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Topic: 2 more stoich chem questions  (Read 4503 times)

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Offline xian

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2 more stoich chem questions
« on: August 07, 2007, 03:45:37 AM »
1. A chemical has a molar weight of 78.3. It is a reducing agent used in lots of enzyme assays to help stabilise an enzyme. Its a liquid at room temp and has a density of 1.12g/ml. You need to make up 50ml of this 5mM chemical, how much would you need to add?

I get 17.5uL or 0.0175ml.  But someone else calculated it as  143.040L.  Anyone let me know which one is correct?
 

2.A reaction is des igned to be carried out in a 5ml total volume. It requires a fi nal A-T-P (mol wt 350) con centration of 5mM. If the prot ocol allows 100 ul for the A-T-P ad.dition, what concentr ation would you make up the st.ock solution?

I get 100 mM.  Is that correct?
« Last Edit: August 07, 2007, 07:29:30 AM by xian »

Offline sdekivit

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Re: 2 more stoich chem questions
« Reply #1 on: August 07, 2007, 09:45:34 AM »
1. i agree with 17.5 uL

2.

i don't agree with 100 mM: you need 5 mL 5 mM ATP --> how many mol ATP is that?
--> that amount of mol ATP must be present in 100 uL solution so what should be you concentration ATP then ?

Offline enahs

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Re: 2 more stoich chem questions
« Reply #2 on: August 07, 2007, 09:47:16 AM »
Quote
I get 17.5uL or 0.0175ml.  But someone else calculated it as  143.040L.  Anyone let me know which one is correct?

143 Liters? Can you get 143 liters in 50 mL? If so, your friend deserves to win the Nobel Prize.

This is an example were I love to point out, do not focus just on just the math! Think about the science and the world.

Quote
2.A reaction is des igned to be carried out in a 5ml total volume. It requires a fi nal A-T-P (mol wt 350) con centration of 5mM. If the prot ocol allows 100 ul for the A-T-P ad.dition, what concentr ation would you make up the st.ock solution?

That is 100 micro liters correct?

If so, no.

When you dilute 100 micro liters into 5 mL, how much do you dilute it by?



*edit* Darn, sdekivit beat me.

Offline xian

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Re: 2 more stoich chem questions
« Reply #3 on: August 07, 2007, 10:00:30 AM »
Oh yes, silly me!  Sometimes you forget logic...

For the 2nd one, I do this:

Concentration (mM) = amount(mmol) / volume (l)

So 5mM = x/0.005

so x = 0.0025mmol.

Is that correct?

To start this problem off I am using M1V1 = M2V2

with M2=5mM
V2=100ml
V1 = 5mL

then M1 ends up equalling 100mM = (100umol/L).  But then we dilute it and get 0.00025umol/1000uL = 0.25mol/L ?

Is that the way to go about it?

Just one more thing, I have never come across g/dl...

How would I calculate how many mg contained in 50uL ?  Youve got the following solutions of compound (ml wt 200).

 x g/dl

Just some tips would help.
« Last Edit: August 07, 2007, 11:17:21 AM by xian »

Offline Yggdrasil

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Re: 2 more stoich chem questions
« Reply #4 on: August 07, 2007, 05:19:34 PM »
Here's how I would do problem #2:

MinitialVinitial = MfinalVfinal

Minitial = x
Vinitial = 100┬ÁL = 0.1mL
Mfinal = 5mM
Vfinal = 5mL
(note: Instead of using M1V1 = M2V2, I find that it is easier to work MinitialVinitial = MfinalVfinal.  That way, it's harder to confuse which volumes go with which molarities).

Therefore:  (0.1mL)x = (5mM)(5mL)
x = (5mM)(5mL)/(0.1mL) = 250mM

g/dl means grams per deciliter.  A deciliter is 1/10 of a liter = 0.1L = 100mL
So, probably the easiest way of thinking about it is that:

1g/dL = 1g/100mL

NB: g/dL is equivalent to (w/v)%, if you are familiar with that measure of concentration.

Offline enahs

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Re: 2 more stoich chem questions
« Reply #5 on: August 07, 2007, 06:00:04 PM »
Here is another way to work it (it is really kinda the same, but you look at it from a different way).

100 Micro liters = 100 x 10-6 L
5 mL = 5 x 10-3 L

  5 x 10-3
   =  50
100 x 10-6

A dilution factor of 50, you therefor need a concentration 50 times greater then what you desire in the end. 5 mM * 50 = 250mM.



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