MOLECULAR WEIGHT CALCULATIONS

MOLECULAR WEIGHT CALCULATIONS Calculating Formula Weight; Significant Figures Page 2 Percentage Composition of a Compound 18 Exponential Notation; Avogadro s Number; Moles 28 Calculating Empirical Formulas 42 OBJECTIVES FOR THIS UNIT This unit will help you learn certain mathematical relationships which are fundamental to chemistry. It is assumed that you can give from memory the names and symbols of common elements. You should know the meaning and significance of atomic weight and atomic number. You should be able to determine the atomic weight of common elements from a periodic table. You should be able to write the names and formulas of many compounds with the aid of a reference table. You should be able to predict whether a compound is primarily ionic or covalent in nature. INSTRUCTIONS TO THE STUDENT Programmed instruction is a method to help you learn better and more easily. You proceed in small steps, check yourself at each step, make few errors, and work at your own speed. The form of programmed instruction may make it look like a test, but this is not a test. This is a method of teaching yourself. You will not be graded on the responses you make while learning. However, you will be held responsible for mastery of the content of this unit at a later time. In addition to a periodic table and this program, you need a sheet of paper, a pen, an uncluttered desk and a slide rule, if you use one. Later you will need your notebook. Your instructor will tell you whether to write your response in the book or on a separate sheet of paper. The sheet may also be used for scratch work. Place the answer sheet at the bar, so it will cover the rest of the page. 1

CALCULATING FORMULA WEIGHT: SIGNIFICANT FIGURES INSTRUCTIONS TO THE STUDENT In all chemical calculations there are two important considerations. One is the units; the other, significant figures. Units are the most important part of any calculation. They describe the physical and mathematical relationships which exist. Units tell what is related and how these things are related. The numbers simply indicate how much is involved. 1. The expression 16 kilometres per hour, the units are and. The numbers are and. Kilometres hour 16 1(understood from the expression per hour ) 2. If your answer is correct, slide the answer sheet to the next bar. If you were i error, read the item again. Draw a line through the incorrect response, and write the correct response several times. On your answer sheet write the units of each of the following expressions: 17 grams Grams 3. 27.5 ml Ml 4. 32.0 grams/mole Grams/mole 5. 0.87 cal/g 2

Cal/g 6. 16.0 u U (u stands for unified atomic mass unit and will be used throughout. It is the mass of one atom of a specific carbon isotope and is equal to 1.66053 X 10-27 kg.) 7. Show all units at every step of a calculation. They provide a built-in check on the method. Let us take as an example the simple arithmetic problem, If an automobile travels at an average rate of 64 kilometres per hour, how far does it travel in 45 minutes? Distance = rate x time D = R x T 64km 1 hr x 45 min D = 1hr x 60 min D = 48 km Kilometres are units of distance and the answer is correct. If by carelessness, you 60 simply combine some numbers, such as 64 x 45, and then tag kilometers onto the numerical answer, you will come up with a ridiculous answer. If you make the same calculation but with the units shown at each step, the 64 km 60min 1 1 hr x 1 hr 45 min Answer will come out in units of km/hr 2. Common sense tells you these units are not units of distance. A solution contains 0.10 gram of salt in each ml of solution. If 12.5 ml of solution are evaporated to dryness, what mass of salt remains in the evaporating dish? Work out this problem on your answer sheet, then check the steps below. Show units with each and every step. 3

0.10 G 1 ml x 12,5 ml = 1.25 g (Grams are units of mass and the problem asks, what mass? ) 8. A problem is not worked satisfactorily unless units are used at each step, even though you may arrive at a correct numerical answer. A correct method is at least as important as the correct answer. In the pervious problem 1.25 g is the correct answer in terms of units but not in terms of significant figures. By definition, significant figures are numbers which express the results of a measurement so that only the last digit is in doubt. If a number has 4 significant figures, there are 4 digits in the number. The first 3 are known exactly, the 4 th is doubtful, and by definition there are 4 significant figures. In 167 there are significant figures and the number is doubtful. 3 7 9. In 1264 there are significant figures and the number is doubtful. 4 4 10. 3927 4 significant figures 7 is doubtful 11. 69 2 significant figures 9 is doubtful 12. 96,438 4

5 significant figures 8 is doubtful 13. Significant figures are concerned with the number of digits (exact and doubtful) in a number, not with the number of decimal places. 123, 12.3, and 0.123 all contain the same number of significant figures. The number of significant figures is, and in each case the is doubtful. 3 3 14. In the next problem you are given a group of numbers. One of the numbers has a different number of significant figures from all the rest. Select this number and write it on your answer sheet. How many significant figures are there in each of the other numbers? 18 63 127 46 127 does not belong with the group because it has 3 significant figures. All the other numbers have 2 significant figures. 15. 16.2 2.734 621 0.962 does not belong with the others because all the rest have significant figures. 2.743 3 significant figures 16. 84.1 73 246 1.87 73 3 significant figures 17. 12.17 9461 6.2 476.3 6.2 4 significant figures 18. 64 0.12 0.6 6.6 5

0.6 2 significant figures 19. Zero is significant or not, depending upon where it occurs in the number. Rule I: If it occurs within the number (not the first or last digit) it is always significant. Example: Zero is significant in 607, 19.06, 104.3. Both zeros are significant in 10.601. Rule II: Zero as the first digit in a number is never significant. Its function as the first digit or digits is to locate the decimal point. In these examples, the underlined zero or zeros are not significant. They locate the decimal point. 0.62.07 0.063.0064 0.601 2 1 2 2 3 20. For each of the next five items, answer the following questions: (1) How may significant figures are there? (2) If the number contains a zero or zeros, are they significant? (apply the rules) (3) Why is the zero significant? Example: 106.7 has 4 significant figures. The 0 is significant (is/is not) because it occurs within the number. Answer the questions for 1206.3 5 significant figures is significant occurs within the number 21. 70.632 5 significant figures is significant occurs within the number 22. 0.62 2 significant figures is not significant locates decimal point 23. 602 6

