CORE Chemistry (9222) | Conservation of mass including the quantitative interpretation of chemical equations

Chemistry Core 9222: Study Notes - Conservation of Mass and Quantitative Equations

Hello future chemist! Welcome to the world of Quantitative Chemistry. This chapter might sound complicated, but it's really just about counting atoms and proving that in chemistry, nothing is ever truly lost. We’re going to learn how to predict exactly how much product we can make from a given amount of starting material. Think of this as the essential "recipe book" for chemistry!

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Section 1: The Law of Conservation of Mass

What is the Law of Conservation of Mass?

This is one of the most fundamental rules in all of science. It’s simple, powerful, and always true:

The total mass of reactants before a chemical reaction must equal the total mass of products after the reaction.

In simpler terms:

Mass Before = Mass After

Atoms are Rearranged, Not Destroyed

A chemical reaction isn't magic; it’s just a rearrangement of atoms. When substances react:

• The atoms break their old bonds.
• They shuffle around.
• They form new bonds to create new substances (the products).

Analogy: Imagine you have a box of LEGO bricks. You build a car (Reactant A) and a plane (Reactant B). When you combine them, you break them down and build a spaceship (Product C). No matter what structure you build, the total number and mass of the original LEGO bricks stays exactly the same. Atoms work just like those bricks!

Key Takeaway: Mass is conserved because atoms themselves are neither created nor destroyed during a chemical reaction.

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Section 2: The Mystery of Changing Mass (Open vs. Closed Systems)

Sometimes, when you do an experiment, it looks like the mass has changed! Don't worry, the Law of Conservation of Mass hasn't been broken. This apparent change is usually due to gases either escaping or being used up.

1. Mass Appears to Decrease

This happens when a reaction produces a gas, and that gas is allowed to escape into the atmosphere.

Example: Heating calcium carbonate (\(CaCO_3\)).

\(CaCO_3 (s) \to CaO (s) + CO_2 (g)\)

If you weigh the solid reactant (\(CaCO_3\)) in an open crucible and then heat it, the product (\(CaO\)) will weigh less. Why? The gaseous carbon dioxide (\(CO_2\)) has escaped! If you could capture and weigh that gas, the total mass would be conserved.

2. Mass Appears to Increase

This happens when a reactant is a gas from the atmosphere (like oxygen) that gets incorporated into a solid or liquid product.

Example: Burning magnesium ribbon (oxidation).

\(2Mg (s) + O_2 (g) \to 2MgO (s)\)

If you weigh the magnesium ribbon and then burn it in an open container, the white powder (magnesium oxide, \(MgO\)) produced will weigh more than the original ribbon. Why? The magnesium has chemically combined with oxygen gas from the air, which adds mass to the solid product.

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Section 3: The Tool for Calculation: Relative Formula Mass (\(M_r\))

To prove conservation of mass mathematically and to calculate masses in equations, we need to know the 'weight' of our substances. This is called the Relative Formula Mass (\(M_r\)).

1. Prerequisite: Relative Atomic Mass (\(A_r\))

Every element has a specific mass (found on the Periodic Table, usually the larger number). This is the Relative Atomic Mass (\(A_r\)). For example, the \(A_r\) of Carbon (C) is 12, and Oxygen (O) is 16.

2. Calculating Relative Formula Mass (\(M_r\))

The \(M_r\) is simply the sum of the \(A_r\) values of all the atoms in the chemical formula.

Step-by-Step Example: Calculating \(M_r\) for Water (\(H_2O\))

Given: \(A_r\) of H = 1, \(A_r\) of O = 16.

1. Identify the atoms and counts: 2 Hydrogen (H) atoms, 1 Oxygen (O) atom.
2. Calculate the total mass for each element:
   • H: \(2 \times 1 = 2\)
   • O: \(1 \times 16 = 16\)
3. Sum the masses: \(M_r(H_2O) = 2 + 16 = 18\)

Don't worry if this seems tricky at first—it’s just careful addition!

