Welcome to Quantitative Chemistry (Unit C1: Section 1.7)
Have you ever baked a cake? If you don't measure your flour, sugar, and eggs accurately, the cake won't turn out right! Chemistry works in the exact same way. Quantitative chemistry is simply the study of chemical measurements, masses, and amounts. It allows chemists to calculate exactly how much reactant they need to make a specific amount of product.
Don't worry if maths in science feels daunting at first! In this chapter, we will break every calculation down into simple, step-by-step recipes so you can tackle any question on your CCEA exam with confidence.
---1. Relative Formula Mass (\(M_r\))
Every element on the Periodic Table has a relative atomic mass (\(A_r\)). You do not need to memorise these numbers because they are always provided on your CCEA Data Leaflet / Periodic Table.
How to Find Relative Formula Mass (\(M_r\)):
The relative formula mass (\(M_r\)) of a compound is calculated by adding together the relative atomic masses (\(A_r\)) of all the atoms shown in the chemical formula.
Step-by-Step Example 1: Water (\(\text{H}_2\text{O}\))
• Look up the \(A_r\) values on the Periodic Table: \(A_r(\text{H}) = 1\), \(A_r(\text{O}) = 16\)
• Count the atoms: 2 Hydrogen atoms and 1 Oxygen atom
• Add them up: \(M_r = (2 \times 1) + (1 \times 16) = 2 + 16 = 18\)
Step-by-Step Example 2: Calcium Hydroxide (\(\text{Ca(OH)}_2\))
• Look up the \(A_r\) values: \(A_r(\text{Ca}) = 40\), \(A_r(\text{O}) = 16\), \(A_r(\text{H}) = 1\)
• Notice the brackets! The subscript \(2\) multiplies everything inside the bracket: 1 Calcium, 2 Oxygen, 2 Hydrogen
• Add them up: \(M_r = (1 \times 40) + (2 \times 16) + (2 \times 1) = 40 + 32 + 2 = 74\)
Common Pitfall to Avoid: On your Periodic Table, each element has two numbers. The smaller number is the atomic number (number of protons). The larger number is the relative atomic mass (\(A_r\)). Always use the larger number (\(A_r\)) for mass calculations!
Key Takeaway: \(M_r\) is simply the sum of all the individual atomic masses (\(A_r\)) in a chemical formula.
---2. Percentage of an Element by Mass
Sometimes you need to find out what fraction of a compound's total mass comes from one specific element. For example, what percentage of water is just oxygen?
The Formula:
\(\text{Percentage of an element by mass} = \left(\frac{\text{total mass of atoms of the element in formula}}{M_r \text{ of compound}}\right) \times 100\)
Worked Example:
Question: Calculate the percentage of oxygen by mass in calcium carbonate (\(\text{CaCO}_3\)).
Given \(A_r\) values: \(\text{Ca} = 40\), \(\text{C} = 12\), \(\text{O} = 16\).
• Step 1: Calculate the \(M_r\) of \(\text{CaCO}_3\):
\(M_r = 40 + 12 + (3 \times 16) = 40 + 12 + 48 = 100\)
• Step 2: Find the total mass of oxygen in the formula:
\(\text{Total mass of oxygen} = 3 \times 16 = 48\)
• Step 3: Apply the formula:
\(\text{Percentage of oxygen} = \left(\frac{48}{100}\right) \times 100 = 48\%\)
Key Takeaway: Divide the mass contribution of the target element by the total \(M_r\) of the compound, then multiply by \(100\).
---3. The Mole Concept
Atoms and molecules are unimaginably tiny. If you tried counting them one by one, it would take forever! Instead, chemists count particles in large bundles called moles (symbol: \(\text{mol}\)).
Think of it like buying eggs: you ask for a "dozen" when you want 12 eggs. A "mole" is just a chemist's counting word for a very large, fixed collection of particles!
The Mass of One Mole:
The mass of 1 mole of any substance in grams (\(\text{g}\)) is numerically equal to its relative atomic mass (\(A_r\)) or relative formula mass (\(M_r\)). This is known as its molar mass (\(\text{g/mol}\)).
• 1 mole of Carbon atoms (\(A_r = 12\)) has a mass of \(12\text{ g}\)
• 1 mole of Water molecules (\(M_r = 18\)) has a mass of \(18\text{ g}\)
• 1 mole of Calcium hydroxide (\(M_r = 74\)) has a mass of \(74\text{ g}\)
Core Conversion Formulae:
\(\text{Moles} = \frac{\text{Mass (g)}}{M_r \text{ (or } A_r \text{)}}\)
\(\text{Mass (g)} = \text{Moles} \times M_r \text{ (or } A_r \text{)}\)
Worked Example 1: Finding Moles from Mass
Question: How many moles are there in \(44\text{ g}\) of carbon dioxide (\(\text{CO}_2\))?
Given \(A_r\) values: \(\text{C} = 12\), \(\text{O} = 16\).
• Step 1: Calculate \(M_r\) of \(\text{CO}_2 = 12 + (2 \times 16) = 44\)
• Step 2: Use the formula: \(\text{Moles} = \frac{\text{Mass}}{M_r} = \frac{44\text{ g}}{44} = 1\text{ mol}\)
Worked Example 2: Finding Mass from Moles
Question: What is the mass of \(0.5\text{ mol}\) of magnesium oxide (\(\text{MgO}\))?
Given \(A_r\) values: \(\text{Mg} = 24\), \(\text{O} = 16\).
