Question

The analysis of a hydrocarbon revealed that it was 85.6281% C and 14.3719% H by mass. When 3.41 g of the gas was stored in a 1.4 L flask at -86.6273 C, it exerted a pressure of 505 Torr. What is the molecular formula of the hydrocarbon?

Answer 1

A good general rule of thumb is that if you get a bunch of information in General Chemistry in one or two sentences, consider checking if you see terms related to the Ideal Gas Law.

`PV = nRT`

In fact, you were given the pressure, volume, and temperature all in one sentence. You should already know or be given that `R = "0.082057 L"cdot"atm/mol"cdot"K"`.

GENERAL PATHWAY

The mass in `"g"` can be used to give you the molar mass in `"g/mol"`, because you can solve for the moles using the Ideal Gas Law.

Once you know the molar/molecular mass, you can use the percentages you were given to determine the atomic contribution to the molar mass. That would give you the molecular formula.

DETERMINING YOUR IDEAL GAS LAW TERMS

A good way to figure out what units you need is to pick a value for `R`, and I picked the one that requires the least work to convert the pressure. `"torr"` to `"atm"` is basically one step, but technically `"1 torr = 1 mm Hg"`, so I included that just so you know it.

The units of `R` basically tell you what units you need for pressure and volume, while you should assume temperature is in `"K"` and `n` is in `"mol"`s.

`P = "505 torr" = "505" cancel"mm Hg" xx ("1 atm")/("760" cancel"mm Hg") = "0.6648 atm"`

`T = -86.6273^@ "C" = -86.6273 + "273.15 K" = "186.5227 K"`

And you know `V` and `R` are in the right units.

NUMBER OF MOLS OF GAS

`color(green)(n) = (PV)/(RT)`

`= (("0.6648" cancel"atm")("1.4" cancel"L"))/(("0.082057" cancel"L"cdotcancel"atm")/("mol"cdotcancel"K")("186.5227" cancel" K"))`

`=` `color(green)("0.0608 mols")`

Now that we have the number of `"mol"`s, we can figure out the molar mass.

MOLAR MASS OF THE MOLECULE

`"MM" = "g molecule"/"mol molecule"`

`=` `"3.41 g"/("0.0608 mol")`

`=` `color(blue)("56.1043 g/mol")`

DETERMINING THE MOLECULAR FORMULA

At this point, we can figure out the molecular formula. Simply multiply by the decimal forms of each percentage to get:

`56.1043 xx 0.856281 = color(green)("48.0410 g/mol C")`

`56.1043 xx 0.143719 = color(green)("8.0632 g/mol H")`

Therefore, carbon constitutes about `"48.04 g"` of the molecular mass, while hydrogen constitutes about `"8.06 g"` of the molecular mass.

When you divide those by the molar mass of `"C"` and `"H"`, you'll get the subscripts in the molecular formula, which are the actual contributions.

`48.0410/12.011 = 3.9998 = color(green)(4 xx "C")`

`8.0632/1.0079 = 8.0000 = color(green)(8 xx "H")`

Therefore, the molecular formula is `color(blue)("C"_4"H"_8)`, which is some sort of butene.