Chapter 7 Chemical Quantities. 7.1 A Mole. Samples of One Mole Quantities. A Mole of Molecules

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1 Chapter 7 Chemical Quantities 7.1 A Mole 7.1 The Mole 7.2 Molar Mass 7.3 Calculations Using Molar Mass 7.4 Percent Composition and Empirical Formulas A mole contains 6.02 x particles (atoms, ions, molecules, formula unit) The number 6.02 x is known as Avogadro s number. One mole of any element contains Avogadro s number of atoms. 1 mole Na = 6.02 x Na atoms 1 mole Au = 6.02 x Au atoms 1 2 A Mole of Molecules Samples of One Mole Quantities One mole of a covalent compound contains Avogadro s number of molecules. 1 mole CO 2 = 6.02 x CO 2 molecules 1 mole H 2 O = 6.02 x H 2 O molecules One mole of an ionic compound contains Avogadro s number of formula units. 1 mole NaCl = 6.02 x NaCl formula units 3 4 1

2 Avogadro s Number Avogadro s number is written as conversion factors x particles and 1 mole 1 mole 6.02 x particles The number of molecules in 0.50 mole of CO 2 molecules is calculated as 0.50 mole CO 2 molecules x 6.02 x CO 2 molecules 1 mole CO 2 molecules = 3.0 x CO 2 molecules 5 A. Calculate the number of atoms in 2.0 moles of Al. B. Calculate the number of moles of S in 1.8 x S Molar Mass A. Calculate the number of atoms in 2.0 moles of Al. 2.0 moles Al x 6.02 x Al atoms 1 mole Al =1.2 x Al atoms B. Calculate the number of moles of S in 1.8 x S. 1.8 x S atoms x 1 mole S 6.02 x S atoms = 3.0 mole S atoms The mass of one mole is called molar mass (g/mole). The molar mass of an element is the atomic mass expressed in grams

3 Give the molar mass to the nearest 0.1 g. A. 1 mole of K atoms = 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 = B. 1 mole of Sn atoms = g 9 10 Molar Mass of CaCl 2 Molar Mass of K 3 PO 4 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 CaCl 2 to the nearest 0.1 g as follows. Formula mass of CaCl 2 = [ (35.45)] = 111.1amu Formula mass = molar mass, so Determine the molar mass of K 3 PO 4 to 0.1 g. Molar Mass = [3(39.1) (16)] = 212.3g/mol 111.1amu = 111.1g/mol

4 One-Mole Quantities A. 1 mole of K 2 O = g B. 1 mole of antacid Al(OH) 3 = g 32.1 g 55.9 g 58.5 g g g A. 1 mole of K 2 O 2 (39.1) + 1 (16.0) = 94.2 g/mol 1mole K 2 O x 94.2g/mol = 94.2g Prozac, C 17 H 18 F 3 NO, is an antidepressant that inhibits the uptake of serotonin by the brain. What is the molar mass of Prozac? 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

5 Prozac, C 17 H 18 F 3 NO, 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) = = 309 g/mole Molar Mass Factors Methane CH 4 known as natural gas is used in gas cook tops and gas heaters. 1 mole CH 4 = 16.0 g The molar mass of methane can be written as conversion factors g CH 4 and 1 mole CH 4 1 mole CH g CH Acetic acid C 2 H 4 O 2 gives the sour taste to vinegar. Write two molar mass conversion factors for acetic acid. Acetic acid C 2 H 4 O 2 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

6 7.3 Calculations with Molar Mass Mole factors are used to convert between the grams of a substance and the number of moles. 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? Grams Mole factor Moles 3.00 moles Al x 27.0 g Al = 81.0 g Al 1 mole Al mole factor for Al The artificial sweetener aspartame (Nutri-Sweet) C 14 H 18 N 2 O 5 is used to sweeten diet foods, coffee and soft drinks. How many moles of aspartame are present in 225 g of aspartame? Calculate the molar mass of C 14 H 18 N 2 O (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) = mole aspartame

