Unit 6 Mole Calculations

The Mole Mole – measurement of the ______________ 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 ______________= 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  22.4 L Gas

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

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

conversions 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

Empirical formula 1. Change percent to grams 11.2% H = 11.2 g H 88.8% O = 88.8 g O

Empirical formula 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

Empirical formula 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

Empirical formula 36.84% N 63.16% O

Empirical formula 35.98% Al 64.02% S

Molecular Formula

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

Molecular formula 1. The problem will give you a molar mass of the compound. 2. Calculate the empirical formula as usual.

Molecular formula 40.68% carbon 5.08% hydrogen 54.24% oxygen Molar mass = 118.1 g/mol

Molecular formula Change % to grams 40.68 g C | 1 mol = 3.387 mol 5.08 g H | 1 mol = 5.030 mol | 1.01 g 54.24 g O | 1 mol = 3.390 mol | 16 g

Molecular formula 3.387 mol / 3.387 mol = 1 x2 = 2 Empirical Formula = C2H3O2

Molecular formula To find the molecular formula: (EFamu)x = MM (molar mass)

Molecular formula C2H3O2 = 59.05 = amu (59.05)x = 118.1 X = 2 (EF)x2 = (C2H3O2)x2 = C4H6O4

Molecular formula 65.45% C 5.45% H 29.09% O MM = 110.0 g/mol

Molecular formula 49.98 g C 10.47 g H MM = 58.12 g/mol

Molecular formula 46.68% N 53.32% O MM = 60.01 g/mol