# How would you determine the molecular formula of a substance that contains 92.3%C and 7.7%H by mass. This same substance when in the gas phase has a density of 0.00454g/ml at stp conditions?

Dec 23, 2015

${\text{C"_8"H}}_{8}$

#### Explanation:

Your strategy here will be to

• use a sample of this compound to help you determine its empirical formula
• use the ideal gas law equation, the gas' density, and the known STP conditions for pressure and temperature to find th gas' molar mass
• use the empirical formula and the gas' molar mass to find its molecular formula

So, you know that you're dealing with a hydrocarbon that contains 92.3% carbon and 7.7% hydrogen by mass. To make the calculations easier, select a $\text{100-g}$ sample of this compound.

You know from the aforementioned percent composition that this sample will contain $\text{92.3 g}$ of carbon and $\text{7.7 g}$ of hydrogen.

Use the molar masses of the two elements to figure out how many moles of each you'd get in this sample

92.3 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = "7.6846 moles C"

7.7 color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.00794color(red)(cancel(color(black)("g")))) = "7.6393 moles H"

Divide both values by the smallest one to get the mole ratio that exists between the two elements in the compound

"For C: " (7.6846 color(red)(cancel(color(black)("moles"))))/(7.6393color(red)(cancel(color(black)("moles")))) = 1.006 ~~1

"For H: " (0.76393color(red)(cancel(color(black)("moles"))))/(0.76393color(red)(cancel(color(black)("moles")))) = 1

The empirical formula for this hydrocarbon is

$\text{C"_1"H"_1 implies "CH}$

Now, STP conditions are characterized by a pressure of $\text{100 kPa}$ and a temperature of ${0}^{\circ} \text{C}$. Keep this in mind.

The ideal gas law equation

$\textcolor{b l u e}{P V = n R T}$

can be rewritten using the definition of $n$, the number of moles of gas. As you know, the number of moles if equal to the ratio between the mass and the molar mass of a compound

$n = \frac{m}{M} _ M$

Plug this into the ideal gas law equation to get

$P V = \frac{m}{M} _ M \cdot R T$

Rearrange to get ${M}_{M}$ isolated on one side of the equation

${M}_{M} = {\overbrace{\frac{m}{V}}}^{\textcolor{b l u e}{= \rho}} \cdot \frac{R T}{P}$

Since density is defined as mass per unit of volume, you can say that

${M}_{M} = \rho \cdot \frac{R T}{P}$

Plug in the STP pressure and temperature and solve for ${M}_{M}$ - do not forget to convert the pressure from kPa to atm and the temperature from degrees Celsius to Kelvin!

Also, convert the density of the gas from grams per milliliter to grams per liter

$0.00454 \text{g"/color(red)(cancel(color(black)("mL"))) * (1000color(red)(cancel(color(black)("mL"))))/"1 L" = "4.54 g/L}$

M_M = 4.54"g"/color(red)(cancel(color(black)("L"))) * (0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 0)color(red)(cancel(color(black)("K"))))/(100/101.325color(red)(cancel(color(black)("atm"))))

${M}_{M} = \text{103.2 g/mol}$

Now, the molecular formula will always be a multiple of the empirical formula. The molar mass of the empirical formula is

$1 \times \text{12./011 g/mol" + 1 xx "1.00794 g/mol" = "13.019 g/mol}$

This means that you have

$13.019 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g/mol"))) xx color(blue)(n) = 103.2 color(red)(cancel(color(black)("g/mol}}}}$

This will get you

$\textcolor{b l u e}{n} = \frac{103.2}{13.019} = 7.93 \approx 8$

Therefore, the molecular formula of the hydrocarbon will be

$\left({\text{CH")_color(blue)(8) implies color(green)("C"_8"H}}_{8}\right)$