3 significant figures is significant occurs within the number 24. 0.07603 4 significant figures (The first 2 zeros are not significant, they locate the decimal point. The 3 rd zero is significant because it occurs within the number.) 25. The only problem left is whether zero is significant as the final digit in a number. Rule III: If the zero is the final digit, and follows the decimal point, the zero is significant. Otherwise there is no reason for it to be there. 17.70 contains 4 significant figures 73.0 contains 3 significant figures 6.020 contains 4 significant figures 0.70 contains 2 significant figures 17.0900 contains 6 significant figures Rule IV: If zero is the final digit or digits and precedes the decimal point, you cannot decide whether its function is to locate the decimal point or whether it is the result of measurement. Exponential notation, which we will not develop here, is required. So that you will not be reduced to guessing, we will state arbitrarily for our purpose that zero as a final digit or digits preceding the decimal point is significant. Example: In 170 the zero is significant. In 60,200 all the zeros are significant. The four rules for zero as a significant figure, thus condense to the statement that zero is a significant figure except at the beginning of a number. Now go back to the beginning of the program and summarize in your notebook the important points about units and significant figures. When you return to this point, practice what you have learned on some additional examples. How many significant figures are there in 16,730? 5 the significant figures are underlined 16.730 26. In 0.071 2 0.071 27. In 17,063.12 7

7 17,063.12 28. 6270 3 6270 29. 709.00 4 709. 00 30. 0.0690 3 0.0690 31. 22.4 3 22.4 32. Why do you need to know how many significant figures a number has? You need to know how many significant figures there are in numbers involved in calculations so that you can tell what degree of accuracy is warranted in your answer. The result of a calculation can be no more precise than the least accurate measurement involved in the calculation. You may know that one measurement of length is 673.21 inches (5 significant figures). You have been able to measure another length, 10.5 inches, only to 3 significant figures. If you subtract 673.21 inches 5 significant figures 010.5 2 significant figures 662.71 inches You are justified in reporting your answer only to 3 significant figures or as 663 inches. (3 significant figures) Another example: You are able to measure the length of a block of wood as 4.62 cm and the width as 9.3 cm. What is the area? 8

4.62 cm 3 significant figures 9.3 2 significant figures 1386 _4159 42.966 cm 2. 5 figures are not justified. The least accurate measurement contains only 2 significant figures, so the answer is 43 cm 2. If the number following the last significant figure is 5 or more, increase the last significant figure by 1. One final refinement. Certain numbers by definition are exact numbers. For instance the density of water at 4 o C is 1 g/ml. The 1 is defined as 1.00000 g to as many places as necessary. Thus in calculation, such as 17.9 ml x g/ml = 17.9 g, the limiting number of significant figures is. 3 (in 17.9 ml) 33. When we talk of 2 atoms or 3 molecules, the 2 and 3 are exact numbers. They have as many significant figures as you need. You should apply these conventions about units and significant figures in all chemical calculations. Refer now to the problem in item 7. How should the answer be reported? Why? 1.3 g ( The least accurate measurement, 0.10 g, is to 2 significant figures, therefore the answer is only justified to 2 significant figures (not 2 decimal places). 1.25 g is rounded to 1.3 g. In all your succeeding work, check units and significant figures.) If this is your first experience with significant figures, you may find this enough new material for one session. You may find it useful to invent some additional problems for yourself and return to this program at a later time. 9

34. In order to compare the atomic weight or atomic mass of one element with the atomic mass of another element, you need a reference standard. The actual masses are minute and cumbersome. In 1961 the International Commission of Atomic Weight selected C 12 as the reference standard. The following material is based on this standard. If the Periodic Table you are using is different from this standard, your numerical answer may be slightly different, but the principles are the same. The atomic masses of all elements are now compared to the isotope of the element,, which weighs atomic mass units. C 12 35. Elements heavier than carbon have masses than 12. Greater higher (or some such expression) 36. Elements lighter than carbon have atomic masses less than. (number) 12 37. Now make use of your Periodic Table. Consult the key to locate the atomic weight or atomic mass, not the atomic number. You should express answers in the following work to 3 significant figures. From the Table, the atomic mass of barium, Ba, is 137.34 amu. Written as 3 significant figures, the atomic weight of Ba is. 137 amu 38. Fluorine s atomic mass is 18.9984 amu. This number has significant figures. 6 39. Fluorine s atomic mass, rounded to 3 significant figures is amu. 10

19.0 40. The element whose mass is approximately twice that of C is. (name and symbol) Magnesium Mg 41. Its atomic mass to 3 significant figures is. 24.3 amu 42. You notice that atomic mass and atomic weight are used interchangeable-ably. However, molecular weight and formula weight should not be used interchangeably. Molecular weight should be applied only to those substances that form by covalent bonding. Substances whose bonds are primarily ionic, are said to have formula weights. You would thus refer to the of NH 3. (molecular weight/formula weight) Molecular weight 43. You calculate the of K 2 CO 3. (molecular weight/formula weight) Formula weight 44. Calculating the molecular weight or formula weight of any substance is a simple process of arithmetic based on the axiom that the whole is equal to the sum of its parts. If you know how many atoms of each element are required you can simply total the weights. Problem: calculate the molecular weight of hydrogen chloride. Report your answer in amu and to three significant figures. The formula for hydrogen chloride is. 11

HC1 45. The formula, HC1, shows that hydrogen chloride contains atom of Hydrogen combined with 1 atom of. 1 chlorine 46. Round the atomic masses to 3 significant figures before calculations, and report your final answer to only 3 significant figures. From the Periodic Table determine that the relative mass of H is amu. The relative mass of C1 is amu. 1.01 35.453 or 35.5 47. To 3 significant figures the mass of 1 H is 1.01 amu; the mass of 1 C1 is 35.5. (The 4 th significant figures in 35.453 is 5; therefore, to 3 significant figures, the mass of 1 C1 is 35.5) The molecular weight of HC1 (1 H = 1.01 amu and 1 C1 = 35.5 amu), is. 36.5 amu (3 significant figures) 48. In referring to sodium chloride, we use the term formula weight because the bonding in sodium chloride is. (covalent/ionic) Ionic 49. The formula weight of sodium chloride is calculated in a similar fashion. The formula is. NaC1 50. The relative mass of Na atom is. 1 23.0 amu 51. The relative mass of one C1 atom is. 12

35.5 amu 52. The formula weight of NaC1 is amu. 58.5 amu 53. Calculate the formula weight of calcium oxide. The formula for calcium oxide is. CaO 54. CaO conatins 1 Ca atom weighing and 1 O atom weighing. The formula weight is. (Ca) 40.1 amu + (O) 16.0 amu = 56.1 amu 55. The formula weight of magnesium sulfide is. Mgs (Mg) 24.3 amu + (S) 32.1 amu = (MgS) 56.4 amu 56. Calculate to 3 significant figures, the molecular weight of water. The formula for water is. H 2 O 57. H 2 O contains atoms of hydrogen. 2 58. Each H atom has a relative mass of amu. 1.01 amu 59. Two H atoms therefore weight 2 x 1.01 amu or. H 2 O contains atom(s) of oxygen with a relative mass of. 13

