# How do you balance tricky chemical equations

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val7322

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#1

can someone tell me their process to solve this

Last edited by val7322; 4 months ago

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TriplexA

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#2

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#2

(Original post by

can someone tell me their process to solve this

**val7322**)can someone tell me their process to solve this

First balance the C's so there's the same number on each side.

Then balance O.

Then balance H.

If you could show me your final equation just as a final check - that'd be great

Last edited by TriplexA; 4 months ago

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username5613084

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#3

(Original post by

can someone tell me their process to solve this

**val7322**)can someone tell me their process to solve this

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ghostwalker

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#4

(Original post by

can someone tell me their process to solve this

**val7322**)can someone tell me their process to solve this

Start by assuming 1 molecule of glucose, becomes a,b,c molecules of propanone, carbon dioxide, and water respectively.

Then consider the no. of C atoms. On the left we have 6, and on the right 3a+b. Giving us one equation:

6=3a+b.

Repeat for H and O.

Solve the 3 simultaneous equations, to get an initial chemical equation.

Your values for a,b,c may be, and in this case are, fractions, so multiply through by the lowest common multiple of the denominators of any fractions to get your desired result.

Last edited by ghostwalker; 4 months ago

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val7322

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#5

(Original post by

Prehaps not the simplest, but:

Start by assuming 1 molecule of glucose, becomes a,b,c molecules of propanone, carbon dioxide, and water respectively.

Then consider the no. of C atoms. On the left we have 6, and on the right 3a+b. Giving us one equation:

6=3a+b.

Repeat for H and O.

Solve the 3 simultaneous equations, to get an initial chemical equation.

Your values for a,b,c may be, and in this case are, fractions, so multiply through by the lowest common multiple of the denominators of any fractions to get your desired result.

**ghostwalker**)Prehaps not the simplest, but:

Start by assuming 1 molecule of glucose, becomes a,b,c molecules of propanone, carbon dioxide, and water respectively.

Then consider the no. of C atoms. On the left we have 6, and on the right 3a+b. Giving us one equation:

6=3a+b.

Repeat for H and O.

Solve the 3 simultaneous equations, to get an initial chemical equation.

Your values for a,b,c may be, and in this case are, fractions, so multiply through by the lowest common multiple of the denominators of any fractions to get your desired result.

6= 3a +b

12 = 6a +2c

6 = a +2b +c

but i dont know how to solve three simultaneous equations

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ran-dumb

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#6

I think it's easier to think this way, the carbon compound on the right has 3 carbons, so looking at the left we have 6 carbons, so its either 1 or 2 of the CH3COCH3. Not that much work to try both, one of them doesn't work and will tell you what to put for the rest.

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KA_P

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#7

(Original post by

I've gotten

6= 3a +b

12 = 6a +2c

6 = a +2b +c

but i dont know how to solve three simultaneous equations

**val7322**)I've gotten

6= 3a +b

12 = 6a +2c

6 = a +2b +c

but i dont know how to solve three simultaneous equations

From first equation we have

b = 6 - 3a. (i)

From the second equation we have

c = 6 - 3a. (ii)

Sub this into the third equation to get

6 = a + 2(6-3a) + (6 - 3a)

6 = a + 12 - 6a + 6 - 3a

-12 = - 8 a

12/8 = a

Lol are these supposed to be integers?

Well must've made a mistake but you get the overall method right?

Then you sub the value of a into (i) then (ii) to find the values of b and c

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ghostwalker

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#8

(Original post by

Lol are these supposed to be integers?

Well must've made a mistake but you get the overall method right?

**KA_P**)Lol are these supposed to be integers?

Well must've made a mistake but you get the overall method right?

val7322 If you're not familiar with solving 3 simulaneous equations, then you can't use this method, though the poster above has given one way to do it. Can't see another easy method, given the oddity of this particular chemical decomposition. Perhaps you can make use of the other posters' suggestions.

Last edited by ghostwalker; 4 months ago

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KA_P

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#9

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#9

Ngl what I got taught to do is to make columns with the number of each element present in that molecule. And then trial by fire.

