1.6.1b describe the formation of ions in terms of electron loss or gain AND 1.6.1c draw electron configuration diagrams of cations and anions using dots or crosses to represent electrons AND 1.6.1e describe ionic bonding as the result of strong net electrostatic attraction between ions

IONIC BONDING 
- ionic bonds form between a metal and a non-metal
- ionic bonding is the complete transfer of electrons between positive and negative ions. The positive ion transfers electrons to the negative ion to empty its outer shell and become stable, forming a strong electrostatic attraction between the ions.

You should know:
- that cations are ions that have lost electrons and that anions have gained electrons
- how to draw electronic configuration diagrams of cations and anions using dots or crosses to represent electrons
- to describe ionic bonding as the result of strong net electrostatic attraction between ions
- that the melting point of ionic substances is higher the stronger the bond and why
----------------------------------------------------------------------------------------------------------------------------------
PRACTICE QUESTIONS
1.i) Name the type of bonding in magnesium chloride.

ii) Draw a diagram (using dots or crosses) to show the bonding in magnesium chloride. Include all of the electrons and the charges present.








iii) Suggest why the melting temperature of magnesium oxide is higher than that of magnesium chloride, even though both are almost 100% ionic.
Magnesium chloride has an equation of MgCl2, as Mg has a charge of 2+ and Cl has a charge of 1-. Magnesium oxide has the equation MgO, as O has a charge of 2-. Therefore, MgCl2 has two individual sets of bonds, while MgO only has one with two pairs within it. This means that MgO has a higher melting point, as more energy is required to break the bond, as it has stronger electrostatic attraction.

2. The bonding in magnesium oxide, MgO, is:
- ionic
- metallic and ionic
- ionic and covalent
- metallic and covalent

3. Element R is in group 1 of the periodic table and element T is in group 6. R and T are not symbols for the elements.
i) The compound of R and T will have the formula:
- RT
- RT6
- RT2
- R2T

ii) The compound of R and T will have bonding which is predominantly:
- ionic
- covalent
- dative covalent
- metallic

iii) In terms of its electrical conductivity, the compound of R and T will:
- conduct when solid and liquid
- conduct when solid but not when liquid
- conduct when liquid but not when solid
- not conduct when solid or liquid

1.6.3a demonstrate an understanding that metals consist of giant lattices of metal ions in a sea of delocalised electrons AND 1.6.3b describe metallic bonding as the strong attraction between metal ions and the sea of delocalised electrons AND 1.6.3c use the models in 1.6.3a and 1.6.3b to interpret simple properties of metals, eg conductivity and melting temperatures

METALLIC BONDING

Metallic bonds are formed when metals give electrons to the sea of delocalized electrons, resulting in positive ions held together by negative electrons.
The strength of metallic bonding depends on two factors:
- the charge of the metal ions
The greater the charge on the metal ions, the greater the attraction between the ions and the delocalized electrons and the stronger the metallic bonds. A higher melting point is evidence of stronger bonds in the substance.
- the size of the metal ions
The smaller the metal ion, the closer the positive nucleus is to the delocalized electrons. This means there is a greater attraction between the two, which creates a stronger metallic bond.
----------------------------------------------------------------------------------------------------------------------------------
SODIUM AND MAGNESIUM
Na: melting point of 371K








Mg: melting point of 922K








----------------------------------------------------------------------------------------------------------------------------------
PRACTICE QUESTIONS
1.i) Describe the bonding in magnesium and explain why it is a good conductor of electricity.
Magnesium is a lattice structure of Mg2+ ions that have lost two electrons each to the sea of delocalized electrons. These delocalized electrons hold all of the magnesium ions together with electrostatic attraction, as opposite charges attract. This is metallic bonding.
Magnesium is a good conductor of electricity because of all of the delocalized electrons that hold it together, which are free to move around and carry electrical charge.

ii) State the type of bonding that exists in solid magnesium.
metallic bonding

iii) Explain fully why the melting point of magnesium is higher than that of sodium.
Magnesium loses 2 electrons to the sea of delocalised electrons to become stable, while sodium only loses 1 electron. Therefore, magnesium has more delocalised electrons between the magnesium ions to hold them together, resulting in greater electrostatic attraction in the lattice structure. As sodium has less delocalised electrons, it has weaker electrostatic attraction, so less energy is needed to break the bonds, so it has a lower melting point. Also, magnesium has a higher charge, so it is more strongly attracted to the delocalised electrons.

2. Metals usually have high melting points and boiling points because there are:
- strong attractions between the ions
- strong attractions between the delocalised electrons
- strong attractions between the ions and the delocalised electrons
- strong intermolecular forces

3. The melting temperatures of the elements of period 3 are given in the table below. Use these values to answer the questions that follow.

i) explain why the melting temperature of sodium is very much less than of magnesium.
The metallic bonds in sodium are weaker as the ions have a lower charge, so they have weaker attraction to the delocalised electrons. Also, there are fewer delocalised electrons in sodium than in magnesium, so there is less electrostatic attraction. As the forces of attraction are weaker in sodium, less energy is needed to break the bonds, so it has a lower melting point.

