# Module 2, Chapter 3

- Created by: Foxswimmer01
- Created on: 18-10-18 18:23

## 3.1 Amount of substance and the mole

**The Mole**

One mole contains 6.02x10^{23 }particles.

Avagadro constant = N_{A }= 6.02x10^{23 }mol^{-1 }

This is the number of particles in each mole of carbon-12.

1 mole of atoms of an element is its relative atomic mass in grams

**Molar Mass**

Molar mass is the mass per mole of a substance. Units are g mol^{-1 }

Amount of substance (n) = mass (m) / molar mass (M)

## 3.2 Determination of Formulae

**Molecular Formulae**

The number of atoms of each element in a molecule.

**Empirical Formulae**

The simplest whole-number ratio of atoms of each element in a compound.

**Relative Molecular Mass**

Compares the mass of a molecule with the mass of an atom of carbon-12.

Relative Formula Mass

Compares the mass of a formula unit with the mass of an atom of carbon-12. Calculated by adding the relative atomic masses of the elements in the empirical formula.

## 3.2 Determination of Formulae

**Finding formulae by experiment**

Empirical formula from mass:

- Convert mass into moles using n = m / M
- Find smallest whole-number ratio, divide by smallest whole number ratio
- Write empirical formula

Determination of a molecular formula:

- Convert % by mass into moles of atoms using n = m / M
- Find smallest whole-number ratio and empirical formula
- Write relative mass of empirical formula
- Find number of units in one molecule
- Write molecular formula

## 3.2 Determination of Formulae

**Hydrated Salts**

Many coloured salts are hydrated – water molecules are part of their crystalline structure.

This water is water of crystallisation.

When the water of crystallisation is removed through heating the compound, anhydrous crystals are left.

## 3.3 Moles and Volume

Volume

Volume is measured in:

· cm^{3} or ml – 1cm^{3} = 1ml

· dm^{3} or litre – 1dm^{3} = 1000cm^{3} = 1000ml = 1l

Moles and solutions

n = c x V

Moles = Concentration x Volume

Or if volume is in cm^{3} divide it by 1000

To convert between mass concentrations, g dm^{-3} , and concentrations in mol dm^{-3}, you need to convert between moles and grams.

## 3.3 Moles and Volume

**Standard solution**

A solution of known concentration. Which is prepared by dissolving an exact mass of the solute in a solvent and making up the solution to an exact volume.

**Moles and gas volumes**

Molar gas volume is the volume per mole of a gas molecule at a stated temperature and pressure.

RTP – Room Temperature Pressure is about 20 degrees Celsius and 101kPa (1 atm).

At RTP, 1 mole of gas molecules has a volume of approximately 24.0 dm^{3}, making the molar gas volume at RTP = 24.0 dm^{3} mol^{-1}.

When the volume is in cm^{3}, divide by 24000.

When the volume is in dm^{3} divide by 24.0.

## 3.3 Moles and Volume

Ideal Gas Equation

**pV = nRT**

- R = ideal gas constant = 8.314 J mol
^{-1}K^{-1} - p = pressure = Pa
- V = volume = m
^{3} - n = amount of gas molecules = mol
- T = temperature = K

Converting between units

- cm
^{3}to m^{3}= x10^{-6} - dm
^{3}to m^{3}= x10^{-3} ^{o}C to K = +273- kPa to Pa = x10
^{3}

## 3.4 Reacting Quantities

Stoichiometry

The ratio of the amount, in moles, of each substance.

Quantities from amounts and equations

1. Work out amount in moles of everything you can.

2. Use the equations to work out the amount of moles of the unknown chemical.

3. Work out the unknown information which is required.

## 3.4 Reacting Quantities

Percentage Yield

The maximum possible amount of product produced is the theoretical yield.

The theoretical yield is unlikely to be achieved because:

· The reaction may not have ‘finished’

· There may be side reactions occurring

· Some of the product may be lost in transferring and purifying

The actual yield is usually lower than the theoretical yield.

Percentage yield = (actual yield / theoretical yield) x 100 %

The limiting reagent – the reactant which is not in excess so will be completely used up first and will cause the reaction to stop.

## 3.4 Reacting Quantities

Atom Economy.

How well atoms have been utilised in a chemical reaction,

Atom economy =

(sum of molar masses of desired products / sum of molar masses of all products) x 100%

Higher atom economies mean that reactions are more efficient and produce less waste.

Atom economy also reflects the sustainability of reactions and processes.

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