21/04/23Atomic massAtomic mass

SYMBOL

PROTON NUMBER = number of protons (obviously)

RELATIVE ATOMIC MASS, Ar

(“Mass number”) = number of protons + number of neutrons

21/04/23Relative formula mass, MRelative formula mass, Mrr

The relative formula mass of a compound is the relative atomic masses of all the elements in the compound added together.

E.g. water H2O:

Therefore Mr for water = 16 + (2x1) = 18

Work out Mr for the following compounds:

1) HCl

2) NaOH

3) MgCl2

4) H2SO4

5) K2CO3

H=1, Cl=35 so Mr = 36

Na=23, O=16, H=1 so Mr = 40

Mg=24, Cl=35 so Mr = 24+(2x35) = 94

H=1, S=32, O=16 so Mr = (2x1)+32+(4x16) = 98

K=39, C=12, O=16 so Mr = (2x39)+12+(3x16) = 138

Relative atomic mass of O = 16

Relative atomic mass of H = 1

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More examplesMore examplesCaCO3 40 + 12 + 3x16 100

HNO3 1 + 14 + 3x16

2MgO 2 x (24 + 16) 80

3H2O 3 x ((2x1) + 16)

4NH3

2KMnO4

3C2H5OH

4Ca(OH)2

Moles – The relative formula mass of a substance, in grams, is known as 1 mole of that substance. E.g. 18g of H2O = 1 mole of H2O

21/04/23Calculating percentage Calculating percentage massmass

If you can work out Mr then this bit is easy…

Calculate the percentage mass of magnesium in magnesium oxide, MgO:

Ar for magnesium = 24 Ar for oxygen = 16

Mr for magnesium oxide = 24 + 16 = 40

Therefore percentage mass = 24/40 x 100% = 60%

Percentage mass (%) =

Mass of element Ar

Relative formula mass Mr

x100%

Calculate the percentage mass of the following:

1) Hydrogen in hydrochloric acid, HCl

2) Potassium in potassium chloride, KCl

3) Calcium in calcium chloride, CaCl2

4) Oxygen in water, H2O

21/04/23Empirical formulaeEmpirical formulaeEmpirical formulae is simply a way of showing how many atoms are in a molecule (like a chemical formula). For example, CaO, CaCO3, H20 and KMnO4 are all empirical formulae. Here’s how to work them out:

A classic exam question:

Find the simplest formula of 2.24g of iron reacting with 0.96g of oxygen.

Step 1: Divide both masses by the relative atomic mass:

For iron 2.24/56 = 0.04 For oxygen 0.96/16 = 0.06

Step 2: Write this as a ratio and simplify:

0.04:0.06 is equivalent to 2:3

Step 3: Write the formula:

2 iron atoms for 3 oxygen atoms means the formula is Fe2O3

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Example questionsExample questions

1) Find the empirical formula of magnesium oxide which contains 48g of magnesium and 32g of oxygen.

2) Find the empirical formula of a compound that contains 42g of nitrogen and 9g of hydrogen.

3) Find the empirical formula of a compound containing 20g of calcium, 6g of carbon and 24g of oxygen.

21/04/23Calculating the Calculating the mass of a productmass of a product

E.g. what mass of magnesium oxide is produced when 60g of magnesium is burned in air?

Step 1: READ the equation:

2Mg + O2 2MgO

IGNORE the oxygen in step 2 – the question

doesn’t ask for it

Step 3: LEARN and APPLY the following 3 points:

1) 48g of Mg makes 80g of MgO

2) 1g of Mg makes 80/48 = 1.66g of MgO

3) 60g of Mg makes 1.66 x 60 = 100g of MgO

Step 2: WORK OUT the relative formula masses (Mr):

2Mg = 2 x 24 = 48 2MgO = 2 x (24+16) = 80

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Work out Mr: 2H2O = 2 x ((2x1)+16) = 36 2H2 = 2x2 = 4

1. 36g of water produces 4g of hydrogen

2. So 1g of water produces 4/36 = 0.11g of hydrogen

3. 6g of water will produce (4/36) x 6 = 0.66g of hydrogen

Mr: 2Ca = 2x40 = 80 2CaO = 2 x (40+16) = 112

80g produces 112g so 10g produces (112/80) x 10 = 14g of CaO

Mr: 2Al2O3 = 2x((2x27)+(3x16)) = 204 4Al = 4x27 = 108

204g produces 108g so 100g produces (108/204) x 100 = 52.9g of Al2O3

1) When water is electrolysed it breaks down into hydrogen and oxygen:

2H2O 2H2 + O2

What mass of hydrogen is produced by the electrolysis of 6g of water?

