# 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.?

##### 1 Answer

#### Explanation:

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

*universal gas constant*, usually given as

Rearrange this equation to solve for

#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

Now, a compound's **molar mass** tells you what the mass of **one mole** of that compound is. In your case, you know 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 *smallest whole number ratio* that exists between the number of atoms of carbon and the number of atoms of hydrogen is

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 e*empirical 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#