All of the elements on the periodic table make up everything in our world!

All of the elements on the periodic table make up everything in our world!

The periodic table consisting of most of the known elements was developed by a Russian man called Dimitri Mendelev around 1860. He managed to put

all the elements in order of weight !

without sophisticated scientific

equipment.

Mendeleevs First Periodic Table

Relative atomic Mass (Ar) numbers

All atoms were measured against Carbon. The carbon atom was assigned an atomic mass of 12

C6

12 This tell us that each carbon atom has a mass of 12

It also tells us the number of protons and neutrons

We say the relative atomic mass (Ar) of Carbon is 12

H1

1 hydrogen has a relative mass of 1 compared to carbon

Relative atomic Mass (Ar) numbers

We say the relative atomic mass (Ar) of hydrogen is 1

O8

16 oxygen has a relative mass of 16 compared to carbon

Relative atomic Mass (Ar) numbers

We say the relative atomic mass (Ar) of oxygen is 16

All of the relative atomic masses for all the atoms are found on the periodic table

Through numerous experiments Avogadro found that one mole (ie 6.02x10 23 atoms ,yes that is a very, very large number)

of any element equalled the relative atomic mass (Ar) in grams of each substance

This means:

6.02x10 23 oxygen (O) atoms = 16 grams

6.02x10 23 carbon (C) atoms = 12 grams

6.02x10 23 hydrogen (H) atoms = 1 gram

6.02x10 23 sulfur (S) atoms = 32 grams

Avogadro's Number

Atomic Mass (Ar) and Molecular Mass (Mr)

We can find the molecular mass (Mr) of a compound by adding the atomic mass (Ar) numbers of the atoms in a compound.

Be aware that the terms atomic mass and molecular mass are also known as molar masses

Ar and Mr are very similiar• Ar = the relative atomic mass and is used for single atoms eg Ca, N, O etc

in gmol-1

Eg The Ar of Ca = 40 gmol -1

Mr = the relative molecular mass of a substance and is the sum of the Ar values

Eg 1 mole of H2 = 1 + 1 = 2 gmol -1

Eg What is the Mr of H2O? (use your PT to find out)

(2 x H = 2 ) + (1 x O = 16) = 18gmol -1

What is the Mr of H2SO4 ?

(2 x H = 2) + (1 x S = 32) + (4 x O = 64)

2 + 32 + 64 = 98 gmol-1

Measuring in ChemistryMeasuring in ChemistryA baker uses a dozen (12) of something

A chemist uses a mole 6.02 x 1023 of something

Units for Mr are gmol-1

Moles, Mass and Mr relationshipMoles, Mass and Mr relationship

Moles/Mass and Mr are all related by the formula

Mrm

n

Where: m = mass in grams Mr = molar mass (g mol-1) n = number of moles

m

n Mr

X

Moles,Mass and Mr relationshipMoles,Mass and Mr relationship

Using this relationship we are able to work out:

1.Moles of a substance given the mass and Mr

2.Mass of a substance given the number of moles and Mr

3.Even the Mr of a substance given the mass and number of moles

Mrm

n

QuestionsQuestions

• Find the number of moles of Ca in 50 grams of the metal.

Mrm

n 1- mol 40g

50gn

n = 1.25 moles

n = 1.25 moles (3sf)

QuestionsQuestions

• Find the number of moles in 8 grams of oxygen gas (O2).

Mrm

n 1- mol 32g

8gn

n = 0.250 moles (3sf)

QuestionsQuestions

• Find the number of moles in 12 grams of carbon dioxide gas (CO2).

Mrm

n 1- mol 44g

12gn

n = 0.273 moles (3sf)

Starter Questions – Using your periodic tableStarter Questions – Using your periodic table

(this means that one mole of CO2 has a mass of 44g)

Find the molar mass (Mr) for CO2

(this means that one mole of Na has a mass of 23g)

Find the atomic mass (Ar) for Na

(2 x 23) + 12 + (3 x 16) = 106 gmol-1

Find the molar mass (Mr) for Na2CO3

11.5 gFind the atomic mass of 0.5 mole of Na metal

2.3 gFind the mass of 0.1 mole of Na metal

12 + (2 x 16) = 44gmol-1

23 gmol-1

Toughie? :How many Na atoms are there in 0.1 mol of Na metal?

