1 Principles of chemistry
2 Inorganic chemistry
3 Physical chemistry
4 Organic chemistry
1 Principles of chemistry
(e) Chemical formulae, equations and calculations Students should: 1.25 write word equations and balanced chemical equations (including state symbols): • for reactions studied in this specification • for unfamiliar reactions where suitable information is provided. 1.26 calculate relative formula masses (including relative molecular masses) (Mr) from relative atomic masses (Ar) 1.27 know that the mole (mol) is the unit for the amount of a substance 1.28 understand how to carry out calculations involving amount of substance, relative atomic mass (Ar) and relative formula mass (Mr) 1.29 calculate reacting masses using experimental data and chemical equations 1.30 calculate percentage yield 1.31 understand how the formulae of simple compounds can be obtained experimentally, including metal oxides, water and salts containing water of crystallisation 1.32 know what is meant by the terms empirical formula and molecular formula 1.33 calculate empirical and molecular formulae from experimental data 1.34C understand how to carry out calculations involving amount of substance, volume and concentration (in mol/dm3) of solution 1.35C understand how to carry out calculations involving gas volumes and the molar volume of a gas (24 dm3 and 24 000 cm3 at room temperature and pressure (rtp)) 1.36 practical: know how to determine the formula of a metal oxide by combustion (e.g. magnesium oxide) or by reduction (e.g. copper(II) oxide)
Notes:
Chemical formulae equations and calculations
1.25: Write Word Equations and Balanced Chemical equations (Including State Symbols): For Reactions Studied in this Specification, For Unfamiliar Reactions Where Suitable Information is Provided
WRITING BALANCED CHEMICAL EQUATIONS:
s
When Balancing Equations, there Needs to be the Same Number of Atoms of Each Element on Either Side of the Equation:
- Work Across the Equation from Left to Right, Checking One Element After Another
- If there is a Group of Atoms (Such as Nitrate Group, NO3), Which has Not Changed from One Side to the Other, then Count the Whole Groups, Rather than Counting the Individual Atoms
s
USING STATE SYMBOLS:
s
State Symbols are Written After formulae in Chemical Equations to Show which Physical State Each Substance is in:
SOLID
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LIQUID
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GAS
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AQUEOUS
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(s)
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(l)
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(g)
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(aq)
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s
Example 1:
s
Aluminium (s) + Copper (II) Oxide (s) → Aluminium Oxide (s) + Copper (s)
UNBALANCED SYMBOL EQUATION: Al + CuO → Al2O3 + Cu
2Al + CuO → Al2O3 + Cu
2Al + 3CuO → Al2O3 + Cu
2Al + 3CuO → Al2O3 + 3Cu
The Equation is Now Balanced
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Example 2:
s
Magnesium Oxide (s) + Nitric Acid (aq) → Magnesium Nitrate (aq) + Water (l)
UNBALANCED SYMBOL EQUATION: MgO + HNO3 → Mg(NO3)2 + H2O
MgO + HNO3 → Mg(NO3)2 + H2O
MgO + HNO3 → Mg(NO3)2 + H2O
MgO + 2HNO3 → Mg(NO3)2 + H2O
The Equation is Now Balanced
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1.26: Calculate Relative Formula Masses (Including Relative Molecular Masses) (Mr) from Relative Atomic Masses (Ar)
RELATIVE ATOMIC MASS: Given by the Symbol Ar
s- Calculated from the Mass Number and Relative Abundances of All the Isotopes of a Particular Element
Example for Isotopes:
s
The Table Shows Information about the Isotopes in a sample of Rubidium
Use Information from the Table to Calculate the Relative Atomic Mass of
this Sample of Rubidium. Give your Answer to One Decimal Place
( 72 x 85 ) + ( 28 x 87 )
____________________________ = 85.6
100
Relative Atomic Mass = 85.6
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s
RELATIVE FORMULA MASS: Given by the Symbol Mr
s- To Calculate the Mr of a Substance, You have to Add Up the Relative Atomic Masses of All the Atoms Present in the Formula
Example:
s
SUBSTANCE
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ATOMS PRESENT
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Mr
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HYDROGEN
( H2 )
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2 x H
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( 2 x 1 ) = 2
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WATER ( H2O )
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( 2 x H ) + ( 1 x O )
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( 2 x 1 ) + 16 = 18
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POTASSIUM CARBONATE
( K2CO3 )
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( 2 x K ) + ( 1 x C ) + ( 3 x O )
