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1 © 2006 Brooks/Cole - Thomson Kull Spring 2007 Chem 105 Lsn 5 Chapter 3 Topics Announcements –All homework quizzes completed on line –Extension for chapter.

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Presentation on theme: "1 © 2006 Brooks/Cole - Thomson Kull Spring 2007 Chem 105 Lsn 5 Chapter 3 Topics Announcements –All homework quizzes completed on line –Extension for chapter."— Presentation transcript:

1 1 © 2006 Brooks/Cole - Thomson Kull Spring 2007 Chem 105 Lsn 5 Chapter 3 Topics Announcements –All homework quizzes completed on line –Extension for chapter 2 until Jan 28, 8 pm to allow for minor computer issues to be resolved –Monday is a review day Topics –Using percent composition »Emperical and molecular formulas –Calculating a formula –Formula of a compound from combining masses –Formula from mass data –Hydrated compounds

2 2 © 2006 Brooks/Cole - Thomson Kull Spring 2007 Chem 105 Lsn 5 Using percent composition Emperical and molecular formulas A pure compound always consists of the same elements combined in the same proportions by weight. Therefore, we can express molecular composition as PERCENT BY WEIGHT Ethanol, C 2 H 6 O 52.13% C 13.15% H 34.72% O

3 3 © 2006 Brooks/Cole - Thomson Kull Spring 2007 Chem 105 Lsn 5 Percent Composition Consider NO 2, Molar mass = ? What is the weight percent of N and of O? What are the weight percentages of N and O in NO?

4 4 © 2006 Brooks/Cole - Thomson Kull Spring 2007 Chem 105 Lsn 5 Calculating a formula In chemical analysis we determine the % by weight of each element in a given amount of pure compound and derive the EMPIRICAL or SIMPLEST formula. PROBLEM : A compound of B and H is 81.10% B. What is its empirical formula?

5 5 © 2006 Brooks/Cole - Thomson Kull Spring 2007 Chem 105 Lsn 5 Because it contains only B and H, it must contain 18.90% H.Because it contains only B and H, it must contain 18.90% H. In 100.0 g of the compound there are 81.10 g of B and 18.90 g of H.In 100.0 g of the compound there are 81.10 g of B and 18.90 g of H. Calculate the number of moles of each constitutent.Calculate the number of moles of each constitutent. A compound of B and H is 81.10% B. What is its empirical formula?

6 6 © 2006 Brooks/Cole - Thomson Kull Spring 2007 Chem 105 Lsn 5 Calculate the number of moles of each element in 100.0 g of sample. A compound of B and H is 81.10% B. What is its empirical formula?

7 7 © 2006 Brooks/Cole - Thomson Kull Spring 2007 Chem 105 Lsn 5 Now, recognize that atoms combine in the ratio of small whole numbers. 1 atom B + 3 atoms H --> 1 molecule BH 3 or 1 mol B atoms + 3 mol H atoms ---> 1 mol BH 3 molecules Find the ratio of moles of elements in the compound. A compound of B and H is 81.10% B. What is its empirical formula?

8 8 © 2006 Brooks/Cole - Thomson Kull Spring 2007 Chem 105 Lsn 5 But we need a whole number ratio. 2.5 mol H/1.0 mol B = 5 mol H to 2 mol B EMPIRICAL FORMULA = B 2 H 5 Take the ratio of moles of B and H. Always divide by the smaller number. A compound of B and H is 81.10% B. What is its empirical formula?

9 9 © 2006 Brooks/Cole - Thomson Kull Spring 2007 Chem 105 Lsn 5 A compound of B and H is 81.10% B. Its empirical formula is B 2 H 5. What is its molecular formula ? Is the molecular formula B 2 H 5, B 4 H 10, B 6 H 15, B 8 H 20, etc.? B 2 H 6 is one example of this class of compounds. B2H6B2H6

10 10 © 2006 Brooks/Cole - Thomson Kull Spring 2007 Chem 105 Lsn 5 A compound of B and H is 81.10% B. Its empirical formula is B 2 H 5. What is its molecular formula ? We need to do an EXPERIMENT to find the MOLAR MASS. Here experiment gives 53.3 g/mol Compare with the mass of B 2 H 5 = 26.66 g/unit = 26.66 g/unit Find the ratio of these masses. Molecular formula = B 4 H 10

11 11 © 2006 Brooks/Cole - Thomson Kull Spring 2007 Chem 105 Lsn 5 Formula of a compound from combining masses

12 12 © 2006 Brooks/Cole - Thomson Kull Spring 2007 Chem 105 Lsn 5 DETERMINE THE FORMULA OF A COMPOUND OF Sn AND I Sn(s) + some I 2 (s) ---> SnI x 4 +, 2 + 1 - Formula of a compound from combining masses

13 13 © 2006 Brooks/Cole - Thomson Kull Spring 2007 Chem 105 Lsn 5 Data to Determine the formula of a Sn—I Compound Reaction of Sn and I 2 is done using excess Sn.Reaction of Sn and I 2 is done using excess Sn. Mass of Sn in the beginning = 1.056 gMass of Sn in the beginning = 1.056 g Mass of Sn remaining (recovered) = 0.601 gMass of Sn remaining (recovered) = 0.601 g Mass of iodine (I 2 ) used = 1.947 gMass of iodine (I 2 ) used = 1.947 g (See p. 125) Convert these masses to moles

14 14 © 2006 Brooks/Cole - Thomson Kull Spring 2007 Chem 105 Lsn 5 Reaction of Sn and I 2 is done using excess Sn.Reaction of Sn and I 2 is done using excess Sn. Mass of iodine (I 2 ) used = 1.947 g Mass of Sn initially = 1.056 g Mass of Sn recovered = 0.601 g Mass of Sn used = 0.455 g Find the mass of Sn that combined with 1.947 g I 2. Find moles of Sn used: Tin and Iodine Compound

15 15 © 2006 Brooks/Cole - Thomson Kull Spring 2007 Chem 105 Lsn 5 Tin and Iodine Compound Now find the number of moles of I 2 that combined with 3.83 x 10 -3 mol Sn. Mass of I 2 used was 1.947 g. How many mol of iodine atoms ? = 1.534 x 10 -2 mol I atoms

16 16 © 2006 Brooks/Cole - Thomson Kull Spring 2007 Chem 105 Lsn 5 Tin and Iodine Compound Now find the ratio of number of moles of moles of I and Sn that combined. Empirical formula is SnI 4

17 17 © 2006 Brooks/Cole - Thomson Kull Spring 2007 Chem 105 Lsn 5 Formula from mass data

18 18 © 2006 Brooks/Cole - Thomson Kull Spring 2007 Chem 105 Lsn 5 Hydrated compounds


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