Friday, April 15th: “A” DayMonday, April 18th: “B” Day

Agenda

Homework questions/problems?Quiz over section 7.2Begin 7.3: “Formulas & Percentage Composition”In-Class Assignments:–Practice pg. 243: #1-4–Practice pg. 245: #1-3

Quiz 7.2: “Relative Atomic Mass and Chemical Formulas”

You can use your book and your guided notes for this walk-talk quiz…

Remember: answer only the question that corresponds to the month you were born in.

If you were born in November, answer question #2. If you were born in December, answer question # 7.

Once everyone has answered their question, get up, walk around, and talk/compare answers with others.

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, the empirical formula for hydrogen peroxide is HO, while the actual formula is H2O2

Determining Empirical Formulas You can use the percentage composition for a

compound to determine its empirical formula.1. Convert the percentage of each element in the

compound to grams.

2. Convert from grams to moles using the molar mass of each element as a conversion factor.

3. Compare these amounts in moles to find the simplest whole-number ratio among the elements.

Determining Empirical FormulasTo find the simplest whole-number ratio,

divide each amount in moles by the smallest of all the amounts of moles.

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 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. Assume that you have a 100 g sample so that each percentage is the same as the amount in grams: for C: 60.0% = 60.0 g C for H: 13.4% = 13.4 g H for O: 26.6% = 26.6 g O

Sample Problem G, continued…2. Use the molar mass to convert each amount in grams to amount in moles:

Sample Problem G, continued…3. Divide each number of moles found by the

smallest number of moles found. (1.66 moles O)

Carbon: 5.00 mol = 3.01 mol C 1.66 mol

Hydrogen: 13.3 mol= 8.01 mol H1.66 mol

Oxygen: 1.66 mol = 1 mol O 1.66 mol

These numbers are within experimental error to be considered whole numbers so the empirical formula is: C3H8O

Additional PracticeFind the empirical formula given the following

composition:26.58% K, 35.35% Cr, and 38.07% O

1.Assume 100 g sample:26.58 g K35.35 g Cr38.07 g O

Additional Practice2. Use molar mass to convert from grams to

moles. 26.58 g K X 1 mole K = .6798 mol K

39.10 g K

35.35 g Cr X 1 mole Cr = .6798 mol Cr52.00 g Cr

38.07 g O X 1 mole O = 2.379 mol O16.00 g O

Additional Practice3. Divide each number of moles found by the

smallest number of moles found (.6798 mol).

.6798 mol K = 1 mol K .6798 mol

.6798 mol Cr = 1 mol Cr.6798 mol

2.379 mol O = 3.5 mol O.6798 mol

Additional Practice4. Since these are not whole numbers, multiply

each one by 2 to get whole numbers. 1 mol K (2) = 2 mol K 1 mol Cr (2) = 2 mol Cr 3.5 mol O (2) = 7 mol O

These ARE whole numbers, so the empirical formula is:

K2Cr2O7

Molecular Formulas are Multiples of Empirical Formulas

The formula for an ionic compound shows the simplest whole-number ratio of the large numbers of ions in a crystal of the compound.

A molecular formula is a whole-number multiple of the empirical formula.

The molar mass of any compound is equal to the molar mass of the empirical formula times a whole number, n.

n (empirical formula) = molecular formula

Comparing Empirical and Molecular Formulas

Compound Empirical Formula

Molecular Formula

Formaldehyde CH2O CH2O

Acetic Acid CH2O C2H4O2

2X the empirical formulan = 2

Glucose CH2O C6H12O6

6X the empirical formulan = 6

Determining a Molecular Formula from an Empirical Formula

(Sample Problem H, pg. 245)The empirical formula for a compound is P2O5.

Its experimental molar mass is 284 g/mol. Determine the molecular formula of the compound.

1. Use the periodic table to find the molar mass of the empirical formula:

For P: 2(30.97) = 61.94 g/molFor O: 5(16.00) = 80.00 g/molMolar mass of P2O5 = 141.94 g/mol

Sample Problem H, continued…2. Find the multiplier, n:n = experimental molar mass of compound

molar mass of empirical formula

n = 284 g/mol = 2 Hint: the bigger # 141.94 g/mol always goes on top!

3. To find the molecular formula, simply multiply the empirical formula by 2:

2 (P2O5) = P4O10

Additional PracticeDetermine the molecular formula for the

following:Molar mass: 232.41 g/molEmpirical formula: OCNCl1.Find molar mass of empirical formula:O = 16.00 g/molC = 12.01 g/molN = 14.01 g/molCl = 35.5 g/mol = 77.52 g/mol

Additional Practice2. Find the multiplier, n: n = experimental molar mass of compound

molar mass of empirical formula

n = 232.41 g/mol = 3 77.52 g/mol

3. To find the molecular formula, simply multiply the empirical formula by 3:

3 (OCNCl) = O3C3N3Cl3

In-Class AssignmentsYou Must SHOW WORK!

Practice pg. 243: #1-4Practice pg. 245: #1-3

We will finish this section next time…