# Moles

## Relative molecular mass and relative formula mass

**Relative Molecular Mass and Relative Formula Mass**

**Relative molecular mass**- The weighted mean mass of a molecule compared with 1/12 of the mass of an atom of carbon-12.**Relative formula mass**- The weighted mean mass of a formula unit compared with 1/12 of the mass of an atom carbon-12.- The relative mass of a simple covalent substance (e.g.H
_{2}O or O_{2}), is called its relative molecular mass. - Relative molecular masses add relative formula masses are calculated by adding up relative atomic masses.

**Example**

Relative formula mass of sodium carbonate, Na_{2}CO_{3}

Na 2 x 23.0 = 46.0

C 1 x 12.0 = 12.0

O 3 x 16.0 = 48.0

Total = 106.0

## Empirical Formulae

**Empirical Formula**

**Empirical Formula**- The simplest whole-number ratio of atoms of each element present in a compound.

**Example**

Find the empirical formula of a compound which is found to contain 1.40g of nitrogen and 0.30g of hydrogen

N H

Composition = 1.40 0.30

Divide by r.a.m. = 1.40/14.0 = 0.1 0.3/1.0 = 0.3

Divide by smallest = 0.1/0.1 = 1 0.3/0.1 = 3

Empirical formula = NH_{3}

## Molecular Formulae

**Molecular Formulae**

**Molecular Formula**- The number of atoms of each element in a molecule.- Molecular formula is a multiple of empirical formula

**Example**

Find the molecular formula of the compound whose empirical formula is CH_{2}O and whose relative molecular mass is 60.0

Mass of empirical formula = (1 x 12.0) + (2 x 1.0) + (1 x 16.0) = 30.0

60/30 = 2 so molecular formula = 2 x empirical formula = C_{2}H_{4}O_{2 }

## The Mole

**The Mole**

**A mole -**contains 6.02 x 10^{23}particles.- In Chemistry amounts of substance are measure in moles.
- There are 4 ways of calculating number of moles:
- Moles = number of particles/6.02 x 10
^{23} - Moles = mass/molar mass
- Moles = volume indm
^{3}/24 - Moles = volume incm
^{3}/24000 - Moles = concentration x volume/1000

- Moles = number of particles/6.02 x 10

**Avogadro Constant**

**Avogadro Constant**- Number of particles present in a mole (6.02 x 10^{23}mol^{-1})

**Concentration**

**Concentrated -**containing a large amount of solute per dm^{3}**Dilute**- containing a small amount of solute per dm^{3}

## Reacting Mass Calculations

**Reacting Mass Calculations**

- Step 1 - Find the number of mole of the thing you are told about.
- Step 2 - Use the equation to find out the moles of the thing you are asked about.
- Step 3 - Find the mass of the thing you are asked about.

**Example**

Work out the mass of HCl formed from 6 g of hydrogen.

**H _{2} + Cl_{2} --->**

**2HCl**

Step 1: Moles H_{2} = 6/2.0 = 3 (mass/molar mass)

Step 2: Moles HCl = 3 x 2 (from equation) = 6

Step 3: Mass HCl = 6 x molar mass = 6 x 36.5 = 219g (moles x molar mass)

## Gas Volume Calculations

**Gas Volume Calculations**

- Step 1 - Find the number of moles of the thing you are told about
- Step 2 - Use the equation to find out the moles of the thing you are asked about
- Step 3 - Find the volume of the thing you are asked about.

**Example**

Work out the volume of CO_{2} formed from 3.99 kg of iron (III) oxide

**Fe _{2}O_{3} + 3CO --->**

**2Fe + 3CO**

_{2}Step 1: Moles Fe_{2}O_{3} = 3990/159.6 = 25 (mass/molar mass)

Step 2: Moles CO_{2} = 25 x 3 (from equation) = 75

Step 3: Volume CO_{2} = 75 x 24 = 1800 dm^{3} (moles x 24)

## Titration Calculations

**Titration Calculations**

- Step 1 - Find the number of moles of the thing you know the concentration and volume of.
- Step 2 - Use the equation to find out the moles of the thing you are asked out.
- Step 3 - Find the unknown concentration or molar mass.

**Example**

25 cm^{3} of NaOH needed 21.5 cm^{3} of 0.1 mol dm^{-3} H_{2}SO_{4} for neutralisation. Calculate the concentration of the NaOH solution.

** H _{2}SO_{4} + 2NaOH -->**

**2NaCl + 2H**

_{2}OStep 1: Moles H_{2}SO_{4} = 0.1 x 21.5/1000 = 2.15 x 10^{-3} (conc x vol/1000)

Step 2: Moles NaOH = 2.15 x 10^{-3} x 2 (from equation) = 4.30 x 10^{-3}

Step 3: Conc NaOH = 4.30 x 10^{-3}/(25/1000) = 0.172 mol dm^{-3} (moles/volume in dm^{3})

## Water of Crystallisation

**Water of Crystallisation**

- Hydrated salts contain water of crystallisation as art of their structure.
- The water of crystalisation is shown in the formula, e.g. CuSO
_{4}.5H_{2}O.

**Example**

- Find mass of water by subtracting the two masses.
- Find the moles of compound and water.
- Make moles into whole number.
- The ratio is the number of water.

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