A 0.2420 sample of an unknown containing C, O, and H is burned completely. Analysis shows that 0.2265 grams of water are produced. How many grams of hydrogen were present in the 0.2420 grams of sample?

Question

anonymous · Answer

$$a C_{x}H_{y}{O_{z}} + b O_{2} \rightarrow c CO_{2} + d H_{2}O$$,
This is the general complete combustion (burning) reaction, where
a, b, c, d represent the coefficients of the reactants and products, and
x, y, z represent the subscripts of the unknown sample molecular formula.

If 0.2265 g of water (molar mass = H+H+O ~18.02) were formed, the amount of moles is:
$$(0.2265)/(18.02) = 0.01257$$
You should see that for every molecule of water, there are two hydrogen atoms.
So, given 0.01257 moles of water molecules, there are 0.02514 moles of hydrogen atoms.
Converting this amount to grams (multiply by hydrogen molar mass, 1.00794) yields:
$$0.02514*1.00794=0.02515$$
Since a nuclear reaction is not happening, we can safely assume the amount of hydrogen in the reactants is equal to the amount of hydrogen in the products (Law of Conservation of Matter).
Your answer should be 0.02515 grams of hydrogen present in the original unknown sample.