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Chemistry: Mixture and Mass

by Tricause
Tags: chemistry, mass, mixture
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Tricause
#1
Sep2-09, 05:45 PM
P: 8
1. The problem statement, all variables and given/known data
A mixture of NaCl and NaBr has a mass of 2.08 g and is found to contain 0.76 g of Na. What is the mass of NaBr in the mixture?

2. Relevant equations
Molar mass equations (possibly), formula mass

3. The attempt at a solution
I've calculated the percent composition of Na in both NaCl and NaBr to be 36.22 and 63.78% respectively. With the given information I have found that sodium makes up 36.54% of the the total 2.08 g.

I am currently stuck on where to go from here since the mass of sodium in the mixture is a different percentage from an ideal mixture where it would be 28.50% of the mixture. Any help in the right direction would surely be appreciated.
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Bohrok
#2
Sep2-09, 06:25 PM
P: 867
I'm not sure how you can do it using percent compositions, at least I can't see it.

How many moles of Na do you have? Since the elements in each compound are 1 to 1, the moles of Na and Br added together equal the total number of moles of Na in the mixture.
What's the mass of the Br and Cl together? The moles of each times their atomic weight, added together will equal that mass.

I would try solving this as a system of two equations with the two variables being the moles of Cl and Br in the mixture.
Tricause
#3
Sep2-09, 07:37 PM
P: 8
Quote Quote by Bohrok View Post
I'm not sure how you can do it using percent compositions, at least I can't see it.

How many moles of Na do you have? Since the elements in each compound are 1 to 1, the moles of Na and Br added together equal the total number of moles of Na in the mixture.
What's the mass of the Br and Cl together? The moles of each times their atomic weight, added together will equal that mass.

I would try solving this as a system of two equations with the two variables being the moles of Cl and Br in the mixture.
I did: [tex]\frac{.76 g Na}{1}[/tex] x [tex]\frac{mol Na}{22.99 g}[/tex] = .033 mol Na

2.08 gmixture - .76 gNa = 1.32 gCl + Br

I then tried: [tex]\frac{.033 mol Na}{1}[/tex] x [tex]\frac{mol Br}{2 mol Na}[/tex] x [tex]\frac{79.909 g}{mol Br}[/tex] = 1.32 g Br

That mass of Br cannot be true, so apparently I am doing something wrong.

Bohrok
#4
Sep2-09, 08:46 PM
P: 867
Chemistry: Mixture and Mass

The last equation is wrong. You won't be using moles of Na, and you need to be adding the masses of Cl and Br.

After you let two variables represent mol of Cl and mol of Br, convert each one to mass with their atomic weights. They need to be added and set equal to the mass of 1.32 g; this mass is the mass of Cl and Br together, not just Br.
Tricause
#5
Sep2-09, 09:19 PM
P: 8
Sorry, I was slightly confused. Now I have found there to be .266 g of Br and 1.05 g of Cl, matching the total mass. Thank you for the help so far, but I have one final question. How do I determine the mass of NaBr from the mass of Na and Br when Na is split into two different molecules?
Bohrok
#6
Sep2-09, 09:36 PM
P: 867
From what you have, find the number of moles of Br. That number will be the same number of moles of Na in NaBr since Na and Br are 1 to 1. Use that number with the atomic mass of Na and add the masses of the two to get the mass of NaBr.
Tricause
#7
Sep2-09, 09:54 PM
P: 8
I've calculated the proper mass; thank you very much for your help.


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