2.02 amu 1 16.0 amu 60. The total weight of 2 H atoms and 1 O atom is + =. 2.02 amu + 16.0 amu = 18.0 amu 61. The molecular weight of water, 18.0 amu, is the of the weight of individual atoms. Sum (total) 62. In order to calculate the molecular weight or formula weight of any substance you need to know the for that substance. Formula 63. From the formula you can tell the number and kinds of elements in the substance. You also need a Periodic Table so you can determine the relative masses of the. Elements 64. From the formulas and the relative masses of the elements, you can calculate the or weight. Molecular formula (either order) 65. Calculate the of calcium chloride. (molecular wt/formula wt) Formula weight 66. Its formula is. CaC1 2 67. Its formula weight is. 14

Ca = 40.1 2C1 = 2 x 35.5 = 71.0 71.0 + 40.1 = 111 amu (significant figures) 68. Calculate the formula weight of sulfuric acid, H 2 SO 4. 2H = 2 x 1.01 = 2.02 S = 1 x 32.1 = 32.1 4O = 4 x 16.0 = 64.0 98.12 or 98.1 amu 69. To calculate the formula weight of copper (II) sulfate, you first need to know its formula,. The formula weight is amu. CuS 4 Cu = 63.5 amu S = 32.1 amu 4 O = 4 x 16.0 64.0 amu 159.6 amu 3 significant figures = 160 amu 70. The formula for cupric nitrate is. Cu(NO 3 ) 2 71. This indicates that cupric nitrate contains atom of copper, Atoms of nitrogen, and atoms of oxygen. 1 2 6 72. Remember that a subscript following the parenthesis affects everything within the parentheses. The formula weight of cupric nitrate is. 1 Cu = 63.5 amu 2 N = 2 x 14.0 = 28.0 amu 6 O = 6 x 16.0 = 96.0 amu 187.5 amu or 188 amu 73. Calculate the formula weight of ammonium phosphate. Its formula is. (NH 4 ) 3 PO 4 74. Its formula weight to 3 significant figures is. 15

3 N = 3 x 14.0 = 42.0 12 H = 12 x 1.01 = 12.12 1 P = 1 x 31.0 = 31.0 4 O = 4 x 16.0 = 64.0 149.12 or 149 amu 75. The molecular weight of sulfur trioxide is. SO 3 1 S = 32.1 3O = 3 x 16.0 = 48.0 80.1 amu 76. The formula for manganese dioxide is and its formula weight is amu. MnO 2 Mn = 54.9 2 O = 2 x 16.0 = 32.0 86.9 amu 77. There is only one further variation. Certain salts build water molecules into their crystals. Such water is called water hydration, or water of crystallization. Blue copper (II) sulfate is an example. Blue copper sulfate has the formula CuSO 4. 5 H 2 O. This is read as copper (II) Sulfur sulfate with 5 molecules of water of hydration. The dot(. ) stands for with and is not the. used in algebra to indicate multiplication. The formula weight is the sum of the weight of the copper (II) sulfate and the 5 molecules of water. You may refer to previous answers in order to calculate that the formula weight of CuSO 4. 5 H 2 O is. CuSO 4 = 160 amu Formula weight = 250 amu 5 H 2 O = 5 x 18.0 = 90 amu 78. The formula for barium chloride with 2 molecules of water of crystallization is. BaC1 2. 2 H 2 O 79. Its formula weight is. 16

Ba = 137.0 2 C1 = 2 x 35.5 = 71.0 2 H 2 O = 2 x 18.0 = 36.0 244.0 or 244 amu 80. The useful household cleaner called washing soda is chemically sodium carbonate with 10 molecules of water crystallization. Its formula weight to 3 significant figures is. Na 2 CO 3. 10 H 2 O 2 Na = 2 x 23.0 = 46.0 C = 12.0 3 O = 3 x 16.0 = 48.0 10 H 2 O = 10 x 18.0 = 180.0 286.0 or 286 amu 81. You now understand the principle and practice of calculating formula or molecular weights. Copy the following questions in your notebook and write your answers following the question in your own words. (1) When do we use the term molecular weight, and when do we use the term formula weight? (2) What is the general principle by which we calculate molecular of formula weight? (3) What information and what reference material do we need? Go back to item 34 and read the program again to check answers. Calculate to 3 significant figures the formula weight of nitric acid, HNO 3. 63.0 amu 82. the molecular weight of phosphorus tribromide is. 271 amu 83. The formula weight of aluminum sulfate is. 342 amu 17

PERCENTAGE COMPOSITION OF A COMPOUND INSTRUCTIONS TO THE STUDENT The purpose of this set is to help you learn to calculate the percentage of an element which is present in a compound. This information is of practical importance in helping you calculate what amount of an element can be obtained from a given quantity of a compound, or conversely, what amount of the element is required to produce a certain quantity of the compound. It is assumed that you can write the correct formula for most compounds and can calculate its molecular, or formula weight. In making these calculations, you will need a Periodic Table. Since answers are required to only 3 significant figures, you are encouraged to use a slide rule. The materials required and the method of procedure are the same as in the previous set. 1. First, let us review. The molecular weight or formula weight of a substance id the of the atomic of the that make up the substance. Sum or total weights elements 2. The molecular weight of carbon dioxide, CO 2, is. Remember to give your answer to 3 significant figures. (C) 12.01 amu + (2 O) 32.0 amu = 44.0 amu 3. The formula weight of sodium sulfate, Na 2 SO 4, is. If you missed any of the first three questions, you are advised to review the set on calculating molecular or formula weight. 2 Na = 2 x 23.0 = 46.0 amu 1 S = 1 x 32.1 = 32.1 amu 4 O = 4 x 16.0 = 64.0 amu 142.1 amu 142 amu to 3 significant figures 18

4. Calculating the percentage composition by weight of the element in a compound Is only a specific application of the general method of figuring percentage. A precentage is the ratio of the part the whole x 100%. Example: A chemistry class has 11 girls and 13 boys. The % of girls in the class to 3 significant figures is. The % of boys is. 11 girl students 24 students x 100% = 45.8% girls 13 boy students 24 students x 100% = 54.2% boys 5. On a recent test, six students made grades between 90 and 100. Ten made grades between 80 and 89. Four made grades between 70 and 79. Two made grades between 60 and 69. Two made failing grades, below 60. % failed the test. % scored 90 or better. 2 students failed 24 students x 100% = 8.3% 6 students earned 90 or better 24 students x 100% = 25% 6. In order to calculate the composition of anything, you must know the part and the whole. % represents the ratio of the part to the x 100%. Whole 7. By exactly the same method, we calculate the % composition by weight of an element in a compound. The whole compound of NaC1 has a formula weight of. 58.5 amu 8. The part of the NaC1 compound that is sodium weighs. 23.0 amu Na the part (Na) 9. The ratio of NaC1 = the whole (NaC1) =. 19

