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Empirical and Molecular Formulas

Empirical and Molecular Formulas. Empirical vs Molecular Formula. The Molecular Formula (MF) gives the actual number of each type of atom present. The Empirical Formula (EF) gives the lowest whole-number ratio of the atoms present. Example: C 2 H 6 and C 3 H 9

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Empirical and Molecular Formulas

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  1. Empirical and Molecular Formulas

  2. Empirical vs Molecular Formula • The Molecular Formula (MF) gives the actual number of each type of atom present. • The Empirical Formula (EF) gives the lowest whole-number ratio of the atoms present. • Example: C2H6 and C3H9 • they have the same EF CH3 yet have very different MF

  3. Types of Formulas The formulas for compounds can be expressed as an empirical formula and as a molecular(true) formula. Empirical Molecular (true) Name CH C2H2 acetylene CH C6H6 benzene CO2 CO2 carbon dioxide CH2O C5H10O5 ribose Timberlake LecturePLUS

  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. Timberlake LecturePLUS

  5. EF vs. MF When you have an ionic compound: the EF = MF For some molecular compounds: the EF = MF, but not always

  6. Determining the EF Remember: The EF is the lowest whole-number ratio of the moles of each atom present. Example: CH4 has 1 mol C atoms 4 mol H atoms

  7. Determining the EF If a compound consists of: 62.1% C 13.8% H 24.1% N The percentages are based on MASS not MOLES We can compare moles, not masses

  8. Determining the EF In order to determine the ratio of C:H:N, we need to know the mole ratio Step 1: Convert % of each into grams Make it easy on yourself, assume a sample size of 100.00g 62.1g C 13.8g H 24.1g N

  9. Determining the EF Now that you know how many grams of each atom you have: Step 2: Convert grams to moles using the molar mass of each 62.1g C 13.8g H 24.1 g N

  10. Determining the EF Now that you have the mol ratio, you need to make them Whole-Numbers. Step 3: Divide each mol by the smallest mol value from step #2 5.17 mol C 13.7 mol H 1.72 mol N

  11. Determining the EF This results in 3 mol C, 8 mol H and 1 mol N therefore the EF = C3H8N

  12. Learning Check EF-1 A. What is the empirical formula for C4H8? 1) C2H4 2) CH2 3) CH B. What is the empirical formula for C8H14? 1) C4H7 2) C6H12 3) C8H14 C. What is a molecular formula for CH2O? 1) CH2O 2) C2H4O2 3) C3H6O3 Timberlake LecturePLUS

  13. Solution EF-1 A. What is the empirical formula for C4H8? 2) CH2 B. What is the empirical formula for C8H14? 1) C4H7 C. What is a molecular formula for CH2O? 1) CH2O 2) C2H4O2 3) C3H6O3 Timberlake LecturePLUS

  14. 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) SN4 3) S4N4 Timberlake LecturePLUS

  15. 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) S4N4 If the actual formula has 4 atoms of N, and S is related 1:1, then there must also be 4 atoms of S. Timberlake LecturePLUS

  16. 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 Timberlake LecturePLUS

  17. EF to MF To go from the EF to the MF, you need two additional pieces of information: 1 – calculate the mass from your EF 2 – You must be given the mass of the MF (X) EFmass = MFmass X = MFmass / EFmass

  18. EF to MF Example: You found the EF to be HO, and the MFmass = 34.02g • Calculate the EFmass = 17.01g • Calculate X = 34.02g / 17.01g • X = 2 • EF = HO MF = H2O2

  19. Learning Check EF-3 A compound has a formula mass of 176.0 and an empirical formula of C3H4O3. What is the molecular formula? 1) C3H4O3 2) C6H8O6 3) C9H12O9 Timberlake LecturePLUS

  20. Solution EF-3 A compound has a formula mass of 176.0 and an empirical formula of C3H4O3. What is the molecular formula? 2)C6H8O6 C3H4O3 = 88.0 g/EF 176.0 g = 2.00 88.0 Timberlake LecturePLUS

  21. 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 C7H6O4? 1) C7H6O4 2) C14H12O8 3) C21H18O12 Timberlake LecturePLUS

  22. Solution EF-4 If there are 192.0 g of O in the molecular formula, what is the true formula if the EF is C7H6O4? 3) C21H18O12 192 g O = 3 x O4 or 3 x C7H6O4 64.0 g O in EF Timberlake LecturePLUS

  23. 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 g C 24.27 g H 4.07 g Timberlake LecturePLUS

  24. 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 Timberlake LecturePLUS

  25. Why moles? Why do you need the number of moles of each element in the compound? Timberlake LecturePLUS

  26. 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 CH2Cl Timberlake LecturePLUS

  27. 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 (CH2Cl)2 = C2H4Cl2 Timberlake LecturePLUS

  28. 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. Timberlake LecturePLUS

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

  30. 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 Timberlake LecturePLUS

  31. 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?_____ Timberlake LecturePLUS

  32. 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?_____ Timberlake LecturePLUS

  33. 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 x2 = 1 (1/3) 0.333 x 3 = 1 (1/4) 0.25 x4 = 1 (3/4) 0.75 x 4 = 3 Timberlake LecturePLUS

  34. Multiply everything x 4 C: 2.25 mol C x 4 = 9 mol C H: 2.0 mol H x 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 C9H8O4 Timberlake LecturePLUS

  35. 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? Timberlake LecturePLUS

  36. 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 = SNCl2 = 117.1 g/EF Mol. Mass/ Empirical mass 351/117.1 = 3 Molecular formula = S3N3Cl6 Timberlake LecturePLUS

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