Help me Chem Gods!

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#### mlchang

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I believe you need the grams per mole of AgBr...

So that you can then figure out how many moles of silver of AgBr you have. If you figure out the number of moles of AgBr, then you know the number of moles of Br from the original sample.

This can tell you the number of moles that were in the original mixture. Once you do that you can subtract out the weight of the Br from the original 0.56 grams. This leaves you with the weight of sodium and potassium. Then you can solve it easily with algebra.

Once you have the remainder after subtracting you set up an equation where the weight of sodium is X and potassium is 1.8X (based on AMUs) and then solve for X

X + 1.8X = remainder

Multiply X back and you will have the weight of potassium. Figure out how many moles that is, then you know how many moles of KBr you had and then divide by 0.56 = answer.

This is the only way that I can figure to do it. Let me try it and I'll post my result.

So that you can then figure out how many moles of silver of AgBr you have. If you figure out the number of moles of AgBr, then you know the number of moles of Br from the original sample.

This can tell you the number of moles that were in the original mixture. Once you do that you can subtract out the weight of the Br from the original 0.56 grams. This leaves you with the weight of sodium and potassium. Then you can solve it easily with algebra.

Once you have the remainder after subtracting you set up an equation where the weight of sodium is X and potassium is 1.8X (based on AMUs) and then solve for X

X + 1.8X = remainder

Multiply X back and you will have the weight of potassium. Figure out how many moles that is, then you know how many moles of KBr you had and then divide by 0.56 = answer.

This is the only way that I can figure to do it. Let me try it and I'll post my result.

#### andrzejpw

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Ok, I got .5571. Can anyone confirm or deny this?

#### Born2bwire

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My guess is .377.

The reaction is going to be:

x*KBr + y*NaBr + (x+y)*AgNO3 --> (x+y)*AgBr + a*KNO3 + b*NaNO3

The KNO3 and NaNO3 may actually be K+, Na+, and NO3-. I forget whether or not they will make precipitates in aqueous solutions (silver bromide definitely will, but that's given in the problem statement).

The main problem is that we only know that all of the bromide ions react with with the silver ions. Thus, the number of moles of bromide ions added must equal the number of moles of silver bromide that are created. Now, taking the values of the atomic masses to be: K=39.10 amu Br=79.90 amu Na=22.99 amu and Ag= 107.9 amu, we can derive a system of equations relating the masses of potassium bromide and sodium bromide added and the resulting mass of the silver bromide.

Thus, we have:

x*(39.10+79.90) + y*(22.99+79.90) = .560 g

(x+y)*(107.9+79.90) = .970 g

I'll leave you to do the calculations, but I derived that the system is solved by:

x = .00177 mol

y = .00339 mol

Please note that I'm totally disregarding proper significant figures here because for speed I used my calculator to find the solutions and this does not preserve proper significant figure notations.

Now, we use the value of x to derive the mass of potassium bromide used.

.00177 mol * 119.00 g/mol = .211 g.

Now find the fraction of mass to the mass of the original sample:

.211g/.560g = .377

Thus, my answer would be .377. Feel free to pick at this the best you can, I took the AP Chem 2 years ago and haven't touched the stuff since.

The reaction is going to be:

x*KBr + y*NaBr + (x+y)*AgNO3 --> (x+y)*AgBr + a*KNO3 + b*NaNO3

The KNO3 and NaNO3 may actually be K+, Na+, and NO3-. I forget whether or not they will make precipitates in aqueous solutions (silver bromide definitely will, but that's given in the problem statement).

The main problem is that we only know that all of the bromide ions react with with the silver ions. Thus, the number of moles of bromide ions added must equal the number of moles of silver bromide that are created. Now, taking the values of the atomic masses to be: K=39.10 amu Br=79.90 amu Na=22.99 amu and Ag= 107.9 amu, we can derive a system of equations relating the masses of potassium bromide and sodium bromide added and the resulting mass of the silver bromide.

