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?

May 9, 2017

You probably have......$\text{ethylene}$

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 $P V = n R T$, then $1 \cdot a t m \times 10 L = \frac{11.6 \cdot g}{\text{Molar mass}} \times 0.0821 \frac{L \cdot a t m}{K \cdot m o l} \times 298 \cdot K$

OR

$\text{Molar mass} = \frac{11.6 \cdot g}{1 \cdot a t m \times 10 \cdot L} \times \times 0.0821 \frac{L \cdot a t m}{K \cdot m o l} \times 298 \cdot K = 28.4 \cdot g \cdot m o {l}^{-} 1$.

Now this molecular mass is consistent with that of $\text{ethylene}$, ${H}_{2} C = C {H}_{2}$ whose molecular mass is $28.0 \cdot g \cdot m o {l}^{-} 1$.

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

Had we calculated as molecular mass of $26 \cdot g \cdot m o {l}^{-} 1$, or $30 \cdot g \cdot m o {l}^{-} 1$, what would be the likely identity of the gas under the given scenario?

May 9, 2017

The molecular formula is ${\text{C"_2"H}}_{2}$.

Explanation:

This is an empirical formula/molecular formula problem.

(a) Calculate the moles of $\text{C}$.

${\text{Moles of C" = 3.38 color(red)(cancel(color(black)("g CO"_2))) × ("1 mol CO"_2)/(44.01 color(red)(cancel(color(black)("g CO"_2)))) × "12.01 g C"/(1 "mol C") = "0.076 80 mol CO}}_{2}$

(b) Calculate the moles of $\text{H}$.

I assume that you meant 0.69 g of water.

$\text{Moles of H" = 0.69 color(red)(cancel(color(black)("g H"_2"O"))) × (1 color(red)(cancel(color(black)("mol H"_2"O"))))/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) × "2 mol H"/(1 color(red)(cancel(color(black)("mol H"_2"O")))) = "0.0766 mol H}$

(c) Find the molar ratios.

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

$\boldsymbol{\text{Element"color(white)(Xll) "Moles"color(white)(m) "Ratio" color(white)(m)"Integers}}$
$\textcolor{w h i t e}{m m} \text{C} \textcolor{w h i t e}{m m m l l} 0.07680 \textcolor{w h i t e}{X} 1.003 \textcolor{w h i t e}{m m m l} 1$
$\textcolor{w h i t e}{m m} \text{H} \textcolor{w h i t e}{m m m l l} 0.0766 \textcolor{w h i t e}{m l l} 1 \textcolor{w h i t e}{X m m m m l} 1$

The empirical formula is $\text{CH}$.

(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).

$p V = n R T = \frac{m}{M} R T$

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 $\text{CH}$ 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.

${\text{MF" = ("EF")_2 = ("CH")_2 = "C"_2"H}}_{2}$