1. Molecular Formula Represents the actual number of atoms of each element in compound – Not necessary for ionic compounds – Necessary for covalent compounds The molecular formula for water is H 2 O, and the empirical formula for water is H 2 O The molecular formula for hydrogen peroxide is H 2 O 2,and the empirical formula is HO. Episode 703

2. The empirical formula for glucose is CH 2 O If the molar mass is 180.0 g/mol, find the molecular formula. – Find the mass of the empirical formula. – Compare the molar mass of molecular formula to molar mass of empirical formula. C 1 x 12 g/mol = 12 g/mol H 2 x 1 g/mol = 2 g/mol O 1 x 16 g/mol = 16 g/mol 30 g/mol 180 g/mol 30 g/mol = 6 x CH 2 O = C 6 H 12 O 6 Episode 703

3. The empirical formula for glucose is CH 2 O If the molar mass is 240.0 g/mol, find the molecular formula. – Find the mass of the empirical formula. – Compare the molar mass of molecular formula to molar mass of empirical formula. C 1 x 12 g/mol = 12 g/mol H 2 x 1 g/mol = 2 g/mol O 1 x 16 g/mol = 16 g/mol 30 g/mol 240 g/mol 30 g/mol = 8 x CH 2 O = C 8 H 16 O 8

4. Problem Set 1 Empirical Formula Molar Mass of molecular formula Molar mass of empirical formula Compare molar masses Actual molecular formula CH26 g/mol NO 2 230 g/mol C3H8C3H8 44 g/mol 13 g/mol 230 g/mol 44 g/mol 2 5 1 C2H2C2H2 N 5 O 10 C3H8C3H8 Episode 703

5. Find the molecular formula for a compound with - 4.04 g N 11.46 g O Molar mass 108 g/mol – Find the empirical formula. – Find the mass of the empirical formula – Compare the molar mass of molecular formula to molar mass of empirical formula. – Empirical formula N 2 O 5 N 2 x 14 g/mol = 28 g/mol O 5 x 16 g/mol = 80 g/mol 108 g/mol = 1 x N 2 O 5 = N 2 O 5 Episode 703

6. Hydrates Crystals with water molecules adhering to the ions or molecules – Na 2 CO 310 H 2 O – Indicates 10 water molecules adhering to each formula unit of sodium carbonate Mass of water = mass of “hydrated” compound minus mass of “dry” compound – Anhydrous means “dry” Episode 703

7. Determine the formula of hydrated barium chloride from this data: Initial mass of hydrated compound = 1.373 g Mass after heating = 1.175 g 1. Determine formula for barium chloride. 2. Determine the mass of water removed from hydrate. 3. Find the ratio between anhydrous compound and water. BaCl 2 1.373 g – 1.175 g = 0.198 g water 1.175g BaCl 2 208g BaCl 2 1 mol BaCl 2 = 0.0056 molBaCl 2 0.198g H 2 O 18g H 2 O 1 mol H 2 O = 0.012 mol H 2 O 0.0056 BaCl 2 2H 2 O Episode 703

8. Determine the formula for the hydrate that is 76.9% CaSO 3 and 23.1% H 2 O. Remember the trick of changing % to grams! Find the ratio between anhydrous compound and water. 76.9g CaSO 3 120g CaSO 3 1mol CaSO 3 = 0.641 mol CaSO 3 23.1g H 2 O 18g H 2 O 1 mol H 2 O = 1.28 mol H 2 O 0.641 CaSO 3 2H 2 O Episode 703