Question

Combustion of a 0.9835-g sample of a compound containing only carbon, hydrogen, and oxygen produced 1.900 g of co2 and 1.070 g of h2o. what is the empirical formula of the compound?

Answer 1

The empirical formula of the compound is determined by calculating moles of carbon and hydrogen from combustion products CO2 and H2O, followed by a stoichiometric conversion to the simplest whole number ratio.

Step-by-step explanation:

To determine the empirical formula of the compound from the given combustion data, we start by calculating the molar amounts of carbon (C) and hydrogen (H) we have in the product. Since we have 1.900 g of CO2, we divide this by the molar mass of CO2 (44.01 g/mol) to find the moles of carbon. A similar calculation for the 1.070 g of H2O yields the moles of hydrogen by dividing by its molar mass (18.015 g/mol). Each mole of CO2 provides one mole of carbon, and each mole of H2O provides two moles of hydrogen.

Answer 2

Find grams of C and H, using molar masses:
12.0 g C x (1.900 g CO2 / 44.0 g CO2) = 0.5182 grams C
2.02 g H x (1.070 g H2O /18 g H2O) = 0.1201 grams H

Since all the C and H in CO2 and H2O will come from the sample except oxygen, because we need to provide oxygen for sample to burn.

0.9835 g sample - 0.5182 g O - 0.1201 g H = 0.3452 grams O


find moles, using molar masses:
0.5182 grams C / 12.0 g/mol C = 0.0432 moles C
0.1201 grams H / 1.01 g/mol H = 0.119 moles H
0.3452 grams O / 16.0 g/mol O = 0.0216 moles O

0.0216 is lesser, so use it to normalize to find the ratio.
find ratios:
0.0432 moles C / 0.0216 = 2 moles C
0.119 moles H / 0.0216 = 5.5 moles H
0.0216 moles O / 0.0216 = 1 mole O


so, the ratio is C2 H{5.5} O1
double the ration to eliminate decimals.

gives your answer is
`C_4 H_(11) O_2`