Ask A Scientist Chemistry Archive

name         Jacqui
status       student
grade        9-12
location     WA

Question -   I have a problem that states: A gasoline pump says that
the gasoline is 70% heptane and 30% octane by volume. It also shows the
densities to be 683.76 g/L and 702.5 g/L at 20 degrees Celsius. What
would be the mass of water produced by the combustion of 7 gallons of
gasoline?

I am not just looking for the answer; instead, if I could at least have a
direction in which to go to solve this problem, that would be wonderful. I
have pummeled my brain and I have no idea how to utilize the information
given.
----------------------------------------------------
 The 70%/30% (v/v) of heptane and hexane, assuming that they mix ideally
(a good assumption here), for the 7 gal of the blend is 4.9 gal (heptane)
and 2.1 gal (octane). There are 3.785 liters / gal, so this blend is 18.55
liters (heptane) and 7.95 liters (octane). Using the two densities you
give:
683.76 gm / liter (heptane)  and 702.5 gm / liter (octane) means that the
18.55 liters (heptane) and 7.95 liters (octane) weigh: 12683.75 gm (heptane)
and 5584.9 gm (octane). The molecular weight of heptane and octane is 100 gm
/ mol heptane and 114 gm / mol octane. So converting the weights to mols
gives:
126.84 mol (heptane) and 49.00 mol (octane).
     Now the combustion reactions are as follows:
1 C7H16 +  22 O2 = 7 CO2 + 8 H20
1 C8H18 +  25 O2=  8 CO2 + 9 H2O

So 126.84 mol (heptane) produces 8 x 126.84 mol H2O = 1015 mol water
And 49.00 mol (octane) produces 9 x mol H2O = 441 mol water
For a total of 1015 + 441 = 1456 mol H2O
At 18 gm water (liquid) / mol water =26208 gm water or about 26.2 liters of
water.

FYI: the amount of heat produced can be calculated as follows:
The heat of combustion (a necessary additional piece of data) of heptane and
octane is: 1150 k-cal / mol (heptane) and 1302.8 k-cal / mol (octane).
Multiplying each molar heat of combustion by the respective number of mols
gives: 145866 k-cal and 63837 k-cal contributed by the two components -- for
a total of 209703 k-cal for the sum of the two components in 7 gal of the
70/30 (v/v) blend.

Vince Calder
=====================================================
Jacqui,

Remember that no matter how complex a problem like this can get, it is 
still a stoichiometry problem and therefore the steps remain the same. (1) 
You need to start from a balanced equation, (2) work with moles, (3) set 
up a ratio using moles and stoichiometric coefficients.

The problem you present needs to be broken down in simpler terms. Isolate 
each chemical reaction: separate the heptane from the octane and write a 
balanced chemical equation for both. I am sure you can do that.

Next you need to know how many moles of each you have in a given sample - 
this usually means working out the number of grams and dividing by the 
molar mass of each reactant. Since you know what is the percent by volume 
of each reactant, you can get the volume of each one found in 7gal. Since 
you know the density, you can get the mass of each one in 7gal (watch your 
unit conversions!).

The rest is just setting up the ratios. Solve how much water the heptane 
alone produces and then how much the octane produces, and then add them up.

Remember the three steps and know that if you can break down the problem 
into those three steps that you can solve any stoichiometry problem.

Greg (Roberto Gregorius)
====================================================
Jacqui,

     You need to:
1        write the chemical equations for combustion of heptane and octane
2        determine the mass of water produced by one mole of heptane and by
one mole of octane (used balanced chemical equations)
3        determine how much heptane and how much octane would be in 7
gallons of gasoline (use the mixture percentages, densities, and molecular
weights)
4        determine the total amount of water produced from combustion of
heptane and octane and add them together (use #2 and #3)

Greg Bradburn
====================================================
Jacqui,

Convert volumes to masses based on given densities of heptane and "octane".

Also you need to decide whether or not you want to assume that the
hydrocarbon + oxygen O(2) combustion equation is carried out with adequate
oxygen for total combustion...RARELY the case.

OR consider the more likely event that there would be other byproducts as
well the would lessen the amount of calculated H(2)O.....namely CO (carbon
monoxide).

Once you have converted from VOLUME => MASS => MOLES...balance the combustion
equation ... recall STOICHIOMETRY?  It gets a little silly with CO as a
byproduct but not impossible to balance the equation.

Solve for moles of water and water is about 18.0152 g / mole....voila !!

Appendix A (below)::

Heptane is easy...read on...or just go to Yahoo and type:  heptane +
structure (probably will get you there).  C(7)H(16) ==> figure MWT from here.

Someone once told me that the "octane" in gasoline was actually in the form
of 2,2,3 - trimethyl pentane.  I have also heard of 2, 4, 4 TMP.  But this is
similar to 2, 2, 3 and identical in all but name to 2, 2, 4 (which is the
correct IUPAC name for AN iso-octane) Which if you add up all the carbons
and hydrogens in this it will still add up to C(n)H(2n+2)...aka the basic
formula for a straight - chained hydrocarbon.

So octane (n=8) or C(8)H(18) is the same as (Me) - C (Me(2)) - CH (Me) -
CH(2) - (Me) ==> adding all C's and H's you get ==>  C(8)H(18)
Where (Me) is a -CH(3) aka methyl group.

The average MWT for C is 12.011 g / mole
That of H is 1.0079 g / mole.

I hope this answers more questions than it creates.
Darin Wagner
====================================================
Jacqui,

The first thing to do is to write two balanced
chemical reactions: one for the complete
combustion of heptane, and another for the
complete combustion of octane.

Now, imagine that you could separate the
gasoline into its components, heptane and octane.
How many gallons of heptane and how many of octane
would you obtain? (Hint: use the volume percentages).

Now you know how many gallons of octane you will burn.
Convert that to liters, and use the density in g/L
to get the number of grams. Now you know how many
grams of octane you will be burning. You can use
the molar mass of octane to figure out how many moles
of octane that would be. Then, the balanced reaction
will tell you the ratio of moles of octane burned to
moles of water produced. The mass of water can be obtained
by using the molar mass of water.

Then do the same thing for heptane, and add up the results.

Good luck!

Dr. Topper
====================================================
Jacqui,

Let me see if I can take a crack at this.  You have a volume of gasoline
and you know the breakdown of components in percent of volume.  This
will give you the volume of each component Yes?  With the densities you
now have the mass.

Look up the formulas for both constituents and write the combustion
equations assuming complete conversion to CO2 and H2O.  Balance the
equations and determine the mass of water produced.

Bob Hartwell
=====================================================
That is a fairly demanding problem; it is straightforward but has a lot of 
steps.
Perceiving these steps is 50% of the problem, but since it's so long,
I think I would like to tell you.

- convert:
   - gallons to liters,
   - total volume to two volumes,
   - two volumes to two masses
- balance:  derive mass ratios  m(water) / m(fuel)  for burning of each 
kind of fuel
- multiply the two masses by respective mass-ratios
- add together.

Jim Swenson
=====================================================