1. Empirical/Molecular Formulas

2. Objective/Warm-Up SWBAT calculate molar mass of compounds. What is the molar mass of each of these elements? Na Cl C H

3. GFM= Gram Formula Mass Find the gram formula mass of C 2 H 6 O C = 12.01 x 2 = 24.02 H = 1.01 x 6 = 6.06 O = 16.00 x 1 = 16.01 Total = 46.09 g/mol Keep 2 decimal places. Unit is g/mol

4. Molar Mass Example: Ca(OH) 2 Ca = 40.01 g/mol x 1 = 40.08 g/mol O = 16.00 g/mol x 2 = 32.00 g/mol H = 1.01 g/mol x 2 = 2.02 g/mol Total = 74.10 g/mol

5. Practice problems

6. Objective/Warm-Up SWBAT convert using molar mass. What is the molar mass of each of these compounds? NaCl CaCl 2 Mg(NO 3 ) 2

8. Objective/Warm-Up SWBAT calculate percent composition by mass. SWBAT distinguish between empirical and molecular formulas. How do you find the percent of something? For example, how would you find the percent of girls or boys in this class?

9. SWBAT calculate percent composition by mass. How do you find the percent of something? For example, how would you find the percent of girls or boys in this class?

11. Calculating Percent Composition Example: Na 2 O Find the mass of each element: Na 2 = 22.99 x 2 = 45.98 g/mol O = 16.00 g/mol Take the part divided by the whole: % Na = (45.98 g/mol) / (61.98 g/mol) = 74.2 % % O = (16.00 g/mol) / (61.98 g/mol) = 25.8 % The total should add up to 100 %

12. Practice Problems

13. Objectives/Warm-Up SWBAT distinguish between empirical and molecular formulas. SWBAT determine the empirical formula of a compound from percent composition. Find the percent of oxygen in Ca(OH) 2

14. Intro to Empirical Formula http://www.chemcollective.org/stoich/empiri calformula.php

15. CH 2 O C 2 H 4 O 2 CH 3 O Empirical Formula A formula that gives the simplest whole-number ratio of the atoms of each element in a compound. Molecular FormulaEmpirical Formula H2O2H2O2 HO C 6 H 12 O 6 CH 2 O

16. Wrap-Up Give three new examples of a molecular formula and give the corresponding empirical formula.  Why is it important to know the difference between molecular and empirical formulas?

17. Determine the empirical formula for a compound containing 2.128 g Cl and 1.203 g Ca. Steps 1.Find mole amounts. 2.Divide each mole by the smallest mole. Steps to Determine Empirical Formula

18. Determine the empirical formula for a compound containing 2.128 g Cl and 1.203 g Ca. 1.Find mole amounts. 2.128 g Cl x 1 mol Cl = 0.0600 mol Cl 35.45 g Cl 1.203 g Ca x 1 mol Ca = 0.0300 mol Ca 40.08 g Ca

19. Determine the empirical formula for a compound containing 2.128 g Cl and 1.203 g Ca. 2.Divide each mole by the smallest mole. Cl = 0.0600 mol Cl = 2.00 mol Cl 0.0300 Ca = 0.0300 mol Ca = 1.00 mol Ca 0.0300 Ratio – 1 Ca: 2 Cl Empirical Formula = CaCl 2

20. A compound weighing 298.12 g consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula? Hint “Percent to mass Mass to mole Divide by small Multiply ‘til whole”

21. A compound weighing 298.12 g consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula? Percent to mass: Mg – (72.2%/100)*298.12 g = 215.24 g N – (27.8%/100)*298.12 g = 82.88 g Mass to mole: Mg – 215.24 g * ( 1 mole ) = 8.86 mole 24.3 g N – 82.88 g * ( 1 mole ) = 5.92 mole 14.01 g Divide by small: Mg - 8.86 mole/5.92 mole = 1.50 N - 5.92 mole/5.92 mole = 1.00 Multiply ‘til whole: Mg – 1.50 x 2 = 3.00 N – 1.00 x 2 = 2.00 Mg 3 N 2

22. Practice If the problem does not give you how many grams, assume 100 grams of the sample. http://www.chemcollective.org/stoich/e f_analysis.php

23. Wrap-Up What is the difference between molecular and empirical formulas? Label as molecular or empirical: C 2 H 4 Na 2 O 2 Na 2 SO 4 Explain how to calculate the empirical formula.

