Formulae and amounts of a substance

Welcome to Further Calculations in Chemistry

Welcome to Section 2.1: Formulae and amounts of a substance (often referred to as Further Calculations) in your CCEA AS 2 chemistry module! Quantitative chemistry is the numerical backbone of the physical sciences. Don't worry if calculations in chemistry have felt intimidating in the past. We will break every concept down into clear, logical, bite-sized steps with relatable analogies, memory aids, and step-by-step methods.

By mastering these principles, you will gain complete confidence in handling gas equations, calculating trace concentrations in solutions, evaluating the efficiency of industrial processes, and working out the exact chemical formulae of unknown substances.

1. Foundational Quantities and Core Definitions

Before diving into advanced calculations, let's review the fundamental definitions defined in the CCEA specification. In the exam, definitions must be stated precisely.

• Relative Atomic Mass (\(A_r\)): The weighted average mass of an atom of an element relative to \(\frac{1}{12}\text{th}\) of the mass of an atom of carbon-12.

• Relative Molecular Mass (\(M_r\)): The average mass of a molecule relative to \(\frac{1}{12}\text{th}\) of the mass of an atom of carbon-12. This is determined by summing the \(A_r\) values of all the atoms present in the molecule.

• Relative Formula Mass (\(M_r\)): Used specifically for giant crystalline or ionic substances (such as \(\text{NaCl}\) or \(\text{CaCO}_3\)), defined as the sum of the relative atomic masses of the atoms in the empirical formula relative to \(\frac{1}{12}\text{th}\) of the mass of an atom of carbon-12.

• The Mole: The amount of substance containing the same number of fundamental entities (atoms, molecules, or ions) as there are carbon atoms in exactly \(12\text{ g}\) of carbon-12. This number is known as Avogadro’s constant (\(L = 6.02 \times 10^{23}\text{ mol}^{-1}\)).

• Molar Mass (\(M\)): The mass of one mole of any substance, expressed in units of \(\text{g mol}^{-1}\).

The Fundamental Mass-Mole Relationship:
\(\text{Amount in moles } (n) = \frac{\text{mass in grams } (m)}{\text{Molar mass } (M_r)}\)

Everyday Analogy: Think of a "mole" just like a "dozen." A dozen always means 12 items (whether eggs or cars), but a dozen cars weighs vastly more than a dozen eggs. Similarly, 1 mole always contains \(6.02 \times 10^{23}\) particles, but 1 mole of lead (\(\text{Pb}\), \(M_r = 207.2\)) weighs much more than 1 mole of helium (\(\text{He}\), \(M_r = 4.0\)).

Key Takeaway: Always refer strictly to the standard of \(\frac{1}{12}\text{th}\) the mass of a carbon-12 atom when defining relative masses in your written answers.

2. Solution Chemistry, Mass Concentrations, and Parts Per Million (ppm)

Molar Concentration and Mass Concentration

Concentration tells us how much solute is dissolved in a given volume of solution.

• Molar concentration formula:
\(\text{Moles } (n) = \text{Concentration } (\text{mol dm}^{-3}) \times \text{Volume } (\text{dm}^3) = \frac{c \times V\text{ (in cm}^3\text{)}}{1000}\)

• Converting between mass concentration and molar concentration:
\(\text{Concentration in g dm}^{-3} = \text{Concentration in mol dm}^{-3} \times M_r\)

Parts Per Million (ppm)

When dealing with extremely dilute solutions — such as pollutants in river water, trace minerals in drinking water, or atmospheric gases — expressing concentration in \(\text{mol dm}^{-3}\) yields inconveniently tiny decimals. We use parts per million (ppm) instead.

Formula for ppm:
\(\text{ppm} = \left(\frac{\text{mass of component}}{\text{total mass of solution or mixture}}\right) \times 10^6\)

For very dilute aqueous solutions, because \(1\text{ dm}^3\) of water has a mass of approximately \(1000\text{ g}\) (\(1,000,000\text{ mg}\)), \(1\text{ ppm}\) is equivalent to:

\(1\text{ ppm} = 1\text{ mg dm}^{-3}\)

Worked Example:
A \(250\text{ cm}^3\) sample of tap water contains \(0.0015\text{ g}\) of dissolved fluoride ions (\(\text{F}^-\)). Calculate the concentration of fluoride in \(\text{ppm}\) (assume the density of the solution is \(1.00\text{ g cm}^{-3}\)).
• Mass of sample \(= 250\text{ cm}^3 \times 1.00\text{ g cm}^{-3} = 250\text{ g}\)
• \(\text{ppm} = \left(\frac{0.0015\text{ g}}{250\text{ g}}\right) \times 10^6 = 6.0\text{ ppm}\)

Key Takeaway: To convert from molarity to mass concentration, multiply by \(M_r\). Use \(\text{ppm}\) for tiny trace quantities by multiplying the mass ratio by \(10^6\).

3. Gas Calculations: RTP vs The Ideal Gas Equation

A. Room Temperature and Pressure (RTP)

Under CCEA specifications, RTP is defined as a temperature of \(293\text{ K}\) (\(20^\circ\text{C}\)) and a pressure of \(1\text{ atm}\) (\(101\text{ kPa}\)).

