1. Percent Composition, Empirical and Molecular Formulas Courtesy www.lab-initio.com

2. Calculating Percent Composition of a Compound  Like all percent problems: part whole 1)Find the mass of each of the components (the elements), 2)Divide by the total mass of the compound; then x 100 x 100 % = percent

3. Calculating Percent Composition Calculate the percent of oxygen in magnesium carbonate, MgCO 3. U: % O K: MgCO 3 P: whole mass = 24.3 g + 12 g + 3 x (16.00 g) = 84.3 g MgCO 3 S: Part oxygen = 3 x (16.00 g) = 48.0 g O x 100 % = percent whole part

4. Calculating Percent composition Calculate the percent of oxygen in magnesium carbonate, MgCO 3. 56.9 % O S: x 100 % = percent whole part x 100 % = 84.3 g 48.0g

5. Examples  Calculate the percent composition of C 2 H 4 ?  U: % C; % H  K: C 2 H 4, assume 1 mole  P: : (part/whole) * 100 whole = 2(12.0g) + 4(1.0g) = 28.0 g C 2 H 4 85.7% C, 14.3 % H 24.0g C 28.0 g C 2 H 4 X 100 = 4.0 g H 28.0 g C 2 H 4 X 100 = Total = 100 % 85.7 % C 14.3 % H

7. What is the percent carbon in C 5 H 8 NO 4 (the glutamic acid used to make MSG monosodium glutamate), a compound used to flavor foods and tenderize meats? a) 8.22 %C b) 24.3 %C c) 41.1 %C Percent Composition

8. What is the percent carbon in C 5 H 8 NO 4 (the glutamic acid used to make MSG monosodium glutamate), a compound used to flavor foods and tenderize meats? a) 8.22 %C b) 24.3 %C c) 41.1 %C Percent Composition

9. Percent composition of Chocolate Chip Cookies Check your percents Cookie Brand Cookies mass Chocolate chip mass “Dough” mass Percent Chocolate Percent Dough Chips AhoyFood LionChips Ahoy - chunkyCostco Lab Group 1 25 %40 % Lab Group 2 18 %27 % Lab Group 3 31 %11 % Lab Group 4 Average

10. GroupChips AhoyFood LionChips Ahoy- Chunky Costco Average percentage

11. How could you determine the percent of sugar in a piece of bubble gum?

12. Formulas  molecular formula = (empirical formula) n  molecular formula = C 6 H 6 = (CH) 6  empirical formula = CH Empirical formula: the lowest whole number ratio of atoms in a compound. Molecular formula: the true number of atoms of each element in the formula of a compound.

13. Formulas (continued ) Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio). Examples: NaCl MgCl 2 Al 2 (SO 4 ) 3 K 2 CO 3

14. Formulas (continued ) Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio). Molecular: H2OH2O C 6 H 12 O 6 C 12 H 22 O 11 Empirical: H2OH2O CH 2 OC 12 H 22 O 11

15. C 6 H 12 O 6 C3H6O3C3H6O3

16. To obtain an Empirical Formula 1.Determine the mass in grams of each element present, if necessary. 2.Calculate the number of moles of each element. 3.Divide each by the smallest number of moles to obtain the simplest whole number ratio. 4.If whole numbers are not obtained * in step 3), multiply through by the smallest number that will give all whole numbers * Be careful! Do not round off numbers prematurely

17. A sample of a brown gas, a major air pollutant, is found to contain 2.34 g N and 5.34g O. Determine a formula for this substance. Formula: 2.34 g N 1 mole N 14.0 g N 5.34 g O 1 mole O 16.0 g O = 0.167 mole N = 0.334 mol O 0.167 mole = 2 NO 2 = 1

18. Empirical Formula Determination Adipic acid contains 49.3% C, 6.9% H, and 43.8% O by mass. What is the empirical formula of adipic acid? 1.Assume there is 100 grams so you can treat % as mass 2. Convert grams to moles

19. Empirical Formula Determination 3. Divide each value of moles by the smallest of the values. Carbon: Hydrogen: Oxygen:

20. Empirical Formula Determination 4. Multiply each number by an integer to obtain all whole numbers. Carbon: 1.50Hydrogen: 2.50 Oxygen: 1.00 x 2 3 5 2 Empirical formula: C3H5O2C3H5O2

21. Empirical Formula Determination A compound contains 75% C and 25% H. What is the empirical formula of this compound? (#1 on worksheet 5) 1.Assume there is 100 grams so you can treat % as mass, and convert grams to moles

22. Empirical Formula Determination 2. 2. Divide each value of moles by the smallest of the values. Carbon: Hydrogen: Since the mole ratios are whole number there is no need to multiply to find whole numbers. Empirical Formula: CH 4

23. Finding the Molecular Formula The empirical formula for adipic acid was just calculated as C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 1. Find the formula mass of C 3 H 5 O 2 3(12.0 g) + 5(1.0) + 2(16.0) = 73.0 g

24. Finding the Molecular Formula The empirical formula for adipic acid was just calculated as C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 3(12.0 g) + 5(1.0) + 2(16.0) = 73.0 g 2. Divide the molecular mass by the mass given by the empirical formula.

25. Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 3. Multiply the empirical formula by this number to get the molecular formula. (C 3 H 5 O 2 ) x 2 = C 6 H 10 O 4

26. Molecular Formula The empirical formula for vitamin C is C 3 H 4 O 3. However, experimental data indicates that the molecular mass of vitamin C is about 180 g/mol. What is the molecular formula of vitamin C?

27. Calculation of the Molecular Formula 2 C6H8O6C6H8O6C6H8O6C6H8O6

28. Calculation of the Molecular Formula Interestingly, the colorless liquid, used in rocket engines has the same empirical formula as the brown air pollutant calculated previously, NO 2. However, this compound has a molar mass of 92.0 g/mole. What is the molecular formula of this substance?

29. Calculation of the Molecular Formula 2 N2O4N2O4N2O4N2O4

30. Empirical Formula from % Composition A substance has the following composition by mass: 60.80 % Na ; 28.60 % B ; 10.60 % H What is the empirical formula of the substance? Consider a sample size of 100 grams This will contain 60.8g Na, 28.60 g B and 10.60 g H Determine the number of moles of each Determine the simplest whole number ratio

31. A substance has the following composition by mass: 60.80 % Na ; 28.60 % B ; 10.60 % H

32. 60.80 g Na1 mole Na 23.0 g Na 28.60 g B1 mole B 10.81 g B = 2.64 mole Na = 2.64 mol B 2.64 mole NaBH 4 10.6 g H1 mole H 1.0 g H = 10.6 mol H 2.64mole = 1 = 4