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Balancing Equations Reactants Products This means that you HAVE to have the same number of atoms on the left and right hand sides of the equation.

Balancing Equations

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Balancing Equations. Reactants  Products This means that you HAVE to have the same number of atoms on the left and right hand sides of the equation. What do the numbers mean?. Subscript numbers refer only to the single element that the number is immediately after. No number = 1 - PowerPoint PPT Presentation

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Page 1: Balancing Equations

Balancing Equations

Reactants Products

This means that you HAVE to have the same number of atoms on the left and right hand sides of the equation.

Page 2: Balancing Equations

What do the numbers mean?

• Subscript numbers refer only to the single element that the number is immediately after. No number = 1

• For example: CO2 1 x C and 2 x O

• Full size numbers (2,3,4) refer to the whole molecule immediately after the number. No number = 1

• For example: 2H2SO4 2 molecules of sulphuric acid

Page 3: Balancing Equations

Changing numbers

• You can only change BIG numbers.

• If you change subscript numbers then you change the chemical!

Page 4: Balancing Equations

An example:

H2 + O2 H2O

Page 5: Balancing Equations

Try this one:

Na + HCl NaCl + H2

Page 6: Balancing Equations

Percentage yields

• When you are doing a chemical reaction you don’t necessarily get as much product as you think you should because not all of the reactants react 100%, or some gas might escape, or you drop a bit… Lots of reasons.

Page 7: Balancing Equations

Step 1: How much should you have?

You work this out by looking at the formula for the equation:

2Mg + O2 2MgO

This tells us that 2 moles of Mg reacts with 1 mole of oxygen to give 2 moles of MgO

Page 8: Balancing Equations

2Mg + O2 2MgO

So, if 48g of Mg was reacted with unlimited oxygen, how much MgO would you expect?

Ar of Mg is 24. Therefore 48g is 2 moles.

This means we must expect 2 moles of MgO

Mr of MgO = 24+16 = 40

Therefore the expected mass of MgO is 80g

Page 9: Balancing Equations

Percentage yield

You can calculate % yield by:

% yield = (actual mass / expected mass ) x 100

In this case, if we got 60g of MgO:% yield = 60/80 = 75%

Page 10: Balancing Equations

1) C + O2 CO2

How many moles are there in 8.8g of CO2? Therefore, how many moles of C and O2 would I

need to produce this much CO2? What masses of C and O2 required for this

reaction?

2)CaCO3 CaO + CO2

How much calcium oxide is produced by heating 25 g of calcium carbonate?

Page 11: Balancing Equations

What is an empirical formula?

• A formula giving the proportions of the elements present in a compound but not the actual numbers or arrangement of atoms.

Page 12: Balancing Equations

What is the ratio of atoms?Question Ratio

Hydrogen to oxygen atoms in H2O

Copper to Oxygen in CuO

Magnesium to Chlorine in MgCl2

Carbon to Hydrogen in Propene C3H6

Carbon to Hydrogen in acetic acid CH3COOH

Can you simplify any of the ratios?

Page 13: Balancing Equations

What is the ratio of atoms?Question Ratio

Hydrogen to oxygen atoms in H2O 2:1

Copper to Oxygen in CuO 1:1

Magnesium to Chlorine in MgCl2 1:2

Carbon to Hydrogen in Propene C3H6 3:6 can be simplified to 1:2

Carbon to Hydrogen in acetic acid CH3COOH 2:4 can be simplified to 1:2

Page 14: Balancing Equations

• The formula that gives the simplest whole number ratio of atoms of each element present in a compound.

• Ratios are in WHOLE NUMBERS ONLY (to represent atoms)

FOR YOUR GLOSSARY: Empirical Formula

Finished? 1 to 10 in your book for a formula challenge!

Page 15: Balancing Equations

Write down the empirical formula for these molecules?

