1. Chapter 3 Molecules, Compounds, & Chemical Equations CHE 123: General Chemistry I Dr. Jerome Williams, Ph.D. Saint Leo University

2. Overview Empirical vs. Molecular Formulas Mass Percent Determining Chemical Formulas

3. Empirical vs. Molecular Formulas Empirical Formula represents the simplest whole number ratio of atoms in a formula. Molecular Formula represents the true formula for a substance and is usually a multiple of the empirical formula.

4. Mass Percent Empirical Formulas are determined from mass percent data. To find the mass percent of an element use the following equation. Mass % of X = (mass X / total mass of compound) x 100% The mass percentages for all elements in a compound must sum to 100.00%.

7. Mass Percent Find the mass % oxygen in sodium nitrate (NaNO 3 ). Molar mass of NaNO 3 is 85.07 g/mole.

9. Mass Percent Given: sodium nitrate NaNO 3 Molar Mass = 85.07 g/mole Unknown: ? Mass % oxygen Relevant Information: 1 mole Na, 1 mole N, and 3 moles O are found in NaNO 3 Mass Percent X = (mass X / total mass of compound) x 100%

10. Mass Percent Solution: mass O = (3 mol x 16.00 g/mol) = 48.00 g Total mass (NaNO 3 ) = (1 mol x 23.00 g/mol) + (1 mol x 14.07 g/mol) + (3 mol x 16.00 g/mol) = 85.07 g Mass % O = (48.00 g / 85.07 g) x 100% = 56.42% O Check: Answer is reasonable.

11. Mass Percent For additional practice, find the % Na and % N in sodium nitrate. Answers: 27.04% Na 16.54% N

12. Determining Chemical Formulas To find an empirical formula, one would use the following method (Works Every Time).

13. Determining Chemical Formulas

14. Example Problem – 71.65% ClMol. Wt. = 98.96 g/mol – 24.27% C – 4.07% H – Find the empirical & molecular formulas for this compound.

15. Determining Chemical Formulas The Traditional Approach – Step 1: Assume 100.00 g sample We pick this sample size so that our percentage data will translate directly into mass in grams. 71.65% Cl71.65 g Cl 24.27% C24.27 g C 4.07% H 4.07 g H

16. Determining Chemical Formulas The Traditional Approach – Step 2: Convert mass values to moles (Axiom #3) 71.65 g Cl x ( 1 mol Cl / 35.45 g Cl) = 2.021 mol Cl 24.27 g C x ( 1 mol C / 12.01 g C) = 2.021 mol C 4.07 g H x ( 1 mol H / 1.008 g H) = 4.04 mol H

17. Determining Chemical Formulas

18. Traditional Approach (Works Every Time) – Step 3: Divide number of moles obtained from step 2 by smallest number of moles present. This gives the mole ratios for each component (look at this value as being the same thing as the number of atoms present).

19. Determining Chemical Formulas The Traditional Approach – Step 3: Convert moles to mole ratios (# atoms) 2.021 mol Cl / 2.021 mol = 1.000 Cl atoms 2.021 mol C / 2.021 mol = 1.000 C atoms 4.04 mol H / 2.021 mol = 2.00 H atoms

20. Determining Chemical Formulas

21. Traditional Approach (Works Every Time) – Step 4: Examine values obtained from step 3. The mole ratios should be whole numbers. If you have whole numbers, then you have the subscripts for the empirical formula directly. If not, proceed to step 5. Note: Values like 2.98, 4.03, etc. translate into 3.00 and 4.00, respectively. Do not round numbers like 2.5, 3.67, 1.33, etc. to whole numbers.

22. Determining Chemical Formulas The Traditional Approach – Step 4: Values are whole numbers, so we can now write the empirical formula 1.000 Cl atoms 1.000 C atomsEmpirical Formula = CH 2 Cl 2.00 H atoms

23. Determining Chemical Formulas

24. Traditional Method (Works Every Time) – Step 5: If mole ratios are not whole numbers, then one must introduce a factor that will convert the mole ratios into whole numbers. – Once you have found the whole number ratios, then the subscripts for the empirical formula are then known. Simply write out the empirical formula.

25. Determining Chemical Formulas

26. To find a molecular formula, one would use the same procedure as the empirical formula with the following additional steps. – Step 6: Once you have empirical formula, calculate the empirical molar mass (EMM).

27. Determining Chemical Formulas The Traditional Approach – Step 6: Calculate Empirical Molar Mass (EMM) Empirical Formula = CH 2 Cl EMM = (1 x 12.01 g/mol) + (2 x 1.01 g/mol) + (1 x 35.45 g/mol) = 49.48 g/mol

28. Determining Chemical Formulas To find a molecular formula – Step 7: Using the molecular weight (this must be given in problem), find the factor (x) that must be multiplied through to get molecular formula. EMM (x) = Mol. Wt. (Empirical Formula) (x) => Molecular Formula

29. Determining Chemical Formulas The Traditional Approach – Step 7: Determine Factor (x) EMM (x) = Mol. Wt. 49.48 g/mol (x) = 98.96 g/mol(x) = 2.000 = 2 Molecular Formula = (CH 2 CL) (2) => C 2 H 4 Cl 2