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( 2 x 39 ) + 12 + ( 3 x 16 ) = 138
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CALCIUM HYDROXIDE
( Ca(OH)2 )
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( 1 x Ca ) + ( 2 x O ) + ( 2 x H )
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40 + ( 2 x 16 ) + ( 2 x 1 ) = 74
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AMMONIUM SULFATE
( (NH4)2SO4 )
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( 2 x N ) + ( 8 x H ) + ( 1 x S ) + ( 4 x O )
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( 2 x 14 ) + ( 8 x 1 ) + 32 +
( 4 x 16 ) = 132
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1.27: Know that the Mole (Mol) is the Unit for the Amount of a Substance
MOLE: Measure of the Amount of a Substance
s- One Mole (1 Mol) is the Amount of a Substance that Contains 6 x 1023 Particles (Atoms, Molecules or Formulae) of the Substance (6 x 1023 is Known as the Avogadro Number
Example:
s
1 Mol of Sodium (Na) contains 6 x 1023 Atoms of Sodium
s
1 Mol of Hydrogen (H2) contains 6 x 1023 Molecules of Hydrogen
s
1 Mol of Sodium Chloride (NaCl) contains 6 x 1023 Formulae of Sodium Chloride
1.28: Understand How to Carry Out Calculations Involving Amount of Substance, Relative Atomic Mass (Ar) and Relative Formula Mass (Mr)
MOLE CALCULATIONS
- CALCULATING MOLES
EQUATION:
s
Amount in Moles = Mass of Substance in Grams ÷ Mr
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Example:
s
SUBSTANCE
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MASS
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Mr
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AMOUNT
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NaOH
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80 g
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40
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( 80 ÷ 40 ) = 2 Mol
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CaCO3
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25 g
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100
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( 25 ÷ 100 ) = 0.25 Mol
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H2SO4
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4.9 g
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98
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( 4.9 ÷ 98 ) = 0.05 Mol
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H2O
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108 g
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18
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( 108 ÷ 18 ) = 6 Mol
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CuSO4.5H2O
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75 g
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250
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( 75 ÷ 250 ) = 0.3 Mol
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- CALCULATING MASS
EQUATION:
s
Mass of Substance (Grams) = Moles x Mr
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Example:
s
SUBSTANCE
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AMOUNT
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Mr
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MASS
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H2O
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0.5 Mol
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18
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( 0.5 x 18 ) = 9 g
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NaCl
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3 Mol
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58.5
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( 3 x 58.5 ) = 175.5 g
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K2CO3
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0.2 Mol
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138
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( 0.2 x 138 ) = 27.6 g
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(NH4)2SO4
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2.5 Mol
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132
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( 2.5 x 132 ) = 330 g
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MgSO4.7H2O
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0.25 Mol
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246
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( 0.25 x 246 ) = 61.5 g
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- CALCULATING RELATIVE FORMULA MASS
EQUATION:
s
Mr = Mass of Substance in Grams ÷ Moles
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Example:
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10 Mol of Carbon Dioxide has a Mass of 440 g. What is the Relative Formula Mass of Carbon Dioxide?
