1.44k likes | 1.93k Views
Unit 1: Stoichiometry. Chemistry 2202. Stoichiometry. Stoichiometry deals with quantities used in OR produced by a chemical reaction. 3 Parts. Mole Calculations (Chp. 2 & 3) Stoichiometry and Chemical Equations (Chp. 4) Solution Stoichiometry (Chp. 6). PART 1 - Mole Calculations.
E N D
Unit 1: Stoichiometry Chemistry 2202
Stoichiometry Stoichiometry deals with quantities used in OR produced by a chemical reaction
3 Parts Mole Calculations (Chp. 2 & 3) Stoichiometry and Chemical Equations (Chp. 4) Solution Stoichiometry (Chp. 6)
PART 1 - Mole Calculations Isotopes and Atomic Mass (pp. 43 - 46) Avogadro’s number (pp. 47 – 49) Mole Conversions (pp. 50 - 74) M, MV, NA, n, m, v, N
Questions p. 45 #’s 1 – 4 p. 46 #’s 1 – 6 p. 75 #’s 9 – 12 p. 51-53 #’s 5 – 15 p. 57 #’s 16 – 19 p. 59,60 #’s 20 – 27 p. 63,64 #’s 28 - 37 p. 54 #’s 5 – 8 p. 65 #’s 2, 4, 5 p. 75 #’s 13, 14 p. 76 #’s 15, 17–19, 21-23 p. 73 #’s 38 – 43 p. 74 #’s 1 – 4 p. 76 #’s 26, 27
PART 1 - Mole Calculations Percent composition: - given mass (p. 79 - 82) - given the chemical formula (p. 83 - 86) Empirical Formulas (pp. 87 - 94) Molecular Formulas (pp. 95 - 98) Lab: Formula of a Hydrate
Questions p. 82 #’s 1 – 4 p. 85 #’s 5 – 8 p. 89 #’s 9 – 12 p. 91 #’s 13 – 16 p. 97 #’s 17 - 20 p. 103 #’s 23 – 24 p. 86 #’s 1, 3 – 6 p. 94 #’s 1 - 7 p. 106 #’s 1 - 3, 6, 7 p. 107 – 109 #’s 5 – 23, 25
Isotopes and Atomic Mass atomic number - the number of protons in an atom or ion mass number - the sum of the protons and neutrons in an atom isotope - atoms which have the same number of protons and electrons but different numbers of neutrons
Isotopes and Atomic Mass 11 % 79 % 10 % not all isotopes are created equal
Isotopes and Atomic Mass atomic mass unit (AMU - p.43) a unit used to describe the mass of individual atoms the symbol for the AMU is u 1 u is 1/12 of the mass of a carbon-12 atom
Isotopes and Atomic Mass average atomic mass (AAM) the AAM is the weighted average of all the isotopes of an element (p. 45) p. 14 # 5 p. 45 #’s 1 – 4 p. 46 #’s 1 – 6 p. 75 #’s 9 - 12
Finding % Abundance eg. Br has two naturally occurring isotopes. Br-79 has a mass of 78.92 u and Br-81 has a mass of 80.92 u. If the AAM of Br is 79.90 u, determine the percentage abundance of each isotope.
Let x = fraction of Br-79 Let y = fraction of Br-81 x + y = 1 78.92x + 80.92y = 79.90 Finding % Abundance y = 1 - x
x + y = 1 78.92x + 80.92y = 79.90
Avogadro’s Number The MOLE is a number used by chemists to count atoms The MOLE is the number of atoms contained in exactly 12 g of carbon-12. In honor of Amedeo Avogadro, the number of particles in 1 mol has been called Avogadro’s number.
How big is Avogadro's number? An Avogadro's number of soft drink cans would cover the surface of the earth to a depth of over 200 miles. Avogadro's number of unpopped popcorn kernels spread across the USA, would cover the entire country to a depth of over 9 miles.
If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years to count the atoms in one mole. How big is Avogadro's number?
