The Language of Chemistry: Formulae, Equations, and Calculations

Hello future chemist! Welcome to one of the most important chapters in IGCSE Chemistry. This chapter is the foundation for almost all calculations you will do, and it helps you understand what happens and how much of it happens during a chemical reaction.

Don't worry if maths isn't your favorite subject; we will break down every calculation into simple, easy-to-follow steps. Think of this chapter as learning the special language (formulae and equations) and the measuring tools (moles and mass) that chemists use every day!

1. Writing Chemical Formulae

A chemical formula is like a passport for a molecule—it tells you exactly which elements are present and how many atoms of each element are bonded together.

a. Valency: The Combining Power

Before writing a formula, we need to know the valency of the elements. Valency is simply the number of bonds an atom can form, or the charge an ion carries.

  • Elements in Group 1 have a valency of 1+ (e.g., Na+).
  • Elements in Group 2 have a valency of 2+ (e.g., Mg2+).
  • Elements in Group 3 have a valency of 3+ (e.g., Al3+).
  • Elements in Group 7 (Halogens) have a valency of 1- (e.g., Cl-).
  • Elements in Group 6 have a valency of 2- (e.g., O2-).

Analogy: Think of valency as the number of "hands" an atom has. When forming a stable compound, all the hands must be held!

b. Writing Formulae for Ionic Compounds (The Criss-Cross Method)

Ionic compounds (metals + non-metals) must be electrically neutral (total charge must be zero). We use the criss-cross method to figure out the ratio needed.

Step-by-step example: Magnesium Chloride

  1. Write down the ions and their charges:
    Magnesium (Group 2) is Mg2+. Chloride (Group 7) is Cl-.
  2. "Criss-Cross" the charge values (ignoring the + or - sign) to become the subscript for the other element.
  3. The 2 from Mg goes to Cl. The 1 from Cl goes to Mg.
  4. The resulting formula is MgCl2. (One Mg2+ ion balances two Cl- ions).

Common Mistake to Avoid: If the subscripts can be simplified (e.g., Ca2O2), you must simplify them (to CaO).

Polyatomic Ions: These are groups of atoms that act as a single unit with an overall charge (e.g., Sulfate \(\text{SO}_4^{2-}\), Nitrate \(\text{NO}_3^{-}\), Carbonate \(\text{CO}_3^{2-}\), Hydroxide \(\text{OH}^{-}\)). If you need more than one polyatomic ion, you must put it in brackets!

Example: Calcium Hydroxide. \(\text{Ca}^{2+}\) and \(\text{OH}^{-}\). Formula: \(\text{Ca(OH)}_2\).

c. Formulae for Covalent Compounds

Covalent compounds (non-metals only) usually use prefixes to indicate the number of atoms (mono=1, di=2, tri=3, tetra=4, etc.).

Example: Carbon dioxide means one carbon and two oxygen atoms: \(\text{CO}_2\).

Key Takeaway: Formulae tell us the exact ratio of atoms. Use the criss-cross method for ionic compounds, ensuring the overall charge is zero.

2. Writing and Balancing Chemical Equations

A chemical equation is a shorthand way of describing a reaction. It follows the fundamental rule: The Law of Conservation of Mass.

Did you know? This law, established by Antoine Lavoisier, means that mass is never created or destroyed in a chemical reaction. Therefore, the total mass of the reactants must equal the total mass of the products!

a. Structure of an Equation

Reactants (starting materials) \(\rightarrow\) Products (what is formed)

We use large numbers (coefficients) in front of the formulae to balance the equation. Subscripts (the small numbers) are part of the molecule and must never be changed.

b. State Symbols

State symbols show the physical state of the substance, which is essential information:

  • \((s)\): Solid
  • \((l)\): Liquid (e.g., water)
  • \((g)\): Gas
  • \((aq)\): Aqueous (dissolved in water)
c. Step-by-Step Guide to Balancing Equations

Balancing means ensuring that the number of atoms of each element is the same on both the reactant and product sides.

Example: Methane combustion (\(\text{CH}_4 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}\))

Step 1: Count the atoms on each side.

Reactants: C=1, H=4, O=2

Products: C=1, H=2, O=3

Step 2: Balance elements other than Hydrogen (H) and Oxygen (O) first. (C is already balanced: 1 and 1).

Step 3: Balance Hydrogen (H). We have 4 H on the left, but only 2 H on the right. Place a coefficient of 2 in front of \(\text{H}_2\text{O}\).

\(\text{CH}_4 + \text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}\)

Step 4: Balance Oxygen (O) last. Now count O on the product side: 2 in \(\text{CO}_2\) plus \(2 \times 1\) in \(2\text{H}_2\text{O}\). Total O = 4. We only have 2 O on the reactant side (\(\text{O}_2\)). Place a 2 in front of the reactant \(\text{O}_2\).