23.0 amu 58.5 amu 23.0 amu 10. The % of Na in NaC1 is, therefore, 58.5 amu x 100% or. 39.3% Part (C1) 11. The % of C1 in NaC1 is whole or 58.5 amu x 100% or %. 35.5 60.7% 12. 39.3% Na + 60.7% C1 = % NaC1. 100% 13. To calculate the % by weight of carbon and of oxygen in carbon dioxide, CO 2, first calculate the weight of the compound to be. Molecular (total) 44.0 amu 14. The % by weight of carbon is the ratio of the weight of carbon to the weight of carbon dioxide x 100%, or the ratio of x 100%. 12.0 amu 44.0 amu 15. 12.0 amu 44.0 amu x 100% =. 27.3% 16. There are atoms of oxygen in CO 2. Each oxygen atom weighs. Therefore the total weight of oxygen in CO 2 is. 20

2 16.0 amu 2 x 16.0 amu = 32.0 amu 17. The % of oxygen by weight in carbon dioxide is x 100% or. 32.0 amu, 44.0 amu 72.7% 18. To determine the % composition by weight of water, H 2 O, you first calculate that the molecular weight of H 2 O is. 18.0 amu 19. Two atoms of hydrogen each weighing, make the weight of hydrogen in water, H 2 O, equal to. 1.01 amu 2.02 amu 20. The ratio of hydrogen to water, the ratio of the part to the whole is. 2.02 amu 18.0 amu oxygen 21. The percentage of oxygen in water is the ratio of water x 100% of %. 100% 11.2% oxygen 22. The percentage of oxygen in water is the ratio of water x 100% or %. 16.0 amu 18.0 amu 88.8% 23. In order to calculate the % of calcium and oxygen in calcium oxide, CaO, you first calculate that the formula weight of CaO is. The compound contains atom(s) of Ca, whose atomic weight is. 21

56.1 amu one 40.1 amu the part (Ca) 24. The % of oxygen of Ca present is the ratio of the whole (CaO) x 100% or X 100% or %. 40.1 amu 56.1 amu 71.5% 25. The % of oxygen in CaO is the ratio of O to CaO x 100% or x 100% or %. 16.0 amu 56.1 amu 28.5% 26. Now calculate the % by weight of sodium, sulfur, and oxygen in sodium sulfate. First, determine that the formula of sodium sulfate is and then that its formula weight is amu. Na 2 SO 4 142 amu 27. Sodium sulfate,, contains atoms of Na. (formula) Na 2 SO 4 2 28. Each atom of sodium weighs, so the total weight of sodium in sodium sulfate is. 23.0 amu 46.0 amu 29. The % of Na in Na 2 SO 4 is x 100% or %. 46.0 amu 142 amu 32.4% 30. Na 2 SO 4 contains atom(s) of sulfur, and the total weight of sulfur in this compound is. 22

1 32.0 amu 31. The % by weight of S in Na 2 SO 4 then is x 100% or %. 32.0 amu 142 amu 22.5% 32. There are atom(s) of oxygen in Na 2 SO 4. Each oxygen atom has an atomic weight of 16.0 amu, so the total weight of oxygen is. 4 64.0 amu 33. The % of oxygen by weight in NaSO4 is therefore x 100% or. 64.0 amu 142 amu 45.1% 34. In the same manner, calculate the % by weight of each element in calcium carbonate, CaCO 3. The formula weight of CaCO 3 is. 100 amu 35. The weight of Ca present is. The % of _Ca_ is x 100% CaCO 3 or %. 40.1 amu 40.1 amu 100 amu 40.1% C 36. The one atom of C present in CaCO 3 weighs. CaCO 3 x 100% = x 100% =. 12.0 amu 12.0 amu 100 amu 12.0% 37. In CaCO 3, there are atom(s) of oxygen, each having an atomic weight of O. The ratio of CaCO 3 is and the % of oxygen is. 23

48.0 amu 3 16.0 amu 100 amu 48.0% 38. To calculate the % by weight of each element in KOH, first calculate the weight of KOH to be. Formula 39.1 amu + 16.0 + 1.01 amu = 56.1 39. In the general case, the percentage of an element is the ratio of the x 100%. part_ whole _K 40. The % of K in KOH is KOH x 100% or %. The % of O in KOH is %. The % of H in KOH is %. K = 69.7% O = 28.5% H = 1.8% 41. Calculate the % by weight of each element in ZnSO 4. There is % of Zn. There is % of S. There is % of O. (formula weight = 162 amu) Zn = 40.7% S = 19.8% O = 39.5% 42. The % each element in copper (II) chloride, CuC1 2 is: copper %, chloride %. (formula weight = 135 amu) Cu = 47.4% C1 = 52.6% 43. The % of each element in potassium chlorate, KC1O 3 is: potassium %, chloride %, oxygen %. 24

(formula weight = 123 amu) K = 31.9% C1 = 29.0% O = 39.1% 44. Calculate the % of each element in barium chloride with two molecules of water of hydration. Its formula and formula weight are and. BaC1 2. H 2 O 244 amu 45. The % of Ba is. The % C1 is. 137 amu _71 amu 244 amu x 100% = 56.1% 244 amu x 100% = 29.1% 46. The formula contains atoms of hydrogen and atoms of oxygen. 4 2 47. Remember that the coefficient 2 before 2 molecules of water, H 2 O affects both the H and the O. The % of H is therefore. The % of O is therefore. 4.04 amu 32.0 amu 244 amu x 100% = 1.66% 244 amu x 100% = 13.1% 48. In BaC1 2. 2 H 2 O the % of water is. 2 H 2 O 36.0 amu BaC1. 2 2 H 2 O 244 amu x 100% = 14.8% 49. Calculate the % of each element in copper (II) sulfate with 5 molecules of water of crystallization. The formula of the compound is, and its formula weight is. CuSO 4. 5 H 2 O 250 amu 50. The % of Cu =. The % of S =. 25

