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Concept 02 / 06Some Basic Concepts of Chemistry

Atomic Mass, Molar Mass and the Mole Concept

The unified atomic mass unit and relative atomic mass, average atomic mass from isotopic abundances, molecular and formula mass, the mole and Avogadro's constant, molar mass, molar volume of a gas, and converting between mass, moles, particles and volume.

Figure 1
Atomic mass. Atoms are far too light to weigh individually in grams, so masses are expressed in unified atomic mass units: 1 u1\ \mathrm{u} (or 11 amu) is exactly 112\frac{1}{12} of the mass of one carbon-12 atom, about 1.66×10−241.66 \times 10^{-24} g. The relative atomic mass of an element is the average mass of its atoms relative to this unit, weighted by natural isotopic abundance: Aˉ=∑(fractional abundance×isotopic mass)\bar{A} = \sum(\text{fractional abundance} \times \text{isotopic mass}). For chlorine, 75%75\% 35Cl^{35}\mathrm{Cl} and 25%25\% 37Cl^{37}\mathrm{Cl} give 0.75×35+0.25×37=35.50.75 \times 35 + 0.25 \times 37 = 35.5. This is why tabulated atomic masses are rarely whole numbers. Molecular mass is the sum of the atomic masses in a molecule (H2O\mathrm{H_{2}O}: 2×1+16=182 \times 1 + 16 = 18 u). For ionic solids, which do not contain discrete molecules, the same sum over the formula unit is called the formula mass (NaCl\mathrm{NaCl}: 58.558.5 u). The mole. One mole is the amount of substance containing exactly NA=6.02214076×1023N_{A} = 6.02214076 \times 10^{23} specified entities (atoms, molecules, ions or electrons). The number is chosen so that the mass of one mole in grams equals the relative mass in u: the molar mass MM of water is 18 g mol−118\ \mathrm{g\ mol^{-1}}. The mole is the chemist's counting unit, like a dozen but enormous, and it links the invisible atomic scale to measurable masses. The central conversions, summarised in the figure: n=mMn = \frac{m}{M} (mass to moles), N=nNAN = nN_{A} (moles to particles), n=VVmn = \frac{V}{V_{m}} for a gas (molar volume Vm=22.4V_{m} = 22.4 L at STP, 0∘0^{\circ}C and 11 atm; 22.722.7 L at 0∘0^{\circ}C and 11 bar), n=Mconc×Vn = M_{\text{conc}} \times V for a solution (molarity times volume in litres). Moles is the hub: convert whatever is given into moles, use the balanced equation, then convert out. Counting sub-particles. One mole of H2SO4\mathrm{H_{2}SO_{4}} contains 22 mol H atoms, 44 mol O atoms, 77 mol atoms in total, and 2+16+32=502 + 16 + 32 = 50 mol electrons (from the atomic numbers 1,16,81, 16, 8). Questions often ask for numbers of atoms, ions, electrons, protons or neutrons in a given mass, and the method is always: mass to moles, then multiply by the count per formula unit and by NAN_{A}. Gram atomic and gram molecular mass are older terms for the molar mass of atoms and molecules. Mass of one atom =MNA= \frac{M}{N_{A}} grams: one carbon-12 atom has mass 126.022×1023=1.993×10−23\frac{12}{6.022 \times 10^{23}} = 1.993 \times 10^{-23} g. Key Takeaways: - 1 u=1121\ \mathrm{u} = \frac{1}{12} mass of a 12C^{12}\mathrm{C} atom ≈1.66×10−24\approx 1.66 \times 10^{-24} g. - Average atomic mass =∑(abundance×mass)= \sum(\text{abundance} \times \text{mass}). - NA=6.022×1023 mol−1N_{A} = 6.022 \times 10^{23}\ \mathrm{mol^{-1}}; molar mass in g/mol equals relative mass in u. - n=mM=NNA=V22.4 Ln = \frac{m}{M} = \frac{N}{N_{A}} = \frac{V}{22.4\ \mathrm{L}} (gas at STP). - Count atoms or electrons per formula unit, then scale by moles. - Pitfall: using 22.422.4 L for liquids or solids, or at conditions other than STP. Brain-Teaser 1: Boron has two isotopes, 10B^{10}\mathrm{B} and 11B^{11}\mathrm{B}, and average atomic mass 10.8110.81. Find the abundance of 10B^{10}\mathrm{B}. Solution: Let the fraction of 10B^{10}\mathrm{B} be xx: 10x+11(1−x)=10.8110x + 11(1 - x) = 10.81, so 11−x=10.8111 - x = 10.81 and x=0.19x = 0.19. So 19%19\% 10B^{10}\mathrm{B} and 81%81\% 11B^{11}\mathrm{B}. The average lies closer to the more abundant isotope, which is a good sanity check. Brain-Teaser 2: How many electrons are present in 1.61.6 g of methane, CH4\mathrm{CH_{4}}? Solution: M=16M = 16 g/mol, so n=0.1n = 0.1 mol. Each molecule has 6+4×1=106 + 4 \times 1 = 10 electrons. Total =0.1×10×6.022×1023=6.022×1023= 0.1 \times 10 \times 6.022 \times 10^{23} = 6.022 \times 10^{23} electrons, exactly one mole of electrons. Counting per molecule first and then scaling keeps the arithmetic clean. Brain-Teaser 3: Which contains the most atoms: 11 g of H2\mathrm{H_{2}}, 1616 g of O2\mathrm{O_{2}}, or 22.422.4 L of CO2\mathrm{CO_{2}} at STP? Solution: 11 g H2\mathrm{H_{2}} is 0.50.5 mol molecules, 11 mol atoms. 1616 g O2\mathrm{O_{2}} is 0.50.5 mol molecules, 11 mol atoms. 22.422.4 L CO2\mathrm{CO_{2}} is 11 mol molecules, 33 mol atoms. So the CO2\mathrm{CO_{2}} has the most atoms (3NA3N_{A}). Always distinguish moles of molecules from moles of atoms.

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