Mr = 24 + 32 + 4(16) = 120

Percentage Composition

Eg. Determine which oxides of iron of Fe₂O₃ or Fe₃O₄ has more iron.

Solution

Mr(Fe₂O₃) = 2(56) + 3(16) = 160

Percentage of Fe in Fe₂O₃
= = (2 x 56)/160 x 100%
= 70%

Mr(Fe₃O₄) = 3(56) + 4(16) = 232
Percentage of Fe in Fe₂O₃ = x 100 %
= x 100%
= 72%

Therefore, Fe₃O₄ has more iron composition than that of Fe₂O₃.

Calculating the Mass of an Element in a Compound

Use the example of Fe₂O₃ in the example above.
The percentage mass of iron in iron(III) oxide is 70%.
Therefore to calculate mass of iron in a 200g compound of iron(III) oxide is (0.7 x 200)g = 140g

Eg. Determine the mass of iron in 200g of Fe₂O₃.

Mr(Fe₂O₃)= 2(56) + 3(16) = 160
Mass of Fe in Fe₂O₃
= = (2 x 56)/160 x 200g
= 140g

Calculating the Mass of Water in a Compound

Compound with water mass is ‘hydrated’ and has H₂O in their formula.

Eg. Calculate water mass in 12.5g hydrated copper sulfate, CuSO₄.5H₂O

Mass of 5H₂O in CuSO₄.5H₂O
= = (5 x 18)/250 x 12.5g
= 4.5g

Mole

Counting Particles

• Unit for particles = mole
• Symbol = mol
• 1 mol = 6 x 10²³ atoms
  

Moles of Particles - Calculating the Number of Moles

Eg 1: How many molecules in 6 x 10²⁴ molecules of water, H₂O?
= 5 mol

Eg 2: Calculate the number of molecules in 0.25 mole of CO₂. Hence, how many atoms are present?

Number of particles = 0.25 mol x 6 x 10²³
= 1.5 x 10²³ molecules

Number of atoms = total number of atoms in CO₂ x number of particles
= 3 x 1.5 x 10²³
= 4.5 x 10²³ atoms

Molar mass

• Molar mass – the mass of one mole of any substances
• For substances consisting of atoms
• It is the Ar of the element in grams. Eg. Ar(C) = 12, molar mass = 12g
• For substances consisting of molecules
• It is the Ar of the substance in grams. Eg. Ar(H₂O) = 18, molar mass = 18g
• For substances consisting of ions
• It is the Ar of substance in grams. Eg. Ar(NaCl)= 58.5, molar mass= 58.5g
  

Eg. Find the mass of 0.4 mol of iron atom.

n = m / Mr
m = n x Mr
m = 0.4 x 56 = 22.4 g

Eg. Argon Fluorohydride gas, HArF, first known noble gas compound, has molar mass of 60g. Find the number of moles Argon atom in 6.66g of HArF.

n (HArF) = 6.66/60
= 0.111 mol

n (Ar) = 0.111 mol x 1 Ar atom in HArF
= 0.111 mol

Different Kinds of Chemical Formulae

• Molecular Formula – shows the actual formula and kinds of atoms present, e.g. C₂H₆
• Empirical Formula – shows the simplest whole number ratio of the atoms present, e.g. C₂H₆, ratio 1:3, therefore C₁H₃, simply CH₃
• Structural Formula – shows how atoms are joined in the molecule. It can be represented by ball-and-stick model or diagrammatically.
  

Calculating the Empirical Formula of a Compound

Find the empirical formula of an oxide of magnesium consisting of 0.32g of oxygen and 0.96g of magnesium.

Step 1: find the number of moles of the 2 elements.

n(Mg) =0.96 / 24
= 0.04 mol

 n(O) = 0.32 / 16
= 0.02 mol

Step 2: Divide the moles by the smallest number.

Mg = 0.04 / 0.02
= 2

O = 0.02 / 0.02
= 1

Therefore, the empirical formula is Mg₂O

Calculating the Empirical Formula from Percentage Composition

An oxide of sulphur consists of 40% sulphur and 60% oxygen.
Take the total 100% to be 100g.

Step 1: find the number of moles of the 2 elements.

n(S) = 40 / 32
= 1.25 mol

n(O) = 60 / 16
= 3.75 mol

Step 2: Divide the moles by the smallest number.
S = 1.25 / 1.25
= 1

O = 3.75 / 1.25
= 3

Therefore, the empirical formula is SO₃

Changing From Empirical formula to Molecular Formula

Find the molecular formula of propene, CH₂, having molecular mass of 42.
Molecular formula will be C_nH_2n

Relative molecular mass = 12n(from carbon Ar) + 2n(2 x hydrogen Ar) = 14n

14 n = 42
n = 42 / 14 = 3

Therefore --> C₃H₆

Molar Volume of Gases

The Avogadro’s Law: 

Equal volume of gases at same temperature and volume contain equal number of particles or molecules.

