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UNIT 4

UNIT 4. Empirical Formulas. Determining the Molecular Formula from the Empirical Formula and the Molecular Mass. The empirical formula of our compound is NO 2 . Suppose its molar mass is 92 g. Then the molecular formula will be an integral number of the empirical formula, (NO 2 ) n

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UNIT 4

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  1. UNIT 4 Empirical Formulas

  2. Determining the Molecular Formula from the Empirical Formula and the Molecular Mass The empirical formula of our compound is NO2 . Suppose its molar mass is 92 g. Then the molecular formulawill be an integral number of the empirical formula, (NO2)n molecular formula = (NO2)n molar mass = 92 g empirical formula = NO2mass of NO2 = 46 g Dividing gives us n: (NO2)n = 92 g gives n = 2 NO2 46 g molecular formula = (NO2)2= N2O4

  3. Determining the Empirical Formula in the Laboratory There are three ways to determine the empirical formula of a compound in the laboratory: • by a combination reaction • by a decomposition reaction • by indirect analysis (often a combustion reaction)

  4. Determining the Empirical Formula from a Combination Reaction Incombinationreactions, two or more reactants combine to form a single product. A + B  AB Example: A student burns 12.3 g of Mg to produce an oxide that weighs 20.4 g. What is theempirical formula of the oxide? Mass of Mg = 12.3 g Mass of O = 20.4 – 12.3 = 8.1 g 12.3 g Mg x 1 mol Mg =0.506 mol Mg8.1 g O x 1 mol O = 0.51 mol O 24.305 g 16.00 g This gives us Mg0.506O0.51 as our formula. Mg0.506O0.51 Mg1O1.01 MgO 0.5060.506

  5. Determining the Empirical Formula from a Decomposition Reaction Indecompositionreactions, a single reactant breaks up into two or more products. AB  A + B Example: A student heats 75.0 g of an unknown compound to produce oxygen and 45.6 g of KCl. What is theempirical formula of the unknown? Mass of KCl = 45.6 g Mass of O = 75.0 – 45.6 = 29.4 g 45.6 g KCl x 1 mol KCl = 0.612 mol KCl29.4 g O x 1 mol O = 1.84 mol O 74.551 g 16.00 g This gives us (KCl)0.612O1.84 as our formula. (KCl)0.612O1.84 (KCl)1O3.01 KClO3 0.6120.612

  6. Determining the Empirical Formula from a Combustion Reaction Incombustionreactions, an organic compound isburned in oxygento produce CO2 and water (and sometimes SO2). This method can be applied to compounds containing only C, H, O, and S. CxHyOz + O2(g)  CO2(g) + H2O(l) CxHySz + O2(g)  CO2(g) + H2O(l) + SO2(g) Note: Neither of these equations can be balanced until we know the values of x,y, and z.

  7. How to Write the Equation for a Combustion Reaction • One reactant is the chemical being combusted. You will either be given the formula for this chemical or you may be expected to know it (from, say, having used it in lab). • The second reactant is always oxygen, O2(g). • If the reactant contains C and H only (hydrocarbon) or C, H, and O, theproducts are CO2(g) and H2O(l). • Once you write the reactants and products, with their states, you may proceed to balance the equation.

  8. Determining the Empirical Formula from a Combustion Reaction Example: Combustion of cyclohexane, a compound containing only C and H, produces 5.86 mg of CO2 and 2.40 mg of water. What is the empirical formula for cyclohexane? CxHy + O2(g)  CO2(g) + H2O(l) (not balanced) Here we must relate the moles of the element in the product to the moles of the free element: 5.86 mg CO2 x 1g CO2 x 1 mol CO2 x 1 mol C = 1000 mg 44.01 g CO2 1 mol CO2 = 1.33 x 10-4 mol C

  9. Determining the Empirical Formula from a Combustion Reaction 2.40 mg H2O x 1 g H2O x 1 mol H2O x 2 mol H = 1000 mg 18.02 g H2O 1 mol H2O = 2.66 x 10-4 mol H This gives us C0.000133H0.000266 as our formula. C0.000133H0.000266 CH2 0.0001330.000133 This is the empirical formula for cyclohexane.

  10. Determining the Empirical Formula from a Combustion Reaction – Indirect Analysis When the combustion is of a compound containing O as well as C and H, the mass of the O is obtained indirectly. Example: The combustion of a 5.048 g sample of a compound of C, H, and O gave 7.406 g CO2 and 3.027 g water. What is the empirical formula of the compound? CxHyOz + O2(g)  CO2(g) + H2O(l) (not balanced!)

  11. Determining the Empirical Formula from a Combustion Reaction – Indirect Analysis Example: The combustion of a 5.048 g sample of a compound of C, H, and O gave 7.406 g CO2 and 3.027 g water. What is the empirical formula of the compound? CxHyOz + O2(g)  CO2(g) + H2O(l) 7.406 g CO2 x 1 mol CO2 x 1 mol C = 0.1683 mol C 44.01 g CO2 1 mol CO2 3.027 g H2O x 1 mol H2O x 2 mol H = 0.3360 mol H 18.02 g H2O 1 mol H2O This gives us C0.1683H0.3360Oz. How do we find z, the moles of O?

  12. Determining the Empirical Formula from a Combustion Reaction – Indirect Analysis We have the mass of the sample. If we can find the mass of C and of H, we can subtract those masses from that of the sample to get the mass of O. The mass of C: 0.1683 mol C x 12.011 g = 2.021 g C 1 mol C The mass of H: 0.3360 mol H x 1.008 g = 0.3386 g H 1 mol H The mass of O: 5.048 – 2.021 – 0.3386 = 2.688 g O

  13. Determining the Empirical Formula from a Combustion Reaction – Indirect Analysis Moles of O 2.688 g x 1 mol O = 0.1680 mol O 16.00 g O The moles of each gives us our formula: C0.1683H0.3360O0.1680 C0.1683H0.3360O0.1680 C1.002H2O1 CH2O 0.1680 0.1680 0.1680

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