1. Please look at the sheets attached to the questions that you completed on Friday.
If the sheet is titled “Progression” please start working.
If the sheet is titled “Revision” you need to pay attention to me right now.

2. If you don’t follow my instructions you will not get the RIGHT answer
Balancing Equations…
The main issues were as follows:
NOT assigning the correct charge
NOT checking your formulae
NOT remembering that water gives a hydroxide
If you don’t follow my instructions you will not get the RIGHT answer

3. Empirical Formulae…
The issue with these questions wasn’t the calculation of the empirical formulae. It was READING the question!

4. 4g of oxygen reacts with an excess of vanadium to form 9.094g of salt
The clue here is “EXCESS of vanadium”
This means that you need to subtract the 4g of oxygen from the total mass of the salt:
Mass V = mass salt – mass oxygen
= – 4
= 5.094g

5. 84. 31g of cadmium reacts with 100g of phosphorus gas, 53
84.31g of cadmium reacts with 100g of phosphorus gas, g of the phosphorus remains
The clue here is “53.545g of phosphorus REMAINS”
This means that you need to subtract the g of phosphorus from the 100g used.
Mass P = total mass P – mass P remaining
= 100 –
= g

6. Homework…
You are to research the way that the atomic model has changed from ancient Greece to the modern day.
You will undertake a task based on this research in Monday’s lesson.
Good sources of information include:
Textbooks (study centre)
Scientific blogs and journals
YouTube videos

7. Homework assessment… Collect Task Sheet A Complete
Check answers with your peers
Bring to me to sign off
Collect the next Task Sheet

8. The Mole

9. The mole is a way to simplify the way we look at quantities in reactions.
“One mole contains the same number of particles as in EXACTLY 12g of carbon-12”
This is given by the AVOGADRO constant.

11. The AVOGADRO constant
This is the number of particles in one mole of a substance.
NA = 6.02 * 1023 mol-1
To find the number of particles we multiply the number of moles by NA

12. The AVOGADRO constant
1) Calculate the number of particles in 1 mole of F-
No particles = 1 x (6.02 * 1023)
= 6.02 * 1023
2) Calculate the number of particles in 1 mole of F2

14. To calculate the number of moles we use one of two equations:
Moles = mass (g) * RFM
Moles = conc * volume (dm-3)
We can calculate any of these values by rearranging the above equations

15. Standard Solutions…
A standard solution contains EXACTLY the expected number of molecules.
It is used to calculate the concentration of unknown substances because its concentration is KNOWN.

16. Standard Solutions…
What is the mass of NaOH required to produce 250.0cm3 of a 0.500M standard solution?
Calculate the number of moles of NaOH
n = c * v
c = 0.5M v = 250/1000
= 0.25dm3
n = 0.5 * 0.25
= 0.125mol
Convert this to a mass
m = Mr * n
m = 40 * = 5.0g

17. A “Limiting Reactant” is the reactant with the smallest molar value.
Theoretical Yield…
The theoretical yield of a reaction is the EXPECTED mass (or volume) of a reaction that had proceeded to COMPLETION
i.e. there are no unreacted limiting reactants remaining
A “Limiting Reactant” is the reactant with the smallest molar value.

18. To calculate the TY you:
Theoretical Yield…
To calculate the TY you:
Write out the balanced equation
Calculate which is the limiting reactant
Calculate the moles of the limiting reactant
Compare the moles of the limiting reactant with the desired product
Convert the moles to a mass

19. A student mixes 5g of Sodium with 10g of water
A student mixes 5g of Sodium with 10g of water. What is the theoretical yield of sodium hydroxide?
Write out the balanced equation
Calculate which is the limiting reactant
2Na + 2H2O  2NaOH + H2
Na: n = 5/23 H2O: n = 10/18
= 0.217mol = 0.556mol
Sodium is the limiting factor as it has the lowest molar value.

20. A student mixes 5g of Sodium with 10g of water
A student mixes 5g of Sodium with 10g of water. What is the theoretical yield of sodium hydroxide?
Compare the moles of the limiting reactant with the desired product
Convert the moles to a mass
2Na + 2H2O  2NaOH + H2
Na : NaOH = 1 : 1
mass(NaOH) = n * Mr
= * 40
= 8.68

21. Percentage Yield…
The reason we calculate the theoretical yield is to determine how EFFICIENT a reaction is.
We can’t know if 5g is a good yield until we know that we were only able to obtain 5.2g maximum!
If we could have made 56g it would be a rubbish yield…

22. Percentage Yield… To calculate percentage yield all you have to do is:
% Yield = Actual Yield * 100
Theoretical Yield

23. Percentage Yield…
A chemist obtain 7.6g of aluminium sulphate in a reaction. The theoretical yield was 12g, calculate the percentage yield.
% Yield = Actual Yield * 100
Theoretical Yield
= (7.6 / 12.0) * 100
= * 100
= 63.3%

24. Gas Volumes…
It’s all very well calculating the mass of a product but if you are forming a gas the chances of you weighing it are unlikely!
In this case we measure volumes.