63.5 amu 32.1 amu 250 amu x 100% = 25.4% 250 amu x 100% = 12.8% 51. CuSO 4. 5 H 2 O contains oxygen atoms. 9 (4 from CuSO 4 ; 5 from 5 H 2 O) 52. The % of O is. The % of H is. 144 amu 10 x 1.01 amu 250 amu x 100% = 57.6% 250 amu x 100% = 4.04% 53. Since the formula weight of washing soda, Na 2 CO 3. 10 H 2 O is, The % of each element is: Na = % C = % O = % H = %. 46.0 amu 12.0 amu 286 amu; 286 amu x 100% = 16.1% 286 amu x 100% = 4.23%; 208 amu 20.2 amu 286 amu x 100% = 72.6%; 286 amu x 100% = 7.06% 54. The % of H 2 O in washing soda is. 180 amu 10 x 18.0 amu = 180 amu 286 amu x 100% = 62.9% If you are making errors in calculating % composition by weight, CHECK: (1) Have you the correct formula for the compound? (2) Have you the correct molecular or formula weight of the compound? (3) Have you the correct number of each atom? (4) Have you read the correct atomic weight for the element from the periodic Table? (5) Have you set up the correct ratio of the part of the whole? (6) Have you made an arithmetic error? You should now be able to calculate the % of any element is a compound. Go back to item 1 and read the program again. In your notebook, list principles and examples. 26

You would be wise to include an example of (1) a compound involving only two elements, (2) a compound containing a radical, (3) a compound containing a radical required more than once such as Zn(NO 3 ) 2, and (4) a compound containing water of crystallization. The check list on page 26 will also be useful. Now apply these principles to a practical problem. 55. Hematite is an iron ore with the composition Fe 2 O 3. What % of each ton of ore is Iron? How many tons of iron could you extract from 700 tons of Hematite, assuming it to be essentially pure Fe 2 O 3? 2 Fe 112 amu Fe 2 O 3 = 160 amu x 100% = 70% 70% Fe in 700 tons of Fe 2 O 3 = 490 tons of Fe. 27

AVOGADRO S NUMBER, GRAM ATOMS, MOLES, EXPONENTIAL NOTATION INSTRUCTIONS TO THE STUDENT This set will help you learn what exponential notation is and how to use it. You will learn how to describe the quantity of an element in several systems and how to convert from one system to another; how to convert from grams to gram atoms to numbers of atoms. It also teaches you how to describe the quantity of a compound in several systems and how to convert from one system to another. Specifically described are how to convert from grams to gram molecules or moles, or to numbers of molecules, and how to convert from grams to gram formulas to numbers of formula units. The chemistry student needs an understanding of these relationships and skills in his calculations because he will use these relationships later in the solution of problems involving weight to weight to volume ratios of substances. Previously you should have mastered the skills and information specified on page 1 of the program and in addition be able to convert quantities from one order of magnitude to another within the metric system. You should be able to perform mathematical operations showing units at each step and report the result to the indicated number of significant figures. 1. Solve the following problem expressing the answer in scientific notation to the correct number of significant figures. What is the volume of a block whose dimensions are 120 cm, 300 cm, and 4,800 cm? Vol = 1 x w x h = 1.2 x 10 2 cm x 3.0 x 10 2 cm x 4.8 x 10 3 cm = 17 x 10 7 cm or 1.7 x 10 8 cm If you solved this problem correctly and know why you are correct, so immediately to item 31. If you did not arrive at this answer by using scientific notation, work through the following items. 2. Scientific notation or exponential notation is a method of expressing a number as the product of a figure with only one digit to the left of the decimal point and multiplied by10 raised to some power. It is assumed that you know, from some previous work in mathematics, how to perform calculations with numbers written exponentially. 28

For example: 10 is written exponentially as 10 1 100 is written exponentially as 10 2 1000 is written exponentially as. 10 3 1_ 3. 10 is written as a power of 10 as 10-1. _1 1000 is written as a power of 10 as. 10-3 4. By definition 10 (and any number to the zero power) is equal to. 1 5. To review further, 10 2 x 10 3 = 10 5 because to multiply numbers expressed as exponents to the same base, you add exponents. 10 4 x 10 3 =. 10 7 6. 10 6 x 10-2 =. 104 [6 + (-2) = 4] 7. 10 9 / 10 7 = 10 2 because to divide numbers expressed as powers of the same base, you subtract exponents. 10 4 / 10 2 =. 10 2 8. 10 5 / 10-3 = 10 8 [5 (-3) = 8] 10 5 / 10 7 =. 29

10-2 9. Scientific notation expresses all numbers as the product of a number with only one Digit to the left of the decimal point and 10 raised to one correct power. For example: 180 = 1.8 x 10 2 (1.8 x 100) 16 = 1.6 x 10 1 (1.6 x 10) 1_ 0.25 = 2.5 x 10-1 (2.5 x 10) 72 = 7.2 x. 7.2 x 10 1 10. 1776 = x 10-1.776 x 10 3 11. 410 = x 10-4.10 x 10 2 12..0121 = x 10-1.21 x 10-2 13. 0.63 = x 10-6.3 x 10-1 14. 7 = x 10-7 x 10 0 (However since 10 0 = 1, the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, are written without 10 0 following them.) 15. There is a mechanical method which will help you arrive at the correct expression for a number is scientific notation. The steps are as follows: 30

(1) Locate the decimal point. (2) Move the decimal point the correct number of spaces to leave only one digit to the left of the decimal point. (3) The number of places the decimal point must be moved is the exponent for 10. The direction in which the decimal point is moved gives you the sign of the exponent. (4) (5) If the decimal point is moved to the left, the exponent is positive. (6) If the decimal point is moved to the right, the exponent is negative. Apply these steps to the following problem. Express 1492 in scientific notation. 1492 = 1.492 x 10 - Since the decimal point must be moved 3 places, the power of 10 is 10 3, and since the direction of the move is to the left, the exponent is positive; therefore 1492 = 1.492 x 10 3. 18.65 = x 10-1.865 x 10 1 (1 place to left) 16..072 = x 10-7.2 x 10-2 (2 places to right) 17. 0.52 = x 10-5.2 x 10-1 (one place to right) 18. 5280 = x 10-5.280 x 10 3 19..00528 = x 10-31

