Chapter 3 - Chapter 3 Stoichiometry Will consider the...

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Chapter 3 Stoichiometry Will consider the quantities of materials consumed or produced in chemical reaction. 2.1 Atomic Masses: The system of atomic masses was made based on 12 C as the standard of exactly 12 atomic mass units (amu). The most accurate method for comparing the masses of atoms is mass spectrometer: (See Figure 3.1) Natural carbon is a mixture of the isotopes 12 C (98.89 %), 13 C (1.11 %) and 14 C (negligibly small percentage at this level of precision): 0836129 . 1 12 13 = C Mass C Mass Mass of 13 C = (1.0836129)(12 amu) = 13.003355 amu The reported atomic mass of carbon is an average value: ) )( 100 % ( ) )( 100 % ( 13 13 12 12 C mass C C mass C + = = (0.9889)(12 amu) + (0.0111)(13.0034 amu) = 12.01 amu
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Example 3.1 : For a sample of copper Cu : Mass of 63 Cu = 62.93 amu (69.09 %) Mass of 65 Cu = 64.93 amu (30.91 %) Calculate average mass of Cu Answer : Average mass of Cu = (0.6909)(62.93) + (0.3091)(64.93) = 63.55 amu 3.2 The mole : One mole of an element is its atomic mass grams and contains avogadro’s number (6.022 x 10 23 ) atoms of this element Mole = (mass g)/(atomic mass g/mol) MM m n = 3.3 Molar Mass: Molar mass of a compound is the sum of the atomic masses of the component atoms n m MM = 2
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Example 3.2 : Compute the mass in grams of a sample of americium (243 g/mole) containing six atoms. Answer : 243 g/mol contains 6.022 x 10 23 atom/mol m g (?) in 6 atoms m = (6 atoms x 243 g mol -1) /(6.022 x 10 23 atoms g -1 ) = 2.42 x 10 -21 g Example 3.5 : a) Calculate number of moles (n) in a sample of cobalt containing 5.00 x 10 20 atoms and b) The mass of the sample in g when the atomic mass is 58.93 g/mol Answer : a) 1 mole contains 6.022 x 10 23 atom/mol n (?) in 5.00 x 10 20 atoms n = (5.00 x 10 20 atom x 1 mol)/(6.022 10 23 atom mol -1 ) = 8.30 10 -4 mol b) MM m n = m = n x MM = 8.30 x 10 -4 mol x 58.93 g/mol = 4.89 x 10 -2 g of Co 3
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Example 3.6 : For C 10 H 6 O 3 calculate: a) Molar mass b) Moles of 1.56 x 10 -2 g Answer : a) MM = (10 x 12.01) + (6 x 1.008) + (3 x 16.00) = 174.10 g/mol b) n = m / MM = 1.56 x 10 -2 g/174.1 g mol -1 = 8.96 x 10 -5 mol Example 3.7 : For calcium carbonate CaCO 3 (calcite) Calculate: a) Molar mass b) Mass of 4.86 moles c) Mass of CO 3 2- ions present Answer : a) MM = (1 x 40.08) + (1 x 12.01) + (3 x 16.00) = 100.09 g/mol b) m = n x MM = 4.86 mol x 100.09 g/mol = 486 g CaCO
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This note was uploaded on 10/25/2010 for the course CHEMISTRY 2010 taught by Professor Thwapiah during the Fall '10 term at Hashemite University.

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Chapter 3 - Chapter 3 Stoichiometry Will consider the...

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