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winter 2011

# winter 2011 - Midterm 1 Chemistry 14A Dr Lavelle 1 Exactly...

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Midterm 1 Chemistry 14A Dr. Lavelle 1. Exactly 4.35 g of compound X are put into a combustion chamber. Combustion yields 9.52 g (ERROR SHOULD READ 10.47 g) carbon dioxide and 4.82 g water. (a) What is the empirical formula of compound X? (3 points) Answer: First, convert grams CO 2 and H 2 O to moles C and H (1 point) 10.47 g CO 2 x(1 mol CO 2 / 44.01 g CO 2 )= 0.2379 moles CO 2 x (1 mol C / 1 mol CO 2 ) = 0.2379 moles C (if multiplied by 12.01 g C mol -1 = 2.857 g C) 4.82 g H 2 Ox(1 mol H 2 O / 18.016 g H 2 O)= 0.2675 moles H 2 O x(2 mol H / 1 mol H 2 O) = 0.5350 moles H (if multiplied by 1.0079 g H mol -1 = 0.5392 g H) Second, find moles of oxygen in compound X (1 point) 0.2379 moles C (12.01 g C mol -1 ) = 2.857 g C 0.5350 moles H (1.0079 g H mol -1 ) = 0.5392 g H Recognize that these masses do not add up to 4.35 and another element is present. Since only CO 2 and H 2 O were reported as products, missing element is oxygen. 4.35 g Compound X – 2.857 g C – 0.5392 g H = 0.9538 g O (1 mol O / 16.0 g O) = 0.0596 mol O Third, find the molar ratios of C, H, and O in whole numbers (i.e. empirical formula) (1 point) C: 0.2379 moles C / 0.0596 = 3.991 ~ 4 H: 0.5350 moles H / 0.0596 = 8.977 ~ 9 O: 0.0596 moles O / 0.0596 = 1.000 Therefore, empirical formula is C 4 H 9 O .

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winter 2011 - Midterm 1 Chemistry 14A Dr Lavelle 1 Exactly...

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