The Masses of chemicals - Learning Together (KS4 …...1 mole of any substance always contains the...

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The Masses of chemicals

WILF•To give a definition of relative formula mass Mr.• To calculate relative formula mass if its formula and the relative atomic mass are given.• To give a full definition of relative atomic mass Ar. • To explain what a mole is.


What’s in a Symbol?

12

6

C is the symbol for carbon

Relative atomic mass(Mass number)

Proton number(Atomic number)

This is equal to the number of protons andneutrons in the nucleus

This is equal to the number of protons in the nucleus

Its also equal to the number of electrons. Why?


Relative Atomic Mass of ElementsDefiniton• The relative atomic mass of an element (Ar ) compares

the mass of atoms of the element with the carbon-12 isotope. It is an average value for the isotope of the element.

• Carbon is given a relative atomic mass (Ar) of 12.• The Ar of other atoms compares them with carbon.• Eg. Hydrogen has a mass of only one twelfth that of

carbon and so has a Ar of 1.Element Symbol Times as heavy as carbon Ar

Helium He one third

Beryllium Be three quarters

Molybdenum Mo Eight

Krypton Kr Seven

4

9

96

84


Relative Formula Mass (Molecular Mass)• For a number of reasons it is useful to use

something called the formula mass sometimes called the relative molecular mass. Mr

• To calculate this we simply add together the atomic masses of all the atoms shown in the formula. (N=14; H=1; Na=23; O=16; Mg=24; Ca=40)

Substance Formula Relative Formula Mass

Ammonia NH3

Sodium oxide Na2O

Magnesium hydroxide Mg(OH)2

Calcium nitrate Ca(NO3)2

14 + (3x1)=17

(2x23) + 16 =62

24+ 2(16+1)=58

40+ 2(14+(3x16))=164


ISOTOPES• Isotopes are virtually identical in their chemical

reactions. (There may be slight differences in speeds of reaction).

• This is because they have the same number of protons and the same number of electrons.

• The uncharged neutrons make no difference to chemical properties but do affect physical properties such as melting point and density.

DefinitionAn isotope is an atom of the same element but with a different number of neutrons


Isotopes: Carbon• Natural samples of elements are often a

mixture of isotopes. About 1% of natural carbon is carbon-13.

Protons

Electrons

Neutrons

C12

699% C

13

61%

6

6

6

6

6

7


Isotopes: Chlorine• About 75% of natural chlorine is 35Cl the rest is

37Cl.

Cl35

1775%

17Protons

Electrons

Neutrons

17

18Protons

Electrons

Neutrons

171720

Cl37

1725%


Isotopes and Relative Atomic Mass

• Many natural elements are a mixture of isotopes.

• This means that when we react atoms of an element we are using a mixture of atoms with different mass numbers.

• The relative atomic mass given in the periodic table takes account of this.

E.g.. For 100 atoms of chlorine:

Mass of 75 atoms of Chlorine 35: 75 x 35 =2625

Mass of 25 atoms of Chlorine 37: 25 x 37 =925

Total = 3550

Average (divide by 100) = 35.5


The Mole

• An Italian count called Amadeo Avogadro worked out that a mole contained 6.02 x1023 particles. This is called the Avogadro Number

• He worked out that one mole of any substance is the relative atomic or formula mass expressed in grams.

• The relative atomic mass AR in grams is known as a MOLE. So a mole of carbon 12 atoms has a mass 12g.

• The relative formula mass MR of a compound in grams is also one mole. 1 mole of CO2 =44g. C+0+0 = 12+16+16.

Amadeo Avogadro


How big is a mole?• 1 mole of marshmallows would be enough

marshmallows to make a 12 mile thick layer of marshmallows covering the entire face of the Earth.

• 1 mole of popcorn kernels could be spread uniformly over the USA if the thickness of the layer was about 9 miles.

• 1 mole of donut holes would cover the earth and be 5 miles deep.