So here goes:

(1) is C6H12O6

(2) is CH3COCH3

(3) is CO2

(4) is H2O

So the total number of each element on each side should be equal therefore: (1) = (2) + (3) + (4)

(1) - C=6 ; H=12 ; O=6

(2) - C=3 ; H=6 ; O=1

(3) - C=1 ; O=2

(4) - H=2 ; O=1

I'm going to multiply (1) by 2.

So (1) - C=12 ; H=24 ; O=12

(2)+(3) [carbon] is 4 carbons. And 12/4 is 3. So let's say that (2) and (3) are potentially each multiplied by 3.

That would mean we have 18 Hydrogens in (2) now but we need 24. 24 - 18 is 6. There's 2 Hydrogens in (4) so (4) must also be multiplied by 3?

So a potential balanced equation could be:

2(1) -> 3(2) + 3(3) + 3(4)

So here goes:

(1) is C6H12O6

(2) is CH3COCH3

(3) is CO2

(4) is H2O

So the total number of each element on each side should be equal therefore: (1) = (2) + (3) + (4)

(1) - C=6 ; H=12 ; O=6

(2) - C=3 ; H=6 ; O=1

(3) - C=1 ; O=2

(4) - H=2 ; O=1

I'm going to multiply (1) by 2.

So (1) - C=12 ; H=24 ; O=12

(2)+(3) [carbon] is 4 carbons. And 12/4 is 3. So let's say that (2) and (3) are potentially each multiplied by 3.

That would mean we have 18 Hydrogens in (2) now but we need 24. 24 - 18 is 6. There's 2 Hydrogens in (4) so (4) must also be multiplied by 3?

So a potential balanced equation could be:

2(1) -> 3(2) + 3(3) + 3(4)

Last edited by KA_P; 4 months ago

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Slx.24

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#10

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#10

(Original post by

First you want a

From first equation we have

b = 6 - 3a. (i)

From the second equation we have

c = 6 - 3a. (ii)

Sub this into the third equation to get

6 = a + 2(6-3a) + (6 - 3a)

6 = a + 12 - 6a + 6 - 3a

-12 = - 8 a

12/8 = a

Lol are these supposed to be integers?

Well must've made a mistake but you get the overall method right?

Then you sub the value of a into (i) then (ii) to find the values of b and c

**KA_P**)First you want a

From first equation we have

b = 6 - 3a. (i)

From the second equation we have

c = 6 - 3a. (ii)

Sub this into the third equation to get

6 = a + 2(6-3a) + (6 - 3a)

6 = a + 12 - 6a + 6 - 3a

-12 = - 8 a

12/8 = a

Lol are these supposed to be integers?

Well must've made a mistake but you get the overall method right?

Then you sub the value of a into (i) then (ii) to find the values of b and c

(Original post by

Ngl what I got taught to do is to make columns with the number of each element present in that molecule. And then trial by fire.

So here goes:

(1) is C6H12O6

(2) is CH3COCH3

(3) is CO2

(4) is H2O

So the total number of each element on each side should be equal therefore: (1) = (2) + (3) + (4)

(1) - C=6 ; H=12 ; O=6

(2) - C=3 ; H=6 ; O=1

(3) - C=1 ; O=2

(4) - H=2 ; O=1

I'm going to multiply (1) by 2.

So (1) - C=12 ; H=24 ; O=12

(2)+(3) [carbon] is 4 carbons. And 12/4 is 3. So let's say that (2) and (3) are potentially each multiplied by 3.

That would mean we have 18 Hydrogens in (2) now but we need 24. 24 - 18 is 6. There's 2 Hydrogens in (4) so (4) must also be multiplied by 3?

So a potential balances equation could be:

2(1) -> 3(2) + 3(3) + 3(4)

**KA_P**)Ngl what I got taught to do is to make columns with the number of each element present in that molecule. And then trial by fire.

