1. 3-1 *See PowerPoint Image Slides for all figures and tables pre-inserted into PowerPoint without notes. CHEMISTRY The Molecular Nature of Matter and Change Third Edition Chapter 3 Lecture Outlines* Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. 3-2 Chapter 3 Stoichiometry

3. 3-3 Mole - Mass Relationships in Chemical Systems 3.5 Fundamentals of Solution Stoichiometry 3.1 The Mole 3.2 Determining the Formula of an Unknown Compound 3.3 Writing and Balancing Chemical Equations 3.4 Calculating the Amounts of Reactant and Product

4. 3-4 mole - the amount of a substance that contains the same number of entities as there are atoms in exactly 12g of carbon-12. This amount is 6.022x10 23. The number is called Avogadro’s number and is abbrieviated as N. One mole (1 mol) contains 6.022x10 23 entities (to four significant figures)

5. 3-5 Counting Objects of Fixed Relative Mass 12 red marbles @ 7g each = 84g 12 yellow marbles @4g each=48g 55.85g Fe = 6.022 x 10 23 atoms Fe 32.07g S = 6.022 x 10 23 atoms S Figure 3.1

6. 3-6 Water 18.02 g Figure 3.2 One Mole of Common Substances CaCO 3 100.09 g Oxygen 32.00 g Copper 63.55 g

7. 3-7 Molecular Mass vs. Molar Mass ( M ) The Molecular mass of a compound expressed in amu is numerically the same as the mass of one mole of the compound expressed in grams. For water: H 2 O Molecular mass = (2 x atomic mass of H ) + atomic mass of O = 2 ( 1.008 amu) + 16.00 amu = 18.02 amu Mass of one molecule of water = 18.02 amu Molar mass = ( 2 x atomic mass of H ) + atomic mass of O = 2 ( 1.008 g ) + 16.00 g = 18.02 g 18.02 g H 2 O = 6.022 x 10 23 molecules of water = 1 mole H 2 O

8. 3-8 Sample Problem 3.1Calculating the Mass and the Number of Atoms in a Given Number of Moles of an Element PROBLEM: PLAN: SOLUTION: amount(mol) of Ag mass(g) of Ag (a) Silver (Ag) is used in jewelry and tableware but no longer in U.S. coins. How many grams of Ag are in 0.0342mol of Ag? (b) Iron (Fe), the main component of steel, is the most important metal in industrial society. How many Fe atoms are in 95.8g of Fe? (a) To convert mol of Ag to g we have to use the the molar mass of Ag, M (b) To convert g of Fe to atoms we first have to find the #mols of Fe and then convert mols to atoms. multiply by M of Ag (107.9g/mol) 0.0342mol Ag x mol Ag 107.9g Ag = 3.69g Ag PLAN: mass(g) of Fe amount(mol) of Fe atoms of Fe SOLUTION: 95.8g Fe x 55.85g Fe mol Fe = 1.72mol Fe 1.72mol Fe x 6.022x10 23 atoms Fe mol Fe = 1.04x10 24 atoms Fe divide by M of Fe (55.85g/mol) multiply by 6.022x10 23 atoms/mol

9. 3-9 Calculating the Number of Moles and Atoms in a Given Mass of Element Problem: Tungsten (W) is the element used as the filament in light bulbs, and has the highest melting point of any element 3680 o C. How many moles of tungsten, and atoms of the element are contained in a 35.0 mg sample of the metal? Plan: ?? Solution: ?? Answer: 1.90 x 10 - 4 mol W 1.15 x 10 20 atoms of Tungsten

10. 3-10 Table 3.1 Summary of Mass Terminology TermDefinitionUnit Isotopic mass Mass of an isotope of an elementamu Atomic mass (atomic weight) Molecular mass (formula mass or molecular weight) Molar mass ( M ) (gram-molecular weight) amu g/mol Average of the masses of the naturally occurring isotopes of an element weighted according to their abundance Sum of the atomic masses of the atoms (or ions) in a molecule (or formula unit) Mass of 1 mole of chemical entities (atoms, ions, molecules, formula units)

