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Empirical/Molecular Formulas

Empirical/Molecular Formulas. Objective/Warm-Up. SWBAT calculate molar mass of compounds. What is the molar mass of each of these elements? Na Cl C H. GFM= Gram Formula Mass. Find the gram formula mass of C 2 H 6 O C = 12.01 x 2 = 24.02 H = 1.01 x 6 = 6.06 O = 16.00 x 1 = 16.01

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Empirical/Molecular Formulas

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  1. Empirical/Molecular Formulas

  2. Objective/Warm-Up • SWBAT calculate molar mass of compounds. • What is the molar mass of each of these elements? • Na • Cl • C • H

  3. GFM= Gram Formula Mass Find the gram formula mass of C2H6O • C = 12.01 x 2 = 24.02 • H = 1.01 x 6 = 6.06 • O = 16.00 x 1 = 16.01 • Total = 46.09 g/mol • Keep 2 decimal places. • Unit is g/mol

  4. Molar Mass • Example: Ca(OH)2 • Ca = 40.01 g/mol x 1 = 40.08 g/mol • O = 16.00 g/mol x 2 = 32.00 g/mol • H = 1.01 g/mol x 2 = 2.02 g/mol • Total = 74.10 g/mol

  5. Practice problems

  6. Objective/Warm-Up • SWBAT convert using molar mass. • What is the molar mass of each of these compounds? • NaCl • CaCl2 • Mg(NO3)2

  7. Objective/Warm-Up • SWBAT calculate percent composition by mass. • SWBAT distinguish between empirical and molecular formulas. • How do you find the percent of something? For example, how would you find the percent of girls or boys in this class?

  8. SWBAT calculate percent composition by mass. • How do you find the percent of something? For example, how would you find the percent of girls or boys in this class?

  9. Calculating Percent Composition • Example: Na2O • Find the mass of each element: • Na2= 22.99 x 2 = 45.98 g/mol • O = 16.00 g/mol • Take the part divided by the whole: • % Na = (45.98 g/mol) / (61.98 g/mol) = 74.2 % • % O = (16.00 g/mol) / (61.98 g/mol) = 25.8 % • The total should add up to 100 %

  10. Practice Problems

  11. Objectives/Warm-Up • SWBAT distinguish between empirical and molecular formulas. • SWBAT determine the empirical formula of a compound from percent composition. • Find the percent of oxygen in Ca(OH)2

  12. Intro to Empirical Formula • http://www.chemcollective.org/stoich/empiricalformula.php

  13. Empirical Formula Empirical Formula A formula that gives the simplest whole-number ratio of the atoms of each element in a compound. CH3O CH3O C2H4O2 CH2O

  14. Wrap-Up • Give three new examples of a molecular formula and give the corresponding empirical formula. • Why is it important to know the difference between molecular and empirical formulas?

  15. Steps to Determine Empirical Formula Determine the empirical formula for a compound containing 2.128 g Cl and 1.203 g Ca. Steps 1. Find mole amounts. 2. Divide each mole by the smallest mole.

  16. Determine the empirical formula for a compound containing 2.128 g Cl and 1.203 g Ca. 1. Find mole amounts. 2.128 g Cl x 1 mol Cl = 0.0600 mol Cl 35.45 g Cl 1.203 g Ca x 1 mol Ca = 0.0300 mol Ca 40.08 g Ca

  17. Determine the empirical formula for a compound containing 2.128 g Cl and 1.203 g Ca. 2. Divide each mole by the smallest mole. Cl = 0.0600 mol Cl = 2.00 mol Cl 0.0300 Ca = 0.0300 mol Ca = 1.00 mol Ca 0.0300 Ratio – 1 Ca: 2 Cl Empirical Formula = CaCl2

  18. A compound weighing 298.12 g consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula? Hint “Percent to mass Mass to mole Divide by small Multiply ‘til whole”

  19. A compound weighing 298.12 g consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula? Percent to mass: Mg – (72.2%/100)*298.12 g = 215.24 g N – (27.8%/100)*298.12 g = 82.88 g Mass to mole: Mg – 215.24 g * ( 1 mole ) = 8.86 mole 24.3 g N – 82.88 g * ( 1 mole ) = 5.92 mole 14.01 g Divide by small: Mg - 8.86 mole/5.92 mole = 1.50 N - 5.92 mole/5.92 mole = 1.00 Multiply ‘til whole: Mg – 1.50 x 2 = 3.00 Mg3N2 N – 1.00 x 2 = 2.00

  20. Practice • If the problem does not give you how many grams, assume 100 grams of the sample. • http://www.chemcollective.org/stoich/ef_analysis.php

  21. Wrap-Up • What is the difference between molecular and empirical formulas? • Label as molecular or empirical: • C2H4 • Na2O2 • Na2SO4 • Explain how to calculate the empirical formula.

