Welcome to Quantitative Chemistry: The Art of Measurement!
Hello future chemist! This chapter, Conservation of Mass and Chemical Equations, is incredibly important. It moves chemistry from just describing reactions to measuring them.
Don't worry if maths isn't your favorite subject—we’re going to break down all the calculations into simple, logical steps. By the end, you will be able to predict exactly how much product you can make from a given amount of starting material!
Let's learn how to weigh a chemical reaction!
Section 1: The Law of Conservation of Mass
The Golden Rule of Chemical Reactions
The most fundamental principle in all of chemistry is the Law of Conservation of Mass.
What does it mean?
In any chemical reaction, mass is neither created nor destroyed. The total mass of the reactants (the starting materials) must equal the total mass of the products (what is formed).
Analogy: The LEGO Principle
Imagine you have a specific number of LEGO bricks. You can take them apart and build something completely different (a chemical reaction), but you will still have the exact same number of bricks and therefore the exact same total mass.
Atoms are like these bricks: they are rearranged, but their number (and total mass) remains constant.
Consequences for Chemical Equations
Because mass must be conserved, all chemical equations must be balanced.
- A balanced equation shows the same number of atoms of each element on both sides (reactant side and product side).
- If the atoms balance, the mass must also balance!
Example: The reaction between Hydrogen and Oxygen
Unbalanced: \(H_2 + O_2 \rightarrow H_2O\) (We have 2 O atoms on the left, but only 1 on the right!)
Balanced: \(2H_2 + O_2 \rightarrow 2H_2O\) (Now we have 4 H atoms and 2 O atoms on both sides.)
Common Pitfall: Reactions Involving Gases
Sometimes, students think mass has been lost in an experiment. This usually happens when gases are involved.
- If a gas is produced and escapes into the air (e.g., burning wood, or reacting an acid with a carbonate), the remaining mass of the solid/liquid materials will seem lower. The mass wasn't lost; the gas just floated away!
- If a gas reactant is used up from the air (e.g., burning Magnesium in Oxygen), the final product mass will be higher than the initial reactant mass. The mass wasn't created; the reactant gas was incorporated!
Key Takeaway: Mass is always conserved. If mass seems to change, it means gases have either escaped or been absorbed.
Section 2: Relative Formula Mass (\(M_r\))
Before we can link the mass of reactants to products, we need a way to measure the "weight" of a single molecule or formula unit.
1. Relative Atomic Mass (\(A_r\))
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.
Don't worry about the definition; just know that the \(A_r\) is the large number shown on the Periodic Table for each element.
Example: Carbon (\(C\)) has an \(A_r\) of 12; Oxygen (\(O\)) has an \(A_r\) of 16.
2. Calculating Relative Formula Mass (\(M_r\))
The Relative Formula Mass (\(M_r\)) (sometimes called Relative Molecular Mass) is the sum of the \(A_r\) values of all atoms shown in the chemical formula.
Step-by-Step Example: Calculating the \(M_r\) of Water (\(H_2O\))
Assume \(A_r\) values: Hydrogen (H) = 1, Oxygen (O) = 16.
- Identify all atoms: 2 Hydrogen atoms, 1 Oxygen atom.
-
Multiply \(A_r\) by the number of atoms:
- H: \(2 \times 1 = 2\)
- O: \(1 \times 16 = 16\)
-
Add them up:
\(M_r\) of \(H_2O = 2 + 16 = 18\)
Important Tip for Brackets: If you see brackets, multiply everything inside the bracket by the subscript outside it.
Example: \(M_r\) of Calcium Hydroxide, \(Ca(OH)_2\) (Assume Ca = 40, O = 16, H = 1)
- Ca: 1 atom = 40
- O: \(2 \times 16 = 32\)
- H: \(2 \times 1 = 2\)
- Total \(M_r = 40 + 32 + 2 = 74\)
Key Takeaway: The \(M_r\) is essential because it tells us the mass ratio of the chemical unit.
Section 3: The Mole Concept – The Chemist's Dozen
Atoms are far too tiny to weigh individually. We need a way to group atoms or molecules so we can measure them easily in the lab. This "grouping" unit is called the Mole.
What is a Mole?
The mole (symbol: \(mol\)) is simply a counting unit, much like a "dozen" (which means 12).
- A dozen eggs = 12 eggs.
- A mole of atoms = \(6.02 \times 10^{23}\) atoms. (This huge number is called Avogadro's Constant.)
Did you know? If you had a mole of pennies, the pile would cover the entire surface of the Earth about 100 meters deep! This shows just how many atoms are in a typical lab sample.
Molar Mass and Grams
The real power of the mole comes when we link it to mass:
The mass of one mole of any substance is called the Molar Mass.
- The Molar Mass is numerically equal to the \(A_r\) or \(M_r\) of the substance.
- The units for Molar Mass are grams per mole (\(g/mol\)).
Example:
- The \(A_r\) of Carbon is 12. Therefore, 1 mole of Carbon atoms has a mass of 12 grams.
