6.1 Counting by Weighing Name: Group: Date: Ch. 6 Chemical Composition - Notetaker (Key) Chem 6 Objective: To understand the concept of average mass and explore how counting can be done by _ weighing Average mass for a jelly bean = Total mass of beans Number of beans 5.1 g + 5.2 g + 5.0 g + 4.8 g + 5.0 g + 5.0 g + 5.1 g + 4.9 g + 5.0 g 10 Average mass for a bean = 50.0 = 5.0 _ g 100 jelly beans x 5 g = 500 _ g of jelly beans 1 Jelly bean If the average mass for a mint is 15 g then the ratio of the mass of mints to the mass of jelly beans is _ 15 g = 3 5 g 10 100 mints x 15 g = 1500 g of mints 1 mint 6.2 Atomic Masses: Counting Atoms by Weighing C(s) + O 2 (g) CO 2 (g) 1 atom reacts with 1 molecule to yield 1 molecule To determine the number of oxygen molecules required we must know how many carbon atoms are present in the pile of carbon. But individuals atoms are far too small to
see. We must learn to count atoms by weighing samples containing large numbers of atoms. The mass of a single carbon atom is 1.99 x 10-23 g but to avoid using terms like 10-23 scientists have defined a smaller unit of mass called the _ atomic mass unit which is abbreviated amu. 1 amu = 1.66 x 10-24 g (Do not memorize or use this number) 12 6 The isotopes of carbon are C, C, and C. Any sample of carbon contains a mixture of these isotopes, always in the same proportions. Because each of these isotopes is has a slightly different mass, we need to use an average mass for the carbon atoms. The average atomic mass for carbon atoms is 12.01 amu. 13 6 14 6 This means that any sample of carbon from nature can be treated as though it were composed of identical carbon atoms, each with a mass of 12.01 amu. Mass of 1000 natural carbon atoms = (1000 atoms) x 12.01 amu _ Atom = 12,010 amu = 1.201 x 10 4 amu Calculate the mass in amu of a sample of aluminum that contains 75 atoms. 1 Al atom = 26.98 amu 75 Al atoms x 26.98 amu = 2024 amu 1 Al atom Calculate the number of sodium atoms present in a sample that has a mass of 1172.49 amu. 1 Na atom = 22.99 amu 1172.49 amu x 1 _ Na atom = 51.00 Na atoms 22.99 amu
6.3 The Mole Objectives: To understand the mole concept and Avogadro s number. To learn to convert among moles, mass and number of atoms in a given sample. What mass of copper contains exactly the same number of atoms as this sample of aluminum? 26.98 g Al ---------------------------------------------------? grams copper Contains the same number of atoms 26.98 g Al = 26.98 amu 63.55 g Cu 63.55 amu The number of atoms present in these samples assumes a special importance in chemistry. It is called the mole. The mole (abbreviated _ mol ) can be defined as the number equal to the number of atoms in 12.01 grams of carbon. The number is called Avogadro s number. One mole of something consists of 6.022 x 10 23 units of that substance. 1 dozen eggs is 12 eggs. 1 mole of eggs is 6.022 x 10 23 eggs. 1 mole of water contains 6.022 x 10 23 H 2 O molecules. 1 mole of marbles is enough to cover the entire earth to a depth of 50 miles. A 12.01 g sample of carbon contains 6.022 x 10 23 carbon atoms. A 1.008 g sample of hydrogen contains 6.022 x 10 23 hydrogen atoms. A 26.98 g sample of aluminum contains 6.022 x 10 23 aluminum atoms.
Summary: A sample of an element with a mass equal to that element s average atomic mass expressed in grams contains 1 mole of atoms. Example 6.3: Aluminum is often used for structures such as high-quality bicycle frames. Compute both the number of moles of atoms and the number of atoms in a 10.0 g sample of aluminum. 10.0 g Al =? moles of Al atoms 1 mol Al = 26.98 g Al 10.0 g Al x 1 mol Al = 0.371 mol Al 26.98 g Al 6.022 x 10 23 Al atoms = 1 mol Al atoms 0.371 mol Al x 6.022 x 10 23 _ Al atoms = _ 2.23 x 10 23 Al atoms 1 mol Al Example 6.4: A silicon chip used in an integrated circuit of a microcomputer has a mass of 5.68 mg. How many silicon (Si) atoms are present in this chip? The average atomic mass for silicon is 28.09 amu. Strategy to solve : Milligrams of Si atoms Grams of Si atoms Moles of Si atoms Number of Si atoms 5.68 mg Si x 1 g Si _ = 5.68 x 10-3 g Si 1000 mg Si 1 mol Si atoms = 28.09 g Si 5.68 g x 10-3 g Si x 1 mol Si _ = 2.02 x 10-4 mol Si 28.09 g Si 2.02 x 10-4 mol Si x 6.022 x 10 23 atoms _ = 1.22 x 10 20 Si atoms 1 mol Si 6.4 Molar Mass Objectives: To understand the definition of molar mass. To learn to convert between moles and mass of a given sample of a chemical compound.
