Name: Unit 7A- The Mole Accelerated Chemistry I Date: The Avogadro Constant and the Mole The number of atoms of an element that have a total mass in grams equivalent to the mass of one atom in amu is called the Avogadro constant, and is equal to 6.02 10 23. One mole (mol) of things equals 6.02 10 23 things. The figure below shows how to convert from one unit (grams, moles, or particles, i.e. atoms, molecules, formula units) to another. 1. Write the missing conversion factors a. b. c. c. b. a. 2. How many moles are in 25.0 g of sodium? Estimate before you solve: is 25 g more or less than one mole? Answ: 1.09 mol Na 3. Calculate the moles of sodium a sample of 2.41 10 48 atoms of sodium. Answ: 4.00x10 24 mol Na 4. Calculate the moles of fluorine in a sample containing 8.6 10 18 atoms of fluorine. Answ: 1.4x10-5 mol 5. How many moles of magnesium are in 4.32 10-4 g magnesium? Answ: 1.78x10-5 mol Mg 1
Complete the following 2-step conversions. SHOW YOUR WORK! 6. 3.68 10 40 atoms C =? g C Answ: 7.34x10 17 g C 7. 284 g Mg =? atoms Mg Answ: 7.03x10 24 atoms Mg 8. 5.22 10 8 atoms B =? g B Answ: 9.37x10-15 g B Molar Mass of a compound = mass of one mole of that compound Find the molar mass of the following compounds: 9. zinc chloride Answ: 136.27 g/mol 10. 2-methylpropane Answ: 58.14 g/mol 11. 1-bromo, 3,4-dimethylpentane Answ: 179.12 g/mol 12. Sulfurous acid Answ: 82.09 g/mol 13. sodium carbonate Answ: 105.98 g/mol Solve the following problems using the factor-label method; SHOW YOUR WORK! 14. 25.0 g zinc chloride =? moles zinc chloride Answ: 0.183 mol 2
15. 2.5 10 24 molecules 2-methylpropane =? moles 2-methylpropane Answ: 4.2mol 16. 6.42 µg 1-bromo, 3,4-dimethylpentane =? moles 1-bromo, 3,4-dimethylpentane Answ: 3.58x10-8 mol 17. 5.32 mg sulfurous acid =? atoms of H? Answ: 7.81x10 19 atoms H 18. 4.38 10 26 formula units sodium carbonate =? g sodium ions Answ: 3.35x10 4 g Na + 19. Calculate the numbers of individual C, H, and O atoms in 1.50 g of glucose (C 6 H 12 O 6 ). Answ: 3.01x10 22 atoms of O and C 6.02x10 22 atoms H 20. Pheromones are a special type of compound secreted by the females of many insect species to attract the males for mating. One pheromone has the molecular formula C 19 H 38 O. Normally, the amount of this pheromone secreted by a female insect is about 1.0 10-12 g. How many molecules are there in this quantity? Answ: 2.1x10 9 molecules 3
21. How many water molecules are present in 2.56 cm 3 of water at standard temperature and pressure? Answ: 8.55x10 22 molecule Percentage Composition Example: Find the percentage composition of arsenic(iii) oxide. knowns: As 2 O 3 = 2 As + 3 O = 2 74.9 amu + 3 16.0 amu = 198 amu or 198 g/mol Percentage of As = Percentage of O = 2 74.9 amu 198 amu 3 16.0 amu 198 amu 100% = 75.8% 100% = 24.2% 22. What is the percentage composition of calcium hydroxide? Answ: 54.10% Ca, 2.73% H, 43.17% O 23. It is determined that in a 500.0-g sample of pitchblende, a uranium ore, there are 2.5 g of uranium, U. What percent of the ore is uranium? Answ: 0.50% U 24. What is the percent of copper and chlorine in a. Copper (I) chloride? Answ: 64.19% Cu, 35.81% Cl 4
b. Copper (II) chloride? Answ: 47.27% Cu, 52.73% Cl 25. A sample of an unknown compound with a mass of 2.876 g has 66.07 percent carbon, 6.71 percent hydrogen, 4.06 percent nitrogen and 23.16 percent oxygen. What is the mass of each element in this compound? Answ: 1.900 g C, 0.195 g H, 0.117 g N, 0.6661 g O 26. Find the percent composition of a compound that contains 2.7369 g of chlorine, 0.4116 g of oxygen and 0.7971 g of phosphorus. Answ: 69.366% Cl, 10.43% O, 20.20 % P 27. A sample of a compound with a mass of 0.432 g is analyzed. The sample is found to contain only fluorine and oxygen. If the compound contains 0.128 g oxygen, calculate the percent composition of the compound. Answ: 29.6% O, 70.4% F 5
