Question

How to find the molecular formula of an unknown substance?

An organic compound contains C, H and O only. When 9.52g of this compound is burned in an atmosphere of oxygen it produces 13.96g of CO2 and 5.72g of H2O. If 0.36 mole of this compound weighs 32.4g, what is the Molecular formula of this compound?

Answer 1

The molecular formula is #"C"_3"H"_6"O"_3

Explanation:

We must calculate the masses of `"C, H"` and `"O"` from the masses given.

`"Mass of C" =13.96 color(red)(cancel(color(black)("g CO"_2))) × "12.01 g C"/(44.01 color(red)(cancel(color(black)("g CO"_2)))) = "3.810 g C"`

`"Mass of H" = 5.72 color(red)(cancel(color(black)("g H"_2"O"))) × "2.016 g H"/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) = "0.6400 g H"`

`"Mass of O" = "Mass of compound -Mass of C - Mass of H" = "(9.52 - 3.810 - 0.6400) g" = "5.070 g"`

Now, we must convert these masses to moles and find their ratios.

From here on, I like to summarize the calculations in a table.

`ulbb("Element"color(white)(X) "Mass/g"color(white)(X) "Amt/mol"color(white)(m) "Ratio"color(white)(m)"Integers")`
`color(white)(mm)"C"color(white)(mmmll)3.810color(white)(mmm)0.3172color(white)(Xml)1.001color(white)(Xmm)1`
`color(white)(mm)"H" color(white)(XXXll)0.6400 color(white)(mmll)0.6349 color(white)(mm)2.003color(white)(Xmm)2`
`color(white)(mm)"O" color(white)(XXXll)5.070 color(white)(mmm)0.3169 color(white)(mml)1color(white)(Xmmmm)1`

The empirical formula is `"CH"_2"O"`.

Calculate the molar mass of the compound

`"Molar mass" = "mass"/"moles" = "32.4 g"/"0.36 mol" = "90.0 g/mol"`

Calculate the molecular formula of the compound

The empirical formula mass of `"CH"_2"O"` is 30.03 u.

The molecular mass of the gas is 90.0 u.

The molecular mass must be an integral multiple of the empirical formula mass.

`"MM"/"EFM" = (90.03 color(red)(cancel(color(black)("u"))))/(30.03 color(red)(cancel(color(black)("u")))) = 3.00 ≈ 3`

The molecular formula must be three times the empirical formula.

`"MF" = ("EF")_3 = ("CH"_2"O")_3 = "C"_3"H"_6"O"_3`

The molecular formula is `"C"_3"H"_6"O"_3`.