The mole, chemical formulas and names

Chemists count atoms by weighing them. A balance cannot see individual atoms, but if we know how much one atom weighs on average, the mass of a sample tells us how many atoms it contains. The unit that makes this practical is the mole. This lesson introduces the mole and molar mass, shows how to move between mass, moles and numbers of particles, and uses those skills to find chemical formulas from experimental data. It also sets out the rules for naming simple compounds, a language you will need for the rest of the program.

The mole and the Avogadro constant

The mole (mol) is the SI unit of amount of substance. One mole contains exactly 6.022 140 76 × 10²³ specified entities, whether atoms, molecules, ions or electrons. This number is the Avogadro constant, N_A, usually rounded to 6.022 × 10²³ mol⁻¹. Its value was chosen so that the mass of one mole of a substance in grams is numerically equal to the mass of one of its entities in atomic mass units. One carbon-12 atom has a mass of exactly 12 u (before 2019, by definition; now to within a few parts in a billion), and one mole of carbon-12 has a mass of 12 g.

The number is almost impossible to picture. A mole of grains of sand would cover Canada to a depth of many metres; a mole of seconds is about 20 million times the age of the universe. Yet a mole of water is only 18 g, about a tablespoon. Atoms are that small.

Here N is the number of particles, n the amount in moles, m the mass and M the molar mass. These two relationships, and the conversion factors they imply, are the core of quantitative chemistry.

Molar mass

The molar mass of an element is its average atomic mass expressed in grams per mole: carbon is 12.011 g/mol, oxygen 15.999 g/mol, chlorine 35.45 g/mol. The molar mass of a compound is the sum of the molar masses of the atoms in its formula. For ionic compounds, which do not consist of discrete molecules, the same calculation gives the formula mass per mole of formula units. Modern periodic tables give atomic masses to four or more significant figures; for most calculations in this course, two decimal places are plenty.

The code below parses chemical formulas, including parentheses and hydrate dots, and calculates molar masses. Programs like this are built into every chemistry software package.

import re

MASS = {"H": 1.008, "C": 12.011, "N": 14.007, "O": 15.999, "Na": 22.990,
        "Mg": 24.305, "Al": 26.982, "P": 30.974, "S": 32.06, "Cl": 35.45,
        "K": 39.098, "Ca": 40.078, "Fe": 55.845, "Cu": 63.546}

def parse(formula):
    """Return a dict of element counts. Handles parentheses and hydrate dots."""
    counts = {}
    for i, part in enumerate(formula.split(".")):
        mult = 1
        if i > 0:
            m = re.match(r"(\d+)(.*)", part)
            if m:
                mult, part = int(m.group(1)), m.group(2)
        for el, n in parse_group(part).items():
            counts[el] = counts.get(el, 0) + n * mult
    return counts

def parse_group(s):
    """Parse a formula with no dots, using a stack for parentheses."""
    stack = [{}]
    tokens = re.findall(r"[A-Z][a-z]?|\d+|\(|\)", s)
    i = 0
    while i < len(tokens):
        tok = tokens[i]
        n = int(tokens[i + 1]) if i + 1 < len(tokens) and tokens[i + 1].isdigit() else 1
        if tok == "(":
            stack.append({}); i += 1; continue
        if tok == ")":
            group = stack.pop()
            for el, c in group.items():
                stack[-1][el] = stack[-1].get(el, 0) + c * n
        else:
            stack[-1][tok] = stack[-1].get(tok, 0) + n
        i += 2 if n != 1 or (i + 1 < len(tokens) and tokens[i + 1] == "1") else 1
    return stack[0]

def molar_mass(formula):
    return sum(MASS[el] * n for el, n in parse(formula).items())

for f in ["H2O", "C6H12O6", "Ca3(PO4)2", "CuSO4.5H2O", "Al2(SO4)3"]:
    print(f"{f:12s} {parse(f)}  M = {molar_mass(f):.2f} g/mol")

H2O          {'H': 2, 'O': 1}  M = 18.02 g/mol
C6H12O6      {'C': 6, 'H': 12, 'O': 6}  M = 180.16 g/mol
Ca3(PO4)2    {'Ca': 3, 'P': 2, 'O': 8}  M = 310.17 g/mol
CuSO4.5H2O   {'Cu': 1, 'S': 1, 'O': 9, 'H': 10}  M = 249.68 g/mol
Al2(SO4)3    {'Al': 2, 'S': 3, 'O': 12}  M = 342.13 g/mol

Converting between mass, moles and particles

Every conversion runs through moles. Mass converts to moles by dividing by molar mass; moles convert to particles by multiplying by the Avogadro constant. The formula then links moles of a compound to moles of each element in it: one mole of H₂O contains two moles of H atoms and one mole of O atoms.

A useful habit is to estimate first. Five grams of methane is about a third of a mole, so it contains about 2 × 10²³ molecules and four times as many hydrogen atoms, roughly 8 × 10²³. The calculated answer agrees.