25. Gas Volumes…
At RTP one mole of gas is 24dm3 or 24000cm3
RTP is:
298K (25oC)
1atm

26. Gas Volumes…
To calculate the gas volume at RTP you simply multiply the moles by 24
Volume of 5mol of He
v = n * 24
= 5 * 24
= 120dm3

27. STP uses m3 instead of dm3 (/1000) STP uses Pa instead of atm
Gas Volumes…
At STP one mole of gas is 22.4dm3
STP is:
273K (0oC)
101325Pa
STP uses m3 instead of dm3 (/1000)
STP uses Pa instead of atm

28. This is the IDEAL GAS EQUATION
Gas Volumes…
There is a relationship between the volume, pressure, temperature and number of moles of a gas.
This is the IDEAL GAS EQUATION
pV=nRT

29. YOU MUST ALWAYS CONVERT TO THE RIGHT UNITS!
Gas Volumes…
pV=nRT
p = pressure (Pa)
V = volume (m3)
N = number of moles
T = temperature (K)
R = gas constant (8.314Jmol-1K-1)
YOU MUST ALWAYS CONVERT TO THE RIGHT UNITS!

30. The value for R never changes 
Gas Volumes…
The ideal gas equation can be used to find the pressure, volume, temperature and number of moles of a gas if you are given (can calculate) the other three terms.
The value for R never changes 

31. A gas occupies 350dm3 at STP. How many moles of gas are present?
Gas Volumes…
A gas occupies 350dm3 at STP. How many moles of gas are present?
Substitute into the ideal gas equation
Rearrange the equation
Solve
pV = nRT
* 0.35 = n * * 273
= n
n = /
n = mol

32. 4moles of gas occupy 2300dm3 at 150Pa. At what temp is this true?
Gas Volumes…
4moles of gas occupy 2300dm3 at 150Pa. At what temp is this true?
Substitute into the ideal gas equation
Rearrange the equation
Solve
pV = nRT
150 * 2.3 = 4 * * T
= * T
T = 345/33.256
n = 10.37K

33. Atom economy…
Percentage yield is useful is showing how well we perform a reaction.
BUT its doesn’t show how efficiently each atom in the reaction is used.

34. Atom economy…
Atom economy shows how well our reactions are using the (often very scarce) raw materials.
If more products are made than we are aiming for we have decreased our atom economy.

35. Atom economy…
The reactions below both produce compound X. Which of the three wastes less atoms?
A + B  X + Y
A  X + Y
A + B  X
If no “by-products” are formed atom economy is 100%

36. Atom economy…
To calculate the atom economy we compare the RFM of the desired product with that of ALL products
Atom economy = RFM desired * 100
RFM all

37. Atom economy…
Calculate the atom economy for the production of NaCl from NaOH and HCl.
Construct balanced equation
Calculate RFM of products
Calculate atom economy
NaOH + HCl  NaCl + H2O
NaCl = H2O = 18
Atom economy = 78.5/( ) * 100
= 81.3%

38. Atom economy…
There are a few rules to help you sense check your atom economies:
Addition reactions have an atom economy of 100%
Substitution/elimination reactions must have an atom economy of <100%

39. Water of Crystallisation…
If a crystal/compound is completely devoid of water it is ANHYDROUS
If a crystal/compound contains any water at all it is HYDRATED
Hydration occurs when water molecules form inside the crystalline structure of a compound during a reaction

40. Water of Crystallisation…
We show the extent of hydration using “dot formulae”
All this involves is showing the number of molecules of water per molecule of compound
Na2SO4.7H2O
7 molecules of water per Na2SO4

41. Co2N3O12H6 Water of Crystallisation…
We can calculate the ratio of the water molecules from the empirical formula.
Hydrated cobalt nitrate
Co2N3O12H6

42. Co2N3O12H6 Water of Crystallisation… Hydrogens = 6
Count the number of hydrogens
Calculate the number of waters
Determine the formula of the salt
Put it all together
Hydrogens = 6
Waters = hydrogens/2 = 3
Nitrate = NO3
Co2(NO3)3
Co2(NO3)3.3H2O

43. Water of Crystallisation…
We can also calculate the dot formula from experimental data:
Salt = magnesium sulphate
Hydrated salt has a mass of 4.312g
After heating it has a mass of 2.107g

44. Water of Crystallisation…
Calculate the mass of water
Calculate the moles of water
Calculate the moles of the salt
Find the ratio of salt to water
Put it all together
m = = g
n = 2.015/18 = mol
n = 2.017/120.4 = mol
0.0175: :7
MgSO4.7H2O