7 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 CO 2 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 CO g O x 100 = 72.7 % O 44.0 g CO % What is the percent carbon in C 5 H 8 NNaO 4 (MSG monosodium glutamate), a compound used to flavor foods and tenderize meats? 1) 7.10 %C 2) 35.5 %C 3) 60.0 %C Types of Formulas 2) 35.5 %C Molar mass = g % = total g C x 100 total g MSG = 60.0 g C x 100 = 35.5 % C g MSG 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. C 5 H 10 O 5 5 = C 1 H 2 O 1 = CH 2 O molecular empirical formula formula

8 Some Molecular and 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. Which is a possible molecular formula for CH 2 O? 1) C 4 H 4 O 4 2) C 2 H 4 O 2 3) C 3 H 6 O A. What is the empirical formula for C 4 H 8? 2) CH 2 C 4 H 8 4 B. What is the empirical formula for C 8 H 14? 1) C 4 H 7 C 8 H 14 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) SN 4 C. Which is a possible molecular formula for CH 2 O? 2) C 2 H 4 O 2 3) C 3 H 6 O 3 3) S 4 N

9 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 molecular formula has 4 atoms of N, and S and N are related 1:1, then there must also be 4 atoms of S. 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 number to determine the molecular formula 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 g/mol = g/mol 4. Multiply the subscripts in CH by Molecular formula = (CH) 6 = C 6 H 6 A compound has a formula mass of and an empirical formula of C 3 H 4 O 3. What is the molecular formula?

10 A compound has a formula mass of and an empirical formula of C 3 H 4 O 3. What is the molecular formula? C 3 H 4 O 3 = 3(12.0) + 4(1.0) + 3(16.0) = 88.0 g/mol g/mol (molar mass) = g/mol (empirical mass) 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 g sample of the compound. C g H 4.07 g Cl g Molecular formula = 2 (empirical formula) (C 3 H 4 O 3 ) 2 = C 6 H 8 O Finding the Molecular Formula Continued Finding the Molecular Formula (continued) 2. Calculate the number of moles of each element 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 3. Divide each by the smallest 2.02 moles C = 1 mole C moles H = 2 moles H moles Cl = 1 mole Cl g Cl x 1 mole Cl = 2.02 moles Cl 35.5 g Cl Empirical formula = C 1 H 2 Cl 1 = CH 2 Cl

11 Finding the Molecular Formula (continued) 4. Calculate empirical mass (EM) empirical mass CH 2 Cl = 49.5 g/mol Aspirin is 60.0% C, 4.5 % H and 35.5 % O. Calculate its empirical (simplest) formula. 5. Divide molar mass by empirical mass Molar mass = 99.0 g/mol = 2 Empirical mass 49.5 g/mol 6. Determine Molecular formula (CH 2 Cl) 2 = C 2 H 4 Cl (continued) (continued) In 100 g of aspirin, there are 60.0 g C, 4.5 g H, and 35.5 g O 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 Divide by the smallest number of moles moles C = 2.25 moles C moles H = 2.0 moles H 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

12 (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 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? Use the whole numbers as subscripts to obtain the simplest formula C 9 H 8 O (continued) In g, there are 27.4 g S, 12.0 g N and 60.6 g Cl g S x 1 mole S = mole S 32.1 g S 12.0 g N x 1 mole N = moles N 14.0 g N Dividing by the smallest number of moles mole S /0.854 = 1.00 mole S mole N/0.854 = 1.00 mole N 1.71 moles Cl/0.854 = 2.00 moles Cl Empirical formula = SNCl 2 = g/mol Molar Mass/ Empirical mass 60.6 g Cl x 1mole Cl = 1.71 moles Cl 35.5 g Cl 351 = Molecular formula = (SNCl 2 ) 3 = S 3 N 3 Cl