Chemistry (9620) Study Notes: 3.1.2 Amount of Substance
Hello future chemist! Welcome to the heart of Physical Chemistry: the "Amount of Substance" chapter. Don't worry if this chapter involves a lot of numbers—it's essentially learning the language chemists use to count atoms and molecules. Understanding this is the foundation for almost every other calculation you will do in A-Level Chemistry!
In simple terms, this chapter teaches you how to measure exactly how much stuff you have, whether it's a solid, a liquid solution, or a gas, so you can make chemical reactions work perfectly.
3.1.2.1 Relative Atomic Mass and Relative Molecular Mass
Atoms are tiny, so we can’t weigh them directly on a balance. Instead, we use a system of comparison. This system relies on the standard definition involving a specific isotope of carbon.
Key Definitions Based on Carbon-12
The standard reference point for all atomic masses is the isotope Carbon-12 (\(^{12}C\)).
- Relative Atomic Mass (\(A_r\)):
Definition: The weighted mean mass of an atom of an element compared with \(1/12\)th the mass of an atom of carbon-12. - Relative Molecular Mass (\(M_r\)):
Definition: The sum of the relative atomic masses of the atoms shown in the molecular formula. This is used for substances that exist as discrete molecules (like \(\text{H}_2\text{O}\) or \(\text{CO}_2\)). - Relative Formula Mass (RFM):
Used for: Ionic compounds (like \(\text{NaCl}\) or \(\text{MgBr}_2\)). Since ionic compounds don't form true molecules, we use the term Relative Formula Mass. The calculation is exactly the same as for \(M_r\): sum up all the \(A_r\) values in the formula.
Think of it like this: If carbon-12 has a standard weight of exactly 12 units, then hydrogen (which is much lighter) has an \(A_r\) of about 1.0, and oxygen (which is heavier) has an \(A_r\) of about 16. We are just comparing their weights to this standard unit.
Quick Review: Relative Mass
- \(A_r\) is the average mass of an atom, relative to \(1/12\)th of a \(^{12}C\) atom.
- \(M_r\) or RFM is the sum of the \(A_r\) values in the compound's formula.
3.1.2.2 The Mole and the Avogadro Constant
Imagine trying to count grains of sugar in a teaspoon. Impossible, right? Chemists deal with billions of particles, so they needed a standard counting unit: the Mole.
The Concept of the Mole (n)
The mole (\(n\)) is the fundamental unit for the amount of substance. It is defined as the amount of substance that contains the same number of specified entities (atoms, molecules, ions, etc.) as there are atoms in exactly 12 g of carbon-12.
Analogy: Just as a "dozen" means 12 eggs, a "mole" means a specific, massive number of particles.
The Avogadro Constant (\(N_A\))
The specific number of particles in one mole is known as the Avogadro Constant (\(N_A\)).
- \(N_A = 6.022 \times 10^{23}\ \text{particles per mole}\)
Important Note: You will not be expected to recall the numerical value of the Avogadro constant for the exams, but you must know how to use it in calculations.
Calculations involving Moles
You need to be able to switch easily between moles, mass, number of particles, and concentration/volume.
1. Moles, Mass, and \(M_r\)
This is the most fundamental relationship:
\[n = \frac{m}{M_r}\]
Where:
\(n\) = Amount of substance (moles, mol)
\(m\) = Mass of substance (grams, g)
\(M_r\) = Relative molecular mass (or formula mass, \(\text{g mol}^{-1}\))
Step-by-Step Trick (The Molar Triangle):
Cover the variable you want to find. If they are side-by-side, multiply; if they are top/bottom, divide.
Mass = Moles \(\times\) \(M_r\)
2. Moles and Number of Particles
To find the actual number of atoms, molecules, or ions:
\[\text{Number of Particles} = n \times N_A\]
Step-by-Step Example:
Question: How many molecules are in 0.5 mol of water?
