👩🏽‍🔬Honors Chemistry Unit 3 Review

Atomic mass and the mole connect the invisible world of atoms to measurable quantities you can actually weigh in a lab. These concepts are the foundation for nearly every calculation you'll do in chemistry, from balancing reactions to determining chemical formulas.

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Atomic Mass

Atomic mass is the weighted average mass of all naturally occurring isotopes of an element, measured in atomic mass units (amu). It's not simply the mass of one isotope; it accounts for both the mass and the relative abundance of each isotope found in nature.

This is the number you see on the periodic table beneath each element's symbol, and it's what you'll use for virtually all calculations going forward.

Why It Matters: Atomic mass lets chemists predict how much of an element will participate in a reaction and form compounds. Without it, there'd be no way to connect what's happening at the atomic scale to what you can measure on a balance.

Isotopes

Isotopes are atoms of the same element (same number of protons) but with different numbers of neutrons. Because neutrons have mass, different isotopes have different masses.

For example, carbon-12 has 6 neutrons while carbon-13 has 7. Both are carbon, but carbon-12 is far more abundant (~98.9%), which is why carbon's atomic mass on the periodic table (12.011 amu) is very close to 12, not halfway between 12 and 13.

Calculating Atomic Mass

To find an element's atomic mass, multiply each isotope's mass by its fractional abundance, then add the results:

$\text{Atomic Mass} = \sum (\text{Isotope Mass} \times \text{Fractional Abundance})$

Practice Calculation:

An element X has two naturally occurring isotopes: X-10 (10 amu, 20% abundance) and X-11 (11 amu, 80% abundance).

$\text{Atomic Mass} = (10 \times 0.20) + (11 \times 0.80) = 2.0 + 8.8 = 10.8 \text{ amu}$

Notice the result is closer to 11 than to 10. That makes sense because X-11 is much more abundant. Always do this quick sanity check on your answer.

The Mole

Avogadro's Number

A mole is simply a counting unit, like a "dozen" means 12. One mole equals $6.022 \times 10^{23}$ particles (atoms, molecules, ions, etc.). This value is called Avogadro's number.

Why this specific number? One mole is defined so that the mass of one mole of an element, in grams, equals that element's atomic mass in amu. Carbon has an atomic mass of 12.011 amu, so one mole of carbon atoms weighs 12.011 grams. This is what makes the mole so useful: it bridges atomic-scale masses and lab-scale masses.

Molar Mass

Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol).

For elements, the molar mass numerically equals the atomic mass from the periodic table.
For compounds, add up the molar masses of all atoms in the formula.

Example: For $\text{NO}_2$ , with $\text{N} = 14.01 \text{ g/mol}$ and $\text{O} = 16.00 \text{ g/mol}$ :

$\text{Molar Mass} = 14.01 + (2 \times 16.00) = 46.01 \text{ g/mol}$

Mole Conversions

These four conversions come up constantly. In each case, you're using dimensional analysis (the factor-label method) to cancel units.

Moles → Particles: Multiply moles by Avogadro's number.

How many formula units are in 3 moles of NaCl?

$\frac{3 \text{ mol}}{1} \times \frac{6.022 \times 10^{23} \text{ formula units}}{1 \text{ mol}} = \boxed{1.807 \times 10^{24} \text{ formula units}}$

(Note: NaCl is ionic, so we say "formula units," not "molecules.")
Particles → Moles: Divide the number of particles by Avogadro's number.
Mass → Moles: Divide the given mass by the molar mass.

How many moles are in 203 g of sodium?

$\frac{203 \text{ g Na}}{1} \times \frac{1 \text{ mol}}{22.990 \text{ g Na}} = \boxed{8.83 \text{ mol Na}}$
Moles → Mass: Multiply moles by the molar mass.

What is the mass of 3 moles of chlorine atoms?

$\frac{3 \text{ mol Cl}}{1} \times \frac{35.45 \text{ g Cl}}{1 \text{ mol Cl}} = \boxed{106 \text{ g Cl}}$

Practice Conversion

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

Solution:

$3 \text{ mol } \text{H}_2\text{O} \times \frac{6.022 \times 10^{23} \text{ molecules}}{1 \text{ mol}} = \boxed{1.807 \times 10^{24} \text{ molecules}}$

Empirical and Molecular Formulas

Empirical formulas give the simplest whole-number ratio of elements in a compound. Molecular formulas give the actual number of each type of atom per molecule.

A molecular formula is always a whole-number multiple of the empirical formula. For glucose, the molecular formula is $\text{C}_6\text{H}_{12}\text{O}_6$ , but the empirical formula is $\text{CH}_2\text{O}$ (each subscript divided by 6).

Determining Formulas from Percent Composition

Convert percentages to grams. If you're given percent composition and a molar mass, multiply each percentage (as a decimal) by the molar mass. If percentages don't add to 100%, the missing element makes up the remainder.
Convert grams to moles by dividing each mass by that element's molar mass.
Find the simplest ratio. Divide all mole values by the smallest one. Round to the nearest whole number. This gives you the empirical formula.
Determine the molecular formula. Calculate the molar mass of the empirical formula. Divide the compound's known molar mass by the empirical formula mass to get a whole-number multiplier. Multiply all subscripts in the empirical formula by that number.

Worked Example

Dichloroethane has a molar mass of 99 g/mol. Analysis shows it contains 24.3% carbon and 4.1% hydrogen. Find its molecular formula.

Step 1: Find the percentage of chlorine and convert to grams.

Chlorine makes up the remainder: $100\% - 24.3\% - 4.1\% = 71.6\%$

Using the molar mass as your sample size (99 g):

Carbon: $99 \times 0.243 = 24.06 \text{ g C}$
Hydrogen: $99 \times 0.041 = 4.06 \text{ g H}$
Chlorine: $99 \times 0.716 = 70.88 \text{ g Cl}$

Step 2: Convert to moles.

$\frac{24.06 \text{ g C}}{12.011 \text{ g/mol}} = 2.00 \text{ mol C}$

$\frac{4.06 \text{ g H}}{1.008 \text{ g/mol}} = 4.03 \text{ mol H}$

$\frac{70.88 \text{ g Cl}}{35.45 \text{ g/mol}} = 2.00 \text{ mol Cl}$

Step 3: Find the simplest ratio.

Divide each by the smallest value (2.00): C = 1, H = 2, Cl = 1. The empirical formula is $\text{CH}_2\text{Cl}$ .

Step 4: Compare to the known molar mass.

Empirical formula mass: $12.01 + 2(1.008) + 35.45 = 49.48 \text{ g/mol}$

$\frac{99}{49.48} = 2$

Multiply all subscripts by 2. The molecular formula is $\text{C}_2\text{H}_4\text{Cl}_2$ .

Real-World Applications

The mole concept shows up everywhere in chemistry:

Stoichiometry: Predicting how much product a reaction will yield (or how much reactant you need).
Formula determination: Identifying unknown compounds from composition data, as in the example above.
Solution preparation: Calculating how much solute to dissolve to make a solution of a specific concentration (molarity).

These skills build directly on each other, so getting comfortable with mole conversions now will pay off in every unit that follows.

👩🏽‍🔬Honors Chemistry Unit 3 Review