A pure organic substance (1.000 g) containing C, H, and O, undergoes combustion analysis. The mass of the water obtained is 0.6662 g. What is the mass percent of hydrogen in the compound? a. 3.699% b. 7.456%

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A pure organic substance (1.000 g) containing C, H, and O, undergoes combustion analysis. The mass of the water obtained is 0.6662 g. What is the mass percent of hydrogen in the compound? a. 3.699% b. 7.456%

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Answer 1 · 2015-02-24T01:32:57

The mass percent of hydrogen in the compound will be $7.456%$ $"7.456%"$ .

You're dealing with an unknown organic compound that weighs $1.000 g$ $"1.000 g"$ . After it undergoes combustion, you collect $0.6662 g$ $"0.6662 g"$ of water, $H_{2} O$ $H_2O$ . The key to this problem is to realize that all the hydrogen that was a part of the organic substance is now a part of the water produced.

Since water is comprised of two hydrogen atoms, you know that for every $18.0153 g$ $"18.0153 g"$ of $H_{2} O$ $H_2O$ , you get $2 \cdot 1.008 = 2.016 g$ $2 * 1.008 = "2.016 g"$ of hydrogen. This means that the mass of water produced will contain

$0.6662 g water \cdot \frac{2.016 g hydrogen}{18.0153 g water} = 0.07455 g$ $"0.6662 g water" * ("2.016 g hydrogen")/("18.0153 g water") = "0.07455 g"$ $H_{2}$ $H_2$

Like I've said, this is also how much hydrogen you had in the organic compound. Therefore, the mass percent of hydrogen in the original $1.000-g$ $"1.000-g"$ sample will be

$\frac{0.07455 g}{1.000 g} \cdot 100 = 7.455 % ≅ 7.456 %$ $"0.07455 g"/"1.000 g" * 100 = "7.455 %" ~= 7.456%$ - close enough to option b.

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A pure organic substance (1.000 g) containing C, H, and O, undergoes combustion analysis. The mass of the water obtained is 0.6662 g. What is the mass percent of hydrogen in the compound? a. 3.699% b. 7.456%

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