5.28 x 10-3 20. To convert numbers in scientific notation to the more usual decimal form, reverse the the procedure. 9.2 x 10 2 =. 920 (2 places to the right) 21. 7.6 x 10 3 =. 7600 (3 places to the right) 22. 9.2 x 10-2 =..092 (2 places to the left) 23. 8.1 x 10-1 =..81 (1 place to the left) 24. What is the volume of a block whose dimensions are 120 cm, 300 cm, and 4800 cm? (2 significant figures) First express the dimensions in scientific notation. 120 cm = x 10 - cm 300 cm = x 10 - cm 4800 cm = x 10 - cm 1.2 x 10 2 cm 3.0 x 10 2 cm 4.8 x 10 3 cm 25. Volume = 1 x w x h, substituting the dimensions gives Volume = 1.2 x 10 2 cm x 3.0 x 10 2 cm 4.8 x 10 3 cm Multiply the numbers together 1.2 x 3.0 x 4.8 =. 32

17.28 (Since the answer is justified only to 2 significant figures, this is written as 17.) 26. Multiply the power of 10 together by adding exponents, 10 2 x 10 2 x 10 3 = 10 -, then calculate the volume of the block. 107 Vol = 1.2 x 10 2 cm x 3.0 x 10 2 cm x 4.8 x 10 3 cm = 17 x 10 7 cm 3. 27. In scientific notation, only 1 number is shown left of the decimal, so 17 x 10 7 cm 3 is written as x 10 - cm 3. 1.7 x 108 cm 3 (Converting 17 to 1.7 is the equivalent of dividing by 10, therefore 10 7 must be multiplied by 10 to equalize the change. 10 7 x 10 1 = 10 8.) 28. Express in scientific notation the product of 186 x 12.2. 1.86 x 10 2 x 1.22 x 10 1 = 1.86 x 1.22 x 10 2 x 10 1 = 2.27 x 10 3 29. Express in scientific notation the product of 1824 x 17. 1.824 x 10 3 x 1.7 x 10 1 = 3.1 x 10 4 30. Express in scientific notation the result of 826 / by 41 =. 8.26 x 10 2 2.01 x 10 1 or 8.26 x 10 2 / 4.1 x 10 1 = 4.1 x 10 1 = 2.0 x 10 1 to 2 significant figures Now go back to the first item of this set and read through the program again. List in your notebook the rules for writing a number in scientific notation and write an example to illustrate each principle. You may want to write additional examples of your own. 31. Let us examine the relationship between grams of an element (gram-atoms) and numbers of atoms of an element. One of the most useful generalizations in chemistry was discovered in 1811 at the University of Turin by an Italian professor named Amadeo Avogardro. He first developed his generalization about the numbers of molecules in a certain volume of gases, but the idea has since been applied widely. 33

The application we use in this set is that the weight in grams of an element which is equal to its atomic weight (mass) contains 6.02 x 10 23 atoms of that element. 6.02 x 10 23 is known as Avogadro s number. If the element is sodium, then 23.0 grams (rounded to 3 significant figures) of sodium contain 6.02 x 10 23 atoms of sodium. If the element is sulfur, grams of sulfur contain an Avogadro s number of atoms of sulfur. 32.1 32. If the element is carbon, 12.0 grams of carbon contain x 10 - atoms of carbon. 6.02 x 10 23 33. The proof of this generalization and a discussion of the experiments by which this number was determined are beyond the scope of this course, but a good encyclopedia or college text should give you more information if you want it. An Avogadro s number of nitrogen atoms weigh grams. 14.0 34. 40.1 grams of calcium contain Avogadro s number of calcium atoms or x 10 - calcium atoms. 1 or an 6.02 x 10 23 35. 80.2 grams of calcium contain Avogadro s number of calcium atoms or x 10 - calcium atoms. 2 2 x 6.02 x 10 23 or 12.04 x 10 23 or 1.20 x 10 24 36. 20.1 grams of calcium contains Avogadro s number of calcium atoms or x atoms of calcium. 34

6.02 x 10 23 1/2 or 0.5 2 = 3.01 x 10 23 37. 4.01 grams of calcium contain Avogadro s number of calcium atoms or x atoms of calcium. 0.1 or 1/10 6.02 x 10 22 38. You can see that using the expression an Avogadro s number of atoms or 6.02 x 10 23 atoms of an element is very cumbersome. Since the absolute number of atoms of an element is seldom important, but the relative number of atoms is useful, the term and relationship of gram-atoms has been defined. A gram atom is the weight of an element which contains an Avogadro s Number of atoms of an element. And conversely, an Avogadro s number of atoms of an element will have a weight in grams equal to the atomic weight of that element. One gram atom of calcium contains one Avogadro s number of calcium atoms and weights grams. 40.1 39. One gram atom of sulfur contains one Avogadro s number of sulfur atoms and weighs grams. 32.0 40. You recall that a mole of any substance contains an Avogadro s number of particle of that substance. A gram atom is therefore a mole of atoms. One gram atom, one mole, of sodium weighs grams. 23.0 41. One mole, a gram atom, of carbon weighs grams. 12 42. A gram atom of any element will contain x atoms of that element. 35

6.02 x 10 23 43. A gram atom of one element will have a different weight from a gram atom of another element because atoms of one element have a average weight from atoms of all other elements. Different 44. The relationship between weight of an element in grams and the number of gram atoms is expressed by the following equation: weight in grams of the element atomic weight of the element = number of gram atoms of that element For example 18 grams of C 12 grams of C/gram atom = 1.5 gram atoms of C How many gram atoms of carbon are 73 grams of carbon? 73 grams of C 12 grams of C/gram atom = 6.1 gram atoms 45. How many gram atoms of sulfur are provided by 24 grams of sulfur? 24 grams of S 32 grams of S/gram atom = 0.75 gram atom of S 46. How many gram atoms, moles, of helium are there in 24 grams of helium? 24 grams 4 grams of He/gram atom = 6 gram atoms of helium 24 grams He_ 4 grams He/mole = 6 moles of He 47. The relationship can also be used to convert gram atoms to grams. 1.5 gram atoms of oxygen weighs how many grams? 1.5 gram atoms oxygen x 16 grams of O/gram atom of O =. 36