1.6.1c draw electron configuration diagrams of cations and anions using dots or crosses to represent electrons AND 1.6.2b draw electron configuration diagrams for simple covalently bonded molecules, including those with multiple bonds and dative covalent bonds, using dots or crosses to represent electrons

BONDING

You should be able to:
- draw dot and cross diagrams for cations and anions with electronic configurations
- describe how ions are formed by transfer of electrons
- construct ionic half equations for electrolysis
- explain why metals form positive ions and non-metals form negative ions
- predict charges of ions in the periodic table

Dative covalent bond: when both electrons in a covalent bond are from the same atom.

If you look at the periodic table, you can see that the group number (along the top) is the same as the number of outer shell electrons.

Group 1 metals all become 1+ ions, group 2 make ions with a 2+ charge, group 3 make ions with 3+ ions, group 7 make ions with a 2- charge, group 7 make ions with a 1- charge.

Transition metals have more complicated shells so they have several different + charges, depending on how oxidised they have become (eg. Fe2+ or Fe3+)

Group 0 or 8 are the noble gases, they are stable and inert and have full outer shells.

----------------------------------------------------------------------------------------------------------------------------------

PRACTICE QUESTIONS

1. Draw a diagram (using dots or crosses) for the ions in magnesium fluoride showing all the electrons and the ionic charges on:

i) the magnesium ion

ii) the fluoride ion

2. Chlorine forms compounds with magnesium and with carbon.

i) draw a dot and cross diagram to show the electronic structure of the compound magnesium chloride (only the outer electrons need to be shown). Include the charges present.

ii) draw a dot and cross diagram to show the electronic structure of the compound tetrachloromethane (only the outer electrons need to be shown).

3.i) Define the term 'covalent bond'.

A covalent bond is formed when electron pairs are shared between non-metal atoms. The bonded atoms form a molecule. Covalent bonds have no free electrons or ions, so they cannot conduct electricity, and they are mainly gases or liquids at room temperature, as they have low melting and boiling points.

ii) Nitrogen forms an oxide called nitrous oxide (N2O). The bonding in nitrous oxide can be represented as: N---N-O

(--- is a triple bond)
Complete the diagram below for the N2O molecule using dots or crosses to represent electrons. Only show all of the outer electrons.

1. 
   

1.3d calculate the amount of substance in a solution of known concentration

You should also be able to use the equation moles = volume x concentration and perform basic titration calculations.
----------------------------------------------------------------------------------------------------------------------------------
CALCULATIONS BASED ON CONCENTRATIONS IN SOLUTION
(These were summer homework so I only did odd questions)

Calculate the number of moles in each given volume of solution.
1. 25cm3 of 1.0mol dm-3 HCl
cm3 into dm3 = 25/1000 = 0.025
moles = volume x concentration = 0.025 x 1.0 = 0.025 moles
3. 250cm3 of 0.25mol dm-3 HCl
cm3 into dm3 = 250/1000 = 0.25
moles = volume x concentration = 0.25 x 0.25 = 0.0625 moles
5. 25cm3 of 1.0mol dm-3 NaOH
cm3 into dm3 = 25/1000 = 0.025
moles = volume x concentration = 0.025 x 1.0 = 0.025 moles
7. 50cm3 of 0.25mol dm-3 HNO3cm3 into dm3 = 50/1000 = 0.050
moles = volume x concentration = 0.050 x 0.25 = 0.0125 moles
9. 25cm3 of 0.05mol dm-3 KMnO4
cm3 into dm3 = 25/1000 = 0.025
moles = volume x concentration = 0.025 x 0.05 = 0.00125 moles

Calculate the mass of material in the given volume of solution.
11. 25cm3 of 1mol dm-3 HCl
cm3 into dm3 = 25/1000 = 0.025
moles = volume x concentration = 0.025 x 1 = 0.025 moles
mass = moles x RFM = 0.025 x (1+35.5) = 0.9125g 
13. 100cm3 of 0.25mol dm-3 NH4NO3
cm3 into dm3 = 100/1000 = 0.1
moles = volume x concentration = 0.1 x 0.25 = 0.025 moles
mass = moles x RFM = 0.025 x (4+48+28) = 2g
15. 25cm3 of 1mol dm-3 BaCl2
cm3 into dm3 = 25/1000 = 0.025
moles = volume x concentration = 0.025 x 1 = 0.025 moles
mass = moles x RFM = 0.025 x (137.3+71) = 5.21g
17. 20cm3 of 0.1mol dm-3 NaOH
cm3 into dm3 = 25/1000 = 0.020
moles = volume x concentration = 0.020 x 0.1 = 0.002 moles
mass = moles x RFM = 0.002 x (23+16+1) = 0.08g
19. 25cm3 of 0.02mol dm-3 KMnO4
cm3 into dm3 = 25/1000 = 0.025
moles = volume x concentration = 0.025 x 0.02 = 0.0005 moles
mass = moles x RFM = 0.0005 x (39.1+55+64) = 0.0791g 

