mass of 2 x 1 + 16 = 18g.
- Hydrogen which has an atomic mass of 1 is therefore equal to 1/12 the mass of a
^{12}C atom. - So one carbon atom has the same mass as 12 hydrogen atoms.

**Mole Calculations**

Specification Point 1.51:

## Calculate the number of:

a) moles of particles of a substance in a given mass of that substance and vice versa

b) particles of a substance in a given number of moles of that substance and vice versa

c) particles of a substance in a given mass of that substance and vice versa

**Moles of Particles in a given Mass**

**Equation:**

**Examples:**

- The mass can be calculated from the given number of moles by rearranging the equation, which becomes:

*Mass = Moles x M*_{r} (or A_{r})

_{r}(or A

_{r})

**Examples:**

**Number of Particles in given Number of Moles**

**Equation:**

*Number of particles = Moles x 6.02 x 10*^{23}

^{23}

**Examples:**

- The moles can be calculated from the given number of particles by rearranging the equation, which becomes:

*Moles = Number of particles *÷* 6.02 x 10*^{23}

^{23}

**Examples:**

**Number of Particles in a given Mass**

**Equation:**

*Number of particles = Moles x 6.02 x 10*^{23}

^{23}

**Example:**

- The mass can be calculated from the given number of particles by calculating the amount of moles from the number of particles and then multiplying the number of moles by Avogadro’s constant.

**Example:**

**Limiting Reagent**

Specification Point 1.52:

## Explain why, in a reaction, the mass of product formed is controlled by the mass of the reactant which is not in excess.

- A chemical reaction stops when one of the reagents is used up.
- The reagent that is used up first is the
**limiting reagent**, as it limits the duration and hence the amount of product that a reaction can produce. - The amount of product is therefore
**directly proportional**to the amount of the limiting reagent added at the beginning of a reaction. - The limiting reagent is the reactant which is
**not present in excess**in a reaction. - In order to determine which reactant is the limiting reagent in a reaction, we have to consider the ratios of each reactant in the balanced equation.
- When performing reacting mass calculations, the limiting reagent is always the number that should be used as it indicates the maximum possible amount of product.

**Example:**

9.2g of sodium is reacted with 8.0g of sulfur to produce sodium sulfide, NaS. Which reactant is in excess and which is the limiting reactant?

Step 1 Calculate the moles of each reactant

Moles = Mass ÷ Ar

Moles Na = 9.2/23 = 0.40

Moles S = 8.0/32 = 0.25

Step 2 Write the balanced equation and determine the molar ratio

2Na + S → Na2S so the molar ratios is 2 : 1

Step 3 Compare the moles. So to react completely 0.40 moles of Na requires 0.20 moles of S and since there are 0.25 moles of S, then S is in excess. Na is therefore the limiting reactant.

**Reaction Stoichiometry**

Specification Point 1.53:

## Deduce the stoichiometry of a reaction from the masses of reactants and products.

- Stoichiometry refers to the numbers in
**front**of reactants and products which must be adjusted to give a balanced equation. - These numbers are called
**coefficients**and from knowledge of the masses of reactants and products, the balanced chemical equation for a given reaction can be found by calculating the coefficients. - First convert the masses of each reactant and product to moles by dividing by the relative atomic/formula masses.
- If the result yields
**uneven numbers**then multiply across to find the**smallest****whole****number**for the coefficient of each species. - Then use the molar ratio to write out the balanced equation.

**Example:**

127g of Cu metal reacts with 32g of oxygen to produce 159g of copper oxide. Work out the balanced equation from the given masses.

**Step 1 **Calculate the moles of each reactant

Moles = Mass ÷ A_{r }

Moles Cu = 127/63.5 = 2

Moles O_{2} = 32/32 = 1

Moles CuO = 159/79.5 = 2

**Step 2 **Write the balanced equation using the moles to indicate the amount of each species

#### 2Cu + O_{2} → 2CuO

#### Edexcel GCSE Chemistry Notes

- 1. Atomic Structure
- 2. The Periodic Table
- 3. Ionic Bonding
- 4. Covalent Bonding
- 5. Types of Substances
- 6. Calculations Involving Masses
- 7. States of Matter and Mixtures
- 8. Methods of Separating & Purifying Substances
- 9. Acids
- 10. Electrolytic Processes
- 11. Obtaining and Using Metals
- 12. Reversible Reactions & Equilibria
- 13. Transition Metals, Alloys & Corrosion
- 14. Quantitative Analysis
- 15. Dynamic Equilibria
- 16. Chemical Cells & Fuel Cells
- 17. Group 1
- 18. Group 7
- 19. Group 0
- 20. Rates of Reaction
- 21. Heat Energy Changes in Chemical Reactions
- 22. Fuels
- 23. Earth & Atmospheric Science
- 24. Qualitative Analysis: Tests for Ions
- 25. Hydrocarbons
- 26. Polymers
- 27. Alcohols & Carboxylic Acids
- 28. Nanoparticles & Bulk & Surface Properties of Matter

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### Author: Morgan

Morgan’s passion for the Periodic Table begun on his 10th birthday when he received his first Chemistry set. After studying the subject at university he went on to become a fully fledged Chemistry teacher, and now works in an international school in Madrid! In his spare time he helps create our fantastic resources to help you ace your exams.

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