0.1 x 6.02 x 1023 = 6.02 x 10 22 Na atoms

Moles, Mass and Mr relationshipMoles, Mass and Mr relationshipMoles/Mass and Mr are all related by what formula?

)Mr(gmol(g) m

n 1-

Where: m = mass inMr =

m

n Mr

Xgrams

molar mass (gmol-1)n = number of

moles

We will use a grid to help in calculations. These become very helpful with complicated questions - just fill what you know then use this to work out your unknown

Question- iron oxide (Fe2O3) is used in the production of iron. How many moles of iron oxide would there be in 200 grams?

Mr

Mrm

n

mass

Fe2O3

?

200 grams

160g200g n = 1.25 moles

(2 x 56) + (3 x 16) = 160 gmol -1

Copy out the grid

In industrial operations large quantities are used

Question - How many moles of iron oxide would there be in 1 tonne (1000kgs) ?

Mr

Mrm

n

mass

Fe2O3

?

1000kg x 1000 = 1 x 106 grams

160gg10 x 1

n6

= 6,250 moles

(2 x 56) + (3 x 16) = 160 gmol -1

Copy out the grid

How to work out the % of an element in a compound

What % of Fe is there in Fe2O3?

32

32

OFe in Fe %70

100 x 160112

OFe Mr2Fe Mr

%Fe

How much iron would you get from 100kgs of Fe2O3?0.7 x 100 = 70kg of iron

Reading Equations using MolesCa + H2SO4 CaSO4 + H2

The above reaction reads:

1 amount of Ca reacts with 1 amount of H2SO4 to give 1 amount of CaSO4 and 1 amount of H2 gas.

(we don’t usually put in the ones – but if we did it would look like this)

The amount we use is the mole

Now the reaction reads:

1 mole of Ca reacts with 1 mole of H2SO4 to give 1 mole of CaSO4 and 1 mole of H2 gas.

1 1 1 1

Reading Equations using MolesCa + H2SO4 CaSO4 + H2

Questions:

1. If we have 1 mole of Ca how many moles of H2SO4 are needed to fully react the Ca?

2. If we had 0.5 mole of Ca how many moles of H2SO4 would be required?

3. If 0.5 mole of Ca reacted with 0.5 mole of H2SO4 how many moles of CaSO4 and how many moles of H2 gas are formed?

Ans: 1mole

Ans: 0.5 mole

Ans : 0.5 mole of H2 gas, 0.5 mole of CaSO4

Reading Equations using MolesCa + H2SO4 CaSO4 + H2

Questions:

If 0.5 mole of Ca reacted with 0.5 mole of H2SO4 how many moles of CaSO4 and how many moles of H2 gas are formed?

Ans : 0.5 mole of H2 gas, 0.5 mole of CaSO4

Given the Ar (atomic mass) values

H = 1gmol-1 Ca = 40gmol-1 S = 32gmol-1 O =16gmol-1

Find the molar masses (Mr) of H2SO4 , CaSO4 and H2

Then use these and the mole (n) formula to work out the mass of H2 gas and the mass of CaSO4 formed

(Hint you will need to rearrange the formula)

Reading Equations using MolesFe2O3 + CO 2Fe + 2CO2

Questions:

If 200kg of Fe2O3 is reacted in the above equation what mass of Fe is produced?

32

32

OFe kg200 in Fe kg140

200 x 160112

OFe Mr2Fe Mr

%Fe

Reaction ratio

Mg + 2 HCl MgCl2 + H2

Work out how many moles of HCl is used to completely react 60grams of Mg metal in the following reaction? Draw grid

2

Mrm

n

mass

Mg

?

60 grams

1-24gmol

60g n = 2.5 moles

24 gmol -1Mr

HCl

36.5 gmol -1

1 Therefore 2 x moles of Mg = moles of HCl

Therefore 2 x 2.5 moles = 5 moles of HCl

What was the mass of HCl used?

