# Moles

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• Created by: Sophie
• Created on: 27-01-14 14:35

## 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.H2O or O2), 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, Na2CO3

Na        2 x 23.0             = 46.0

C          1 x 12.0             = 12.0

O         3 x 16.0             = 48.0

Total                              = 106.0

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## 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 =                                       NH3

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## 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 CH2O 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 = C2H4O

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## The Mole

The Mole

• A mole - contains 6.02 x  1023 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 1023
• Moles = mass/molar mass
• Moles = volume indm3/24
• Moles = volume incm3/24000
• Moles = concentration x volume/1000

• Avogadro Constant - Number of particles present in a mole (6.02 x 1023 mol-1)

Concentration

• Concentrated - containing a large amount of solute per dm3
• Dilute - containing a small amount of solute per dm3
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## 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.

H2 + Cl2 ---> 2HCl

Step 1: Moles H2 = 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)

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## 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 CO2 formed from 3.99 kg of iron (III) oxide

Fe2O3 + 3CO ---> 2Fe + 3CO2

Step 1: Moles Fe2O3 = 3990/159.6 = 25                        (mass/molar mass)

Step 2: Moles CO2 = 25 x 3 (from equation) = 75

Step 3: Volume CO2 = 75 x 24 = 1800 dm3                     (moles x 24)

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## 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 cm3 of NaOH needed 21.5 cm3 of 0.1 mol dm-3 H2SO4 for neutralisation. Calculate the concentration of the NaOH solution.

H2SO4 + 2NaOH --> 2NaCl + 2H2O

Step 1: Moles H2SO4 = 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 dm3)

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## 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. CuSO4.5H2O.

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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