Announcement. Quiz # 1 and 2 on Friday, 28 November. Please bring a blue book!

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1 Announcement Quiz # 1 and 2 on Friday, 28 November. Please bring a blue book! 1

2 Borax is the common name of an industrial cleaner, sodium tetraborate, Na 2 B 4 O 7. Suppose you have 20.0 g of borax. Calculate the ff: (a) formula mass of borax; (b) # of moles of borax; (c) # of moles of boron; (d) how many grams of boron; (e) how many atoms of O are present. 2

3 Formula Mass of Na 2 B 4 O 7 : (2 x 22.99) + (4 x 10.81) + (7 x 16.00) = amu # of moles of borax in 20.0 g sample: (20.0 g) / (201.2 g mol -1 ) = mol borax # of moles of boron in 20.0 g sample: (20.0 g) / (201.2 g mol -1 ) x (4 mol B/1 mol borax) = mol B How many grams of boron in 20.0 g sample: mol B x (10.81 g mol -1 ) = 4.30 g B How many atoms of O are present: (20.0 g) / (201.2 g mol -1 ) x (7 mol O/1 mol borax) x (6.022x10 23 atoms O/1 mol O) = 4.19 x atoms O 3

4 Learning Objectives 1. Connect the dots between amu, grams, and the periodic table via the mole. 2. Differentiate mass terms: isotopic, atomic, molecular, and molar mass. 3. Compute the molar mass of a substance from a chemical formula. 4. Use factor label method (dimensional analysis) to convert between grams moles molecules. 5. % mass to empirical formula and vice versa. 6. Balancing equations and mastering stoichiometry. 7. Limiting reagent, yields, and solution stoichiometry. 4

5 Chemical Formula Types of chemical formula [X m Y n ] Molecular Empirical Structural Actual # of atoms of each element Relative # of atoms of each element Actual # of atoms and the bonds between them H 2 O 2 HO H-O-O-H 5

6 Composition by Mass Determine the % mass of each element in a compound n molar mass of element molar mass of compound 100% Where n is the number of moles of an element in 1 mol of the compound 6

7 Practice Problem Determine the % mass of each element in C 3 H 6 O 3. Determine the % mass of each element in C 5 H 10 O 5. 7 Molar Mass = = g/mol % C = % H = % O = % = 40.00% 100% = 6.714% 100% = 53.29% Molar Mass = = g/mol % C = % = 40.00% % H = 100% = 6.714% % O = % = 53.29%

8 Empirical Formula The simplest formula for a compound that gives rise to the smallest set of whole numbers of atoms. Compounds with the same % mass of its elements have the same empirical formula. 8

9 Elemental Analysis A lab technique that can determine % mass of each element in a compound. Using this technique, you can generate the empirical formula of any unknown compound. Mass Percent %C %H %O % of element atomic mass Empirical Formula C x H y O z ratio of moles 9

10 How to get Empirical or Molecular Formula If we know the % mass of elements in a compound, we can determine the empirical formula. With molar mass, we can get molecular formula. mass % of elements Assume 100 g of sample grams of each element Use atomic weight molecular formula Use molar mass empirical formula Calculate mole ratio moles of each element 10

11 Practice Problem Elemental analysis of a sample of an ionic compound gave the following results: 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. Determine the empirical formula and the name of this compound. 11

12 Practice Problem Elemental analysis of a sample of an ionic compound gave the following results: 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. Determine the empirical formula and the name of this compound. 12 mol Na 2.82 g Na = mol Na g Na mol Cl 4.35 g Cl = mol Cl g Cl mol O 7.83 g O = mol O g O Divide by the smallest # of moles: Na 1 Cl 1 O 3.98 NaClO 4 sodium perchlorate

13 Practice Problem Dibutyl succinate is an insect repellent used against 13 household ants and roaches. Elemental analysis indicates that the % mass of the composition is 62.58% C, 9.63% H, and 27.79% O. Its experimentally determined molecular mass is 230 amu. What are the empirical and molecular formula of dibutyl succinate? The empirical formula is C 6 H 11 O 2. The molecular formula is C 12 H 22 O 4.

14 STEP 1. Determine the mass of each element. Assuming a 100 g sample: g C, 9.63 g H, and g O. STEP 2. Convert masses to amount in moles. 1 mol C n C = g C = mol C g C 1 mol H n H = 9.63 g H = 9.55 mol H g H 1 mol O n O = g O = mol O g O STEP 3. Write a tentative empirical formula: C 5.21 H 9.55 O STEP 4. Convert to whole numbers by dividing by smallest number of moles: C 2.99 H 5.49 O 14

15 STEP 5. Convert to small whole number ratio. Multiply by 2 to get: C 5.98 H O 2 The empirical formula is C 6 H 11 O 2. Empirical formula mass: (6x12.011) + (11x1.008) + (2x15.99) = 115 amu STEP 6. Compare empirical formula mass and molecular mass to determine the molecular formula. The empirical formula mass is 115 amu. The molecular formula mass is 230 amu. molecular mass 230 amu n = = empirical mass 115 amu = 2 The molecular formula is C 12 H 22 O 4. 15

16 Practice Problem 1. [3.31] A sample of mol of a metal M reacts completely with excess fluorine to form 46.8 g of MF 2. a. How many moles of F are in the sample of MF 2 that forms? b. How many grams of M are in this sample of MF 2? c. What element is represented by the symbol M? 16

17 Practice Problem 2. [3.33] Cortisol (MW = g/mol) is a steroid hormone involved in protein synthesis. Medically, it has a major use in reducing inflammation from rheumatoid arthritis. Cortisol is 69.6% C, 8.34% H, and 22.1% O by mass. What is its molecular formula? 17

18 Practice Problem 3. [3.29] Determine the empirical formulas of the following compounds: a mol of chlorine atoms combined with 0.22 mol of oxygen atoms. b g of silicon combined with 12.4 g of chlorine. c mass % carbon and 72.7 mass % oxygen. 18

19 Practice Problem 4. [Follow-up problem 3.5] One of the most widespread environmental carcinogens is benzo[a]pyrene (MW= g/mol). It is found in coal dust, in cigarette smoke, and even in charcoal-grilled meat. Analysis of this hydrocarbon shows mass % C and 4.79 mass % H. What is the molecular formula of benzo[a]pyrene? 19

Chapter 3. Mass Relationships, Stoichiometry and Chemical Formulas. Announcements. Learning Objectives MOLE

Chapter 3. Mass Relationships, Stoichiometry and Chemical Formulas. Announcements. Learning Objectives MOLE Announcements HOUR EXAM 1 --Critical To Understand Chapter 3 July 18 6-7:30PM --Skip Combustion Analysis & Isomers (p.82-83 in Principles of Chemistry Text) See me if you donʼt understand! Chapter 3 Relationships,