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Objectives

Section 4 Determining Chemical Formulas. Chapter 7. Objectives. Define empirical formula, and explain how the term applies to ionic and molecular compounds. Determine an empirical formula from either a percentage or a mass composition.

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Objectives

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  1. Section 4 Determining Chemical Formulas Chapter 7 Objectives • Define empirical formula, and explain how the term applies to ionic and molecular compounds. • Determine an empirical formula from either a percentage or a mass composition. • Explain the relationship between the empirical formula and the molecular formula of a given compound. • Determine a molecular formula from an empirical formula.

  2. Section 4 Determining Chemical Formulas Chapter 7 Empirical and Actual Formulas

  3. Section 4 Determining Chemical Formulas Chapter 7 • An empirical formulaconsists of the symbols for the elements combined in a compound, with subscripts showing the smallest whole-number mole ratio of the different atoms in the compound. • For an ionic compound, the formula unit is usually the compound’s empirical formula. • For a molecular compound, however, the empirical formula does not necessarily indicate the actual numbers of atoms present in each molecule. • example: the empirical formula of the gas diborane is BH3, but the molecular formula is B2H6.

  4. Section 4 Determining Chemical Formulas Chapter 7 Calculation of Empirical Formulas • To determine a compound’s empirical formula from its percentage composition, begin by converting percentage composition to a mass composition. • Assume that you have a 100.0 g sample of the compound. • Then calculate the amount of each element in the sample. • example: diborane • The percentage composition is 78.1% B and 21.9% H. • Therefore, 100.0 g of diborane contains 78.1 g of B and 21.9 g of H.

  5. Section 4 Determining Chemical Formulas Chapter 7 Calculation of Empirical Formulas, continued • Next, the mass composition of each element is converted to a composition in moles by dividing by the appropriate molar mass. • These values give a mole ratio of 7.22 mol B to 21.7 mol H.

  6. Section 4 Determining Chemical Formulas Chapter 7 Calculation of Empirical Formulas, continued • To find the smallest whole number ratio, divide each number of moles by the smallest number in the existing ratio. • Because of rounding or experimental error, a compound’s mole ratio sometimes consists of numbers close to whole numbers instead of exact whole numbers. • In this case, the differences from whole numbers may be ignored and the nearest whole number taken.

  7. Section 4 Determining Chemical Formulas Chapter 7 Calculation of Empirical Formulas, continued Sample Problem L Quantitative analysis shows that a compound contains 32.38% sodium, 22.65% sulfur, and 44.99% oxygen. Find the empirical formula of this compound.

  8. Section 4 Determining Chemical Formulas Chapter 7 Calculation of Empirical Formulas, continued Sample Problem L Solution Given:percentage composition: 32.38% Na, 22.65% S, and 44.99% O Unknown:empirical formula Solution: percentage composition mass composition composition in moles smallest whole-number mole ratio of atoms

  9. Section 4 Determining Chemical Formulas Chapter 7 Calculation of Empirical Formulas, continued Sample Problem L Solution, continued

  10. Section 4 Determining Chemical Formulas Chapter 7 Calculation of Empirical Formulas, continued Sample Problem L Solution, continued Smallest whole-number mole ratio of atoms: The compound contains atoms in the ratio 1.408 mol Na:0.7063 mol S:2.812 mol O. Rounding yields a mole ratio of 2 mol Na:1 mol S:4 mol O. The empirical formula of the compound is Na2SO4.

  11. Section 4 Determining Chemical Formulas Chapter 7 Calculation of Molecular Formulas • The empirical formulacontains the smallest possible whole numbers that describe the atomic ratio. • The molecular formulais the actual formula of a molecular compound. • An empirical formula may or may not be a correct molecular formula. • The relationship between a compound’s empirical formula and its molecular formula can be written as follows. x(empirical formula) = molecular formula

  12. Section 4 Determining Chemical Formulas Chapter 7 Calculation of Molecular Formulas, continued • The formula masses have a similar relationship. x(empirical formula mass) = molecular formula mass • To determine the molecular formula of a compound, you must know the compound’s formula mass. • Dividing the experimental formula mass by the empirical formula mass gives the value of x. • A compound’s molecular formula mass is numerically equal to its molar mass, so a compound’s molecular formula can also be found given the compound’s empirical formula and its molar mass.

