Theory Exercises

What is a Mole?

The mole (symbol: mol) is the SI unit for the amount of substance.

Just as a "dozen" always means 12 items (like eggs or donuts), a "mole" always means a specific number of particles: Avogadro's Number.

\[1 \text{ mol} = 6.022 \times 10^{23} \text{ particles}\]

These particles can be atoms, molecules, ions, or any other microscopic entity.

How to Convert Between Moles and Particles

To convert between moles and particles (atoms or molecules), we use Avogadro's Number.

  • Moles → Particles: Multiply the moles by \(6.022 \times 10^{23}\).
  • Particles → Moles: Divide the particles by \(6.022 \times 10^{23}\).
\[\text{Moles} \xrightarrow{\times 6.022 \times 10^{23}} \text{Particles (Atoms/Molecules)}\]
\[\text{Particles (Atoms/Molecules)} \xrightarrow{\div 6.022 \times 10^{23}} \text{Moles}\]

Why do we use the Mole?

Atoms and molecules are incredibly small. If you were to count the atoms in a glass of water one by one, it would take you millions of years!

The mole allows chemists to:

  1. Work with manageable numbers: instead of saying \(602,200,000,000,000,000,000,000\) atoms, we just say "1 mole".
  2. Connect the microscopic to the macroscopic: It bridges the gap between the tiny mass of a single atom and the grams we measure in a laboratory.

Solved Examples

When solving mole problems, we use conversion factors (fractions) or direct proportions (the rule of three).

Example: Converting Molecules to Moles
Question: Calculate how many moles are in \(1.806 \times 10^{24}\) molecules of Carbon Dioxide (\(CO_2\)). Solution: We use Avogadro's number as a fraction. Since we want to find "moles", we put "1 mol" on top:
\[1.806 \times 10^{24} \text{ molecules } CO_2 \times \frac{1 \text{ mol } CO_2}{6.022 \times 10^{23} \text{ molecules } CO_2} = 3 \text{ mol } CO_2\]
Example: Finding Atoms of an element in a compound
Question: How many atoms of Hydrogen are in \(2\) moles of Methane (\(CH_4\))? Solution:
  1. Each molecule of \(CH_4\) contains 4 atoms of Hydrogen (the subscript).
  2. Therefore, \(1\) mole of \(CH_4\) contains 4 moles of Hydrogen atoms.
\[2 \text{ mol } CH_4 \times \frac{4 \text{ mol H}}{1 \text{ mol } CH_4} \times \frac{6.022 \times 10^{23} \text{ atoms H}}{1 \text{ mol H}}\]

>

\[= 8 \text{ mol H} \times 6.022 \times 10^{23} \text{ atoms/mol} = 4.817 \times 10^{24} \text{ atoms of Hydrogen}\]

Example: Atoms in a complex molecule (Glucose)
Question: How many atoms of Oxygen are in \(0.5\) moles of Glucose (\(C_6H_{12}O_6\))? Solution:
\[0.5 \text{ mol } C_6H_{12}O_6 \times \frac{6 \text{ mol O}}{1 \text{ mol } C_6H_{12}O_6} \times \frac{6.022 \times 10^{23} \text{ atoms O}}{1 \text{ mol O}} = 1.806 \times 10^{24} \text{ atoms O}\]

Mole-to-Mole Calculations (Proportions)

In a balanced chemical equation, the coefficients (the numbers in front of the formulas) tell us the mole ratio between the substances. We can think of these as "recipes".

Example: \(2 H_2 + O_2 \rightarrow 2 H_2O\) This recipe says that for every 2 moles of \(H_2\), we need 1 mole of \(O_2\) to produce 2 moles of \(H_2O\).

We can use direct proportions (also known as the rule of three) or conversion factors to calculate how much of one substance we need if we know the amount of another.

Example: Calculating Moles Produced
Reaction: \(N_2 + 3 H_2 \rightarrow 2 NH_3\) Question: How many moles of Ammonia (\(NH_3\)) are produced if we react \(9\) moles of Hydrogen (\(H_2\))? Solution (using a Proportion): \( \frac{3 \text{ mol } H_2}{2 \text{ mol } NH_3} = \frac{9 \text{ mol } H_2}{x \text{ mol } NH_3} \) \( x = \frac{9 \times 2}{3} = 6 \text{ mol } NH_3 \)
Example: Calculating Moles Needed
Reaction: \(N_2 + 3 H_2 \rightarrow 2 NH_3\) Question: How many moles of Nitrogen (\(N_2\)) are needed to react completely with \(12\) moles of Hydrogen (\(H_2\))? Solution (using a Conversion Factor): \( 12 \text{ mol } H_2 \times \frac{1 \text{ mol } N_2}{3 \text{ mol } H_2} = 4 \text{ mol } N_2 \)
Example: Calculating Total Atoms
Question: How many total atoms are in \(0.1\) moles of Water (\(H_2O\))? Solution:
  1. Each molecule of \(H_2O\) has 3 atoms (\(2\) Hydrogen + \(1\) Oxygen).
  2. Moles of atoms:
\( n_{\text{atoms}} = 0.1 \text{ mol } H_2O imes 3 \text{ atoms/molecule} = 0.3 \text{ moles of atoms} \)
  1. Number of atoms (\(N\)):
\( N = 0.3 \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol} = 1.8066 imes 10^{23} \text{ atoms} \)
Example: Large Scale Molecules to Moles
Question: Calculate how many moles are in \(1.8066 imes 10^{25}\) molecules of Methane (\(CH_4\)). Solution: \( n = \frac{1.8066 \times 10^{25}}{6.022 \times 10^{23}} = 0.3 \times 10^2 = 30 \text{ moles} \)