Determine the molecular formula of a hydrocarbon if the combustion of 5.3 mg obtained 16.6 mg #"CO"_2#. The density of the gas at standard conditions is #"2.504 g/dm"^3#. What is the molar mass of the gas and its molecular and empirical formula?
Molecular formula is
Molar mass is
Empirical formula is
Original question stated
student mentioned that the question used 6.48 mg of carbon dioxide, as opposed to 8.48 mL.
Solution with revised inputs from student.
All things same except
"if the combustion of 5.3 mg obtained 8.48 mg CO2"
Both values of carbon dioxide gave erroneous results for the working out empirical formula of the hydrocarbon.
from the original question this value for calculation purpose modified original value of
It has been shown that a value taken as above gives a possible solution.
If we assume that in the original question there was a typo and "obtained 16.48mg of carbon dioxide" is the correct text, it will also lead to same empirical formula.
Given the sample is a hydrocarbon. As such its constituent elements are Carbon
We know that 1 GMW of any gas occupies
Therefore calculating mass of
Molar mass of hydrocarbon
We also know that 1 GMW of
Percent of carbon in sample
To find out the empirical formula of hydrocarbon we divide the above percentages by respective Average atomic masses.
We obtain Empirical formula of Hydrocarbon as
Let molecular formula be
As we already have Molar mass
Equating the two we have
Hence molecular formula of the hydrocarbon is
Molar mass of the gas
Molecular formula of the gas
Empirical formula of the gas
Given that the density of hydrocarbon at STP is
Let the molecular formula of HC be
And the balanced equation of the combustion reaction of the HC in oxygen is
So by this equation the stochiometric ratio of HC and the
But by the given data this ratio is
So equating these two we get
Now calculated value of molar mass of the HC (using atomic mass of carbon as
Inserting the value of
Hence the molecular formula of HC is
And obviously the empirical formula of HC is