  13. Section 4 Determining Chemical Formulas Chapter 7 Comparing Empirical and Molecular Formulas

  14. Section 4 Determining Chemical Formulas Chapter 7 Calculation of Molecular Formulas, continued Sample Problem N In Sample Problem M, the empirical formula of a compound of phosphorus and oxygen was found to be P2O5. Experimentation shows that the molar mass of this compound is 283.89 g/mol. What is the compound’s molecular formula?

  15. Section 4 Determining Chemical Formulas Chapter 7 Calculation of Molecular Formulas, continued Sample Problem N Solution Given:empirical formula Unknown:molecular formula Solution: x(empirical formula) = molecular formula

  16. Section 4 Determining Chemical Formulas Chapter 7 Calculation of Molecular Formulas, continued Sample Problem N Solution, continued Molecular formula mass is numerically equal to molar mass. molecular molar mass = 283.89 g/mol molecular formula mass = 283.89 amu empirical formula mass mass of phosphorus atom = 30.97 amu mass of oxygen atom = 16.00 amu empirical formula mass of P2O5 = 2  30.97 amu + 5 × 16.00 amu = 141.94 amu

  17. Section 4 Determining Chemical Formulas Chapter 7 Calculation of Molecular Formulas, continued Sample Problem N Solution, continued Dividing the experimental formula mass by the empirical formula mass gives the value of x. (P2O5) = P4O10 The compound’s molecular formula is therefore P4O10.

  18. Chapter 7 Standardized Test Preparation Multiple Choice 1. Which of the following compounds does not contain a polyatomic ion? A. sodium carbonate B. sodium sulfate C. sodium sulfite D. sodium sulfide

  19. Chapter 7 Standardized Test Preparation Multiple Choice 1. Which of the following compounds does not contain a polyatomic ion? A. sodium carbonate B. sodium sulfate C. sodium sulfite D. sodium sulfide

  20. Chapter 7 Standardized Test Preparation Multiple Choice 2. The correct formula for ammonium phosphate is A. (NH4)3PO4. B. (NH4)2PO4. C. NH4PO4. D. NH4(PO4)2.

  21. Chapter 7 Standardized Test Preparation Multiple Choice 2. The correct formula for ammonium phosphate is A. (NH4)3PO4. B. (NH4)2PO4. C. NH4PO4. D. NH4(PO4)2.

  22. Chapter 7 Standardized Test Preparation Multiple Choice 3. When writing the formula for a compound that contains a polyatomic ion, A. write the anion’s formula first. B. use superscripts to show the number of polyatomic ions present. C. use parentheses if the number of polyatomic ions is greater than 1. D. always place the polyatomic ion in parentheses.

  23. Chapter 7 Standardized Test Preparation Multiple Choice 3. When writing the formula for a compound that contains a polyatomic ion, A. write the anion’s formula first. B. use superscripts to show the number of polyatomic ions present. C. use parentheses if the number of polyatomic ions is greater than 1. D. always place the polyatomic ion in parentheses.

  24. Chapter 7 Standardized Test Preparation Multiple Choice 4. The correct name for NH4CH3COO is A. ammonium carbonate. B. ammonium hydroxide. C. ammonium acetate. D. ammonium nitrate.

  25. Chapter 7 Standardized Test Preparation Multiple Choice 4. The correct name for NH4CH3COO is A. ammonium carbonate. B. ammonium hydroxide. C. ammonium acetate. D. ammonium nitrate.

