1. Moles Number of atoms/molecules ; Relative Molecular Mass ;
Volume occupied.
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2. Avogadro’s Number of particles = 6.0x1023
I mole of a substance is the amount of that substance which contains particles of that 6.0 x 1023 substance.
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Avogadro’s Number of particles = 6.0x1023

3. 1 mole of oxygen = 16g

4. Definitions
RMM in grams – Relative Molecular Mass The RMM of a compound = Sum of the molecular masses of the atoms in the compound.

5. RMM in Grams What is the RMM of each of the following
H2O = (2 x 1) + 16 = 18
(NH4)2SO4 = (14+1*4)* *4 = 132
What is the RMM of each of the following
H3PO4 ; Al2(SO4)3 ; CuSO4.5H2O?
H3PO4 = 1 * * 4 = 98
Al2(SO4)3 = 27 * 2 + (32 +16*4)*3 = 342
CuSO4.5H2O = *4+ 5(1*2 +16) = 250

6. Questions H3PO4 ; Al2(SO4)3 ; CuSO4.5H2O
How many moles in 10 grams of each?
What mass of each is needed to make 0.75 mole?
H3PO4 =
Al2(SO4)3 =
CuSO4.5H2O =

7. Questions H3PO4 ; Al2(SO4)3 ; CuSO4.5H2O
How many moles in 10 grams of each?
What mass of each is needed to make 0.75 mole?
H3PO4 = 10/98 = 0.1 mole
Al2(SO4)3 = 10/342 = mole
CuSO4.5H2O 10/250 = mole
H3PO4 = 0.75 of 98 = 73.5 g
Al2(SO4)3 = 0.75 of 342 = g
CuSO4.5H2O = 0.75 of 250 = g

8. Ammonium nitrate [NH4NO3] Calcium carbonate [CaCO3]
Avogadro’s Number = 6 x 1023
How many molecules are there in 5 g of each of the following?
Hydrogen gas [H2]
Methane [CH4]
Water [H2O]
Ammonium nitrate [NH4NO3]
Calcium carbonate [CaCO3]

9. Grams H2 1 mole 2 g 1 g 5 g
Molecules 6 * * 1023 / 2 5 * 6 * 1023/ * 1024

10. Ammonium nitrate [NH4NO3] Calcium carbonate [CaCO3]
Avogadro’s Number = 6 * 1023
How many atoms are there in 5 g of each of the following?
Hydrogen gas [H2]
Methane [CH4]
Water [H2O]
Ammonium nitrate [NH4NO3]
Calcium carbonate [CaCO3]

11. H2 = 1 * 2 = 2g 5/2 = 2.5 moles = 2.5 * 6 * 1023 = 15 * 1023 * 2 = 3.0 *1024
CH4 = 12 + (1*4) = 16 g5/16 = 0.31 mole = 0.31 * 6 * 1023 = 1.88 * 1023 * 5 = 9.4 * 1023
H2O = (1*2) + 16 = 18g 5/18 = 0.28 mole = 0.28 * 6 * 1023 = 1.68 * 1023 * 3 = 5.04 * 1023
NH4NO3 = 14 + (1*4) (16*3) = 80 g 5/80 = mole = * 6 * = * 1023 = 3.75 * 1022 * 9 = 3.38 * 1023
CaCO3 = (16*3) =100 = 5/100 = mole 0.05*6*1023 = 0.3*1023 = 3.0*1022*5 = 1.5 * 1023

12. 22.4 L of any gas at STP Volume occupied by each of the following
8g of Oxygen
32g methane
20g of neon
7.1g of chlorine
0.01g of hydrogen

13. O2 = 16 * 2 = 32 g  8/32 = 0.25 moles of 22.4 L = 5.6 L
CH4 = 12+ (1*4) = 16g  32/16 = 2 mole = 2 * 22.4 l = 44.8 L
Ne = 20 = 20g 20/20 = 1 mole = 22.4 L
Cl2 = 35.5 * 2 = 71 g  7.1/71 = 0.1 moles 0.1 of 22.4 L = 2.24 L

14. General rule for questions
Write down the two units
Put the one you are given the amount of first and the one you are asked the amount of second
Give the value of 1 mole of each
Bring the one you know the amount of down to 1 unit and divide the other by the molar amount
Multiply both by the amount given
Watch out for the number of atoms / molecules

15. Example How many atoms in 2.24L of methane at STP?
22.4L = 6 * 1023 molecules
1 L = 6 * 1023 / 22.4 molecules
2.24L = 2.24 * 6 * 1023 / 22.4
Methane is CH4 so 5 atoms per molecule
Atoms = 5 * 2.24 * 6 * 1023 / 22.4 = 3 * 1023

16. Race Find the number of grams of: 3.1 moles of sodium chloride
2.5 moles of MgSO4.7H2O
1.5mol. of chlorine gas (remember that chlorine is diatomic)
0.25mol. of sodium carbonate

17. Christmas Test Revision!
Structure = 10 Short Questions
3 Long Questions
Exam = 90%
Essay = 5%
Presentation = 5%
Atomic Structure – Calculating Protons, Neutrons and Electrons
Balancing Chemical Equations
Physical and Chemical Change
Isotopes – Calculations
Electronic Configuration
Mole Calculation