11. 3-11 Interconverting Moles, Mass, and Number of Chemical Entities Mass (g) = no. of moles x no. of grams 1 mol No. of moles = mass (g) x no. of grams 1 mol No. of entities = no. of moles x 6.022x10 23 entities 1 mol No. of moles = no. of entities x 6.022x10 23 entities 1 mol

12. 3-12 Sample Problem 3.2Calculating the Moles and Number of Formula Units in a Given Mass of a Compound PROBLEM: PLAN: SOLUTION: Ammonium carbonate is white solid that decomposes with warming. Among its many uses, it is a component of baking powder, first extinguishers, and smelling salts. How many formula unit are in 41.6g of ammonium carbonate? mass(g) of (NH 4 ) 2 CO 3 number of (NH 4 ) 2 CO 3 formula units amount(mol) of (NH 4 ) 2 CO 3 After writing the formula for the compound, we find its M by adding the masses of the elements. Convert the given mass, 41.6g to mols using M and then the mols to formula units with Avogadro’s number. The formula is (NH 4 ) 2 CO 3. M = (2 x 14.01g/mol N)+(8 x 1.008g/mol H) +(12.01g/mol C)+(3 x 16.00g/mol O) = 96.09g/mol 41.6g (NH 4 ) 2 CO 3 x 2.61x10 23 formula units (NH 4 ) 2 CO 3 divide by M multiply by 6.022x10 23 formula units/mol mol (NH 4 ) 2 CO 3 96.09g (NH 4 ) 2 CO 3 6.022x10 23 formula units (NH 4 ) 2 CO 3 mol (NH 4 ) 2 CO 3 x=

13. 3-13 Mass % of element X = atoms of X in formula x atomic mass of X (amu) molecular (or formula) mass of compound(amu) x 100 Mass % of element X = moles of X in formula x molar mass of X (amu) molecular (or formula) mass of compound (amu) x 100

14. 3-14 Sample Problem 3.3Calculating the Mass Percents and Masses of Elements in a Sample of Compound PLAN: SOLUTION: amount(mol) of element X in 1mol compound mass(g) of X in 1mol of compound mass % X in compound PROBLEM:Glucose (C 6 H 12 O 6 ) is the most important nutrient in the living cell for generating chemical potential energy. (a) What is the mass percent of each element in glucose? (b) How many grams of carbon are in 16.55g of glucose? mass fraction of X multiply by M (g/mol) of X divide by mass(g) of 1mol of compound multiply by 100 We have to find the total mass of glucose and the masses of the constituent elements in order to relate them. There are 6 moles of C per mole glucose; 12 moles H; 6 moles O. 6 mol C12 g C mol C = 72 g C

15. 3-15 Empirical and Molecular Formulas Empirical Formula - Molecular Formula - The simplest formula for a compound that agrees with the elemental analysis and gives rise to the smallest set of whole numbers of atoms. The formula of the compound as it exists, it may be a multiple of the empirical formula.

16. 3-16 Information Contained in the Chemical Formula of Glucose C 6 H 12 O 6 ( M = 180.16 g/mol) Oxygen (O) Mass per mole of compound 6 atoms 96.00 g Table 3.2 Carbon (C)Hydrogen (H) Atoms per molecule of compound Moles of atoms per mole of compound Atoms per mole of compound Mass per molecule of compound 6 atoms12 atoms 6 moles of atoms 12 moles of atoms 6 moles of atoms 6(6.022 x 10 23 ) atoms 12(6.022 x 10 23 ) atoms 6(6.022 x 10 23 ) atoms 6(12.01 amu) =72.06 amu 12(1.008 amu) =12.10 amu 6(16.00 amu) =96.00 amu 72.06 g12.10 g