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Relative Formula Mass = Mass ÷ Number of Moles
Relative Formula Mass = 440 ÷ 10 =44
Relative Formula Mass of Carbon Dioxide = 44
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1.29: Calculate Reacting Masses Using Experimental Data and Chemical Equations
REACTING MASSES: Chemical Equations can be Used to Calculate the Moles or Masses of Reactants and Products
s- Use Information from the Question to Find the Mole of a Substance / Reactant
- Identify the Ratio of the Substance and Reactants and Find the Moles of Others
- Apply Mole Calculations to Find Answer
Example 1:
s
Calculate the Mass of Magnesium Oxide that can be made by Completely Burning 6 g of Magnesium in Oxygen
s
Magnesium (s) + Oxygen (g) → Magnesium Oxide (s)
SYMBOL EQUATION: 2Mg + O2 → 2MgO
RELATIVE FORMULA MASS: Magnesium : 24 Magnesium Oxide : 40
STEP 1 - Calculate the Mol of Magnesium Used in Reaction
Moles = Mass ÷ Mr Moles = 6 ÷ 24 = 0.25
STEP 2 - Find the Ratio of Magnesium to Magnesium Oxide using Chemical Equation
Magnesium Magnesium Oxide
Mol 2 2
Ratio 1 1
Mol 0.25 0.25
Mol of Magnesium Oxide = 0.25
STEP 3 - Find the Mass of Magnesium Oxide
Moles of Magnesium Oxide = 0.25
Mass = Moles x Mr Mass = 0.25 x 40 = 10
Mass of Magnesium Oxide Produced = 10 g
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Example 2:
s
Calculate the Mass, in Tonnes, of Aluminium that can be Produced from 51 Tonnes of Aluminium Oxide
s
Aluminium Oxide (s) → Aluminium (s) + Oxygen (g)
SYMBOL EQUATION: Al2O3 → 2Al + 3O2
RELATIVE FORMULA MASS: Aluminium : 27 Oxygen : 16 Aluminium Oxide : 102
1 Tonne = 106 g
STEP 1 - Calculate the Mol of Aluminium Oxide Used
Mass of Aluminium Oxide in Grams = 51 x 106 = 51000000
Moles = Mass ÷ Ar Moles = 51000000 ÷ 102 = 1000000
STEP 2 - Find the Ratio of Aluminium Oxide to Aluminium using Chemical Equation
Aluminium Oxide Aluminium
Mol 1 2
Ratio 1 2
Mol 1000000 2000000
Mol Aluminium = 2000000
STEP 3 - Find the Mass of Aluminium
Moles of Aluminium = 2000000
Mass = Moles x Mr Mass = 2000000 x 27 = 54000000
Mass in Tonnes = 54000000 ÷ 106 = 54 Tonnes
Mass of Aluminium Produced = 54 Tonnes
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1.30: Calculate Percentage Yield
PERCENTAGE YIELD: Calculation of the Percentage of Yield Obtained from the Theoretical Yield
s
EQUATION:
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Percentage Yield = ( Yield Obtained / Theoretical Yield ) x 100
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Example:
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In an Experiment to Displace Copper from Copper Sulfate, 6.5 g of Zinc was Added to an Excess of Copper (II) Sulfate Solution. The Copper was Filtered Off, Washed and Dried. The Mass of Copper Obtained was 4.8 g. Calculate the Percentage Yield of Copper
s
EQUATION OF REACTION: Zn (s) + CuSO4 (aq) → ZnSO4 (aq) + Cu (s)
STEP 1: Calculate the Amount, in Moles of Zinc Reacted
Moles of Zinc = 6.5 ÷ 65 = 0.10 Mol
STEP 2: Calculate the Maximum Amount of Copper that could be Formed
Maximum Moles of Copper = 0.10 Mol
STEP 3: Calculate the Maximum Mass of Copper that could be Formed
Maximum Mass of Copper = ( 0.110 x 64 ) = 6.4 g
STEP 4: Calculate the Percentage of Yield of Copper
Percentage Yield = ( 4.8 ÷ 6.4 ) x 100 = 75%
Percentage Yield of Copper = 75%
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1.31: Understand How the Formulae of Simple Compounds can be Obtained Experimentally, including Metal Oxides, Water and Salts containing Water of Crystallisation
METAL OXIDES
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Diagram Showing the Apparatus Needed to find the Formulae of a Metal Oxide
