What is the relative formula mass (M_{r}) of Li_{2}O (lithium oxide)? **Write down your workings.**

A_{r }(relative atomic mass) of Li = 7

A_{r }(relative atomic mass) of O = 16

Li_{2} = 7 x 2 = 14

O = 16

Li_{2} + O = 14 + 16 = 30

What is the relative formula mass (M_{r}) of C_{3}H_{8} (propane)? **Write down your workings.**

A_{r }(relative atomic mass) of C = 12

A_{r }(relative atomic mass) of H = 1

C_{3} = 12 x 3 = 36

H_{8} = 1 x 8 = 8

C_{3} + H_{8} = 36 + 8 = 44

What is the relative formula mass (M_{r}) of Ca(OH)_{2} (calcium hydroxide)? **Write down your workings.**

A_{r} of Ca = 40

A_{r }of O = 16

A_{r }of H = 1

Ca = 40

O + H = 16 + 1 = 17

(OH)_{2} = 17 x 2 = 34

Ca + (OH)_{2} = 40 + 34 = 74

What is the formula for calculating the percentage mass of an element in a compound?

**A _{r}** (relative atomic mass of element)

**x number of atoms**

**x 100**

**M _{r}** (relative formula mass of compound)

What is the percentage mass of lithium in lithium oxide (Li_{2}O)? **Write down your workings.**

A_{r} of Li = 7

A_{r }of O = 16

**M _{r }of Li_{2}O**

- Li
_{2}= 7 x 2 = 14 - O = 16
- Li
_{2}+ O = 14 + 16 = 30

**Percentage mass of lithium in lithium oxide**

- Li
_{2 }÷ Li_{2}O x 100 = - (7x2) ÷ 30 x 100 =
- 14 ÷ 30 x 100 =
**47%**

What is the percentage mass of hydrogen in propane (C_{3}H_{8})? **Write down your workings.**

Ar of C = 12

Ar of H = 1

**M _{r} of C_{3}H_{8}**

- C
_{3}= 12 x 3 = 36 - H
_{8}= 1 x 8 = 8 - C
_{3}+ H_{8}= 36 + 8 = 44

**Percentage mass of hydrogen in propane**

- H
_{8 }÷ C_{3}H_{8}x 100 = - (1x8) ÷ 44 x 100 =
- 8 ÷ 44 x 100 =
**18%**

What is the percentage mass of oxygen in calcium hydroxide (Ca(OH)_{2})?**Write down your workings.**

Ar of Ca = 40

Ar of O = 16

Ar of H = 1

**M _{r} of Ca(OH)_{2}**

- Ca = 40
- O + H = 16 + 1 = 17
- (OH)
_{2}= 17 x 2 = 34 - Ca + (OH)
_{2}= 40 + 34 = 74

**Percentage mass of hydrogen in calcium hydroxide**

- O
_{2}÷ Ca(OH)_{2 }x 100 = - (16x2) ÷ 74 x 100 =
- 32 ÷ 74 x 100 =
**43%**

If 6g of magnesium reacts completely with 4g of oxygen what mass of magnesium oxide is formed?

10g

When calcium carbonate is hated it breaks down to give calcium oxide and carbon dioxide.

calcium carbonate →calcium oxide + carbon dioxide

6.00g of calcium carbonate is heated, this gives 3.36g of calcium oxide. What mass of carbon dioxide was made?

**Extension: **Explain how you work this out.

6.00g - 3.36g = **2.64g**

**Extension: **

The total mass of products must equal the total mass of reactants. So to work out the mass of carbon dioxide yuo take away the mass of calcium oxide (product) from the mass of calcium carbonate (reactant).

Liam adds two reactants to an open conical flask on a mass balance. The starting mass is 10g. At the end of the reaction the mass is 8.95g. What might have caused the decrease in mass during the reaction?

###
- One of the products was a gas
- The gas will have left the conical flask decreasing the mass of the flask.

Sodium + Hydrochloric acid →Sodium chloride + Hydrogen

Jack reacts sodium with hydrochloric acid in an open (unsealed) conical flask. What will happen to the mass of the flask as the reaction takes place? Explain why this will happen.

###
- The mass of the flask will decrease
- Because, hydrogen is a gass and will be lost from the flask.

Jamie reacts 6g of magnesium with oxygen (from the air) by heating it. At the end of the reaction he finds he has 9.3g of magnesium oxide.

How much oxygen was used in the reaction?

mass of magnesium oxide produced - mass of magnesium used = mass of oxygen used

9.3 - 6 = 3.3 g

**HIGHER:**

How many **atoms** are in 1 mole of lithium (Li)?

6.02 x 10^{23 }(Avogadro's constant)

**HIGHER:**

How many **molecules** are in 1 mole of oxygen (O_{2})?

6.02 x 10^{23 }(Avogadro's constant)

**HIGHER:**

How many **atoms** are in 1 mole of oxygen (O_{2})?

2 x (6.02 x 10^{23}): Each oxygen molecule has two oxygen atoms

**1.204 x 10 ^{24}**

**HIGHER:**

How many **molecules** are in 1 mole of carbon dioxide (CO_{2})?

6.02 x 10^{23}

**HIGHER:**

How many **atoms** are there in 1 mole of carbon dioxide (CO_{2})?

3 x (6.02 x 10^{23}): Each carbon dioxide molecule is made from 3 atoms, 1 carbon atom and 2 oxygen atoms.

