Atomic Mass and the Mole

In chemistry, understanding the atomic mass and the concept of a mole is fundamental to analyzing and predicting reactions. This study guide will help you master these essential topics and apply them in various chemical calculations.

🏷️ Atomic Mass

Atomic mass represents the average mass of atoms of an element, taking into account the different masses of isotopes and their relative abundance on Earth. Each isotope has a different number of neutrons, which affects their mass.

🧮 Calculating Atomic Mass

To calculate atomic mass, we use isotopic masses and their natural abundances:

❓Practice Calculation

Suppose an element X has two naturally occurring isotopes: X-10 (10 amu, 20% abundance) and X-11 (11 amu, 80% abundance). The atomic mass would be:

🔄 Isotopes

Isotopes are atoms with the same number of protons but different numbers of neutrons. They are significant because:

1. They contribute to the average atomic mass.
2. Natural variations in isotopic composition can lead to slight differences in atomic mass between samples from different sources.

📊 The Mole

6️⃣ Avogadro’s Number

Avogadro's number, $6.023 \times 10^{23}$, refers to the number of particles (atoms, molecules, ions) in one mole of substance.

⚖️ Molar Mass

Molar mass is defined as the mass in grams of one mole of particles—its unit is grams per mole (g/mol). An element’s atomic mass is also its molar mass.

✖️Calculating Molar Mass:

For compound $\text{XY}_2$ with atomic masses $\text{X}=14\ \text{amu}$ and $\text{Y}=16 \ \text{amu}$:

🧮 Using Moles in Calculations

To convert between moles, number of particles, and mass:

1. Moles to Particles:

   • Multiply moles by Avogadro’s number
   • ex. If you have 3 moles of NaCl, how many molecules of NaCl do you have? $$\frac{3\ \text{mol}}{1}\cdot \frac{6.023 \cdot 10^{23}\ \text{molecules}}{\text{mol}}=\boxed{18.069\cdot10^{23}\ \text{molecules}}$$

2. Particles to Moles:

   • Divide the number of particles by Avogadro’s number.

3. Mass to Moles:

   • Divide given mass by molar mass.
   • ex. If you have 203 grams of sodium, how many moles do you have? $$\frac{203\ g\ \text{Na}}{1}\cdot\frac{1 \ \text{mol}}{22.9898\ g\ \text{Na}}=\boxed{8.83\ \text{mol Na}}$$

4. Moles to Mass:

   • Multiply moles by molar mass.
   • ex. If you have 3 moles of chlorine, how many grams do you have? $$\frac{3\ \text{mol Cl}}{1}\cdot\frac{35.453 \ g \text{ Cl}}{1 \text{mol Cl}}=\boxed{106\ g\text{ Cl}}$$

❓Practice Conversion

Convert 3 moles of $\text{H}_2\text{O}$ to particles.

Solution:

ℹ️ Formulas

Empirical formulas represent the simplest whole-number ratio of elements in a molecule. Molecular formulas represent the actual number of each element in the compound. For example, the the empirical formula for glucose, $\text{C}_6\text{H}_{12}\text{O}_6$, is $\text{C}\text{H}_{2}\text{O}$.

✖️Determining Formulas

To determine a formula, you can use data from percent composition or combustion analyses.

1. First, you can convert the percentages or masses given into moles.
2. Then, you should find the simplest ratio between the moles (aka the empirical formula!).
3. Finally, you need to check if this matches the molecular formula. You do this by comparing the empirical formula’s mass to the known mass. If they aren’t the same, you’ll have to multiply all the elements in the empirical formula by a single value to get the correct mass.

Let’s look at an example:

Dichloroethane, a compound that is often used for dry cleaning, contains carbon, hydrogen, and chlorine. It has a molar mass of 99 g/mol. Analysis of a sample shows that it contains 24.3% carbon and 4.1% hydrogen. What is its molecular formula?

Solution:

1. Convert percentages to masses.

   Carbon: $99\text{ g/mol}\times.243=24.057\text{ g C}$

   Hydrogen: $99\text{ g/mol}\times.041=4.059\text{ g H}$

   Chlorine: $99\text{ g/mol}\times.716=70.884\text{ g Cl}$

2. Convert masses to moles.

   $$\frac{24.057\text{ g C}}{1}\times\frac{1 \text{ mol}}{12.011\text{ g C}}=2\text{ mol C}$$ $$\frac{4.059\text{ g H}}{1}\times\frac{1 \text{ mol}}{1.008\text{ g H}}=4\text{ mol H}$$ $$\frac{70.884\text{ g Cl}}{1}\times\frac{1 \text{ mol}}{35.453\text{ g Cl}}=2\text{ mol Cl}$$

3. Find simplest ratio for empirical formula.

   The simplest ratio between these three is $\text{CH}_2\text{Cl}$.

4. Compare empirical formula’s molar mass with compound's known molar mass for molecular formula.

   The molar mass of $\text{CH}_2\text{Cl}$ is 49.48 amu, which doesn’t match our given mass. We just need to multiply it by two to get our given mass of 99 amu, so the molecular formula is $\text{C}_2\text{H}_4\text{Cl}_2$.

🌍 Real-World Applications

The concept of the mole is vital throughout chemistry. It can help us in stoichiometry, the practice of determining how much and what type of product will be formed in reaction, determining unknown chemical formulas based on composition analysis, and preparing reagents and solutions of a particular concentration.

Using this guide as your foundation will enable you to confidently face calculations involving atomic masses and the extensive applications involving moles—from basic conversions to complex stoichiometric problems!