Calculate the concentration in mol dm-3 of the following.
21. 3.65g of HCl in 1000cm3 of solution
cm3 into dm3 = 1000/1000 = 1
moles = mass/RFM = 3.65/(1+35.5) = 0.1 moles
concentration = moles/volume = 0.1/1 = 0.1mol dm-3
23. 6.624g of Pb(NO3)2 in 250cm3 of solution
cm3 into dm3 = 250/1000 = 0.25
moles = mass/RFM = 6.624/(207.2+28+96) = 0.02
concentration = moles/volume = 0.02/0.25 = 0.08mol dm-3 (I'm not sure why this is wrong)
25. 1.962g of H2SO4 in 250cm3 of solution
cm3 into dm3 = 250/1000 = 0.25
moles = mass/RFM = 1.962/(2+32.1+64) = 0.02 moles
concentration = moles/volume = 0.02/0.25 = 0.08mol dm-3 (I'm not sure why this is wrong either, even though they're identical)
27. 25g of Na2SO3.5H2O in 250cm3 of solution
cm3 into dm3 = 250/1000 = 0.25
moles = mass/RFM = 25/(46+32.1+10+128) = 0.116 moles
concentration = moles/volume = 0.116/0.25 = 0.464mol dm-3 (This one is wrong too, for some reason)
29. 4.80g of (COOH)2.2H2O in 250cm3 of solution
cm3 into dm3 = 250/1000 = 0.25
moles = mass/RFM = 4.80/(6+64+12) = 0.059 moles
concentration = moles/volume = 0.059/0.25 = 0.236mol dm-3 (also wrong, if anyone has any idea why, please comment)
----------------------------------------------------------------------------------------------------------------------------------
SIMPLE VOLUMETRIC CALCULATIONS

Calculate the molarity (the number of moles) of the first named solution.
1. 25cm3 of sodium hydroxide reacts with 21cm3 of 0.2 mol dm-3 HCl
NaOH + HCl --> NaCl + H2O
cm3 into dm3 = 21/1000 = 0.021
moles = concentration x volume = 0.021 x 0.2 = 0.0042 moles
- ratio
NaOH:HCl, 1:1, 0.0042
cm3 into dm3 = 25/1000 = 0.025
concentration = moles/volume = 0.0042/0.025 = 0.168 mol dm-3 
3. 20cm3 of hydrochloric acid reacts with 23.6cm3 of 0.1 mol dm-3 NaOH
HCl + NaOH --> NaCl + H2O
cm3 into dm3 = 23.6/1000 = 0.0236
moles = concentration x volume = 0.0236 x 0.1 = 0.00236 moles
- ratio
HCl : NaOH, 1:1, 0.00236 
cm3 into dm3 = 20/1000 = 0.020
concentration = moles/volume = 0.00236/0.020 = 0.118 mol dm-3 
5. 25cm3 of nitric acid reacts with 15cm3 of a solution of 0.2 mol dm-3  NH4OH
HNO3 + NH4OH --> NH4NO3 + H2O
cm3 into dm3 = 15/1000 = 0.015
moles = concentration x volume = 0.015 x 0.2 = 0.003 moles
- ratio
HNO3 : NH4OH, 1:1, 0.003 
cm3 into dm3 = 25/1000 = 0.025
concentration = moles/volume = 0.003/0.025 = 0.12 mol dm-3 
7. 25cm3 of a solution of NaCl reacts with 10cm3 of a 0.02 mol dm-3 silver nitrate
NaCl + AgNO3 --> NaNO3 + AgCl
cm3 into dm3 = 10/1000 = 0.010
moles = concentration x volume = 0.010 x 0.02 = 0.0002 moles
- ratio
NaCl : AgNO3, 1:1, 0.0002 
cm3 into dm3 = 25/1000 = 0.025
concentration = moles/volume = 0.0002/0.025 = 0.008 mol dm-3 
9. 25cm3 of HxA reacts with 25cm3 of 0.2 mol dm-3 NaOH to give Na2A
HxA + NaOH --> Na2A + H2O
cm3 into dm3 = 25/1000 = 0.025
moles = concentration x volume = 0.025 x 0.2 = 0.005 moles
- ratio
HxA : NaOH, 1:1, 0.005 
cm3 into dm3 = 25/1000 = 0.025
concentration = moles/volume = 0.005/0.025 = 0.02 mol dm-3 (this is probably due to me not working out the x)

1.3b write balanced equations (full and ionic) for simple reactions, including the use of state symbols

An important principle in chemical reactions is that matter cannot be created or destroyed. Therefore it is important that symbol equations are balanced.
A balanced equation has the same number of each type of atom of each side of the equation.
eg.
unbalanced: Na + Cl2 --> NaCl
balanced: 2Na + Cl2 --> 2NaCl
This shows that two moles of sodium react with one mole of chlorine to make two moles of sodium chloride.

