# Question ae4f9

Jun 6, 2015

Your compound's empirical formula will be ${C}_{4} {H}_{11} {O}_{2}$.

So, you know that your compound contains only carbon, hydrogen, and oxygen. Its combustion reaction will only produce carbon dioxide, $C {O}_{2}$, and water, ${H}_{2} O$.

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" $C {O}_{2}$

SInce every 1 mole of $C {O}_{2}$ 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" ${H}_{2} O$

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

${m}_{\text{sample}} = {m}_{C} + {m}_{O} + {m}_{H}$

${m}_{O} = 0.9835 - 0.5182 - 0.1199 = \text{0.3454 g O}$

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#

${C}_{2} {H}_{5.5} {O}_{1}$
${C}_{4} {H}_{11} {O}_{2}$ $\to$ your compound's empirical formula.