Stoichiometry and the Mole Concept Notes

Stoichiometry and The Mole Concept

Symbols and Formulae

A symbol is used to represent one atom of an element. The symbol can be a two letter, the first of which must be a capital. Examples are H for hydrogen and Li for Lithium.

An integer in front of a symbol indicates the number of that atom. Examples are 3H represents three atoms of hydrogen, 5Cu represents five atoms of copper.

The formula of a substance shows the relative numbers of atoms or ions that have combined together. Examples are as follow:

Substance is an element

e.g

N2

The formula represents one molecule of nitrogen

One molecule of nitrogen contains two nitrogen atoms

2N2 represent two molecules of nitrogen. Two molecules of nitrogen contain a total of four N atoms

Substance is a covalent compound

e.g

H2

O

The formula represents one molecule of water

One molecule of water contains two hydrogen atoms and one oxygen atom

2H2O represent two molecules of water that contain a total of four H atoms and two O atoms

Substance is an ionic compound

e.g

Na2O

Na2O represents one unit of sodium oxide

One unit of sodium oxide contains two Na+ ions and one O2- ion

2Na2O represents two units of sodium oxide that contain four Na+

ions and two O2- ions

Writing Formulae of Simple Compounds

The valency of an element is needed to be able to write the formulae of a compound.

The valency of an element is a number which shows its combining power and in ionic compounds, the valency of an ion is equal to the charge

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POSITIVE ION VALENCY NEGATIVE ION VALENCYH+ Hydrogen 1 F- Fluoride 1Na+ Sodium 1 Cl- Chloride 1K+ Potassium 1 Br- Bromide 1

Ba2+ Barium 2 I- Iodide 1Ca2+ Calcium 2 S2- Sulphide 2Mg2+ Magnesium 2 O2- Oxide 2Al3+ Aluminium 3 OH- Hydroxide 1Ag+ Silver (I) 1 NO3

- Nitrate 1Cu2+ Copper (II) 2 NO2

- Nitrite 1Fe2+ Iron (II) 2 HCO3

- Hydrogen carbonate

1

Pb2+ Lead (II) 2 CH3COO-

Ethanoate 1

Zn2+ Zinc 2 SO32- Sulphite 2

Ni2+ Nickel (II) 2 SO42- Sulphate 2

Fe3+ Iron (III) 3 CO32- Carbonate 2

Cr3+ Chromium (III)

3 PO43- Phosphate 3

In writing formula of ionic compounds note that:

Ionic compounds are made up of positive and negative ions In an ionic compound, the total positive charges must equal the

total negative charges. In other words, the total charges in an ionic compound must equal zero

Example 1Method A Method B

Sodium chloride

Formula of ions

Na+ Cl-

Valency of ions

1 1

Simplest ratio of combining

ions

1 1

Formula NaClMagnesium

oxideFormula of

ionsMg2+ O2-

Valency of ions

2 2


ions

1 1

Formula MgOCalcium nitrate

Formula of ions

Ca2+ NO3-

Valency of ions

2 1

Simplest ratio 1 2

3

of combining ions

Formula Ca(NO3)2

Sodium sulphate

Formula of ions

Na+ SO42-

Valency of ions

1 2


ions

2 1

Formula Na2SO4

Iron (II) phosphate

Formula of ions

Fe2+ PO43-

Valency of ions

2 3


ions

3 2

Formula Fe3(PO4)2

Worked Example:

1. Iron (III) oxide is a chemical compound found in an iron ore called haematite

(a) Give the symbol of all the elements in iron (III) oxide

(b) Give the formula of all the ions in iron (III) oxide

(c) What is the formula of iron (III) oxide?

In writing the formula of covalent compounds, the valency of an element is the number of covalent bonds which it can form with hydrogen atom.

Relative Molecular Mass (Mr)

By definition, Mr of a substance is defined as the average mass of a molecule of a compound, compared with the mass of a 12

6C atom which is taken as 12 units.

To obtain the relative molecular mass, we add together the individual relative atomic masses. These masses can be found from the Periodic Table.

