Relative Atomic Mass (Ar)

- Relative atomic mass (Ar) is the weighted average mass of an atom of an element relative to 1/12th the mass of a carbon-12 atom.

- All relative atomic masses in the periodic table consider that element's abundance. That's why sometimes you end up with a decimal, such as in chlorine, where the relative atomic mass is 35.5.

Formula:

Examples:

1. Copper has 2 isotopes; 69% of copper atoms are Cu⁶³, and the rest are Cu⁶⁵. Calculate the Ar of copper.

2. Element X has 3 isotopes; calculate the Ar, given to 1 d.p.

Mass no.	% abundance
24	79.00
25	10.00
26	11.00

Relative Formula Mass (Mr)

- To calculate the relative formula mass (Mr), we add up all the relative atomic masses (Ar) of the atoms in the substance using a periodic table.

Examples:

1. NaCl : 23 + 35.5 = 58.5

2. Ca(OH)₂ = 40 + (2 x 16) + (2 x 1) = 74

3. MgSO₄·7H₂O = 24 + 32 + (4 x 16) + (7 x 2) + (7 x 16) = 246

4. Mg(NO₃)₂ = 24 + (2 x 14) + (2 x 16 x 3) = 180

5. Sodium sulfate = Na₂SO₄ = (23 x 2) + 32 + (16 x 4) = 142

Percentage by Mass

- To calculate the % by mass of an element in a compound:

Examples:

1) % by mass of C in Cr(CO)₆

52 + (6 x 12) + (6 x 16) = 220

2) % by mass of O in CuSO₄·5H₂O

16 x 9 = 144

3. % by mass of Na in Na₂CO₃

23 x 2 = 46

The Mole

- In chemistry, the mole is a measure of the amount of substance.
- The unit of the mole is mol:
- 1 mol = 6.02 x 10²³ particles (e.g., atoms, molecules, ions)
- It's easier to count in moles than individual particles.
E.g.
1 mole = 6.022 x 10²³ particles
2 moles = 6.022 x 10²³ x 2 = 1.2044 x 10²⁴ particles
3 moles = 6.022 x 10²³ x 3 = 1.8066 x 10²⁴ particles
- You can use expressions such as:
→ A mole of oxygen gas.
→ 3 moles of copper (II) sulfate crystals.

Examples:

1. How many moles of water molecules are present in 54 g of water?

Mr(H₂O) = (1 x 2) + 16 = 18,

2. How many moles in 102 tonnes of O₂?

3. Mass of 0.4 moles of C₂H₅OH.

Mass = Mr x Moles, 0.4 x 46 = 18.4 g

4. 0.200 mol of a compound has a mass of 1.64 g. What's the Mr.?

Reacting Masses

Method: Using Moles to Calculate Reacting Masses in Equation.
1. Write out the information from the question in the balanced equation.
2. Calculate the number of moles for the known substance (n = mass/Mr).
3. Use the mole ratio from the balanced equation to calculate the number of moles of the target substance from the number of moles of the known substance.
4. Find the mass of the target substance (mass = n x Mr).

Examples:

1. What mass of calcium oxide will be formed by the decomposition of 200 g of calcium carbonate?

CaCO₃ → CaO + CO₂ irrelevant

200g ?

CaCO₃ : CaO

2 mol : 2 mol : n(CaO) = 2 mol

m(CaO) = Mr x n = 56 x 2 = 112 g

2. What mass of ethanol could burn in 100g of oxygen?

C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O irrelevant

? 100g

C₂H₅OH : 3O₂

1 : 3

n : 1 : 0.4 : 3.125

m(C₂H₅OH) = 1.04 mol x 46.0g/mol = 47.9 g

3. How many g of Na₂SO₄ are formed when 49 g of H₂SO₄ are dissolved/ neutralised by NaOH solution?

H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)

49g ?

H₂SO4 : Na₂SO4

1 : 1

m(Na₂SO4) = 0.5 x (46 + 32 + 64) = 71 g

4. What mass of C do you need to reduce 15.9 g of Cu(II) oxide to copper?

CuO(s) + C(s) → Cu(s) + CO(g)

15.9 ?

CuO : C

1 : 1

m(C) = 0.2 x 12 = 2.4 g

Avogadro's Number

- 1 mole = 6.02 x 1023 (Avogadro's number) of particles.
- In 0.5 moles of H2 there are 6.02 x 1023 molecules = 3.01 x 1023

Examples:

- E.g., carbon has an Ar of 12, so 1 mole of carbon weighs 12 g.
Similarly, CO2 has an Mr of 44 g, so 1 mol of CO2 weighs 44 g.