How can we calculate the mass of 1 mol of methane (CH 4 )? That is, what is the mass of 6.022 x 10 23 CH 4 molecules? 1 mol of CH 4 molecules (6.022 x 10 23 CH 4 molecules) contains 1 mol C atoms (6.022 x 10 23 C atoms) 4 mol H atoms (6.022 x 10 23 H atoms) Mass of 1 mol of C = 1 x 12.01 g = 12.01 g Mass of 4 mol of H = 4 x 1.008 g = + 4.032 g Mass of 4 mol of CH 4 = 16.04 g 16.04 g/mol is called the _ molar mass for methane: The mass of 1 mol of CH 4 molecules. Example 6.5: Calculating Molar Mass Calculate the molar mass of sulfur dioxide. Mass of 1 mol of S = 1 x 32.07 = 32.07 g Mass of 2 mol of O = 2 x 16.00 = + 32.00 g Mass of 1 mol of SO 2 = 64.07 g = molar mass The molar mass of SO 2 is 64.07 g/mol. It represents the mass of 1 mol of SO 2 Some substances exist as a collection of ions rather than as separate molecules. For example, NaCl is composed of an array of Na+ and Cl- ions. There are no molecules. In some books the term _ formula weight is used. We will continue to use the term molar mass instead. Molar mass of NaCl = Mass of 1 mol of Na+ = 22.99 g Mass of 1 mol of Cl- = Mass of 1 mol of NaCl = Example 6.6: Calculating Mass from Moles + 35.45 g A. Calculate the molar mass of calcium carbonate. 58.44 g = molar mass Mass of 1 mol of Ca 2+ = 1 x 40.08 g = 40.08 g Mass of 1 mol CO 3 2- (contains 1 mol of C and 3 mol of O): 1 mol C = 1 x 12.01 = 12.01 g 3 mol O = 3 x 16.00 = 40.08 g Mass of 1 mol of CaCO 3 100.09 g/mol = molar mass B. A certain sample of calcium carbonate contains 4.86 mol. What is the mass in grams of this sample? 4.86 mol CaCO 3 x 100.09 g CaCO 3 = 486 g CaCO 3
1 mol CaCO 3 Example 6.7: Calculating Moles from Mass A. Calculate the molar mass of juglone (C 10 H 6 O 3 ). Mass of 10 mol of C = 10 x 12.01 g = 120.1 g Mass of 6 mol of H = 6 x 1.008 g = 6.048 g Mass of 3 mol of O = 3 x 16.00 g = 48.00 g Mass of 1 mol of C 10 H 6 O 3 = 174.1 g/mol = molar mass B. A sample of 1.56 g of pure juglone was extracted from black walnut husks. How many moles of juglone does this sample represent? 1 mol = 174.1 g juglone 1.56 g juglone x 1 mol juglone = 0.00896 mol juglone 174.1 g juglone = 8.96 x 10-3 mol juglone Example 6.8: Calculating Number of Molecules Isopentyl acetate, C 7 O 2, the compound responsible for the scent of bananas, can be produced commercially. Interestingly, bees release about 1 μg (1 x 10-6 g ) of this compound when they sting. This attracts other bees which then join the attack. How many molecules of isopentyl acetate are released in a typical bee sting? 7 mol C x 12.01 g = 84.07 g C 1 mol 14 mol H x 1.008 g = 14.11 g H 1 mol 2 mol O x 16.00 g = + 32.00 g O 1 mol 130.18 g/mol 1 mol of isopentyl acetate (6.022 x 10 23 molecules) has a mass of 130.18 g. 1 x 10-6 g C 7 O 2 x 1 mol C 7 O 2 = 8 x 10-9 mol C 7 O 2
130.18 g C 7 O 2 8 x 10-9 mol C 7 O 2 x 6.022 x 10 23 molecules = 5 x 10 15 molecules 1 mol C 7 O 2 This very large number of molecules is released in each bee sting.