# How do you balance tricky chemical equations

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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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4 months ago
#2
(Original post by val7322)
can someone tell me their process to solve this
Hi there.

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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4 months ago
#3
(Original post by val7322)
can someone tell me their process to solve this
You will need an old style weighing scale
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4 months ago
#4
(Original post by val7322)
can someone tell me their process to solve this
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.
Last edited by ghostwalker; 4 months ago
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#5
(Original post by 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.
I've gotten
6= 3a +b
12 = 6a +2c
6 = a +2b +c

but i dont know how to solve three simultaneous equations
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4 months ago
#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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4 months ago
#7
(Original post by val7322)
I've gotten
6= 3a +b
12 = 6a +2c
6 = a +2b +c

but i dont know how to solve three simultaneous equations
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
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4 months ago
#8
(Original post by KA_P)
Lol are these supposed to be integers?
Well must've made a mistake but you get the overall method right?
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.
Last edited by ghostwalker; 4 months ago
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4 months ago
#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) Last edited by KA_P; 4 months ago
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4 months ago
#10
(Original post by 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 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) Wtf ur a mad brainiac too Can I fall harder?
0
4 months ago
#11
(Original post by Slx.24)
Wtf ur a mad brainiac too Can I fall harder? it'll be more difficult to catch you 🥺
Spoiler:
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Spoiler:
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Spoiler:
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it's been way too long since I did chem/maths so it could be wrong 🙈

0
4 months ago
#12
(Original post by KA_P) it'll be more difficult to catch you 🥺
Spoiler:
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Spoiler:
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it's been way too long since I did chem/maths so it could be wrong 🙈

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

Spoiler:
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I didn't even bother reading that shi I just saw the essay of maths and was like "Wao"
0
4 months ago
#13
(Original post by Slx.24)
Ur right about that Tbf, I weigh a feather so might not be that hard

Spoiler:
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I didn't even bother reading that shi I just saw the essay of maths and was like "Wao"
😂😂 I'll go to all lengths to catch you dw Spoiler:
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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
0
4 months ago
#14
(Original post by KA_P)
😂😂 I'll go to all lengths to catch you dw Spoiler:
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My brain feels like a hackers computer sometimes - even I don't understand how it works Hahahaha you better get rdy cus I'm comingggg

Spoiler:
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Me who does coding 0
4 months ago
#15
(Original post by Slx.24)
Hahahaha you better get rdy cus I'm comingggg

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Me who does coding Slxxxxxx 🙈 - let's leave this thread railed shall we Spoiler:
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see just for you 0
4 months ago
#16
(Original post by KA_P)
Slxxxxxx 🙈 - let's leave this thread railed shall we Spoiler:
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see just for you Damn it I nearly did it again why do I do this everytime Spoiler:
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Pfft 🦋🦋
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4 months ago
#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.
0
4 months ago
#18
(Original post by 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.
Twas unintentional 0
4 months ago
#19
(Original post by 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.
Frr 😂 Sleeps da best
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#20
(Original post by 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.
I've done it !!!!!!!!
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

2C6H12O6 = 3CH3COCH3 + 3CO2 + 3H2O
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