Under these specific conditions, 1 mole of any gas occupies a volume of \(24.0\text{ dm}^3\) (\(24,000\text{ cm}^3\)).

\(\text{Moles of gas at RTP} = \frac{\text{Volume in dm}^3}{24.0} = \frac{\text{Volume in cm}^3}{24000}\)

B. The Ideal Gas Equation: \(pV = nRT\)

What happens when a gas is not at standard room conditions? In AS 2, you use the Ideal Gas Equation to calculate properties at any given temperature and pressure:

\(pV = nRT\)

Where:
• \(p = \text{pressure measured in Pascals (Pa or N m}^{-2}\text{)}\)
• \(V = \text{volume measured in cubic metres (m}^3\text{)}\)
• \(n = \text{amount of gas in moles (mol)}\)
• \(R = \text{molar gas constant} = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\) (provided on the CCEA Data Leaflet)
• \(T = \text{temperature in Kelvin (K)}\)

Crucial Unit Conversions (The #1 Examiner Trap!)

The vast majority of student errors in \(pV = nRT\) calculations come from incorrect units. Memorise these conversion rules:

1. Pressure (\(p\)):
• From \(\text{kPa}\) to \(\text{Pa}\): multiply by \(1000\) (e.g., \(101\text{ kPa} = 101,000\text{ Pa}\))
• From \(\text{MPa}\) to \(\text{Pa}\): multiply by \(10^6\)

2. Volume (\(V\)):
• From \(\text{dm}^3\) to \(\text{m}^3\): divide by \(1000\) (\(\times 10^{-3}\))
• From \(\text{cm}^3\) to \(\text{m}^3\): divide by \(1,000,000\) (\(\times 10^{-6}\))

3. Temperature (\(T\)):
• From Celsius (\(^\circ\text{C}\)) to Kelvin (\(\text{K}\)): add \(273\) (e.g., \(20^\circ\text{C} = 20 + 273 = 293\text{ K}\))

Step-by-Step Worked Example:
Calculate the mass of a sample of carbon dioxide gas (\(\text{CO}_2\), \(M_r = 44.0\)) that occupies \(450\text{ cm}^3\) at a pressure of \(150\text{ kPa}\) and a temperature of \(27^\circ\text{C}\).

Step 1: Convert all values to SI units
• \(p = 150\text{ kPa} = 150 \times 10^3\text{ Pa} = 150,000\text{ Pa}\)
• \(V = 450\text{ cm}^3 = 450 \times 10^{-6}\text{ m}^3 = 0.000450\text{ m}^3\)
• \(T = 27^\circ\text{C} + 273 = 300\text{ K}\)
• \(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\)

Step 2: Rearrange \(pV = nRT\) to solve for moles (\(n\))
\(n = \frac{pV}{RT} = \frac{150,000 \times 0.000450}{8.31 \times 300} = \frac{67.5}{2493} \approx 0.02708\text{ mol}\)

Step 3: Calculate the mass
\(\text{Mass} = n \times M_r = 0.02708\text{ mol} \times 44.0\text{ g mol}^{-1} = 1.19\text{ g}\) (to 3 significant figures)

Key Takeaway: Always check your units before plugging numbers into \(pV = nRT\): Pressure in \(\text{Pa}\), Volume in \(\text{m}^3\), and Temperature in \(\text{K}\).

4. Reaction Efficiency: Percentage Yield and Atom Economy

In chemical manufacturing, chemical engineers want to produce products as efficiently, cheaply, and cleanly as possible. We use two complementary metrics to evaluate this: Percentage Yield and Atom Economy.

A. Percentage Yield

Percentage Yield compares the actual mass of product obtained in a practical reaction against the theoretical maximum mass that could be formed based on balanced stoichiometry:

\(\% \text{ Yield} = \left(\frac{\text{Actual Mass of Product Obtained}}{\text{Theoretical Maximum Mass of Product}}\right) \times 100\)

Why is percentage yield rarely \(100\%\)?
• Incomplete reaction: The reaction may not have gone to completion, or it may be a reversible reaction that reached dynamic equilibrium.
• Side reactions: Unexpected competing reactions may form unwanted side-products.
• Physical losses during purification: Product is inevitably lost on glassware, during filtration, transfer between containers, or recrystallisation.

B. Atom Economy

Atom Economy is a core principle of Green Chemistry. It measures the proportion of reactant atoms that are successfully converted into the desired useful product, rather than ending up as wasteful by-products.

\(\% \text{ Atom Economy} = \left(\frac{M_r\text{ of Desired Product}}{\sum M_r\text{ of All Reactants}}\right) \times 100 = \left(\frac{M_r\text{ of Desired Product}}{\sum M_r\text{ of All Products}}\right) \times 100\)

Note: In calculating the denominator, you must include stoichiometric coefficients (the balancing numbers in the equation).

