Molecular Formula Represents the actual number of atoms

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

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. 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 – Compare the molar mass of molecular formula to molar mass of empirical formula. 180 g/mol 30 g/mol = 6 x CH 2 O = C 6 H 12 O 6 Episode 703

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. 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 – Compare the molar mass of molecular formula to molar mass of empirical formula. 240 g/mol 30 g/mol = 8 x CH 2 O = C 8 H 16 O 8

Problem Set 1 Empirical Formula CH NO 2 C 3 H 8 Molar Mass of Molar mass of molecular empirical formula 26 g/mol 13 g/mol 230 g/mol 44 g/mol Compare molar masses Actual molecular formula 2 5 1 C 2 H 2 N 5 O 10 C 3 H 8 Episode 703

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

Hydrates • Crystals with water molecules adhering to the ions or molecules – Na 2 CO 3 • 10 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

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. Ba. Cl 2 2. Determine the mass of water removed from hydrate. 1. 373 g – 1. 175 g = 0. 198 g water 3. Find the ratio between anhydrous compound and water. 1. 175 g Ba. Cl 2 1 mol Ba. Cl 2 = 0. 0056 mol. Ba. Cl 2 208 g Ba. Cl 2 0. 0056 0. 198 g H 2 O 1 mol H 2 O = 0. 012 mol H O 2 18 g H 2 O 0. 0056 Ba. Cl 2 • 2 H 2 O Episode 703

Determine the formula for the hydrate that is 76. 9% Ca. SO 3 and 23. 1% H 2 O. Remember the trick of changing % to grams! Find the ratio between anhydrous compound and water. 76. 9 g Ca. SO 3 1 mol Ca. SO 3 120 g Ca. SO 3 23. 1 g H 2 O 1 mol H 2 O 18 g H 2 O = 0. 641 mol Ca. SO 3 0. 641 = 1. 28 mol H 2 O 0. 641 Ca. SO 3 • 2 H 2 O Episode 703