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Chapter 7 Chemical Quantities

Chapter 7 Chemical Quantities. 7.1 The Mole 7.2 Molar Mass 7.3 Calculations Using Molar Mass 7.4 Percent Composition and Empirical Formulas. 7.1 A Mole. A mole contains 6.02 x 10 23 particles (atoms, ions, molecules, formula unit)

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Chapter 7 Chemical Quantities

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  1. Chapter 7 Chemical Quantities 7.1 The Mole 7.2 Molar Mass 7.3 Calculations Using Molar Mass 7.4 Percent Composition and Empirical Formulas

  2. 7.1 A Mole • A mole contains 6.02 x 1023 particles (atoms, ions, molecules, formula unit) • The number 6.02 x 1023 is known as Avogadro’s number. • One mole of any element contains Avogadro’s number of atoms. 1 mole Na = 6.02 x 1023 Na atoms 1 mole Au = 6.02 x 1023 Au atoms

  3. A Mole of Molecules • One mole of a covalent compound contains Avogadro’s number of molecules. 1 mole CO2 = 6.02 x 1023CO2 molecules 1 mole H2O = 6.02 x 1023H2O molecules • One mole of an ionic compound contains Avogadro’s number of formula units. 1 mole NaCl = 6.02 x 1023 NaCl formula units

  4. Samples of One Mole Quantities

  5. Avogadro’s Number • Avogadro’s number is written as conversion factors. 6.02 x 1023 particles and 1 mole 1 mole 6.02 x 1023 particles • The number of molecules in 0.50 mole of CO2 molecules is calculated as 0.50 mole CO2 molecules x 6.02 x 1023 CO2 molecules 1 mole CO2 molecules = 3.0 x 1023 CO2 molecules

  6. Learning Check A. Calculate the number of atoms in 2.0 moles of Al. B. Calculate the number of moles of S in 1.8 x 1024 S.

  7. Solution A. Calculate the number of atoms in 2.0 moles of Al.2.0 moles Al x 6.02 x 1023 Al atoms 1 mole Al =1.2 x 1024 Al atoms B. Calculate the number of moles of S in 1.8 x 1024 S. 1.8 x 1024 S atoms x 1 mole S 6.02 x 1023 S atoms = 3.0 mole S atoms

  8. 7.2 Molar Mass • The mass of one mole is called molar mass (g/mole). • The molar mass of an element is the atomic mass expressed in grams.

  9. Learning Check Give the molar mass to the nearest 0.1 g. A. 1 mole of K atoms = ________ B. 1 mole of Sn atoms = ________

  10. Solution Give the molar mass to the nearest 0.1 g. A. 1 mole of K atoms = 39.1 g B. 1 mole of Sn atoms = 118.7 g

  11. Molar Mass of CaCl2 • For a compound, the molar mass is the sum of the molar masses of the elements in the formula. We calculate the molar mass of CaCl2 to the nearest 0.1 g as follows. • Formula mass of CaCl2 = [40.1 + 2(35.45)] = 111.1amu • Formula mass = molar mass, so 111.1amu = 111.1g/mol

  12. Molar Mass of K3PO4 Determine the molar mass of K3PO4 to 0.1 g. Molar Mass = [3(39.1) + 31.0 + 4(16)] = 212.3g/mol

  13. One-Mole Quantities 32.1 g 55.9 g 58.5 g 294.2 g 342.3 g

  14. Learning Check A. 1 mole of K2O = ______g B. 1 mole of antacid Al(OH)3 = ______g

  15. Solution A. 1 mole of K2O 2 (39.1) + 1 (16.0) = 94.2 g/mol 1mole K2O x 94.2g/mol = 94.2g B. 1 mole of antacid Al(OH)3 1 (27.0) + 3 (16.0) + 3 (1.0) = 78.0 g/mol 1mole Al(OH)3 x 78.0 g/mol = 78.0g

  16. Learning Check Prozac, C17H18F3NO, is an antidepressant that inhibits the uptake of serotonin by the brain. What is the molar mass of Prozac?