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METHOD:
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WORKING OUT EMPIRICAL FORMULA / FORMULAE:
Mass of Metal: Subtract Mass of Crucible from Metal and Mass of Empty Crucible
Mass of Oxygen: Subtract Mass of Metal used from the Mass of Magnesium Oxide
STEP 1 - Divide Each of the Two Masses by the Relative Atomic Masses of Elements
STEP 2 - Simplify the Ratio
Metal Oxygen
Mass x y
Mole x / Mr y / Mr
= a = b
Ratio a : b
STEP 3 - Represent the Ratio into the ‘ Metal O ‘ E.g, MgO
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WATER AND SALTS CONTAINING WATER OF CRYSTALLISATION
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Diagram Showing the Apparatus Needed to find the Formulae of Crystals
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METHOD:
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WORKING OUT EMPIRICAL FORMULA / FORMULAE:
Mass of White Anhydrous Salt: Measure Mass of White Anhydrous Salt Remaining
Mass of Water: Subtract Mass of White Anhydrous Salt Remaining from the Mass of Known Hydrated Salt
STEP 1 - Divide Each of the Two Masses by the Relative Atomic Masses of Elements
STEP 2 - Simplify the Ratio of Water to Anhydrous Salt
Anhydrous Salt Water
Mass a b
Mole a / Mr b / Mr
= y = x
Ratio 1 : x
STEP 3 - Represent the Ratio into ‘ Salt.xH2O ’
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1.32: Know What is Meant by the Terms Empirical Formula and Molecular Formula
EMPIRICAL FORMULA: Gives the Simplest Whole Number Ratio of Atoms of Each Element in the Compound
s- Calculated from Knowledge of the Ratio of Masses of Each Element in the Compound
Example:
s
A Compound that Contains 10 g of Hydrogen and 80 g of Oxygen has an Empirical Formula of H2O. This can be Shown by the Following Calculations:
s
Amount of Hydrogen Atoms = Mass in Grams ÷ Ar of Hydrogen = (10 ÷ 1) = 10 Mol
Amount of Oxygen Atoms = Mass in Grams ÷ Ar of Oxygen = (80 ÷ 16) = 5 Mol
The Ratio of Moles of Hydrogen Atoms to Moles of Oxygen Atoms:
Hydrogen Oxygen
Moles 10 : 5
Ratio 2 : 1
Since Equal Numbers of Moles of Atoms contain the Same Number of Atoms, the Ratio of Hydrogen Atoms to Oxygen Atoms is 2: 1
Hence the Empirical Formula is H2O
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MOLECULAR FORMULA: Gives the Exact Numbers of Atoms of Each Element Present in the Formula of the Compound
s- Divide the Relative Formula Mass of the Molecular Formula by the Relative Formula Mass of the Empirical Formula
- Multiply this to Each Number of Elements
RELATIONSHIP BETWEEN EMPIRICAL AND MOLECULAR FORMULA:
s
NAME OF COMPOUND
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EMPIRICAL FORMULA
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MOLECULAR FORMULA
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METHANE
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CH4
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CH4
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ETHANE
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CH3
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C2H6
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ETHENE
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CH2
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C2H4
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BENZENE
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CH1
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C6H6
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Example:
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The Empirical Formula of X is C4H10S1 and the Relative Formula Mass of X is 180
s
What is the Molecular Formula of X?
