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Recall that in combustion analysis, a compound reacts with excess O2 to form products.
For a compound composed of C, H, and O, the reaction looks like this:
CxHyOz + O2 (excess) → x CO2 + y H2O
Calculate the mass of C and H.
molar mass of CO2 = 44.01 g/mol
There is 1 mole of C in 1 mole of CO2. Finding the mass of C:
mass of C = 0.4367 g C
molar mass of H2O = 18.02 g/mol
There are 2 moles of H in 1 mole of H2O. Finding the moles of H:
mass of H = 0.0734 g H
Calculate the mass of O:
We have 0.8009 g of the sample compound. Subtracting the masses of C and H from this will give us the mass of O:
mass of O = 0.8009 g – (0.4367 g C + 0.0734 g H) = 0.2908 g O
Calculate the moles of each element and determine the lowest whole number ratio.
moles of C = 0.0363 mol C
Upon combustion, a 0.8009 g sample of a compound containing only carbon, hydrogen, and oxygen produces 1.6003 g CO2 and 0.6551 g H2O. Find the empirical formula of the compound.
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Based on our data, we think this problem is relevant for Professor Altomare's class at UCF.