Friday, April 15th: “A” Day Monday, April 18th: “B” Day Agenda Homework questions/problems? Quiz over section 7.2 Begin 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 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, 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. Convert the percentage of each element in the compound to grams. Convert from grams to moles using the molar mass of each element as a conversion factor. Compare these amounts in moles to find the simplest whole-number ratio among the elements.

Determining Empirical Formulas To 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… 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 H Oxygen: 1.66 mol = 1 mol O These numbers are within experimental error to be considered whole numbers so the empirical formula is: C3H8O

Additional Practice Find the empirical formula given the following composition: 26.58% K, 35.35% Cr, and 38.07% O Assume 100 g sample: 26.58 g K 35.35 g Cr 38.07 g O

Additional Practice 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 Cr 52.00 g Cr 38.07 g O X 1 mole O = 2.379 mol O 16.00 g O

Additional Practice 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 2.379 mol O = 3.5 mol O

Additional Practice K2Cr2O7 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 Acetic Acid C2H4O2 2X the empirical formula n = 2 Glucose C6H12O6 6X the empirical formula n = 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/mol For O: 5(16.00) = 80.00 g/mol Molar mass of P2O5 = 141.94 g/mol

Sample Problem H, continued… 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 Practice Determine the molecular formula for the following: Molar mass: 232.41 g/mol Empirical formula: OCNCl Find molar mass of empirical formula: O = 16.00 g/mol C = 12.01 g/mol N = 14.01 g/mol Cl = 35.5 g/mol = 77.52 g/mol

Additional Practice 2. 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 Assignments You Must SHOW WORK! Practice pg. 243: #1-4 Practice pg. 245: #1-3 We will finish this section next time…