Molar Volume of Gas: 

• volume occupied by one mole of gas
• All gases at room temperature and pressure (r.t.p.) = 24dm³
• 1dm³ = 1000cm³

Formulae:
Eg. What is the number of moles of 240cm³ of Cl₂ at r.t.p.?

        = 0.01 mol

Molar Volume and Molar Mass

Gases have same volume but not necessarily same mass
Example: Hydrogen -> 2g, Carbon Dioxide -> 44g

Eg. Find the volume of 7g of N₂ at r.t.p.

Step 1: Find the number of moles from the mass of nitrogen
n = 7 / 28
= 0.25 mol

Step 2: Find the volume of nitrogen, now with formula of gas
0.25 mol = volume of gas / 24

Volume of gas = 0.25 mol x 24
= 6 dm³ (or 6000cm³)

Concentration of Solutions

Concentration of solution tells the number of solute in a volume of solution

Calculating the Amount of Solute

Moles of solute (n) = Concentration (mol/dm³ ) x Volume of solution (dm³)

Eg. What is the mass of solute in 600cm³ of 1.5 NaOH solution?

Volume of solution in dm3 = 0.60 dm³
n = 1.5 x 0.60
= 0.9 mol

Number of moles of NaOH = m / Mr
0.9 = m / 40

m = 0.9 x 40
= 36g

Calculations using Chemical Equations

Constructing Chemical Equations

Eg. 1: Reaction Between Hydrogen and Oxygen

Word Equation: Oxygen + Hydrogen --> Water

To write the chemical equation, we use symbols of atoms/molecules:
O₂ + H₂ --> H₂O

Balance the equation!
O₂ + 2H₂ --> 2H₂O

Calculations from Equations

Reacting Masses

• In every equation, each atom is rational to each other.
  
• Suppose we want to find moles of X atoms that reacted to form 0.25 mole of Y atoms.
  
• We always put the atom we want to find as numerator and the denominator being the atom we know.

Eg. X + 2Z --> 2Y

1. Find the ratio first:

2. Then multiply the ratio by no. of moles of Y to find the reacting mole of X.
1/2 x 0.25 = 0.125 mole

Therefore 0.125 mole of X reacted with 0.25 mole of Y.

3. To find the reacting mass of X, e.g. Y is given as 35g, we just multiply the mole by the mass of Y as they are always in ratio:

0.125 x 35 = 4.375 g

Reacting Masses and Volumes

First, find the ratio of moles and multiply the mole of the gas volume you want to find with the volume of gas at room temperature (24dm³)

Example

MgCl₂ is formed by reacting Mg and HCl according to equation:
Mg (s) + 2HCl (aq) --> MgCl₂ (s) + H₂ (g)

Find the amount of hydrogen gas, in cm3, formed when 14.6g of HCl is reacted.

m(HCl) = 14.6/36.5
=0.4 mol

Multiply ratio by mole of HCl = 1/2 x 0.4 = 0.2 mol

Multiply mole by molar volume of gas at r.t.p. = 0.2 x 24 dm³= 4.8 dm³
1dm³ = 1000cm³

Therefore, 4.8dm³ x 1000 = 4800 cm³

4800cm³ of gas is formed

Example Questions

1 mole of CO₂ is equal to

- 6.02 x 10²³ molecules
- 2 moles of O
- 1 mole of C
- 44.0095 grams
- 22.4 L at r.t.p.
- 31.9988 grams of O
- 12.0107 grams of C

MCQ Questions

1. One mole of each of the following compounds is burnt in excess oxygen. Which compound will produce three moles of carbon dioxide and three moles of steam only?
a. C₃H₈
b. C₃H₇OH
c. C₃H₇CO₂H
d. CH₃CO₂CH₃

2. 20 cm³ of carbon monoxide are reacted with 10 cm³ of oxygen. The equation for the reaction is shown:
2CO + O₂ --> 2CO₂
Which volume of carbon dioxide will be produced?

a. 10 cm³
b. 20 cm³
c. 30 cm³
d. 40 cm³

3. What has a mass equal to that of one mole of water?
a. 24  dm³ of water
b. one mole of steam
c. one molecule of water
d. two moles of hydrogen molecules and one mole of hydrogen molecules

4. An 8 g sample of oxygen contains the same number of atoms as 16 g of element X. What is the relative atomic mass, Ar of X?
a. 4
b. 8
c. 16
d. 32

5. Which quantity is the same for one mole of ethanol and one mole of ethane?
a. mass
b. number of atoms
c. number of molecules
d. volume at r.t.p.