\( \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \)

Step 5: Check the final count.

Reactants: C=1, H=4, O=4

Products: C=1, H=4, O=4. (Balanced!)

Quick Review: Never change the subscripts. Use only coefficients (the big numbers) to balance. Balance O and H last!

3. Relative Mass and The Mole

To do calculations, we need a way to measure the 'amount' of substance. Atoms are too small to weigh individually, so we use relative masses and the concept of the mole.

a. Relative Atomic Mass (\(A_r\)) and Relative Formula Mass (\(M_r\))

The Relative Atomic Mass (\(A_r\)) is the average mass of an atom of an element compared to 1/12th the mass of a carbon-12 atom. This is the larger number found on the Periodic Table.

  • Example: \(A_r\) of Oxygen is 16. \(A_r\) of Sodium is 23.

The Relative Formula Mass (\(M_r\)) is the sum of the \(A_r\) values of all the atoms shown in the chemical formula.

Step-by-step example: Calculate \(M_r\) of \(\text{H}_2\text{SO}_4\)

  1. H: \(A_r = 1\). We have 2 H atoms: \(2 \times 1 = 2\)
  2. S: \(A_r = 32\). We have 1 S atom: \(1 \times 32 = 32\)
  3. O: \(A_r = 16\). We have 4 O atoms: \(4 \times 16 = 64\)
  4. Total \(M_r\): \(2 + 32 + 64 = \mathbf{98}\)

Note: \(M_r\) has no units, as it is a relative measure. However, when we use it in mole calculations, we give it units of grams per mole (\(\text{g/mol}\)).

b. The Mole: The Chemist's Dozen

The mole (abbreviated \(\text{mol}\)) is the unit for the amount of substance. Just like a "dozen" means 12, a "mole" means a specific, huge number of particles (atoms, molecules, or ions). This number is the Avogadro Constant (\(6.02 \times 10^{23}\)).

Analogy: A mole of marshmallows and a mole of bowling balls both contain the same number of items, but they weigh vastly different amounts!

The key connection is that:

1 mole of any substance has a mass in grams equal to its \(M_r\).

  • Example: Since the \(M_r\) of water (\(\text{H}_2\text{O}\)) is 18, 1 mole of water weighs 18 grams.
c. The Essential Mole Calculation Formula

This formula is vital. Memorize it!

$$ \text{Moles} = \frac{\text{Mass (g)}}{\text{Relative Formula Mass } (M_r)} $$ $$ \text{or } n = \frac{m}{M_r} $$

Example Calculation: How many moles are in 88 g of \(\text{CO}_2\)? (Given: \(M_r\) of \(\text{CO}_2\) is 44)

$$ \text{Moles} = \frac{88 \text{ g}}{44 \text{ g/mol}} = 2.0 \text{ mol} $$

Key Takeaway: \(A_r\) is mass from the periodic table, \(M_r\) is the sum of \(A_r\text{s}\). The mole links the mass (grams) to the formula mass (\(M_r\)).

4. Calculations Involving Formulae and Equations

These calculations allow us to figure out the exact composition of compounds and the required amounts in chemical reactions.

a. Empirical and Molecular Formulae

The Empirical Formula is the simplest whole-number ratio of atoms in a compound.

The Molecular Formula is the actual number of atoms of each element in a molecule.

Example: Ethane's molecular formula is \(\text{C}_2\text{H}_6\). Its empirical formula is \(\text{CH}_3\).

Step-by-step: Calculating the Empirical Formula

Example: A compound contains 40.0% Carbon, 6.7% Hydrogen, and 53.3% Oxygen.

  1. Assume 100g sample: Convert percentages to mass (C=40.0 g, H=6.7 g, O=53.3 g).
  2. Convert Mass to Moles: Divide the mass by the \(A_r\) (C=12, H=1, O=16).
    • C: \(40.0 / 12 = 3.33 \text{ mol}\)
    • H: \(6.7 / 1 = 6.7 \text{ mol}\)
    • O: \(53.3 / 16 = 3.33 \text{ mol}\)
  3. Find the Simplest Ratio: Divide all mole values by the smallest number (3.33).
    • C: \(3.33 / 3.33 = 1\)
    • H: \(6.7 / 3.33 \approx 2\)
    • O: \(3.33 / 3.33 = 1\)
  4. Write the Formula: The empirical formula is \(\text{CH}_2\text{O}\).

If you are then told the Molecular Mass (\(M_r\)) of the compound is 180, you can find the Molecular Formula:

Empirical \(M_r\) of \(\text{CH}_2\text{O}\) = \(12 + (2 \times 1) + 16 = 30\).

