Calculating Moles in C1V1=C2V2 Reactions

[Procrastinate]

Just a quick question, when calculating C₁V₁ = C₂V₂, do I need to account for Mole Rato?

Say that C1 had 1 moles required in reaction as opposed to C2 which required 2 moles, would I just multiply C2 by 1/2?

 

[QuarkCharmer]

I'm not sure I understand your question exactly.

You can't use the mole ratios given by C1V1 = C2V2 because the equation is unbalanced. It's also a sort of identity!

Let's assume:
$$2KOH + H_2SO_4 \rightarrow K_2SO_4 + 2H_2O$$
as an example. What are you asking exactly?

 

[Procrastinate]

I'm not sure I understand your question exactly.

You can't use the mole ratios given by C1V1 = C2V2 because the equation is unbalanced. It's also a sort of identity!

Let's assume:
$$2KOH + H_2SO_4 \rightarrow K_2SO_4 + 2H_2O$$
as an example. What are you asking exactly?

Okay, say I want to calculate the Concentration of K2SO4 with the known concentration and volumes of 2KOH.

I plug both of these numbers into the C1V1=C2V2 calculation. Would I need to divide the concentration of 2KOH by 2 in the equation like I would when doing other similar equations i.e. finding the moles of a particular substance?

 

[QuarkCharmer]

It's easier to think of these problems in terms of "Dimensional Analysis"

Let's say you had 15 grams of KOH and you wanted the resulting mass of K2SO4. (first you should determine the limiting reactant, but let's assume for now that there is an infinite supply of H2SO4.

The first thing you have to do is figure out how many moles of KOH 15 grams is.
$$15g KOH (\frac{1 mol KOH}{(39+16+1)g KOH})$$

Now, looking at the balanced formula, you know that the ratio of KOH to K2SO4 is 2/1 (The coefficients). You simply put that into the analysis.
$$15g KOH (\frac{1 mol KOH}{(39+16+1)g KOH})(\frac{1 mol K2SO4}{2 mol KOH})$$
(grams of KOH and moles of KOH should cancel out in the multiplication leaving a number and the desired unit, moles in this example)

This produces the max possible number of moles that the 15 grams of KOH can produce. You can convert that to grams by using the mass numbers from the periodic table, like I did for the first unit factor in the Dimensional Analysis.

You have to pay close attention to what units you are using, and be sure that before you multiply in a unit factor converting the mols of one thing to another you are actually dealing with moles. Also be sure that the top and bottom of the equation cancels out the unwanted units, that is always a good sign you are on the right track.

The result is something like this:
$$15g KOH (\frac{1 mol KOH}{(39+16+1)g KOH})(\frac{1 mol K2SO4}{2 mol KOH})(\frac{(39*2+32+16*4)g K2SO4}{1 mol K2SO4}) = 23.304 g K2SO4$$
Edit: I just re-read your first post! I didn't catch that you were talking about dilutions. You don't need to account for any mol ratio in the equation. The molarity is already given as the initial and final variable.

 

[Norfonz]

Yes, you have to account for the mole ratio because you are often going to have stoicheometries which are not one-to-one.

Saying c1v1 = c2v2 is a statement of the conservation of mass by the way, moles in equal moles out. These moles are a reflection of one species technically, so if you have something reacting, you need the coefficient which corresponds to the species you are looking at in the balance.

 

[Borek]

C₁V₁ = C₂V₂

That's the worst way of remembering how to do this type of calculation. For some reason most students remember simplified version that works only in some cases, but they have no idea what the general approach is.

http://www.titrations.info/titration-calculation