Equations containing ions should have the same overall charge on each side in order to be balanced. This can be achieved by balancing in the normal way.
eg.
unbalanced: Ca2+ + Cl- --> CaCl2
(overall charge of +1 --> overall charge of 0)
balanced: Ca2+ + 2Cl- --> CaCl2
(overall charge of 0 --> overall charge of 0)

State symbols are letters that are added to a formula to indicate what state each reactant and product is in. The four state symbols are: s, solid; l, liquid; g, gas; aq, aqueous
These are added after the formula in brackets and subscript.
eg. 2H2(g) + O2(g) --> 2H2O(g)

----------------------------------------------------------------------------------------------------------------------------------
IONIC EQUATIONS
Turning a word equation into a symbol equation, adding state symbols, and balancing (charges are in bold and italic):
iron (II) sulphate + sodium hydroxide --> iron (II) hydroxide + sodium sulphate
Fe2+SO42- + Na+OH- --> Fe2+OH- + Na+SO42-
FeSO4 + NaOH --> Fe(OH)2 + Na2SO4
FeSO4 + 2NaOH --> Fe(OH)2 + Na2SO4
FeSO4(aq) + 2NaOH(aq) --> Fe(OH)2(s) + Na2SO4(aq)
Turning everything into an ionic equation:
Fe2+(aq)SO42-(aq) + 2Na+(aq)2OH-(aq) --> Fe2+(s)2OH-(s)+ 2Na+(aq)SO42-(aq)
Remove spectator ions (ions which did not change):
Fe2+(aq) + 2OH-(aq) --> Fe2+(OH-)2(s)

A step-by-step breakdown of how to write an ionic equation:
1. write the word equation out as a symbol equation
2. add charges
3. balance the equation
4. add state symbols (remember your solubility rules)
5. separate the ions
6. remove spectator ions
7. rewrite and "tidy up" the equation
----------------------------------------------------------------------------------------------------------------------------------
PRACTICE QUESTIONS
They were supposed to be from pg. 82 of the student workbook but the document was not included.

1.3i analyse and evaluate the results obtained from finding a formula or confirming an equation by experiment, eg the reaction of lithium with water and deducing the equation from the amounts in moles of lithium and hydrogen

CALCULATION OF FORMULAE FROM EXPERIMENTAL DATA
Calculate the empirical formula of the compound from the given data. This may be as a percentage composition or as the masses of materials found in an experiment. (This was my summer homework, and i was only required to do questions 5, 6, and 10 in each section)

5. Pb 90.66%, O 9.34%
- assume that 100g of the compound is present
Pb 90.66g, O 9.34g
- convert masses to moles
moles = mass/RAM = 90.66g/207 = 0.438 moles
                                = 9.34g/16 = 0.584 moles
- divide both by the lowest moles
0.438/0.438 = 1
0.584/0.438 = 1.333
Pb3O4

6. H 3.66%, P 37.8%, O 58.5%
H 3.66g, P 37.8g, O 58.5g
- convert masses to moles
moles = mass/RAM = 3.66/1 = 3.66 moles
                                = 37.8/31 = 1.22 moles
                                = 58.5/16 = 3.66 moles
- divide by the lowest moles
3.66/1.22 = 3
1.22/1.22 = 1
3.66/1.22 = 3
H3PO3

10. H 5.88%, O 94.12%
H 5.88g, O 94.12
- convert masses into moles
moles = mass/RFM = 5.88/1 = 5.88
                               = 94.12/16 = 5.88
- divide by the lowest amount of moles
5.88/5.88 = 1
5.88/5.88 = 1
HO (H2O2)

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

5. 50cm3 of a solution of citric acid, Mr = 192, containing 19.2dm-3 reacted with 50cm3 of a solution of sodium hydroxide containing 12gdm-3. Citric acid can be represented by the formula HxA, where x represents the number of hydrogen atoms in the molecule. Use the data above to calculate the number of moles of sodium hydroxide that react with one mole of citric acid and hence find the value of x.

6. When 12.475g of hydrated copper (II) sulphate, CuSO4.xH2O, was heated, 7.980g of anhydrous salt was produced. Use the data to find the value of x and write out the equation for the reaction.
- calculate the masses of the copper (II) sulphate, the anhydrous salt, and the water
hydrated copper (II) sulphate: 12.475g
anhydrous copper (II) sulphate: 7.980g
water: 4.495g
- calculate the moles 
moles = mass/RFM = 7.980/(63.5+32.1+64) = 0.05 moles
                               = 4.495/(2+16) = 0.25 moles
- divide by the smallest moles
0.05/0.05 = 1
0.25/0.05 = 5
x = 5
CuSO4.5H2O --> CuSO4 + 5H2O

10. When 13.9g of FeSO4.xH2O is heated, 4g of solid iron (III) oxide is produced, together with the loss of 1.6g of sulphur dioxide and 2g of sulphur trioxide. The rest of the mass loss being due to the water of crystallisation lost. Use the data to write out the full equation for the action of heat.
FeSO4.xH2O: 13.9g
Fe: 4g
SO3: 2g
SO2: 1.6g
H2O: 6.3g
13.9g - 6.3 = 7.6g
- calculate the moles 
moles = mass/RFM = 7.6/(55.8+32.1+64) = 0.05 moles
                               = 6.3/(2+16) = 0.35 moles
- divide by the smallest amount of moles
0.05/0.05 = 1
0.35/0.05 = 7
x = 7
FeSO4.7H2O --> Fe2O3 + SO3 + SO2 + 14H2O

