When 0.72 g of a liquid is vaporized at 110 C and 0.967 atm, the gas occupies a volume of 0.559 L. The empirical formula of the gas is CH2. What is the molecular formula of the gas? Please give easy to follow, step by step instruction and the answer.?

The idea here is that you need to use the ideal gas law equation to determine how many moles of gas you have in that sample.

Once you know how many moles of gas were vaporized, you can use the sample's mass to determine the compound's' molar mass.

So, the ideal gas law equation looks like this

#color(blue)(PV = nRT)" "#, where

#P# - the pressure of the gas
#V# - the volume it occupies
#n# - the number of moles of gas
#R# - the universal gas constant, usually given as #0.082("atm" * "L")/("mol" * "K")#
#T# - the temperatue of the gas, expressed in Kelvin

Rearrange this equation to solve for #n# and plug in your values - don't forget to convert the temperature to Kelvin!@

#PV = nRT implies n = (PV)/(RT)#

#n = (0.967color(red)(cancel(color(black)("atm"))) * 0.559color(red)(cancel(color(black)("L"))))/(0.082(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 110)color(red)(cancel(color(black)("K")))) = "0.01721 moles"#

So, your sample contained#0.01721# moles of gas. Assuming that all the moles that were present in the liquid were vaporized, you can say that the initial sample contained #0.01721# moles as well.

Now, a compound's molar mass tells you what the mass of one mole of that compound is. In your case, you know that #0.01721# moles have a mass of #0.72# grams, which means that one mole will have a mass of

#M_"M" = m/n#

#M_"M" = "0.72 g"/"0.01721 moles" = "41.84 g/mol"#

You know that the empirical formula of the compound is #"CH"_2#. This tells you that the smallest whole number ratio that exists between the number of atoms of carbon and the number of atoms of hydrogen is #1:2#.

In order to determine the compound's molecular formula, you need to determine exactly how many atoms of each element you get per molecule.

This is of course equivalent to finding how many moles of carbon and moles of hydrogen you get per mole of compound.

The molecular formula is actually a multiple of the empirical formula. The molar mass of th eempirical formula will be

#"12.011 g/mol" + 2 xx "1.00794 g/mol" = "14.029 g/mol"#

So, how many empirical formulas are needed to make the molecular formula?

#14.029color(red)(cancel(color(black)("g/mol"))) xx color(blue)(n) = 41.84color(red)(cancel(color(black)("g/mol")))#

#color(blue)(n) = 41.84/14.029 = 2.982 ~~ 3#

Therefore, the molecular formula of the compound is

#("CH"_2)_color(blue)(3) implies "C"_3"H"_6#