c161l7_09_17

# c161l7_09_17 - Combustion analysis Combustion analysis W...

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Unformatted text preview: Combustion analysis Combustion analysis W Remember empirical formula? The masses used for that come from combustion analysis W Mainly used for organic compounds Products of combustion CO 2 , H 2 O and N 2 We will mostly ignore the N 2 Empirical formula from Empirical formula from combustion analysis combustion analysis 1.621g of a compound is analyzed by combustion analysis. 1.902 g of water and 3.095 g of CO 2 were found. Determine the empirical formula. 3.095g CO 2 × 1mol/44.01g × 1 mol C/1 mol CO 2 = 0.07032 mol C 1.902g H 2 O × 1 mol/18.02g × 2 mol H/1 mol H 2 O = 0.2111 mol H Oxygen must be found by difference (mass sample - mass of C - mass of H) 1.621g 1.621g - (0.07032 (0.07032 ×12.01) 12.01) - (0.2111 (0.2111 ×1.008) = 0.564g 1.008) = 0.564g mols of O = 0.564/16 = 0.0352 W C:H:O::0.0703:0.2111:0.0352 W Divide by smallest number 2:6:1 W Empirical formula C 2 H 6 O Symbolizing chemical reactions Symbolizing chemical reactions We use chemical equations to symbolize reactions.reactions....
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c161l7_09_17 - Combustion analysis Combustion analysis W...

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