  17. Solution Prozac, C17H18F3NO, is a widely used antidepressant that inhibits the uptake of serotonin by the brain. What is the molar mass of Prozac? 17C (12.0) + 18H (1.0) + 3F (19.0) + 1N (14.0) + 1 O (16.0) = 204 + 18 + 57.0 + 14.0 + 16.0 = 309 g/mole

  18. Molar Mass Factors Methane CH4 known as natural gas is used in gas cook tops and gas heaters. 1 mole CH4 = 16.0 g The molar mass of methane can be written as conversion factors. 16.0 g CH4 and 1 mole CH4 1 mole CH4 16.0 g CH4

  19. Learning Check Acetic acid C2H4O2 gives the sour taste to vinegar. Write two molar mass conversion factors for acetic acid.

  20. Solution Acetic acid C2H4O2 gives the sour taste to vinegar. Write two molar mass factors for acetic acid. 2(12.0) + 4(1.0) + 2(16) = 60.0g/mol 1 mole of acetic acid = 60.0 g acetic acid 1 mole acetic acid and 60.0 g acetic acid 60.0 g acetic acid 1 mole acetic acid

  21. 7.3 Calculations with Molar Mass • Mole factors are used to convert between the grams of a substance and the number of moles. Mole factor Grams Moles

  22. Calculating Grams from Moles Aluminum is often used for the structure of lightweight bicycle frames. How many grams of Al are in 3.00 moles of Al? 3.00 moles Al x 27.0 g Al = 81.0 g Al 1 mole Al mole factor for Al

  23. Learning Check The artificial sweetener aspartame (Nutri-Sweet) C14H18N2O5 is used to sweeten diet foods, coffee and soft drinks. How many moles of aspartame are present in 225 g of aspartame?

  24. Solution Calculate the molar mass of C14H18N2O5. 14 (12.0) + 18 (1.0) + 2 (14.0) + 5(16.0) = 294 g/mole Set up the calculation using a mole factor. 225 g aspartame x 1 mole aspartame 294 g aspartame mole factor(inverted) = 0.765 mole aspartame

  25. 7.4 Percent Composition • In a compound, the percent composition is the percent by mass of each element in the formula. • In one mole of CO2 there are 12.0 g of C and 32.0 g of O (molar mass 44.0 g/mol), 12.0 g C x 100 = 27.3 % C 44.0 g CO2 32.0 g O x 100 = 72.7 % O 44.0 g CO2100.0 %

  26. Learning Check What is the percent carbon in C5H8NNaO4 (MSG monosodium glutamate), a compound used to flavor foods and tenderize meats? 1) 7.10 %C 2) 35.5 %C 3) 60.0 %C

  27. Solution 2) 35.5 %C Molar mass = 169.1 g % = total g C x 100 total g MSG = 60.0 g C x 100 = 35.5 % C 169.1 g MSG

  28. Types of Formulas • The molecular formula is the true or actual number of the atoms in a molecule. • The empirical formula is the simplest whole number ratio of the atoms. • The empirical formula is calculated by dividing the subscripts in the molecular formula by a whole number to give the lowest ratio.C5H10O5  5 = C1H2O1 = CH2Omolecular empirical formulaformula

  29. Some Molecular and Empirical Formulas

  30. Learning Check 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. Which is a possible molecular formula for CH2O? 1) C4H4O4 2) C2H4O2 3) C3H6O3

  31. Solution A. What is the empirical formula for C4H8? 2) CH2C4H8  4 B. What is the empirical formula for C8H14? 1) C4H7C8H14 2 C. Which is a possible molecular formula for CH2O? 2) C2H4O2 3) C3H6O3

  32. Learning Check 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

  33. Solution 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 molecular formula has 4 atoms of N, and S and N are related 1:1, then there must also be 4 atoms of S.

  34. Relating Empirical and Molecular Formulas • A molecular formula is equal to or a multiple of the empirical formula. • Thus, the molar mass is equal to or a multiple of the empirical mass. molar mass = a whole number empirical mass • Multiply the empirical formula by the whole numberto determine the molecular formula.

