# empirical formulas & molecular formulas empirical formulas molecular formulas

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- Slide 1
- Empirical Formulas & Molecular Formulas Empirical Formulas Molecular Formulas
- Slide 2
- 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
- Slide 3
- Empirical Formulas Write your own one-sentence definition for each of the following: Empirical formula Molecular formula
- Slide 4
- 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.
- Slide 5
- Empirical Formulas 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
- Slide 6
- Solution 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
- Slide 7
- Molecular Formula 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
- Slide 8
- Solution 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.
- Slide 9
- Empirical and Molecular Formulas To obtain the molecular formula, you must first have an empirical formula. You must also have the molar mass or be given enough information to determine the molar mass. With these two values, you can determine the molecular formula.
- Slide 10
- Empirical and Molecular Formulas molar mass = a whole number = n EF 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
- Slide 11
- Empirical Formula Empirical Mass Molecular Formula Molecular Mass
- Slide 12
- Molecular Formula 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
- Slide 13
- Solution 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
- Slide 14
- Molecular Formula If there are 192.0 g of O in the molecular formula, what is the true formula if the Empirical Formula 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
- Slide 15
- Solution 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
- Slide 16
- 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
- Slide 17
- 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
- Slide 18
- Why moles? Why do you need the number of moles of each element in the compound?
- Slide 19
- Why moles? Why do you need the number of moles of each element in the compound? Recall Avogadros number! A mole contains equal numbers of particles. That means we are comparing atoms on an equal basis.
- Slide 20
- 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
- Slide 21
- 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
- Slide 22
- Empirical Formula 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.
- Slide 23
- Solution 60.0 g C x ___________= ______ mol C 4.5 g H x ___________ = _______mol H 35.5 g O x ___________ = _______mol O
- Slide 24
- Solution 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
- Slide 25
- 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?_____
- Slide 26
- 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 the results whole numbers?_____
- Slide 27
- 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
- Slide 28
- 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 (2/3)0.667x 3=2 (3/4)0.75 x 4 = 3
- Slide 29
- Molecular Formula 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?
- Slide 30
- Solution 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