1. Water – H2O2. Sulfur – S8

3. Hexane – C6H14

4. Benzene – C6H6

5. Carbon dioxide - CO2

6. Oxygen - O2

7. Propene - C3H6

8. Octene - C4H8

9. Nitrogen - N2

10.Calcium Carbonate - CaCO3

YOU HAVE 5 MINUTES

Page 16: Balancing Equations

Write down the empirical formula for these molecules

1. Water – H2O H2O2. Sulfur – S8 S3. Hexane – C6H14 C3H7

4. Benzene – C6H6 CH5. Carbon dioxide – CO2 CO2

6. Oxygen - O2 O7. Propene - C3H6 CH2

8. Octene - C4H8 CH2

9. Nitrogen - N2 N10.Calcium Carbonate - CaCO3 CaCO3

Mark these now!What was your score out of 10?

Page 17: Balancing Equations

Step 1: Divide the mass or % of each element by its RAM (find MOLES of each).

Step 2: Divide both mole amounts by the smaller number.

Step 3: Work out the smallest possible whole number ratio = simplest ratio of atoms.

Step 4: Express as a formula.

Calculating Empirical Formula-MethodSample question: If 22.55% of phosphorus chloride is phosphorous, what

is the empirical formula of the compound? (P=31, Cl=35.5)

Page 18: Balancing Equations

Calculating empirical formula from a percentage composition…

Weights or Weight percentages

22.55% = P

77.45=Cl

Convert into moles(Divide by Relative Atomic Mass Ar)

22.55/31

= 0.73

77.45/35.5

= 2.18Write as a mole ratio(this gives a ratio between elements)

0.73 : 2.18

Convert to whole number ratio(by dividing by smallest number)

/0.73

1:3 Express as formula PCl3

If 22.55% of phosphorus chloride is phosphorous, what is the empirical formula of the compound? (P=31, Cl=35.5)

1.A compound which contains 11.6% nitrogen and 88.4% chlorine.

2. A compound which contains 40% sulphur and 60% oxygen.

3. A compound which contains 37.21% carbon, 7.75% hydrogen and 55.04% chlorine.

Page 19: Balancing Equations

Questions

1. A compound which contains 11.6% nitrogen and 88.4% chlorine.

2. A compound which contains 40% sulphur and 60% oxygen.

Page 20: Balancing Equations

Calculating empirical formula from a percentage composition…

Weights or Weight percentages

11.6 % = N 88.4%=Cl

Convert into moles(Divide by Relative Atomic Mass Ar)

11.6 / 14

= 0.828571

88.4/35.5

= 2.4901Write as a mole ratio(this gives a ratio between elements)

0.8285:2.4901

Convert to whole number ratio(by dividing by smallest number)

1:3

1:3 Express as formula NCl3

A compound which contains 11.6% nitrogen and 88.4% chlorine.

Page 21: Balancing Equations

Calculating empirical formula from a percentage composition…

Weights or Weight percentages

40% 60%

Convert into moles(Divide by Relative Atomic Mass Ar)

40/32

= 1.25

60/16

= 3.75Write as a mole ratio(this gives a ratio between elements)

1.25:3.75

Convert to whole number ratio(by dividing by smallest number)

1:3

Express as formula SO3

A compound which contains 40% sulphur and 60% oxygen? (S= 32, 0=16)

Page 22: Balancing Equations

Gas Chromatography/Mass Spec

Page 23: Balancing Equations
Page 25: Balancing Equations

Instrumental methods are accurate, sensitive and rapid and are particularly useful when the amount of a sample is very small.

Page 26: Balancing Equations

There are pros and cons though!

Advantages Disadvantages

Highly accurate and sensitive Expensive

Relatively quick results Specialist training required

Can analyse very small samples Results can be interpreted only by comparison with known substances

You need to know the pros and cons for your exam!

Page 27: Balancing Equations

Gas Chromatography

• Gas chromatography is often used with mass spectrometry. A sample made of a mixture of chemicals is injected into a gas chromatography machine in order to separate the mixture. The separate samples then go into a mass spectrometer for identification.

• A mass spectrometer is a device for finding the mass of atoms and groups of atoms.

• The mass spectrometer can also give the relative molecular mass of each of the substances separated in the column.

Page 28: Balancing Equations

Gas Chromatography

• Used to separate a mixture of compounds, like paper chromatography, but the sample is vaporised first

• Can detect and identify compounds in tiny amounts of a sample, e.g. can be used to detect the presence of banned drugs in samples from athletes.

