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# Yet another Chemistry question! PART 4 added 7/14/10

Registered User regular
edited July 2010
Another Chemistry question for you smart people out there.
A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4 , for a laboratory experiment. How many grams of Na3PO4 will be needed to produce 750mL of a solution that has a concentration of Na+ ions of 1.50 ?

I've spent about an hour on this problem and can't figure out how it got the answer "61.5g" Can someone PLEASE point me in the right direction? The book is entirely worthless.
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Part 2:

How many grams of HCl are formed from reaction of 3.56g of H2 with 8.98g of Cl2?

Reaction: H2 + Cl2 => HCl

It seems like a simple question but I can't think of anyway to do this. Do I convert H2 and Cl2 to moles? How would this help me at all?

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Part 3:

Give the quantum numbers for the outer most electron (by filling order) in Co

Co
n = 3
l = 2
ml = -2,-1,0,1,2
ms = +- 1/2

My question is how they arrived at n = 3 and l = 2. I did the electronic configuration for Co, and it's:

1s2 2s2 2p6 3s2 3p6 3d7 4s2

Does that make sense? How do they arrive at n = 3, l = 2?
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Part 4:

This one is hard to type out. Basically I am trying to find the formal charge of an atom, with a given structure.

The formula I have is: # valence electrons - # lone pair electrons - # bonds

One structure has Sulfur with two pairs of lone electrons and two bonds. The options are:

a. 0 b. 1 c. -1 d. -2 e. 2

I said e, because Sulfur has 6 valence electrons, so I subtract 4 from that ( 2 bonds and 2 lone pair electrons) to get 2. But his answer key says: "ba". I assume this is a typo, but neither b or a make sense. Am I doing something wrong?

The structure looks something like this:

H ---- S(two dots above, and two dots below)
C ---(3 bars)---- N:

The reason why I'm confused is that even without the typo the rest of the answers for this part don't make any sense. Am I right or is he?

urahonky on

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San DiegoRegistered User regular
edited June 2010
The concentration unit on 1.50 is apparently M, which is important. So:

1. Find the amount of Na3PO4 needed (moles)

1a. Find target concentration of Na+ in 750 mL
1.50 mol Na+ / 1,000 mL solution = x mol Na+ / 750 mL
x = 1.125 mol Na+

1b. Convert to Na3PO4
3 Na+ : 1 Na3PO4, so
1.125 mol Na+ : x mol Na3PO4
x = 0.375 mol Na3PO4

2. Calculate molecular weight of Na3PO4
Na = 23, Na3 = 69
P = 31
O = 16, O4 = 64
Na3PO4 = 164 g/mol

3. Calculate amount of Na3PO4 needed (g)
0.375 mol Na3PO4 * 164 g/mol = 61.5 g Na3PO4

Orogogus on
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Registered User regular
edited June 2010
Wow yeah... I was close all the way to... 1a, then I lost it. So how did you know the ratio of Na+ to the entire formula? I'm not sure I get how it was a 3:1 ratio... I know it's N3, but how... Does that mean every time this formula is used, there are 3 Na+'s? Is that the easiest way to remember it?

Also i didn't realize that you can just make M == mol like that.

urahonky on
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Registered User regular
edited June 2010
The ratio of 3:1 is because of the formula. That is to say, one mole of Na3PO4, when fully ionised, will yield 3 moles of Na+. Another example would be say if you were using NaCl instead. Now in this case, one mole of NaCl will yield one mole of Na+ when fully ionised, so it's a 1:1 ratio.

Iso on
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San DiegoRegistered User regular
edited June 2010
It's been a while, so I could have it wrong, but to the best of my recollection M is mol/L. M is a measure of concentration, while mol is a count (like how doz would mean 12).

The formula tells you how many of each ion you'll end up with. Since it's Na3PO4, assuming it dissolves entirely you end up with 3 Na+ ions and 1 (PO4)--- ion for each molecule of Na3PO4 you start with.

Like, if the formula was X57Y21, you'd know that each molecule of X57Y21 would dissolve into 57 Xs and 21 Ys.

Orogogus on
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Registered User regular
edited June 2010
Ah okay that makes sense then. That is why you divide the mol of Na by 3, to get the mol of the entire formula. That was the framework needed for me to understand how to get this answer. I really wish something like that was written in the book since, you know, this question CAME FROM THE BOOK. Ugh.

urahonky on
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Registered User regular
edited June 2010
Updated OP with new question... I'm sure it's an easy one. I can't wait till these 5 weeks are done...... Just two to go.

urahonky on
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San DiegoRegistered User regular
edited June 2010
urahonky wrote: »
Part 2:

How many grams of HCl are formed from reaction of 3.56g of H2 with 8.98g of Cl2?

Reaction: H2 + Cl2 => HCl

It seems like a simple question but I can't think of anyway to do this. Do I convert H2 and Cl2 to moles? How would this help me at all?

This is a limiting reactant question. If you have an infinite amount of Cl2 and a little H2, then your H2 supply determines how much HCl you can make. So you figure out if you have more moles of H or of Cl and then calculate how many g of HCl that would be.

The reaction should be H2 + Cl2 -> 2 HCl, though.

Orogogus on
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Registered User regular
edited June 2010
So if I found out H => 1.78 mol, and Cl => .1266573 mol, the limiting reactant is Cl. From there... I'm not sure how to proceed. So assume .1266573 is the total mol in HCl? Use that to find the weight?

urahonky on
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Registered User regular
edited June 2010
Hi again.