• 1 mole of sheets of paper could form a million stacks from the surface of the earth, all that would pass the sun.


The Mole (cont’d)1 mole of any substance always contains the same number of particles – 6 x 1023

1 mole of carbon weighs differently to 1 mole of oxygen

1 mole of carbon, C, atoms weighs 12g

1 mole of oxygen, O, atoms weighs 16g. Use a periodic able and work out the mass of each of the following:

1 mole of sodium Na 2 moles carbon C

1 mole of nitrogen N 10 moles chlorine Cl

1 mole of calcium Ca 0.5 moles oxygen O

1 mole of hydrogen H 0.1 moles sulphur S

1 mole of sodium Na 23g 2 moles carbon C 24g

1 mole of nitrogen N 14g 10 moles chlorine Cl 355g

1 mole of calcium Ca 40g 0.5 moles oxygen O 8g

1 mole of hydrogen H 1g 0.1 moles sulphur S 3.2g

Worksheet

skl_massofmole.doc


The Mole (cont’d)Changing grams to moles

No of moles= mass of substance (g)

relative atomic (Ar) or formula mass (Mr)

e.g. How many moles in 12g of magnesium?

No moles = mass of substance (g)

relative atomic mass (Ar)

= 12 = 0.5 moles (Ar Mg = 24)

24

e.g. How many moles in 4g of oxygen? Trick – oxygen exists as molecules = O2

No moles = mass of substance (g)

relative formula mass (Mr)

= 4

32 (2x16) Mr O2 = 1

8


Check Up

How many moles in the following:

1. 46g Sodium Na

2. 3.5g lithium Li

3. 3.2g oxygen O2

4. 71g Chlorine gas Cl25. 4.4g carbon dioxide CO2

6. 3.6g water H2O?

ANSWERS


1. 46g sodium Na.

No moles = mass ÷ Ar

= 46 ÷ 23

= 2 moles

2. 3.5g Lithium Li

No moles = mass ÷ Ar

= 3.5 ÷ 7

= 0.5 moles

3. 3.2g oxygen O2

No moles = mass ÷ Mr

= 3.2 ÷ (2x16)

= 3.2 ÷ 32

= 0.1 moles

4. 71g chlorine gas Cl2.


= 71 ÷ (2 x 35.5)

= 71 ÷ 71

= 1 mole

5. 4.4g carbon dioxide CO2


= 4.4 ÷ 44

= 0.1 moles

6. 3.6g water H2O


= 3.6 ÷ 18

= 0.2 moles


Changing Moles to grams:

Mass of substance = no of moles x Ar or Mr

E.G What is the mass of 4 moles of magnesium oxide MgO?

Mass = no. moles x Mr

= 4 x (24 + 16) MgO

= 4 x 40

= 160g


What is the mass of the following?

1. 10 moles of carbon C

2. 5 moles of beryllium Be

3. 0.2 moles oxygen O2

4. 0.1 moles magnesium chloride MgCl25. 0.25 moles hydrogen H2

6. 0.2 moles aluminium oxide Al2O3

Check Up

ANSWERS


1. 10 moles of carbon Cmass = No moles x Ar

mass = 10 x 12 = 120g

2. 5 moles Bemass = No moles x Ar

mass = 5 x 9 = 45g

3. 0.2 moles O2

mass = No moles x Mr

Mass = 0.2 x (2 x 16)= 0.2 x 32 = 6.4g

4. 0.1 moles MgCl2mass = No moles x Mr

Mass = 0.1 x (24 +(2 x 35.5)= 0.1 x (24 + 71) = 9.5g

5. 0.25 moles H2


Mass = 0.25 x 2 = 0.5g

6. 0.2 moles A12O3


Mass = 0.2 x ((2x27) + (3x16))

= 0.2 x (54 + 48)

=0.2 x 102 = 20.4g


Percentage Composition by Mass

• It is sometimes useful to know how much of a compound

is made up of some particular element.

• This is called the percentage composition by mass – the

percentage of an element in a compound.