**1.806 x 10 ^{24}**

**HIGHER:**

What is the mass of 5 moles of lithium oxide (Li_{2}O)? **Write down your workings.**

Ar of Li = 7

Ar of O = 16

**Mass of 1 mole of lithium oxide**

- Li
_{2}= 7 x 2 = 14 - O = 16
- Li
_{2}+ O = 14 + 16 =**30g**

**Mass of 5 moles lithium oxide**

- 5 x 30g =
**150g**

**HIGHER:**

What is the mass (in grams) of 3C_{3}H_{8} (propane)? **Write down your workings.**

Ar (relative atomic mass) of C = 12

Ar (relative atomic mass) of H = 1

**Mass of 1 mole of propane (C _{3}H_{8})**

- C
_{3}= 12 x 3 = 36 -
H

_{8}= 1 x 8 = 8 -
C

_{3}+ H_{8}= 36 + 8 = 44g

**Mass of 3 moles propane**

- 3 x 44g =
**132g**

**HIGHER:**

How many moles of water (H_{2}O) are there in 36g?

Ar (relative atomic mass) of H = 1

Ar (relative atomic mass) of O = 16

**Mass of 1 mole of water (H _{2}O)**

- H
_{2}= 1 x 2 = 2 - O = 16
- H
_{2}+ O = 2 + 16 = 18g

**Moles of water in 36g**

- number of moles = total mass ÷ mass of 1 mole
- 36 ÷ 18 = 2 moles

** HIGHER:**

How many moles of ammonia (NH_{3}) are there is 8.5g.

Ar (relative atomic mass) of N = 14

Ar (relative atomic mass) of H = 1

**Mass of 1 mole of water (NH _{3})**

- N = 14
- H
_{3}= 1 x 3 = 3 - N + H
_{3}= 14 + 3 = 17

**Moles of ammonia in 8.5g**

- number of moles = total mass ÷ mass of 1 mole
- 8.5 ÷ 17 = 0.5 moles

**HIGHER: **

3.5 g of Li reacts completely with 4g of O_{2} to produce 7.5 g of Li_{2}O. Use this to work out a balanced equation for the reaction.

Ar (relative atomic mass) of Li = 7

Mr (relative formula mass) of O_{2} = 32 (16 x2)

Mr (relative formula mass of Li_{2}O = 30 ( 14 x 2 + 16)

**Write you answer out.**

###
- Li + O
_{2} → Li_{2}O
- Moles of Li in 3.5 g = 3.5 ÷ 7 = 0.5
- Moles of O
_{2} in 4g = 4 ÷ 32 = 0.125
- Moles of Li
_{2}O in 7.5 g = 7.5 ÷ 30 = 0.25
- ratio of atoms/molecules = 0.5: 0.125: 0.25
- To get whole number divide each by the smallest (0.125)
- ratio of atoms/molecules = 4:1:2
- Balanced equation = 4Li + 1O
_{2} → 2Li_{2}O

_{2}→ Li_{2}O_{2}in 4g = 4 ÷ 32 = 0.125_{2}O in 7.5 g = 7.5 ÷ 30 = 0.25_{2}→ 2Li_{2}O4Al + 3O_{2} → 2Al_{2}O_{3}

Calculate the number of moles of aluminium oxide formed when 135g of aluminum is burned in air.

###
- 4Al + 3O
_{2} → 2Al_{2}O_{3}
- Number of moles of aluminium = mass of aluminium ÷ relative formula mass
- 135 ÷ 27 = 5
- Ratio of aluminium to aluminium oxide 4:2
- There are half the number of moles of aluminium oxide as there are of aluminium
- There will therefore be 5 x 0.5 =
**2.5 moles of aluminium oxide**

_{2}→ 2Al_{2}O_{3}**2.5 moles of aluminium oxide**##
- What volume does a mole of gas occupy at toom temperature (20
^{o}C) and pressure (1 atm)

^{o}C) and pressure (1 atm)###
- 24 dm
^{3}

^{3}If you were given the mass of a gas, how would you calculate the volume it takes up?

Volume of gas (in dm^{3}) = number of moles x 24, therefore..

Volume of gas (in dm^{3}) = (mass ÷ relative formula mass) x 24

What is the volume of 80 g of oxygen at room temp and pressure?

Volume of gas (in dm^{3}) = number of moles x 24

Volume of gas (in dm^{3}) = (mass ÷ relative formula mass) x 24

= (80 ÷ 16) x 24 = 5 x 24

= 120 dm^{3}

What volume of ammonia is formed when 100 dm^{3} of nitrogen reacts with hydrogen.

N_{2} + 3H_{2} → 2NH_{3}

N_{2} + 3H_{2} → 2NH_{3}

The ratio of nitrogen to ammonia is 1:2

Therefore if you use 100 dm^{3} of nitrogen you will get 200 dm^{3 }of ammonia.

What is the formula for working out the concentration of a solution in g/dm^{3}?

Concentration (g/dm^{3}) = mass of solute (g) ÷ volume of solvent (dm^{3})

What is the formula for working out the concentration of a solution in mol/dm^{3}?

Concentration (mol/dm^{3}) = moles of solute ÷ volume of solvent (dm^{3})

If you were given the concentration of a solution (mol/dm^{3}) and the volume (dm^{3}), how would you calculate the number of moles of solute present?

moles of solute = Concentration (mol/dm^{3}) x volume of solvent (dm^{3})