1.3e use chemical equations to calculate reacting masses and vice versa using the concepts of amount of substance and molar mass AND 1.3f use chemical equations to calculate volumes of gases and vice versa using the concepts of amount of substance and molar volume of gases, eg calculation of the mass or volume of CO2 produced by combustion of a hydrocarbon (given a molar volume for the gas)

You should know how to calculate reacting masses using this equation. For example:
What mass of magnesium oxide would be produced from 16g of magnesium in the reaction between magnesium and oxygen?
The equation is 2Mg + O2 --> 2MgO
You know that you have 16g of magnesium, so you can use this to calculate the moles of magnesium.
Moles = mass/RAM = 16g/24 gmol-1 = 0.67 moles
Next, you find out the ratio:
Mg : MgO, 2 : 2, 1:1, therefore both have 0.67 moles.
Therefore, the mass of MgO can be calculated with moles x RAM
Mass = moles x RAM = 0.67 x (24+16) = 26.67g

With the same question, you should be able to calculate the volume of O2 required. 

You already know how many moles of Mg you have (0.67 moles).

However, the ratio is different as you are trying to work out volume of oxygen, not magnesium oxide. Therefore, the ratio of Mg : O is 2 : 1, so the amount of moles of oxygen is half of 0.67, 0.034.

1 mole of a gas is 24dm3, so to work out the volume of oxygen, you have to do 0.034 x 24dm3, which equals 0.804dm3.

----------------------------------------------------------------------------------------------------------------------------------
CALCULATIONS OF PRODUCTS/REACTANTS BASED ON EQUATIONS
1. What mass of barium sulphate would be produced from 10g of barium chloride in the following reaction?
BaCl2 + H2SO4 --> BaSO4 + 2HCl
- calculate moles of barium chloride
moles = mass/RFM = 10g/(137+71) = 0.048 moles
- ratio
BaCl2 : BaSO4, 1 : 1, so both have 0.048 moles
- calculate mass of barium sulphate
mass = moles x RFM = 0.048 x (137+32+64) = 11.2g

2. What mass of potassium chloride would be produced from 20g of potassium carbonate?

3. What masses of ethanol and ethanoic acid would need to react together to give 1g of ethyl ethanoate?

4. What mass of iron (III) oxide would need to be reduced to produce 100 tonnes of iron in a blast furnace?

5. What mass of silver nitrate as a solution in water would need to be added to 5g of sodium chloride to ensure complete precipitation of the chloride?
AgNO3(aq) + NaCl(aq) --> AgCl(s) + NaNO3(aq)
- calculate the moles of sodium chloride
moles = mass/RFM = 5g/(23+35) = 0.086 moles
- ratio
NaCl : AgNO3, 1:1, therefore they have the same amount of moles
- calculate the mass of silver nitrate
mass = moles x RFM = 0.086 x (108+14+48) = 14.62g

6. A solution of copper sulfate reacts with sodium hydroxide solution to produce a precipitate of copepr hydroxide according to the following equation:
CuSO4(aq) + 2NaOH(aq) --> Cu(OH)2(s) + Na2SO4(aq)
What mass of sodium hydroxide would be needed to convert 15.96g of copper sulfate to copper hydroxide and what mass of copper hydroxide would be produced?
- calculate the moles of copper sulphate
moles = mass/RFM = 15.96g/(64+32+64) = 0.093 moles
- ratio
CuSO4 : 2NaOH, 1:2, 0.093 : 0.186
- calculate the mass of sodium hydroxide
mass = moles x RFM = 0.186 x (23+16+1) = 7.44g


7. What volume of ammonia gas would be needed to produce 40g of ammonium nitrate in the following reaction?
NH3(g) + HNO3(aq) --> NH4NO3(aq)
- calculate moles of ammonium nitrate
moles = mass/RFM = 40g/(28+4+48) = 0.5 moles
- ratio
NH3 : NH4NO3, 1:1, 0.5 moles
- calculate volume
moles x 24dm3 = 0.5 x 24dm3 = 12dm3 or 12000cm3

8. In the reaction between calcium carbonate and nitric acid what mass of calcium nitrate and what volume of carbon dioxide would be produced from 33.3g of calcium carbonate?

9. What would be the total volume of gas produced by the action of heat on 33.12g of lead (II) nitrate?

10. Magnesium reacts with sulfuric acid to produce a solution of magnesium sulphate. If this is allowed to crystallize out the solid produced has the formula MgSO4.7H20. Write out the equation for this reaction and calculate the mass of magnesium sulphate heptahydrate that could be produced from 4g of magnesium.
Mg(s) + H2SO4(aq) --> MgSO4(aq) + H2(g)
Mg(s) + H2SO4(aq) + 7H20(l)--> MgSO4.7H2O(s) + H2(g)
- calculate the moles of magnesium
moles = mass/RFM = 4g/24 = 0.167 moles
- ratio
Mg : MgSO4.7H2O, 1:1, 0.167 moles for both
- calculate the mass of magnesium sulphate heptahydrate
mass = moles x RFM = 0.167 x (24+32+176+14) = 41.082g