Empirical and Molecular Formulas Example: Determine the empirical formula for the compound with a composition of 46.56% Fe and 53.44% S. Assume you have 100 g of compound; there are 46.56 g Fe and 53.44 g S 46.56 g Fe 1mol Fe 53.44 g s 1mol S Fe: = 0.834 mol Fe S: = 1.66 mol S 55.8 g Fe 32.1g S Ratio of moles: Fe 0.834 = S 1. 66 = 0.834 / 0.834 1.66 / 0.834 = 1.00 1.99 so FeS 2 (divide through by the smaller number) or Fe 0.834 S 1.66 = Fe (0.834/0.834) S 1.66/0.834 = FeS 2 28. Determine the empirical formula of a compound containing 2.644 g of gold and 0.476 g chlorine. Answ: AuCl 29. Determine the empirical formula of a compound containing 0.928 g of gallium and 0.412 g phosphorus. Answ: GaP 30. What is the empirical formula for a compound containing zinc and oxygen if a 5.09 g sample contains 4.09 g of zinc? Answ: ZnO 31. Determine the empirical formula of a compound containing 2.16 g of aluminum, 3.85 g of sulfur and 7.68 g oxygen. Answ: Al 2 S 3 O 12 = Al 2 (SO 4 ) 3 Unit 7 Student Handout The Mole 6
32. Determine the molecular formula of a compound that is 30.45% nitrogen and 69.55% oxygen. The molar mass of the compound is 92 g/mol. Answ: N 2 O 4 33. Determine the molecular formula of a compound containing 56.36 g of oxygen and 43.64 g of phosphorus. The molar mass of the compound is 283.9 g/mol. Answ: P 4 O 10 34. Ascorbic acid, aka vitamin C, has 40.9% carbon, 4.58% hydrogen, and 54.5% oxygen. It s molar mass is 176.1 g/mol. What is its molecular formula? Answ: C 6 H 8 O 6 Hydrates Example: A 10.0 g sample of hydrated copper(ii) sulfate is heated to drive off the water. The dry sample has a mass of 6.4 g. What is the mass of the water driven off? What is the formula of the hydrate? 10.0 g CuSO 4 hydrate 6.4 g CuSO 4 dry = 3.6 g water driven off Convert the masses of dry CuSO 4 and water to moles 6.4g CuSO4 1 mol CuSO4 160 g CuSO4 = 0.040 mol CuSO 4 3.6 g H2O 1 mol H2O = 0.20 mol H 2 O 18.0 g H2O The ratio between moles of CuSO 4 and H 2 O is 1 to 5, so CuSO 4 5H 2 O. Unit 7 Student Handout The Mole 7
Find the formulas for the hydrates containing the following masses of components. Name the hydrates. 35. 3.71 g sodium carbonate, 6.29 g water Answ: Na 2 CO 3 10H 2 O 36. Two hydrates of cobalt(ii) chloride exist. One is 21.68% water the other is 45.43% water. What are their empirical formulas? Answ: CoCl 2 2H 2 O and CoCl 2 6H 2 O Molarity Molarity is a measure of a solution s concentration. It is equal to the ratio between the moles of dissolved substance (solute) and the volume of a solution in L. mol of solute M = L of solution 37. Complete the following table by filling in the correct values. Concentration Moles of solute Volume of solution a. 10.0M 3.00 L b. 0.062M 0.651 mol c. 6.3 mol 5.9 L d. 6.0M 1.00 L e. 0.0555 mol 500 ml 38. What is the molarity of a solution in which 90.0 g of C 6 H 12 O 6, glucose, is dissolved in enough water to make 250.0 ml of solution? Answ: 2.00 M C 6 H 12 O 6 Unit 7 Student Handout The Mole 8
39. If 0.0100 kg of sodium hydroxide is dissolved in enough water to make 250.0 ml of solution, what is the molarity? Answ: 1.00 M NaOH 40. Describe the preparation of 250.0 ml of 0.2 M calcium nitrate. 41. What mass of potassium phosphate is needed to prepare 4.00 liters of 1.50 M solution? Answ: 1.27 kg 42. What is the molarity of a solution that contains 85.6 grams of phosphorous acid in 0.385 L of solution? Answ: 2.70 M Unit 7 Student Handout The Mole 9
