Welcome to Reacting Masses and Volumes!
Welcome, future chemists! This chapter is the backbone of quantitative chemistry. If you can master these skills, you can predict exactly how much product you will get, or how much starting material you need, for virtually any reaction.
Think of it as the ultimate recipe book for chemistry. Just like a chef needs to know how many grams of flour match up with how many millilitres of milk, you need to know how many moles of one chemical react with another.
Don't worry if the calculations look complex at first. We will break down every type of calculation—mass, volume, and concentration—into simple, manageable steps. Let's build your stoichiometric confidence!
1. The Foundation: Moles and Molar Mass
1.1 Defining Key Terms
To perform any calculation in this chapter, we must be comfortable using the concept of the mole (\(n\)).
- Relative Atomic Mass (\(A_r\)): The average mass of an atom of an element relative to $\frac{1}{12}$ the mass of a carbon-12 atom.
- Relative Molecular Mass (\(M_r\)): The sum of the relative atomic masses of all atoms shown in the molecular formula (for covalent compounds).
- Relative Formula Mass (\(M_r\)): Used specifically for ionic compounds (or when the structure is unknown). It is calculated in the same way as \(M_r\).
- The Mole (mol): The amount of substance that contains the same number of elementary particles as there are atoms in exactly 12 g of carbon-12. This number is the Avogadro constant (\(L\)): \(6.02 \times 10^{23}\) particles per mole.
1.2 Mass-to-Mole Conversion (The Key Formula)
The most fundamental tool in stoichiometry is converting mass (in grams) to moles, and vice versa.
$$ n = \frac{m}{M_r} $$
Where:
- \(n\) = number of moles (mol)
- \(m\) = mass of substance (g)
- \(M_r\) = relative molecular/formula mass (\(\text{g} \, \text{mol}^{-1}\))
Memory Tip: If you want to find the number of moles (n), put the smaller unit (mass, m) over the bigger unit (Molar Mass, \(M_r\)).
Key Takeaway for Section 1: Moles link the measurable world (mass, volume) to the world of atoms and equations. Always start by converting known quantities into moles!
2. Reacting Masses and Percentage Yield
2.1 Stoichiometry: The Mole Bridge (2.4.1a, 2.4.1e)
Stoichiometry simply means calculating the quantities of reactants and products involved in a chemical reaction. The crucial piece of information comes from the balanced chemical equation.
Step-by-Step Reacting Mass Calculation:
Let's find the mass of product B formed when a mass of reactant A is used.
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Balance the Equation: Ensure the equation is balanced. This gives you the mole ratio (the Stoichiometric Relationship, 2.4.1e).
Example: $2\text{A} + 1\text{B} \rightarrow 3\text{C}$. The mole ratio A:C is 2:3. - Mass $\rightarrow$ Moles (Reactant A): Convert the given mass of A into moles using $n = m/M_r$.
- Moles $\rightarrow$ Moles (Mole Bridge): Use the stoichiometric ratio from the balanced equation to find the moles of the unknown substance (B or C).
- Moles $\rightarrow$ Mass (Product C): Convert the calculated moles of C back into a mass using $m = n \times M_r$.
Common Mistake to Avoid: Never use the masses or the concentration values directly in the mole ratio step. The ratio only applies to moles.
2.2 Calculating Percentage Yield (2.4.1a)
In a perfect world, all reactants turn into products, giving the theoretical yield (what you calculated in step 4 above). In reality, mass is lost (spillage, side reactions, incomplete reactions), resulting in the actual yield.
The percentage yield tells us how efficient the reaction was.
$$ \text{Percentage Yield} = \frac{\text{Actual Yield (mass or moles)}}{\text{Theoretical Yield (mass or moles)}} \times 100\% $$
Did you know? Industrial chemists strive for high percentage yields because even a small increase can save millions of dollars in raw materials!
Quick Review: Reacting Masses
- Mass (g) $\rightarrow$ Moles (mol)
- Use the Mole Ratio (from the balanced equation).
- Moles (mol) $\rightarrow$ Mass (g)
3. Reactions in Solution: Concentration and Volume (2.4.1c)
When reactions happen in liquids (aqueous solutions), we use concentration to measure the amount of substance.
3.1 Concentration Definitions and Formulas
Concentration ($C$) is the amount of solute (in moles) dissolved per unit volume of solution.
The standard unit for concentration in AS Chemistry is moles per cubic decimetre (\(\text{mol} \, \text{dm}^{-3}\)), also known as molarity (\(\text{M}\)).
$$ n = C \times V $$
Where:
- \(n\) = number of moles (mol)
- \(C\) = concentration (\(\text{mol} \, \text{dm}^{-3}\))
- \(V\) = volume (\(\text{dm}^{3}\))
3.2 The Importance of Volume Units
This is where most students make a calculation error! You must convert volume units before using $n = C \times V$.
- $$ 1 \text{ dm}^3 = 1000 \text{ cm}^3 $$
- To convert from $\text{cm}^3$ to $\text{dm}^3$, divide by 1000.
Analogy: A $\text{dm}^3$ is about the size of a large carton of milk (1 L). A $\text{cm}^3$ is tiny (like a sugar cube).
3.3 Calculations Involving Titrations
A titration is a key experimental technique used to determine the exact concentration of a solution (often an acid or alkali).
Titration Calculation Steps:
- Balance the Equation: Find the mole ratio (e.g., $1 \text{ mol}$ acid reacts with $2 \text{ mol}$ alkali).
- Find Moles of Known Solution: Use the known concentration ($C_1$) and the measured volume ($V_1$) of the standard solution to calculate moles ($n_1$). Remember to convert $V_1$ to $\text{dm}^3$!
