# The analysis of a hydrocarbon revealed that it was 85.6281% C and 14.3719% H by mass. When 3.41 g of the gas was stored in a 1.4 L flask at -86.6273 C, it exerted a pressure of 505 Torr. What is the molecular formula of the hydrocarbon?

##### 1 Answer

A good general rule of thumb is that if you get a bunch of information in General Chemistry in one or two sentences, consider checking if you see terms related to the Ideal Gas Law.

#PV = nRT#

In fact, **you were given the pressure, volume, and temperature all in one sentence**. You should already know or be given that

**GENERAL PATHWAY**

The mass in **molar mass** in

Once you know the molar/molecular mass, you can use the percentages you were given to determine the **atomic contribution** to the molar mass. That would give you the molecular formula.

**DETERMINING YOUR IDEAL GAS LAW TERMS**

A good way to figure out what units you need is to pick a value for ** least** work to convert the pressure.

The units of

#P = "505 torr" = "505" cancel"mm Hg" xx ("1 atm")/("760" cancel"mm Hg") = "0.6648 atm"#

#T = -86.6273^@ "C" = -86.6273 + "273.15 K" = "186.5227 K"#

And you know

**NUMBER OF MOLS OF GAS**

#color(green)(n) = (PV)/(RT)#

#= (("0.6648" cancel"atm")("1.4" cancel"L"))/(("0.082057" cancel"L"cdotcancel"atm")/("mol"cdotcancel"K")("186.5227" cancel" K"))#

#=# #color(green)("0.0608 mols")#

Now that we have the number of

**MOLAR MASS OF THE MOLECULE**

#"MM" = "g molecule"/"mol molecule"#

#=# #"3.41 g"/("0.0608 mol")#

#=# #color(blue)("56.1043 g/mol")#

**DETERMINING THE MOLECULAR FORMULA**

At this point, we can figure out the molecular formula. Simply multiply by the decimal forms of each percentage to get:

#56.1043 xx 0.856281 = color(green)("48.0410 g/mol C")#

#56.1043 xx 0.143719 = color(green)("8.0632 g/mol H")#

Therefore, carbon constitutes about

When you divide those by the molar mass of

#48.0410/12.011 = 3.9998 = color(green)(4 xx "C")#

#8.0632/1.0079 = 8.0000 = color(green)(8 xx "H")#

Therefore, the molecular formula is **butene**.