A0.100 g sample of a carbon–hydrogen–oxygen compound is combusted in a stream of pure oxygen and produces 0.220 g co2 and 0.239 g h2o. what is the mass percent of hydrogen in the sample? a.13.3

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

answer is: empirical formula is tl₂o. if we use 100 grams of compound, tha:

m(tl) = 96.2%  · 100 g ÷ 100%.

m(tl) = 96.2 g.

n(tl) = m(tl)  ÷ m(tl). n(tl) = 96.2 g  ÷ 204.4 g/mol.

n(tl) = 0.47 mol. m(o) = 3.77 g. n(o) = m(o)  ÷ m(o). n(o) = 3.77 g  ÷ 16 g/mol. n(ag) = 0.235 mol. n(tl) . n(o) = 0.47 mol : 0.235 mol /0.235. n(tl) . n(o) = 2 : 1.

m(tl₂o) = 204.4 g/mol  · 2 + 16 g/mol = 424.8 g/mol.

m(compound) = 456. 8 g/mol - 424.8 g/mol = 32 g/mol ( 2 oxygens); tl₂o₃.