# Chemical Forums

## Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: nectarines on December 19, 2012, 02:00:06 PM

Title: Balancing really difficult chemical equations
Post by: nectarines on December 19, 2012, 02:00:06 PM
I know this seems extremely simple, but I've been taking practice tests for my gen chem final, and the balancing equations problems have been the most challenging for me. Here are examples:

?P + ?HNO3 + ?H2O -> ?H3PO4 + ?NO

?Cu + ?HNO3 -> ?Cu(NO3)2 + ?NO + ?NO2 + ?H2O, where NO and NO2 are formed in a 2:3 ratio

Can someone explain the process of solving these? I can do most balancing problems, but these are just ridiculous.
Title: Re: Balancing really difficult chemical equations
Post by: curiouscat on December 19, 2012, 06:10:27 PM
Have you been taught how to solve simultaneous linear equations?

Then, write atom balances. Say, in 2nd example for Cu, H, N, O.
4 linear eq. in 6 unknowns. Solve.

Matrix based solving might speed up the process.

It isn't the way Chem. classes typically teach but I find it neater. If you're lucky maybe your Scientific Calculator will have a linear eq. solving function too.

Title: Re: Balancing really difficult chemical equations
Post by: curiouscat on December 19, 2012, 06:35:33 PM
a Cu + b HNO3   :rarrow: c Cu(NO3)2 + d NO + e NO2 + f H2O

(Cu)  a = c
(H)   b  =  2f
(N)  b = 2c +d + e
(O) 3b = 6c +d + 2e + f

3d=2e

Solve.
Title: Re: Balancing really difficult chemical equations
Post by: AWK on December 19, 2012, 07:11:36 PM
Quote
?Cu + ?HNO3 -> ?Cu(NO3)2 + ?NO + ?NO2 + ?H2O, where NO and NO2 are formed in a 2:3 ratio
It will be more simple to split this eqution into two equations with NO and NO2 formed separately.

correction of printing error (quote instead of subscript)
Title: Re: Balancing really difficult chemical equations
Post by: Borek on December 19, 2012, 09:26:00 PM
As AWK wrote, these are two separate processes and each has to be balanced separately.
Title: Re: Balancing really difficult chemical equations
Post by: AWK on January 03, 2013, 01:52:00 AM
In this case the other possibility is joining 2NO + 3NO2 into artificial molecule N5O8
?Cu + ?HNO3 -> ?Cu(NO3)2 + ?N5O8 + ?H2O
Title: Re: Balancing really difficult chemical equations
Post by: Ben Cohen on January 09, 2013, 07:22:36 PM
It's all about making connections between the elements and using variables efficiently
Title: Re: Balancing really difficult chemical equations
Post by: Big-Daddy on January 10, 2013, 06:45:41 AM
a Cu + b HNO3   :rarrow: c Cu(NO3)2 + d NO + e NO2 + f H2O

(Cu)  a = c
(H)   b  =  2f
(N)  b = 2c +d + e
(O) 3b = 6c +d + 2e + f

3d=2e

Solve.

6 variables in 4 equations? This cannot be solved numerically by the standard method of solving simultaneous linear equations.
Title: Re: Balancing really difficult chemical equations
Post by: curiouscat on January 10, 2013, 07:03:30 AM
a Cu + b HNO3   :rarrow: c Cu(NO3)2 + d NO + e NO2 + f H2O

(Cu)  a = c
(H)   b  =  2f
(N)  b = 2c +d + e
(O) 3b = 6c +d + 2e + f

3d=2e

Solve.

6 variables in 4 equations? This cannot be solved numerically by the standard method of solving simultaneous linear equations.

5 equations. Also have "3d=2e"?
Title: Re: Balancing really difficult chemical equations
Post by: Borek on January 10, 2013, 07:35:05 AM
6 variables in 4 equations? This cannot be solved numerically by the standard method of solving simultaneous linear equations.

Quite often we do miss one equation when balancing reactions, but we also have additional condition that the solution should consist of the smallest integers - and that's enough to find the unique solution.
Title: Re: Balancing really difficult chemical equations
Post by: curiouscat on January 10, 2013, 08:30:50 AM
6 variables in 4 equations? This cannot be solved numerically by the standard method of solving simultaneous linear equations.

This cannot be solved numerically uniquely by the standard method of solving simultaneous linear equations.

An under-determined system of simultanous linear eqns. is not the same as an inconsistent system.
Title: Re: Balancing really difficult chemical equations
Post by: Big-Daddy on January 10, 2013, 09:13:53 AM
6 variables in 4 equations? This cannot be solved numerically by the standard method of solving simultaneous linear equations.

This cannot be solved numerically uniquely by the standard method of solving simultaneous linear equations.

An under-determined system of simultanous linear eqns. is not the same as an inconsistent system.

What is the process to force a solution out of something like this?

(When I say force a solution I mean find numerical values by actual number calculation rather than guesswork.)
Title: Re: Balancing really difficult chemical equations
Post by: Borek on January 10, 2013, 09:56:08 AM
What is the process to force a solution out of something like this?

(When I say force a solution I mean find numerical values by actual number calculation rather than guesswork.)

I told you - there is an additional condition that can be used. Depending on how you approach the problem you either need some guesswork, or you can do it algorithmically. In a way it is not much different from finding the empirical formula - you have numbers that have to be converted to integers without changing ratios. It is not that difficult, as you can do Gauss elimination using rational numbers, so in the end it is just a matter of finding GCD of all coefficients.

Title: Re: Balancing really difficult chemical equations
Post by: curiouscat on January 10, 2013, 11:21:57 AM
What is the process to force a solution out of something like this?

(When I say force a solution I mean find numerical values by actual number calculation rather than guesswork.)