  22. Warm-Up/Objective • SWBAT calculate molecular formulas from empirical formulas or percent composition. • Label as molecular or empirical: • C2H4 • Na2O2 • Na2SO4 • What are the steps to calculate the empirical formula?

  23. You are a Forensic Scientist •             The victim in the following case is a 35-year old white male named Tony DeMoy.  Initial investigators say they found several signs around the death site that suggest foul play.  Four possible causes of his untimely death have been suggested by his wife who has been ruled out as a suspect because of a proven alibi.  Your task is to identify who and what killed Tony DeMoy.       

  24. Molecular Formula The molecular formula gives the actual number of atoms of each element in a molecular compound. Steps 1. Find the empirical formula. 2. Calculate the Empirical Formula Mass. 3. Divide the molar mass by the “EFM”. 4. Multiply empirical formula by factor. Find the molecular formula for a compound whose molar mass is ~124.06 and empirical formula is CH2O3. 2. “EFM” = 62.03 g 3. 124.06/62.03 = 2 4. 2(CH2O3) = C2H4O6

  25. Find the molecular formula for a compound that contains 4.90 g N and 11.2 g O. The molar mass of the compound is 92.0 g/mol. Steps 1. Find the empirical formula. 2. Calculate the Empirical Formula Mass. 3. Divide the molar mass by the “EFM”. 4. Multiply empirical formula by factor.

  26. Find the molecular formula for a compound that contains 4.90 g N and 11.2 g O. The molar mass of the compound is 92.0 g/mol. Empirical formula. A. Find mole amounts. 4.90 g N x 1 mol N = 0.350 mol N 14.01 g N 11.2 g O x 1 mol O = 0.700 mol O 16.00 g O

  27. Find the molecular formula for a compound that contains 4.90 g N and 11.2 g O. The molar mass of the compound is 92.0 g/mol. B. Divide each mole by the smallest mole. N = 0.350 = 1.00 mol N 0.350 O = 0.700 = 2.00 mol O 0.350 Empirical Formula = NO2 Empirical Formula Mass = 46.01 g/mol

  28. Find the molecular formula for a compound that contains 4.90 g N and 11.2 g O. The molar mass of the compound is 92.0 g/mol. Molecular formula Molar Mass = 92.0 g/mol = 2.00 Emp. Formula Mass 46.01 g/mol Molecular Formula = 2 x Emp. Formula = N2O4

  29. Solving the Crime • With a partner, analyze each piece of evidence in the lab area. • There are 4 suspected compounds. • Find the molecular formula of each compound, then see the teacher for possible identity of those compounds.

  30. Wrap-Up • Summarize how to find the molecular formula from the empirical formula.

  31. A 528.39 g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~222.25 g/mol. What is its molecular formula?

  32. mole C - 255.64 g * ( 1 mole ) = 21.29 mol 12.01 g mole H – 42.91 g * ( 1 mole ) = 42.49 mol 1.01 g mole O – 229.85 g * ( 1 mole ) = 14.37 mol 16.00 g A 528.39 g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~222.25 g/mol. What is its molecular formula? g C – (48.38/100)*528.39 g = 255.64 g g H – (8.12/100)*528.39 g = 42.91 g g O – (43.5/100)*528.39 g = 229.85 g

  33. A 528.39 g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~222.25 g/mol. What is its molecular formula? From last slide: 21.29 mol C, 42.49 mol H, 14.27 mol O C – 21.29/14.27 = 1.49 H – 42.49/14.27 = 2.98 (esentially 3) O – 14.27/14.27 = 1.00 C – 1.49 x 2 = 3 H – 3 x 2 = 6 O – 1 x 2 = 2 C3H6O2

  34. A 528.39 g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~222.25 g/mol. What is its molecular formula? From last slide: Empirical formula = C3H6O2 “EFM” = 74.09 Molar mass = 222.24 = ~3 EFM 74.09 3(C3H6O2) = C9H18O6

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