Question

A 10 g sample of a compound contains 4.00g `C`, 0.667g `H`, and 5.33g `O`. What is the molecular formula?

The molecular mass is 180.156 g/mol.

Answer 1

The molecular formula is `"C"_6"H"_12"O"_6`.

Explanation:

We must first determine the empirical formula of the compound.

We must figure out the moles of `"C"`, `"H"`, and `"O"` and then calculate their ratio.

Step 1. Calculate the moles of each element.

`"Moles of C" = 4.00 color(red)(cancel(color(black)("g C"))) × "1 mol C"/(12.01 color(red)(cancel(color(black)("g C")))) = "0.3331 mol C"`

`"Moles of H" = 0.667 color(red)(cancel(color(black)("g H"))) × "1 mol H"/(1.008 color(red)(cancel(color(black)("g H")))) = "0.6617 mol C"`

`"Moles of O" = 5.33 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)("g O")))) = "0.3331 mol O"`

Step 2. Calculate the empirical formula.

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

`"Element"color(white)(Agll) "Mass/g"color(white)(Xm) "Moles"color(white)(Xm) "Ratio"color(white)(m)color(white)(l)"Integers"`
`stackrel(———————————————————)(color(white)(m)"C" color(white)(XXXmm)4.00 color(white)(Xmm)0.3331 color(white)(Xm)1color(white)(mmmmml)1)`
`color(white)(m)"H" color(white)(XXXXm)0.667 color(white)(mml)0.6617 color(white)(Xm)1.987 color(white)(mmml)2`
`color(white)(m)"O" color(white)(XXXXm)5.33 color(white)(mmm)0.3331 color(white)(Xm)1.000 color(white)(mmml)1`

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

Step 3. Calculate the empirical formula mass

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

Step 4. Calculate the molecular mass.

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

`"MM" = n × "EFM"`

`n = "MM"/"EFM" = (180.156 color(red)(cancel(color(black)("u"))))/(30.03 color(red)(cancel(color(black)("u")))) = 5.999 ≈ 6`

Step 5. Calculate the Molecular Formula

The molecular formula is 6 times the empirical formula.

The molecular formula is `("CH"_2"O")_6 = "C"_6"H"_12"O"_6`.