Question

What is the molecular formula of a material for which complete combustion of a unknown mass gives `3.38*g` of `CO_2(g)`, and `0.69*g` of water. An `11.6*g` mass of this gas gives a volume of `10.0*L` at `0` `""^@C` of `10*L`?

Answer 1

You probably have......`"ethylene"`

Explanation:

We can only interrogate the molecular formula on the basis of the Ideal Gas equation. The combustion data that you gave us were non-kosher (neither the mass of the combusted gas, nor the water content of the combustions were included).

If `PV=nRT`, then `1*atmxx10L=(11.6*g)/"Molar mass"xx0.0821(L*atm)/(K*mol)xx298*K`

OR

`"Molar mass"=(11.6*g)/(1*atmxx10*L)xxxx0.0821(L*atm)/(K*mol)xx298*K=28.4*g*mol^-1`.

Now this molecular mass is consistent with that of `"ethylene"`, `H_2C=CH_2` whose molecular mass is `28.0*g*mol^-1`.

Anyway, I don't think you have included all of the data for this question.

Had we calculated as molecular mass of `26*g*mol^-1`, or `30*g*mol^-1`, what would be the likely identity of the gas under the given scenario?

Answer 2

The molecular formula is `"C"_2"H"_2`.

Explanation:

This is an empirical formula/molecular formula problem.

(a) Calculate the moles of `"C"`.

`"Moles of C" = 3.38 color(red)(cancel(color(black)("g CO"_2))) × ("1 mol CO"_2)/(44.01 color(red)(cancel(color(black)("g CO"_2)))) × "12.01 g C"/(1 "mol C") = "0.076 80 mol CO"_2`

(b) Calculate the moles of `"H"`.

I assume that you meant 0.69 g of water.

`"Moles of H" = 0.69 color(red)(cancel(color(black)("g H"_2"O"))) × (1 color(red)(cancel(color(black)("mol H"_2"O"))))/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) × "2 mol H"/(1 color(red)(cancel(color(black)("mol H"_2"O")))) = "0.0766 mol H"`

(c) Find the molar ratios.

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

`bb("Element"color(white)(Xll) "Moles"color(white)(m) "Ratio" color(white)(m)"Integers")`
`color(white)(mm)"C" color(white)(mmmll)0.07680color(white)(X)1.003color(white)(mmml)1`
`color(white)(mm)"H" color(white)(mmmll)0.0766 color(white)(mll)1 color(white)(Xmmmml)1`

The empirical formula is `"CH"`.

(d) Calculate the molar mass of the gas

You don’t give the pressure or temperature of the gas, so I will assume STP (1 bar and 0 °C).

`pV = nRT = m/MRT`

`M = (mRT)/(pV) = ("11.6 g" × "0.083 14" color(red)(cancel(color(black)("bar·L·K"^"-1")))"mol"^"-1" × 273.15 color(red)(cancel(color(black)("K"))))/(1 color(red)(cancel(color(black)("bar"))) × 10 color(red)(cancel(color(black)("L")))) = "26.3 g/mol"`

(d) Calculate the formula of the gas

The empirical formula mass of `"CH"` is 13.02 u.

The molecular mass of the gas is 26.3 u.

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

`"MM"/"EFM" = (26.3 color(red)(cancel(color(black)("u"))))/(13.02 color(red)(cancel(color(black)("u")))) = 2.02 ≈ 2`

The molecular formula must be twice the empirical formula.

`"MF" = ("EF")_2 = ("CH")_2 = "C"_2"H"_2`