11. Copper (II) oxide reacts with sulphuric acid to produce copper (II) sulphate. If this is allowed to crystallize the formula of the crystals is CuSO4.5H2O.
What mass of copper oxide would be needed to produce 100g of crystals?
CuO + H2SO4 + 4H2O --> CuSO4.5H2O
- calculate moles of crystals
moles = mass/RFM = 100g/(64+32+144+10) = 0.4 moles
- ratio
CuSO4.5H2O : CuO, 1:1, 0.4 moles
- calculate the mass of copper oxide
mass = moles x RFM = 0.4 x (64+16) = 32g

12. Sulphur dioxide can be removed from the waste gases of a power station by passing it through a slurry of calcium hydroxide. The equation for this reaction is:
SO2(g) + Ca(OH)2(aq) --> CaSO3(aq) + H2O(l)
What mass of calcium hydroxide would be be needed to deal with 1000dm3 of sulphur dioxide?

13. In a fermentation reaction, glucose is converted to alcohol and carbon dioxide according to the following equation:
C6H12O6 --> 2C2H5OH + 2CO2
What mass of alcohol and what volume of carbon dioxide would be produced from 10g of glucose?
- calculate moles of glucose
moles = mass/RFM = 10g/(72+12+96) = 0.056 moles
- ratio
C6H12O6 : 2C2H5OH, 1:2, 0.056:0.111
- calculate mass of alcohol
mass = moles x RFM = 0.111 x (24+6+16) = 5.106g
- ratio 
C6H12O6 : 2CO2, 1:2, 0.056:0.111
- calculate volume
volume = moles x 24dm3 = 0.111 x 24dm3 = 2.664dm3

14. In the following reactions calculate the mass of precipitate formed from 20g of the metal salt in each case.
i) ZnSO4 + 2NaOH --> Zn(OH)2 + Na2SO4
- calculate the moles of ZnSO4
moles = mass/RFM = 20g/(65+32+64) = 0.124 moles
-ratio
ZnSO4 : Zn(OH)2, 1:1, 0.124 moles
- calculate the mass of Zn(OH)2
mass = moles x RFM = 0.124 x (65+2+32) = 12.276g
ii) Al2(SO4)3 + 6NaOH --> 2Al(OH)3 + 3Na2SO4
- calculate the moles in Al2(SO4)3
moles = mass/RFM = 20g/(54+96+192) = 0.058 moles
- ratio
Al2(SO4)3 : 2Al(OH)3, 1:, 0.058 moles:0.117
- calculate the mass of 2Al(OH)3
mass = moles x RFM = 0.117 x (27+3+48) = 9.126g
iii) MgSO4 + 2NaOH --> Mg(OH)2 + Na2SO4
- calculate the moles of MgSO4
moles = mass/RFM = 20g/(24+32+64) = 0.167 moles
- ratio
MgSO4 : Mg(OH)2, 1:1, 0.167 moles
- calculate the mass of Mg(OH)2
mass = moles x RFM = 0.167 x (24+2+32) = 9.686g

15. What volume of hydrogen would be produced by 1g of calcium in its reaction with water?

16. What mass of magnesium would be needed to produce 100cm3 of hydrogen?

17. Chlorine reacts with sodium hydroxide as follows:
Cl2 + 6NaOH --> 5NaCl + NaClO3 + 3H2O
What mass of sodium chloride and what mass of sodium (V) chlorate would be produced from 240cm3 of chlorine gas?
- calculate moles of chlorine
moles x 24dm3 = volume of a gas, so moles = volume of a gas/24dm3
moles = volume of a gas/24dm3 = 2.4dm3/24dm3 = 0.1 moles
- ratio 
Cl2 : 5NaCl, 1:5, 0.1:0.5
- calculate mass of 5NaCl
mass = moles x RFM = 0.5 x (23+35) = 29g (the answer should be 2.9g, I think that this is a mistake a made when I was calculating the moles from the volume. Instead of calculating it all in cm3, I calculated it all in dm3)
- ratio
Cl2 : NaClO3, 1:1, 0.1 moles
- calculate the mass of NaClO3
mass = moles x RFM = 0.1 x (23+35+48) = 10.6g (again, this answer is almost right)

18. When nitrogen reacts with hydrogen in the Haber process only 17% of the nitrogen is converted to ammonia. What volume of nitrogen and what volume of hydrogen would be needed to produce 1 tonne of ammonia? (1 tonne = 1x106g)

19. Nitric acid is produced by the following series of reactions:
4NH3 + 5O2 --> 4NO + 3H2O
4NO + 02 --> 4NO2
4NO2 + O2 + 2H2O --> 4HNO3
What mass of nitric acid would be produced from 17 tonnes of ammonia and what volume of oxygen would be needed in the reaction?
- calculate the moles of ammonia
moles = mass/RFM = 17t/(14+3) = 1 mole
- ratio
4NH3 : 4HNO3, 4:4, 1:1, 1 mole
- calculate the mass of nitric acid
mass = moles x RFM = 1 x (1+14+48) = 720 tonnes (this is completely wrong, the answer should be 63 tonnes, but I'm not sure what went wrong)
- ratio
4NH3 : O2, 4:1, 1:0.25 moles
- calculate the volume of oxygen
volume = moles x 24dm3 = 0.25 x 24dm3 = 6dm3 (this is also extremely wrong)

20. Hardness in water is caused by dissolved calcium compounds. When heated, some of these break down and deposit calcium carbonate as follows:
Ca(HCO3)2 --> CaCO3 + H2O +CO2
This builds up as "fur" on the inside of boilers. It can be removed by reaction with hydrochloric acid.
What mass of calcium carbonate would be produced from 10,000dm3 of water containing 0.356g of calcium hydrogen carbonate per dm3 of water and what volume of 10 mol dm-3 hydrochloric acid solution would be needed to remove the solid calcium carbonate from the inside of the boiler?