% mass of an

element in a

compound

=

Relative atomic

mass of the

element Ar

No. of atoms of

the element in

the formulaX

Relative formula mass of

the compound Mr

%


Percentage Composition by Mass cont’d

0

20

40

60

80

%

Carbon OxygenE.g. % of oxygen in carbon dioxide (Atomic Masses: C=12. O=16)

Formula = Number oxygen atoms =

Rel. Atomic Mass of O =

Rel. Formula Mass CO2 =

% oxygen =

CO2

2

12 +(2x16)=44

(2 x 16 ) x 100 = 72.7%44

Now work out the percentage composition of potassium K in Potassium nitrate, - KNO3

K= 39, N = 14, O=16

16


Formula Atoms

of O

Mass of

O

Formula

Mass

%age Oxygen

MgO 1

K2O 1

NaOH 1

SO2 2

• Calculate the percentage of oxygen in the compounds shown below

32+(2x16)=

6432

23+16+1

=4016

(2x39)+16

=9416

24+16=4016 16x100/40=40%

16x100/94=17%

16x100/40=40%

32x100/64=50%

% Z = (Number of atoms of Z) x (atomic Mass of Z)

Formula Mass of the compound

Activity


• Nitrogen is a vital ingredient of fertiliser that is

needed for healthy leaf growth.

• But which of the two fertilisers ammonium nitrate

or urea contains most nitrogen?

• To answer this we need to calculate what

percentage of nitrogen is in each compound

Activity


Formula Atoms

of N

Rel. At

Mass of

N

Relative Formula

Mass

%age Nitrogen

NH4NO3 2 28

CON2H4 2 28

• Formulae: Ammonium Nitrate NH4NO3: Urea CON2H4

28x100 /80 =

35%

28x100 /60 =

46.7%

14+(1x4)+14+(3x16)=

80

12+16+(2x14)+(4x1)=

60

And so, in terms of % nitrogen

urea is a better fertiliser than

ammonium nitrate 0

10

20

30

40

50

1st Qtr

Amm.Nitrate UreaAtomic masses H=1: C=12: N=14: O=16

Activity


Working out a formula from the

percentage composition.

This is just working backwards from the percentage

composition. If we know the %age composition we can work

out the ratio of the atoms of the elements in the compound.

This is called the empirical formula. This is the simplest

whole number ratio. Sometime it is the same as the actual

number – this is called the molecular formula.

e.g. Empirical formula water – H2O

Molecular formula water – H2O

Empirical formula hydrogen peroxide – HO

Molecular formula hydrogen peroxide – H2O2




Example

9g of aluminium react with 35.5g chlorine gas.

What is the empirical formula of the compound formed?

Important bit – you have to convert these masses into

moles (number of atoms). How do we do that?

Divide mass by relative atomic mass.

No of moles in 9g of Aluminium = 9 ÷ 27 = 1/3

No moles in 35.5g chlorine = 35.5 ÷ 35.5 = 1

Ratio = 1Al : 3Cl

Empirical Formula = AlCl31 mole Al react with 3 moles Cl

So 1 atom Al reacts with 3 atoms Cl




Problems

1. 1.2 g of carbon react with 3.2 g of oxygen.

What is the empirical formula of Carbon Dioxide?

2. 80g of calcium react with 32g of oxygen.

What is the empirical formula of Calcium oxide ?


Representing Chemical reactions:Equations.

WILF1. To balance symbol equations.2. To interpret how many moles of

reactants/products are shown in balanced equation.

3. To use a balanced symbol equation to calculate the mass of reactants or products.


Word Equations• All equations take the general form:

Reactants ProductsWord equations simply replace “reactants andproducts” with the names of the actual reactants and products.

E.g

Reactants Products

Magnesium + oxygen

Sodium + water

Magnesium + lead nitrate

Nitric acid + calcium

hydroxide

Magnesium oxide

Magnesium nitrate + lead

Sodium hydroxide + hydrogen

Water + calcium nitrate


• Write the word equations for the descriptions below.

