Question

Question #b183e

Answer 1

Empirical formula: `"CH"`
Molecular formula: `"C"_6"H"_6`

Explanation:

Your strategy here will be to

• use the percent composition of water and of carbon dioxide to determine how much hydrogen and carbon were produced by the reaction

• use the molar mass of carbon and hydrogen to determine how many moles were present in the `"5.50-g"` sample

• use the molar mass of benzene to determine the molecular formula of the compound

Benzene is actually a hydrocarbon, i.e. a compound that contains only carbon and hydrogen, but let's assume that you don't know that.

You can prove it by adding up the mass of carbon and the mass of hydrogen produced by the reaction. If this number turns out to be equal to the initial mass of the sample, then you can say for a fact that benzene is a hydrocarbon.

So, every mole of water, `"H"_2"O"`, contains two moles of hydrogen and one mole of oxygen. This means that the percent composition of hydrogen in water will be

`(2 xx 1.00794 color(red)(cancel(color(black)("g/mol"))))/(18.015color(red)(cancel(color(black)("g/mol")))) xx 100 = "11.19% H"`

This means that every `"100 g"` of water will contain `"11.19 g"` of hydrogen.

In your case, `"3.81 g"` of water will contain

`3.81 color(red)(cancel(color(black)("g H"_2"O"))) * "11.19 g H"/(100color(red)(cancel(color(black)("g H"_2"O")))) = "0.4263 g H"`

Do the same for carbon dioxide, `"CO"_2`, for which every mole contains one mole of carbon and two moles of oxygen

`(1 xx 12.011 color(red)(cancel(color(black)("g/mol"))))/(44.01 color(red)(cancel(color(black)("g/mol")))) xx 100 = "27.29% C"`

In your case, `"18.59 g"` of carbon dioxide will contain

`18.59 color(red)(cancel(color(black)("g CO"_2))) * "27.29 g C"/(100color(red)(cancel(color(black)("g CO"_2)))) = "5.073 g"`

The two masses add up to

`m_"total" = "0.4263 g" + "5.073 g" = "5.4993 g" ~~ "5.50 g"`

So, you know that your compound is a hydrocarbon. Use the molar masses of the two elements to figure out how many moles of each you had in the initial sample

`"For C: " 5.073color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011 color(red)(cancel(color(black)("g")))) = "0.42236 moles C"`

`"For H: " 0.4263 color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.00794 color(red)(cancel(color(black)("g")))) = "0.42294 moles H"`

Divide these values by the smallest one to get the mole ratio that exists between the two elements in the compound

`"For C: " (0.42236 color(red)(cancel(color(black)("moles"))))/(0.42236 color(red)(cancel(color(black)("moles")))) = 1`

`"For H: " (0.42294 color(red)(cancel(color(black)("moles"))))/(0.42236 color(red)(cancel(color(black)("moles")))) = 1.004 ~~ 1`

The empirical formula of the compound, which tells you what the smallest whole number ratio that exists between the two elements is, will thus be

`"C"_1"H"_1`

Now, in order to determine the molecular formula, you need to first find the molar mass of the empirical formula. To do that, add the molar masses of each atom that makes up the empirical formula

`1 xx "12.011 g/mol" + 1 xx "1.00794 g/mol" = "13.019 g/mol"`

The molecular formula will always be a multiple of the empirical formula

`13.019 color(red)(cancel(color(black)("g/mol"))) xx color(blue)(n) = 78 color(red)(cancel(color(black)("g/mol")))`

`color(blue)(n) = 78/13.019 = 5.9912 ~~ 6`

The molecular formula of benzene will thus be

`("C"_1"H"_1)_color(blue)(6) = color(green)("C"_6"H"_6)`