For example;

1. Relative molecular mass of sulphuric acid

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The formula of sulphuric acid = H2SO4

Sulphuric acid contains:2 Hydrogens 2 x Ar of H = 2 x 1 = 21 Sulphur 1 x Ar of S = 1 x 32 = 324 Oxygens 4 x Ar of O = 4 x 16 = 64

Therefore Mr = 2 + 32 + 64 = 98

2. Relative molecular mass of hydrated sodium carbonate

The formula of hydrated sodium carbonate = Na2CO3 . 10H2O

Hydrated sodium carbonate contains:2 Sodiums 2 x Ar of Na = 2 x 23 = 461 Carbons 1 x Ar of C = 1 x 12 = 123 Oxygens 3 x Ar of O = 3 x 16 = 4810 Water molecules

10 x (Mr of H2O)10 x (2 x Ar of H + 1 x Ar of O)10 x (2 x 1 + 1 x 16)10 x (18) = 180Therefore Mr = 46 + 12 + 48 + 180 = 286

Test Your Understanding

Using the Ar found in your periodic table, find the Mr of the following substances:Calcium hydroxide Ca(OH)2

Sodium oxide Na2O

Ammonium sulphate

Calcium hydrogen carbonate

Ca(HCO3)2

Nitrogen trichloride

Hydrogen sulphide

Sodium peroxide Na2O2

Calcium hydrogen sulphate

Ca(HSO4)2

The Percentage Composition of Elements in a Molecule

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If we know the chemical formula of a substance, we can calculate its percentage composition by mass, without doing as experiment. The rules are:

1. Write down the chemical formula of the substance2. Find out its Mr

3. Express the total mass of each element as a percentage of the total relative molecular mass

For example;

1. Percentage composition of sulphuric acidChemical formula H2SO4

Relative molecular mass

98

Percentage composition % of hydrogen =

% of sulphur =

% of oxygen =

6

2. Percentage composition of hydrated sodium carbonate crystalsChemical formula Na2CO3 . 10H2ORelative molecular mass

286

Percentage composition % of sodium =

% of carbon =

% of oxygen =

% of water =

The percentage composition of pure substances does not change. For example water molecule will always contain 11.11% hydrogen and 88.89% oxygen whether its in New York, New Delhi, New Zealand or the Moon (it never varies)


What is the percentage composition of the elements, carbon and hydrogen, in the hydrocarbon molecules listed below?

a) Methane (CH4)

b) Ethane (C2H6)

c) Propane (C3H8)

d) Butane (C4H10)

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The Emperical Formula: The Formula from the Percentage Composition

By definition, the empirical formula is the simplest formula which shows the relative numbers of the atoms of the different elements present.

The empirical formula can be determined once the percentage or mass of each element is known. The rules are:

1. Divide the percentage or mass of each element by its Ar

2. Divide by the smallest number to convert to the simplest ratio3. The number of atoms of the different elements is the empirical

formula

For example:Given that a molecule contain 88.89% oxygen and 11.11% hydrogen, what is its empirical formula?

H ODivide % by Ar 11.11% 1 = 11.11 88.89% 16 = 5.55Simplest ratio 11.11 5.55 = 2 5.55 5.55 = 1Empirical formula H2 O


Calculate the empirical formula of the hydrocarbon molecules listed below using the percentage composition given

a) 75% C, 25% H

b) 80% C, 20% H

c) 81.8% C, 18.2% H

d) 82.7% C, 17.3% H

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THE MOLE

What is the mole?

It is a term in chemistry to describe an amount of atoms, ions and molecules. It enables chemists to count these particles by weighing.

Definition of the mole:The amount of substance which contains the Avogadro Number of particles. The Avogadro Number is 6.02 X 1023 (or 602 000 000 000 000 000 000 000)

Definition of Avogadro Number:The number of atoms in 12g of the carbon-12 isotope

Substances Unit NumberDrawing Pins

Gross 144

Football boots

Pair 2

Eggs

Dozen 12

Playing cards

Pack 52

Paper

Ream 480 sheets

CARBON (12g)

MOLE 6.02 x 1023 PARTICLES

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Moles of Atoms: Relative Atomic Mass in Grams

If we weigh 12 g of carbon, this would contain 6.02 x 1023 atoms of carbon. In the same way, so would 23 g of sodium, or 27 g of aluminium, or 16g of oxygen, or 1 g of hydrogen.

