# Question 579cc

Feb 3, 2015

You need to know two things to solve this problem:

1. That 1 mole of any ideal gas occupies $\text{22.4 L}$ at STP;
2. The relationship between moles and molar mass

"moles" = ("mass of substance")/("molar mass")

So, you know that the vapor weighs $\text{0.96 g}$ at STP. STP conditions imply a temperature of $\text{273.15 K}$ and a pressure of $\text{1.0 atm}$. This means that the number of moles you have is

n=V/V_("molar") = (250 * 10^(-3)"L")/("22.4 L") = "0.01116 moles"

SIde note - you can use the ideal gas law equation, $P V = n R T$, to double check this result;

Since $n = \text{m"/"MM}$, you get that

$\text{n" = "m"/"MM" = "0.01116 moles}$

Since $m = \text{0.96 g}$, the value of the molar mass will be

$\text{molar mass" = m/n = ("0.96 g")/("0.01116 moles") = "86.02 g/mol}$

Your compound's empirical formula is ${\left(C {H}_{2}\right)}_{x}$, or ${C}_{x} {H}_{2 x}$. The value of $x$ is determined by dividing the molar mass of the compound by the molar mass of the empirical formula

x = ("86.02 g/mol")/("14.0 g/mol") = "6.14"

Ideally, $x$ should be as close to an integer as possible; in this case, the closest integer would be $6$, which would make the compound's molecular mass equal to $\text{84.0 g/mol}$ and the molecular formula

${C}_{6} {H}_{12}$.

Since this problem describes the Dumas method of molecular weight determination, an experimental method used to determine molar mass, you could calculate the percent error for the result

"%error" = |"accepted value - experimental value"|/("accepted value") * 100#

$\text{%error" = |84.0 - 86.02|/84.0 * 100 = "2.4%}$

which is a relatively small percent error $\to$ the actual molecular formula is ${C}_{6} {H}_{12}$.