Dealing with Brackets and Coefficients

If a formula has brackets, the number outside the bracket multiplies everything inside.

Example: Calculate \(M_r\) for Calcium Hydroxide, \(Ca(OH)_2\).

Given: \(A_r\) Ca = 40, O = 16, H = 1.

• Ca: \(1 \times 40 = 40\)
• O: \(2 \times 16 = 32\) (The 2 multiplies the O)
• H: \(2 \times 1 = 2\) (The 2 multiplies the H)
• Total \(M_r\): \(40 + 32 + 2 = 74\)

Quick Review: \(M_r\) allows us to assign a specific mass 'weight' to every compound, which is crucial for proving mass conservation.

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Section 4: Quantitative Interpretation of Chemical Equations (Stoichiometry by Mass)

A balanced chemical equation tells us two things:

1. The ratio of particles (e.g., 2 molecules of H and 1 molecule of O).
2. The ratio of total mass needed for the reaction.

We use the \(M_r\) values and the balancing numbers (coefficients) to find this exact mass ratio.

Step-by-Step: Using Equations for Mass Calculation

Let's use the reaction where hydrogen reacts with oxygen to form water:

\(2H_2 + O_2 \to 2H_2O\)

Assume \(A_r\): H=1, O=16.

Step 1: Calculate the \(M_r\) for each substance.

• \(H_2\): \(1 \times 2 = 2\)
• \(O_2\): \(16 \times 2 = 32\)
• \(H_2O\): \(18\) (calculated in Section 3)

Step 2: Calculate the Total Mass Ratio using the balancing numbers.

This is where students often make mistakes! You must multiply the \(M_r\) by the big number in front (the coefficient).

• \(2H_2\): \(2 \times M_r(H_2) = 2 \times 2 = 4\)
• \(1O_2\): \(1 \times M_r(O_2) = 1 \times 32 = 32\)
• \(2H_2O\): \(2 \times M_r(H_2O) = 2 \times 18 = 36\)

Proof of Conservation of Mass:

Reactants total mass: \(4 + 32 = 36\) units.
Products total mass: \(36\) units.
Mass is conserved! (36 = 36)

The resulting mass ratio is: 4 : 32 : 36 (Hydrogen : Oxygen : Water).

(We can simplify this ratio by dividing by 4: 1 : 8 : 9)

Step 3: Solving Mass Problems (The Proportionality)

Once you have the mass ratio, you can solve any mass problem for that reaction using scaling or proportionality.

Question: If 64 g of oxygen (\(O_2\)) reacts fully with hydrogen, what mass of water (\(H_2O\)) is produced?

1. Use the calculated ratio from Step 2:

   Oxygen : Water = 32 : 36

2. Set up the relationship:

   Known Mass (Oxygen) / Ratio Mass (Oxygen) = Unknown Mass (Water) / Ratio Mass (Water)

   \(64 / 32 = x / 36\)

3. Calculate the scaling factor:

   \(64 \div 32 = 2\)

   (This means the actual reaction is 2 times bigger than our ratio calculation.)

4. Apply the scaling factor to the unknown:

   \(x = 36 \times 2 = 72 g\)

   Answer: 72 g of water is produced.

Did you know? This ability to precisely calculate amounts is why chemists can scale up reactions safely in industrial manufacturing!

Common Pitfalls to Avoid

• Forgetting the Coefficient: Always multiply the \(M_r\) by the big number (coefficient) in the balanced equation. If you miss this, your entire mass ratio will be wrong.
• Mixing up subscripts and coefficients: Subscripts (the small numbers, like the 2 in \(H_2\)) are used to calculate the \(M_r\). Coefficients (the big numbers, like the 2 in \(2H_2\)) are used to find the overall mass ratio.
• Not using a balanced equation: If the equation isn't balanced, the ratio of masses will be incorrect, and the Law of Conservation of Mass will appear to be broken.

You have successfully mastered the foundation of quantitative chemistry! Keep practicing these ratio calculations—they are the key to success in this section.