The mass of one mole of atoms is its relative atomic mass in grams

Element Relative atomic mass

Mass of 1 mole Number of atoms in 1 mole

Hydrogen 1 1 g 6.02 x 1023

Carbon 12 12 g 6.02 x 1023

Oxygen 16 16 g 6.02 x 1023

Aluminium 27 27 g 6.02 x 1023

Calcium 40 40 g 6.02 x 1023

Silver 108 108 g 6.02 x 1023

Moles of Molecules: Relative Molecular Mass in Grams

If we weigh 44 g of carbon dioxide, this would contain 6.02 x 1023

molecules of carbon dioxide. In the same way, so would 18 g of water, or 32 g of oxygen gas, or 2 g of hydrogen gas.

The mass of one mole of molecules is its relative molecular mass in grams

Molecule Relative Molecular

mass

Mass of 1 mole Number of atoms in 1 mole

Carbon dioxide (CO2)

44 44 g 6.02 x 1023

Water (H2O) 18 18 g 6.02 x 1023

Oxygen (O2) 32 32 g 6.02 x 1023

Hydrogen (H2) 2 2 g 6.02 x 1023

Moles of Gases: Molar Gas Volumes

Gases are made up of particles and are usually molecules except for the inert gases (Group 8) which composed of atoms.

Accordingly, as one mole contain the same number of particles, one mole of different gases would also occupy identical volume (under the same temperature and pressures).

One mole of any gas at room temperature and pressure occupies a volume of 24000 cm3 or 24dm3

This is sometimes called the molar gas volume.

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In all chemical calculations, gas volumes will be measured at room temperature and pressure (r.t.p). Room temperature is taken as 25oC (298K) and room pressure as one atmosphere (760 mm of mercury).

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Molecule Relative Molecular

mass

Mass of 1 mole

Number of atoms in 1

mole

Volume occupied at r.t.p

Carbon dioxide (CO2)

44 44 g 6.02 x 1023 24 000 cm3

Chlorine (Cl2) 71 71 g 6.02 x 1023 24 000 cm3

Oxygen (O2) 32 32 g 6.02 x 1023 24 000 cm3

Hydrogen (H2) 2 2 g 6.02 x 1023 24 000 cm3

Calculations with Moles

Formula A:

Example:

How many moles are there in(a) 88 g of carbon dioxide?(b) 64 g of oxygen molecules?

Solution:

(a)Mr of carbon dioxide molecule = 44

Number of moles = = 2 moles

(b)Mr of oxygen molecule = 32

Number of moles = = 2 moles


How many moles are there in

(a) 25 g of calcium carbonate CaCO3?

(b) 49 g of sulphuric acid H2SO4?

(c) 64 g of methane gas?

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Formula B:

Mass in grams = Number of moles x Ar or Mr of substance

Example:

How many grams are there in(a)10 moles of water?(b)0.25 moles of oxygen atoms?

Solution:

(a)Mr of water molecules H2O = 18

Mass in grams = 10 x 18 = 180 g

(b)Mr of oxygen atoms O = 16

Mass in grams = 0.25 x 16 = 4 g


How many grams are there in

(a)5 moles of carbon dioxide CO2 ?

(b)0.5 moles of ethene C2H4 ?

(c) 0.1 moles of calcium carbonate?

(d)0.2 moles of bromine gas?

(e)2 moles of sulphur, S8 ?

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Formula C:

Number of particles = Number of moles x Avogadro Number

Example

How many particles are there in(a)16 g of oxygen molecules?(b)11 g of carbon dioxide molecules?

Solution

(a)Number of moles of oxygen molecules = Mass / Mr

= 16 / 32 = 0.5 moles

Number of particles = 0.5 x 6 x 1023

= 3 x 1023 molecules

(b)Number of moles of carbon dioxide molecules = Mass / Mr

= 11 / 44 = 0.25 moles

Number of particles = 0.25 x 6 x 1023

= 1.5 x 1023 molecules


How many particles are there in

(a)0.1 moles of ethene?

(b)5 moles of bromine?

(c) 64 g of methane gas?

