Notes on Nomenclature, Moles, Percent Composition, and Empirical vs Molecular Formulas
Nomenclature, Moles, Percent Composition, and Empirical vs. Molecular Formulas
Nomenclature overview
Binary acids naming rule (as introduced): use the hydro- prefix, the root name of the element, and end with -ic acid (e.g., hydrofluoric acid).
Distinguish between covalently bonded vs ionic compounds; atoms can be ions or molecules.
Nomenclature summary to study and master: you should be able to name any compound once you know whether it is covalent or ionic.
The mole concept and Avogadro's number
A mole is a counting unit for atoms, ions, formula units, etc.
Avogadro's number: N_A = 6.022 \times 10^{23} particles per mole.
Conversion factors:
From moles to particles (atoms, ions, or formula units): multiply by N_A.
From particles to moles: divide by N_A.
Example hint: For every one mole of carbon tetrachloride (CCl₄), there is 1 mole of carbon and 4 moles of chlorine.
When converting number of atoms to moles, multiply moles by the appropriate stoichiometric coefficients in the formula.
Molar mass and mass-to-moles conversions
Molar mass is the mass per mole of a species, expressed in g/mol.
To go from moles to grams: m = n \times Mm where Mm is the molar mass.
To go from grams to moles: n = \dfrac{m}{M_m}.
The molar mass of a compound is found by summing the molar masses of all atoms in the formula.
Example: calculate the molar mass of acetic acid by adding masses of all constituent atoms (H, C, O).
The total molar mass is used as a conversion factor between moles and grams.
Percent composition (mass percent of components)
Definition: the percent composition of an element X in a compound is the mass fraction of X in one mole of the compound, expressed as a percentage.
Formula:
\%X = \left( \dfrac{mX}{M{compound}} \right) \times 100\%
where mX is the mass of element X in one mole of the compound and M{compound} is the molar mass of the compound.
Example: copper in copper sulfate pentahydrate (CuSO₄·5H₂O).
For this compound, determine the masses of Cu, S, O, and H in one mole:
Cu: 63.546 g/mol
S: 32.066 g/mol
O: 16.00 g/mol (nine O atoms total in CuSO₄·5H₂O: 4 from sulfate + 5 from water)
H: 1.008 g/mol (10 H atoms total)
Molar mass of CuSO₄·5H₂O ≈ 249.68 g/mol
Percent Cu: \%Cu = \dfrac{63.546}{249.68} \times 100\% \approx 25.45\%
For a 20 g sample, grams of Cu ≈ 0.2545 \times 20 \text{ g} = 5.090\text{ g}
Empirical vs molecular formulas
Empirical formula: the simplest whole-number ratio of atoms in a compound (not necessarily the actual number of atoms).
Molecular formula: the actual number of each type of atom in a molecule; it is not just a ratio.
Relationship: the molecular formula is a multiple of the empirical formula.
Examples:
Glucose has empirical formula CH₂O; its molecular formula is C₆H₁₂O₆ (a sixfold multiple of CH₂O).
Acetylene (C₂H₂) and benzene (C₆H₆) have the same empirical formula CH, but benzene has molecular formula C₆H₆ (a sixfold multiple of CH) and acetylene has C₂H₂ (a twofold multiple of CH).
How to determine the empirical formula from percent composition (step-by-step)
1) Start from the percent composition of the compound (assume a 100 g sample so that the percent values directly give grams of each element).Example percentages: Carbon 60.0%, Hydrogen 4.48%, Oxygen 35.52% (sum = 100%).
In 100 g of compound: C = 60.0 g, H = 4.48 g, O = 35.52 g.
2) Convert each mass to moles using the respective molar masses:C: n_C = \dfrac{60.0\text{ g}}{12.01\text{ g/mol}} \approx 4.996\text{ mol}
H: n_H = \dfrac{4.48\text{ g}}{1.008\text{ g/mol}} \approx 4.444\text{ mol}
O: n_O = \dfrac{35.52\text{ g}}{16.00\text{ g/mol}} \approx 2.22\text{ mol}
3) Determine the simplest whole-number ratio by dividing all mole values by the smallest number of moles among them (here, 2.22):C: 4.996 / 2.22 ≈ 2.25
H: 4.444 / 2.22 ≈ 2.00
O: 2.22 / 2.22 = 1.00
4) Convert to whole numbers: since 2.25 is not a whole number, multiply all by a factor to get whole numbers. Here, multiply by 4 to get integers:C: 2.25 × 4 = 9
H: 2.00 × 4 = 8
O: 1.00 × 4 = 4
Empirical formula: \mathrm{C9H8O_4}
5) Compute the molar mass of the empirical formula:M{empirical} = 9(12.01) + 8(1.008) + 4(16.00) \approx 180.16\ \text{g/mol} 6) Determine the molecular formula by comparing the molar mass of the actual compound ($M{compound}$) to $M_{empirical}$:
n = \dfrac{M{compound}}{M{empirical}}
If $M{compound}$ = 180.2 g/mol, then n \approx \dfrac{180.2}{180.16} \approx 1.00, so the molecular formula is the same as the empirical formula (C₉H₈O₄). 7) In a different example (fructose), the empirical formula CH₂O with $M{empirical} = 30.03$ g/mol, and if the compound’s molar mass is $M_{compound} = 180.2$ g/mol, then
n = \dfrac{180.2}{30.03} \approx 6
Molecular formula = $(CH2O) \times 6 = \mathrm{C6H{12}O6}$ (glucose).
Quick concept check: which has more atoms for the same mass, 10 g of Mg or 10 g of Ca?
Mg has a smaller molar mass (24.305 g/mol) than Ca (40.78 g/mol), so for the same mass, Mg has more moles and therefore more atoms.
Intuition: If you compare two samples by mass with different molar masses, the one with the smaller molar mass will consist of more moles and thus more atoms for the same mass.
Recap of vocabulary and key relationships
Atomic mass: average mass of an atom, as listed on the periodic table (in amu) and used to compute molar masses of elements.
Molecular mass / formula mass: mass of a molecule or formula unit in amu; molar mass in g/mol is numerically equal to this mass in amu.
Molar mass: grams per mole; used to convert between grams and moles.
Percent composition: mass fraction of each element in a compound expressed as a percentage.
Empirical formula: simplest whole-number ratio of atoms in a compound.
Molecular formula: actual numbers of atoms in a molecule; is a whole-number multiple of the empirical formula.
Quick reminders for future sections
You will frequently flip conversion factors when moving between species (moles of A to moles of B, or moles to grams, etc.).
In stoichiometry, you’ll also move between moles of substances in a reaction using coefficients from the balanced equation.
Practice both directions of conversions (grams → moles and moles → grams) and become fluent with empirical-to-molecular formula calculations.