(http://i.imgur.com/Ur3MC.gif)

(http://i.imgur.com/if011.gif)

Title: Re: Balancing really difficult chemical equations
Post by: Big-Daddy on January 10, 2013, 11:43:18 AM
What is the process to force a solution out of something like this?

(When I say force a solution I mean find numerical values by actual number calculation rather than guesswork.)

I told you - there is an additional condition that can be used. Depending on how you approach the problem you either need some guesswork, or you can do it algorithmically. In a way it is not much different from finding the empirical formula - you have numbers that have to be converted to integers without changing ratios. It is not that difficult, as you can do Gauss elimination using rational numbers, so in the end it is just a matter of finding GCD of all coefficients.

You specified only the condition before, not the process for calculating this. I had seen that algebraic method and it is the one I was referring to. But I suppose with reference to curiouscat's example that whole number guesswork becomes trivial at this stage however complicated the original problem.

I suppose that due to my lack of knowledge of what a Gaussian elimination is that this is probably what I was asking for when I asked what process was needed to force the solution.
Title: Re: Balancing really difficult chemical equations
Post by: Borek on January 10, 2013, 09:53:46 PM
Gaussian elimination is just a step on the numerical approach to solving system of equations - that is, it is a tool that will help you find a solution like the one curiouscat posted:

(http://i.imgur.com/if011.gif)

Now as everything is expressed using a, assume a=1 and see if all coefficients are small integers. If they are not - see what you can do to make them integers (that is, find the lowest number that will make them all integers once you multiply them by this number). Using rational numbers from the start helps enormously.
Title: Re: Balancing really difficult chemical equations
Post by: curiouscat on January 10, 2013, 10:07:04 PM
Quite often we do miss one equation when balancing reactions, but we also have additional condition that the solution should consist of the smallest integers - and that's enough to find the unique solution.

Not "quite often" but always I think. Because we have a homogeneous system (Since we always have a+b=c etc. and never a+b=c+2) of linear equations and such a system would always admit all zeros as a trivial equation. So a unique solution would mean all coefficients zero.

So the fact that we get a non-unique solution (before applying the smallest integers constraint) is a feature and not a bug. If you ever got a full rank matrix, be worried! ;)
Title: Re: Balancing really difficult chemical equations
Post by: Borek on January 10, 2013, 10:38:01 PM
Not "quite often" but always I think.

Sometimes we miss even more than one equation.

C7H6O3 + C4H6O3  :rarrow: C9H8O4 + HC2H3O2

(salicylic acid + acetic anhydride -> aspirin + acetic acid)

4 coefficients, but only two independent equations.

Quote
Because we have a homogeneous system (Since we always have a+b=c etc. and never a+b=c+2) of linear equations and such a system would always admit all zeros as a trivial equation. So a unique solution would mean all coefficients zero.

So the fact that we get a non-unique solution (before applying the smallest integers constraint) is a feature and not a bug. If you ever got a full rank matrix, be worried! ;)

Good point :) But I am not convinced - just because there exist trivial solution doesn't mean there is no other, non-trivial solution. I have no time to investigate now, but I have a feeling I have seen reactions with a full rank matrix.
Title: Re: Balancing really difficult chemical equations
Post by: curiouscat on January 11, 2013, 12:55:25 AM

Sometimes we miss even more than one equation.

Ah! True. No oubt.

Quote
Good point :) But I am not convinced - just because there exist trivial solution doesn't mean there is no other, non-trivial solution. I have no time to investigate now, but I have a feeling I have seen reactions with a full rank matrix.

I'd love to see one if you come up with one.
Title: Re: Balancing really difficult chemical equations
Post by: kingkev101 on January 05, 2018, 10:27:51 AM
aCu + bHNO3 -> cCu(NO3)2 + dNO + eNO2 + fH2O

Completing the solution for this:-

(Cu)  a = c....1)
(H)   b  =  2f......2)
(N)  b = 2c +d + e ..3)
(O) 3b = 6c +d + 2e + f....4)

Also Remember that in the initial post we are told that d:e must be 2:3 and so d/e = 2/3 which when removing denominators leaves 3d = 2e (basic maths).

3d = 2e...5)

We can derive one more simple equation. Notice that the LHS of eq 4) is 3 times that of eq 3)
so we can immediately say that 6c + 3d + 3e = 6c + d + 2e + f. Substituting 3d = 2e gives:
6c + 2e + 3e = 6c + d +2e +f. Next you can remove 6c and 2e from both sides giving 3e = d + f but we know d in relation e i.e. d = 2e/3 so 3e = 2e/3 +f ie

7e/3 = f.....6)

Now we have 2 simple equations with f in each so lets try f = 1 this => b =2 (eq 2) and e=3/7 (eq 6) and knowing e gives us d ie d = 2e/3 = 2/3 * 3/7 = 2/7 ie d = 2/7

knowing f,b,d & e then eq 4)...3 * 2 = 6c + 2/7 + 6/7 + 1 => 6 = 6c + 15/7 ie 6c = 27/7 => c = 9/14 and so eq 1 gives us a = 9/14

so a:b:c:d:e:f   =   9/14:2:9/14:2/7:3/7:1
Multiply throughout by 14 gives 9:28:9:4:6:14

9Cu + 28HNO3 -> 9Cu(NO3)2 + 4NO + 6NO2 + 14H2O
Double checking if it balances we see
Cu:   9 on LHS and 9 on RHS
H:   28 on LHS and 14*2 = 28 on RHS
N:   28 on LHS and 18+ 4 + 6 = 28 on RHS
O:   3 * 28 = 84 on LHS and 9 * 3* 2 + 4 + 12 + 14 = 54 + 4 + 12 + 14 = 84 on RHS
Ie balances …..!