Factor needed = (Molecular \(M_r\)) / (Empirical \(M_r\)) = \(180 / 30 = 6\).

Multiply the empirical formula subscripts by 6: \(\text{C}_6\text{H}_{12}\text{O}_6\).

b. Reacting Mass Calculations (Stoichiometry)

This is where we use the coefficients in a balanced equation to predict how much product we can make or how much reactant we need. The coefficients give us the mole ratio.

Balanced Equation: \( \text{N}_2 (g) + 3\text{H}_2 (g) \rightarrow 2\text{NH}_3 (g) \)

Mole Ratio: 1 mole of \(\text{N}_2\) reacts with 3 moles of \(\text{H}_2\) to produce 2 moles of \(\text{NH}_3\).

The 4-Step Calculation Process (The Moles Path):

Question: What mass of ammonia (\(\text{NH}_3\)) is produced from 42 g of nitrogen gas (\(\text{N}_2\))? (\(A_r\): N=14, H=1)

Step 1: Find \(M_r\) of known substance (\(\text{N}_2\)) and unknown substance (\(\text{NH}_3\)).

\(M_r (\text{N}_2)\) = \(2 \times 14 = 28\).
\(M_r (\text{NH}_3)\) = \(14 + (3 \times 1) = 17\).

Step 2: Convert Mass of Known Substance to Moles.

$$ \text{Moles } \text{N}_2 = \frac{42 \text{ g}}{28 \text{ g/mol}} = 1.5 \text{ mol} $$

Step 3: Use the Mole Ratio to find Moles of Unknown Substance (\(\text{NH}_3\)).

Ratio \(\text{N}_2 : \text{NH}_3\) is 1 : 2.

Moles \(\text{NH}_3 = 1.5 \times 2 = 3.0 \text{ mol}\)

Step 4: Convert Moles of Unknown Substance back to Mass.

$$ \text{Mass } \text{NH}_3 = \text{Moles} \times M_r = 3.0 \text{ mol} \times 17 \text{ g/mol} = \mathbf{51 \text{ g}} $$

Quick Tip: If you get stuck, remember the path: Mass A \(\rightarrow\) Moles A \(\rightarrow\) Moles B \(\rightarrow\) Mass B.

c. Limiting Reactants (Optional but Important)

In most real experiments, one reactant runs out before the other. This substance is the limiting reactant (or limiting reagent). Once it's gone, the reaction stops, limiting the amount of product formed.

Analogy: Making cheese sandwiches. If you have 10 slices of bread and 1 slice of cheese, the cheese is the limiting reactant. You can only make 1 sandwich, even though you have lots of bread left over.

To identify the limiting reactant, calculate the number of moles available for each reactant, and then compare these moles using the mole ratio from the balanced equation.

5. Efficiency of Reactions: Yield and Atom Economy

In industry, chemists need reactions to be safe, fast, and, crucially, efficient. Efficiency is measured using percentage yield and atom economy.

a. Percentage Yield

The theoretical yield is the maximum mass of product that could be formed (calculated using stoichiometry, as in Section 4b).

The actual yield is the mass of product actually obtained in the lab (always less than the theoretical yield).

Percentage yield tells us how successful the reaction was:

$$ \text{Percentage Yield} = \frac{\text{Actual Mass Yield}}{\text{Theoretical Mass Yield}} \times 100\% $$

Yield is often less than 100% because:

  • The reaction may be incomplete or reversible.
  • Some product may be lost when separated or purified (e.g., sticking to glassware).
  • Side reactions may occur, producing unwanted by-products.
b. Atom Economy

Atom economy is a measure of how efficiently the atoms from the reactants are incorporated into the desired product. A high atom economy is better for the environment (less waste).

$$ \text{Atom Economy} = \frac{M_r \text{ of Desired Product}}{\text{Sum of } M_r \text{ of ALL Reactants}} \times 100\% $$

Example: If a reaction produces 100 g of desired product and 50 g of waste, the total mass of reactants used was 150 g. The atom economy is \( (100 / 150) \times 100\% = 66.7\% \).

Key Takeaway: High percentage yield means you didn't lose much product. High atom economy means you didn't make much waste.

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Quick Review Checklist

  • Can I write the formula for an ionic compound using valency?
  • Can I balance a chemical equation by using coefficients?
  • Can I calculate the \(M_r\) of any compound?
  • Do I know the relationship: \(\text{Moles} = \text{Mass} / M_r\)?
  • Can I use the mole ratio from a balanced equation to find reacting masses?
  • Can I calculate Percentage Yield and Atom Economy?

You have tackled the biggest calculation chapter in IGCSE Chemistry! Mastering the mole concept takes practice, so keep practicing those 4-step calculations. You’ve got this!