- Use the Mole Bridge: Use the stoichiometric ratio to find the moles of the unknown solution ($n_2$).
- Find Concentration of Unknown Solution: Use the calculated moles ($n_2$) and the measured volume ($V_2$) of the unknown solution to find its concentration ($C_2 = n_2 / V_2$). Convert $V_2$ to $\text{dm}^3$.
Key Takeaway for Section 3: When dealing with solutions, always ensure volume is in $\text{dm}^3$ before calculating moles.
4. Reactions Involving Volumes of Gases
Gases are special because, under the same conditions, equal volumes of gases contain the same number of molecules (Avogadro's Law). We have two main ways to calculate moles for gases depending on the conditions.
4.1 Molar Gas Volume (MGV) (2.4.1b)
At standard temperature and pressure (or room temperature and pressure, RTP, which is often $298 \text{ K}$ and $101 \text{ kPa}$), we can assume a fixed volume for one mole of any ideal gas.
- The most commonly used value at RTP is $24.0 \text{ dm}^3 \text{mol}^{-1}$.
$$ n = \frac{V (\text{dm}^3)}{24.0} $$
Note: Just like with solutions, volumes must be in $\text{dm}^3$. If given in $\text{cm}^3$, divide by 1000.
4.2 Using the Ideal Gas Equation (\(pV = nRT\)) (4.1)
When the reaction is NOT at RTP (or the conditions are not specified), you must use the Ideal Gas Equation.
$$ pV = nRT $$
Where:
- \(p\) = Pressure (must be in Pascals, Pa)
- \(V\) = Volume (must be in cubic metres, $\text{m}^3$)
- \(n\) = moles (mol)
- \(R\) = Ideal Gas Constant ($8.31 \text{ J} \, \text{K}^{-1} \, \text{mol}^{-1}$)
- \(T\) = Temperature (must be in Kelvin, K)
Crucial Unit Conversions for \(pV = nRT\):
This is the trickiest part! You MUST use SI units:
- Pressure: $\text{kPa} \times 1000 = \text{Pa}$
- Volume: $\text{dm}^3 \times 10^{-3} = \text{m}^3$ (or $\text{cm}^3 \times 10^{-6} = \text{m}^3$)
- Temperature: $\text{Temp } ({}^{\circ}\text{C}) + 273 = \text{Temp } (\text{K})$
Tip for $pV=nRT$: If you are calculating the $M_r$ of a gas, you can combine this with the mole formula: $n = m/M_r$. Substituting gives: $pV = (m/M_r)RT$. You can then solve for $M_r$.
Key Takeaway for Section 4: Check the conditions! If it's standard conditions (RTP), use $24.0 \text{ dm}^3$. If temperature and pressure are given specifically, use the $pV = nRT$ equation, ensuring all units are converted to SI units (Pa, $\text{m}^3$, K).
5. Limiting and Excess Reagents (2.4.1d)
Chemical reactions stop when one of the reactants is completely used up. This reactant is the limiting reagent.
5.1 Defining the Reagents
- Limiting Reagent: The reactant that is completely consumed during the reaction. It determines (limits) the maximum amount of product that can be formed (the theoretical yield).
- Excess Reagent: The reactant that is left over once the reaction has stopped.
Analogy: If you are making 10 sandwiches and you have 20 slices of bread and 3 kg of cheese. The bread (limiting reagent) will run out first, and you will have leftover cheese (excess reagent). The maximum number of sandwiches you can make is 10, limited by the bread.
5.2 Identifying the Limiting Reagent (The Critical Step)
You must identify the limiting reagent whenever quantities of *two or more* reactants are provided.
- Convert all known reactants to moles.
- Choose one reactant (A) and use the mole ratio (from the balanced equation) to calculate the moles of the other reactant (B) that would be needed to react completely with A.
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Compare: Compare the moles of B calculated as "needed" with the moles of B actually "available."
- If "Available" > "Needed," then B is in excess, and A is limiting.
- If "Available" < "Needed," then B is limiting, and A is in excess.
- Perform all subsequent calculations (e.g., finding product mass/volume) using the moles of the LIMITING REAGENT.
5.3 Calculating the Amount of Excess Reagent Remaining
The question might ask how much excess reagent is left over.
- Use the moles of the limiting reagent to calculate the moles of the excess reagent that were used up.
- Moles remaining = (Moles available) - (Moles used up).
- Convert the moles remaining back into the requested unit (mass or volume).
Key Takeaway for Section 5: If you have quantities for two reactants, you MUST determine the limiting reagent first. All product calculations rely solely on the limiting reagent.
Summary and Problem-Solving Strategy
Stoichiometry is about connecting different measurements via the mole concept. Here is your universal roadmap for solving nearly every quantitative chemistry problem:
The Universal Stoichiometry Triangle
(How to get to moles)
- For Solids (Mass): \(n = m / M_r\)
- For Solutions: \(n = C \times V\) ($\text{V}$ in $\text{dm}^3$)
- For Gases (at RTP): \(n = V / 24.0\) ($\text{V}$ in $\text{dm}^3$)
- For Gases (at non-standard conditions): \(n = pV / RT\) (Use SI units: Pa, $\text{m}^3$, K)
The Overall Plan of Attack
- Write the balanced chemical equation.
- Convert all known quantities (masses, volumes, concentrations) into moles.
- If two reactants are given, determine the limiting reagent.
- Use the mole ratio from the balanced equation (using the limiting reagent) to find the moles of the required substance (product or excess reagent remaining).
- Convert the final answer from moles back into the required unit (mass, volume, or concentration).
Practice, practice, practice! By systematically applying these steps and diligently checking your units, you will master reacting masses and volumes!