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Caclulations involving masses

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Relative Atomic Mass

Different elements have different numbers of protons, neutrons and

electrons.

The mass of an element comes from the neutrons and protons in its

nucleus (an electron’s mass is negligible).

The relative atomic mass is the largest number in the element’s box

on the periodic table.

It is usually at the top of the box.

It is often written as Ar.

It tells us how heavy an element is compared to other elements.

Because it is a way of comparing elements it doesn’t have any units.

Relative Atomic Mass Questions

Question 1:

Write down the Ar for the following elements:

Osmium Lithium

Sodium Nitrogen

Platinum Helium

Titanium Carbon

Chlorine Hydrogen

Argon Sulphur

Question 2:

Write down the elements with the following Ar values:

Ar Element

1

12

16

23

222

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Relative Molecular Mass

The relative molecular mass can be worked out by adding up

the relative atomic masses of the elements in the molecule.

It can also be written as Mr or relative formula mass

Example 1: Calculating the relative molecular mass of CH4

CH4 contains 1 x C and 4 x H atoms (remember the small number applies

to the H only)

Ar for C = 12

Ar for H = 1

So Mr = (1 x 12) + (4 x1)

= 16

Calculate the Mr of the following substances:

(the first 2 have been done for you)

(a) H2O = 1 + 1 + 16

= 18

(b) CO2 = 12 + 16 + 16

= 44

(c) NaCl =

(d) O2 =

(e) H2 = (f) HCl =

(g) NaOH =

(h) KOH = (i) MgO =

(j) CO =

(k) C2H4 = (l) C2H6 =

(m)CH4 =

(n) C4H10 = (o) C5H12 =

(p) C6H6 =

(q) S8 = (r) Al2O3 =

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Example 2: Calculating the relative molecular mass of Al(OH)3

Al(OH)3 contains 1 x Al, 3 x O and 3 x H (remember the small number

applies to everything inside the bracket)

Ar for Al = 27

Ar for O = 16

Ar for H = 1

So Mr = (1 x 27) + (3 x 16) + (3 x 1)

= 78 .

Now try these examples – the first one has been done for you.

(a) Ca(OH)2 = 40 + (16 + 1)2 = 74

(b) (NH4)2SO4 =

(b) (NH4)2Cr2O7 =

(d) Fe(OH)2 =

(e) Fe(OH)3 =

(f) (NH4)2S203 =

(g) (NH4)3PO4 =

(h) C6H2CH3(NO2)3 =

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Calculating the Empirical Formula

You can use relative atomic masses (Ar) to calculate the empirical formula

of compounds when you know the mass of reactants you used.

Example:

3.2g of sulphur reacts with oxygen to produce 6.4g of sulphur oxide.

What is the formula of the oxide?

Step 1 – Write down the masses of reactants:

S = 3.2g O = 6.4 - 3.2 = 3.2g

Step 2 – Divide Masses by Ar

3.2 = 0.1 3.2 = 0.2

32 16

Step 3 – Divide everything by the smallest number to get the ratio

0.1 = 1 0.2 = 2

0.1 0.1

Ratio S : O = 1:2

Step 4 - Write out the formula

* If you are

given % just use

the % as the

mass in grams.

Answer these questions to calculate the formula of simple compounds:

1. 10.2g of an oxide of aluminium contains 5.4g of aluminium. Calculate

the formula of the oxide (i.e. empirical formula)

Step Process

1 Write down masses of reactants*

2 Divide by Ar

3 Divide by smallest number to get ratio

4 Write down formula

SO2

Use this table as a

prompt until you are

confident with the

method.

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2. A compound contains 29.11% sodium, 40.51% sulphur and 30.38%

oxygen by mass. Calculate its simplest formula (empirical formula)

3. 14.2g of a compound that contains 3 elements (sodium, sulphur and

oxygen). The mass of sodium is 4.6g and sulphur is 3.2g of sulphur.

The rest is oxygen. Calculate the mass of oxygen, and then

calculate the simplest formula of the compound.

4. A compound contains 40.21% of potassium, 26.80% chromium and

32.99% oxygen by mass. Calculate its empirical formula

5. A sample of a compound containing 3 elements (potassium,

chromium, oxygen) weighs 22.62g. This amount contains 6g of

potassium and 8g of chromium and the rest is oxygen. Calculate the

mass of oxygen and then calculate the simplest formula (empirical

formula)

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Calculating the Mass of Products from a Reaction Equation

You can use the reaction equation and information about relative atomic

mass to calculate the mass of products you would expect to get.

Example

N2 + 3H2 → 2NH3

If we have 7g of N2, how much NH3 can we make?

Step 1

Write down the Mr or Ar underneath each chemical and multiply by any

balancing number

N2 + 3H2 → 2NH3

28 3x2 = 6 2x17 = 34

This means that for every 28g of N2 we would expect to get 34g of NH3.

