Honors Chemistry Unit 5 Review: Chemical Quantities & Stoichiometry

A mole contains exactly $6.02214076 \times 10^{23}$ specified particles, called elementary entities. The Avogadro constant is $N_A = 6.02214076 \times 10^{23}\ \text{mol}^{-1}$, and for most class calculations you round its numerical value to $6.022 \times 10^{23}$. Before May 20, 2019, the mole was defined as the amount of substance containing as many entities as there are atoms in exactly 0.012 kg (12 g) of carbon-12. Both definitions give the same value for your calculations.

The particles can be atoms, molecules, ions, or formula units, so always say what you're counting. One mole of $O_2$ is $6.022 \times 10^{23}$ molecules, but it contains twice that many oxygen atoms.

To go from moles to particles, multiply by $N_A$. To go from particles to moles, divide by $N_A$. Set it up with dimensional analysis so the unit you start with cancels:

$3.00 \text{ mol He} \times \dfrac{6.022 \times 10^{23} \text{ atoms}}{1 \text{ mol}} = 1.81 \times 10^{24} \text{ atoms He}$

The constant is named after Amedeo Avogadro, whose work on gases led to it. He never measured the number himself, and the term "mole" came decades after he died.

To find the mass of a formula, multiply each element's atomic mass by its subscript and add the results. For glucose, $C_6H_{12}O_6$:

$6(12.01) + 12(1.008) + 6(16.00) = 180.16 \text{ g/mol}$

A molecular compound like glucose gets a molecular weight. An ionic compound like NaCl has no separate molecules, so its sum is called a formula weight ($22.99 + 35.45 = 58.44$). The math is identical. In amu the number describes one molecule or formula unit, and in g/mol the same number describes one mole.

Percent composition tells you what fraction of a compound's mass comes from each element. Oxygen in glucose is $96.00 / 180.16 \times 100 = 53.3\%$.

To go backward from percents to an empirical formula, assume a 100 g sample, convert each element's grams to moles, and divide by the smallest mole value. A compound that is 40.0% C, 6.7% H, and 53.3% O gives 3.33 mol C, 6.65 mol H, and 3.33 mol O, a 1 : 2 : 1 ratio, so the empirical formula is $CH_2O$. If the molar mass is 180.16 g/mol, divide by the empirical mass (30.03) to get 6, and the molecular formula is $C_6H_{12}O_6$.

Every conversion in this topic follows one map, grams ↔ moles ↔ particles. Molar mass links grams and moles, and Avogadro's number links moles and particles. You can't skip from grams to particles directly, because moles sit in the middle.

To go from grams to moles, divide by molar mass. To go from moles to grams, multiply by it. Writing each factor as a fraction with units keeps you from flipping one by accident.

Worked example. How many molecules are in 50.0 g of water? First find the molar mass, $2(1.008) + 16.00 = 18.02$ g/mol. Then convert grams to moles:

$50.0 \text{ g} \times \dfrac{1 \text{ mol}}{18.02 \text{ g}} = 2.775 \text{ mol}$

Then convert moles to molecules:

$2.775 \text{ mol} \times \dfrac{6.022 \times 10^{23}}{1 \text{ mol}} = 1.67 \times 10^{24} \text{ molecules}$

A quick sense check helps here. 50 g is a bit less than three moles of water, so the answer should be a bit less than three times $6 \times 10^{23}$, and it is.

Later units add a third bridge. At STP (0°C and 1 atm), one mole of an ideal gas takes up about 22.4 L.

Atoms aren't created or destroyed in a reaction, so a balanced equation has the same count of each element on both sides. Change only coefficients, never subscripts, since changing $H_2O$ to $H_2O_2$ makes a different substance. It often helps to balance elements that appear in only one compound on each side first. H and O often get balanced later, but the best order depends on the equation.

The coefficients are mole ratios. The general plan is given amount → moles → mole ratio → moles of target → final unit.

Worked example. For $2H_2 + O_2 \rightarrow 2H_2O$, you start with 4.0 g $H_2$ and 16.0 g $O_2$. Converting to moles gives $4.0 / 2.016 = 2.0$ mol $H_2$ and $16.0 / 32.00 = 0.500$ mol $O_2$. Divide each amount by its coefficient, which gives about 1.0 for $H_2$ and 0.500 for $O_2$. Oxygen has the smaller value, so it is the limiting reactant. The theoretical yield is

$0.500 \text{ mol } O_2 \times \dfrac{2 \text{ mol } H_2O}{1 \text{ mol } O_2} \times 18.02 \text{ g/mol} = 18.0 \text{ g } H_2O$

If the lab actually collects 15.3 g, the percent yield is $15.3 / 18.0 \times 100 = 85.0\%$. Hydrogen is the excess reactant.