Mole Calculations

The Mole

• Mole – measurement of the amount of a substance.– We know the amount of different

substances in one mole of that substance.

Atomic Mass Unit

• Mass of 1 mole of compound• Found by adding the atomic weights

of each atom of each element that makes up a compound.

• H2O

• There are 2 atoms of hydrogen and 1 atom of oxygen.

AMU

• H = 2 x 1.01 = 2.02• O = 1 x 16 = 16 _____

18.02 = amu

1 mole of any substance = the amu of that substance

AMU

• Molar mass = the sum of the molar masses of atoms of the elements in the formula

• Calculated the same way

The Mole

• One mole of any substance has Avogadro’s number.

• 6.022 x 1023 atoms, molecules, ions

The Mole

• For gases only – 1 mole of a gas occupies 22.4 L

Converting Between Units

amu 1 mole 6.022 x 1023 atoms, mlc, ions

(expressed in grams)

Conversions

• Change 5.0 grams of sodium chloride to moles of sodium chloride.

• Change 0.45 moles of barium chlorate to grams of barium chlorate.

• Determine the number of atoms in 15 grams of water.

Empirical Formulas

Empirical Formulas

• The simplest whole number ratio of moles of each element in the compound.

• H2O

• NaCl

Empirical Formulas

• Usually given in percentages of each element in the compound.

• Based on 100% of the compound.• Can be compared to a 100 gram

sample of the substance.

• 11.2% Hydrogen• 88.8% Oxygen

• 1. Change percents to grams

11.2% H = 11.2 g H88.8% O = 88.8 g O

• 2. Convert grams to moles. (We know an empirical formula is the lowest whole number ratio of moles)

11.2 g H | 1 mol = 11.089 mol | 1.01 g

88.8g O | 1 mol = 5.550 mol | 16 g

Cannot have decimals, the numbers must be whole numbers. Divide by the lowest.

11.2 g H | 1 mol = 11.089 mol / 5.550 = 2 | 1.01 g

88.8g O | 1 mol = 5.550 mol / 5.550 = 1 | 16 g

• H2O = empirical formula

• 36.84% N• 63.16% O

• 35.98% Al• 64.02% S

Molecular Formula

• Specifies the actual number of atoms of each element in one molecule or formula unit of the substance.

• The molar mass of acetylene is 26.04 g/mol and the mass of the empirical formula, CH, is 13.02 g/mol

• 1. The problem will give you a molar mass of the compound.

• 2. Calculate the empirical formula as usual.

• 40.68% carbon• 5.08% hydrogen• 54.24% oxygen

• Molar mass = 118.1 g/mol

• Change % to grams

• 40.68 g C | 1 mol = 3.387 mol | 12.01 g

• 5.08 g H | 1 mol = 5.030 mol| 1.01 g

• 54.24 g O | 1 mol = 3.390 mol | 16 g

• 3.387 mol / 3.387 mol = 1 x2 = 2• 5.030 mol / 3.387 mol = 1.5 x2 = 3• 3.390 mol / 3.387 mol = 1 x2 = 2

• Empirical Formula = C2H3O2

• To find the molecular formula:

• (EFamu)x = MM (molar mass)

• C2H3O2 = 59.05 = amu

• (59.05)x = 118.1• X = 2

• (EF)x2 = (C2H3O2)x2 = C4H6O4

• 65.45% C• 5.45% H• 29.09% O

• MM = 110.0 g/mol

• 49.98 g C• 10.47 g H

• MM = 58.12 g/mol

• 46.68% N• 53.32% O

• MM = 60.01 g/mol