s
RELATIVE FORMULA MASS: Carbon : 12 Hydrogen : 1 Sulfur : 32
STEP 1 - Calculate Relative Formula Mass of Empirical Formula
( C x 4 ) + ( H x 10 ) + ( S x 1) = ( 12 x 4 ) + ( 1 x 10 ) + ( 32 x 1) = 90
STEP 2 - Divide Relative Formula Mass of X by Relative Formula Mass of Empirical
Formula
180 / 90 = 2
STEP 3 - Multiply Each Number of Elements by 2
( C 4 x 2 ) + ( H 10 x 2 ) + ( S 1 x 2 ) = ( C 8 ) + ( H 20 ) + ( S 2 )
Molecular Formula of X = C8H20S2
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1.33: Calculate Empirical and Molecular Formulae from Experimental Data
METAL OXIDES
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Diagram Showing the Apparatus Needed to find the Formulae of a Metal Oxide
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METHOD:
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WORKING OUT EMPIRICAL FORMULA / FORMULAE:
Mass of Metal: Subtract Mass of Crucible from Metal and Mass of Empty Crucible
Mass of Oxygen: Subtract Mass of Metal used from the Mass of Magnesium Oxide
STEP 1 - Divide Each of the Two Masses by the Relative Atomic Masses of Elements
STEP 2 - Simplify the Ratio
Metal Oxygen
Mass x y
Mole x / Mr y / Mr
= a = b
Ratio a : b
STEP 3 - Represent the Ratio into the ‘ Metal O ‘ E.g, MgO
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WATER AND SALTS CONTAINING WATER OF CRYSTALLISATION
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Diagram Showing the Apparatus Needed to find the Formulae of Crystals
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METHOD:
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WORKING OUT EMPIRICAL FORMULA / FORMULAE:
Mass of White Anhydrous Salt: Measure Mass of White Anhydrous Salt Remaining
Mass of Water: Subtract Mass of White Anhydrous Salt Remaining from the Mass of Known Hydrated Salt
STEP 1 - Divide Each of the Two Masses by the Relative Atomic Masses of Elements
STEP 2 - Simplify the Ratio of Water to Anhydrous Salt
Anhydrous Salt Water
Mass a b
Mole a / Mr b / Mr
= y = x
Ratio 1 : x
STEP 3 - Represent the Ratio into ‘ Salt.xH2O ’
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1.34C: Understand How to Carry Out Calculations involving Amount of Substance, Volume and Concentration (in Mol / dm3 ) of Solution
CALCULATIONS INVOLVING SUBSTANCES
GENERAL EQUATION:
s
- CALCULATING MOLES
EQUATION:
s
Amount of Substance (Mol) = Concentration x Volume of Solution (dm3)
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Example:
s
Calculate the Moles of Solute Dissolved in 2 dm3 of a 0.1 mol / dm3 Solution
Concentration of Solution : 0.1 mol / dm3
Volume of Solution : 2 dm
Moles of Solute = 0.1 x 2 = 0.1
Amount of Solute = 0.2 Mol
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2. CALCULATING CONCENTRATION
EQUATION:
s
Example:
s
25.0 cm3 of 0.050 mol / dm3 Sodium carbonate were Completely Neutralised by 20.00 cm3 of Dilute Hydrochloric Acid. Calculate the Concentration, in Mol / dm3 of the Hydrochloric Acid
s
STEP 1 - Calculate the Amount, in Moles, of Sodium Carbonate Reacted
Amount of Na2CO3 = ( 5.0 x 0.050 ) ÷ 1000 = 0.00125 Mol
STEP 2 - Calculate the Amount, in Moles, of Hydrochloric Acid Reacted
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
1 Mol of Na2CO3 Reacts with 2 Mol of HCl
0.00125 Mol of Na2CO3 Reacts with 0.00250 Mol of HCL
STEP 3 - Calculate the Concentration, in Mol / dm3, of the Hydrochloric Acid
1 dm3 = 1000 cm3
Concentration ( Mol / dm3 ) = 0.00250 ÷ ( 20 ÷ 1000 ) = 0.125
Concentration of Hydrochloric Acid = 0.125 Mol / dm3
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3. CALCULATING VOLUME
EQUATION:
s
Example:
s
Calculate the Volume of Hydrochloric Acid of Concentration 1.0 mol / dm3 that is Required to React Completely with 2.5g of Calcium Carbonate
s
STEP 1 - Calculate the Amount, in Moles, of Calcium Carbonate that Reacts
Mr of CaCO3 is 100
Amount of CaCO3 = ( 2.5 ÷ 100 ) = 0.025 Mol
STEP 2 - Calculate the Moles of Hydrochloric Acid Required
CaCO3 + 2HCl → CaCl2 + H2O + CO2
1 Mol of CaCO3 Requires with 2 Mol of HCl
0.025 Mol of CaCO3 Requires 0.05 Mol of HCL
STEP 3 - Calculate the Volume of HCl Required