Mrm

n

m = 5 x Mr = 5 mol x 36.5 gmol-1

= 182.5 g

= 182 g (3sf)

SherbetSherbet

NaHCO3 + C4 H6O6 + C6 H7 O7 Na tartarate + H2O + CO2

Sherbet balanced reaction

Tartaricacid

Citricacid

Sherbet

Given 1 gram of NaHCO3 work out how much of each ingredient

NaHCO3 Tartaric acid Citric acid

Mr

Molar ratio 1 1 1

184gmol

1g

Mr

m n = 0.0119 mol

84 gmol -1 150 gmol -1 191 gmol -1

mass = n x Mr 0.0119 x 150= 1.785g

0.0119 x 191= 2.273g

0.0119 mol 0.0119 mol

1g

Reaction ratio

CH4 + 2O2 CO2 + 2H2O

Work out how many moles of carbon dioxide is created when 100grams of oxygen is reacted in the following reaction? Draw grid

1

Mrm

n

mass

?

100 grams

1-32gmol100g

n = 3.125 moles

32 gmol -1Mr 44 gmol -1

2 Therefore ½ x moles of O2 = moles of CO2

Therefore ½ x 3.125 moles = 1.5625 moles of CO2

CO2O2

Would you give this answer? 1.56 moles of CO2 (3 sig fig)

Empirical Formula

This is the simplest ratio of atoms in a compound

For example the empirical formula for ethane C2H6

is CH3

the empirical formula for butane C4H10 is C2H5

the empirical formula for methane CH4 is CH4

the empirical formula for hydrogen peroxide H2O2 is HO

Molecular Formula

The molecular formula of a compound is the formula giving the actual number of atoms in a compound

We will be asked to find the molecular formula given % mass of a compound

For example find the molecular formula for a compound which was found to contain 80% carbon and 20% hydrogen

Example: finding the empirical formula for a compound that was found to contain 80% carbon and 20% hydrogen

C H

80g 20g

Step two :

divide each element by its Ar value

simply convert each % to grams

Step one :

666.612

80 20

1

20

Step three :

What is the smaller number?Now divide each of these bythis smaller number

16.6666

6.6666 00.3

6.66666

20

Now this is the empirical formula for the compound just write out the formula

Step four :

1 x C 3 x H

CH3

Example : Now find the molecular formula for the compound that was found to contain 80% carbon and 20% hydrogen.Given that the compound was found to have a molar mass of 30 gmol-1

Step five: use the empirical formula you just found ie CH3 and follow the method belowStep six : divide your given molar mass by the molar mass of CH3

massmolar formula empirical

massmolar given 2

15gmol

30gmol

1-

-1

Step seven: now multiply your empirical formula by 2 to get the molecular formula - 2 x CH3 = C2H6

A compound was found to have a % composition of 2.0% Hydrogen, 32.7% Sulfur, 65.3% Oxygen find the empirical formula

OH65.3g2g

Step two :

divide each element by its Ar value

simply convert each % to grams

Step one :

Step three :

What is the smaller number? Now divide each of these bythis smaller number

Now round and just write out the empirical formula

Step four :

08125.416

65.3 2

1

2

11.02187

1.02187 9571

1.02187

2 .

A compound was found to have a % composition of 2.0% Hydrogen, 32.7% Sulfur, 65.3% Oxygen find the empirical formula

S

32.7g

02187132

32.7 .

9939.31.02187

4.08125

412

SH2 O4

A sample with composition 40% C, 6.7% H and 53.3% O had a molar mass 60 g mol-1 find the Empirical formula and also the molecular formula

OC53.3g40g

Step two :

divide each element by its Ar value

simply convert each % to grams

Step one :

Step three :

What is the smaller number? Now divide each of these bythis smaller number

Now round and just write out the empirical formula

Step four :

H6.7g

121

H2C O

3333312

40 . 76

1

6.7 . 33123

16

53.3 .

000613.3312

3.3333 . 0112

3.3312

6.7 . 1

3.3312

3.3312

A sample with composition 40% C, 6.7% H and 53.3% O had a molar mass 60 g mol-1 find the Empirical formula and also the molecular formula

Step five: use the empirical formula you just found ie CH2O and follow the method below

Step six : divide your given molar mass (60 gmol-1) by the molar mass of CH2O (30 gmol-1)

Step seven: now multiply your empirical formula by 2 to get the molecular formula - 2 x CH2O = C2H4O2

massmolar formula empirical

massmolar given 2

30gmol

60gmol

1-

-1

Water of Crystallisation

• Some hydrated salts have H2O molecules in their structure eg CuSO4 5H2O (hydrated means with water)

