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Friday, April 15 th : “A” Day Monday, April 18 th : “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

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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.

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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.

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Percent Composition of Iron Oxides

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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 H 2 O 2

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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.

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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.

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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

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

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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 H 1.66 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: C 3 H 8 O

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Additional Practice Find 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 K 35.35 g Cr 38.07 g O

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Additional Practice 2.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

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Additional Practice 3.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

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Additional Practice 4.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: K 2 Cr 2 O 7

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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

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Comparing Empirical and Molecular Formulas CompoundEmpirical Formula Molecular Formula FormaldehydeCH 2 O Acetic AcidCH 2 OC 2 H 4 O 2 2X the empirical formula n = 2 GlucoseCH 2 OC 6 H 12 O 6 6X the empirical formula n = 6

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Determining a Molecular Formula from an Empirical Formula (Sample Problem H, pg. 245) The empirical formula for a compound is P 2 O 5. 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 P 2 O 5 = 141.94 g/mol

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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 (P 2 O 5 ) = P 4 O 10

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Additional Practice Determine the molecular formula for the following: Molar mass: 232.41 g/mol Empirical formula: OCNCl 1.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

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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) = O 3 C 3 N 3 Cl 3

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

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