Answer: \(\text{0.5 mol} \times (6.022 \times 10^{23}\ \text{mol}^{-1}) = 3.011 \times 10^{23}\ \text{molecules}\)
3. Concentration of Solutions
Concentration tells us how much substance (solute) is dissolved in a specific volume of solution. It is measured in mol \(\text{dm}^{-3}\) (moles per cubic decimetre).
\[c = \frac{n}{V}\]
Where:
\(c\) = Concentration (\(\text{mol dm}^{-3}\))
\(n\) = Amount of substance (mol)
\(V\) = Volume of solution (\(\text{dm}^3\))
Crucial Unit Conversion Reminder:
Volume must be in \(\text{dm}^3\)!
\(1\ \text{dm}^3 = 1000\ \text{cm}^3\ \text{(or 1 Litre)}\)
If you are given volume in \(\text{cm}^3\), remember to divide by 1000 to convert to \(\text{dm}^3\).
Key Takeaway
The mole links the microscopic world (atoms) to the macroscopic world (grams and litres), allowing us to measure and use chemicals in the laboratory accurately.
3.1.2.3 The Ideal Gas Equation
Gases behave very differently from solids or liquids, so we need a special equation to relate their amount (moles) to their physical conditions (pressure, volume, temperature).
The Ideal Gas Equation
The equation that describes the behaviour of an ideal gas (an approximation of real gases) is:
\[pV = nRT\]
Where:
- \(p\) = Pressure (Pascals, Pa)
- \(V\) = Volume (Cubic metres, \(\text{m}^3\))
- \(n\) = Amount of substance (moles, mol)
- \(R\) = The Gas Constant (This value will be provided in exams)
- \(T\) = Absolute Temperature (Kelvin, K)
Danger! Unit Alert!
This equation requires SI units. If you use the wrong units, your answer will be wildly wrong!
- Pressure (p): Often given in \(\text{kPa}\). Convert to \(\text{Pa}\) by multiplying by 1000. (\(1\ \text{kPa} = 1000\ \text{Pa}\)).
- Volume (V): Often given in \(\text{dm}^3\) or \(\text{cm}^3\). Convert to \(\text{m}^3\).
\(\text{dm}^3 \rightarrow \text{m}^3\): Divide by 1000.
\(\text{cm}^3 \rightarrow \text{m}^3\): Divide by 1,000,000 (\(10^6\)). - Temperature (T): Often given in \({}^{\circ}\text{C}\). Convert to \(\text{K}\) by adding 273. (\(T(\text{K}) = T({}^{\circ}\text{C}) + 273\)).
Did you know? Real gases only behave "ideally" at high temperatures and low pressures. At other conditions, the volume of the particles themselves and the forces between them become significant.
Key Takeaway
The Ideal Gas Equation is a powerful tool, but mastering the unit conversions (Pa, \(\text{m}^3\), K) is essential for success.
3.1.2.4 Empirical and Molecular Formula
When you analyse a compound, you might find its elemental composition by mass or percentage. From this data, you can determine its formula.
Definitions
- Empirical Formula: The simplest whole number ratio of atoms of each element in a compound.
- Molecular Formula: The actual number of atoms of each element in a molecule of the compound.
Example: Ethane has the molecular formula \(\text{C}_2\text{H}_6\). The simplest whole number ratio is \(\text{C}\text{H}_3\). Therefore, \(\text{C}\text{H}_3\) is the empirical formula.
Step-by-Step: Calculating the Empirical Formula
To find the simplest ratio, you must first convert the mass data into moles.
- Write down the mass (g) or percentage (%) for each element provided in the data.
- Convert Mass/Percentage to Moles (\(n\)): Divide the mass (or percentage) of each element by its Relative Atomic Mass (\(A_r\)).
\[n = \frac{\text{Mass}}{\text{A}_r}\] - Find the simplest whole number ratio: Divide all the mole values calculated in Step 2 by the smallest mole value obtained.
- Round (if necessary): If the numbers are not whole (e.g., you get 1.5, 2.5, 3.33), multiply all numbers by a common factor (e.g., multiply by 2 for .5 ratios, or by 3 for .33 ratios) until you get whole numbers.
Step-by-Step: Calculating the Molecular Formula
Once you have the Empirical Formula, you need the Relative Molecular Mass (\(M_r\)) of the compound to find the molecular formula.