1. The copper oxide was added to hot sulphuric acid and it

reacted to give a blue solution of copper sulphate and

water.

water+copper

sulphate

sulphuric

acid

+Copper

oxide

2. The magnesium was added to hot sulphuric acid and it

reacted to give colourless magnesium sulphate solution

plus hydrogen

hydrogen+Magnesium

sulphate

sulphuric

acid

+Magnesium

Activity


• Write the word equations for the descriptions below.

3. The methane burned in oxygen and it reacted to give

carbon dioxide and water.

water+Carbon

dioxide

oxygen+methane

4. The copper metal was placed in the silver nitrate solution.

The copper slowly disappeared forming blue copper

nitrate solution and needles of silver metal seemed to

grow from the surface of the copper

silver+Copper

nitrate

Silver nitrate+copper

Activity


Chemical Equations

• Step 1: Write down the word equation.

• Step 2: Replace words with the chemical formula .

• Step 3: Check that there are equal numbers of each type

of atom on both sides of the equation. If not, then balance

the equation by using more than one.

• Step 4: Write in the state symbols (s), (l), (g), (aq).

2MgO(s)2Mg(s) +O2(g)

2MgO2Mg + O2

Oxygen doesn’t balance.Need 2 MgO and so need 2 Mg

MgOMg + O2

magnesium oxidemagnesium + oxygen

ProductsReactants








Reactants Products

sodium + water hydrogen + sodium hydroxide

+ +

+ +

+ +

Na H2O H2 NaOH

2Na 2H2O 2NaOHH2

2Na(s) 2H2O(l) H2(g) 2NaOH(aq)

Hydrogen doesn’t balance. Use 2 H2O, NaOH, 2Na

Chemical Equations








Reactants Products

magnesium + lead nitrate magnesium nitrate + lead

+ +

+ +

Mg Mg(NO3)2 Pb

Mg(s) Pb(NO3)2(aq) Mg(NO3)2(aq) Pb(s)

Already balances. Just add state symbols

Pb(NO3)2

Chemical Equations


• Below are some chemical equations where the formulae are correct but the balancing step has not been done. Write in appropriate coefficients (numbers) to make them balance.

Reactants Products

AgNO3(aq) + CaCl2(aq) Ca(NO3)2(aq) + AgCl(s)

CH4(g) + O2(g) CO2(g) + H2O(g)

Mg(s) + Ag2O(s) MgO(s) + Ag(s)

NaOH + H2SO4(aq) Na2SO4(aq) + H2O(l)

22

2 2

2

2 2

Activity


Reacting Masses


Conservation of Mass

• New substances are made during chemical reactions

• However, the same atoms are present before and after

reaction. They have just joined up in different ways.

• Because of this the total mass of reactants is always equal

to the total mass of products.

• This idea is known as the Law of Conservation of Mass.

Reaction

but no

mass change


• There are examples where the mass may seem to change

during a reaction.

• Eg. In reactions where a gas is given off the mass of the

chemicals in the flask will decrease because gas atoms

will leave the flask. If we carry the same reaction in a

strong sealed container the mass is unchanged.

Mg

HCl

Gas given off.

Mass of

chemicals in flask

decreases

11.71

Same reaction in

sealed container:

No change in

mass

Conservation of Mass


Reacting Mass and formula mass

• The formula mass in grams of any substance contains the same number of particles. We call this amount of substance 1 mole.

Atomic Masses: H=1; Mg=24; O=16; C=12; N=14

1 mole of methane molecules12 + (1x4)CH4

1 mole of magnesium oxide24 + 16MgO

1 mole of hydrogen molecules1x2H2

1 mole of nitric acid1+14+(3x16)HNO3

ContainsFormula MassSymbol


Reacting Mass and Equations

• By using the formula masses in grams ( moles) we can deduce what masses of reactants to use and what mass of products will be formed.

carbon + oxygen carbon dioxide

C + O2 CO2

12 + 2 x 16 12+(2x16)

12g 32g 44g

So we need 32g of oxygen to react with 12g of carbon and

44g of carbon dioxide is formed in the reaction.

