# Question 18b16

Jun 3, 2015

Styart by determining your compound's empirical formula.

You know the percent composition of your substance to be 54.5% carbon, 9.1% hydrogen, and 36.4% oxygen. This means that a 100-g sample will contain 54.5 g of carbon, 9.1 g of hydrogen, and 36.4% g of oxygen.

Use each element's molar mass to figure out how many moles of each you'd get in that sample

$\text{For C": (54.5cancel("g"))/(12.0cancel("g")/"mol") = "4.542 moles C}$

$\text{For H": (9.1cancel("g"))/(1.01cancel("g")/"mol") = "9.01 moles H}$

$\text{For O": (36.4cancel("g"))/(16.0cancel("g")/"mol") = "2.275 moles O}$

Divide these numbers by the smallest one to get the mole ratios of the elements in the compound.

"For C": (4.542cancel("moles"))/(2.275cancel("moles")) = 2

"For H": (9.01cancel("moles"))/(2.275cancel("moles")) = 4

"For O": (2.275cancel("moles"))/(2.275cancel("moles")) = 1#

Your compound's empirical formula will thus be

${C}_{2} {H}_{4} O$

Now for the molecular mass. You know that your compound's relative density compared with steam is equal to 5. Since no mention of pressure or temperature was made, you're going to have to find a relationship that will allow you to solve for the molar mass of the compound without knowing pressure and temperature.

Use the ideal gas law equation to find a relationship between density and molar mass

$P V = n R T$

Since number of moles is equal to mass divided by molar mass, you can write

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

Rearrange this equation to get

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

This means that density will be equal to

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

Now, the relative density of the compound can be written like this

${\rho}_{\text{relative" = rho_"compound"/rho_"steam}} = 5$

Assuming that you have the same conditions for pressure and temperature for the compound and for the steam, and knowing that the molar mass of steam is 18.02 g/mol, you can write

$\left(\left(\cancel{\text{P") * M_M)/(cancel("RT")))/((cancel("P") * 18.02)/(cancel("RT}}\right)\right) = {M}_{M} / 18.02 = 5$.

Therefore,

${M}_{M} = 5 \cdot 18.02 = \textcolor{g r e e n}{\text{90.1 g}}$

SIDE NOTE Your compound's molecular formula will be

${\left({C}_{2} {H}_{4} O\right)}_{\textcolor{b l u e}{\text{n}}} = 90.1$

$\textcolor{b l u e}{\text{n}} = \frac{90.1}{2 \cdot 12.0 + 4 \cdot 1.01 + 16.0} = 2.05 \cong 2$

${C}_{4} {H}_{8} {O}_{2}$ $\to$ molecular formula.