For this question, you first have to find out how many moles of HCl can be formed. And to find that, you need to find out how many moles of H and Cl you have. What they give you is the mass of the H and Cl, which is a measure of quantity, but worthless in terms of chemical reactions, so you need to convert.

Number of moles of H:
3.56 / 1.00794 = 3.532 moles

Number of moles of Cl:
8.98 / 35.453 = 0.25329 moles

Now, H and Cl combine to form HCl in a 1:1 ratio, so it seems that we have more H atoms and Cl atoms. You need one H for every Cl, so in this case, we can only form 0.25329 moles of HCl. The mass of HCl formed is therefore:

0.25329 x (1.00794 + 35.453) = 9.235g

Iso on
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Registered User regular
edited June 2010
So do I look at the right hand side in order to get my moles or the left? I got H as 1.78 moles, which is half of what you've got because I put the MW as 2, instead of 1.

urahonky on
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Registered User regular
edited June 2010
urahonky wrote: »
So if I found out H => 1.78 mol, and Cl => .1266573 mol, the limiting reactant is Cl. From there... I'm not sure how to proceed. So assume .1266573 is the total mol in HCl? Use that to find the weight?

In this case, you're mistaking H2 for H. With 3.56g of H2, you have 1.78 moles of H2, but 3.56 moles of H, and it's important to make that distinction in your calculations.

The formula for the reaction is H2 + Cl2 => 2HCl. You could perform the calculations like you did, e.g. find out the number of moles of H2, and Cl2, then convert, but it's far easier to just find out how many moles of H and Cl you have, because that's your final product. And referring to the final product makes things easier when the product is something like Fe2O3, and you just find out how many 'blocks' of Fe2 and O3 you have to build the molecule with.

Iso on
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Registered User regular
edited June 2010
urahonky wrote: »
So do I look at the right hand side in order to get my moles or the left? I got H as 1.78 moles, which is half of what you've got because I put the MW as 2, instead of 1.

You can look at either side, but it's important to note that if you look at the left side, you get 1.78 moles because the MW of H2 is 2. This means you have 1.78 moles of H2. However, for the purpose of this calculation, you don't really care how much H2 you have, what you care about is how much H you have.

Iso on
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Registered User regular
edited June 2010
Hmm, I think an easier way to look at it is say for example, you have 10 boxes of burgers containing 2 burgers each, and 20 boxes of chips containing 2 servings of chips each. You want to find out how many set meals you can make (a set meal being one burger + one serving of chips).

Referring to your earlier calculations, you have calculated the number of boxes of burgers (e.g moles of H2) you have. But in this calculation, what's important isn't how many boxes you have, but how many burgers you have. So, you then see that you have 20 burgers, and 40 servings of chips. Now, one meal requires one burger and a serving of chips, so you can only make 20 set meals. And then it's a matter of finding out how much 20 set meals weigh.

Iso on
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Registered User regular
edited June 2010
Ah, that would be why I got the wrong answer then. :P

urahonky on
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Registered User regular
edited June 2010
Looks like I'm getting the hang of this. Seriously guys you are the best and I really appreciate all the help.

urahonky on
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Registered User regular
edited July 2010
Part 3 added to the OP, if anyone would be so kind as to help me out I would love them forever! I have the rest of this sample test figured out except this one!

urahonky on
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San DiegoRegistered User regular
edited July 2010
urahonky wrote: »
Part 3:

Give the quantum numbers for the outer most electron (by filling order) in Co

Co
n = 3
l = 2
ml = -2,-1,0,1,2
ms = +- 1/2

My question is how they arrived at n = 3 and l = 2. I did the electronic configuration for Co, and it's:

1s2 2s2 2p6 3s2 3p6 3d7 4s2

Does that make sense? How do they arrive at n = 3, l = 2?

You're going after 3d7. The 3 in 3d7 is where n = 3 comes from, and the d is where l = 2 comes from -- s=1, p=2, d=3, f=4.

You don't figure for 4s2 because 4s comes before 3d in the electron filling order.

Orogogus on
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Registered User regular
edited July 2010
So if l = 2, then it's d? But you wrote d=3. So is it +1? Or is s = 0, p = 1, d = 2, etc?

urahonky on
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San DiegoRegistered User regular
edited July 2010
Bah, sorry. Yes, s=0 etc.

Orogogus on
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Registered User regular
edited July 2010
Aufbau's rule will guide you in all your electron filling needs
1s
2s 2p
3s 3p 3d
4s 4p 4d 4f

Just make sure to draw arrows to remind you how they fill.
There should be an easy pattern!

Fuzzy Cumulonimbus Cloud on
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Registered User regular
edited July 2010
Thanks for the help thus far guys. It seriously has made this quarter a lot easier. Another question posted that I need help with.

Only two more days of this class left! Tomorrow is the Chapter 6/7 exam, and Thursday is the final. This is going to be a terrible week, but it'll be worth it in the end.

urahonky on
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When the last moon is cast over the last star of morning And the future has past without even a last desperate warningRegistered User, Moderator mod
edited July 2010
Despite my being in organic chemistry II right the fuck now, I'm a little shaky on things like formal charge because my genchem 10 years ago spent more time balancing equations than talking about formal charge.

However, I believe the correct answer is 0. If S has two loan PAIRS, that's 4 electrons right there. Plus the two bonds and you get 6. 6-6=0.

ceres on
And it seems like all is dying, and would leave the world to mourn
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Registered User regular
edited July 2010
Ah shit. You're right. I thought it was asking the number of pairs, not the number of electrons. That makes much more sense. Thank you ceres. Now I won't be missing 20% of my grade on the exam tomorrow.

urahonky on