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### Topic: Calculating the number of particles per milliliter of CdSe  (Read 22241 times)

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#### fudawala

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« Reply #15 on: July 01, 2008, 03:01:48 PM »
Hi sjb,

The concentration of the sample is 10 mg/mL.

#### enahs

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« Reply #16 on: July 01, 2008, 03:05:39 PM »
Outline:

You use the density to convert volume to mass.
You use the mass and Molecular Weight of the compound to convert to # of moles.
You use the # of moles and Avagrado's Number/Constant to convert to # of molecules/particles.

You now have the # of molecules/particles in that volume.

Do the math. Follow the units and cancel them out.

#### fudawala

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« Reply #17 on: July 01, 2008, 03:16:48 PM »
The only problem that I have is to convert the volume to mass using the density. Do you know how to do that. Thanks.

#### enahs

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« Reply #18 on: July 01, 2008, 03:28:48 PM »
Use your units.

This is not the numbers for your problem, just some made up.

Density: mass / volume (example, g/mL)

So, if I have volume * density I am left with mass. Or mass / density I am left with volume.

:
density of 33.3g/mL

55 mL * 33.3 g =    1,831.5 g
1 mL

density of 10g/mL

6 g * 1 mL    = 0.6 mL
10 g

#### fudawala

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« Reply #19 on: July 01, 2008, 03:35:31 PM »
Can I ask you a question. On the sample it says the concentration which is 10 mg per mL. It also labels that the nanomole/mL is 69.4. I just want to know which one is the mass of the substance of 1 mL. Thanks again.

#### enahs

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« Reply #20 on: July 01, 2008, 03:46:07 PM »
10mg/mL implies there are 10 mg of CdSe per mL of solvent.

Where as 69.4 n moles/mL implies there are 13.28 μG (micro grams) per mL of solvent.

Those contradict each other.

Perhaps you should start over fresh. What is the question (word for word) and what are you given.

#### fudawala

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« Reply #21 on: July 01, 2008, 05:19:07 PM »
My question is to calculate the number of molecules per milliliter of a solution?

The solution is a Core Shell Evidot in Toluene of a Hops Yellow color. The molecular weight is given to me which is 20 micrograms/nmol. This value is given to me from the table.

Another thing that is given to me is listed on the solution bottle itself. It labels the absorption and emission peaks in wavelength. The absorption peak is 565 nanometers. The emission peak is 578 nanometers.

The concentration is given to me which is 10 mg/mL.
The nmol/mL is also given to me which is 69.4.

I'm being asked to find the number of particles per milliliter of this substance? I don't know how to solve the problem from this given to me?

Thanks Again. I appreciate your help. Please I need your help on this one.

#### Borek ##### Re: Calculating the number of particles per milliliter of CdSe
« Reply #22 on: July 01, 2008, 05:30:57 PM »
Either you are given wrong information, or you are misreading it. 20 μg/nmol is at least expressed in strange units, but it means molar mass of 2 x 105 g/mol. Then you are given information that in each mL there is 10 mg of substance and it is 69.4 nmol. This in turn means molar mass in the range of 1.4 x 105. Molar mass of CdSe is 191.37 g/mol. As nothing fits, I doubt anyone will be able to help.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### fudawala

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« Reply #23 on: July 01, 2008, 06:22:14 PM »
Can you tell me how did you get 2 X 10^5 of one molar mass. The other molar mass of 1.4 X 10^5.

The molecular weight is not 20 micrograms per nanomole. Gave wrong info. But the molecular weight is 23 micrograms per nanomole.

#### Borek ##### Re: Calculating the number of particles per milliliter of CdSe
« Reply #24 on: July 01, 2008, 06:25:36 PM »
20 x 10-6/10-9 & 0.01/69.4 x 10-9.

And sorry, typo, first one should be 20k.

Molar mass is mass of the sample divided by the number of moles in the sample.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### fudawala

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« Reply #25 on: July 01, 2008, 06:35:13 PM »
.01 is the density right because the concentration was 10 mg/mL.

#### fudawala

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« Reply #26 on: July 01, 2008, 06:57:36 PM »
Can someone explain me what is a molar extinction coefficient at first exciton up to 10^5. The Core Shell Evidot of the Hops Yellow color is about 0.98 (molar extinction coefficient at first exciton [10^5]).

#### fudawala

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« Reply #27 on: July 01, 2008, 07:08:59 PM »
Hi enahs,

I just want to know how did you get 13.28 micrograms from 69.4 nanomoles.

Thanks

#### enahs

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« Reply #28 on: July 01, 2008, 07:32:54 PM »
You said the compound was Cadmium Selenide, which as a molecular weight of 191.4 g/mol.
You said it says 69.4 nanomoles / 1mL

69.4 nanomoles = 69.4 (10-9)moles

69.4 (10-9) moles * 191.4 g  = 13.2 (10-6) g = 13.2 μg
1 mole

The molar extinction coefficient is the older term for the molar absorptivity coefficient.  It is a constant for a specific substance in a specific solvent. It relates the absorbance of the compound to its concentration and path length in the Beer-Lambert law.

Your material is not juse Cadmium Selenide though as you previously said. Based on the "Core Shell Evidot" comment, it is actually a more complicated mixture with Zinc Sulfide and such, not just the CdSe. But who knows what proportions the ZnS and CdSe are in, so the molecular weight can not be added up from constitute parts.

The best you can do is take the bottles word.

You say the bottle says 69.4 nmol/mL and 10mg/mL, which means there are 10mg/69.4 nmol
10 (10-3) g     *         1 mL          = 144,092 g/mol
1 mL                   69.4 (10-9)mol

But your quantum dots also probably have amines in them as well to "cap" them.

But back to your first question:
Quote
My question is to calculate the number of molecules per milliliter of a solution?
You are told there is 69.4 nmol's of stuff in 1 mL. Therefor, multiplying it by avagrados number gives you how many molecules of stuff in 1mL.

This is a very poor question and answer though.

#### fudawala

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« Reply #29 on: July 01, 2008, 07:54:58 PM »
Can I ask a question about the mixture between CdSe and Zinc sulfide. You said that when you mix these two compounds the molecular weight does not add. Can you tell me why?