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ChemProblem

ChemProblem - 2 = R N If N = C we know immediately that β...

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Given: Lavoisier measured: H 2 = 0.02394 O 2 = 0.34211 Lavoisier H 2 /O 2 ratio (R L ): 0.069979538 However, when we use current values to reconstruct this ratio where: H 2 = (2 moles)(1.00794 g/mole) = 2.01588 O 2 = (2 moles)(15.9994 g/mole) = 31.9988 current H 2 /O 2 ratio (R C ): 0.062998612 We compare the two ratios, and find R L > R C . Either Lavoisier's H 2 is too heavy, or his O 2 is too light. The explanation for Lavoisier's error is his H 2 sample was not pure hydrogen; that the heavier air would account for the inflated R L value. So we know that Lavoisier's H 2 was really (H 2 + air). An equation demonstrating what we know: [α (H 2 ) + β (air)]/(O
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Unformatted text preview: 2 ) = R N If N = C, we know immediately that β = 0, and we have an expression for Rc . But, if N = R, we cannot be sure, on inspection, what α and β are (and this is the crux of the problem). So, we express one as a function of the other: = 1 – x α = amount of H 2 = x β = amount of air Now, equation reads: [(1 – x )(H 2 ) + (x)(air)]/(O 2 ) = R L = 0.069979538 Substituting values (H 2 ) = (2)*(1.00794) = 2.01588 (O 2 ) = (2)*(31.9988) = 31.9988 (air) = 29 Solving x yields, x = 0.008278248, which is 0.828% of air contaminant in Lavoisier's H 2 sample....
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