- The \(M_r\) of \(H_2O\) is 18. Therefore, 1 mole of \(H_2O\) molecules has a mass of 18 grams.
The Central Equation for Quantitative Chemistry
The relationship between mass, moles, and molar mass must be memorized and understood perfectly!
$$n = \frac{m}{M_r}$$
Where:
\(n\) = number of moles (\(mol\))
\(m\) = mass in grams (\(g\))
\(M_r\) = Relative Formula Mass (or Molar Mass in \(g/mol\))
Memory Aid: The Triangle
Imagine a triangle split into three parts:
(Mass (m) on top, Moles (n) and \(M_r\) on the bottom.)
- To find Mass (m), cover 'm': \(m = n \times M_r\)
- To find Moles (n), cover 'n': \(n = \frac{m}{M_r}\)
- To find \(M_r\), cover '\(M_r\)': \(M_r = \frac{m}{n}\)
Quick Review Calculation: How many moles are in 80 g of Sodium Hydroxide (\(NaOH\))? (Na = 23, O = 16, H = 1)
1. Calculate \(M_r\): \(23 + 16 + 1 = 40\) g/mol.
2. Apply the formula: \(n = \frac{80 g}{40 g/mol} = 2 moles\)
Key Takeaway: The mole is the link between the tiny world of atoms (\(M_r\)) and the measurable world of the lab (mass in grams).
Section 4: Quantitative Interpretation of Chemical Equations (Stoichiometry)
This is where everything comes together! We use the mole ratio from a balanced equation to calculate the mass of chemicals involved.
The Importance of Ratios
In a balanced chemical equation, the large numbers (coefficients) placed in front of the compounds tell us the exact mole ratio needed for the reaction.
Consider the equation for ammonia production:
$$N_2 (g) + 3H_2 (g) \rightarrow 2NH_3 (g)$$
This equation tells us the ratio is 1 : 3 : 2.
- 1 mole of Nitrogen reacts with 3 moles of Hydrogen to produce 2 moles of Ammonia.
- This is a mole ratio, not a mass ratio!
The 3-Step Stoichiometry Calculation
If you are asked to calculate the mass of a product from the mass of a reactant, use this universal three-step method (often called "Mass-to-Mole-to-Mass"):
Example Problem:
What mass of iron (III) oxide (\(Fe_2O_3\)) is produced if 11.2 g of iron (\(Fe\)) reacts completely with oxygen? (Relative Atomic Masses: Fe = 56, O = 16)
Step 0: Ensure the equation is balanced!
$$4Fe + 3O_2 \rightarrow 2Fe_2O_3$$
Step 1: Convert Known Mass to Moles
We know the mass of \(Fe\) (11.2 g). We need its moles.
1. Calculate \(M_r\) of \(Fe\): \(M_r = 56\).
2. Calculate moles of \(Fe\):
$$n_{Fe} = \frac{m}{M_r} = \frac{11.2 g}{56 g/mol} = 0.2 mol$$
Step 2: Use the Mole Ratio (The Bridge)
Look at the balanced equation coefficients to link what you know (\(Fe\)) to what you want to find (\(Fe_2O_3\)).
The ratio \(Fe\) : \(Fe_2O_3\) is 4 : 2 (which simplifies to 2 : 1).
If we have 0.2 mol of \(Fe\), we will produce half that amount of \(Fe_2O_3\).
$$n_{Fe_2O_3} = 0.2 mol \times \frac{2}{4} = 0.1 mol$$
(We have successfully crossed the bridge from reactant moles to product moles!)
Step 3: Convert Target Moles back to Mass
We now know we made 0.1 mol of \(Fe_2O_3\). We need its mass in grams.
1. Calculate \(M_r\) of \(Fe_2O_3\):
Fe: \(2 \times 56 = 112\)
O: \(3 \times 16 = 48\)
\(M_r = 112 + 48 = 160\) g/mol.
2. Calculate mass:
$$m = n \times M_r = 0.1 mol \times 160 g/mol = 16.0 g$$
Answer: 16.0 g of iron (III) oxide is produced.
Encouragement: Don't worry if this feels like a long process. With practice, these three steps become second nature. Always write down your units and \(M_r\) calculations neatly!
Summary of Stoichiometry Steps
- Balance the equation.
- Calculate \(M_r\) for the substances you need.
- Convert Mass of the Known substance to Moles (\(n = m/M_r\)).
- Use the equation coefficients to find the Moles of the Unknown substance.
- Convert Moles of the Unknown back to Mass (\(m = n \times M_r\)).
Key Takeaway: The balanced chemical equation provides the essential mole ratio needed for all quantitative calculations.
Final Quick Review: Conservation of Mass & Stoichiometry
Core Concepts You Must Know:
- Law of Conservation of Mass: Mass of reactants = Mass of products.
- \(M_r\): Sum of \(A_r\) values. Necessary to convert between grams and moles.
- Mole Formula: \(n = m/M_r\).
- Quantitative Link: Coefficients in a balanced equation provide the mole ratio.
You have now mastered the foundation of quantitative chemistry! Keep practicing those 3-step problems, and you will ace this section!