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Formula D:

Volume of gas = Number of moles x molar gas volume

Example

What volume (at r.t.p) would the following gases occupy?(a)2 moles of oxygen gas?(b)0.1 mole of hydrogen gas?

(Molar Gas Volume at r.t.p = 24 dm3)

Solution

(a)Volume of oxygen gas = 2 x 24 = 48 dm3

(b)Volume of hydrogen gas = 0.1 x 24 = 2.4 dm3


What volume (at r.t.p) would the following gases occupy?

(a)0.2 moles of carbon dioxide gas

(b)3 moles of nitrogen gas

(c) 4 g of oxygen gas

(d)4 g of sulphur dioxide gas


What would be the mass of 12 dm3 of the following gases at r.t.p?

(a) chlorine gas (b) sulphur dioxide gas

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Moles and Equations

Stoichiometry is the study of the quantitative composition of chemical substances and quantitative changes that occur in chemical reactions.

Consider a car manufacturer uses liquid hydrogen as fuel. He may have to determine how much fuel is necessary for a particular journey. When hydrogen burns in oxygen it forms water. The balanced chemical equation is:

2H2 (g) + O2 (g) 2H2O (g)

In the reaction, 2 hydrogen molecules react with one oxygen molecule to form 2 water molecules.

If we use moles, then 2 moles of hydrogen gas react with one mole of oxygen gas to form two moles of steam. A balanced chemical equation shows the stoichiometric ratio and this gives the exact amount of reactant and product.

Lets use the equation again to find out how much water would be produced if 10g of hydrogen were burnt in the air.

2H2 (g) + O2 (g) 2H2O (g) 2 moles 1 mole 2 moles

Mole Ratio:

No of moles of water = 2 x No of moles of hydrogen

No of moles of water = 2 x

No of moles of water = 2 x = 5 moles

In the reaction, 5 moles of water or 90g of water [5 x 18] are produced.

Now, lets look at the same problem from a different point of view.

How much oxygen is necessary to form 54g of water by burning hydrogen?

Taking the same chemical equation:

2H2 (g) + O2 (g) 2H2O (g) 2 moles 1 mole 2 mole

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Mole Ratio:

No of moles of oxygen = x No of moles of water

No of moles of oxygen = x

No of moles of oxygen = x = 1.5 moles

In the reaction, 1.5 moles of oxygen or 48g of oxygen [1.5 x 32] are required.


For each of these equations, state how many moles of each reactant and product there are. Convert the moles into grams.

S (s) + O2 (g) SO2 (g) CaCO3 (s) CaO (s) + CO2 (g)

2Cu (s) + O2 (g) 2CuO (s) 2NaOH (aq) + H2SO4 (aq) Na2SO4 (aq) + 2H2O (l)

Calculating Percentage Purity and Yield

Percentage Purity

Most chemicals that we use are never 100% pure. When comparing chemical’s purity, we refer to its percentage purity. For example, crystals of potassium nitrate can be sold as at least 99% pure.

e.g potassium nitrateminimum assay 99%maximum limits of impurity:chloride 0.02%

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sulphate 0.02%sodium 0.05%

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Percentage Yield

In any reaction, the amount of product produced experimentally is always less than produced theoretically (by calculation from equation). This can be expressed in terms of percentage yield

Percentage yield =

Example:

Calculate the percentage yield if, on heating 50g of limestone (CaCO3), 21g of lime (CaO) were obtained.

Calcium carbonate calcium oxide + carbon dioxide

CaCO3 (s) CaO (s) + CO2 (g)

Mr = 100 Mr = 56 Mr = 44(1 mole) (1 mole) (1 mole)

Mole Ratio:

Theoretically, 0.5 moles or 28g (0.5 x 56) of CaO is produced when 50g of CaCO3 is heated

Therefore, percentage yield =

=

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Moles and Solution

What is a solute? solvent? and solution?

Solute: a substance that dissolves in a liquidSolvent: the liquid in which the solute dissolvesSolution: the resulting mixture

+

(g) (dm3) (g/dm3)

The volume of the solution is measured in dm3. The concentration of the solution tells you how much solute is

dissolved in 1 dm3 of solution. The amount of solute can be measured in grams or moles.