24 grams 48. What is the weight of 0.600 gram atom of magnesium? 0.600 gram atom Mg x 24.3 grams Mg/gram atom Mg = 14.6 grams Mg 49. How many atoms of zinc are there in 50 grams of zinc? Since 1 gram atom zinc.76 gram atom 6.02 x 10 23 atoms zinc = x x = 4.6 x 10 23 atoms of zinc Or since 65 grams of zinc contains = 50 grams zinc 6.02 x 10 23 atoms zinc x x = 4.6 x 10 23 atoms zinc 50. Now return to item 38 and copy in your notebook the definition of a gram atom. From item 44 copy the equation for converting grams of an element to gram atoms. Work the next 3 examples. After you have checked your solutions with the response, copy these 3 examples into your notebook for reference. Example 1: you have 80 grams of iron, (1) How many gram atoms of iron do you have? (2) How many atoms of iron? (a) 80 grams of Fe 55.8 grams of Fe/gram atom = 1.4 gram atoms (b) 1.4 gram atoms x 6.02 x 10 23 atoms/gram atom = 8.6 x 10 23 atoms 52. Example 2: 0.620 gram atom of aluminum represents (a) What is the weight of 9.03 x 10 22 atoms of silver? (b) How many gram atoms of silver is this? 37

(a) 6.02 x 10 23 atoms silver weigh 9.03 x 10 22 atoms silver 107.9 grams = x x = 16.2 grams (b) 1 gram atom contains x 6.02 x 10 23 atoms = 9.03 x 10 22 atoms x = 0.15 gram atom 53. Avogadro s principle can also be applied to substances which occur in molecules or formula units. A gram molecule is that weight of a covalently bonded substance which contains an Avogadro s Number, 6.02 x 10 23, of molecules of that substance. One gram molecule Has a weight in grams equal to the molecular weight. A gram molecule is a mole of molecules. Example: 18 grams of water contain 6.02 x 10 23 molecules of water. One gram Molecule of water weighs 18 grams. Equation: _weight in grams of a substance Molecular weight of the substance = number of gram molecules A gram formula is that weight of an ionically bonded substance which contains an Avogadro s Number, 6.02 x 10 23, of formula units of the substance. One gram Formula has a weight in grams equal to the formula weight of the substance. A gram formula is a mole of formula units. Example: 6.02 x 10 23 formula units of sodium chloride weigh 58.8 grams and constitute one gram formula of sodium chloride. Equation: weight in grams of a substance formula weight of a substance = number of gram formulas Gram atoms, gram molecules, and gram formulas are all calculated by the same process. The terms refer to elements, covalently bonded substances, and ionically bonded substances respectively. The shortened term, the mole is used to stand for any of them. A mole is that quantity of a substance which contains an Avogadro s number of units of that substance. Since you now understand the principle and the process, work out a few examples for practice. 38

How many gram molecules of CO 2 are 80 grams of CO 2? Work out your answer, then check it. 80 g CO 2 44 g CO 2 /molecular weight = 1.8 gram molecules of 1.8 moles 54. How many molecules are there in 1.8 gram molecules of CO 2? 10 23 molecules 1.8 gram molecules x 6.02 gram molecules = 10.8 x 10 23 or 1.1 x 10 24 molecules 55. What is the weight of 2.5 gram formulas of NaC1? 58.5 grams 2.5 gram formulas of NaC1 x gram formula NaC1 = 146 grams 56. How many formula units are there in 146 grams of NaC1? 6.02 x 10 23 formula units 2.5 gram formulas x gram formula = 15 x 10 23 or 1.5 x 10 24 formula units 57. How many gram atoms of nitrogen are there in 50 grams of nitrogen? 50 grams N 14 grams N/gram atom = 3.6 gram atoms of nitrogen 58. How many gram molecules of nitrogen are there in 50 grams of nitrogen? 50 grams of nitrogen 28 grams/gram molecular weight = 1.8 gram formula or 1.8 moles Remember, nitrogen is diatomic, N 2. 59. How many atoms of nitrogen do 50 grams of nitrogen contain? 39

50 grams = 3.6 gram atoms (from item 57) 6.02 x 10 23 atoms 3.6 gram atoms x gram atom = 22 x 10 23 or 2.2 x 10 24 atoms 60. How many molecules of nitrogen do 50 grams of nitrogen contain? This can be solved in several ways but the simplest at this point is to reason that since each nitrogen molecule contains 2 atoms of nitrogen, 2.2 x 10 24 atoms of N form 1.1 x 10 24 molecules of nitrogen. 61. Return to item 53. Copy in your notebook the definition, example, and equation for gram formulas. Then return to this point and work out the next few problems. When you have checked them against the responses, add the solutions to your notebook as applications of this process. How many moles--that is, how many gram molecules are there in 100 grams of hydrogen chloride? How many molecules? 100 grams HC1 36.5 grams HC1/gram molecules = 2.7 gram molecules or moles 6.02 x 10 23 molecules 2.7 moles x mole = 16 x 10 23 or 1.6 x 10 24 molecules 62. What is the weight of 5.7 moles of zinc sulfide? A mole of ZnS is a gram formula of ZnS and weighs 97.5 grams. 5.7 moles x 97.5 grams/mole = 556 grams. 63. 8.2 x 10 23 molecules of water (a) weigh how many grams? (b) constitute how many moles of water? 40

(a) 6.02 x 10 23 molecules of H 2 O 8.2 x 10 23 molecules H 2 O 18 grams = x (b) 24.5 grams H 2 O 18 grams/mole = 1.4 moles x = 24.5 grams 64. How many grams do 4.9 gram molecules of oxygen weigh? 32 gram O 2 4.9 gram molecules x gram molecule = 157 grams You will use these relationships in determining the simplest formula of a substance and in other chemical calculations. CONTINUE ON 41

CALCULATING EMPIRICAL FORMULAS Instructions to the Student The purpose of this set is to help you learn how to arrive at the simplest formula for a compound, either from the relative weights of the elements, or from its percentage composition. It is assumed that you can write formulas for most substances, and that you understand that a formula describes the elements in a substance and the ratios of its elements to one another. You should also be able to calculate the number of gram atoms in a given weight of an element. It will be a great help if you can use a slide rule accurately and rapidly. You will also need a Periodic Table. The method and materials required are the same as in the previous set. 1. The simplest formula or empirical formula of a compound indicates the elements which make up the compound and the ratios of the elements to one another. The simplest formula does not indicate the total number of atoms in the formula; a molecular formula does. The formula for sulfuric acid, H 2 SO 4, indicates that sulfuric acid is composed of the elements,,, and in the ratios of to to. H, S, O, or hydrogen, sulfur, oxygen 2 to 1 to 4 2. The formula for potassium carbonate,, indicates that this compound Contains the elements,, and in the ratios of to to,. K 2 CO 3 K, C, O, or potassium, carbon, oxygen 2 to 1 to 3 3. There is a family of hydrocarbons, some of which have the formulas C 2 H 4, C 3 H 6, C 4 H 8 and C 5 H 10. In each case, the compound consists of the elements and in the ratio of to. 42

