Question #ae4f9

1 Answer
Jun 6, 2015

Your compound's empirical formula will be #C_4H_11O_2#.

So, you know that your compound contains only carbon, hydrogen, and oxygen. Its combustion reaction will only produce carbon dioxide, #CO_2#, and water, #H_2O#.

This means that all the carbon that the compound contained will now be a part of the carbon dioxide, and all the hydrogen that the compound contained is now a part of the water.

Use carbon dioxde's molar mass to determine how many moles your reaction produced

#1.900cancel("g") * "1 mole"/(44.0cancel("g")) = "0.04318 moles"# #CO_2#

SInce every 1 mole of #CO_2# contains 1 mole of carbon, you compound contained 0.04318 moles of carbon. Use carbon's molar mass to see how many grams you had

#0.04318cancel("moles C") * "12.0 g"/(1cancel("mole C")) = "0.5182 g C"#

Do the same for water.

#1.070cancel("g") * "1 mole"/(18.02cancel("g")) = "0.95938 moles"# #H_2O#

This time, 1 mole of water contains 2 moles of hydrogen so you have

#0.5938cancel("moles"H_2O) * "2 moles H"/(1cancel("mole"H_2O)) = "0.1188 moles H"#

Use hydrogen's molar to see how many grams you had

#0.1188cancel("moles H") * "1.01 g"/(1cancel("mole H")) = "0.1199 g H"#

Use the total mass of the sample to determine how much oxygen you had

#m_"sample" = m_(C) + m_(O) + m_(H)#

#m_(O) = 0.9835 - 0.5182 - 0.1199 = "0.3454 g O"#

Since you already know how many moles of carbon and hydrogen the sample contained, use oxygen's molar masses to calculate how many moles you had

#0.3454cancel("g") * "16.0 g"/(1cancel("mole")) = "0.02159 moles O"#

Divide the number of moles of each element by the smallest of the group to determine the mole ratios that existed in the sample

#"For C": (0.04318cancel("moles"))/(0.02159cancel("moles")) = 2#

#"For H": (0.1188cancel("moles"))/(0.02159cancel("moles")) = 5.5#

#"For O": (0.02159cancel("moles"))/(0.02159cancel("moles')) = 1#

Your sample looked like this

#C_2H_5.5O_1#

Since you can't have fractional subscripts in the empirical formula, multiply all by 2 to get

#C_4H_11O_2# #-># your compound's empirical formula.