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An unknown compound contains only C, H, and O.  Combustion of 2.50 g of this compound produced 6.10 g of CO2 and 2.50 g of H2O.  What is the empirical formula of the unknown compound?

Oct 2nd, 2014

The combustion yields the products, CO2 and H2O. Since all of the carbon in the unknown compound will go to CO2, we can calculate the amount of carbon in the 2.5g sample:

m(Carbon) = 12/44 * 2.5 = 0.6818g

Since all of the hydrogen in the unknown compound will go to H2O, we can calculate the amount of hydrogen in the 2.5g sample:

m(Hydrogen) = 1.01/18.02 * 2.5 = 0.1401g

Therefore, the mass of Oxygen in the sample must be equal to 2.5 - 0.6818 - 0.1401 = 1.6781g.

Now to calculate mole ratios:

Carbon : 0.6818/12 = 0.05682    Hydrogen: 0.1401/1.01 =0.1387  Oxygen: 1.6781/16 = 0.10488

Dividing each by 0.05682, ration is C:H:O = 1: 2.44: 2 which is approx 1:2.5:2

So empirical formula is C2H5O4

Oct 2nd, 2014

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Oct 2nd, 2014
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Oct 2nd, 2014
Sep 20th, 2017
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