Thread: Chemical Equasion help... Please!

Hi, I was doing chemical equasions in class, and the puzzled me.

We/ve been doing them for a few lessons now, but then the teacher asked us to this one:

Glucose > Alcohol + Carbon Dioxide + Water
C.6 H.12 O.6 > C.2 H.5 OH + C O.2 + H.2 O

[C.6 means C sub-script six]

I couldn't figure out how to make the equasion work, even after anout 20 mins of trying [whilke listening to the teacher drone on about things I had already learnt]

What is the solution to this? And are there any easy ways to work out chemical equasions like this? Any hints for long ones with molecules that contain more than 3 elements?

Thanks in advance!

he law of conservation of mass dictates the quantity of each element does not change in a chemical reaction. Thus, each side of the chemical equation must represent the same quantity of any particular element. Similarly, the charge is conserved in a chemical reaction. Therefore, the same charge must be present on both sides of the unbalanced equation.

One balances a chemical equation by changing the scalar number for each molecular formula. Simple chemical equations can be balanced by inspection, that is, by trial and error. Another technique involves solving a system of linear equations.

Example #1: Na + O2 → Na2O

In order for this equation to be balanced, there must be an equal amount of Na on the left hand side as on the right hand side. As it stands now, there is 1 Na on the left but 2 Nas on the right. This is solved by putting a 2 in front of the Na on the left hand side:

2Na + O2 → Na2O

In this there are 2 Na atoms on the left and 2 Na atoms on the right. In the next step the oxygen atoms are balanced as well. On the left hand side there are 2 O atoms and the right hand side only has one. This is still an unbalanced equation. To fix this a 2 is added in front of the Na2O on the right hand side. Now the equation reads:

2Na + O2 → 2Na2O

Notice that the 2 on the right hand side is "distributed" to both the Na2 and the O. Currently the left hand side of the equation has 2 Na atoms and 2 O atoms. The right hand side has 4 Nas total and 2 Os. Again, this is a problem, there must be an equal amount of each chemical on both sides. To fix this 2 more Nas are added on the left side. The equation will now look like this:

4Na + O2 → 2Na2O

This equation is a balanced equation because there is an equal number of atoms of each element on the left and right hand sides of the equation.

Example #2:

P4 + O2 → 2P2O5

This equation is not balanced because there is an unequal amount of Os on both sides of the equation. The left hand side has 4 Ps and the right hand side has 4 Ps. So the P atoms are balanced. The left hand side has 2 Os and the right hand side has 10 Os.

To fix this unbalanced equation a 5 in front of the O2 on the left hand side is added to make 10 Os on both sides resulting in

P4 + 5O2 → 2P2O5

The equation is now balanced because there is an equal amount of substances on the left and the right hand side of the equation.

Thanks for that. I know most of the basics... it was just the equation in the original post that threw me off.

C.6 H.12 O.6 > C.2 H.5 OH + C O.2 + H.2 O

When you have an equation like this, where there are three products, each of the products having at least two different elements in it. I want to know how to solve these, or if there is an easy trick to equalling out the sides.

After my 20 minutes of trying, I had found ways to make the Hydrogen and Oxygen atoms balanced, the Hydrogen and carbon atoms balanced, and the Oxygen and Carbon atoms balanced, but no way to make the equation as a whole balanced. I can do simple equations like:

C.2 H.5 O H + O.2 > C O.2 + H.2 O

C.2 H.5 O H + 3 O.2> 2 C O.2 + 3 H.2 O

That's the chemical equation for complete combustion in alcohols, but the feramntation of yeasts above threw me off, as there was no way that I could solve it.

Hello,
I think that your problem is coming from the water. If you look at the structure of glucose you’ll see that the fermentation of it should produce. 2 molecules of ethanol and 2 molecules of carbon dioxide…. This is why you are having trouble balancing this. : )

The correct answer should be… Glucouse ---> 2C2H50H + 2CO2
See chapter 14.1 in the Lehniinger principles of biochemistry book for more help with this question.. I think it’s around page 520 or something if I remember correctly.

