First we assume that the compound containing only C,H,and O is combusted completely in the presence of excess oxygen, so that the only things that can be produced are water and carbon dioxide.
From there we should back calculate the amount of Hydrogen that is in the original sample by taking all of the hydrogen in the 0.239g to came from the organic compound.
And since we know that the original mass of the sample was .100g, we can also easily get a mass % H by taking the mass Hydrogen calculated over the total original mass (.100 g)
So that:
0.239g H2O / (18.01 g/mol) = .01327 moles H20
.01327 Moles H20 * 2.02g H (per every mole H2O) = .0268g H initially present in the sample
.0268g H / .100g sample = 26.8% H by mass
The mass percent of hydrogen in the 0.1 g sample of compound containing carbon–hydrogen–oxygen is 26.6%
We'll begin by calculating the mass of the hydrogen in the compound. This can be obtained as follow:
Molar mass of H₂O = (2×1) + 16 = 18 g/mol
Molar mass of H₂ = 2 × 1 = 2 g/mol
Mass H₂O produced = 0.239 g
Mass of H = [tex]\frac{2}{18} * 0.239[/tex]
Finally, we shall determine the mass percent of Hydrogen in the compound. This can be obtained as follow:
Mass of H = 0.0266 g
Mass of compound = 0.1
[tex]Percentage = \frac{mass}{mass of compound} * 100\\\\= \frac{0.0266}{0.1} * 100[/tex]
Thus, the mass percent of hydrogen in the sample is 26.6%
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