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# Chapter #3 Lecture Part 1[1]

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### Chapter #3 Lecture Part 1[1]

1. 1. Stoichiometry <ul><li>Stoichiometry = “Chemical Arithmetic” </li></ul><ul><li>Stoichiometry of Elements </li></ul><ul><li>Stoichiometry of Compounds </li></ul><ul><li>Stoichiometry of Chemical Reactions </li></ul>
2. 2. Atomic mass is the mass of an atom in atomic mass units (amu) Micro World atoms & molecules Macro World grams
3. 3. The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12.00 grams of 12 C
4. 4. Molar mass is the mass of 1 mole of in grams eggs shoes marbles atoms 1 mole 12 C atoms = 6.022 x 10 23 atoms = 12.00 g 1 12 C atom = 12.00 amu 1 mole 12 C atoms = 12.00 g 12 C 1 mole lithium atoms = 6.941 g of Li
5. 6. Lecture Problem <ul><li>Calculate the mass, in grams, of 4.50 moles of calcium atoms. </li></ul>
6. 7. Lecture Problem <ul><li>Calculate the number of moles of calcium atoms in 612 grams of calcium. </li></ul><ul><li>Calculate the number of calcium atoms in 612 grams of calcium. </li></ul>
7. 8. Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule. SO 2
8. 9. Molar Mass Problem <ul><li>Calculate the molar mass of sucrose, C 12 H 22 O 11 . </li></ul>
9. 10. Percent composition of an element in a compound = n is the number of moles of the element in 1 mole of the compound 52.14% + 13.13% + 34.73% = 100.0% n x molar mass of element molar mass of compound x 100% C 2 H 6 O %C = 2 x (12.01 g) 46.07 g x 100% = 52.14% %H = 6 x (1.008 g) 46.07 g x 100% = 13.13% %O = 1 x (16.00 g) 46.07 g x 100% = 34.73%
10. 11. Empirical Formula <ul><li>Identifies the elements present and shows the simplest whole-number ratio of the atoms in a substance. </li></ul>
11. 12. Data Utilized To Determine An Empirical Formula <ul><li>Weight of each element present in a given sample of the compound. </li></ul><ul><li>Weight percentage of each element in a compound. </li></ul><ul><li>Weight of products generated from a reaction involving the compound of interest. </li></ul>
12. 13. Empirical Formula Problem <ul><li>Determine the empirical formula of a compound containing 11.66 grams of iron and 5.01 grams of oxygen. </li></ul>
13. 14. Empirical Formula Problem <ul><li>Determine the empirical formula of a compound containing 62.1% carbon, 5.21% hydrogen, 12.1% nitrogen, and 20.7% oxygen. </li></ul>
14. 15. Molecular Formula <ul><li>Identifies the elements present and shows the exact number of atoms of each element present per molecule. </li></ul>
15. 16. Data Utilized To Determine A Molecular Formula <ul><li>Empirical Formula </li></ul><ul><li>& </li></ul><ul><li>Molar Mass of Compound </li></ul>
16. 17. Molecular Formula Problem <ul><li>Cadaverine, a foul smelling substance produced by the action of bacteria on meat, contains 58.55% carbon, 13.81% hydrogen and 27.40% nitrogen. The molar mass of the compound is 102.2 grams/mole. Determine the molecular formula of cadaverine. </li></ul>