1. An 8.00 mg sample of an organic compound was
subjected to quantitative combustion to give 15.33 mg
of carbon dioxide and 9.36 mg of water. The
molecular weight was found to be 46. Calculate the %
composition, empirical and molecular formula of the
organic compound.
Let's say we had a 100 gram sample of this compound. The sample would therefore contain 73.9 grams of mercury and 26.1 grams of chlorine. How many moles of each atom do the individual masses represent?
For Mercury:
(73.9g)×(1mol
200.59g
)=0.368moles
(73.9g)×(1mol200.59g)=0.368moles
For Chlorine:
(26.1g)×(1mol
35.45g
)=0.736mol
(26.1g)×(1mol35.45g)=0.736mol
What is the molar ratio between the two elements?
0.736molCl
0.368molHg
=2.0
0.736molCl0.368molHg=2.0
Thus, we have twice as many moles (i.e. atoms) of Cl
Cl as Hg
Hg. The empirical formula would thus be (remember to list cation first, anion last):
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