Chem June Exam outline

Chemistry June Exam Notes
Quantities in Chemical Reactions
 Molecular and formula mass o The mass of one unit of a compound (a molecule or a formula unit) o The sum of the mass of all the atoms in a compound o With knowledge of the mass of each individual atom, the percentage composition by mass can be determined
 The Mole (mol) o A counting unit, one mole refers to 6.02 x 1023 particles of any given substance o Known as Avogadro’s Constant and given the symbol NA
 Molar Mass o The molar mass of a substance is the mass of one mole of a particular substance
 Mass of one particle x 6.02 x 1023 o Coincidentally, the molar mass of a substance has the same numerical value as the atomic, molecular, or formula mass of the substance, but with different units
 If one single unit of a substance is x u, then one mol of the same substance is x g/mol
 Equations o n = m/M
 n = number of moles (mol)
 m = specific mass of substance (g)
 M = molar mass of substance (g/mol) o n = N/NA
 n = number of moles (mol)
 N = specific number of particles
 NA = Avogadro’s Constant; 6.02 x 1023
 Empirical (Simplest) Formula o Represents smallest whole number ratio of the atoms in a compound
 E.g. empirical formula of H2O2 is HO o To find empirical formula
 Determine mass of each element in the compound, or use percentage composition and assume a 100 g sample
 Find the number of moles of each element (n = m/M)
 Find the ratio among the mole values by dividing the mole value for each element by the smallest mole value
 If these ratios are not whole numbers, multiply by an appropriate factor until whole numbers are obtained (some rounding is necessary)
 In short: Element  Mass  Moles  Ratio  Whole number ratio

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Molecular Formula o Shows the exact number of each atom in a compound
 Multiple of the empirical formula
 E.g. C6H6 = (CH)6 o Molecular formula = (empirical formula)x
 X = molar mass / empirical formula mass o To