Sciencemadness Discussion Board - How can this be a balanced equation?

How can this be a balanced equation?

CHRIS25 - 14-4-2012 at 11:08

Balanced Equation: C6H8O7 + 0H2O = C6H8O7

I tried working out myself, got stuck with trying to balance the oxygen side of things. So I looked it up on a calculator and got this???

I have citric acid granules and want to make concentrated citric acid. I presume? that the 192 g/moles for citric acid is to be taken as 192 grams per litre of water? I actually have learned the gram mole thing but suddenly hit the problem of how to work out how to know when a solution is concentrated.

Always a thought when you have just asked a question. Is the answer 10H2O (192 divided by 18)? 10 grams of water to 192 grams of acid?

[Edited on 14-4-2012 by CHRIS25]

[Edited on 14-4-2012 by CHRIS25]

barley81 - 14-4-2012 at 14:25

You want to make a solution of citric acid?

There is no chemical change, so you don't really need an equation.
You could write (citric acid)(s) -> (citric acid)(aq) but that's not really necessary. How strong exactly do you want the solution? Citric acid is really soluble in water (73g/100mL according to wiki). I suppose that would be concentrated, but "concentrated" doesn't imply any specific concentration for citric acid.

[Edited on 14-4-2012 by barley81]

Polverone - 14-4-2012 at 14:31

Quote: Originally posted by CHRIS25

Balanced Equation: C6H8O7 + 0H2O = C6H8O7

I tried working out myself, got stuck with trying to balance the oxygen side of things. So I looked it up on a calculator and got this???

I have citric acid granules and want to make concentrated citric acid. I presume? that the 192 g/moles for citric acid is to be taken as 192 grams per litre of water? I actually have learned the gram mole thing but suddenly hit the problem of how to work out how to know when a solution is concentrated.

Always a thought when you have just asked a question. Is the answer 10H2O (192 divided by 18)? 10 grams of water to 192 grams of acid?

You appear to have two problems here. The first equation is perfectly balanced but "0H₂O" is redundant. Perhaps it was changed during editing?

The second problem is that of making a citric acid solution of known concentration. You are on the right track: one mole of citric acid in one liter of water gives a solution of one molar concentration. That is, 192 grams of anhydrous citric acid in 1 liter of water gives a one-molar solution of citric acid. If you have citric acid monohydrate, instead of the anhydrous acid, you will have to adjust the mass up correspondingly. Whether the acid you have is anhydrous or a hydrate can only be answered by the product packaging or analysis. You can prepare a smaller quantity of one-molar citric acid solution by scaling down both the water and citric acid quantities. For example, multiply both quantities by 0.2 if you wanted only 200 mL of one-molar citric acid solution. You can also of course adjust the water:acid ratio up or down to prepare solutions of lesser or greater molarity.

There is no hard and fast line to draw where a "concentrated" solution begins, though a saturated solution should qualify. The quantity of a substance in solution at saturation can only be found by experiment. You can look up tabulated experimental results for many organic and inorganic materials in books like Lange's Handbook of Chemistry or the CRC Handbook of Chemistry and Physics. Much of this data is online too, though beware of figures without citations. Solubilities are often expressed in grams of solute per 100 mL of solvent, so you will have to use the formula weight to calculate molarity of the saturated solution.

Pyridinium - 14-4-2012 at 16:04

Quote: Originally posted by Polverone

192 grams of anhydrous citric acid in 1 liter of water gives a one-molar solution of citric acid.

Been on here long enough to know that Polverone knows better (this is what happens sometimes when typing out a response quickly)... but just thought I'd point out that a 1 molar solution would actually be 192 grams citric acid in 1 liter of *solution*, rather than 1 liter of water. Add the citric acid to a little bit of water in a 1000 ml volumetric flask, swirl it around to dissolve, then make up the total volume to 1000 ml.

Dissolving a solid in water creates a volume change. At very low concentrations this is sometimes disregarded.

But yeah, other than that... what he said.

He even recommended Lange's Handbook. When I first saw the post I thought wait a minute, did I forget and type a post on here?

watson.fawkes - 14-4-2012 at 17:49

Quote: Originally posted by barley81

There is no chemical change, so you don't really need an equation.
You could write (citric acid)(s) -> (citric acid)(aq) but that's not really necessary.

You need an equation here if you're tracking ΔG for thermodynamic analysis. It seems to be a pretty common error in such calculations to leave out the phase changes.

barley81 - 14-4-2012 at 18:52

Quote: Originally posted by watson.fawkes

You need an equation here if you're tracking ΔG for thermodynamic analysis. It seems to be a pretty common error in such calculations to leave out the phase changes.

True!

CHRIS25 - 14-4-2012 at 23:23

Barley , Polverone and Pyridinium. Very much appreciated. Yes, thankyou for that reference, I shall hunt it done on the web.

CHRIS25 - 14-4-2012 at 23:27

""How strong exactly do you want the solution? Citric acid is really soluble in water (73g/100mL according to wiki)""

I must have missed that on wiki, thought I read it properly - whoops. Yes I am experimenting with a saturated solution, heaqting copper and then applying a pure brass rub and then brushing on citric acid by accident produced the most incredible metallic rainbow display of colour, something I thought only possible with my cold potassium pyrosulphate treatments. So I want to see if the reaction deepens colours through a very concentrated solution.

arsphenamine - 15-4-2012 at 10:37

Pardon, but that 192 g/mol is a molecular mass or formula weight, not a solubility figure.