Mastering Mass Calculations
Moles, Molecules, and Mass

Understanding the Fundamentals

In chemistry, we often need to bridge the gap between microscopic counts (atoms/molecules) and macroscopic quantities (mass in grams). This is achieved using the Mole Concept and Molar Mass.

Key Concepts:

• Avogadro's Number ($N_A$): $6.022 \times 10^{23} \text{ particles/mol}$. This is the number of units in one mole of any substance.
• Molar Mass ($M$): The mass of one mole of a substance, typically expressed in g/mol. It is calculated by summing the atomic masses from the periodic table.
• Formula: $\text{Mass} (m) = n \times M$, where $n$ is the number of moles.

Solving Part (a): Mass of $1 \times 10^{23}$ molecules of Methane ($CH_4$)

Step 1: Calculate Molar Mass of $CH_4$

• Carbon (C) $\approx 12.01 \text{ g/mol}$
• Hydrogen (H) $\approx 1.01 \text{ g/mol}$
• Molar Mass = $12.01 + 4(1.01) = 16.05 \text{ g/mol}$

Step 2: Convert molecules to moles ($n$)

• $n = \frac{\text{Number of molecules}}{N_A} = \frac{1 \times 10^{23}}{6.022 \times 10^{23}} \approx 0.166 \text{ mol}$

Step 3: Calculate mass

• $m = 0.166 \text{ mol} \times 16.05 \text{ g/mol} \approx 2.66 \text{ g}$

Solving Part (b): Mass of 2.5 moles of Chlorine gas ($Cl_2$)

Step 1: Calculate Molar Mass of $Cl_2$

• Chlorine (Cl) $\approx 35.45 \text{ g/mol}$
• Since chlorine gas exists as a diatomic molecule ($Cl_2$):
• Molar Mass = $2 \times 35.45 = 70.90 \text{ g/mol}$

Step 2: Calculate mass

• $m = n \times M$
• $m = 2.5 \text{ mol} \times 70.90 \text{ g/mol} = 177.25 \text{ g}$

Summary Table