3. Monday, April 21 st : “A” Day Tuesday, April 22 nd : “B” Day Agenda  Go over Sec. 7.2 quiz  Begin 7.3: “Formulas & Percentage Composition”  In-Class Assignment: Practice pg. 243: #1-4  Homework: Pg. 56 Worksheet: 1 a-d Concept Review Must SHOW WORK!

4. Sec. 7.2 Quiz: “Relative Atomic Mass and Chemical Formulas”  This quiz gave some of you trouble, so I wanted to go over it before we continue with section 7.3…

5. 7.3: “Formulas and Percentage Composition”  The percentage composition is the percentage by mass of each element in a compound.  Percentage composition helps verify a substance’s identity.  Percentage composition can also be used to compare the ratio of masses contributed by the elements in two different substances.

6. Percent Composition of Iron Oxides

7. Empirical Formula  An actual formula shows the actual ratio of elements or ions in a single unit of a compound.  Empirical formula: a chemical formula that shows the simplest ratio for the relative numbers and kinds of atoms in a compound.  For example, consider the empirical formula and actual formulas for hydrogen peroxide: HOH 2 O 2 Empirical Formula Actual formula

8. Rules for Determining Empirical Formulas  You can use the percentage composition of a compound to determine its empirical formula. 1.Change the percentage of each element in the compound to grams. % grams 2.Use the molar mass to change grams moles 3.Compare the amounts in moles to find the simplest whole-number ratio.

9. Rules for Determining Empirical Formulas  To find the simplest whole-number ratio, divide each amount of moles by the smallest number of moles you found.  This will give a subscript of 1 for the atoms present in the smallest amount.  Finally, you may need to multiply all of the amounts of moles by a number to convert all subscripts to small, whole numbers.  The final numbers of moles you get are the subscripts in the empirical formula.

10. Determining an Empirical Formula form Percentage Composition (Sample Problem G, pg. 242) Chemical analysis of a liquid shows that it is 60.0% C, 13.4% H, and 26.6% O by mass. Calculate the empirical formula of this substance. 1.Change % grams: Assume that you have a 100 g sample so that each percentage is the same as the amount in grams: C: 60.0% = 60.0 g C H: 13.4% = 13.4 g H O: 26.6% = 26.6 g O

11. Sample Problem G, continued… 2.Use the molar mass to change grams moles: (remember sig figs!)

12. Sample Problem G, continued… 3.Divide each number of moles by the smallest number of moles found. (1.66 mol O) Carbon: 5.00 mol = 3.01 mol C 1.66 mol Hydrogen: 13.3 mol= 8.01 mol H 1.66 mol Oxygen: 1.66 mol = 1.00 mol O 1.66 mol  These numbers are within experimental error to be considered whole numbers and become the subscripts, so the empirical formula is: C 3 H 8 O

13. Example #1  Find the empirical formula given the following percentage composition: 32.37% Na, 22.58% S, 45.05% O. 1.Assume 100 g sample and change % grams: 32.37 g Na 22.58 g S 45.05 g O

14. Example #1 cont… 2.Use the molar mass to change grams moles: Na: 32.37 g Na X 1 mol Na = 1.408 mol Na 22.99 g Na S: 22.58 g S X 1 mol S=.7041 mol S 32.07 g S O: 45.05 g O X 1 mol O = 2.816 mol O 16.00 g O

15. Example #1 cont… 3.Divide each number of moles by the smallest number of moles found (.7041 mol) Na: 1.408 mol Na= 2.000 mol Na.7041 mol S:.7041 mol S= 1.000 mol S.7041 mol O: 2.816 mol O= 3.999 mol O.7041 mol These ARE whole numbers, so the empirical formula is: Na 2 SO 4

16. Additional Practice Find the empirical formula given the following percentage composition: 26.58% K, 35.35% Cr, and 38.07% O 1.Assume 100 g sample and change % grams: 26.58 g K 35.35 g Cr 38.07 g O

17. Additional Practice, cont… 2.Use the molar mass to change grams moles: K: 26.58 g K X 1 mole K =.6798 mol K 39.10 g K Cr: 35.35 g Cr X 1 mole Cr =.6798 mol Cr 52.00 g Cr O: 38.07 g O X 1 mole O = 2.379 mol O 16.00 g O (remember sig figs!)

18. Additional Practice, cont… 3.Divide each number of moles by the smallest number of moles found (.6798 mol) K:.6798 mol K = 1 mol K.6798 mol Cr:.6798 mol Cr = 1 mol Cr.6798 mol O: 2.379 mol O = 3.5 mol O.6798 mol

19. Additional Practice, cont… 4.Since 3.5 mol of oxygen is not a whole number, multiply each number of moles by 2 to get whole numbers: K: (2) 1 mol K = 2 mol K Cr: (2) 1 mol Cr = 2 mol Cr O: (2) 3.5 mol O = 7 mol O  These ARE whole numbers, so the empirical formula is: K 2 Cr 2 O 7

20. In-Class Assignment/Homework  In-Class Assignment: Practice pg. 243: 1-4  Homework: Worksheet pg. 56: 1a-d (side 56 only) Concept Review: “Formulas Percentage Composition”: #1-5 Must Show Work! Next time: You will have a sub Good luck to all juniors taking the ACT!