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abruzzese [7]
3 years ago
13

Complete combustion of 7.20 g of a hydrocarbon produced 22.1 g of co2 and 10.5 g of h2o. what is the empirical formula for the h

ydrocarbon?
Chemistry
1 answer:
jek_recluse [69]3 years ago
5 0
<span>C3H7 The unknown compound consists of only carbon and hydrogen. The oxygen comes from the air. So you need to first determine the relative moles of hydrogen and carbon that are present in the CO2 and H2O. First, look up the molar masses of Carbon, Hydrogen, and Oxygen. Then determine the molar masses of CO2 and H2O. Carbon = 12.0107 Hydrogen = 1.00794 Oxygen = 15.999 Molar mass of CO2 1 * 12.0107 + 2 * 15.999 = 44.0087 Molar mass of H2O 2 * 1.00794 + 1 * 15.999 = 18.01488 Number of moles of CO2 22.1 g / 44.0087 g/mol = 0.502173 Number of moles of H2O 10.5 g / 18.01488 g/mol = 0.582852 Since there's 1 carbon atom per CO2 molecule, there are 0.502173 moles of carbon. Since there's 2 hydrogen atoms per H2O molecule, there are 2 * 0.582852 = 1.165703 moles of hydrogen. Now we need to find a simple ratio of small integers that comes close to the ratio of 0.502173 / 1.165703 = 0.43079 to determine the empirical formula. 3/7 = 0.428571, an error of only 0.002219. The next closest ratio has an error of 0.013654, over 6 times larger. So the empirical formula is C3H7</span>
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Ludmilka [50]

Answer: 466.67 ml

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}   (At constant temperature and number of moles)

Thus

P_1V_1=P_2V_2  [according to Boyle's law]

742torr\times 500ml=795torr\times V_2

V_2=466.67ml


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