Page 2: Activity 3.1
What happens in the reaction flask?
When the solutions X and Y are mixed, they react with each other. For example, if you mix copper sulphate (blue) and sodium carbonate, a chemical reaction happens and new substances are formed.
Do you think that a chemical reaction has taken place?
Yes, a chemical reaction has taken place. We can often tell this because there might be a change in colour or a solid substance (called a precipitate) might form.
Why should we put a cork on the mouth of the flask?
We put a cork on the flask to make it a closed system. This ensures that no matter (like a gas that might be produced) can escape and no air from outside can get in. This is important for measuring the mass accurately.
Does the mass of the flask and its contents change?
No, the mass of the flask and its contents does not change. The total mass before the chemical reaction is exactly equal to the total mass after the reaction.
Page 2 & 3: Questions Block 1
1. In a reaction, 5.3 g of sodium carbonate reacted with 6 g of acetic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium acetate. Show that these observations are in agreement with the law of conservation of mass.
sodium carbonate + acetic acid → sodium acetate + carbon dioxide + water
The law of conservation of mass states that mass is not created or destroyed in a chemical reaction. This means the total mass of the reactants (the starting chemicals) must equal the total mass of the products (the new chemicals).
- Total mass of reactants:
Mass of sodium carbonate = 5.3 g
Mass of acetic acid = 6.0 g
Total mass = 5.3 g + 6.0 g = 11.3 g - Total mass of products:
Mass of sodium acetate = 8.2 g
Mass of carbon dioxide = 2.2 g
Mass of water = 0.9 g
Total mass = 8.2 g + 2.2 g + 0.9 g = 11.3 g
Since the total mass of reactants (11.3 g) is equal to the total mass of products (11.3 g), these observations agree with the law of conservation of mass.
2. Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?
The ratio of hydrogen to oxygen by mass is 1:8. This means 1 g of hydrogen reacts with 8 g of oxygen.
To find out how much oxygen is needed for 3 g of hydrogen, we can set up a simple calculation:
If 1 g of Hydrogen needs 8 g of Oxygen
Then 3 g of Hydrogen will need 3 × 8 g of Oxygen
3 × 8 = 24
So, 24 g of oxygen gas would be required to react completely with 3 g of hydrogen gas.
3. Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?
The postulate is: “Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction.”
Because atoms cannot be created or destroyed, they are just rearranged during a reaction. This is why the total mass stays the same.
4. Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
The postulate is: “The relative number and kinds of atoms are constant in a given compound.”
Because a compound like water always has the same fixed number and kinds of atoms (e.g., 2 hydrogen atoms and 1 oxygen atom), the proportion of their masses is also always constant.
Page 5: Questions Block 2
1. Define the atomic mass unit.
One atomic mass unit (u) is a mass exactly equal to one-twelfth (1/12th) the mass of one atom of carbon-12. It is the standard unit used for measuring the mass of atoms.
2. Why is it not possible to see an atom with naked eyes?
It is not possible to see an atom with naked eyes because atoms are very, very small. They are smaller than anything we can imagine. Millions of atoms stacked on top of each other would only be as thick as a single sheet of paper.
Page 6: Activity 3.2
Refer to Table 3.4 for ratio by mass of atoms present in molecules and Table 3.2 for atomic masses of elements. Find the ratio by number of the atoms of elements in the molecules of compounds given in Table 3.4.
- For Water (H₂O):
- Elements: Hydrogen (H) and Oxygen (O)
- Ratio by Mass: 1:8
- Atomic Mass: H = 1 u; O = 16 u
- Mass / Atomic Mass:
- H: 1 / 1 = 1
- O: 8 / 16 = 0.5
- Simplest Ratio (by number):
- H: 1 / 0.5 = 2
- O: 0.5 / 0.5 = 1
- Ratio by number (H:O) = 2:1
- For Ammonia (NH₃):
- Elements: Nitrogen (N) and Hydrogen (H)
- Ratio by Mass: 14:3
- Atomic Mass: N = 14 u; H = 1 u
- Mass / Atomic Mass:
- N: 14 / 14 = 1
- H: 3 / 1 = 3
- Simplest Ratio (by number):
- N: 1 / 1 = 1
- H: 3 / 1 = 3
- Ratio by number (N:H) = 1:3
- For Carbon Dioxide (CO₂):
- Elements: Carbon (C) and Oxygen (O)
- Ratio by Mass: 3:8
- Atomic Mass: C = 12 u; O = 16 u
- Mass / Atomic Mass:
- C: 3 / 12 = 0.25
- O: 8 / 16 = 0.5
- Simplest Ratio (by number):
- C: 0.25 / 0.25 = 1
- O: 0.5 / 0.25 = 2
- Ratio by number (C:O) = 1:2
Page 9: Questions Block 3
1. Write down the formulae of
(i) sodium oxide
(ii) aluminium chloride
(iii) sodium sulphide
(iv) magnesium hydroxide
(i) Sodium oxide: Na₂O
(ii) Aluminium chloride: AlCl₃
(iii) Sodium sulphide: Na₂S
(iv) Magnesium hydroxide: Mg(OH)₂
2. Write down the names of compounds represented by the following formulae:
(i) Al₂(SO₄)₃
(ii) CaCl₂
(iii) K₂SO₄
(iv) KNO₃
(v) CaCO₃
(i) Al₂(SO₄)₃: Aluminium sulphate
(ii) CaCl₂: Calcium chloride
(iii) K₂SO₄: Potassium sulphate
(iv) KNO₃: Potassium nitrate
(v) CaCO₃: Calcium carbonate
3. What is meant by the term chemical formula?
A chemical formula is a way of writing down the composition of a substance using symbols. It shows which elements are in the compound and how many atoms of each element are present. For example, the chemical formula for water is H₂O.
