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Timberlake LecturePLUS1 Chapter 9 Empirical Formulas

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Timberlake LecturePLUS2 Types of Formulas The formulas for compounds can be expressed as an empirical formula and as a molecular(true) formula. Empirical Molecular (true)Name CHC 2 H 2 acetylene CHC 6 H 6 benzene CO 2 CO 2 carbon dioxide CH 2 OC 5 H 10 O 5 ribose

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Timberlake LecturePLUS3 Empirical Formulas Write your own one-sentence definition for each of the following: Empirical formula Molecular formula

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Timberlake LecturePLUS4 An empirical formula represents the simplest whole number ratio of the atoms in a compound. The molecular formula is the true or actual ratio of the atoms in a compound.

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Timberlake LecturePLUS5 Learning Check EF-1 A. What is the empirical formula for C 4 H 8 ? 1) C 2 H 4 2) CH 2 3) CH B. What is the empirical formula for C 8 H 14 ? 1) C 4 H 7 2) C 6 H 12 3) C 8 H 14 C. What is a molecular formula for CH 2 O? 1) CH 2 O2) C 2 H 4 O 2 3) C 3 H 6 O 3

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Timberlake LecturePLUS6 Solution EF-1 A. What is the empirical formula for C 4 H 8 ? 2) CH 2 B. What is the empirical formula for C 8 H 14 ? 1) C 4 H 7 C. What is a molecular formula for CH 2 O? 1) CH 2 O2) C 2 H 4 O 2 3) C 3 H 6 O 3

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Timberlake LecturePLUS7 Learning Check EF-2 If the molecular formula has 4 atoms of N, what is the molecular formula if SN is the empirical formula? Explain. 1) SN 2) SN 4 3) S 4 N 4

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Timberlake LecturePLUS8 Solution EF-2 If the molecular formula has 4 atoms of N, what is the molecular formula if SN is the empirical formula? Explain. 3) S 4 N 4 If the actual formula has 4 atoms of N, and S is related 1:1, then there must also be 4 atoms of S.

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Timberlake LecturePLUS9 Empirical and Molecular Formulas molar mass = a whole number = n simplest mass n = 1 molar mass = empirical mass molecular formula = empirical formula n = 2 molar mass = 2 x empirical mass molecular formula = 2 x empirical formula molecular formula = or > empirical formula

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Timberlake LecturePLUS10 Empirical Formula Empirical Mass Molecular Formula Molecular Mass

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Timberlake LecturePLUS11 Learning Check EF-3 A compound has a formula mass of 176.0 and an empirical formula of C 3 H 4 O 3. What is the molecular formula? 1) C 3 H 4 O 3 2) C 6 H 8 O 6 3) C 9 H 12 O 9

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Timberlake LecturePLUS12 Solution EF-3 A compound has a formula mass of 176.0 and an empirical formula of C 3 H 4 O 3. What is the molecular formula? 2)C 6 H 8 O 6 C 3 H 4 O 3 = 88.0 g/EF 176.0 g = 2.00 88.0

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Timberlake LecturePLUS13 Learning Check EF-4 If there are 192.0 g of O in the molecular formula, what is the true formula if the EF is C 7 H 6 O 4 ? 1) C 7 H 6 O 4 2) C 14 H 12 O 8 3) C 21 H 18 O 12

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Timberlake LecturePLUS14 Solution EF-4 If there are 192.0 g of O in the molecular formula, what is the true formula if the EF is C 7 H 6 O 4 ? 3) C 21 H 18 O 12 192 g O = 3 x O 4 or 3 x C 7 H 6 O 4 64.0 g O in EF

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Timberlake LecturePLUS15 Finding the Molecular Formula A compound is Cl 71.65%, C 24.27%, and H 4.07%. What are the empirical and molecular formulas? The molar mass is known to be 99.0 g/mol. 1. State mass percents as grams in a 100.00-g sample of the compound. Cl 71.65 gC 24.27 g H 4.07 g

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Timberlake LecturePLUS16 2. Calculate the number of moles of each element. 71.65 g Cl x 1 mol Cl = 2.02 mol Cl 35.5 g Cl 24.27 g C x 1 mol C = 2.02 mol C 12.0 g C 4.07 g H x 1 mol H = 4.04 mol H 1.01 g H

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Timberlake LecturePLUS17 Why moles? Why do you need the number of moles of each element in the compound?

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Timberlake LecturePLUS18 3. Find the smallest whole number ratio by dividing each mole value by the smallest mole values: Cl: 2.02 = 1 Cl 2.02 C: 2.02 = 1 C 2.02 H: 4.04 = 2 H 2.02 4. Write the simplest or empirical formula CH 2 Cl

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Timberlake LecturePLUS19 5. EM (empirical mass) = 1(C) + 2(H) + 1(Cl) = 49.5 6. n = molar mass/empirical mass Molar mass = 99.0 g/mol = n = 2 E M 49.5 g/EM 7.Molecular formula (CH 2 Cl) 2 = C 2 H 4 Cl 2

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Timberlake LecturePLUS20 Learning Check EF-5 Aspirin is 60.0% C, 4.5 % H and 35.5 O. Calculate its simplest formula. In 100 g of aspirin, there are 60.0 g C, 4.5 g H, and 35.5 g O.

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Timberlake LecturePLUS21 Solution EF-5 60.0 g C x ___________= ______ mol C 4.5 g H x ___________ = _______mol H 35.5 g O x ___________ = _______mol O

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Timberlake LecturePLUS22 Solution EF-5 60.0 g C x 1 mol C = 5.00 mol C 12.0 g C 4.5 g H x 1 mol H = 4.5 mol H 1.01 g H 35.5 g O x 1mol O= 2.22 mol O 16.0 g O

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Timberlake LecturePLUS23 Divide by the smallest # of moles. 5.00 mol C = ________________ ______ mol O 4.5 mol H = ________________ ______ mol O 2.22 mol O = ________________ ______ mol O Are are the results whole numbers?_____

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Timberlake LecturePLUS24 Divide by the smallest # of moles. 5.00 mol C = ___2.25__ 2.22 mol O 4.5 mol H = ___2.00__ 2.22 mol O 2.22 mol O = ___1.00__ 2.22 mol O Are are the results whole numbers?_____

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Timberlake LecturePLUS25 Finding Subscripts A fraction between 0.1 and 0.9 must not be rounded. Multiply all results by an integer to give whole numbers for subscripts. (1/2) 0.5 x 2 = 1 (1/3)0.333 x 3 = 1 (1/4)0.25 x 4 = 1 (3/4)0.75 x 4 = 3

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Timberlake LecturePLUS26 Multiply everything x 4 C: 2.25 mol C x 4 = 9 mol C H: 2.0 mol Hx 4 = 8 mol H O: 1.00 mol O x 4 = 4 mol O Use the whole numbers of mols as the subscripts in the simplest formula C 9 H 8 O 4

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Timberlake LecturePLUS27 Learning Check EF-6 A compound is 27.4% S, 12.0% N and 60.6 % Cl. If the compound has a molar mass of 351 g/mol, what is the molecular formula?

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Timberlake LecturePLUS28 Solution EF 6 0.853 mol S /0.853 = 1 S 0.857 mol N /0.853 = 1 N 1.71 mol Cl /0.853 = 2 Cl Empirical formula = SNCl 2 = 117.1 g/EF Mol. Mass/ Empirical mass 351/117.1 = 3 Molecular formula = S 3 N 3 Cl 6

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