Atomic masses: C=12; O=16


aluminium + chlorine aluminium chloride

2Al + 3Cl2 2AlCl3

2 x 27 + 6 x 35.5 2x (27+(3x35.5)

54g 213g 267g

So 54g of aluminium react with 213g of chlorine to give 267g

of aluminium chloride.

Atomic masses: Cl=35.5; Al=27

• What mass of aluminium and chlorine react together?

Activity


magnesium + oxygen

+

+

Atomic masses: Mg=24; O=16

• What mass of magnesium and oxygen react together?

Magnesium oxide

Mg O2 MgO22

2 x 24 2x16 2x(24+16)

48g 32g 80g

So 48g of magnesium react with 32g of oxygen to give 80g

of magnesium oxide.

Activity


Sodium + hydrochloric +

hydroxide + acid

+ +

Atomic masses: Na = 23 O = 16 H = 1 Cl = 35.5

• What mass of sodium chloride is formed when sodium hydroxide and hydrochloric acid react together?

Sodium

chloride

NaOH HCl NaCl

23+1+16 1+35.5 23+35.5

40g 36.5g 58.5g

So 40g of sodium hydroxide react with 36.5g of

hydrochloric acid to give 58.5g of sodium chloride.

H2O

water

(2x1)+16

18g

Activity


Step 1 Word Equation

Step 2 Replace words with correct formula.

Step 3 Balance the equation.

Step 4 Write in formula masses.

Remember: where the equation shows more than 1

molecule to include this in the calculation.

Step 5 Add grams to the numbers.

• It is important to go through the process in the correct order to avoid mistakes.


Reacting Mass and Scale Factors

• We may be able to calculate that 48g of magnesium gives

80g of magnesium oxide – but can we calculate what

mass of magnesium oxide we would get from burning

1000g of magnesium? There are 3 extra steps:

Step 1 Will 1000g of Mg give more or

less MgO than 48g?

Step 2 I need to scale ? the 48g

to 1000g. What scale factor

does this give?

Step 3 If 48g Mg gives 80g of MgO

What mass does 1000g give?

Answer

more

up1000 = 20.83

48

20.83 x 80

1667g

Activity


• Mg + CuSO4 MgSO4 + Cu

• 24 64+32+(4x16) 24+32+(4x16) 64

• 24g 160g 120g 64g

What mass of copper will I get when 2 grams of magnesium is

added to excess (more than enough) copper sulphate?

Step 1 Will 2g of Mg give more or less Cu

than 24g?

Step 2 I need to scale ? the 24g to

2g. What scale factor does this

give?

Step 3 If 24g Mg gives 64g of Cu


Answer

less

down2 = 0.0833

24

0.0833 x 64

5.3

Activity


• CaCO3 CaO + CO2

• 40+12+(3x16) 40+16 12+(2x16)

• 100g 56g 44g

• What mass of calcium oxide will I get when 20 grams of

limestone is decomposed?

Step 1 Will 20g of CaCO3 give more or less

CaO than 100g?

Step 2 I need to scale ? the 100g to

20g. What scale factor does this

give?

Step 3 If 100g CaCo3 gives 56g of CaO


Answer

less

down20 = 0.20

100

0.20 x 56

11.2g

Activity


Elements Carbon Hydrogen

Amount in question (% or mass) 75 25

Atomic mass (periodic table) 12 1

Number of moles = Amount in questionAtomic mass

7512=6.25

251=25

Mole ratio (divide each number of moles by the smallest number of moles)

6.256.25=1

256.25= 4

Empirical formula C H4

Calculation of Empirical Formulae.Example1.A compound contains 75% carbon and 25% hydrogen. What is its empirical formula?