C and H or carbon and hydrogen 1 to 2 4. Since CH 2 describes the elements and their ratios for each of this family of compounds, (CH 2 ) n is the simplest formula, or the empirical formula, for each compound in this group. The value of n must be determined by methods we will not describe now, before one can decide which specific compound is indicated. The formula for potassium chloride,, indicates that in any weight of this substance the ratio of K to C1 is to. KC1 1 to 1 5. 1 atom of K combines with atom(s) of C1. 1 6. 17 atoms of K combine with atoms of C1. 17 7. 6.02 x 10 23 atoms of K combine with atoms of C1. 6.02 x 10 23 (6.02 x 10 23 atoms of any element = 1 gram atom = 1 mole) 8. Therefore, 1 gram atom of K combines with gram atom(s) of C1. 1 9. In H 2 S 2 gram atoms of hydrogen combine with gram atoms(s) of sulfur. 1 10. In aluminum bromide,, gram atom of A1 combines (formula) (how many) with gram atoms of bromide. 43

A1Br 3 1 3 11. In barium chlorate,, composed of the element,, and, the ratios of the numbers of gram atoms are to to respectively. Ba(C1O 3 ) 2 Ba, C1, O 1 to 2 to 6 (the subscript affects all the elements within the parentheses) 12. Conversely, if we can determine the ratios of the gram atoms in any compound, we can determine the simplest formula for the compound. Problem 1: In a compound of carbon, hydrogen, and chlorine, it was determined that the numbers of gram atoms of the elements were 1 gram atom of C, 2 gram atoms of H, and 2 gram atoms of C1. The simplest formula for this compound is C H C1. C 1 H 2 C1 2 or CH 2 C1 2 since we omit a subscript of 1 13. Problem 2: If analysis showed that there was 0.1 gram atom of C, 0.2 gram atoms of H, and 0.2 gram atoms of C1, the simplest formula is C H C1. C 0.1 H 0.2 C1 0.2 14. Since subscripts are always expressed as whole numbers, not fractions or decimals, we multiply each subscript by the same factor (in this case 10), in order to clear the formula of decimals. C 0.1 x 10 = C 1 H 0.2 x 10 = H 2 C1 0.2 x 10 = C1 2 Remember the subscript 1 is understood, so the simplest formula is. CH2C12 15. The simplest formulas are the same for problems 1 and 2 since the ratios of gram Atoms, but not the numbers of gram atoms, are identical. Go back and check this statement for yourself. The problem of determining the simplest formula of a compound consists of two operations: 44

(1) Calculation of the ratios of gram atoms of each element (2) Conversion of the ratios of gram atoms to the smallest whole number ratios Perform these operations in solving problem 3. Problem 3: Chemical analysis shows that a compound contains 6.1 grams of phosphorus and 8.0 grams of oxygen. What is the simplest formula for this compound? First, calculate the number of gram atoms of P. 6.1 grams 31 grams/ gram atom = gram atom of P 0.2 16. How many gram atoms of oxygen are present? 8.0 grams 16 grams/ gram atom = 0.5 gram atom O 17. What factor is required to clear the formula of decimals? 10 18. 10 x 0.2 gram atom of P = gram atoms of P. 10 x 0.5 gram atom of O = gram atoms of O. 2 gram atoms of P 5 gram atoms of O 19. The simplest formula for this compound is. P2O5 20. Problem 4: Analysis of a compound shows that it consists of 10 grams of calcium, 3 grams of carbon, and 12 grams of oxygen. What is the simplest formula? First, how many gram atoms of each element are there? Ca = C = O = 45

10 grams Ca = 40 grams/ gram atom =.25 gram atom Ca 3 grams C = 12 grams/ gram atom =.25 gram atom C O = 12 grams 16 grams/ gram atom =.75 gram atom O 21. Multiply each ratio by? 4 22. 4 x.25 gram atom Ca = gram atom Ca 4 x.25 gram atom C = gram atom C 4 x.75 gram atom O = gram atom O 1 gram atom Ca, 1 gram atom C, 3 gram atoms O 23. The simplest formula is. CaCO 3 24. In some cases it is not so easy to see the factor required to convert the decimal ratios to whole number ratios. In that case, divide each ratio by the smallest. In the above problem of.25 gram atom of Ca,.25 gram atom of C, and.75 gram atom of O, the smallest number of gram atoms is. 0.25 25. Dividing each ratio by 0.25 gives..25.25.75.25 = 1 Ca.25 = 1 C.75 = 3 O 26. The formula is. 46

CaCO 3 27. Problem 5: The quantity of each element may be expressed as a percentage, rather than in units of weight. The method of determining the empirical formula is the same. A compound contains 41% of sulfur and 59% of sodium. This compound can also be considered as a 100 gram sample containing 59 grams of Na and 41 grams of S. What is the simplest formula? There are gram atoms of S There are gram atoms of Na 41% 59% 32 = 1.28 gram atoms S 23 = 2.56 gram atoms Na 28. Reduce the ratios to the simplest terms; divide each ratios by the smallest. 1.28 gram atom S 1.28 = gram atom S 2.56 gram atoms Na 1.28 = gram atoms Na 1 gram atom S, 2 gram atom Na 29. The formula is. Na 2 S (The more electropositive element comes first in the formula.) 30. Problem 6: Analysis shows the following %: K = 31.9% C1 = 28.9% and O = 39.2% There are gram atoms of K There are gram atoms of C1 There are gram atoms of O.815 gram atom K,.815 gram atom C1, 2.45 gram atoms O 31. Dividing each by the smallest gives the formula. 47

KC1O 3 32. Problem 7: A sample of leas oxide contains 10 grams of lead and 1.6 grams of oxygen. Which lead oxide is this?. 10 grams 207 = 0.05 gram atom of lead 1.6 grams 16 = 0.10 gram atom of oxygen 0.05 1 0.10 = 2 Therefore the compound is PbO 2 33. Problem 8: A sample contains 23.6 grams of sulfur, 17.8 grams of oxygen, and 17.0 grams of sodium. What is its simplest formula? 23.6 grams sulfur 32 = 0.74 gram atom S 17.8 grams oxygen 16 = 1.11 gram atoms O 17 grams sodium 23 = 0.74 gram atom Na The ratio is not too obvious, so divide each ratio by the smallest. The new ratios are S = 1, O = 1.5, and Na = 1. The ratios can now be cleared of decimals by multiplying each by 2. The formula is Na 2 S 2 O 3. In your notebook copy the method and examples for calculating the simplest formula of a compound. 48