I am having a problem wrapping my mind around the idea that in $230\:\mathrm{mL}$ of $0.615\:\mathrm{M}$ $\ce{KCl}$ there is $0.141\:\mathrm{mol}$ of $\ce{Cl^-}$ ions. I guess I still do not understand moles.

Is this calculation saying that there $0.141\:\mathrm{mol}$ of $\ce{K}$ and $\ce{Cl}$ in one mole of $\ce{KCl}$?

I know this may be a silly question, but I just want to make sure I am following the correct way of thinking when it comes to this concept.


Your concentration is 0.615 M. The M stands for molar concentration, which is the number of moles per liter of solution. Thus, your concentration of 0.615 M (mol/L) means that in one liter of solution there will be 0.615 moles of solute (KCl). Similarly, a solution with concentration of 0.615 g/L would have 0.615 grams of solute per 1 liter of solution.

A mole is just like any other unit for amount, except that it is built on a very large number. Since, for example, there at $6.02\times 10^{23}$ atoms of chlorine in 35.45 grams of chlorine, we generally do not discuss numbers of atoms (or molecules) as a useful measurable amount. How much would a dozen atoms of chlorine weigh? Much less than any of our balances or scales could measure!

One mole of stuff is an amount of stuff equal to the number of carbon atoms in 12 grams of $\ce{^{12}C}$, most common isotope of carbon. This number happens to be Avogadro's Number - $6.02\times 10^{23}$, which is really large! This is also the number of atoms or molecules present in a number of grams equal to one formula weight of that substance, so it is a convenient number. We know that when we weigh out 12.01 grams of carbon, 1.008 grams of hydrogen, 15.99 grams of oxygen, etc., we know that we always have the same number of atoms in each sample, because each is one mole. We tend not to worry too much about the large number hiding in the definition of one mole.

Now, if you have one mole of a compound, like $\ce{KCl}$, then you know that there is one mole of the $\ce{KCl}$ formula unit in that sample. Since each $\ce{KCl}$ formula unit has one $\ce{K+}$ and one $\ce{Cl-}$, then each mole of $\ce{KCl}$ has one mole of $\ce{K+}$ ions and one mole of $\ce{Cl-}$ ions.

Now, you have your 0.615 M solution, and you only have 230 mL of it. How much $\ce{KCl}$, $\ce{K+}$, and $\ce{Cl-}$ is in that sample? Well, you would have 0.615 moles of each in 1 liter, but you do not have one liter. You have 230 mL. Since there are 1000 mL in a liter, we can do a little maths:

$$230\text{ mL}\times \dfrac{1 \text{ L}}{1000\text{ mL}}\times\dfrac{0.615\text{ mol KCl}}{1\text{ L}}=0.141 \text{ mol KCl}\equiv 0.141 \text{ mol}\,\ce{K+}\equiv 0.141 \text{ mol}\,\ce{Cl-}$$

| improve this answer | |

One thing that may help is to keep in mind that a "mole" is simply a name for a number, like "dozen" (12) or "gross" (144) or "score" (20).

The difference is that a "mole" is a very large number (approximately 6.02x1023) - one that is typically only reached when you're dealing with things the size of atoms.

So 0.141 moles of KCl breaks into 0.141 moles of K+ and 0.141 moles of Cl-, just as 2 dozen eggs can be broken down into 2 dozen yolks and 2 dozen shells (and two dozen egg whites).

"Molar" simply means "moles per liter", and can be treated similarly to eggs per carton. If there are two dozen eggs per carton and you have three quarters of a carton left, you have 2 dozen eggs/carton * 0.75 carton = 1.5 dozen eggs left. You also have 1.5 dozen yolks and 1.5 dozen eggshells, because each egg has both a yolk and a shell. Likewise, having 750 mL of a 2 Molar solution of KCl means that you have 2 moles KCl/liter * 0.75 liters = 1.5 moles KCl. You also have 1.5 moles of K+ and 1.5 moles of Cl-

Keep in mind that just like "dozen" and "score", you have to keep track of what you're measuring. A dozen cars and a dozen wheels are two different things. Two boxes with 1.5 dozen toy cars per box have 3 dozen cars total but 12 dozen wheels, because there are 4 dozen wheels per 1 dozen cars (which is entirely equivalent to saying there are 4 wheels per 1 car). Likewise, for every mole of CaCl2 there are 2 moles of Cl- (which is equivalent to saying there are two chlorines for each "molecule" of CaCl2), so you'll have to work that factor of two into your calcuations if you're going from quantities of CaCl2 to quantities of chloride ions.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.