• This means that each particle of CuSO4 is surrounded by 5H2O molecules

• These H2O molecules are called the water of crystallisation

• The water in the structure can be removed by heating to form the anhydrous (without water) salt ie CuSO4

Finding the Empirical Formula for a Hydrated salt by Gravimetric

Analysis

• To find the mass of H2O in a hydrated (with water) salt

• Firstly find the mass of the hydrated (with water) salt

• Heat strongly for 10 minutes cool then reweigh

• Then heat again and reweigh to check the salt is at constant mass

• Then subtract the mass of the anhydrous(dry) salt from the hydrated(wet) salt

• This will give you the water mass

• Then use your anhydrous salt mass and water mass to workout the empirical formula of the hydrated salt using grid

Find the empirical formula for a hydrated salt Na2CO3XH2OGiven:

Mass of hydrated Na2CO3 salt = 2.58g

Mass of salt after heating to constant weight = 0.95g

Therefore: 2.58g – 0.95g = 1.63g (mass of water)

Calculation for finding the empirical formula for a hydrated salt Na2CO3XH2O

106gmol-1 18gmol-1

Step two :

Given masses find moles of each

Find the Mr ‘s

Step one :

0905555.018

1.63 0089622.0

106

0.95

Step three :

What is the smaller number?Now divide each of these bythis smaller number

10.00896226

0.00896226 1.10

0.00896226

0.0905555

Make small adjustments to give whole numberTo give empirical formula

Step four :

Na2CO3 H2O

1 x Na2CO3 10H2O

Na2CO310H2O

Concentration

Particles per volume

Can be in grams per litre but chemists usually express concentration in moles per litre

This are related by the formula:

Where:n = molesV = volume in litresc = concentration in molL-1

Vn

c

Write the formula for:

1. Moles using mass and Molar mass

2. Concentration using Volume and Moles

)Mr(gmol(g) m

n 1-

Vn

c

Concentration Example15 grams of NaCl is dissolved in 1 litre of water, what is the concentration of the solution in moles per litre? (M NaCl = 58.5 gmol-1)

Find moles of NaCl

Divide moles by volume in litres

mol 0.2564 58.515

Mm

n

1molL256.0litre 1

0.2564mol

Vn

c

Concentration Example20 grams of NaOH is dissolved in 500mls of water, what is the concentration of the solution in moles per litre? (M NaOH = 40gmol-1)

Find moles of NaOH

Divide moles by volume in litres

mol 0.5 40gmol

20g

Mm

n1-

1molL 10.5litre0.5mol

Vn

c

If solution A (25mls of 0.23 molL-1 AgNO3) is mixed withsolution B (82 mls of O.15molL-1 of AgNO3 ) What is the final concentration of the solution when they are mixed

Find total moles by adding moles of A to moles of B

Divide total moles by total volume in litres

moles5.75x10 n

0.025L x 0.23molL n

xV c n

Vn

c

moles A Solution

3-

1-

)SF3(molL 0.169 molL 16869.0s0.107litre

0.01805mol

Vn

c 1-1

moles0.0123 n

0.082L x molL 0.15 n

xV c n

Vn

c

moles B Solution

1-

es0.01805mol

s0.0123mole moles 10x75.5

B molesA moles

moles Total

3

Reaction ratio

CH4 + 2O2 CO2 + 2H2O

2

Mrm

n

mass

?

500 grams

1-16gmol500g

n = 31.25 moles

16 gmol -1Mr 18 gmol -1

1 Therefore 2 x moles of CH4 = moles of H2O Therefore 2 x 31.25 moles = 62.5 moles of H2O

H2OCH4

How many grams of water vapour H2O(g) will be produced in the complete combustion of 500g methane, CH4(g)?

62.5x18 =1,125g

= 1.13kg H2O

Homework for Friday - finish off all worksheet pages in booklet except worksheet 4

Test preparation for MondayRead Unit 11 page 41 in your pathfinder textComplete all questions on page 42 and 43 before Monday

See me before Monday if you are having any trouble

How many grams of water vapour H2O(g) will be produced in the complete combustion of 500g methane, CH4(g)?

Equation: CH4(g) + 2O2(g) CO2(g) + 2H2O(g)