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.

Monday, April 21st: “A” DayTuesday, April 22nd: “B” Day

AgendaGo over Sec. 7.2 quizBegin 7.3: “Formulas & Percentage Composition”In-Class Assignment:

Practice pg. 243: #1-4Homework:

Pg. 56 Worksheet: 1 a-dConcept Review

Must SHOW WORK!

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…

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.

Percent Composition of Iron Oxides

Empirical FormulaAn 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:

HO H2O2 Empirical Formula Actual formula

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.

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.

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 CH: 13.4% = 13.4 g HO: 26.6% = 26.6 g O

Sample Problem G, continued…2. Use the molar mass to change grams moles:

(remember sig figs!)

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: C3H8O

Example #1Find 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 Na22.58 g S45.05 g O

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

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 molThese ARE whole numbers, so the empirical

formula is: Na2SO4

Additional PracticeFind 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 K35.35 g Cr38.07 g O

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!)

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

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 KCr: (2) 1 mol Cr = 2 mol CrO: (2) 3.5 mol O = 7 mol O

These ARE whole numbers, so the empirical formula is: K2Cr2O7

In-Class Assignment/HomeworkIn-Class Assignment:

Practice pg. 243: 1-4Homework:

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!