1. 6.10 Calculating Empirical Formulas The empirical formula gives the lowest whole- number ratio of the atoms of the elements in a compound. The empirical formula gives the lowest whole- number ratio of the atoms of the elements in a compound. It tells the kinds and relative count (atoms or moles of atoms) in molecules or formula units of a compound. It tells the kinds and relative count (atoms or moles of atoms) in molecules or formula units of a compound. Empirical formulas may be interpreted at the microscopic (atoms) or macroscopic level (moles of atoms). Empirical formulas may be interpreted at the microscopic (atoms) or macroscopic level (moles of atoms). Example: the empirical formula of ethyne, C 2 H 2, and benzene, C 6 H 6, is CH. Example: the empirical formula of ethyne, C 2 H 2, and benzene, C 6 H 6, is CH.

2. 6.10 Calculating Empirical Formulas CO 2 (empirical and molecular formula for carbon dioxide) CO 2 (empirical and molecular formula for carbon dioxide) Microscopic interpretation: 1 CO 2 molecule is composed of 1 carbon atom and 2 oxygen atoms. Microscopic interpretation: 1 CO 2 molecule is composed of 1 carbon atom and 2 oxygen atoms. Macroscopic interpretation: 1 mol of CO 2 is composed of 1 mol of carbon atoms and 2 mol oxygen atoms. Macroscopic interpretation: 1 mol of CO 2 is composed of 1 mol of carbon atoms and 2 mol oxygen atoms.

3. 6.10 Calculating Empirical Formulas An empirical formula may or may not be the same as the molecular formula An empirical formula may or may not be the same as the molecular formula Example: CO 2 Example: CO 2 If they are different the molecular formula is a simple multiple of the empirical formula If they are different the molecular formula is a simple multiple of the empirical formula Dinitrogen tetrahydride Dinitrogen tetrahydride Molecular Formula: N 2 H 4 Molecular Formula: N 2 H 4 Empirical Formula: NH 2 Empirical Formula: NH 2

4. 6.10 Calculating Empirical Formulas 1) Find the % composition of each element in the compound 2) Assume that 100g of sample of compound so % = mass 3) Convert mass to mole (x 1mol/gfm = moles of each) 4) Divide all the moles by the smallest – if all numbers come out be whole numbers use the numbers for the subscripts in empirical formula 5) If the numbers are not whole numbers but are close round to whole numbers (1.95 would be 2 – 1.10 would be 1) 6) If numbers end in.25 x 4,.33 x 3,.5 x 2,.67 x3,.75 x 4,.8 x 5 to get whole numbers – all numbers must be multiplied by same number

5. 6.10 Example 15 What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen? What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen? Step 1: Done Step 2/3: 25.9 g N x = 1.85 mol N 74.1 g O x = 4.63 mol O 1 mol N 14.0 gN 1 mol O 16.0 g O

6. 6.10 Example 15 Step 4: = 1 mol N = 1 mol N = 2.50 mol O = 2.50 mol O Step 5: Skip Step 6: 1 mol N x 2 = 2 mol N 2.5 mol O x 2 = 5 mol O 1.85 mol N 1.85 1.85 4.63 mol O 1.85 1.85 Final Answer: Empirical Formula – N 2 O 5

7. 6.11 Calculating Molecular Formulas The molecular formula of a compound is either the same as its empirical formula or a simple whole-number multiple of it. The molecular formula of a compound is either the same as its empirical formula or a simple whole-number multiple of it. You can determine the molecular formula of a compound if you know its empirical formula and its gram formula mass You can determine the molecular formula of a compound if you know its empirical formula and its gram formula mass

8. 6.11 Calculating Molecular Formulas Steps of calculating Molecular formula Steps of calculating Molecular formula 1. Calculate the empirical formula mass (efm) – this is just the gfm of the empirical formula. 2. Divide (efm) into the gram formula mass (gfm) – it takes this many of the empirical formula units to make up the molecular formula of the compound 3. Multiple each subscript in the empirical formula by the multiple that you get.

9. 6.11 Example 16 Calculate the molecular formulas of the following compounds. Calculate the molecular formulas of the following compounds. a. gfm = 60 gempirical formula – CH 4 N Step 1: Calculate efmStep 2: C: 12.0= 2 H: 4 x 1.0 N: 14.0Step 3: efm = 30.0 gC 2 H 8 N 2 60 g 30 g

10. 6.11 Example 16 Calculate the molecular formulas of the following compounds. Calculate the molecular formulas of the following compounds. b. gfm = 78 gempirical formula – NaO c. gfm = 181.5 gempirical formula – C 2 HCl