4. How many atoms are present in a
(i) H₂S molecule and
(ii) PO₄³⁻ ion?
(i) H₂S molecule:
* Hydrogen (H): 2 atoms
* Sulphur (S): 1 atom
* Total = 2 + 1 = 3 atoms
(ii) PO₄³⁻ ion (Phosphate ion):
* Phosphorus (P): 1 atom
* Oxygen (O): 4 atoms
* Total = 1 + 4 = 5 atoms
Page 10: Questions Block 4
1. Calculate the molecular masses of H₂, O₂, Cl₂, CO₂, CH₄, C₂H₆, C₂H₄, NH₃, CH₃OH.
(Atomic masses: H=1 u, C=12 u, O=16 u, Cl=35.5 u, N=14 u)
- H₂: 1 × 2 = 2 u
- O₂: 16 × 2 = 32 u
- Cl₂: 35.5 × 2 = 71 u
- CO₂: 12 + (16 × 2) = 12 + 32 = 44 u
- CH₄: 12 + (1 × 4) = 12 + 4 = 16 u
- C₂H₆: (12 × 2) + (1 × 6) = 24 + 6 = 30 u
- C₂H₄: (12 × 2) + (1 × 4) = 24 + 4 = 28 u
- NH₃: 14 + (1 × 3) = 14 + 3 = 17 u
- CH₃OH: 12 + (1 × 3) + 16 + 1 = 32 u
2. Calculate the formula unit masses of ZnO, Na₂O, K₂CO₃, given atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, and O = 16 u.
- ZnO: 65 + 16 = 81 u
- Na₂O: (23 × 2) + 16 = 46 + 16 = 62 u
- K₂CO₃: (39 × 2) + 12 + (16 × 3) = 78 + 12 + 48 = 138 u
Exercise Questions (Page 11)
1. A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.
Total mass of the compound = 0.24 g
Mass of boron = 0.096 g
Mass of oxygen = 0.144 g
- Percentage of boron: (Mass of boron / Total mass of compound) × 100 = (0.096 g / 0.24 g) × 100 = 0.4 × 100 = 40%
- Percentage of oxygen: (Mass of oxygen / Total mass of compound) × 100 = (0.144 g / 0.24 g) × 100 = 0.6 × 100 = 60%
The compound is 40% boron and 60% oxygen by weight.
2. When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?
- Answer: 11.00 g of carbon dioxide will be formed.
- Law: This is governed by the Law of Constant Proportions (or Definite Proportions).
- Explanation: The law states that elements in a compound always combine in the same fixed ratio by mass. The first part of the question tells us that 3.0 g of carbon reacts with 8.00 g of oxygen to make 11.00 g of carbon dioxide. This means the ratio is always 3:8 for carbon and oxygen. In the second part, even though we have 50.00 g of oxygen, the 3.00 g of carbon will still only react with 8.00 g of oxygen. The extra oxygen (50.00 g – 8.00 g = 42.00 g) will be left over unreacted. The amount of product formed depends on the reactant that runs out first (in this case, carbon).
3. What are polyatomic ions? Give examples.
- What they are: Polyatomic ions are groups (or clusters) of atoms that are joined together and act as a single unit with a fixed charge on them (either positive or negative).
- Examples:
- Ammonium ion (NH₄⁺)
- Hydroxide ion (OH⁻)
- Sulphate ion (SO₄²⁻)
- Carbonate ion (CO₃²⁻)
4. Write the chemical formulae of the following.
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate.
(a) Magnesium chloride: MgCl₂
(b) Calcium oxide: CaO
(c) Copper nitrate: Cu(NO₃)₂
(d) Aluminium chloride: AlCl₃
(e) Calcium carbonate: CaCO₃
5. Give the names of the elements present in the following compounds.
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate.
(a) Quick lime (which is Calcium oxide, CaO): Calcium and Oxygen
(b) Hydrogen bromide (HBr): Hydrogen and Bromine
(c) Baking powder (which is Sodium hydrogencarbonate, NaHCO₃): Sodium, Hydrogen, Carbon, and Oxygen
(d) Potassium sulphate (K₂SO₄): Potassium, Sulphur, and Oxygen
6. Calculate the molar mass of the following substances.
(a) Ethyne, C₂H₂
(b) Sulphur molecule, S₈
(c) Phosphorus molecule, P₄ (Atomic mass of phosphorus = 31)
(d) Hydrochloric acid, HCl
(e) Nitric acid, HNO₃
(Atomic masses: H=1 u, C=12 u, S=32 u, P=31 u, Cl=35.5 u, O=16 u, N=14 u)
Note: Molar mass (in g/mol) is numerically the same as molecular mass (in u).
(a) Ethyne, C₂H₂: (12 × 2) + (1 × 2) = 24 + 2 = 26 u (or 26 g/mol)
(b) Sulphur molecule, S₈: 32 × 8 = 256 u (or 256 g/mol)
(c) Phosphorus molecule, P₄: 31 × 4 = 124 u (or 124 g/mol)
(d) Hydrochloric acid, HCl: 1 + 35.5 = 36.5 u (or 36.5 g/mol)
(e) Nitric acid, HNO₃: 1 + 14 + (16 × 3) = 1 + 14 + 48 = 63 u (or 63 g/mol)
Page 12: Group Activity
Now, write the formula of sodium phosphate.
- The sodium ion has a valency (charge) of 1+ (Na⁺).
- The phosphate ion is a polyatomic ion with a charge of 3- (PO₄³⁻).
- To balance the charges, you need three sodium ions (total charge of 3+) to balance the one phosphate ion (charge of 3-).
- Formula: Na₃PO₄