• Step 1: Calculate \(M_r\) of \(\text{MgO} = 24 + 16 = 40\)
• Step 2: Use the formula: \(\text{Mass} = \text{Moles} \times M_r = 0.5 \times 40 = 20\text{ g}\)
Common Pitfall to Avoid: Always ensure mass is in grams (\(\text{g}\)) before using the formula. If an exam question gives mass in kilograms (\(\text{kg}\)), convert it by multiplying by \(1000\) (e.g. \(2\text{ kg} = 2000\text{ g}\)).
Key Takeaway: \(\text{Moles} = \frac{\text{Mass}}{M_r}\). Mass must always be in grams.
---4. Reacting Mass Calculations & Limiting Reactants
The 3-Step Method for Reacting Mass Calculations:
A balanced chemical equation tells you the ratio of moles reacting together. You can use this to calculate the mass of product made or reactant needed using 3 reliable steps:
• Step 1: Calculate the moles of the substance with the known mass: \(\text{Moles} = \frac{\text{Mass}}{M_r}\)
• Step 2: Use the balanced equation ratio to find the moles of the unknown substance.
• Step 3: Convert those moles into mass: \(\text{Mass} = \text{Moles} \times M_r\)
Worked Example:
Question: What mass of magnesium oxide (\(\text{MgO}\)) is produced when \(12\text{ g}\) of magnesium (\(\text{Mg}\)) is completely burned in oxygen?
\(2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}\)
Given \(A_r\) values: \(\text{Mg} = 24\), \(\text{O} = 16\).
• Step 1 (Find moles of known):
\(\text{Moles of Mg} = \frac{\text{Mass}}{A_r} = \frac{12\text{ g}}{24} = 0.5\text{ mol}\)
• Step 2 (Use equation ratio):
The equation shows: \(2\text{ mol of Mg} \rightarrow 2\text{ mol of MgO}\) (a \(1:1\) ratio).
Therefore, \(0.5\text{ mol of Mg}\) produces \(0.5\text{ mol of MgO}\).
• Step 3 (Convert to mass of unknown):
\(M_r\text{ of MgO} = 24 + 16 = 40\)
\(\text{Mass of MgO} = \text{Moles} \times M_r = 0.5 \times 40 = 20\text{ g}\)
Common Pitfall to Avoid: When calculating \(M_r\) in Step 1 or Step 3, do not multiply by the big balancing number in the equation! The \(M_r\) of \(\text{MgO}\) is \(40\), not \(80\). The big balancing numbers are only used for the mole ratio in Step 2.
Limiting Reactants and Excess
In a real chemical reaction, reactants are rarely mixed in exact stoichiometric proportions.
• Limiting Reactant: The reactant that gets completely used up first. Because it runs out first, it limits and determines the maximum amount (theoretical yield) of product that can form.
• Reactant in Excess: The reactant that is left over after the reaction has finished. There is more of it than needed.
Everyday Analogy: Imagine you are making toasties. Each toastie requires 2 slices of bread and 1 slice of cheese. If you have 10 slices of bread but only 2 slices of cheese, you can only make 2 toasties! The cheese runs out first, so cheese is the limiting reactant. Bread is in excess (6 slices left over).
Key Takeaway: The limiting reactant controls how much product forms because the reaction stops as soon as it is entirely consumed.
---5. Theoretical Yield and Percentage Yield
Key Definitions:
• Theoretical Yield: The maximum possible mass of product that could be formed, calculated from the balanced equation assuming 100% of reactants convert into products.
• Actual Yield: The mass of product actually collected and weighed when you carry out the experiment in the laboratory.
The Percentage Yield Formula:
\(\text{Percentage Yield} = \left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right) \times 100\)
Worked Example:
Question: A student calculates that the theoretical yield of copper(II) sulfate crystals should be \(25\text{ g}\). After carrying out the experiment, purifying, and drying the crystals, the actual yield is \(20\text{ g}\). Calculate the percentage yield.
• Apply formula:
\(\text{Percentage Yield} = \left(\frac{20\text{ g}}{25\text{ g}}\right) \times 100 = 80\%\)
Common Pitfall to Avoid: Your percentage yield can never be greater than \(100\%\) in theory! If you get a value over \(100\%\), check if you accidentally divided the theoretical yield by the actual yield instead of actual by theoretical.
Why is the Percentage Yield Almost Always Less Than 100%?
In school experiments and industrial manufacturing, yields rarely reach \(100\%\). The four main reasons required by the CCEA specification are:
1. Reaction is reversible or incomplete: Some reactions do not go to completion because products reform reactants.
2. Loss of product during physical separation: Product is lost when transferring liquids between beakers, filtering (trapped on filter paper), or during crystallisation.
3. Unexpected side reactions: Unintended reactions may occur, producing unwanted side products instead of the desired product.
4. Impurities in reactants: Starting materials may not be 100% pure, reducing the amount of actual reactant available.
Key Takeaway: \(\text{Percentage Yield} = \left(\frac{\text{Actual}}{\text{Theoretical}}\right) \times 100\). It is less than \(100\%\) due to physical loss, incomplete reactions, side reactions, or impurities.
---Quick Revision Summary Checklist
• Relative Formula Mass (\(M_r\)): Sum of all \(A_r\) values in the formula (use the larger number on the Periodic Table).
• Percentage by Mass: \(\left(\frac{\text{mass of element in formula}}{M_r}\right) \times 100\)
• Mole Calculations: \(\text{Moles} = \frac{\text{Mass (g)}}{M_r}\) and \(\text{Mass (g)} = \text{Moles} \times M_r\)
• Reacting Masses: 1. Moles of known \(\rightarrow\) 2. Mole ratio \(\rightarrow\) 3. Mass of unknown.
• Limiting Reactant: Used up first; dictates the amount of product formed.
• Percentage Yield: \(\left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right) \times 100\)