  35. Finding the Molecular Formula Determine the molecular formula of a compound that has a molar mass of 78.0 and an empirical formula of CH. 1. Empirical mass of CH = 13.0 g/mol 2. Divide the molar mass by the empirical mass. 3. 78.0 g/mol = 6.00 13.0 g/mol 4. Multiply the subscripts in CH by 6. 5. Molecular formula = (CH)6 = C6H6

  36. Learning Check A compound has a formula mass of 176.0 and an empirical formula of C3H4O3. What is the molecular formula?

  37. Solution A compound has a formula mass of 176.0 and an empirical formula of C3H4O3. What is the molecular formula? C3H4O3 = 3(12.0) + 4(1.0) + 3(16.0) = 88.0 g/mol 176.0 g/mol (molar mass) = 2.00 88.0 g/mol (empirical mass) Molecular formula = 2 (empirical formula) (C3H4O3 )2 = C6H8O6

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

  39. Finding the Molecular FormulaContinued 2. Calculate the number of moles of each element. 24.27 g C x 1 mole C = 2.02 moles C 12.0 g C 4.07 g H x 1 mole H = 4.03 moles H 1.01 g H 71.65 g Cl x 1 mole Cl = 2.02 moles Cl 35.5 g Cl

  40. Finding the Molecular Formula(continued) 3. Divide each by the smallest 2.02 moles C = 1 mole C 2.024.03 moles H = 2 moles H 2.022.02 moles Cl = 1 mole Cl 2.02 Empirical formula = C1H2Cl1 = CH2Cl

  41. Finding the Molecular Formula(continued) 4. Calculate empirical mass (EM)empirical mass CH2Cl = 49.5 g/mol 5. Divide molar mass by empirical mass Molar mass = 99.0 g/mol = 2Empirical mass 49.5 g/mol 6. Determine Molecular formula(CH2Cl)2 = C2H4Cl2

  42. Learning Check Aspirin is 60.0% C, 4.5 % H and 35.5 % O. Calculate its empirical (simplest) formula.

  43. Solution (continued) In 100 g of aspirin, there are 60.0 g C, 4.5 g H, and 35.5 g O. 60.0 g C x 1 mole C = 5.00 moles C 12.0 g C 4.5 g H x 1 mole H= 4.5 moles H 1.01 g H 35.5 g O x 1mole O= 2.22 moles O 16.0 g O

  44. Solution (continued) Divide by the smallest number of moles. 5.00 moles C = 2.25 moles C 2.22 4.5 moles H = 2.0 moles H 2.22 2.22 moles O = 1.00 mole O 2.22 Note that the results are not all whole numbers. To obtain whole numbers, multiply by a factor to give whole numbers.

  45. Solution (continued) Multiply each number of moles by 4 C: 2.25 moles C x 4 = 9 moles C H: 2.0 moles H x 4 = 8 moles H O: 1.00 mole O x 4 = 4 moles O Use the whole numbers as subscripts to obtain the simplest formula C9H8O4

  46. Learning Check A compound is 27.4% S, 12.0% N and 60.6 % Cl. If the compound has a molar mass of 351 g, what is the molecular formula?

  47. Solution In 100.0 g, there are 27.4 g S, 12.0 g N and 60.6 g Cl. 27.4 g S x 1 mole S =0.854 mole S 32.1 g S 12.0 g N x 1 mole N= 0.857moles N 14.0 g N 60.6 g Cl x 1mole Cl = 1.71 moles Cl 35.5 g Cl

  48. Solution (continued) Dividing by the smallest number of moles 0.854 mole S /0.854 = 1.00 mole S 0.857 mole N/0.854 = 1.00 mole N 1.71 moles Cl/0.854 = 2.00 moles Cl Empirical formula = SNCl2 = 117.1 g/mol Molar Mass/ Empirical mass 351 = 3117.1 Molecular formula = (SNCl2)3 = S3N3Cl6

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