1) 5 moles of Au¹⁹⁷ atoms have a mass of 984.8 g. Calculate the mass of 1 Au¹⁹⁷ atom.

2. One Cu⁶³ atom has a mass of 1.045 x 10^-22. Calculate the mass of 3.2 x 10²² Cu⁶³ atoms.

(1.045 x 10^-22) x (6.02 x 10²³) x 3 = 189g x 3 = 189 g

- We can also work out the number of particles in a given mass of substance by using n: Mr. Avogadro's number.

1) How many calcium atoms are in 20g of calcium?

= 0.5 mol → 0.5 x 6.02 x 10²³ = 3.01 x 10²³ atoms

2) What mass do 3.01 x 10²⁴ silver atoms have?

3) If 3.01 x 10²³ molecules of a molecule have a mass of 16g what's the Mr of the molecule?

Molar Ratios

- We can use normal ratios in everyday life in balanced equations, too. For example:
- E.g., if there are 2 moles of Ca, how many moles of O₂ and CaO are there? → Ignore subscripts, look at big numbers.
2Ca + O₂ → 2CaO → 2CaO
2 mol : 1 mol : 2 mol
1 mol + 2mol = 3mol of O₂ and CaO and CaO

Examples:

1) TiCl4 + 4 Na → Ti + 4NaCl
If there are 2.5 mol of TiCl₄, how many moles would there be of:
Na → 10 mol
Ti → 2.5 mol
There's 4x as much NaCl + Na than TiCl₄, so we times 2.5 by 4.
NaCl → 10 mol

2. C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O
If is there 0.25 mol of O₂, how many moles would there be of:
C₂H₅OH → ⅛
mol CO₂ → ⅙ mol
H₂O → ¼ mol

Formulae

Molecular Formula: The atoms that make up a molecule, e.g., H₂O.
Empirical Formula: The simplest ratio of the atoms present in a compound.

Molecular formula	Empirical formula
H2O	H2O
C2H4	CH2
CH2	CH
CH3CO2H	CH2O (add carbons)

E.g., 1) 0.795 g of copper oxide was reduced to 0.635 g of copper when heated. What is the empirical formula of copper oxide?

	Cu	O
mass	0.795	0.16	0.16 = 0.795 - 0.635
Ar	63.5	16	divide mass by Ar to get moles.
no. of moles	0.01	0.01
ratio	1	1	ratio is 1:1 so formula = CuO

= CuO

E.g., 2) 10.01 g of a white solid contains 4.01 g of calcium, 1.2 g of carbon, and 4.8 g of oxygen. What's the empirical formula?

	Ca	C	O
mass	4.01	1.2	4.8
Ar	40	12	16
no. of moles	0.1	0.1	0.2
ratio	1	1	2

= CaCO₂

E.g., 3) What's the empirical formula of a compound that contains 85.7% of carbon and 14.3% of hydrogen?

	C	H
mass	85.7	14.3	use % as mass
Ar	12	1
no. of moles	7.14	14.3
ratio	1	2

= CH₂

b) The molar mass of this molecule is 56. What's the molecular formula?

Mr(CH₂) = 12 + (2 x 1) = 14

= CH₂ x 4 = C₄H₈

Water of Crystallisation

- Heat drives off the water of crystallisation from a salt, leaving behind the anhydrous salt.
- You can calculate the mass of the substance before and after heating so you have the mass of water lost.

Examples:

1. BaCl₂.nH₂O

	BaCl2	H2O
mass	2.08	0.36
Ar	2.08	18
no. of moles	0.01	0.02
ratio	1	2

→ BaCl₂.2H₂O

2. CaSO₄.nH₂O has an Mr of 172; find n.

CaSO₄ : 40 + 32 + 64 = 136
172 - 136 = 36
nH₂O = 136
Mr of H₂O = 18 → 36 ÷ 18 = 2 → n = 2

= CaSO₄.2H₂O

3. A hydrated carbonate of a metal has formula X₂CO₃.10H₂O and has a Mr of 286. What's the metal?

10H₂O Mr = 180 → 286-180 = 106
CO₃ Mr (C + O) → 106 - (12 + 48) = 46
X₂ = 46 → 46 ÷ 2 = 23
Mr of 23 = Lithium

4. MgSO₄. x H₂O has a mass of 13.52 g. MgSO₄ anhydrous has a mass of 6.60 g. Calculate x.

	MgSO₄	H2O
mass	6.60	6.92	(13.52 - 6.60)
Ar	120	18
no. of moles	0.055	0.38
ratio	1	7	(round)

x = 7

% Yield

- The percentage yield of a reaction is the product obtained compared to the predicted theoretical maximum yield as calculated from the balanced equation.