Empirical formula CH4


Calculation of Empirical Formulae.Example2. An oxide of carbon contains 27% carbon. What is its empirical formula?

Elements Carbon Oxygen

Amount in question (% or mass) 27

Rel. Atomic mass (periodic table) 12

Number of moles = Amount in questionRel. Atomic mass

2712= 2.25

Mole ratio (divide each number of moles by the smallest number of moles)

= 2.252.25

= 1

Empirical formula C

73

16

7316

=4.56

= 4.562.25

= 2

O2


Moles and solutions

When 1 mole of solute is dissolved in 1 litre of water we say it is a 1 Molar or 1M solution.

A 1M solution of sodium hydroxide (NaOH) solution would contain 1 mole ( 23+16+1 = 40g) of sodium hydroxide solid dissolved in 1 litre of water.

Number of moles = volume(l) x Moles per litre (M)

e.g. How many moles in 250cm3 of a 2M solution?

No of moles = 0.25 l x 2m = 0.5 moles.


Gases are measured by volume not by mass.

At room temperature and pressure 1 mole of any gas occupies 24 litres.

e.g. 0.2 moles of hydrogen H2 occupies what volume?

Volume = 0.2 x 24 = 4.8 litres.

Moles and gases


Formula from Composition by mass.

WILF

To calculate the formula of a compound from the mass of the reactants.


GCSE QuestionAn exothermic reaction takes place when

nitrogen reacts with hydrogen to make ammonia.

The reaction can be represented by this equation.

N2 (g) + 3H2 (g) 2NH3 (g)

(a) Calculate the maximum mass of ammonia that could be made from 1000 g of nitrogen.

Relative atomic masses: H = 1; N = 14


Relative Formula Mass Mr

• When a new compound is discovered we have to

deduce its formula.

• This always involves getting data about the

masses of elements that are combined together.

• What we have to do is work back from this data to

calculate the number of atoms of each element

and then calculate the ratio.

• In order to do this we divide the mass of each

atom by its atomic mass.

• The calculation is best done in 5 stages:


Calculating Empirical Formula

• We found 3.2g of copper reacted with 0.8g of oxygen. What is the formula of the oxide of copper that was formed? (At. Mass Cu=64: O=16)

Substance Copper oxide

1. Elements Cu O

2. Mass of each element

in question (g)

3. Mass ÷ Atomic Mass

(No. of moles)

4. Ratio

5. Formula

3.2 0.8

3.2÷64 =0.05 0.8÷16 =0.05

1:1

CuO


• We found 5.5g of manganese reacted with 3.2g of oxygen. What is the formula of the oxide of manganese formed? (Atomic. Mass Mn=55: O=16)

Substance Manganese oxide

1. Elements Mn O


(g)

3. Mass / Atomic Mass

(No. of moles)

4. Ratio

5. Formula

5.5 3.2

5.5/55 =0.10 3.2/16 =0.20

1:2

MnO2


• A chloride of silicon was found to have the following %

composition by mass: Silicon 16.5%: Chlorine 83.5%

(Atomic. Mass Si=28: Cl=35.5)

Substance Silicon Chloride

1. Elements Si Cl


(g per 100g)

3. Mass / Atomic Mass

4. Ratio

5. Formula

16.5 83.5

16.5/28 =0.59 83.5/35.5 =2.35

Cl÷Si = (2.35 ÷ 0.59) = (3.98)

Ratio of Cl:Si =4:1

SiCl4

Divide biggest by

smallest

Activity


• Calculate the formula of the compounds formed when the

following masses of elements react completely:

(Atomic. Mass Si=28: Cl=35.5)

Element 1 Element 2 Atomic Masses Formula

Fe = 5.6g Cl=106.5g Fe=56 Cl=35.5

K = 0.78g Br=1.6g K=39: Br=80

P=1.55g Cl=8.8g P=31: Cl=35.5

C=0.6g H=0.2g C=12: H=1

Mg=4.8g O=3.2g Mg=24: O=16

FeCl3

KBr

PCl5

CH4

MgO

Activity

masses


Formula from Charges on ions

WILF

To calculate the formula of a compound from the charge(s) on the ions


Charges on ions.