A concentration of 10 gdm-3 means there is 10 g of solute dissolved in 1 dm3 of solution

A concentration of 2 moldm-3 means there are 2 moles of solute dissolved in 1 dm3 of solution

How to convert concentration in gdm3 to moldm3? (Concentration Conversion)

Concentration (gdm-3) = concentration (moldm-3) x mass of 1 mole of solute

Consider dissolving sodium hydroxide NaOH in water:

1 mole (40g) in 1000 cm3 (1 dm3) = 1 moldm-3

2 moles (80g) in 1000 cm3 (1 dm3) = 2 moldm-3

Twice the amount of solute in the same volume of solvent

0.5 moles (20g) in 1000 cm3 (1 dm3) = 0.5 moldm-3

Half the amount of solute in the same volume of solvent

1 mole (40g) in 500 cm3 (0.5 dm3) = 2 moldm-3

Same amount of solute in half the volume of solvent so twice as concentrated

1 mole (40g) in 2000 cm3 (2 dm3) = 0.5 moldm-3

Same amount of solute in twice the volume of solvent so half as concentrated

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SOLUTE SOLUTIONSOLVENT

Concentration of Solutions

It is necessary for a chemist to try to work out the number of moles of solute in a certain volume of solution. For example how much sodium hydroxide is contained in 200 cm3 of 2 mol dm-3 sodium hydroxide solution? This can be calculated as follows:

1000 cm3 of 2 mol dm-3 contains 2 moles of NaOH

1 cm3 of 2 mol dm-3 contains mole of NaOH

200 cm3 of 2 mol dm-3 contain mole of NaOH = 0.4 mole of

NaOH

In general

Number of moles of solute = Concentration in moldm-3

Moles and Titration

What is titration? Quantitative experimental technique Involves reaction between acids and alkalis (Neutralisation

Reaction) How is titration carried out?

o Acid (or alkali) of unknown concentration is added from burette into a conical flask containing alkali (or acid)

o Exact volume of acid or alkali to be placed in the conical flask is measured by using a pipette

o An indicator is added to find the end-point of the titration

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To find the concentration of the unknown acid, the following acid and alkali relationship is required:

where M = molarity or concentration inn mol dm-3

V = volume in cm3

n = number of moles shown in the chemical equation

Example 1:

What volume of 0.1 mol dm-3 hydrochloric acid will be required to neutralize 25cm3 of 0.2 mol dm-3 sodium hydroxide?

The equation for the reaction is:Hydrochloric acid + sodium hydroxide sodium chloride + water

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

From the equation 1 mole of acid reacts with 1 mole of sodium hydroxide

Here;Macid = 0.1 mol dm-3 Malkali = 0.1 mol dm-3

Vacid = unknown Valkali = 25 cm3

nacid = 1 nalkali = 1

Therefore;

Thus;

Vacid =

Example 2:

If 25 cm3 of potassium hydroxide is neutralized by 20 cm3 of 2.5 mol dm-3 sulphuric acid, what is the molarity of the alkali?

The equation for the reaction is:Sulphuric acid + potassium hydroxide potassium sulphate +

water

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What volume of 2 mol dm-3 sulphuric acid would be required to neutralize

(a)25 cm3 of 4 mol dm-3 sodium hydroxide?

(b)25 cm3 of 2 mol dm-3 potassium hydroxide?

(c) 25 cm3 of 2 mol dm-3 sodium hydroxide?

What is the concentration in mol dm-3 of a sodium hydroxide solution if 25 cm3 of it is neutralized by

(a)25 cm3 of 2 mol dm-3 hydrochloric acid?

(b)25 cm3 of 1 mol dm-3 sulphuric acid?

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Further Molar Calculations from Equation

Problem 1Calculate how much carbon dioxide gas would be evolved if 50 g of chalk (calcium carbonate) were dissolved in excess acid. You may assume that the reaction takes place at room temperature and pressure (r.t.p).

Molar gas volume = 24 000 cm3

The equation for the reaction is;

Calcium carbonate + hydrochloric acid calcium chloride + water + carbon dioxide

Problem 2Calculate the volume of

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Stoichiometry and the Mole Concept Notes

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