- Calculate the Mass of the Empirical Formula (\(M_{\text{Empirical}}\)).
- Find the Multiplier (\(x\)): Divide the known \(M_r\) of the actual molecule by the \(M_{\text{Empirical}}\).
\[x = \frac{\text{Known } M_r}{\text{Mass of Empirical Formula}}\] - Calculate the Molecular Formula: Multiply all the subscripts in the empirical formula by the multiplier (\(x\)).
Common Mistake to Avoid: Students sometimes forget to multiply the final empirical formula by the integer derived in Step 3. Remember, the molecular formula is always an integer multiple of the empirical formula.
Key Takeaway
Empirical calculations rely entirely on the mole concept: convert mass to moles, then find the simplest ratio.
3.1.2.5 Balanced Equations and Associated Calculations
A balanced chemical equation is more than just symbols; it's a quantitative recipe. The coefficients (big numbers) tell us the mole ratios of the reactants and products.
Using Mole Ratios (Stoichiometry)
If you have a reaction like: \(\text{A} + 2\text{B} \rightarrow \text{C}\)
This means 1 mole of A reacts with 2 moles of B to produce 1 mole of C. You can use these ratios to calculate masses, volumes, or concentrations of any reactant or product.
The Core Calculation Path:
If you know the mass of substance A and want to find the mass of product C:
Mass A \(\xrightarrow{/ M_r \text{ of A}}\) Moles A \(\xrightarrow{\text{Mole Ratio}}\) Moles C \(\xrightarrow{\times M_r \text{ of C}}\) Mass C
I. Percentage Yield
In the lab, reactions rarely go perfectly; we often lose product (spillage, incomplete reaction, side reactions). Percentage yield measures how successful the reaction was.
- Theoretical Mass: The mass of product that should be made, calculated using stoichiometry (the core calculation path above).
- Actual Mass: The mass of product actually obtained in the experiment.
Formula:
\[\text{Percentage Yield} = \frac{\text{Actual Mass of Product}}{\text{Theoretical Mass of Product}} \times 100\]
II. Percentage Atom Economy
In industry, it's not enough to get a high yield; we also want to be efficient and green. Atom economy measures how many atoms from the reactants end up in the desired product, rather than in waste by-products.
Formula:
\[\text{Percentage Atom Economy} = \frac{\text{Molecular Mass of Desired Product}}{\text{Sum of Molecular Masses of All Reactants}} \times 100\]
Economic, Ethical and Environmental Advantages:
Chemical processes with a high atom economy are preferred because:
- Economic: Less raw material is wasted, saving money.
- Environmental: Less waste is produced, reducing disposal costs and environmental impact (making the process "greener").
III. Solution Calculations (Titrations)
This links back to concentration (\(c = n/V\)). You must be able to use balanced equations to link the moles reacting in solution.
Calculation Steps in Titration:
- Calculate moles of the known substance: Use the known concentration and volume (\(n = c \times V\)).
- Use the mole ratio: Look at the balanced equation to find the moles of the unknown substance.
- Calculate the unknown quantity: Use the moles of the unknown substance to find the concentration (if volume is known) or mass (if \(M_r\) is known).
Key Takeaway
Stoichiometry is about using the mole ratio from the balanced equation to link the quantities (masses, volumes, concentrations) of everything involved in the reaction.
Required Practical 1: Titration
Practical application of "Amount of Substance" is demonstrated in Required Practical 1: making a volumetric solution and carrying out a simple acid-base titration.
This practical skill confirms your understanding of concentration and mole ratios by allowing you to find the exact amount of acid needed to neutralise an unknown amount of base (or vice versa).
- Volumetric Solution: Preparing a solution of precisely known concentration (usually involves weighing a primary standard and dissolving it in a volumetric flask).
- Titration: Accurately measuring the volume of one solution (from a burette) needed to react completely with a measured volume of another solution (in a conical flask), using an indicator to find the equivalence point.
Mastering this practical proves you can handle the concepts of moles and concentration in a real laboratory setting!