• Many elements form ions with some definite

charge (E.g. Na+, Mg2+ and O2-). It is often

possible to work out the charge using the Periodic

Table.

• If we know the charges on the ions that make up

the compound then we can work out its formula.

• This topic is covered in more detail in the Topic on

Bonding but a few slides are included here on how

to work out the charges on ions and use these to

deduce the formula of simple ionic compounds.


Charges and Metal ions

• Metals usually lose electrons to empty this outer shell.

• The number of electrons in the outer shell is usually

equal to the group number in the Periodic Table.

• Eg. Li =Group 1 Mg=Group2 Al=Group3

Mg

2.8.2

Mg2+

Al

2.8.3

Al3+

Li

2.1

Li+


Charges and non-metal ions

• Elements in Groups 4 onwards generally gain electrons and the number of electrons they gain is equal to the Group Number.

• Oxygen (Group 6) gains (8-6) =2 electrons to form O2-

• Chlorine (Group 7) gains (8-7)=1 electron to form Cl-

ClO

2.62.8

O O2-2.8.7 2.8.8

Cl Cl-


• Copy out and fill in the Table below showing

what charge ions will be formed from the

elements listed.

H He

Li

Na

K

Be

Sc Ti

Mg

V Cr Mn Fe Co Ni Cu Zn Ga Ge Se BrCa Kr

Al P

N O

S Cl

F Ne

ArSi

B C

As

Mg

C

Cl

K

Symbol Li N Cl Ca K Al O Br Na

Group No

Charge

1 5 7 2 1 3 6 7 1

1+ 3- 1- 2+ 1+ 3+ 2- 1- 1+

1 2 3 4 5 6 7 0

Activity


The formulae of ionic compounds

This is most quickly done in 5 stages.

Remember the total + and – charges must =zero

• Eg. The formula of calcium bromide.

1. Symbols: Ca Br

2. Charge on ions 2+ 1-

3. Need more of Br

4. Ratio of ions 1 2

5. Formula CaBr2

Br

Ca

Br

Ca2+

Br-

Br-

2 electrons


• Eg. The formula of aluminium bromide.

1. Symbols: Al Br


3. Need more of Br

4. Ratio of ions 1 3

5. Formula AlBr3

BrAl

Br

Br

3 electrons

Al3+ Br-

Br-

Br-



• Eg. The formula of aluminium oxide.

1. Symbols: Al O


3. Need more of O

4. Ratio of ions 2 3 (to give 6 e-)

5. Formula Al2O3

OAl

O

OAl

2e-

2e-

2e-

Al3+

O2-

O2-

O2-

Al3+



• Eg. The formula of magnesium chloride.

1. Symbols: Mg Cl2. Charge on ions3. Need more of4. Ratio of ions5. Formula

2+ 1-

Cl1:2

MgCl2

Cl

MgCl

1e-

1e-

Cl-Mg2+

Cl-

The formulae of ionic compoundsActivity


• Eg. The formula of sodium oxide.

1. Symbols: Na O2. Charge on ions3. Need more of4. Ratio of ions5. Formula

ONa

Na 1e-

1e- Na+

O2-

Na+

1+ 2+

Na

2 : 1

Na2O

ActivityThe formulae of ionic compounds


• Using the method shown on the last few slides,

work out the formula of all the ionic compounds

that you can make from combinations of the

metals and non-metals shown below:

•Metals: Li Ca Na Mg Al K

•Non-Metals: F O N Br S Cl

Activity


Reacting Mass Industrial Processes

• Industrial processes use tonnes of reactants not grams.

• We can still use equation and formula masses to calculate

masses of reactants and products.

• We simply swap grams for tonnes.

• E.g. What mass of CaO does 200 tonnes of CaCO3 give?