Types of chemical formulas

A molecular formula gives the actual number of each kind of atom in a molecule: hydrogen peroxide is H₂O₂, glucose is C₆H₁₂O₆. An empirical formula gives only the simplest whole-number ratio: HO for hydrogen peroxide, CH₂O for glucose. Different compounds can share an empirical formula: formaldehyde (CH₂O), acetic acid (C₂H₄O₂) and glucose all have the empirical formula CH₂O. A structural formula shows how atoms are connected, which molecular formulas cannot: ethanol (CH₃CH₂OH) and dimethyl ether (CH₃OCH₃) both have the molecular formula C₂H₆O but are quite different substances. Ionic compounds are always written as empirical formulas, because they consist of extended arrays of ions rather than molecules.

Percent composition

The mass percent of an element in a compound is its share of the molar mass. Percent composition is used to check the purity of a compound, to compare fertilizers or ores, and, run in reverse, to determine formulas.

Empirical formulas from experimental data

To find an empirical formula from mass data, convert the mass of each element into moles, then find the simplest whole-number ratio. A practical sequence: assume a 100 g sample so that percentages become grams; convert grams to moles; divide every mole value by the smallest; and if the results are not all close to whole numbers, multiply by a small integer (2, 3, 4) until they are. A ratio such as 1 : 1.5 becomes 2 : 3; 1 : 1.33 becomes 3 : 4.

Combustion analysis

For organic compounds, the classic method is combustion analysis. A weighed sample is burned completely in excess oxygen. All the carbon ends up in carbon dioxide and all the hydrogen in water; these are trapped in separate absorbers and weighed. From the masses of CO₂ and H₂O we calculate the masses of C and H in the original sample. If the compound also contains oxygen, its mass is found by difference, because the oxygen in the products came partly from the sample and partly from the supplied oxygen. Modern CHN analyzers automate the process, measuring the gases instrumentally, but the arithmetic is unchanged.

# Combustion analysis of a compound containing C, H and O
M_C, M_H, M_O = 12.011, 1.008, 15.999
M_CO2, M_H2O = M_C + 2 * M_O, 2 * M_H + M_O

sample = 0.2500     # g of compound burned
co2 = 0.3664        # g of CO2 collected
h2o = 0.1500        # g of H2O collected

mass_C = co2 * M_C / M_CO2
mass_H = h2o * 2 * M_H / M_H2O
mass_O = sample - mass_C - mass_H
moles = {"C": mass_C / M_C, "H": mass_H / M_H, "O": mass_O / M_O}
smallest = min(moles.values())
print({el: round(m / smallest, 2) for el, m in moles.items()})
for mult in range(1, 5):
    ratio = {el: m / smallest * mult for el, m in moles.items()}
    if all(abs(r - round(r)) < 0.08 for r in ratio.values()):
        print("multiplier", mult, "->", {el: round(r) for el, r in ratio.items()})
        break

{'C': 1.0, 'H': 2.0, 'O': 1.0}
multiplier 1 -> {'C': 1, 'H': 2, 'O': 1}

Molecular formulas

An empirical formula combined with the molar mass gives the molecular formula. The molar mass, measured by mass spectrometry or from gas density (lesson 11), must be a whole-number multiple of the empirical formula mass. Divide to find that multiple and multiply every subscript by it.

Hydrates

Many ionic compounds crystallize with water molecules built into their crystal structure. Copper(II) sulfate pentahydrate, CuSO₄·5H₂O, is a striking blue; heating it drives off the water and leaves white anhydrous CuSO₄. The number of water molecules per formula unit can be found by weighing a sample before and after heating.

Naming simple compounds

Chemical names follow rules set by the International Union of Pure and Applied Chemistry (IUPAC). Three categories cover most compounds met in this course.

Binary ionic compounds are named cation first, then anion with the ending -ide: NaCl is sodium chloride, CaO calcium oxide, Al₂S₃ aluminum sulfide. For metals that form more than one ion, mostly transition metals, a Roman numeral gives the charge: FeCl₂ is iron(II) chloride and FeCl₃ iron(III) chloride. Polyatomic ions keep their own names, which must be learned.

Binary molecular compounds, formed between two nonmetals, use Greek prefixes for the number of each atom: CO is carbon monoxide, CO₂ carbon dioxide, N₂O₄ dinitrogen tetroxide, PCl₅ phosphorus pentachloride. The prefix mono- is dropped for the first element. The prefixes are mono (1), di (2), tri (3), tetra (4), penta (5), hexa (6), hepta (7), octa (8), nona (9) and deca (10).

Acids in water are named from their anions. Binary acids take the prefix hydro- and the ending -ic acid: HCl(aq) is hydrochloric acid. Oxoacids change -ate to -ic acid and -ite to -ous acid: H₂SO₄ is sulfuric acid (from sulfate), HNO₂ is nitrous acid (from nitrite). CHEM-102 develops the chemistry of acids; organic nomenclature waits for CHEM-261.