  26. Chapter 7 Standardized Test Preparation Multiple Choice 5. Which of the following is the correct formula for iron(III) sulfate? A. Fe3SO4 B. Fe3(SO4)2 C. Fe2(SO4)3 D. 3FeSO4

  27. Chapter 7 Standardized Test Preparation Multiple Choice 5. Which of the following is the correct formula for iron(III) sulfate? A. Fe3SO4 B. Fe3(SO4)2 C. Fe2(SO4)3 D. 3FeSO4

  28. Chapter 7 Standardized Test Preparation Multiple Choice 6. The molecular formula for acetylene is C2H2. The molecular formula for benzene is C6H6. The empirical formula for both is A. CH. B. C2H2. C. C6H6. D. (CH)2.

  29. Chapter 7 Standardized Test Preparation Multiple Choice 6. The molecular formula for acetylene is C2H2. The molecular formula for benzene is C6H6. The empirical formula for both is A. CH. B. C2H2. C. C6H6. D. (CH)2.

  30. Chapter 7 Standardized Test Preparation Multiple Choice 7. Which of the following shows the percentage composition of H2SO4? A. 2.5% H, 39.1% S, 58.5% O B. 2.1% H, 32.7% S, 65.2% O C. 28.6% H, 14.3% S, 57.1% O D. 33.3% H, 16.7% S, 50% O

  31. Chapter 7 Standardized Test Preparation Multiple Choice 7. Which of the following shows the percentage composition of H2SO4? A. 2.5% H, 39.1% S, 58.5% O B. 2.1% H, 32.7% S, 65.2% O C. 28.6% H, 14.3% S, 57.1% O D. 33.3% H, 16.7% S, 50% O

  32. Chapter 7 Standardized Test Preparation Multiple Choice 8. Which of the following compounds has the highest percentage of oxygen? A. CH4O B. CO2 C. H2O D. Na2CO3

  33. Chapter 7 Standardized Test Preparation Multiple Choice 8. Which of the following compounds has the highest percentage of oxygen? A. CH4O B. CO2 C. H2O D. Na2CO3

  34. Chapter 7 Standardized Test Preparation Multiple Choice 9. The empirical formula for a compound that is 1.2% H, 42.0% Cl, and 56.8% O is A. HClO. B. HClO2. C. HClO3. D. HClO4.

  35. Chapter 7 Standardized Test Preparation Multiple Choice 9. The empirical formula for a compound that is 1.2% H, 42.0% Cl, and 56.8% O is A. HClO. B. HClO2. C. HClO3. D. HClO4.

  36. Chapter 7 Standardized Test Preparation Short Answer 10. When a new substance is synthesized or is discovered experimentally, the substance is analyzed quantitatively. What information is obtained from this typical analysis, and how is this information used?

  37. Chapter 7 Standardized Test Preparation Short Answer 10. When a new substance is synthesized or is discovered experimentally, the substance is analyzed quantitatively. What information is obtained from this typical analysis, and how is this information used? Answer:The percentage composition is obtained and used to determine the substance’s empirical formula.

  38. Chapter 7 Standardized Test Preparation Short Answer 11. An oxide of selenium is 28.8% O. Find the empirical formula. Assuming that the empirical formula is also the molecular formula, name the oxide.

  39. Chapter 7 Standardized Test Preparation Short Answer 11. An oxide of selenium is 28.8% O. Find the empirical formula. Assuming that the empirical formula is also the molecular formula, name the oxide. Answer:SeO2; selenium dioxide

  40. Chapter 7 Standardized Test Preparation Extended Response 12. What is an empirical formula, and how does it differ from a molecular formula?

  41. Chapter 7 Standardized Test Preparation Extended Response 12. What is an empirical formula, and how does it differ from a molecular formula? Answer:The empirical formula is the simplest whole-number ratio of the atoms in a chemical formula. The molecular formula is the actual number of atoms of each element in a chemical formula. The two formulas differ by an integer factor.

  42. Chapter 7 Standardized Test Preparation Extended Response 13. What are Stock system names based on?

  43. Chapter 7 Standardized Test Preparation Extended Response 13. What are Stock system names based on? Answer:The Stock system is based on the oxidation number of the cation.

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