43. What volume of 0.025M magnesium nitrate would contain 4.50 g of dissloved nitrate ions? Answ: 1.5 L THE MOLE VOCABULARY REVIEW Match the term with the correct description. a. molecular mass b. Avogadro constant c. molarity d. empirical formula e. formula mass f. molar mass g. percentage composition h. molecular formula 44. the sum of the atomic masses of all atoms in the formula unit of an ionic compound 45. the simplest ratio of the elements in a compound 46. the sum of all the atomic masses in a molecule 47. the mass of 6.02 10 23 molecules, atoms, ions, or formula units of a species 48. the ratio between the moles of dissolved substance and the volume of solution in liters 49. shows the actual number of atoms in a molecule 50. 6.02 10 23 51. a statement of the relative mass each element contributes to the mass of a compound as a whole Challenge Problems 52. A typical virus particle is 5.00 10-6 cm in diameter. If Avogadro s number of these virus particles were laid in a row, how many kilometers long would the line be? Answ: 3.01x10 13 km Unit 7 Student Handout The Mole 10
53. How many molecules are in a sample of water that requires 8.40 kcal of heat energy to raise the temperature by 34.0 C? Answ: 8.25x10 24 molecules 54. Nitroglycerine contains 60% as many carbon atoms as hydrogen atoms; three times as many oxygen atoms as nitrogen atoms; and the same number of carbon and nitrogen atoms. The number of moles in a gram of nitroglycerine is 0.00441. What is the molecular formula of nitroglycerin? Answ: C 3 H 5 N 3 O 9 55. A compound of hydrogen and carbon has a molar mass 114 g/mol. If one mole of the compound contains 18.17 g of hydrogen, what is the compound s molecular formula? Answ: C 8 H 18 Unit 7 Student Handout The Mole 11
56. The element gold has properties that have made it much sought after through the ages. A cubic meter of ocean water contains 6.00 10-6 g gold. If the total mass of the water in the oceans of the world is 4.00 10 20 kg, how many kilograms of gold are distributed throughout the oceans? (Assume the density of sea water is 1.00 g/ml). How many liters of seawater would have to be processed to recover a kilogram of gold (which had a value of about $11,000 at 1989 prices)? Do you think this recovery operation is feasible? Answ: 2.40x10 12 g Au distributed throughout the oceans, 1.67x10 11 L seawater to process 57. What volume of 2.00M aluminum nitrate solution contains 0.250 moles of nitrate ions? Answ: 41.7 ml of solution 58. A solution of ammonium sulfate is made that is 2.25 M for ammonium ions. What volume of this solution can be produced using only 50.0 g of ammonium sulfate? Answ: 336 ml of solution Unit 7 Student Handout The Mole 12
59. You have 250. ml of 0.136 M HCl. Using a volumetric pipet, you take 25.00 ml of that solution and dilute it to 100.00 ml in a volumetric flask. Now you take 10.00 ml of the second solution, using a volumetric pipet, and dilute it to 100.00 ml in a volumetric flask. What is the concentration of HCl in the final solution? (This process is called a serial dilution.) Answ: 0.0034 M HCl 60. Iron metal has density of 7.874 g/cm 3. You have a piece of metal in the shape of a rectangular prism that is 5.500 x 10 8 pm by 1.250 x 10 8 Mm by 1.345 x 10 9 nm. How many atoms of iron do you have? (7 points) Answ: 7.848x10 23 atoms Unit 7 Student Handout The Mole 13