6. Which ion is present in the highest concentration in a 2 mol/dm³ aqueous solution of sodium sulphate?
a. hydrogen ions
b, hydroxide ions
c. sodium ion
d. sulphate ion

7. Magnesium reacts with hydrochloric acid. Which solution would give the fastest initial rate of reaction?
a. 40 g of HCl in 1000 cm³ of water
b. 20 g of HCl in 1000 cm³ of water
c. 10 g of HCl in 100 cm³ of water
d.  4 g of HCl in 50 cm³ of water

8. One mole of hydrated copper(II) sulphate, CuSO₄.5H₂O is dissolved in water. How many moles of ions does the solution contain?
a. 1
b. 2
c. 6
d. 7

9. What is the ratio of the volume of 2 g of hydrogen to the volume of 16 g of methane, both volumes at r.t.p.?
a. 1 to 1
b. 1 to 2
c. 1 to 8
d. 2 to 1

10. Which of the following contains the same number of molecules as 9 g of water?
a. 2 g of hydrogen gas
b. 14 g of nitrogen gas
c. 32 g of oxygen gas
d. 44 g of carbon dioxide gas

11. Calcium reacts with water as shown.
Ca (s) + 2H₂O (l) --> Ca(OH)₂ (aq) + H₂ (g)

What is the total mass of the solution that remains when 40g of calcium reacts with 100 g of water?
a. 58 g
b. 74 g
c. 138 g
d. 140 g

12. 20 cm³ of oxygen are reacted with 20 cm³ of carbon monoxide. What are the volumes of the gases remaining, at the original temperature and pressure?

	oxygen/cm3	carbon monoxide/cm3	carbon dioxide/cm3
a	0	0	20
b	0	0	40
c	10	0	20
d	10	10	20

13. What is the mass of oxygen contained in 72 g of pure water?
a. 16 g
b. 32 g
c. 64 g
d. 70 g

14. A volume of ethane, C₂H₆, at r.t.p. has a mass of 20 g. What is the mass of an equal volume of propene, C₃H₆ at r.t.p.?
a. 20 g
b. 21 g
c. 28 g
d. 42 g

15. Sodium reacts with water according to the equation below.
2Na + 2H₂O --> 2NaOH + H₂
Which volume of hydrogen is produced at r.t.p. when 0.2 mol of sodium reacts?
a. 1.2 dm³
b. 2.4 dm³
c. 4.8 dm³
d. 9.6 dm³

a. 1

b. 2

c. 3

d. 4

a. 15

b. 14

c. 10

d. 7

a. X = 3, Y = 8

b. X = 4, Y = 3

c. X = 3, Y = 4

d. X = 4, Y = 8

a. 4%

b. 12%

c. 40%

d. 44%

a. NH₄NO₃

b. NH₄Cl₂

c. N₂H₂

d. (NH₂)₂CO

a. CH₃

b. CH₄

c. C₂H₆

d. C₂H₄

a. XO

b. XO₂

c. X₂O₃

d. X₂O

a. n

b. n/5

c. 5n/2

d. 2n/5

a. 4g of H₂

b. 8g of O₂

c. 71g of Cl₂

d. 72dm³ of argon gas at rtp

a. 32g of O₂

b. 18g of water

c. 28g of N₂

d. 4g of H₂

a. Mg(OH)₂

b. KF

c. Al₂O₃

d. Ca₂Cl

a. 38g

b. 48g

c. 52g

d. 152g

a. 6.02 x 10²³

b. 4.05 x 10²³

c. 3.01 x 10²²

d. 6.02 x 10²¹

6a. Mr of BaCl₂ = 208

no. of mol of BaCl₂ = 62.1/208 = 0.2985 mol --> 0.3mol (2 dp)

6b. no. of mol of H₂SO₄ = concentration x volume = 0.25 x 55/1000 = 0.01375 --> 0.01 mol (2 dp)

6c. Mr of N₂ gas = 28

no. of mol of N₂ gas = 17/28 = 0.6071 mol --> 0.61 mol (2 dp)

7a. Mr of KOH = 56

no. of mol of KOH = 32/56 = 0.5714mol --> 0.57mol (2 dp)

conc of KOH = no. of mol/(500/1000) = 1.14 mol/dm³

7b. Mr of Fe₂O₃ = 160

Mass of Fe extracted = (56 x 2)/160 x 14.9 = 10.43g

7c. Mr of CO₂ = 44

no. of mol of CO₂ = 100/44 = 2.272 mol --> 2.27 mol (2 dp)

vol of CO₂ = 2.27mol x 24dm³ = 54.48dm³

8. Conc of ethanol in g/dm³ = 22.3 / (250/1000) = 89.2 g/dm³

Mr of ethanol = 46

conc of ethanol in mol/dm³ = 89.2/46 = 1.94 mol/dm³