- Determining % yield allows the efficiency of a manufacturing process to be gauged.
- A 100% yield means no product has been lost, and a 0% yield means no product has been gained.
The actual percentage yield from a manufacturing process will significantly impact its profitability.
- In any chemical process, it's almost impossible to get 100% yield because:
→ The reactants may be impure.
→ The reaction may not fully go to completion/may not have been lost long enough for the reaction to take place.
→ Losses of the product as it's separated from the reaction mixture.
- All separation techniques result in small amounts of material being left behind at each stage.
- Side reactions too.

Examples

Worked Example:

1. 12 g of Mg + excess HCl acid were reacted together to make MgCl₂ what mass of MgCl₂ can be made?

Mg + 2HCl → MgCl₂ + H₂

Mg : MgCl₂
1 : 1 → n(MgCl) = 0.5 mol
m(MgCl₂) = n x Mr(MgCl₂) = 0.5 x (24 + 35.5 x 2) = 47.5 g → theoretical yield

At the end of this experiment, only 4.7 g of MgCl₂ crystals were recovered. What is the % yield of the reaction?

a) 15 g of CH₃COOH reacts with excess C₂H₅OH to make CH₃COOC₂H₅. Calculate theoretical yield.

1 : 1 → (CH₃COOC₂H₅) (g) = 0.25 x 88 = 22g) (g) = 0.25 x 88 = 22 g

b) 8.28 g were collected (of CH₃COOC₂H₅) - calculate % yield.

2.

2) 20 g of methane was completely combusted with air, CH₄ + 2O₂ → CO₂ + 2H₂O, the gases were cooled and collected to condense to water vapour; only 35cm³ water was collected, % yield.

Limiting Reactants

- During a chemical reaction, the limiting reagent/reaction exhausts itself completely. When the limiting reactant is used up, the reaction stops and no more product is made; the number of moles controls product production.
- Chemical reactions do not entirely use up reactants in excess.

Showing Which Reactant is in Excess/Limiting:

Worked example:

1. Cu(NO₃)₂ + Mg → Mg(NO₃)₂ + Cu (There are 0.8 moles of Cu(NO₃)₂ + 0.25 moles of Mg. Show Cu(NO₃)₂ is in excess.

Cu(NO₃)₂ : Mg
0.8 : 25
1: 1
3.2 : 1 (0.8 ÷ 0.25 = 3.2)
3.2 > 1∴ Cu(NO3)2 is in excess.
→ This also means only 0.25 moles of Mg(NO₃)₂ can be made.

2. When mixing 0.3 mol of Zn with 0.4 mol of HCl, determine the limiting reagent by reactant in excess. Zn + 2HCl → ZnCl₂ + H₂

Zn : HCl
0.3 : 0.4
1 : 2 → (0.4 ÷ 0.3)
1 : 1.3 ∴ HCl is limiting + Zn is in excess → 1.3 < 2

3. 9 mol of PCl₅ and 24 mol of H₂O

PCl₅ + 3H₂O → H3PO3 + 3HCl
PCl₅ : 3H₂O
9 : 24
1 : 3
1 : 6 ∴3H₂O is limiting + PCl5 is in excess → 2.6 < 3

4. CuO + Mg → MgO + Cu

24g of Mg + 48g of CuO
24g of Mg + Mgo of CuO

CuO : Mg
1 : 1
0.3 : 1 ∴ CuO is limiting as 0.3 < 1

5. 2SO₂ + O₂ → 2SO₃
96g of SO₂ + 38g O₂

2SO₂: O₂
2 : 1
1.5 : 1.2
1.25 : 1 ∴SO2 is limiting as 1.25 < 2.

6) a) 1.5 g of CaCO₃ reacts with 0.73 g of HCl.
CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O Which reagent is limiting?

CaCO₃ : 2HCl
0.015 : 0.02
1 : 2
1 : 1.3 → HCl is limiting as 1.3 < 2

b) Use part a) answer to determine theoretical yield of CaCl₂.

HCl : CaCl₂
2 : 1
0.02 : 0.01 = m(CaCl₂) = 0.01 x 111 = 1.11g

7) Sulfur is roasted in excess oxygen to form sulfur dioxide. What mass of sulfur dioxide is formed if 32 g of sulfur is burnt?

S + O₂ → SO₂

S : SO₂

1:1 → M(SO₂) 1 x 64 = 64 g

Harder Examples:

8) How many grams of K₂O will be produced from 0.50 g of K and 0.10 g of O?

Determine the limiting reactant first to determine the production capacity.

4K: O₂

4 : 1

4 : 1

4.1: 1 O₂ is limiting

O₂ : 2K₂O

9. How many grams of CH₄ are made when 10 g of H₂ react with 5 g of C?