Step 2

Decide what fraction or multiple of the Mr the reacting amount is:

By examination 7 is 28/4

Step 3

Work out how much product you would get

34 / 4 =

Excess means you have more than enough of that reactant so it won’t

affect your calculation

8.5g

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Calculating the Mass of Reactants from a Reaction Equation

The previous method can also be used to work out the mass of reactants

used in a chemical reaction.

Example

The reaction between iron (Fe) and sulphur (S) to make iron (II) sulphide

(FeS).

Fe + S FeS

How much Fe is needed to make 176g of FeS?

Step 1

Write down the Mr or Ar underneath each chemical and multiply by any

balancing number

Fe + S FeS

56 32 88

This means that for every 88g of FeS made, we need 56g of Fe and 32g

of S

Step 2

Decide what fraction or multiple of the Mr the reacting amount is:

By examination: 176 is 2 x 88

Step 3

Work out how much reactant you would need

56 x 2 =

Q. How much S would be needed to make 176g of FeS?

Now try these questions yourself. Remember to start by writing the Mr or Ar under

each formula in the equation, and multiplying by the balancing number.

5.6g of iron react with excess sulphur to make WHAT MASS of iron(II) sulphide?

Fe + S FeS

112g

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3.65g of HCl react with excess sodium hydroxide. What mass of water is formed?

HCl + NaOH NaCl + H2O

8g of hydrogen are exploded in excess oxygen. What mass of water is formed?

2H2 + O2 2H2O

If 3.2 g of oxygen is reacted with excess hydrogen, what mass of water is formed?

2H2 + O2 2H2O

What mass of hydrogen and oxygen are needed to make exactly 3.6g of water (2

answers needed)

2H2 + O2 2H2O

When sulphur (found in coal) burns it forms acid rain. What mass of sulphur dioxide

is formed from 32 tonnes of sulphur and excess oxygen?

S + O2 SO2

When a hydrocarbon burns in excess oxygen, 2 products are formed (water and

carbon dioxide). What mass of the greenhouse gas (CO2) forms when 10g of

heptane (C7H16), found in petrol, are burned in excess O2?

C7H16 + 11O2 7CO2 + 8H2O

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Extension Questions

Try these questions…

123g of sulphur are burned in excess oxygen. What mass of sulphur dioxide forms?

S + O2 SO2

29 tonnes of iron (III) oxide are reacted in the blast furnace with excess carbon

monoxide. What mass of iron is produced?

Fe2O3 + 3CO 2Fe + 3CO2

2.567g of pure sodium hydroxide are reacted with excess dilute sulphuric acid.

What mass of water is formed?

2NaOH + H2SO4 Na2SO4 + 2H2O

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What mass of lead nitrate has to be thermally decomposed to give 50g of lead

oxide?

2Pb(NO3)2 2PbO + 4NO2 + O2

What mass of water is formed when 1kg (1000g) of octane (equivalent to just over

a litre of petrol) is burned in excess oxygen?

2 C8H18 + 25 O2 16 CO2 + 18 H2O

What mass of carbon dioxide is formed on a typical 25 mile car journey? (Assume

the petrol consists of 2kg {2000g} of pure heptane and that half a gallon of it [2.25

litres] reacts with excess oxygen in the air).

C7H16 + 11 O2 7 CO2 + 8 H2O

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Calculating concentration (gdm-3)

Concentration is the amount of solute dissolved in a stated volume of a

solution. The units are “grams per cubic decimetre” (gdm-3). 1dm3 is the

same volume as 1 litre or 1000cm3.

Calculate the concentration, in gdm-3, of the solute in these solutions.

1 A sample of sea water with a volume of 2 dm3 contains 60 g of salt

dissolved in it. What is the concentration of salt in the sea water?

2 A bottle of ginger beer contains 28 g of sugar in a 0.25 dm3 portion.

What is the concentration of sugar in the ginger beer?

3 The laboratory technician made up some limewater for use in

chemistry experiments. She dissolved 15 g of calcium hydroxide in

water to make up 2 l of solution. What is the concentration of the

limewater?

4 Kyle dissolved a cube of sugar with a mass of 3 g in a 100 cm3 cup of

coffee. What is the concentration of sugar in Kyle’s coffee?

5 A 1 l bottle of mineral water contains 28 mg of calcium. What is the

concentration of calcium

in g dm–3?

6 An Olympic-sized swimming pool has a volume of 2.5 × 106 dm3. 10 kg of

chlorine is needed to disinfect the pool. What is the concentration of

chlorine in the water in g dm–3?

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Moles

In chemistry a mole is another way of measuring an amount of a substance.

One mole of any chemical contains 6.02X1023 particles, this is known as

Avogadro’s number.

The unit for mole is “mol”.

The mass of one mole of a substance is the relative atomic mass (Ar) or

relative formula mass (Mr) in grams.

For example, the relative atomic mass of magnesium is 24, so 24g of

magnesium contains 6.02X1023 atoms of magnesium.

Calculating moles from masses

Try the following:

1 A square sheet of aluminium kitchen foil has a mass of 3g. How many moles

of aluminium atoms are there in the sheet?

