Takenote9 - 1.5279 g H 100%= 10.413%H 14.673 g — 100...

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Unformatted text preview: - 1.5279 g H * 100%= 10.413 %H 14.673 g — 100 — (62.040 + 10.413) = 27.57 % 0 Step 5. Assume the sample size (usually 100 grams) Step 6. Convert grams of each element to moles - 62.040 g C (11101 C) 2 5.1653 mol C 12.011 g — 10413 g H (mol H) = 10-331 molH 1.00794 g - 27.547 g 0 (mol 0) = 1.7218 mol 0 15.994 g Step 7. Write the chemical formula using the calculated amounys as subscripts. Divide each by the smallest mole amount. If necessary multiply each subscript by a small integer to achieve the Whole numbers. The result is the empirical formula 95.1653 Emmi 01.7218 1.7218 1.7218 1.7218 : C2.9999H6.00010 : C3H50 (Empirical formula for acetone) 11. Oxidation reactions: A. (Assigning oxidation numbers) , Rule 1. An atom in its elemental state has oxidation number of 0. Rule 2. For monotonic ions, the oxidation number is equal to the charge. Rule 3. Fluorine always has oxidation number as -1. Rule 4. Oxygen has oxidation number -2 except in peroxides where it is -1. Rule 5. Hydrogen has oxidation number of +1 except in hydride where it is —1. Rule 6. The sum of the oxidation numbers is equal to the charge of the molecule or ion. For example: 1. CO (carbon monoxide) O has —2. For sum to be zero C has to have + III. Nomenclature A. Ionic compounds. 1. a. LiF — lithium fluoride b. K20 — potassium oxide 2. For elements with different oxidation states a. CuBr - cuprous oxide b. (3ng — cupric bromide ...
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