Comparison: Percentage Yield vs Atom Economy

• Focus: Percentage yield measures practical reaction efficiency and laboratory technique. Atom economy measures theoretical atom efficiency and green sustainability.
• Addition Reactions: Have an atom economy of \(100\%\) because all reactant atoms combine into a single product (e.g., \(\text{CH}_2=\text{CH}_2 + \text{H}_2\text{O} \to \text{CH}_3\text{CH}_2\text{OH}\)).
• Substitution or Elimination Reactions: Often have lower atom economy because they produce stoichiometric by-products (e.g., salts, water, halides).

Worked Example:
Consider the extraction of iron in the blast furnace:
\(\text{Fe}_2\text{O}_3 + 3\text{CO} \to 2\text{Fe} + 3\text{CO}_2\)
Calculate the percentage atom economy for the production of pure iron (\(\text{Fe}\), \(A_r = 56.0\)).
• \(M_r\) of desired product \(= 2 \times 56.0 = 112.0\)
• Total \(M_r\) of all products \(= (2 \times 56.0) + 3 \times [12.0 + (2 \times 16.0)] = 112.0 + 3(44.0) = 112.0 + 132.0 = 244.0\)
• \(\% \text{ Atom Economy} = \left(\frac{112.0}{244.0}\right) \times 100 = 45.9\%\)

Key Takeaway: A reaction can have a \(100\%\) experimental yield but still have a low atom economy if the reaction pathway inherently generates large quantities of waste by-products.

5. Formula Determinations: Hydrated Salts and Combustion Analysis

A. Water of Crystallisation in Hydrated Salts

Many ionic compounds incorporate a fixed ratio of water molecules into their crystalline lattice, represented as \(\text{CuSO}_4 \cdot x\text{H}_2\text{O}\) or \(\text{MgSO}_4 \cdot x\text{H}_2\text{O}\).

Experimental Procedure:
1. Weigh an empty crucible and lid.
2. Add the hydrated salt and re-weigh.
3. Heat strongly with a Bunsen burner to drive off the water vapor.
4. Allow to cool and re-weigh.
5. Heat to constant mass: Repeat heating, cooling, and weighing until two consecutive mass measurements are identical. This guarantees all water of crystallisation has been completely expelled.

Calculating \(x\):
• \(\text{Mass of water lost} = \text{Mass of hydrated salt} - \text{Mass of anhydrous residue}\)
• \(\text{Moles of anhydrous salt} = \frac{\text{mass of anhydrous residue}}{M_r\text{ of anhydrous salt}}\)
• \(\text{Moles of water} = \frac{\text{mass of water lost}}{18.0}\)
• \(x = \frac{\text{moles of }\text{H}_2\text{O}}{\text{moles of anhydrous salt}}\)

B. Combustion Analysis of Organic Compounds

When an unknown organic compound containing carbon, hydrogen, and oxygen is burned completely in excess oxygen, all the carbon is converted to \(\text{CO}_2\) and all the hydrogen is converted to \(\text{H}_2\text{O}\).

Step-by-step determination of empirical formula:
Step 1: Calculate the mass of carbon in the sample from the mass of \(\text{CO}_2\) collected:
\(\text{Mass of C} = \text{mass of CO}_2 \times \frac{12.0}{44.0}\)

Step 2: Calculate the mass of hydrogen in the sample from the mass of \(\text{H}_2\text{O}\) collected:
\(\text{Mass of H} = \text{mass of H}_2\text{O} \times \frac{2.0}{18.0}\)

Step 3: Determine the mass of oxygen by subtraction from the initial total mass of the sample:
\(\text{Mass of O} = \text{total sample mass} - (\text{mass of C} + \text{mass of H})\)

Step 4: Convert masses to moles (\(\div A_r\)) and find the simplest whole-number molar ratio to determine the empirical formula.

Key Takeaway: In hydrated salt questions, failure to heat to constant mass leaves residual water, causing the calculated value of \(x\) to be artificially low.

6. Common Exam Pitfalls and Revision Checklist

Review this checklist before your AS 2 exam to ensure you don't drop avoidable marks:

• Significant Figures: Keep all intermediate numbers in your calculator memory throughout multi-step calculations. Round only your final answer to the appropriate number of significant figures (matching the data in the question with the fewest significant figures). Premature rounding is penalised on CCEA mark schemes.

• Balancing Numbers in Atom Economy: Don't forget to multiply the \(M_r\) of each substance by its stoichiometric balancing coefficient when finding total reactant or product masses.

• Temperature Conversions: Always convert Celsius to Kelvin (\(T = ^\circ\text{C} + 273\)). A temperature of \(0^\circ\text{C}\) is \(273\text{ K}\), not \(0\text{ K}\)!

• Volume Conversions in \(pV=nRT\): Remember that \(1\text{ m}^3 = 1000\text{ dm}^3 = 1,000,000\text{ cm}^3\). Double-check your powers of 10.

• Standard Gas Volume: Remember that at CCEA RTP (\(293\text{ K}\), \(101\text{ kPa}\)), \(1\text{ mole}\) of gas occupies \(24.0\text{ dm}^3\).