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Molecular Formula

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

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Molecular Formula

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  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 H2O, and the empirical formula for water is H2O • The molecular formula for hydrogen peroxide is H2O2,and the empirical formula is HO. Episode 703

  2. The empirical formula for glucose is CH2O • 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 = C6H12O6 = 6 x CH2O 30 g/mol Episode 703 Episode 703

  3. The empirical formula for glucose is CH2O • 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 = C8H16O8 = 8 x CH2O 30 g/mol

  4. Problem Set 1 13 g/mol 2 C2H2 230 g/mol 5 N5O10 1 44 g/mol C3H8 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 N2O5 N 2 x 14 g/mol = 28 g/mol O 5 x 16 g/mol = 80 g/mol 108 g/mol 108 g/mol = N2O5 = 1 x N2O5 108 g/mol Episode 703

  6. Hydrates • Crystals with water molecules adhering to the ions or molecules • Na2CO3•10 H2O • 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. BaCl2 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. 1.175g BaCl2 1 mol BaCl2 = 0.0056 molBaCl2 208g BaCl2 0 .0056 BaCl2•2H2O 0.198g H2O 1 mol H2O = 0.012 mol H2O 18g H2O 0 .0056 Episode 703

  8. Determine the formula for the hydrate that is 76.9% CaSO3 and 23.1% H2O. Remember the trick of changing % to grams! Find the ratio between anhydrous compound and water. 76.9g CaSO3 1mol CaSO3 = 0.641 mol CaSO3 120g CaSO3 0.641 23.1g H2O 1 mol H2O = 1.28 mol H2O 18g H2O 0.641 CaSO3•2H2O Episode 703

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