## G5- moles of Na [ENDORSED]

William Cryer 1L
Posts: 12
Joined: Fri Sep 29, 2017 7:05 am

### G5- moles of Na

In exercise G5 on the homework, I'm having trouble finding the moles of Na to solve the problem. Here's the problem:

A student prepared a solution of sodium carbonate by
adding 2.111 g of the solid to a 250.0-mL volumetric fl ask and
adding water to the mark. Some of this solution was transferred
to a buret. What volume of solution should the student transfer
into a fl ask to obtain
(a) 2.15 mmol Na

Now, the solutions manual says that to find the moles of Na, you multiply the moles of the notal solution (Na2CO3) by 2, since there are 2 Na atoms in Sodium Carbonate. My question is, why would the concetration of Na be DOUBLE that of the whole molecule? Shouldn't it be 2/3's of the molarity?

I need some help understanding this concept, I understand how to solve the problem otherwise.

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Joined: Thu Jul 13, 2017 3:00 am
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### Re: G5- moles of Na

You are correct to say there are 2 atoms of Na in NA2(C03). However, I would think about that a bit differently. What this really means is that for every mole of sodium carbonate, you have 2 mols of Na; the ratio of Na to sodium carbonate is 2:1. So, to get a certain number of moles of Na, you need half the number of moles of sodium carbonate.
This is why in the solutions manual they divided by 2 (did you mean divide when you said multiply?); you only need half the moles of sodium carbonate to get the required moles of Na. The molarity you calculated is of sodium carbonate not sodium, which means you have to divide this number by the number of moles of sodium carbonate required (which is why you divided the number of mols of Na required by 2).
Hope this is clear.

Leah Savage 2F
Posts: 74
Joined: Fri Sep 29, 2017 7:06 am

### Re: G5- moles of Na  [ENDORSED]

Molarity is the number of moles per liter. There are 2 atoms of Na for every 1 atom of Na2CO3, meaning that the molar ratio between Na and Na2CO3 is 2:1. This means that the molarity, also, should have a ratio of 2:1 if the volume is constant. A way to think about this problem is to find the molarity of the Na2CO3 (.0199 moles/.25 L), which would be .0797 M, then put the values they give you (2.15 mmol Na+, 4.98 mmol CO3 2-, etc.) into moles of Na2CO3 using molar ratios. Then, set up an equation where .0797 M is equal to the moles of Na2CO3 found from 2.15 mmol Na+ or 4.98 mmol CO3 2- divided by x liters. Then, just solve for liters. The molarity has to stay equal as the moles change, so there should be a new value for liters.

Jesse Jimenez 1L
Posts: 2
Joined: Sat Jul 22, 2017 3:00 am

### Re: G5- moles of Na

When looking at this problem, the main thing is the ration between sodium and sodium carbonate is 2:1. The mistake that is throwing you off is that you have to divide by two rather than multiply. Should the ration be 1:2, then its completely different.

Alexandria Weinberger
Posts: 51
Joined: Fri Sep 29, 2017 7:03 am

### Re: G5- moles of Na

I was stuck on this problem too. Very helpful

Humza_Khan_2J
Posts: 56
Joined: Thu Jul 13, 2017 3:00 am

### Re: G5- moles of Na

With regards to this problem, you have to use the molar ratio between Sodium ions and Sodium carbonate, which is 2:1. You can appropriate calculations based on this ratio.