deduce the formula of the hydrocarbon

Question

10cm3 of hydrocarbon were expose with an excess of oxygen. When cooled to room temperature there was a contraction in volume of 30cm3. on treatment with NaOH there was a further contraction of 30cm3. Deduce the formula of the hydrocarbon.

Answer 1

When you say "exposed", do you mean fully oxidized?  Let's assume you do.  And assume that it a gas to begin with.

 

First of all, let's write out the equation, using unknowns for the numbers of C and H in the original hydrocarbon:

 

C_xH_y + O₂ ---> CO₂ + H₂O.

 

Balancing this equation gives us:

 

C_xH_y + (x + y/4) O₂ ---> x CO₂ + y/2 H₂O

 

we don't know if y is divisible by 4, so this may multiply up to:

 

4 C_xH_y + (4x + 1) O₂ ---> 4x CO₂ + 2y H₂O

 

After oxidation, there was none of the hydrocarbon left, replaced by its oxidation products, H₂O and CO₂.  Cooling to room temperature would cause of the H₂O to condense as a liquid.  So let's say that meant that there were 30 cm³ of H₂O in gaseous form after oxidation, and it condensed after cooling.

 

I also figure that treating with NaOH would then cause all of the CO₂ to be combined with NaOH to form Na₂CO_3.  So that means that there were 30 cm³ of CO₂ after oxidation, the same volume as there was of H₂O.  Since one of the gas laws says that the volume of an ideal gas is proportional to the number of moles, that means that the number of moles of CO₂ and H₂O have to be equal, so 4x = 2y.

 

By the same reasoning about volume, we know that each mole of the hydrocarbon produced three moles each of CO₂ and H₂O.  So 4x and 2y are both 3 times the coefficient of the hydrocarbon in the equation, which is 4.  So 4x and 2y are both 12.

 

So x = 3 and y = 6.

 

So the original hydrocarbon is C₃H₆.

 

Let's see if that works:

 

Start with

 

C₃H₆ + O₂ ---> CO₂ + H₂O

 

...and balance:

 

C₃H₆ + 4.5 O₂ ---> 3CO₂ + 3H₂O

2 C₃H₆ + 9O₂ ---> 6CO₂ + 6H₂O

 

That looks good to me.