----------------------------------------------------------------------------------------------------------------------------------
REACTIONS INVOLVING GASES
Assume that you have 10cm3 of the first named reactant and then calculate the volumes of all the gases involved in the equation. In these examples, the reactions are being carried out at above 100 degrees Celsius and you should assume the water is present as a gas and therefore has a volume.

1. CH4 + 2O2 --> CO2 + 2H2O
CH4: 10cm3 
O2: 20cm3 
CO2: 10cm3
H2O: 20cm3

2. C2H4 + 3O2 --> 2CO2 +2H2O
C2H4: 10cm3 
O2: 30cm3 
CO2: 20cm3 
H2O: 20cm3 

3. 2C2H2 +5O2 --> 4CO2 + 2H2O
C2H2: 10cm3 
O2: 25cm3 
CO2: 20cm3 
H2O: 10cm3 

4. 2C8H18 + 25O2 --> 16CO2 + 18H2O
C8H18: 10cm3 
O2: 125cm3 
CO2: 80cm3 
H2O: 90cm3 

5. N2 + 3H2 --> 2NH3
N2: 10cm3 
H2: 30cm3 
NH3: 20cm3 

1.3h. Demonstrate an understanding of, and be able to perform, calculations using the Avogadro constant

The value for Avogadro's Number is 6.022 x 10²³ mol¯¹.

USING THIS DIAGRAM
in one direction, and then the other, with one step, and with two steps
Example questions:

1. 0.450 mole (or gram) of Fe contains how many atoms?
2. 0.200 mole (or gram) of H₂O contains how many molecules?

Example #1: 0.450 mole of Fe contains how many atoms?

Solution: You have the amount of moles (0.450), so you start from the "moles of substance" box, and skip right to the "number of atoms or molecules" box. Therefore, you need to do: 0.450 mol x 6.022 x 10²³ mol¯¹ which is equal to 

Example #2: 0.200 mole of H₂O contains how many molecules?

Solution: Again, you know the amount of moles (0.200), so you start from the "moles of substance" box and skip right to the "number of atoms or molecules" box, so to calculate the molecules you need to do: 0.200 mol x 6.022 x 10²³ mol¯¹ which gives you 

Here are the same two problems as before, but with gram replacing mole:

1. 0.450 gram of Fe contains how many atoms?
2. 0.200 gram of H₂O contains how many molecules?

Look at the solution steps and you'll see we have to do two steps of working to go from grams (on the left) across to the right through moles and then to the number of atoms or molecules. 
So, the first one it would be like this:

Step One: 0.450 g divided by 55.85 g/mol = 0.00806 mol

Step Two: 0.00806 mol x 6.022 x 10²³ atoms/mol, giving us 

and for the second, we have:

Step One: 0.200 g divided by 18.0 g/mol = 0.0111 mol

Step Two: 0.0111 mol x 6.022 x 10²³ molecules/mol which would give us

the opposite direction

Now, let's see how well you can do the opposite direction. The first two are the one-step type, the second two are the two-step type, and the last two are slightly more challenging, to test you. (I have already attempted these questions)

1) Calculate the number of molecules in 1.058 mole of H₂O

1.058 x Avogadro's number = 6.371 x 10²³ molecules/mol

2) Calculate the number of atoms in 0.750 mole of Fe

0.750 x Avogadro's number = 4.52 x 10²³ atoms/mol

3) Calculate the number of molecules in 1.058 gram of H₂O

1.058g divided by 18g/mol = 0.059 mol
0.059 mol x Avogadro's number = 
molar mass is calculated by atomic mass of hydrogen x 2, as there are two atoms of hydrogen, + atomic mass of oxygen)

4) Calculate the number of atoms in 0.750 gram of Fe
0.750g divided by 55.85g/mol = 0.0134 mol
0.0134 x Avogadro's number =

5) Which contains more molecules: 10.0 grams of O₂ or 50.0 grams of iodine, I₂?

O₂
10g divided by 31.9g/mol = 0.313 mol
0.313 x Avogadro's number =
I₂
50g divided by 253.8g/mol = 0.197 mol
0.197 x Avogadro's number =
Therefore, ... contains more molecules.

6) A solution of ammonia and water contains 2.10 x 10²⁵ water molecules and 8.10 x 10²⁴ ammonia molecules. How many total hydrogen atoms are in this solution?
water = H20
ammonia = NH3

----------------------------------------------------------------------------------------------------------------------------------

UNITS

The unit on Avogadro's Number is mol¯¹ and you would say "per mole" out loud. The question then is WHAT per mole?