Hope this helps. : )

So, My science teacher was wrong in saying that glucose produces Water when it ferments? I actually looked at the equasion after he told us to do it, and saw that you could easily make it work with no water. It would also be easier to solve if you added some oxygen to the reactants' side.

Thanks for your help, chemist and Emmanuel. But I'm still looking for an easy tactic to balance these harder equasions. I've got the "look at how many atoms on each side, and then multiply the molecules' atoms first" teechnique down, but it throws me off when you have an equation which doesn't have an atomic molecule in it [O.2, H.2] because you can;t just throw the extras onto there.

Originally Posted by asxz

So, My science teacher was wrong in saying that glucose produces Water when it ferments? I actually looked at the equasion after he told us to do it, and saw that you could easily make it work with no water. It would also be easier to solve if you added some oxygen to the reactants' side.

Thanks for your help, chemist and Emmanuel. But I'm still looking for an easy tactic to balance these harder equasions. I've got the "look at how many atoms on each side, and then multiply the molecules' atoms first" teechnique down, but it throws me off when you have an equation which doesn't have an atomic molecule in it [O.2, H.2] because you can;t just throw the extras onto there.

Hello,

If the method you described above is not working then your products are likely wrong. You just can't create matter so when doing these just remember that the
ratio of the molecular equivalents must remain the same - otherwise you are creating or destroying matter.

I'm not sure what your teacher meant when giving you the equation but when I TAed the biochemistry course glucose fermentation always produced 2 ethanols and 2 carbon dioxides... this can be seen when observing the structure of glucose.. since matter cannot be created or destroyed without an outside reactant one cannot have water in the fermentation of glucose because there are not enough hydrogens to account for both the ethanol and the water.

Your first example using just glucose was an example of anerobic respiration which produces only carbon dioxide and ethanol.
If your teacher is saying that there must be water in the products then that is an example of aerobic respiration giving water as a product: see link: http://www.purchon.com/chemistry/chemical.htm

Hope that helps..
here are some other links that you might find helpful.. good luck. : )

You can also check out that biochem book I quoted for a reference or

http://dwb.unl.edu/Teacher/NSF/C11/C...mentation.html

http://www.chemcases.com/alcohol/alc-03.htm

Thank you chemist!

Hi asxz,
concerning your initial question, Emmanuel gave you a hint about linear equations, and that's indeed the way to go if you want a method that never fails.
Then of course if you hate maths, it's a different problem.

I will show you the method applied to your original problem.

You wrote:

C.6 H.12 O.6 > C.2 H.5 OH + C O.2 + H.2 O

Now, rewrite this with symbolic coefficients before each molecular formula except for one of them, which means that its coefficient is arbitrarily chosen to be 1. In this case I chose glucose because it's the biggest molecule in the equation:

C.6 H.12 O.6 > a C.2 H.5 OH + b C O.2 + c H.2 O

The conservation of individual atoms requires that the same number of the same atoms be present on either side. If you apply this to all types of atoms (C, H and O in this case) you'll have 3 equations in 3 unknowns (a, b and c):

So, for carbon:

6 = 2*a + b

For hydrogen:

12 = 5*a + a + 2*c

And for oxygen:

6 = a + 2*b + c

Now you can solve the system of these 3 equations by your preferred method, and you'll get:

a=2, b=2, c=0

Which, plugged back into your reactions gives:

C.6 H.12 O.6 > 2 C.2 H.5 OH + 2 C O.2

So as you see the method also tells you that there is no water generated in this reaction.

Sometimes you'll get rational coefficients (fractions) as results. If you want integer coefficients, just multiply the whole reaction by the least common multiple of the denominators.

I'm sure this works for most reactions; I guess that when you write reactions that are impossible to balance, you'll get an impossible system of equations.

I'd say give it a try on the most horrendous-looking equation in your exercise book, and if it works, there you go, you have your answer.

Good luck.

Thanks lavosier. I'm pretty good at maths, so I'll give it a try. It seems to be an easier way to figure it out than trial and error. More solid anyway.