THE MOLE

Atomic and molecular mass• Masses of atoms, molecules, and formula units are given

in amu (atomic mass units).• Example:• Sodium chloride:• (22.99 amu + 35.45 amu =) 58.44 amu• Sulfur dioxide:• (32.07 amu + 2 x 16.00 amu =) 64.07 amu

The Mole• The mole is a counting unit.• The number of particles in 1 mole is called Avogadro’s

Number

(6.02 x 1023)• This is the magic number that allows us to turn atomic

masses (amu) into grams

Molar mass• The molar mass is the mass in grams of one mole of a

compound• The molecular mass can be calculated from atomic

masses:

water = H2O = 2(1.01 amu) + 16.00 amu = 18.02 amu

• 1 mole of H2O will weigh 18.02 g, therefore the molar mass of H2O is 18.02 g

Keep in mind

The masses of individual atoms or molecules are measured

in

AMU

The mass of a mole of atoms or molecules are measured in

GRAMS

Molar Volume• At STP, 1 mol (6.02 x 1023 rep. part.) of ANY gas occupies

a volume of 22.4 L.• The quantity 22.4 L is called the molar volume of a gas.• STP = standard temperature and pressure

• 0o C and 1 atm

• What volume will 0.375 mol of O2 gas occupy at STP?

Mass percent• Measures the mass (by percent) of a single atom in a

compound.

Mass Percent = Element mass x 100 %

Total mass

Percent composition

• Percentage of each element in a compound• By mass

• Can be determined from:

-the formula of the compound or

-the experimental mass analysis of the

compound• The percentages may not always total to 100%

due to rounding

Formulas•Empirical formula: formula in its simplest form.

•Molecular formula: actual formula of a compound.

Examples

• C2H6: Empirical formula = CH3

• N2O4: Empirical formula = NO2

• C6H12O6: Empirical formula = CH2O

• Molecular formula = (empirical formula)n where n is integer

• C2H6= (CH3)2 N2O4= (NO2)2

• C6H12O6= (CH2O)6

Calculations

• 1. Determine the amount in grams of all the elements in the compound

• 2. Convert those grams into moles for each element

• 3. Divide the number of moles by the smallest and round up or down

• 4. Turn those rounded numbers into whole numbers when necessary

Example

•An unknown compound contains 0.426 grams of carbon and 0.107 grams of hydrogen. Determine the empirical formula.

• C: 0.426 g / 12.01 g/mol = 0.0355 mol• H: 0.107 g / 1.008 g/mol = 0.106 mol

• C: 0.0355 / 0.0355 = 1.00; round to 1• H: 0.106 / 0.0355 = 2.99; round to 3

• Empirical formula: CH3

Percentages

•Convert percentages in grams assuming 100 % = 100 grams

•Treat as grams problems

Calculations molecular formulas

• 1. Convert empirical formula to empirical formula mass by using the molar mass

• 2. Determine n by dividing the actual molar mass by the empirical formula molar mass

• 3. Molecular formula = n x empirical formula

Example problem

• The empirical formula of a compound is C2H5 and its molar mass is 58.12 g/mol. What is the molecular formula of this compound?

• C2H5: empirical molar mass = 2 x 12.01 + 5 x 1.008 = 29.06 g/mol

• n = actual molar mass / empirical molar mass =58.12 g/mol / 29.06 g/mol = 2

Molecular formula = 2 x empirical formula = C4H10

Percentages

• When percentages are given, first calculate the empirical formula.

• Use this empirical formula to calculate the molecular formula.