e) Chemical Formulae and Chemical Equations ​

1.21 Write Word Equations and Balanced Chemical Equations to Represent the Reactions Studied in This Specification ​

A chemical equation which is expressed by using words is called a word equation.

Example: Carbon + oxygen → carbon dioxide

A chemical equation which is expressed by using symbols is called a chemical equation.

Example: $C(s) + O_2(g) → CO_2(g)$

1.22 Use the State Symbols (s), (l), (g) and (aq) in Chemical Equations to Represent Solids, Liquids, Gases and Aqueous Solutions Respectively ​

1.23 Understand How the Formulae of Simple Compounds Can Be Obtained Experimentally, Including Metal Oxides, Water and Salts Containing Water of Crystallisation ​

Example: Finding the $n$ in $BaCl_2 \cdot nH_2O$ ​

Experimental Data: ​

• Mass of crucible: $30.00 , \text{g}$
• Mass of crucible + barium chloride crystals ($BaCl_2 \cdot nH_2O$): $32.44 , \text{g}$
• Mass of crucible + anhydrous barium chloride ($BaCl_2$): $32.08 , \text{g}$

Calculations: ​

1. Mass of $BaCl_2$:

   $$ 32.08 - 30.00 = 2.08 , \text{g} $$

2. Mass of water ($H_2O$):

   $$ 32.44 - 32.08 = 0.36 , \text{g} $$

3. Number of moles:

   • Moles of $BaCl_2$: $$ \frac{\text{Mass of } BaCl_2}{\text{Molar Mass of } BaCl_2} = \frac{2.08}{208} = 0.01 , \text{mol} $$
   • Moles of $H_2O$: $$ \frac{\text{Mass of } H_2O}{\text{Molar Mass of } H_2O} = \frac{0.36}{18} = 0.02 , \text{mol} $$

4. Simplest Ratio:

   • Divide the moles of each substance by the smallest number of moles: $$ \text{Ratio of } BaCl_2 : H_2O = \frac{0.01}{0.01} : \frac{0.02}{0.01} = 1 : 2 $$

5. Empirical Formula: $$ BaCl_2 \cdot 2H_2O $$

Thus, the formula of the hydrated salt is $BaCl_2 \cdot 2H_2O$.

1.24 Calculate Empirical and Molecular Formulae from Experimental Data ​

Converting Empirical Formula to Molecular Formula ​

The relative molecular mass of the compound is $56$.
The empirical formula mass of $CH_2$ is:

$$ 12 + (2 \times 1) = 14 $$

To find the molecular formula:

$$ \frac{\text{Relative Molecular Mass}}{\text{Empirical Formula Mass}} = \frac{56}{14} = 4 $$

Thus, the molecular formula is:

$$ C_4H_8 $$

1.25 Calculate Reacting Masses Using Experimental Data and Chemical Equations ​

Calculate the mass of magnesium oxide that can be made by completely burning $6 , \text{g}$ of magnesium in oxygen.

Equation for reaction: $2Mg + O_2 → 2MgO$

1. Calculate the amount, in moles, of magnesium reacted:
   $Ar$ of $Mg$ is $24$

   Amount of magnesium = $\frac{6}{24} = 0.25 , \text{mol}$

2. Calculate the amount of magnesium oxide formed:
   The equation tells us that $2 , \text{mol}$ of $Mg$ reacts to form $2 , \text{mol}$ of $MgO$, hence the amount of $MgO$ formed is the same as the amount of $Mg$ reacted.

   Amount of $MgO$ formed is $0.25 , \text{mol}$

3. Calculate the mass of $MgO$ formed:
   $Mr$ of $MgO = 24 + 16 = 40$

   Mass of magnesium oxide = $0.25 \times 40 = 10 , \text{g}$

1.26 Calculate Percentage Yield ​

The actual yield is the amount of product that's actually there to be used at the end of the manufacturing process.

The predicted yield is the amount that might have been expected if nothing had got lost along the way.

In practice, some product will be lost during the process when purifying the product by filtration or evaporation or when transferring a liquid or when heating.

The percentage yield is a way of comparing the actual yield with the predicted yield. It's calculated using a formula:

$$\text{Percentage yield} = \frac{\text{actual yield} \times 100%}{\text{predicted yield}}$$

1.27 Carry Out Mole Calculations Using Volumes and Molar Concentrations ​

$$\text{Mole} = \frac{\text{Volume (cm}^3)}{1000} \times \text{Concentration (mol/dm}^3)$$

$$\text{Concentration in g/dm}^3 = \text{Concentration in mol/dm}^3 \times \text{Mr}$$