The answer depends on the problem. In the first example, I used iron, an element. Almost all elements come in the form of individual atoms, so the correct numerator with most elements is "atoms", so mol¯¹ atoms or atoms/mol.

So, doing the calculation and rounding off to three significant figures, we get 2.71 x 10²³ mol¯¹ atoms. Notice "atoms" never gets written until the end. It is assumed to be there in the case of elements. If you wrote Avogadro's Number with the unit atoms/mole in the problem, you would be correct.

The same type of discussion applies to substances which are molecular in nature, such as water. So the numerator I use here is "molecule" and the problem answer is 1.20 x 10²³ molecules.

Once again, the numerator part of Avogadro's Number depends on what is in the problem. Other possible numerators include "formula units," ions, or electrons. These, of course, are all specific to a given problem. When a general word is used, the most common one is "entities," as in 6.022 x 10²³ entities/mol.

----------------------------------------------------------------------------------------------------------------------------------

MORE COMPLEX PROBLEMS 

Problem #7: How many atoms of chlorine are in 16.50 g of iron(III) chloride?

Solution:

1) Determine moles of FeCl₃:

16.50 g / 162.204 g mol¯¹ = 0.101723755 mol

2) Determine how many formula units of iron(III) chloride are in 0.1017 mol:

0.101723755 mol x 6.022 x 10²³ = 6.1258 x 10²²

3) Determine number of Cl atoms in 6.1258 x 10²² formula units of FeCl₃:

6.1258 x 10²² x 3 = 1.838 x 10²³ (to 4 sig fig)

---

Problem #8: How much does 100 million atoms of gold weigh?

Solution:

1) Determine moles of gold in 1.000 x 10⁸ (we'll assume four sig figs):

1.000 x 10⁸ atoms divided by 6.022 x 10²³ atoms/mol = 1.660577881 x 10¯¹⁶ mol

2) Determine grams in 1.66 x 10¯¹⁶ mol of gold:

1.660577881 x 10¯¹⁶ mol times 196.97 g/mol = 3.271 x 10¯¹⁴ g (to four sig fig)

----------------------------------------------------------------------------------------------------------------------------------
PRACTICE QUESTIONS 
(these were my summer homework, so I have attempted them)

1)       The mass of one proton is 1.6725 x 10^-24 g.  Calculate the mass of one mole of ¹H⁺ ions given that the Avogadro constant = 6.0225 x 10²³.
6.0225x10²³ x 1.6725 x 10^-24 = 1.0073g

2)       The mass of one atom of ¹⁹F is 3.1546 x 10^-23 g.  Use this and the Avogadro constant (6.0225 x 10²³) to find the mass of one mole of ¹⁹F atoms.
6.0225x10²³ x 3.1546x10^-23g = 18.999g

3)       The mass of one mole of ¹²⁷I atoms is 126.9045 g.  Given that the Avogadro constant equals 6.0225 x 10²³ and the mass of one electron equals 9.1091 x 10^-28 g, calculate the mass of one mole of ¹²⁷I^- ions.
mass of 1 mole of ¹²⁷I^- ions + mass of one electron = 126.9045 g + (9.1091 x 10^-28 g)(6.0225 x 10^{23)
= 126.905g}

4)  a)  Calculate the Avogadro constant given that the mass of one mole of ¹²C atoms is 12.000 g while the mass of one ¹²C atom is 1.9925 x 10^-23 g.
12/1.9925 x 10^-23 g = 60225 x 10^-23

     b)  The relative atomic mass of carbon is 12.011 and not 12.000.  Explain this difference.
Carbon occurs in isotopes such as carbon-12, carbon-13 and carbon-14, so the RAM is an average of these.

5)       Calculate the expected mass of one mole of a particles (⁴He²⁺ ions) given that the mass of one proton is 1.6725 x 10^-24 g, the mass of one neutron is 1.6748 x 10^-24 g and the Avogadro constant is 6.0225 x 10²³.
one particle (⁴He²⁺ ions) = 2 neutrons and 2 protons
mass of 1 mole of particles = 6.0225 x 10²³ x 2(1.6725 x 10^-24 g + 1.6748 x 10^-24 g) = 4.0138g

6)  a)  The mass of one mole of ²⁷Al atoms is 26.9185 g.  Given that the Avogadro constant equals 6.0225 x 10²³ and the mass of one electron equals 9.1091 x 10^-28 g, calculate the mass of one mole of ²⁷Al³⁺ ions.
²⁷Al³⁺ ions have 3 less electrons than ²⁷Al atoms
mass of 1 mole of ²⁷Al³⁺ ions = mass of ²⁷Al atoms - 3(mass of electron) = 26.9185 - (6.0225 x 10^{23 x 9.1091 x} 10^{-28 x 3) = 26.9169g}

     b)  Calculate the percentage difference between the mass of ²⁷Al atoms and ²⁷Al³⁺ ions, and use this to show why the mass of electrons are negligible in normal laboratory use of aluminium and other chemicals.
difference in mass = 26.9185 - 26.9169 = 0.0016g
percentage difference = 0.0016/26.9185 x 100 = 0.00597%