S1.4.3 Molar mass

Understanding Molar Mass: The Key to Counting Particles by Mass

You’re baking a cake.
The recipe calls for 2 cups of flour, 1 cup of sugar, and 3 eggs. But what if you didn’t have measuring cups? Instead, you’d need to weigh the ingredients.
Chemistry works in a similar way.
Atoms and molecules are far too small to count individually, so chemists use a recipe based on mass to measure substances.

This is where the concept of molar mass comes in—it’s the bridge between the microscopic world of particles and the macroscopic world of grams and kilograms.

What Is Molar Mass?

Definition

Molar mass

Molar mass, denoted as $M$ , is the mass of one mole of a substance.

It is expressed in units of grams per mole ( ${g mol}^{- 1}$ ). But what exactly is a mole?

Definition

Mole

The mole (mol) is the SI unit for the amount of substance, defined as $6.02 \times 10^{23}$ particles (atoms, molecules, or ions).

The number $6.02 \times 10^{23}$ is known as Avogadro’s constant.

The molar mass tells you how much one mole of a substance weighs in grams.

Example

The molar mass of carbon ( $C$ ) is $12.01 {g mol}^{- 1}$ . This means 1 mole of carbon atoms weighs $12.01$ grams.
The molar mass of water ( $H_{2} O$ ) is $18.02 {g mol}^{- 1}$ . This means 1 mole of water molecules weighs $18.02$ grams.

The Relationship Between Mass, Moles, and Molar Mass

The relationship between mass, moles, and molar mass is captured by the formula:

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

Where:

$n$ = number of moles (mol)
$m$ = mass of the substance (g)
$M$ = molar mass ( ${g mol}^{- 1}$ )

This formula allows you to calculate any one of the three variables if the other two are known.

Rearranging the Formula

To find mass: $m = n \times M$
To find molar mass: $M = \frac{m}{n}$

Example question

How many moles of water corresponds to $36.04 g$ ?

Solution

Write down the known values:
- $m = 36.04, g$
- $M = 18.02, {g mol}^{- 1}$
Use the formula $n = \frac{m}{M}$ : $n = \frac{36.04}{18.02} = 2.00 mol$

So, $36.04 g$ of water is equivalent to $2.00$ moles of water molecules.

Problem-Solving Applications: Particles, Moles, and Mass

Now that you understand the relationship between mass, moles, and molar mass, let’s apply it to solve problems.
These problems often involve converting between particles, moles, and mass.

Step-by-Step Problem Solving

Identify what is given and what is required.
Use the appropriate formula. If particles are involved, use Avogadro’s constant to convert between particles and moles.
Perform the calculation, ensuring units are consistent.

Example question

How much does $1.204 \times 10^{24}$ molecules of water weigh?

Solution

Step 1: Convert particles to moles. Use Avogadro’s constant:
$n = \frac{Number of particles}{Avogadro's constant}$ $n = \frac{1.204 \times 10^{24}}{6.022 \times 10^{23}} = 2.00 mol$
Step 2: Convert moles to mass. Use the formula $m = n \times M$ : $m = 2.00 mol \times 18.02 {g mol}^{- 1} = 36.04 g$

So, $1.204 \times 10^{24}$ molecules of water weigh $36.04 g$ .

Example question

Converting Mass to Number of Particles

How many molecules are in a 5.0 g sample of water $(H_{2} O)$ ?

Solution

Step 1: Identify Given and Required Information

Mass of water, $m = 5.0 g$
Molar mass of $H_{2} O$ :
$M = (2 \times 1.008) + 16.00 = 18.016 g/mol$
Avogadro's constant, $N_{A} = 6.022 \times 10^{23} {mol}^{- 1}$

Required: Number of molecules $N$ .

Step 2: Convert Mass to Moles
Use the formula:
$n = \frac{m}{M}$

Substitute the values:
$n = \frac{5.0}{18.016} = 0.2776 mol$

Step 3: Convert Moles to Number of Particles
Use the formula:
$N = n \times N_{A}$

Substitute the values:
$N = 0.2776 \times 6.022 \times 10^{23}$

Result:
$N \approx 1.67 \times 10^{23} molecules$

Answer:
There are approximately $1.67 \times 10^{23}$ molecules of water in a 5.0 g sample.

Example question

Converting Particles to Mass

A sample contains $3.0 \times 10^{24}$ molecules of carbon dioxide $(C O_{2})$ . What is the mass of the sample?

Solution

Step 1: Identify Given and Required Information

Number of molecules, $N = 3.0 \times 10^{24}$
Molar mass of $C O_{2}$ :
$M = 12.01 + (2 \times 16.00) = 44.01 g/mol$
Avogadro's constant, $N_{A} = 6.022 \times 10^{23} {mol}^{- 1}$

Required: Mass $m$ of the sample.

Step 2: Convert Particles to Moles
Use the formula:
$n = \frac{N}{N_{A}}$

Substitute the values:
$n = \frac{3.0 \times 10^{24}}{6.022 \times 10^{23}}$

Result:
$n \approx 4.98 mol$

Step 3: Convert Moles to Mass
Use the formula:
$m = n \times M$

Substitute the values:
$m = 4.98 \times 44.01 \approx 219.2 g$

Answer:
The mass of the carbon dioxide sample is approximately $219.2 g$ .

Common Challenges and Tips for Success

Common Mistake

Forgetting to use the periodic table: Always double-check the molar masses of elements.
Mixing up units: Ensure mass is in grams and molar mass is in ${g mol}^{- 1}$ .
Skipping unit conversions: When dealing with particles, always use Avogadro’s constant to convert to moles.

Tip

Practice, practice, practice: The more problems you solve, the more intuitive these calculations will become.
Use dimensional analysis: This method ensures your units cancel out correctly, leading to the right answer.

Reflection and Further Exploration

Theory of Knowledge

How does the concept of molar mass help us understand the scale of chemical reactions in everyday life?
What are the limitations of using the mole concept in real-world applications, such as measuring very small quantities of substances?

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S1.4.3 Molar mass

Understanding Molar Mass: The Key to Counting Particles by Mass

What Is Molar Mass?

The Relationship Between Mass, Moles, and Molar Mass

Rearranging the Formula

Problem-Solving Applications: Particles, Moles, and Mass

Step-by-Step Problem Solving

Converting Mass to Number of Particles

Converting Particles to Mass

Common Challenges and Tips for Success

Reflection and Further Exploration

You've read 2/2 free chapters this week.

Introduction to Molar Mass