Molar Mass

Theory Exercises

1. What is Molar Mass?

The Molar Mass ($M$) is the mass of exactly one mole ($6.022 \times 10^{23}$ particles) of a substance. It is the essential "bridge" between the mass we can measure in a lab and the number of atoms or molecules.

Units: Grams per mole (g/mol).
Periodic Table: For any element, the number you see in the periodic table is its molar mass.

\[M_{\text{molar mass}} = \frac{m_{\text{mass}}}{n_{\text{moles}}}\]

2. Finding Molar Mass for Elements (Isotopes)

Why is the molar mass in the periodic table often a decimal (like $35.45$)? This is because it is the weighted average of all the naturally occurring isotopes of that element.

An isotope is an atom with the same number of protons but a different number of neutrons (making it heavier or lighter).

The Calculation

If you know the mass and abundance (%) of the isotopes, you calculate the average molar mass:

\[M_{\text{ molar atomic mass}} = \frac{(M_{\text{molar mass of isotope 1}} \times \%_{\text{percentage of isotope 1}}) + \dots}{100}\]

Example: Average Molar Mass of Chlorine

Chlorine has two main isotopes:
Cl-35 ($35 \text{ g/mol}$) with an abundance of 75%.
Cl-37 ($37 \text{ g/mol}$) with an abundance of 25%.
Calculation:

\[M = \frac{(35 \times 75) + (37 \times 25)}{100} = \frac{2625 + 925}{100} = 35.5 \text{ g/mol}\]

3. Molar Mass of Compounds

To find the molar mass of a molecule, simply add up the molar masses of every atom in its chemical formula.

\[M_{\text{ compound atomic mass}} = \sum (n_{\text{ atoms}} \times M_{\text{ molar atomic mass}})\]

Example: Molar Mass of Water ($H_2O$)

$H = 1.0 \text{ g/mol}$
$O = 16.0 \text{ g/mol}$

$ M(H_2O) = (2 \times 1.0) + (1 \times 16.0) = 18.0 \text{ g/mol} $

4. Mass-to-Mass Calculations in Reactions

Molar mass is also used in chemical reactions when you know the mass of one substance and want to find the mass of another reactant or product.

The idea is always the same:

Convert the given mass into moles using molar mass.
Use the balanced equation to convert moles of one substance into moles of the other.
Convert those moles back into mass.

\[\text{mass} \rightarrow \text{moles} \rightarrow \text{stoichiometric ratio} \rightarrow \text{moles} \rightarrow \text{mass}\]

If the exercise says a substance reacts completely and does not mention a limiting reagent, we assume the reaction happens in the exact stoichiometric ratio given by the balanced equation.

General Method

For a reaction $aA + bB \rightarrow cC + dD$:

Find moles of the known substance:

$ n = \frac{m}{M} $

Use the coefficients of the balanced equation:

$ n(\text{unknown}) = n(\text{known}) \times \frac{\text{coefficient of unknown}}{\text{coefficient of known}} $

Convert back to mass:

$ m = n \times M $

Solved Examples

Example 1: Mass to Moles

Question: How many moles are in $54 \text{ g}$ of Water ($H_2O$)? (Given: $M_{H_2O} = 18 \text{ g/mol}$) Solution: $ n = \frac{54 \text{ g}}{18 \text{ g/mol}} = 3 \text{ mol} $

Example 2: Moles to Mass

Question: What is the mass of $0.5 \text{ moles}$ of Carbon Dioxide ($CO_2$)? (Given: $M_{CO_2} = 44 \text{ g/mol}$) Solution: $ m = 0.5 \text{ mol} \times 44 \text{ g/mol} = 22 \text{ g} $

Example 3: Two-step Conversion (Mass to Molecules)

Question: How many molecules are in $9 \text{ g}$ of Water ($H_2O$)? (Given: $M_{H_2O} = 18 \text{ g/mol}$) Solution:

Find moles ($n$):

$$n = \frac{m}{M} = \frac{9 \text{ g}}{18 \text{ g/mol}} = 0.5 \text{ mol} $

Find molecules ($N$):

$ N = n \times 6.022 \times 10^{23} = 0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23} \text{ molecules} $

Example 4: Moles to Mass (Complex Compound)

Question: What is the mass of $0.2 \text{ moles}$ of Calcium Carbonate ($CaCO_3$)? (Given: $Ca=40.1, C=12.0, O=16.0$) Solution:

Find Molar Mass ($M$):

$ M = 40.1 + 12.0 + (3 \times 16.0) = 40.1 + 12.0 + 48.0 = 100.1 \text{ g/mol} $

Find Mass ($m$):

$ m = n \times M = 0.2 \text{ mol} \times 100.1 \text{ g/mol} = 20.02 \text{ g} $

Example 5: Particles to Mass (Two-step)

Question: What is the mass of $1.2044 \times 10^{23}$ molecules of Oxygen gas ($O_2$)? (Given: $M_{O_2} = 32 \text{ g/mol}$) Solution:

Find moles ($n$):

$ n = \frac{1.2044 \times 10^{23}}{6.022 \times 10^{23}} = 0.2 \text{ mol} $

Find mass ($m$):

$ m = 0.2 \text{ mol} \times 32 \text{ g/mol} = 6.4 \text{ g} $

Example 6: Mass-to-Mass in a Reaction

Question: In the reaction $2H_2 + O_2 \rightarrow 2H_2O$, what mass of $O_2$ is needed to react completely with $8 \text{ g}$ of $H_2$? Assume the substances are mixed in the exact stoichiometric ratio. Solution:

Find moles of $H_2$:

$ n(H_2) = \frac{8 \text{ g}}{2 \text{ g/mol}} = 4 \text{ mol} $

Use the mole ratio $2:1$:

$ n(O_2) = 4 \times \frac{1}{2} = 2 \text{ mol} $

Convert moles of $O_2$ to mass:

$ m(O_2) = 2 \text{ mol} \times 32 \text{ g/mol} = 64 \text{ g} $