Thus, we have:

x*(39.10+79.90) + y*(22.99+79.90) = .560 g

(x+y)*(107.9+79.90) = .970 g

I'll leave you to do the calculations, but I derived that the system is solved by:

x = .00177 mol

y = .00339 mol

Please note that I'm totally disregarding proper significant figures here because for speed I used my calculator to find the solutions and this does not preserve proper significant figure notations.

Now, we use the value of x to derive the mass of potassium bromide used.

.00177 mol * 119.00 g/mol = .211 g.

Now find the fraction of mass to the mass of the original sample:

.211g/.560g = .377

Thus, my answer would be .377. Feel free to pick at this the best you can, I took the AP Chem 2 years ago and haven't touched the stuff since.

#### mlchang

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Okay so I figured...

AgBr = 188g/mole

thus in 0.97g there are 0.0052moles of AgBr.

So that means there are 0.0052 moles of Br.

0.0052 moles of Br weighs 0.416g (80g/mole Br x 0.0052 moles)

So that means there was 0.416g of Br in the original sample as well.

So that means there was 0.144g of material that is made up of Na and K. This is where I'm not sure if I'm right, but...

If sodium weighs X amount then potassium weighs roughly 1.8X (by molar weight).

So in theory X + 1.8X will tell you the amount of each to allow you to get 0.144

With all my rounding etc, I get X to be 0.055. So then 1.8X (the weight of potassium in the original sample) = 0.099g of potassium

Then this equals roughly 0.0025 moles of potassium in the original sample which equals the moles of KBr in the original sample as well which gives you 0.186g of KBr which is 33% of the original sample.

AgBr = 188g/mole

thus in 0.97g there are 0.0052moles of AgBr.

So that means there are 0.0052 moles of Br.

0.0052 moles of Br weighs 0.416g (80g/mole Br x 0.0052 moles)

So that means there was 0.416g of Br in the original sample as well.

So that means there was 0.144g of material that is made up of Na and K. This is where I'm not sure if I'm right, but...

If sodium weighs X amount then potassium weighs roughly 1.8X (by molar weight).

So in theory X + 1.8X will tell you the amount of each to allow you to get 0.144

With all my rounding etc, I get X to be 0.055. So then 1.8X (the weight of potassium in the original sample) = 0.099g of potassium

Then this equals roughly 0.0025 moles of potassium in the original sample which equals the moles of KBr in the original sample as well which gives you 0.186g of KBr which is 33% of the original sample.

#### mlchang

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Ah, Born2bwire did it the proper with way with a system.

That should be the right way. My way was just some rough guestimation, so I would disregard it.

Sorry. It's been a long time.

That should be the right way. My way was just some rough guestimation, so I would disregard it.

Sorry. It's been a long time.

#### Born2bwire

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The problem is that we can't assume that there are equal amounts of potassium bromide and sodium bromide, but interestingly enough, you got close.

#### mlchang

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Quote:

Yeah I realized that I made that assumption too late after I posted, so anyway...actually I only got close because I rounded off a bunch of numbers on my way through and so the answer was only close by chance. Heh. I feel quite stupid now. And then I worked through the system by hand and got the same answer as Born2bwire

Originally posted by Born2bwire |

Yeah I realized that I made that assumption too late after I posted, so anyway...actually I only got close because I rounded off a bunch of numbers on my way through and so the answer was only close by chance. Heh. I feel quite stupid now. And then I worked through the system by hand and got the same answer as Born2bwire

#### CaptBubba

##### Not dumb enough fora custom title...so he thought.

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Quote:

Chemistry can do that to just about anybody. At least you gave it a shot, I tried to read the question, but my brain detected chemistry and instinctivly shut down as a defense mechanism.

Originally posted by mlchang I feel quite stupid now. |

Chemistry can do that to just about anybody. At least you gave it a shot, I tried to read the question, but my brain detected chemistry and instinctivly shut down as a defense mechanism.

#### dhwilkin

##### Headphone audiophiles are practically the stuff of legend.

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Darn straight... chemistry was the bane of my academic existence.

CaptBubba said... Chemistry can do that to just about anybody. At least you gave it a shot, I tried to read the question, but my brain detected chemistry and instinctivly shut down as a defense mechanism. |

Darn straight... chemistry was the bane of my academic existence.

