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Empirical and Molecular Formulas

Empirical and Molecular Formulas. Empirical Formula. If we are given an unknown substance how could we determine its chemical formula? We can determine its percent composition Then we can determine its empirical formula: the simplest whole-number ratio of atoms in a compound.

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Empirical and Molecular Formulas

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  1. Empirical and Molecular Formulas

  2. Empirical Formula • If we are given an unknown substance how could we determine its chemical formula? • We can determine its percent composition • Then we can determine its empirical formula: the simplest whole-number ratio of atoms in a compound

  3. However, the empirical formula does not necessarily give information about the number of atoms in a molecule (molecular formula). • The empirical formula gives the combining ratio in its simplest form • The molecular formula gives the same ratio but with the actual number of atoms Note: The empirical formula and the molecular formula can be the same in some cases. E.g. H2O, CH4, CO2

  4. Divide each element by the lowest quantity 3.33 mol C = 1 6.63 mol H = 1.99 3.33 mol O = 1 3.33 3.33 3.33 Mole ratio of C:H:O = 1:2:1 Convert mass to moles nC = 40.0 g X 1 mol 12.01 g nC = 3.33 mol nO = 53.3 g X 1 mol 16.00 g nO = 3.33 mol nH = 6.70 g X 1 mol 1.01 g nH = 6.63 mol Calculating Empirical Formula E.g. What is the empirical formula of a substance that is 40.0% carbon, 6.70% hydrogen, and 53.3% oxygen by mass? Given: Assume 100 g mC = 40.0 g mH = 6.70 g mO = 53.3 g  The empirical formula of the substance is CH2O

  5. Calculating Molecular Formula • The empirical formula tells us the simplest ratio of atoms in a molecule, but it does not tell us the actual number of atoms in a molecule • However, if we know molar mass and the empirical formula of a compound we can determine its molecular formula • Divide the molar mass of the compound by the molar mass obtained from the empirical formula • Multiply the subscripts by this number

  6. Molecular Formula = MMcmpd MMHO = 34 g/mol 17.01 g/mol = 2 Given: Empirical Formula: HO MMcmpd = 34 g/mol MMHO = 1.01 + 16.00 = 17.01 g/mol E.g. What is the molecular formula of a compound that has a molar mass of 34 g/mol and the empirical formula HO?  The molecular formula of the compound is H2O2

  7. Molecular Formula = MMcmpd MMCH3 = 45.00 g/mol 15.04 g/mol = 3 Given: Empirical Formula: CH3 MMcmpd = 45.00 g/mol MMCH3 = 12.01 + 3(1.01) = 15.04 g/mol E.g. What is the molecular formula of a compound that has a molar mass of 45.00 g/mol and the empirical formula CH3?  The molecular formula of the compound is C3H9

  8. Calculating Molecular Formula • We can calculate molecular formula if empirical formula and molar mass is given • We can also calculate molecular formula if percent composition and molar mass is given.

  9. mH = 0.175 X 58.0 g = 10.2 g nH = 10.2 g X 1 mol 1.01 g = 10.1 mol mc = 0.825 X 58.0 g = 47.8 g nc = 47.8 g X 1 mol 12.01 g = 3.98 mol Given: % C = 82.5% % H = 17.5% MM = 58.0 g/mol m = 58.0 g (Assume 1 mole) E.g. What is the molecular formula of a compound with a molar mass of 58.0 g/mol and a percent composition of 82.5% C and 17.5% H? Mole ratio C:H = 4:10  The molecular formula is C4H10

  10. Homework • P. 186 #2-6 • P. 188 #1-6 • P. 193 # 8,9

  11. Hydrates • Solid ionic compounds are formed from dehydrating ionic solutions • Some of the liquid water molecules may remain between molecules of the solids, creating a hydrate • When a hydrate is heated and water is evaporated it is referred to as anhydrous

  12. Hydrate Example A 50.0 g sample of a hydrate of barium hydroxide was heated. The anhydrous solid weighed 27.2 g. • Determine the percent composition of water in the hydrate. • Calculate the molecular formula of the hydrate. Name the hydrate. Hint: Start with a BCE Ba(OH)2xH2O Solve for x

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