So here goes:

(1) is C6H12O6

(2) is CH3COCH3

(3) is CO2

(4) is H2O

So the total number of each element on each side should be equal therefore: (1) = (2) + (3) + (4)

(1) - C=6 ; H=12 ; O=6

(2) - C=3 ; H=6 ; O=1

(3) - C=1 ; O=2

(4) - H=2 ; O=1

I'm going to multiply (1) by 2.

So (1) - C=12 ; H=24 ; O=12

(2)+(3) [carbon] is 4 carbons. And 12/4 is 3. So let's say that (2) and (3) are potentially each multiplied by 3.

That would mean we have 18 Hydrogens in (2) now but we need 24. 24 - 18 is 6. There's 2 Hydrogens in (4) so (4) must also be multiplied by 3?

So a potential balances equation could be:

2(1) -> 3(2) + 3(3) + 3(4)

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KA_P

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#11

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#11

(Original post by

Wtf ur a mad brainiac too Can I fall harder?

**Slx.24**)Wtf ur a mad brainiac too Can I fall harder?

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it's been way too long since I did chem/maths so it could be wrong 🙈

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Slx.24

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#12

(Original post by

it'll be more difficult to catch you 🥺

**KA_P**)it'll be more difficult to catch you 🥺

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it's been way too long since I did chem/maths so it could be wrong 🙈

Tbf, I weigh a feather so might not be that hard

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I didn't even bother reading that shi I just saw the essay of maths and was like "Wao"

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#13

(Original post by

Ur right about that

Tbf, I weigh a feather so might not be that hard

**Slx.24**)Ur right about that

Tbf, I weigh a feather so might not be that hard

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I didn't even bother reading that shi I just saw the essay of maths and was like "Wao"

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My brain feels like a hackers computer sometimes - even I don't understand how it works

Last edited by KA_P; 4 months ago

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Slx.24

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#14

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😂😂 I'll go to all lengths to catch you dw

**KA_P**)😂😂 I'll go to all lengths to catch you dw

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My brain feels like a hackers computer sometimes - even I don't understand how it works

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Me who does coding

I'm in your head dun dun dunn

I'm in your head dun dun dunn

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#15

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Hahahaha you better get rdy cus I'm comingggg

**Slx.24**)Hahahaha you better get rdy cus I'm comingggg

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Me who does coding

I'm in your head dun dun dunn

I'm in your head dun dun dunn

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see just for you

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#16

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Slxxxxxx 🙈 - let's leave this thread railed shall we

**KA_P**)Slxxxxxx 🙈 - let's leave this thread railed shall we

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see just for you

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Pfft 🦋🦋

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username5794123

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#17

Why would you guys derail this thread? It’s just a guy asking help on a chem question.

Also, I condone everything that’s in post #3.

Also, I condone everything that’s in post #3.

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#18

(Original post by

Why would you guys derail this thread? It’s just a guy asking help on a chem question.

Also, I condone everything that’s in post #3.

**0ptics**)Why would you guys derail this thread? It’s just a guy asking help on a chem question.

Also, I condone everything that’s in post #3.

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#19

**0ptics**)

Why would you guys derail this thread? It’s just a guy asking help on a chem question.

Also, I condone everything that’s in post #3.

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val7322

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#20

(Original post by

Read my previous post - in full.

val7322 If you're not familiar with solving 3 simulaneous equations, then you can't use this method, though the poster above has given one way to do it. Can't see another easy method, given the oddity of this particular chemical decomposition. Perhaps you can make use of the other posters' suggestions.

**ghostwalker**)Read my previous post - in full.

val7322 If you're not familiar with solving 3 simulaneous equations, then you can't use this method, though the poster above has given one way to do it. Can't see another easy method, given the oddity of this particular chemical decomposition. Perhaps you can make use of the other posters' suggestions.

6= 3a +b

12 = 6a +2c

6 = a +2b +c

b = 6-3a

c=6-3a

a+ 2(6-3a) + 6-3a = 6

then

-8a +18 =6

then

8a =12

a= 12/8 = 3/2

subsitute it in b and c, which both equal 3/2

multiply everything by 2 so it would be

**2**C6H12O6 =

**3**CH3COCH3 +

**3**CO2 +

**3**H2O

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