Avogadro’s Number 1 mole = 6.02214199 x 1023 particles 1 mol = 6.022 x 1023 particles NA = 6.022 x 1023 particles/mol
Avogadro’s Number 5 mol 3.011 x 1024 atoms 0.01 mol 6.022 x 1021 atoms 7.72 mol 4.65 x 1024 atoms 0.0133 mol 8.01 x 1021 atoms
Avogadro’s Number Formulas: N = n x NA n = # of moles N = # of particles (atoms, ions, molecules, or formula units) NA = Avogadro’s #
How many moles are contained in the following? 2.56 x 1028 Pb atoms 7.19 x 1021 CO2 molecules Avogadro’s Number
Avogadro’s Number eg. Calculate the number of moles in 4.98 x 1025 atoms of Al. eg. How many formula units of Na2SO4 are in 5.69 mol of Na2SO4? # of Na ions? # of Oxygen atoms?
Avogadro’s Number 1. How many molecules of glucose are in 0.435 mol of C6H12O6? How many carbon atoms? 2. Calculate the number of moles in a sample of glucose that has 3.56 x 1022 hydrogen atoms.
Avogadro’s Number pp. 51 – 53: #’s 5 – 15 p. 54: #’s 4 - 8
Molar Mass The mass of one mole of a substance is called the molar mass of the substance eg. 1 mole of Pb has a mass of 207.19 g 1 mole of Ag has a mass of 107.87 g
Molar Mass The symbol for molar mass is M and the unit is g/mol eg. MPb = 207.19 g/mol MAg = 107.87 g/mol
Molar Mass The molar mass of a compound is the sum of the molar masses of the elements in the compound eg. Calculate the molar mass of: a) H2O b) C6H12O6 c) Ca(OH)2
Molar Mass H2O has 2 hydrogens and 1 oxygen 2 x 1.01 = 2.02 1 X 16.00 = 16.00 18.02 g/mol
Molar Mass C6H12O6 6 x 12.01 = 72.06 12 x 1.01 = 12.12 6 x 16.00 = 96.00 180.18 g/mol
Molar Mass Ca(OH)2 1 x 40.08 = 40.08 2 x 16.00 = 32.00 2 x 1.01 = 2.02 74.10 g/mol Your calculator may not show the zeroes. There should be 2 digits after the decimal when adding molar masses
Molar Mass p. 57: #’s 16 – 19 & Molar Masses Handout 1. 151.92 g/mol 7. 58.44 g/mol 2. 120.38 g/mol 8. 100.09 g/mol 3. 286.19 g/mol 9. 44.02 g/mol 4. 100.40 g/mol 10. 248.22 g/mol 5. 74.44 g/mol 11. 115.04 g/mol 6. 78.01 g/mol
Molar Mass Calculations mass Avogadro’s # molar mass N = n x NA m = n x M
Molar Mass Calculations m = n x M N = n x NA
Molar Mass Calculations eg. How many moles are in 25.3 g of NO2? m = 25.3 g MNO2 = 46.01 g/mol
Molar Mass Calculations eg. What is the mass of 4.69 mol of water? n = 4.69 mol Mwater = 18.02 g/mol m = n x M = (4.69 mol)(18.02 g/mol) = 84.5 g
Molar Mass Calculations Practice: p. 59 #’s 20 - 23 p. 60 #’s 24 - 27
Molar Mass Calculations Practice: p. 63 #’s 28 - 33 p. 64 #’s 34 – 37 p. 76 # 15 x NA x M particles(N) moles (n) mass (m) ÷ M ÷ NA
Molar Mass Calculations eg. How many molecules are in 26.9 g of water? m = 26.9 g Mwater= 18.02 g/mol NA = 6.022 x 1023 molecules/mol Find N
Molar Mass Calculations N = n x NA = 1.493 X 6.022 x 1023 = 8.99 x 1023 molecules = 1.493 mol H2O
Molar Mass Calculations eg. How many molecules are in 4.78 g of glucose? m = 4.78 g Mwater= 180.18 g/mol NA = 6.022 x 1023 molecules/mol Find N 4:57 AM
Molar Mass Calculations N = n x NA = 1.493 X 6.022 x 1023 = 8.99 x 1023 molecules = 1.493 mol H2O 4:57 AM
Molar Mass Calculations eg. A sample of Sn contains 4.69 x 1028 atoms. Calculate its mass. N = 4.69 x 1028 NA = 6.022 x 1023 molecules/mol MSn = 118.69 g/mol Find m
Molar Mass Calculations m = n x M = 77881 mol x 118.69 g/mol = 9.24 x 10 6 g = 77,881 mol
Molar Mass Calculations Practice: p. 63 #’s 28 - 33 p. 64 #’s 34 – 37 p. 76 # 15