Chapter3Lecture

Chapter3Lecture - Chapter 3: Finding an Empirical Formula...

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Unformatted text preview: Chapter 3: Finding an Empirical Formula More difficult is finding the compound when knowing the masses or percentages. If given percentages, just assume 100 g total. Given: 43.40% Na 11.32% C 45.28% O 43.40 g Na or 43.40 g 11.32 g 45.28 g 1 mole Na 1.888 mole Na 22.99 g Na 1mole C 11.32 g C 0.9433 mole C 12.01g C 45.28 g O 1 mole O 2.830 mole O 16.00 g O Therefore: Na 1.888 C0.9433 O2.830 We must have integers. All seem to be a bit low. So: Na 2 C O3 or Na2(CO3) We have sodium carbonate. Copyright: 2010 Prof. Magde 6 8.377 g Fe 3.600 g O 1 mole Fe 0.1500 mole Fe 55.847 g Fe 1mole O 0.2250 mole O 16.00 g O ———————————————————————— Another: Cumere is a hydrocarbon that is 89.94% C by mass, and has a molar mass of 120 g Find the formula. Assume 100 g: 89.94 g C and 10.06 g of H Convert to moles: 7.49 mole C and 9.98 mole H Looks like 7.5 of C and 10 of H, but we need whole numbers, so it must be 3 C + 4 H or a multiple of that. The Mm of C3 H4 would be 3(12.01) + 4(1.008) = 40 g mol-1 So cumene has 3 units and is C9 H12 !! ...
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This note was uploaded on 01/12/2011 for the course CHEM 6A taught by Professor Pomeroy during the Fall '08 term at UCSD.

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Chapter3Lecture - Chapter 3: Finding an Empirical Formula...

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