CaCO3 CaO + CO2

100 56 44

So 100 tonnes would give ? tonnes

And 200 tonnes will give

Scale factor =

So mass of CaO formed = ? tonnes =

56

more

200/100 =2

2 x 56 112 tonnes


• Iron is extracted from iron oxide Fe2O3

• E.g. What mass of Fe does 100 tonnes of Fe2O3

give?

Fe2O3 + 3CO 2Fe + 3CO2

160 84 112 + 132


And 100 tonnes will give

Scale factor =

So mass of Fe formed = ? =

112

less

100/160 =0.625

0.625 x 112 70 tonnes

Activity


• Ammonia is made from nitrogen and hydrogen

• E.g. What mass of NH3 is formed when 50 tonnes of N2 is completely converted to ammonia?

N2 + 3H2 2NH3

28 6 34


And 50 tonnes will give than 28 tonnes

Scale factor =

So mass of NH3 formed = ? =

34

more

50/28 =1.786

1.786 x 34 60.7 tonnes

Activity


Na is the symbol for?

1. Nitrogen

2. Nickel

3. Neodynium

4. Sodium


Which of these does NOT exist as a diatomic

molecule (2 bonded atoms)?

1. Nitrogen

2. Oxygen

3. Calcium

4. Chlorine


How many oxygen atoms are represented in the

formula Pb(NO3)2?

1. One

2.Two

3.Three

4.Six


What is the formula mass of MgCl2 ?

Mg=24 Cl=35.5

1. 59.5

2. 83.5

3. 95

4. 119


What is the formula mass of Mg(OH)2 ?

Mg=24 O=16 H = 1

1. 41

2. 42

3. 57

4. 58


What is the percentage nitrogen in ammonium

sulphate (NH4)2SO4?

1. 21%

2. 42%

3. 63%

4. 84%


What is the formula of a compound containing

1.4g nitrogen and 3.2g of oxygen? (N=14

O=16)

1. N2O

2. NO

3. NO2

4. N2O3


What is the formula of a compound containing

6.5g zinc and 1.6g oxygen?

(Zn=65 O=16)

1. ZnO

2. Zn2O3

3. ZnO2

4. Zn2O


What is the formula of a compound formed

between Cr3+ ions and O2- ions?

1. CrO

2. Cr2O3

3. CrO2

4. Cr3O2


What is the formula of a compound formed

between Cr3+ ions and OH- ions?

1.CrOH3

2.Cr3OH

3.Cr(OH)3

4.Cr2OH3


What is the word equation for the reaction

described below?

A small piece of strontium metal was added to

water. It fizzed giving off hydrogen gas leaving an

alkaline solution of strontium hydroxide.

1.Strontium + water hydrogen + strontium hydride

2.Strontium + water oxygen + strontium hydroxide

3.Strontium + water hydrogen + strontium hydrate

4.Strontium + water hydrogen + strontium hydroxide


What numbers a - d are needed to balance the

equation?

Strontium + water hydrogen + strontium hydroxide

a Sr + b H2O c H2 + d Sr(OH)2

1 a=1 b=1 c=1 d=1

2 a=1 b=2 c=1 d=1

3 a=1 b=1 c=2 d=1

4 a=1 b=1 c=1 d=2


What is the mass of 2 moles of magnesium

nitrate Mg(NO3)2?

1. 86g

2. 134g

3. 148g

4. 296g


How many moles of iron atoms is 280g of iron?

(Fe=56)

1. One mole

2. Two moles

3. Four moles

4. Five moles


When iron rusts it forms the iron oxide Fe2O3.

What mass of oxygen reacts with 112g of iron?

(Fe=56 O=16)

1. 1g

2. 16g

3. 48g

4. 168g


Hydrogen reacts with chlorine to form hydrogen chloride HCl.

H2 + Cl2 2HCl

What mass of hydrogen chloride will be obtained from 4g of hydrogen gas?

(H=1 Cl=35.5)

1 36.5g

2 73g

3 109.5g

4 146g

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