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Vocabulary-style flashcards covering Avogadro's number, atomic/molar masses, isotopes, percent composition, and how to derive empirical and molecular formulas from data.
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Avogadro's number (N_A)
6.022 × 10^23 particles per mole; the number of atoms or molecules in one mole of a substance.
Molar mass
Mass of one mole of a substance, in grams per mole; numerically equal to the formula mass (in amu) for the substance.
Atomic mass unit (amu)
Unit of mass used for atoms and molecules; 1 amu is 1/12 the mass of a carbon-12 atom.
Atomic number (Z)
Number of protons in the nucleus; identifies the element.
Mass number (A)
Total number of protons and neutrons in an atom; the isotope's mass in amu (e.g., hydrogen-1 has A=1).
Isotopes
Atoms of the same element (same Z) with different mass numbers (A) due to differing numbers of neutrons.
Average atomic mass
Weighted average of isotopic masses based on natural abundances; listed on the periodic table.
Molecular mass
Sum of the atomic masses (in amu) of all atoms in one molecule.
Molar mass (distinction from molecular mass)
Mass of one mole of a substance (g/mol); numerically equal to the molecular/formula mass in amu.
SO2 molecular mass
64.07 amu (32.072 for S + 2×16.00 for O); the mass of one SO2 molecule.
Molar mass of SO2
64.07 g/mol; mass of one mole of SO2.
Mass to moles relationship
n (mol) = mass (g) / molar mass (g/mol). Units cancel to give moles.
Moles to molecules relationship
Number of molecules = moles × Avogadro's number (N_A).
Atoms in a molecule from moles
Total atoms of an element = (atoms per molecule) × (number of molecules).
Percent composition by mass
Mass percent of an element in a compound = (moles of element × molar mass of element) / (molar mass of compound) × 100.
Empirical formula
The simplest whole-number ratio of atoms in a compound, obtained from percent composition by converting to moles and reducing to smallest integers.
Molecular formula
The actual formula of a compound; equals the empirical formula multiplied by an integer factor.
Ascorbic acid empirical formula example
From percent composition (e.g., C 40.92%, H 4.58%, O 54.50%), the empirical formula is C3H4O3.
Molar mass from percent example: MnO
77% Mn and 23% O lead to an empirical formula MnO.
P2O5 empirical formula from percent
43.64% P and 56.36% O yield empirical formula P2O5; if molar mass is 283.88 g/mol, the molecular formula is P4O10.
C2H4Br empirical vs molecular formula
Empirical formula C2H4Br; with molar mass 215.9 g/mol, the molecular formula is C4H8Br2.
Caffeine molar mass
Molar mass of caffeine = 194.2 g/mol (194.2 amu per molecule).