Understanding Molar Mass: The Key to Counting Particles by Mass
What Is Molar Mass?
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 ($\text{g mol}^{-1}$).
- But what exactly is a mole?
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.
- The molar mass of carbon ($C$) is $12.01 \, \text{g mol}^{-1}$.
- This means 1 mole of carbon atoms weighs $12.01$ grams.
- The molar mass of water ($H_2O$) is $18.02 \, \text{g mol}^{-1}$.
- 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 ($\text{g mol}^{-1}$)
- 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: $$M = \frac{m}{n}$$
How many moles of water corresponds to $36.04 \, \text{g}$?
Solution
- Write down the known values:
- $m = 36.04 , \text{g}$
- $M = 18.02 , \text{g mol}^{-1}$
- Use the formula $n = \frac{m}{M}$: $$n = \frac{36.04}{18.02} = 2.00 \, \text{mol}$$
- So, $36.04 \text{g}$ 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.
How much does $1.204 \times 10^{24}$ molecules of water weigh?
Solution
- Step 1: Convert particles to moles. Use Avogadro’s constant:
$$ n = \frac{\text{Number of particles}}{\text{Avogadro's constant}}$$ $$n = \frac{1.204 \times 10^{24}}{6.022 \times 10^{23}} = 2.00 \, \text{mol} $$ - Step 2: Convert moles to mass. Use the formula $m = n \times M$: $$m = 2.00 \, \text{mol} \times 18.02 \, \text{g mol}^{-1} = 36.04 \, \text{g}$$
- So, $1.204 \times 10^{24}$ molecules of water weigh $36.04 \, \text{g}$.
How many molecules are in a 5.0 g sample of water $(H_2O)$?
Solution
Step 1: Identify Given and Required Information
- Mass of water, $m = 5.0 \, \text{g}$
- Molar mass of $H_2O$:
$M = (2 \times 1.008) + 16.00 = 18.016 \, \text{g mol}^{-1}$ - Avogadro's constant, $N_A = 6.022 \times 10^{23} \, \text{mol}^{-1}$
Required: Number of molecules $N$.
Step 2: Convert Mass to Moles
- Use the formula:
$$n = \frac{m}{M}$$ - Substitute the values:
$$n = \frac{5.0}{18.016} = 0.2776 \, \text{mol}$$
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:
$$N \approx 1.67 \times 10^{23} \, \text{molecules}$$
Answer: There are approximately $1.67 \times 10^{23}$ molecules of water in a 5.0 g sample.
A sample contains $3.0 \times 10^{24}$ molecules of carbon dioxide $(CO_2)$. 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 $CO_2$:
$M = 12.01 + (2 \times 16.00) = 44.01 \, \text{g mol}^{-1}$ - Avogadro's constant, $N_A = 6.022 \times 10^{23} \, \text{mol}^{-1}$
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:
$$n \approx 4.98 \, \text{mol}$$
Step 3: Convert Moles to Mass
- Use the formula:
$$m = n \times M$$ - Substitute the values:
$$m = 4.98 \times 44.01 \approx 219.2 \, \text{g}$$
Answer: The mass of the carbon dioxide sample is approximately $219.2 \, \text{g}$.
- 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 $\text{g mol}^{-1}$.
- Skipping unit conversions: When dealing with particles, always use Avogadro’s constant to convert to moles.
- 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.