Volume = ( Mol of Substance ÷ Concentration )
= 0.05 ÷ 1.0
= 0.05 Mol
Volume of Hydrochloric Acid = 0.05 Mol
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1.35C: Understand How to Carry Out Calculations Involving Gas Volumes and the Molar Volume of a Gas ( 24 dm3 and 24 000 cm3 at Room Temperature and Pressure (rtp))
CALCULATIONS INVOLVING GASES
GENERAL EQUATION:
1. CALCULATING VOLUME
s
EQUATION:
Volume of Gas ( dm3 ) = Amount of Gas ( Mol ) x 24
OR
Volume of Gas ( cm3 ) = Amount of Gas ( Mol ) x 24000
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Example:
s
NAME OF GAS
|
AMOUNT OF GAS
|
VOLUME OF GAS
|
HYDROGEN
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3 Mol
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( 3 x 24 ) = 72 dm3
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CARBON DIOXIDE
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0.25 Mol
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( 0.25 x 24 ) = 6 dm3
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OXYGEN
|
5.4 Mol
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( 5.4 x 24000 ) = 129600 cm3
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AMMONIA
|
0.02 Mol
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( 0.02 x 24 ) = 0.48 dm3
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2. CALCULATING MOLES
EQUATION:
s
Example:
s
NAME OF GAS
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VOLUME OF GAS
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MOLES OF GAS
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METHANE
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225.6 dm3
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( 225.6 ÷ 24 ) = 9.4 mol
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CARBON MONOXIDE
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7.2 dm3
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( 7.2 ÷ 24 ) = 0.3 mol
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SULFUR DIOXIDE
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960 dm3
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( 960 ÷ 24 ) = 40 mol
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OXYGEN
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1200 cm3
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( 1200 ÷ 24000 ) = 0.05 mol
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1.36: Practical: Know How to Determine the Formula of a Metal Oxide by Combustion (E.g. Magnesium Oxide) or by Reduction (E.g. Copper (II) Oxide)
METAL OXIDE: When a Metal Reacts With and Gains Oxygen
s
COMBUSTION OF METAL OXIDES
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Diagram Showing the Apparatus Needed to find the Formulae of a Metal Oxide by Combustion
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METHOD:
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WORKING OUT EMPIRICAL FORMULA / FORMULAE:
Mass of Metal: Subtract Mass of Crucible from Metal and Mass of Empty Crucible
Mass of Oxygen: Subtract Mass of Metal used from the Mass of Magnesium Oxide
STEP 1 - Divide Each of the Two Masses by the Relative Atomic Masses of Elements
STEP 2 - Simplify the Ratio
Metal Oxygen
Mass a b
Mole a / Mr b / Mr
= x = y
Ratio x : y
STEP 3 - Represent the Ratio into the Form ‘ xO ‘ E.g, MgO
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s
REDUCTION OF METAL OXIDES
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Diagram Showing the Apparatus Needed to find the Formulae of a Metal Oxide by Reduction
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METHOD:
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WORKING OUT EMPIRICAL FORMULA / FORMULAE:
Mass of Metal: Measure Mass of the Remaining Metal Powder
Mass of Oxygen: Subtract Mass of the Remaining Metal Powder from the Mass of Metal Oxide
STEP 1 - Divide Each of the Two Masses by the Relative Atomic Masses of Elements
STEP 2 - Simplify the Ratio
Metal Oxygen
Mass a b
Mole a / Mr b / Mr
= x = y
Ratio x : y
STEP 3 - Represent the Ratio into the Form ‘ xO ‘ E.g, MgO
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