1. Empirical Formulas & Percent Composition CO 2 Carbon dioxide BCl 3 boron trichloride

2. Molar Mass

3. How many moles of alcohol are there in a “standard” can of beer if there are 21.3 g of C 2 H 6 O? (a)Molar mass of C 2 H 6 O = 46.08 g/mol (b)Calc. moles of alcohol

4. How many molecules of alcohol are there in a “standard” can of beer if there are 21.3 g of C 2 H 6 O? = 2.78 x 10 23 molecules We know there are 0.462 mol of C 2 H 6 O.

5. How many atoms of C are there in a “standard” can of beer if there are 21.3 g of C 2 H 6 O? = 5.57 x 10 23 C atoms There are 2.78 x 10 23 molecules. Each molecule contains 2 C atoms. Therefore, the number of C atoms is

6. Empirical & 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

7. Percent Composition Consider some of the family of nitrogen- oxygen compounds: NO 2, nitrogen dioxide Structure of NO 2 What is the weight percent of N and of O?

8. Percent Composition Consider NO 2, Molar mass = Percent of N and of O? What are percentages of N and O in NO? 46 g/ mol

9. Determining Formulas 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 : What is its empirical formula of a compound of B and H if 81.10% is boron?

10. B %B + %H = 100% 100% - 81.10% B = 18.90% H. If there were 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. Determination of empirical formulas: mass of each

11. Calculate the number of moles of each element assuming 100.0 g total. Determination of empirical formulas: moles of each Determination of empirical formulas: moles of each

12. Take the ratio of moles of B and H. atoms combine in small whole number ratios Determination of empirical formulas: moles ratio Always divide by the smaller number. 18.75 mol H and 7.502 mol B 7.502 mol 2.5 mol H to 1.0 mol B

13. . But we need a whole number ratio. 2.5 mol H or 5 mol H 1.0 mol B 2 mol B EMPIRICAL FORMULA = B 2 H 5

14. If the empirical formula is B 2 H 5, what is its molecular formula ? We need to do an EXPERIMENT to find the MOLAR MASS. MS gives 53.3 g/mol Compare empirical mass = 26.66 g/unit Find the ratio of these masses. Molecular formula = B 4 H 10

15. Determine the formula of a compound of Sn and I using the following data. Reaction of Sn and I 2 is done using excess Sn. Mass of Sn in the beginning = 1.056 g Mass of Sn remaining = 0.601 g Mass of iodine (I 2 ) used = 1.947 g

16. Find the mass of Sn that combined with 1.947 g I 2. Mass of Sn initially = 1.056 g Mass of Sn recovered = 0.601 g Mass of Sn used = 0.455 g Find moles of Sn used: Tin and Iodine Compound

17. 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

18. 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

19. Practice Formula Determination Homework: Worksheets pgs 17-20 Pg 13-14 good review for mole calculations Prelab questions