24. Warm-Up/Objective SWBAT calculate molecular formulas from empirical formulas or percent composition. Label as molecular or empirical: C 2 H 4 Na 2 O 2 Na 2 SO 4 What are the steps to calculate the empirical formula?

25. You are a Forensic Scientist The victim in the following case is a 35-year old white male named Tony DeMoy. Initial investigators say they found several signs around the death site that suggest foul play. Four possible causes of his untimely death have been suggested by his wife who has been ruled out as a suspect because of a proven alibi. Your task is to identify who and what killed Tony DeMoy.

26. Molecular Formula The molecular formula gives the actual number of atoms of each element in a molecular compound. Steps 1.Find the empirical formula. 2.Calculate the Empirical Formula Mass. 3.Divide the molar mass by the “EFM”. 4.Multiply empirical formula by factor. Find the molecular formula for a compound whose molar mass is ~124.06 and empirical formula is CH 2 O 3. 2.“EFM” = 62.03 g 3.124.06/62.03 = 2 4.2(CH 2 O 3 ) = C 2 H 4 O 6

27. Find the molecular formula for a compound that contains 4.90 g N and 11.2 g O. The molar mass of the compound is 92.0 g/mol. Steps 1.Find the empirical formula. 2.Calculate the Empirical Formula Mass. 3.Divide the molar mass by the “EFM”. 4.Multiply empirical formula by factor.

28. Find the molecular formula for a compound that contains 4.90 g N and 11.2 g O. The molar mass of the compound is 92.0 g/mol. Empirical formula. A.Find mole amounts. 4.90 g N x 1 mol N = 0.350 mol N 14.01 g N 11.2 g O x 1 mol O = 0.700 mol O 16.00 g O

29. B.Divide each mole by the smallest mole. N = 0.350 = 1.00 mol N 0.350 O = 0.700 = 2.00 mol O 0.350 Empirical Formula = NO 2 Empirical Formula Mass = 46.01 g/mol Find the molecular formula for a compound that contains 4.90 g N and 11.2 g O. The molar mass of the compound is 92.0 g/mol.

30. Molecular formula Molar Mass = 92.0 g/mol = 2.00 Emp. Formula Mass 46.01 g/mol Molecular Formula = 2 x Emp. Formula = N 2 O 4 Find the molecular formula for a compound that contains 4.90 g N and 11.2 g O. The molar mass of the compound is 92.0 g/mol.

31. Solving the Crime With a partner, analyze each piece of evidence in the lab area. There are 4 suspected compounds. Find the molecular formula of each compound, then see the teacher for possible identity of those compounds.

32. Wrap-Up  Summarize how to find the molecular formula from the empirical formula.

33. A 528.39 g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~222.25 g/mol. What is its molecular formula?

34. g C – (48.38/100)*528.39 g = 255.64 g g H – (8.12/100)*528.39 g = 42.91 g g O – (43.5/100)*528.39 g = 229.85 g mole C - 255.64 g * ( 1 mole ) = 21.29 mol 12.01 g mole H – 42.91 g * ( 1 mole ) = 42.49 mol 1.01 g mole O – 229.85 g * ( 1 mole ) = 14.37 mol 16.00 g

35. A 528.39 g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~222.25 g/mol. What is its molecular formula? From last slide: 21.29 mol C, 42.49 mol H, 14.27 mol O C – 21.29/14.27 = 1.49 H – 42.49/14.27 = 2.98 (esentially 3) O – 14.27/14.27 = 1.00 C – 1.49 x 2 = 3 H – 3 x 2 = 6 O – 1 x 2 = 2 C3H6O2C3H6O2

36. A 528.39 g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~222.25 g/mol. What is its molecular formula? From last slide: Empirical formula = C 3 H 6 O 2 “EFM” = 74.09 Molar mass = 222.24 = ~3 EFM 74.09 3(C 3 H 6 O 2 ) = C 9 H 18 O 6