#### Born2bwire

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LOL, I somehow picture you as looking something like Spunky does in your avatar.

On a side note, mucho kudos on the Rocko's Modern Life avatar, I loved watching that show while growing up and I really miss it. *Sigh*

Originally posted by CaptBubba Chemistry can do that to just about anybody. At least you gave it a shot, I tried to read the question, but my brain detected chemistry and instinctivly shut down as a defense mechanism. |

LOL, I somehow picture you as looking something like Spunky does in your avatar.

On a side note, mucho kudos on the Rocko's Modern Life avatar, I loved watching that show while growing up and I really miss it. *Sigh*

#### mlchang

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No, no you don't understand...I was a BIOCHEMISTRY major in college. Heh. Quantum mechanics my ass. Oh and ask me if I remember anything about x-ray crystallography. I erased that part of my memory to make room for more useful things, like different kinds of amber ale beers that I like. Yay me.

Originally posted by dhwilkin Darn straight... chemistry was the bane of my academic existence. |

No, no you don't understand...I was a BIOCHEMISTRY major in college. Heh. Quantum mechanics my ass. Oh and ask me if I remember anything about x-ray crystallography. I erased that part of my memory to make room for more useful things, like different kinds of amber ale beers that I like. Yay me.

#### Born2bwire

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Oh you poor man. And to think I would prefer to do AP Chem work than my Intermediate Mechanics and Special Relativity Physics homework. Now if someone could save me the trouble and post an answer to this question...

For a critically damped oscillator, we have (beta)^2-w0^2 = 0 and our solution is of the form: x(t)=(A+Bt)exp^9-(beta*t).

a. Given that x(0)=x0 and v(0)=v0, determine constants A and B and write the complete solution.

b. Verify that this solution satisfies our differential equation by expanding x(t) to the 2nd order in t to confirm that it has the correct initial position and velocity.

c. Find the initial acceleration of the mass and show that it is consistent with our expectations based upon Newton's 2nd law.

This looks like a shorter (and thus

Lord I hate mechanics, but I'm looking forward to more quantum physics, which means I'm really messed up
.

For a critically damped oscillator, we have (beta)^2-w0^2 = 0 and our solution is of the form: x(t)=(A+Bt)exp^9-(beta*t).

a. Given that x(0)=x0 and v(0)=v0, determine constants A and B and write the complete solution.

b. Verify that this solution satisfies our differential equation by expanding x(t) to the 2nd order in t to confirm that it has the correct initial position and velocity.

c. Find the initial acceleration of the mass and show that it is consistent with our expectations based upon Newton's 2nd law.

This looks like a shorter (and thus

**easier**) than the normal problems we get.Lord I hate mechanics, but I'm looking forward to more quantum physics, which means I'm really messed up

#### dhwilkin

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Ouch, I'm sorry. Believe it or not, I briefly considered biochemistry as a major, since at the time I was very interested in understanding exactly how various sports supplements and hormones worked. Then I figured out that I sucked at basic chemistry (I'm talking Chem 101), and I already had a good grasp of how the supplements and body reacted together in an applicational sense. And really... I'm convinced that the only chemistry one needs is to know what H20, O2, and NaCl stand for... anything else is gravy.

mlchang said... No, no you don't understand...I was a BIOCHEMISTRY major in college. Heh. Quantum mechanics my ass. Oh and ask me if I remember anything about x-ray crystallography. I erased that part of my memory to make room for more useful things, like different kinds of amber ale beers that I like. Yay me. |

Ouch, I'm sorry. Believe it or not, I briefly considered biochemistry as a major, since at the time I was very interested in understanding exactly how various sports supplements and hormones worked. Then I figured out that I sucked at basic chemistry (I'm talking Chem 101), and I already had a good grasp of how the supplements and body reacted together in an applicational sense. And really... I'm convinced that the only chemistry one needs is to know what H20, O2, and NaCl stand for... anything else is gravy.

#### Born2bwire

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Yeah, well, you'll be eating those words the next time you accidentally down a bottle of H2SO4.