3) What mass of aluminium is produced from 100g of aluminium oxide?

2Al2O3 4Al + 3O2

2) What mass of calcium oxide is produced when 10g of calcium burns?

2Ca + O2 2CaO

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Actual YieldActual Yield

Even though no atoms are ever gained or lost in a chemical reaction, it is not always possible to obtain the calculated amount of product. Because:

• The reaction may not totally finish – it may be reversible

•Some of the product may be lost when it is separated from the reaction mixture – filtered

• Some of the reactants may react in different ways to the expected reaction

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Percentage yieldPercentage yieldThe amount of product obtained is known as the yield. When compared to the maximum theoretical (calculated) amount as a percentage, it is called percentage yield.

Percentage yield (%) =

Actual yield made

Maximum yield possiblex100%

E.g. What mass of aluminium is produced from 100g of aluminium oxide?

2Al2O3 4Al + 3O2Mr: 2Al2O3 = 2x((2x27)+(3x16)) = 204 4Al = 4x27 = 108

204g produces 108g so 100g produces (108/204) x 100 = 52.9g of AlHowever, only 35.6g of Al was actually obtained during the experiment. What is the

percentage yield.Percentage yield (%) =

35.6

52.9x100%

= 67.3%

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Atom economyAtom economy

This is simply a measure of the amount of starting materials that end up as useful products.

It is important for sustainable development and economical reasons that industrial reactions have High Atom Economy.

Atom Economy (%) = Total Mr of useful products

Total Mr of reactants x100%

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Reversible ReactionsReversible ReactionsSome chemical reactions are reversible. In other words, they can go in either direction:

A + B C + D

NH4Cl NH3 + HCl

e.g. Ammonium chloride

Ammonia + hydrogen chloride

When a reversible reaction occurs in a closed system (Where nothing can escape), equilibrium is reached when both reactions occur at exactly the same rate in each direction.The relative amounts of all the reacting substances at equilibrium depend on the conditions of the reaction.

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Making AmmoniaMaking Ammonia

Nitrogen + hydrogen Ammonia N2 + 3H2 2NH3

•High pressure

•450O C

•Iron catalystRecycled H2 and N2

Nitrogen

Hydrogen

Mixture of NH3, H2 and N2. This is cooled causing NH3 to liquefy.

Fritz Haber, 1868-1934

Guten Tag. My name is Fritz Haber and I won the Nobel Prize for chemistry. I am going to tell you

how to use a reversible reaction to produce ammonia, a very important chemical. This is called

the Haber Process.

To produce ammonia from nitrogen and hydrogen you have to use three conditions:

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Uses of AmmoniaUses of Ammonia

Nitrogen monoxide

Hot platinum catalyst

Ammonia gas

Oxygen

Cooled

Water and oxygen

Nitrogen monoxide

Nitric acid

Ammonia + nitric acid Ammonium nitrate NH3 + HNO3 NH4NO3

Ammonia is a very important chemical as it can be used to make plant fertilisers and

nitric acid:

More ammonia can then be used to neutralise the nitric acid to produce AMMONIUM NITRATE (a fertiliser rich in

nitrogen).

The trouble with nitrogen based fertilisers is that they can also create problems – they could contaminate our drinking

water.

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Haber Process SummaryHaber Process Summary

•200 atm pressure

•450O C

•Iron catalyst

Recycled H2 and N2

Nitrogen

Hydrogen

Mixture of NH3, H2 and N2. This is cooled causing NH3 to liquefy.

To compromise all of these factors, these conditions are used to make a reasonable Yield of ammonia, quickly:

A low temperature increases the yield of ammonia but is too slow

A high temperature improves the rate of reaction but decreases the yield too much

A high pressure increases the yield of ammonia but costs a lot of money