The mole and molar mass
Chemists count atoms by weighing them. The mole links the tiny scale of atoms to the grams you measure in the lab.
Why we need the mole
Atoms are far too small to count individually: a drop of water contains about \(10^{21}\) molecules. The mole is a counting unit like a dozen, but enormous: one mole is \(6.022 \times 10^{23}\) particles, Avogadro's number. It lets chemists measure out known numbers of atoms by mass.
N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}Molar mass
The molar mass of an element in grams per mole equals its atomic mass in u: carbon is 12.01 g/mol. For a compound, add the atoms' molar masses: water \(\text{H}_2\text{O}\) is \(2(1.008) + 16.00 = 18.02\) g/mol. So 18.02 g of water contains one mole of molecules.
Converting between grams, moles, and particles
moles \(= \frac{\text{mass}}{\text{molar mass}}\); particles \(= \text{moles} \times N_A\). How many molecules are in 36.0 g of water? \(\frac{36.0}{18.02} \approx 2.00\) mol, times \(6.022 \times 10^{23}\) gives about \(1.20 \times 10^{24}\) molecules.
n = \frac{m}{M}Percent composition
The mass percent of an element in a compound is its mass in one mole divided by the molar mass. In water, hydrogen is \(\frac{2.016}{18.02} \approx 11.2\%\) and oxygen \(88.8\%\). Percent composition helps identify unknown compounds and check purity.
Empirical and molecular formulas
The empirical formula gives the simplest whole-number ratio of atoms; the molecular formula gives actual numbers. Glucose is \(\text{C}_6\text{H}_{12}\text{O}_6\) with empirical formula \(\text{CH}_2\text{O}\). To find an empirical formula from percentages, assume 100 g, convert each element to moles, and divide by the smallest number of moles.
Worked example: an empirical formula
A compound is 40.0% C, 6.7% H, and 53.3% O. In 100 g: \(\frac{40.0}{12.01} = 3.33\) mol C, \(\frac{6.7}{1.008} = 6.65\) mol H, \(\frac{53.3}{16.00} = 3.33\) mol O. Dividing by 3.33 gives 1 : 2 : 1, so the empirical formula is \(\text{CH}_2\text{O}\). If its molar mass is 180 g/mol, the molecular formula is \(\text{C}_6\text{H}_{12}\text{O}_6\).
Molarity
Solution concentration is usually given as molarity: moles of solute per litre of solution. Dissolving 0.50 mol of NaCl to make 2.0 L gives 0.25 M. Diluting follows \(M_1V_1 = M_2V_2\), since the moles of solute stay the same.
M = \frac{n}{V},\quad M_1V_1 = M_2V_2Common mistakes
Forgetting to multiply by subscripts when computing molar mass. Confusing grams and moles. Rounding mole ratios too early when finding empirical formulas. Using the volume of solvent instead of the total solution volume for molarity.
Key takeaways
A mole is \(6.022 \times 10^{23}\) particles. Molar mass converts grams to moles: \(n = m/M\). Percent composition and empirical formulas come from mole ratios. Molarity is moles per litre, and dilution keeps moles constant: \(M_1V_1 = M_2V_2\).
Flashcard
6.022 \times 10^{23} \text{ particles}
Check your understanding
How many moles are in 88.0 g of CO2 (molar mass 44.0 g/mol)?