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CH3%20additional%20problem%20solutions

# CH3%20additional%20problem%20solutions - Additional...

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Additional problems for Chang CH 3 material on analysis of mixtures and general questions on Stoichiometry K4.36 (analysis of mixtures) The aluminum in a 0.764 g sample of an unknown material was precipitated as aluminum hydroxide, Al(OH) 3 , which was then converted to Al 2 O 3 by heating strongly. If 0.127 g of Al 2 O 3 is obtained from the 0.764 g sample, what is the mass percent of aluminum in the sample?

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Additional problems for Chang CH 3 material on analysis of mixtures and general questions on Stoichiometry K4.36 (analysis of mixtures) The aluminum in a 0.764 g sample of an unknown material was precipitated as aluminum hydroxide, Al(OH) 3 , which was then converted to Al 2 O 3 by heating strongly. If 0.127 g of Al 2 O 3 is obtained from the 0.764 g sample, what is the mass percent of aluminum in the sample? 0.127 g Al 2 O 3 · 1 mol Al 2 O 3 102.0 g · 2 mol Al 1 mol Al 2 O 3 · 26.98 g 1 mol Al = 0.0672 g Al 0.0672 g 0.764 g · 100% = 8.79% Al
K4.61 (general questions on Stoichiometry) Menthol, from oil of mint, contains only C, H, and O. If 95.6 mg of menthol burns completely in O 2 , and gives 269 mg of CO 2 and 110 mg of H 2 O, what is the empirical formula of menthol?

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K4.61 (general questions on Stoichiometry) Menthol, from oil of mint, contains only C, H, and O. If 95.6 mg of menthol burns completely in O 2 , and gives 269 mg of CO 2 and 110 mg of H 2 O, what is the empirical formula of menthol? 269 mg CO 2 · 1 g 10 3 mg · 1 mol CO 2 44.01 g · 1 mol C 1 mol CO 2 = 0.00611 mol C 0.00611 mol C · 12.01 g 1 mol C = 0.0734 g C 110 mg H 2 O · 1 g 10 3 mg · 1 mol H 2 O 18.0 g · 2 mol H 1 mol H 2 O = 0.012 mol H 0.012 mol H · 1.01 g 1 mol H = 0.012 g H mass of O = sample mass – mass of C – mass of H = (95.6 mg · 1 g 10 3 mg ) – 0.0734 g – 0.012 g
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CH3%20additional%20problem%20solutions - Additional...

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