17. 3-17 mass(g) of each element Sample Problem 3.4Determining the Empirical Formula from Masses of Elements PROBLEM: PLAN: SOLUTION: amount(mol) of each element empirical formula Elemental analysis of a sample of an ionic compound gave the following results: 2.82g of Na, 4.35g of Cl, and 7.83g of O. What are the empirical formula and name of the compound? preliminary formula change to integer subscripts use # of moles as subscripts divide by M (g/mol) Once we find the relative number of moles of each element, we can divide by the lowest mol amount to find the relative mol ratios (empirical formula). 2.82g Na mol Na 22.99g Na = 0.123 mol Na 4.35g Cl mol Cl 35.45g Cl = 0.123 mol Cl 7.83g O mol O 16.00 O = 0.489 mol O Na 1 Cl 1 O 3.98 NaClO 4 Na 1 Cl 1 O 3.98 NaClO 4 NaClO 4 is sodium perchlorate.

18. 3-18 assume 100g lactic acid and find the mass of each element Sample Problem 3.5Determining a Molecular Formula from Elemental Analysis and Molar Mass PROBLEM: PLAN: amount(mol) of each element During physical activity. lactic acid (M=90.08g/mol) forms in muscle tissue and is responsible for muscle soreness. Elemental anaylsis shows that it contains 40.0 mass% C, 6.71 mass% H, and 53.3 mass% O. (a) Determine the empirical formula of lactic acid. (b) Determine the molecular formula. preliminary formula empirical formula divide each mass by mol mass(M) molecular formula use #mols as subscripts convert to integer subscripts divide mol mass by mass of empirical formula to get a multiplier

19. 3-19 Sample Problem 3.5Determining a Molecular Formula from Elemental Analysis and Molar Mass continued SOLUTION:Assuming there are 100.g of lactic acid, the constituents are 40.0g C6.71g H53.3g Omol C 12.01g C mol H 1.008g H mol O 16.00g O 3.33mol C6.66mol H3.33mol O C 3.33 H 6.66 O 3.33 3.33 CH 2 Oempirical formula mass of CH 2 O molar mass of lactate90.08g 30.03g 3 C 3 H 6 O 3 is the molecular formula

20. 3-20 difference (after-before) = mass of oxidized element Sample Problem 3.6Determining a Molecular Formula from Combustion Analysis PLAN: find the mass of each element in its combustion product molecular formula PROBLEM:Vitamin C (M=176.12g/mol) is a compound of C,H, and O found in many natural sources especially citrus fruits. When a 1.000-g sample of vitamin C is placed in a combustion chamber and burned, the following data are obtained: mass of CO 2 absorber after combustion=85.35g mass of CO 2 absorber before combustion=83.85g mass of H 2 O absorber after combustion=37.96g mass of H 2 O absorber before combustion=37.55g What is the molecular formula of vitamin C? find the mols preliminary formula empirical formula

21. 3-21 Combustion Train for the Determination of the Chemical Composition of Organic Compounds. Figure 3.4 C n H m + (n+ ) O 2 = n CO (g) + H 2 O (g) m 2 m 2

22. 3-22 Flow Chart of Mass Percentage Calculation moles of X in one mol of compound Mass % of XMass fraction of X mass (g) of X in one mol of compound M = (g/mol) of X Divide by mass(g) of one mol of compound Multiply by 100

23. 3-23 SOLUTION: CO 2 85.35g-83.85g = 1.50gH2OH2O37.96g-37.55g = 0.41g There are 12.01g C per mol CO 2. 1.50g CO 2 12.01g CO 2 44.01g CO 2 = 0.409g C 0.41g H 2 O 2.016g H 2 O 18.02g H 2 O = 0.046g H There are 2.016g H per mol H 2 O. O must be the difference: 1.000g - (0.409 + 0.049) = 0.545 0.409g C 12.01g C 0.046g H 1.008g H 0.545g O 16.00g O = 0.0341mol C= 0.0461mol H= 0.0341mol O C 1 H 1.3 O 1 C3H4O3C3H4O3 176.12g/mol 88.06g = 2.000 C6H8O6C6H8O6 Sample Problem 3.6Determining a Molecular Formula from Combustion Analysis continued