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CHEM 111 (69)
Lecture

Determining empirical formulae, molecular formulae, combustion analysis, electronegativity

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Department
Chemistry
Course
CHEM 111
Professor
Dana Johnson
Semester
Fall

Description
8 October Review from Wednesday Information in a formula (e.g. AlCl )3 Ratios of atoms to each other (1:3) Number of moles of Cl in one mole of AlCl 3 Can determine mass percent of Cl Mass of Cl / mass of AlCl3 * 100 Use to determine Cl mass of any sample of AlCl 3 Clicker challenge A 0.371 mole sample of a metal oxide (M O ) w2i3hs 55.4 g. What element is represented by M? M 2 =30.371 moles O = 0.371 (3) = 1.11 mol M = 0.371 (2) = 0.742 mol 1.11 mol of O x 16.00 g / 1 mole = 17.76 g 55.4 g – 17.76 g = 37.64 g 37.64 g / 0.742 mol = 50.72 g / mol M is V Determining empirical formulae Remember, compounds do not vary in composition Formula for glucose → C H O6 12 6 In 1 molecule of glucose, there are 6 carbon atoms In 1 mole of glucose, there are 12 moles of hydrogen atoms Subscripts tell us number (not weight) Empirical formulae can be determined from elemental analysis Compound is decomposed into elements 1. Convert weights to moles 2. Insert moles into a preliminary formula 3. Divide all moles by the lowest number 4. Multiply all values (if necessary) by an integer to get whole numbers Zn 13.74 g x (1 mol / 65.39 g) = 0.21 mol P 4.34 g x (1 mol / 30.97 g) = 0.14 mol O 8.96 g x (1 mol / 16.00 g) = 0.56 mol Zn 0.21 / 00.14 / 00.56 / 0.14 Zn 1.51x 4 Zn 3 2 =8Zn
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