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 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 molecular formula = (NO 2 ) n molar mass = 92 g empirical formula = NO 2 mass of NO 2 = 46 g Dividing gives us n: (NO 2 ) n = 92 g gives n = 2 NO 2 46 g molecular formula = (NO 2 ) 2 = N 2 O 4

3. Determining the Empirical Formula in the Laboratory There are three ways to determine the empirical formula of a compound in the laboratory: 1. by a combination reaction 2. by a decomposition reaction 3. by indirect analysis (often a combustion reaction)

4. Determining the Empirical Formula from a Combination Reaction In combination reactions, 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 the empirical formula of the oxide? Mass of Mg = 12.3 gMass of O = 20.4 – 12.3 = 8.1 g 12.3 g Mg x 1 mol Mg = 0.506 mol Mg 8.1 g O x 1 mol O = 0.506 mol O 24.305 g 16.00 g This gives us Mg 0.506 O 0.51 as our formula. Mg 0.506 O 0.506  Mg 1 O 1  MgO 0.506 0.506

5. Determining the Empirical Formula from a Decomposition Reaction In decomposition reactions, 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 the empirical formula of the unknown? Mass of KCl = 45.6 gMass of O = 75.0 – 45.6 = 29.4 g 45.6 g KCl x 1 mol KCl = 0.612 mol KCl 29.4 g O x 1 mol O = 1.84 mol O 74.551 g 16.00 g This gives us (KCl) 0.612 O 1.84 as our formula. (KCl) 0.612 O 1.84  (KCl) 1 O 3.01  KClO 3 0.612 0.612

6. Determining the Empirical Formula from a Combustion Reaction In combustion reactions, an organic compound is burned in oxygen to produce CO 2 and water (and sometimes SO 2 ). This method can be applied to compounds containing only C, H, O, and S. C x H y O z + O 2 (g)  CO 2 (g) + H 2 O(l) C x H y S z + O 2 (g)  CO 2 (g) + H 2 O(l) + SO 2 (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 1.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). 2.The second reactant is always oxygen, O 2 (g). 3.If the reactant contains C and H only (hydrocarbon) or C, H, and O, the products are CO 2 (g) and H 2 O(l). 4.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 CO 2 and 2.40 mg of water. What is the empirical formula for cyclohexane? C x H y + O 2 (g)  CO 2 (g) + H 2 O(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 CO 2 x 1g CO 2 x 1 mol CO 2 x 1 mol C = 1000 mg 44.01 g CO 2 1 mol CO 2 = 1.33 x 10 -4 mol C

9. Determining the Empirical Formula from a Combustion Reaction 2.40 mg H 2 O x 1 g H 2 O x 1 mol H 2 O x 2 mol H = 1000 mg 18.02 g H 2 O 1 mol H 2 O = 2.66 x 10 -4 mol H This gives us C 0.000133 H 0.000266 as our formula. C 0.000133 H 0.000266  CH 2 0.000133 0.000133 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 CO 2 and 3.027 g water. What is the empirical formula of the compound? C x H y O z + O 2 (g)  CO 2 (g) + H 2 O(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 CO 2 and 3.027 g water. What is the empirical formula of the compound? C x H y O z + O 2 (g)  CO 2 (g) + H 2 O(l) 7.406 g CO 2 x 1 mol CO 2 x 1 mol C = 0.1683 mol C 44.01 g CO 2 1 mol CO 2 3.027 g H 2 O x 1 mol H 2 O x 2 mol H = 0.3360 mol H 18.02 g H 2 O 1 mol H 2 O This gives us C 0.1683 H 0.3360 O z. 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: C 0.1683 H 0.3360 O 0.1680 C 0.1683 H 0.3360 O 0.1680  C 1.002 H 2 O 1  CH 2 O 0.1680 0.1680 0.1680