2 Calculate the number of moles of the named particles in the following

substances.

a 24g of carbon atoms

b 6g of magnesium atoms

c 3.2g of sulfur dioxide molecules (SO2)

d 1kg of octane molecules (C8H18)

3 What is the mass of the following quantities?

a 10 moles of hydrogen molecules (H2)

b 0.1 moles of gold atoms

c 0.5 moles of sodium chloride (NaCl)

d 1.5 moles of calcium carbonate (CaCO3)

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The Avogadro constant is 6.02 × 1023 mol–1.

Calculate the number of moles of:

a water molecules, H2O, in 9 g of water b ethanol molecules, C2H5OH, in 9.2 g

of ethanol.

Calculate the mass of:

a 2.5 mol of potassium iodide, KI b 0.125 mol of calcium sulfate, CaSO4.

Calculate the number of molecules in:

a 0.5 mol of carbon dioxide, CO2 b 2 mol of oxygen, O2.

Calculate the number of moles in:

a 1.505 × 1023 atoms of sodium b 1.806 × 1024 atoms of copper

Calculate the number of molecules in:

a 9 g of hydrogen, H2 b 48 g of oxygen, O2.

Calculate the mass of the following number of molecules:

a 6.02 × 1022 molecules of methane, CH4 b 1.806 × 1025 molecules of ammonia, NH3.

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Further mole calculations

Number of moles of a substance = mass of substance in grams

relative atomic or formula mass

Concentration of solution, mol dm-3 = number of moles of solute

volume of solution in dm3

Relative atomic masses: H = 1, C = 12, O = 16, Na = 23, Cl = 35.5

1 The recommended daily amount of salt, sodium chloride, NaCl, in an adult

diet is 6g. How many moles does this represent?

2 Harry has two sugar cubes in a mug of tea. Each cube weighs 0.4g and he has

four cups of tea each day. (Sugar is sucrose, C12H22O11.)

a How many moles of sugar does Harry consume each day?

b If a mug has a volume of 250cm3, what is concentration of sugar in the

tea in mol dm–3?

3 The human body maintains the amount of glucose in the blood at a

concentration of about

0.004 mol dm–3. An adult male has about 5dm3 of blood.

a How many moles of glucose does an adult male have in his blood?

b What mass of glucose does an adult male have in his blood (the relative

formula mass of glucose is 180)?

4 The concentration of hydrochloric acid (HCl) in the stomach is about 0.01 mol

dm–3. What is the concentration in g dm–3?

5 22.5 cm3 of sodium hydroxide solution reacted with 25.0 cm3 of

0.100mol/dm3 hydrochloric acid.

NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)

a) Calculate the concentration of the sodium hydroxide solution in mol/dm3 .

Give your answer to 3 significant figures

b) Calculate the concentration of the sodium hydroxide solution in g/dm3 .

Give your answer to 3 significant figures.

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Calculating limiting factors

In a chemical reaction, one of the reactants is often added in excess and is

not completely used up in the reaction. The amount of product formed is

determined by the reactant not in excess.

This is called the limiting factor.

Step 1:

How much NH3 is formed from 1.5g of ammonium chloride?

Moles = mass/Mr

Moles of ammonium chloride = 1.5/53.5 = 0.028

2:2 molar ratio, therefore moles of NH3 = 0.028

Mass = moles X Mr = 0.028 X 17 = 0.47g

Step 2:

How much NH3 is formed from 4.00g of calcium hydroxide?

Moles = mass/Mr

Moles of calcium hydroxide = 4.00/74 = 0.54

1:2 molar ratio, therefore moles of NH3 = 0.054 X 2 = 0.11

Mass = moles X Mr = 0.11 X 74 = 8g

Step 3:

Which reactant made the least product? This is your limiting factor

As less product is made from the NH4Cl, this is the limiting factor.

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Try the following:

1. A solution containing 8.0 g of sodium hydroxide was added to a

solution containing 4.38 g hydrogen chloride, HCl. The reaction

taking place is:

HCl + NaOH → NaCl + H2O

Calculate which reactant is the limiting factor.

2. A solution containing 9.8 g of sodium hydroxide was added to a solution

containing 9.8 g sulfuric acid, H2SO4. The reaction taking place is

H2SO4 + 2NaOH → Na2SO4 + 2H2O

Calculate which reactant is the limiting factor

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Working out the stoichiometry of a reaction from masses

Stoichiometry is another name for the molar ratio of an equation

or the balancing.

Al Cl2

Calculate the

number of moles

(Mass/Mr)

10.8

27= 0.4

42.6

71= 0.6

Divide by the

smaller number

0.4

0.4= 1

0.6

0.4= 1.5

Simplest whole

number ratio

1 × 2 = 2 1.5 × 2 = 3

Work out the stoichiometry of the following.

1. 15g of hydrogen gas reacts exactly with 70g of nitrogen to produce

ammonia, NH3. Deuce the balanced equation for the reaction.