The Mole 1 atom or 1 molecule is a very small entity not convenient to operate with The masses we usually encounter in chemical experiments vary from milligrams to kilograms Just like one dozen = 12 things One mole = 6.022 x 10 23 things Avogadro s number: N A = 6.022 x 10 23 1
The Mole N A = 6.022 x 10 23 Why 6.022 x 10 23? This is the number of carbon atoms found in 12 g of the carbon-12 isotope Molar mass mass of one mole of atoms, molecules, ions, etc. Numerically equal to the atomic or molecular weight of the substance in grams: m (1 mole H 2 ) = M r (H 2 ) = m (1 mole Fe) = M r (Fe) = 2
The Mole: Example 1 Example: Calculate the mass of a single Mg atom in grams to 3 significant figures. 3
The Mole: Example 2 Example: How many C 6 H 14 molecules are contained in 55 ml of hexane (d = 0.78 g/ml). 4
The Mole: Example 3 Example: Calculate the number of O atoms in 26.5 g of lithium carbonate, Li 2 CO 3. Solution: 1) Calculate the molar mass of Li 2 CO 3 : M r (Li 2 CO 3 ) = 73.89 g/mol 2) Calculate the amount of Li 2 CO 3 in moles: n (Li 2 CO 3 ) = 0.359 mol 3) Calculate the amount of O atoms in moles: n (O) = (0.359x3) mol = 1.077 mol 4) Multiply by the Avogadro s number to obtain the answer: N (O) = 1.077 mol x 6.022 10 23 atom/mol = = 6.50 10 23 atoms 5
Percent Composition and Formulas of Compounds If the formula of a compound is known, its chemical composition can be expressed as the mass percent of each element in the compound (percent composition), and vice versa. When solving this kind of problems, we can use masses expressed in a.m.u. or in g/mol 6
Percent Composition: Example 1 What is the percent composition of each element in sodium chloride, NaCl? 7
Percent Composition: Example 2 Calculate the percent composition of iron(iii) sulfate, Fe 2 (SO 4 ) 3, to 3 significant figures Solution: 1) Calculate the molar mass of Fe 2 (SO 4 ) 3: M r (Fe 2 (SO 4 ) 3 ) = 399.88 g/mol 2) Calculate the percent composition for the elements: ω(fe) = 2 Mr (Fe) 2 55 85. g/mol 100% = 100% = M (Fe (SO ) ) 399 88. g/mol r 2 4 3 27.93% ω(s) = 3 Mr (S) 3 32 07. g/mol 100% = 100% = M (Fe (SO ) ) 399 88. g/mol r 2 4 3 24.06% ω(o) = 12 Mr (O) 12 16 00. g/mol 100% = 100% = M (Fe (SO ) ) 399 88. g/mol r 2 4 3 48.01% 8
Simplest (Empirical) Formula The smallest whole-number ratio of atoms present in the compound Molecular formula, on the other hand, indicates the actual number of atoms present in a molecule of the compound water hydrogen peroxide 9
Empirical Formula: Example 1 The first high-temperature superconductor, prepared by Bednorz and Müller in 1986, contained 68.54% lanthanum, 15.68% copper, and 15.79% oxygen by mass. What was the simplest formula of this compound? 10
Empirical Formula: Example 2 A sample of a compound contains 6.541g of Co and 2.368g of O. What is its empirical formula? 11
Elemental Composition A combustion train for carbon-hydrogen analysis percent composition is determined experimentally magnesium perchlorate sodium hydroxide 12
Empirical Formula: Example 3 0.1172 g of a pure hydrocarbon was burned in a C-H combustion train to produce 0.3509 g of CO 2 and 0.1915 g of H 2 O. Determine the masses of C and H in the sample, the percentage of these elements in this hydrocarbon, and the empirical formula of the compound. Solution: 1) First we calculate the molar weights of CO 2 and H 2 O: M r (CO 2 ) = 44.01 g/mol; M r (H 2 O) = 18.02 g/mol 2) Now we can calculate the amount of both CO 2 and H 2 O in moles: n (CO 2 ) n (H O) 2 = = m(co M (CO r 2 2 ) ) m(h O) M (H O) r 2 2 = = 0.3509 44 01. 0.1915 g g/mol g 18.02 g/mol = = 0.007973 0.01063 mol = 7.973 10 mol = 1.063 10 2 3 mol mol 13
Example 3 (continued) 3) Notice that 1 mole of CO 2 molecules contains 1 mole of C atoms and 1 mole of H 2 O molecules contain 2 moles of H atoms Therefore, n (C) = 7.973 10-3 mol; n (H) = 2 x 1.063 10-2 mol = 2.126 10-2 mol 4) Now we can answer the first question determine the masses of carbon and hydrogen in the starting compound using the formula: = M m n r m = n M r m (C) = 7.973 10-3 mol x 12.01 g/mol = 0.09576 g m (H) = 2.126 10-2 mol x 1.008 g/mol = 0.02143 g Now, if you add these masses, they should give you the mass of the starting compound: 0.1172 g 14
Example 3 (continued) 5) Now we can easily find the mass percentage of the elements in the hydrocarbon by dividing the mass of the element by the total mass of the hydrocarbon and multiplying by 100%: 0.09576 g ω(c) = 100% = 0 1172. g 81.71% 0.02143 g ω(h) = 100% = 18.28% 0 1172. g We can also find the percentage of hydrogen by subtracting the percentage of carbon from 100%: ω(h) = 100% - ω(c) = 18.29% The slight difference between two results in caused by rounding errors. 15
Example 3 (continued) 6) Finally, to find the empirical formula we use the amounts of C and H in moles found in step 3 and divide them by the smallest of both numbers: -3 7.973 10 C : -3 7.973 10 mol mol = 1.000-2 2 126. 10 H : -3 7.973 10 mol mol = 2.666 We know that the fraction like.666 is in fact 2/3. Therefore, to convert both results to whole numbers, we need to multiply them by 3: C: 3.000 H: 7.998 These result tells us that the empirical formula for the hydrocarbon in question is C 3 H 8 16
Empirical Formula: Example 4 0.1014 g sample of purified glucose was burned in a C-H combustion train to produce 0.1486 g of CO 2 and 0.0609 g of H 2 O. An elemental analysis showed that glucose contains only carbon, hydrogen, and oxygen. Determine the empirical formula of the compound. This example is worked out in your textbook 17
Molecular Formula Indicates the actual number of atoms present in a molecule of the compound To determine the molecular formula for a molecular compound, both its empirical formula and its molecular weight must be known The molecular formula for a compound is either the same as, or an integer of, the empirical formula 18
Molecular Formula: Example A compound is found to contain 85.63% C and 14.37% H by mass. In another experiment its molar mass is found to be 56.1 g/mol. What is its molecular formula? I ll work on this example in the beginning of the next lecture. 19
Other Examples What mass of ammonium phosphate, (NH 4 ) 3 PO 4, would contain 15.0 g of N? You do it! The answer is: 53.2 g 20
Reading Assignment Go through Lecture 3 notes and learn again the problems we solved in class Read Chapter 2 to the end Learn Key Terms (p. 82) Take a look at Lecture 4 notes (already posted on the web) If you have time, read Chapter 3 Homework #1 due by 9/13 21