Page 29: Balancing Equations

Gas Chromatography- why use it?

• Gas chromatography separates mixtures better than paper and thin-layer.

• Has high separating power.• Can be used to trace the source of pollution incidents

by measuring exact characteristics of oil spill• Very small amounts of sample are needed for gas

chromatography• Can detect and identify compounds in trace amounts

(tiny amounts)• Can be used to detect the presence of banned drugs

in samples from athletes.

Page 30: Balancing Equations

How it works

1. A small amount of sample is injected into one end of the heated column, where it turns to vapour (gas)

Page 31: Balancing Equations

How it works

2: The substance is carried by the carrier gas through the tube

Page 32: Balancing Equations

How it works

3: The carrier gas carries the sample through the stationary phase (the liquid)

Page 33: Balancing Equations

How it works

4: The detector records the arrival of each chemical from the column.

Page 34: Balancing Equations

How it works

5: Each arrival registers as a peak of the chromatogram.

Page 35: Balancing Equations

OH– ion + H+ ions H2O = Water

Alkali + Acid Salt + Water

Sodium + HydrochloricHydroxide Acid

Sodium Chloride + Water

For example

Neutralisation of Acids and Alkalis

Acids are a source of H+ ions.Alkalis are a source of OH– ions.

Na + OH – + H + Cl – – Na + Cl – + H2O

Page 36: Balancing Equations

Acid Alkali Salt

Hydrochloric Acid Potassium Hydroxide

Sulfuric Acid Sodium Hydroxide

Nitric Acid Silver Hydroxide

Barium Hydroxide Barium Sulfate

Hydrobromic Acid Calcium Bromide

Name that Salt

Page 37: Balancing Equations

Acid Alkali Salt

Hydrochloric AcidHCl

Potassium HydroxideKOH

Potassium ChlorideKCl

Sulfuric AcidH2SO4

Sodium HydroxideNaOH

Sodium SulfateNa2SO4

Nitric AcidHNO3

Silver HydroxideAgOH

Silver NitrateAgNO3

Sulfuric AcidH2SO4

Barium HydroxideBa(OH)2

Barium SulfateBaSO4

Hydrobromic AcidHBr

Calcium HydroxideCa(OH)2

Calcium BromideCaBr2

Name that Salt answers

Page 38: Balancing Equations

Making and isolating salts

Neutralisation is a very clean process…

We take an acid and alkali…

And just get a salt and water.

Let the water evaporate and we’ll just be left with the crystallised salt…

…as long as we start with just the right amount of acid and alkali so neither is left over. An indicator could help – we’ll know when neutral.

Page 39: Balancing Equations

Ammonium + NitricHydroxide Acid

Ammonium + Water Nitrate

Making and isolating ammonium nitrate

NH4OH + HNO3 NH4NO3 + H2O

Ammonium nitrate is a very important fertiliser. It is highly soluble in water and easily taken up from the soil by plant roots. It is an essential source of nitrogen for plants to be able to grow.

We can cheat this time – ammonia is volatile so we can use an excess and let it evaporate off with the water.

Ammonia dissolves in water to form alkaline ammonium hydroxide

NH3 + H2O NH4+OH--

Page 40: Balancing Equations

Making insoluble salts

Insoluble salts don’t dissolve in water.

When formed they precipitate out of solution…

This process is called precipitation.

Lead iodide, barium sulfate and silver chloride are all examples of insoluble salts.

PbI2

BaSO4

AgCl

Page 41: Balancing Equations

Removing ions from solution.

Filtration is a very easy, cheap and reliable method of separating solids from liquids.

To remove an ion from solution we just need to form an insoluble salt of it.

The precipitate can be then filtered off.

For example, silver ions can be precipitated when combined with chloride ions.

Page 42: Balancing Equations

Industry and the environment.Chemical Industry uses a lot of metal ions.These could harm wildlife if the effluent (waste water) is discharged into rivers untreated.Precipitation is used to remove toxic metal ions before they enter the environment.Raising the pH of the water causes insoluble metal hydroxides to precipitate out.Very heavy fines are imposed on any companies that release effluent that exceeds legal limits.

Page 43: Balancing Equations