Question

Question #ae4f9

Answer 1

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.