1. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-1 Stoichiometry of Formulas and Equations Chapter 3

2. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-2 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

3. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-3 mole(mol) - the amount of a substance that contains the same number of entities as there are atoms in exactly 12 g of carbon-12. This amount is 6.022x10 23. The number is called Avogadro’s number and is abbreviated as N. One mole (1 mol) contains 6.022x10 23 entities (to four significant figures)

4. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-4 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 Counting objects of fixed relative mass.

5. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-5 Water 18.02 g CaCO 3 100.09 g Oxygen 32.00 g Copper 63.55 g One mole of common substances. Figure 3.2

6. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-6 Information Contained in the Chemical Formula of Glucose C 6 H 12 O 6 ( M = 180.16 g/mol) Oxygen (O) Mass/mole of compound 6 atoms 96.00 g Table 3.1 Carbon (C)Hydrogen (H) Atoms/molecule of compound Moles of atoms/ mole of compound Atoms/mole of compound Mass/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

7. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-7 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 gM

8. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-8 MASS(g) of element MASS(g) of element AMOUNT(mol) of element AMOUNT(mol) of element ATOMS of element ATOMS of element Summary of the mass-mole- number relationships for elements. M (g/mol) Avogadro’s number (atoms/mol) Figure 3.3

9. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-9 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 #g Ag/mol Ag, the molar mass 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 107.9 g 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

10. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-10 MASS(g) of compound MASS(g) of compound AMOUNT(mol) of compound AMOUNT(mol) of compound MOLECULES (or formula units) of compound MOLECULES (or formula units) of compound AMOUNT(mol) of elements in compound AMOUNT(mol) of elements in compound Avogadro’s number (molecules/mol) M (g/mol) chemical formula Summary of the mass-mole- number relationships for compounds. Figure 3.4

11. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-11 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 41.6 g 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, 41.6 g 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 14.01 g/mol N)+(8 x 1.008 g/mol H) +(12.01 g/mol C)+(3 x 16.00 g/mol O) = 96.09 g/mol 41.6 g (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 96.09 g (NH 4 ) 2 CO 3 6.022x10 23 formula units (NH 4 ) 2 CO 3 mol (NH 4 ) 2 CO 3 x=

12. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-12 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 Mass percent from the chemical formula

13. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-13 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. Per mole glucose there are 6 moles of C 12 moles H 6 moles O (a)

14. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-14 Sample Problem 3.3Calculating the Mass Percents and Masses of Elements in a Sample of Compound continued 6 mol C x 12.01 g C mol C = 72.06 g C 12 mol H x 1.008 g H mol H = 12.096 g H 6 mol O x 16.00 g O mol O = 96.00 g O M = 180.16 g/mol (b) mass percent of C = 72.06 g C 180.16 g glucose = 0.3999x 100 = 39.99 mass %C mass percent of H = 12.096 g H 180.16 g glucose = 0.06714x 100 = 6.714 mass %H mass percent of O = 96.00 g O 180.16 g glucose = 0.5329 x 100 = 53.29 mass %O

15. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 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. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-16 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 showed 2.82 g of Na, 4.35 g of Cl, and 7.83 g 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). 2.82 g Na mol Na 22.99 g Na = 0.123 mol Na 4.35 g Cl mol Cl 35.45 g Cl = 0.123 mol Cl 7.83 g O mol O 16.00 g 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.

17. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-17 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 =90.08 g/mol) forms in muscle tissue and is responsible for muscle soreness. Elemental analysis shows that this compound 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

18. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-18 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: 40.0 g C6.71 g H53.3 g Omol C 12.01g C mol H 1.008 g H mol O 16.00 g O 3.33 mol C6.66 mol H3.33 mol O C 3.33 H 6.66 O 3.33 3.33 CH 2 Oempirical formula mass of CH 2 O molar mass of lactate90.08 g 30.03 g 3 C 3 H 6 O 3 is the molecular formula

19. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-19 Combustion apparatus for determining formulas of organic compounds. Figure 3.5 C n H m + (n+ ) O 2 = n CO (g) + H 2 O (g) m 2 m 2

20. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 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. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-21 SOLUTION: CO 2 85.35 g-83.85 g = 1.50 g H2OH2O 37.96 g-37.55 g = 0.41 g There are 12.01 g C per mol CO 2 1.50 g CO 2 12.01 g CO 2 44.01 g CO 2 = 0.409 g C 0.41 g H 2 O 2.016 g H 2 O 18.02 g H 2 O = 0.046 g H There are 2.016 g H per mol H 2 O. O must be the difference: 1.000 g - (0.409 + 0.046) = 0.545 0.409 g C 12.01 g C 0.046 g H 1.008 g H 0.545 g O 16.00 g O = 0.0341 mol C= 0.0456 mol H= 0.0341 mol O C 1 H 1.3 O 1 C3H4O3C3H4O3 176.12 g/mol 88.06 g = 2.000 C6H8O6C6H8O6 Sample Problem 3.6Determining a Molecular Formula from Combustion Analysis continued

22. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-22 Table 3.2 Constitutional Isomers of C 2 H 6 O Property EthanolDimethyl Ether M (g/mol) Boiling Point Density at 20 0 C Structural formulas 46.07 78.5 0 C 0.789 g/mL (liquid) 46.07 -25 0 C 0.00195 g/mL (gas) Space-filling models

23. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-23 The formation of HF gas on the macroscopic and molecular levels. Figure 3.6

24. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-24 A three-level view of the chemical reaction in a flashbulb. Figure 3.7

25. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-25 translate the statement balance the atoms specify states of matter adjust the coefficients check the atom balance

26. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-26 translate the statement Sample Problem 3.7 Balancing Chemical Equations PROBLEM: PLAN:SOLUTION: balance the atomsspecify states of matter Within the cylinders of a car’s engine, the hydrocarbon octane (C 8 H 18 ), one of many components of gasoline, mixes with oxygen from the air and burns to form carbon dioxide and water vapor. Write a balanced equation for this reaction. adjust the coefficients check the atom balance C 8 H 18 + O 2 CO 2 + H 2 O 825/29 2C 8 H 18 + 25O 2 16CO 2 + 18H 2 O 2C 8 H 18 ( l ) + 25O 2 ( g ) 16CO 2 ( g ) + 18H 2 O ( g )

27. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-27 MASS(g) of compound A MASS(g) of compound A AMOUNT(mol) of compound A AMOUNT(mol) of compound A MOLECULES (or formula units) of compound A MOLECULES (or formula units) of compound A Avogadro’s number (molecules/mol) M (g/mol) of compound A molar ratio from balanced equation Summary of the mass-mole-number relationships in a chemical reaction. Figure 3.8 MASS(g) of compound B MASS(g) of compound B AMOUNT(mol) of compound B AMOUNT(mol) of compound B MOLECULES (or formula units) of compound B MOLECULES (or formula units) of compound B Avogadro’s number (molecules/mol) M (g/mol) of compound B

28. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-28 Sample Problem 3.8 Calculating Amounts of Reactants and Products PROBLEM: In a lifetime, the average American uses 1750 lb(794 g) of copper in coins, plumbing, and wiring. Copper is obtained from sulfide ores, such as chalcocite, or copper(I) sulfide, by a multistep process. After an initial grinding, the first step is to “roast” the ore (heat it strongly with oxygen gas) to form powdered copper(I) oxide and gaseous sulfur dioxide. (a) How many moles of oxygen are required to roast 10.0 mol of copper(I) sulfide? (b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(I) sulfide is roasted? (c) How many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide? PLAN: write and balance equation find mols O 2 find mols SO 2 find g SO 2 find mols Cu 2 O find mols O 2 find kg O 2

29. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-29 SOLUTION: Sample Problem 3.8 Calculating Amounts of Reactants and Products continued 2Cu 2 S( s ) + 3O 2 ( g ) 2Cu 2 O( s ) + 2SO 2 ( g ) 3mol O 2 2mol Cu 2 S = 15.0mol O 2 = 641g SO 2 = 0.959kg O 2 kg O 2 10 3 g O 2 = 20.0mol Cu 2 O 20.0mol Cu 2 O 3mol O 2 2mol Cu 2 O 32.00g O 2 mol O 2 10.0mol Cu 2 S(a) 10.0mol Cu 2 S 2mol SO 2 2mol Cu 2 S 64.07g SO 2 mol SO 2 (b) 2.86kg Cu 2 O 10 3 g Cu 2 O kg Cu 2 O mol Cu 2 O 143.10g Cu 2 O (c) (a) How many moles of oxygen are required to roast 10.0 mol of copper(I) sulfide? (b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(I) sulfide is roasted? (c) How many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide?

30. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-30 Table 3.3 Information Contained in a Balanced Equation Viewed in Terms of Reactants C 3 H 8 (g) + 5O 2 (g) Products 3CO 2 (g) + 4H 2 O(g) molecules3 molecules CO 2 + 4 molecules H 2 O1 molecule C 3 H 8 + 5 molecules O 2 amount (mol) 1 mol C 3 H 8 + 5 mol O 2 3 mol CO 2 + 4 mol H 2 O44.09 amu C 3 H 8 + 160.00 amu O 2 mass (amu) 132.03 amu CO 2 + 72.06 amu H 2 O mass (g) 44.09 g C 3 H 8 + 160.00 g O 2 132.03 g CO 2 + 72.06 g H 2 O total mass (g) 204.09 g

31. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-31 Sample Problem 3.9Using Molecular Depictions to Solve a Limiting- Reactant Problem PROBLEM: Nuclear engineers use chlorine trifluoride in the processing of uranium fuel for power plants. This extremely reactive substance is formed as a gas in special metal containers by the reaction of elemental chlorine and fluorine. PLAN: (a) Suppose the box shown at left represents a container of the reactant mixture before the reaction occurs (with chlorine colored green). Name the limiting reactant, and draw the container contents after the reaction is complete. (b) When the reaction is run again with 0.750 mol of Cl 2 and 3.00 mol of F 2, what mass of chlorine trifluoride will be prepared? Write a balanced chemical equation. Compare the number of molecules you have to the number needed for the products. Determine the reactant that is in excess. The other reactant is the limiting reactant.

32. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-32 Sample Problem 3.9 Using Molecular Depictions to Solve a Limiting- Reactant Problem SOLUTION: continued Cl 2 (g) + 3F 2 (g) 2ClF 3 (g) F2F2 Cl 2 (a) You need a ratio of 2 Cl and 6 F for the reaction. You have 6 Cl and 12 F. 6 Cl would require 18 F. 12 F need only 4 Cl (2 Cl 2 molecules). There isn’t enough F, therefore it must be the limiting reactant. You will make 4 ClF 2 molecules (4 Cl, 12 F) and have 2 Cl 2 molecules left over. (b) We know the molar ratio of F 2 /Cl 2 should be 3/1. 3.00 mol F 2 0.750 mol Cl 2 = 4 1 Since we find that the ratio is 4/1, that means F 2 is in excess and Cl 2 is the limiting reactant. 0.750 mol Cl 2 2 mol ClF 3 1 mol Cl 92.5 g ClF 3 1 mol ClF 3 =139 g ClF 3

33. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-33 Sample Problem 3.10Calculating Amounts of Reactant and Product in a Limiting-Reactant Problem PROBLEM: A fuel mixture used in the early days of rocketry is composed of two liquids, hydrazine(N 2 H 4 ) and dinitrogen tetraoxide(N 2 O 4 ), which ignite on contact to form nitrogen gas and water vapor. How many grams of nitrogen gas form when 1.00x10 2 g of N 2 H 4 and 2.00x10 2 g of N 2 O 4 are mixed? PLAN: We always start with a balanced chemical equation and find the number of mols of reactants and products which have been given. In this case one of the reactants is in molar excess and the other will limit the extent of the reaction. mol of N 2 divide by M molar ratio mass of N 2 H 4 mol of N 2 H 4 mass of N 2 O 4 mol of N 2 O 4 limiting mol N 2 g N 2 multiply by M

34. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-34 Sample Problem 3.10Calculating Amounts of Reactant and Product in a Limiting-Reactant Problem continued SOLUTION: N 2 H 4 ( l ) + N 2 O 4 ( l ) N 2 ( g ) + H 2 O( l ) 1.00x10 2 g N 2 H 4 = 3.12mol N 2 H 4 mol N 2 H 4 32.05g N 2 H 4 3.12mol N 2 H 4 = 4.68mol N 2 3 mol N 2 2mol N 2 H 4 2.00x10 2 g N 2 O 4 = 2.17mol N 2 O 4 mol N 2 O 4 92.02g N 2 O 4 2.17mol N 2 O 4 = 6.51mol N 2 3 mol N 2 mol N 2 O 4 N 2 H 4 is the limiting reactant because it produces less product, N 2, than does N 2 O 4. 4.68mol N 2 mol N 2 28.02g N 2 = 131g N 2 243

35. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-35 A + B (reactants) C (main product) D (side products) The effect of side reactions on yield. Figure 3.9

36. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-36 Sample Problem 3.11 Calculating Percent Yield PROBLEM: Silicon carbide (SiC) is an important ceramic material that is made by allowing sand(silicon dioxide, SiO2) to react with powdered carbon at high temperature. Carbon monoxide is also formed. When 100.0 kg of sand is processed, 51.4 kg of SiC is recovered. What is the percent yield of SiC from this process? PLAN: write balanced equation find mol reactant & product find g product predicted percent yield actual yield/theoretical yield x 100 SOLUTION: SiO 2 ( s ) + 3C( s ) SiC( s ) + 2CO( g ) 100.0 kg SiO 2 mol SiO 2 60.09 g SiO 2 10 3 g SiO 2 kg SiO 2 = 1664 mol SiO 2 mol SiO 2 = mol SiC = 1664 1664 mol SiC 40.10 g SiC mol SiC kg 10 3 g = 66.73 kg x 100=77.0% 51.4 kg 66.73 kg

37. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-37 Sample Problem 3.12 Calculating the Molarity of a Solution PROBLEM: Glycine (H 2 NCH 2 COOH) is the simplest amino acid. What is the molarity of an aqueous solution that contains 0.715 mol of glycine in 495 mL? mol of glycine concentration(mol/mL) glycine molarity(mol/L) glycine SOLUTION: PLAN: Molarity is the number of moles of solute per liter of solution. 0.715 mol glycine 495 mL soln 1000mL 1 L divide by volume 10 3 mL = 1L = 1.44 M glycine

38. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-38 Summary of mass-mole-number-volume relationships in solution. Figure 3.10 MASS (g) of compound in solution AMOUNT (mol) of compound in solution MOLECULES (or formula units) of compound in solution VOLUME (L) of solution M (g/mol) Avogadro’s number (molecules/mol)

39. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-39 Sample Problem 3.13Calculating Mass of Solute in a Given Volume of Solution PROBLEM: A “buffered” solution maintains acidity as a reaction occurs. In living cells phosphate ions play a key buffering role, so biochemistry often study reactions in such solutions. How many grams of solute are in 1.75 L of 0.460 M sodium monohydrogen phosphate? volume of soln moles of solute grams of solute multiply by M PLAN: SOLUTION: Molarity is the number of moles of solute per liter of solution. Knowing the molarity and volume leaves us to find the # moles and then the # of grams of solute. The formula for the solute is Na 2 HPO 4. 1.75 L0.460 moles 1 L 0.805 mol Na 2 HPO 4 141.96 g Na 2 HPO 4 mol Na 2 HPO 4 = 0.805 mol Na 2 HPO 4 = 114 g Na 2 HPO 4

40. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-40 Converting a concentrated solution to a dilute solution. Figure 3.11

41. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-41 Sample Problem 3.14Preparing a Dilute Solution from a Concentrated Solution PROBLEM: “Isotonic saline” is a 0.15 M aqueous solution of NaCl that simulates the total concentration of ions found in many cellular fluids. Its uses range from a cleaning rinse for contact lenses to a washing medium for red blood cells. How would you prepare 0.80 L of isotomic saline from a 6.0 M stock solution? PLAN: It is important to realize the number of moles of solute does not change during the dilution but the volume does. The new volume will be the sum of the two volumes, that is, the total final volume. volume of dilute soln moles of NaCl in dilute soln = mol NaCl in concentrated soln L of concentrated soln multiply by M of dilute solution divide by M of concentrated soln M dil xV dil = #mol solute = M conc xV conc SOLUTION: 0.80 L soln = 0.020 L soln 0.15 mol NaCl L soln 0.12 mol NaCl L soln conc 6 mol = 0.12 mol NaCl

42. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-42 Sample Problem 3.15Calculating Amounts of Reactants and Products for a Reaction in Solution PROBLEM: Specialized cells in the stomach release HCl to aid digestion. If they release too much, the excess can be neutralized with antacids. A common antacid contains magnesium hydroxide, which reacts with the acid to form water and magnesium chloride solution. As a government chemist testing commercial antacids, you use 0.10M HCl to simulate the acid concentration in the stomach. How many liters of “stomach acid” react with a tablet containing 0.10g of magnesium hydroxide? PLAN: Write a balanced equation for the reaction; find the grams of Mg(OH) 2 ; determine the mol ratio of reactants and products; use mols to convert to molarity. mass Mg(OH) 2 divide by M mol Mg(OH) 2 mol ratio mol HCl L HCl divide by M

43. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-43 Sample Problem 3.15Calculating Amounts of Reactants and Products for a Reaction in Solution SOLUTION: continued Mg(OH) 2 ( s ) + 2HCl( aq ) MgCl 2 ( aq ) + 2H 2 O( l ) 0.10g Mg(OH) 2 mol Mg(OH) 2 58.33g Mg(OH) 2 = 1.7x10 -3 mol Mg(OH) 2 1.7x10 -3 mol Mg(OH) 2 2 mol HCl 1 mol Mg(OH) 2 = 3.4x10 -3 mol HCl 3.4x10 -3 mol HCl 1L 0.10mol HCl = 3.4x10 -2 L HCl

44. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-44 Sample Problem 3.16Solving Limiting-Reactant Problems for Reactions in Solution PROBLEM: Mercury and its compounds have many uses, from fillings for teeth (as an alloy with silver, copper, and tin) to the industrial production of chlorine. Because of their toxicity, however, soluble mercury compounds, such mercury(II) nitrate, must be removed from industrial wastewater. One removal method reacts the wastewater with sodium sulfide solution to produce solid mercury(II) sulfide and sodium nitrate solution. In a laboratory simulation, 0.050L of 0.010M mercury(II) nitrate reacts with 0.020L of 0.10M sodium sulfide. How many grams of mercury(II) sulfide form? PLAN: As usual, write a balanced chemical reaction. Since this is a problem concerning a limiting reactant, we proceed as we would for a limiting reactant problem. Find the amount of product which would be made from each reactant. Then choose the reactant that gives the lesser amount of product.

45. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-45 SOLUTION: Sample Problem 3.16Solving Limiting-Reactant Problems for Reactions in Solution continued L of Na 2 S mol Na 2 S mol HgS multiply by M mol ratio L of Hg(NO 3 ) 2 mol Hg(NO 3 ) 2 mol HgS multiply by M mol ratio Hg(NO 3 ) 2 ( aq ) + Na 2 S( aq ) HgS( s ) + 2NaNO 3 ( aq ) 0.050L Hg(NO 3 ) 2 x 0.010 mol/L x 1mol HgS 1mol Hg(NO 3 ) 2 0.020L Hg(NO 3 ) 2 x 0. 10 mol/L x 1mol HgS 1mol Na 2 S = 5.0x10 -4 mol HgS= 2.0x10 -3 mol HgS Hg(NO 3 ) 2 is the limiting reagent. 5.0x10 -4 mol HgS 232.7g HgS 1 mol HgS = 0.12g HgS