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### #169840Calculating concentration

Posted by on 04 August 2014 - 04:47 AM in Chemistry

yes, you are correct now.

out of curiosity, did you control the spread of the compound on the filter paper? or did you just allow the compound to spread as far as the volume took it?

Thank you. we used split petri dishes. In the control plate, one side of the petri plate is placed with plug of fungi. As soon it reaches the partition of the plate, we took readings.

### #169835Calculating concentration

Posted by on 04 August 2014 - 03:52 AM in Chemistry

Thank you. phage434 and Pito. I simply took 200, 100 and 50 microlitres and used directly for my experiment (3 different volumes of same pure compound, but I would like to express as concentration). I did not diluted it. Coming to my calculation if it is 3.97 mol/L, then can it be a 0.397 M in 100 microlitres? I am not sure, if I calculated it right. I calculated based on : 3.97 mol in 1 Litre. So 1 litre = 1000 ml. So in 100microlitres it is 0.0397 M . Am I right now?

But the concentration is the same than!

If you did not dilute it you would just have added the same concentration to your sample...

The sample itself of course would have an effect on the end concentration.

I will give you a simple example:

if you have 10 moles per liter for a certain solution than this concentration does not change whether you have taken 100 microliter or 50 microliter.

The concentration is the same!

The amount of moles is different of course: for 100 microliter you have twice the amount as for 50 microliter.

1 liter = 10 moles

100 microliter = 0,1 ml = 0,1ml x 10^-3 liter , this meaning 10/10 000 moles!

50 microliter is than the half of 10/10 000 moles..

Now the end concentration of course depends on to what you add this 100 microliter or 50 microliter.

Thank you. The result of my experiment is like this, as the volume increases from 50 to 200, the results were better. I did not get good results when I used 50 microlitre. Does this mean concentration has an effect on my sample? So you are saying, even though the volume increases, concentration remain same? Am I right? Do I need say as the volume increases the results were better, (instead of saying as concentration increases the results were better)?

But what did you do with the "volumes" ?

Did you add the 200 microliter to something else?

The concentration stays the same! In the 200 microliter , 50 microliter and 100 microliter the concentration is the same!

The total amount of your substance is different of course..... because 200 microliter has more moles in it than 50 microliter!

Sorry I am totally confused with the term concentration and amount. I did not diluted 200 microlitres. I know, you understood my point. But I am confused. I am working on fungal inhibition with a pure chemical compounds described above. I took 3 volumes and tested on fungal inhibition. When I used 200 microlitre the inhibition was higher, but could not see inhibition at 50 microlitre. So If I understood correctly, I used only one concentration (is conc expressed in %?) or  (I think concentration is expressed in moles, millimoles, micromoles - if I dilute from 1 mole to 1milli mole, the concentration usually decreases because of dilution?). So in my experiment, is the quantity/amount or concentration (do not know the right term here to be used) have an effect on fungal growth. Also in one paper, authors used 1 ng  1 μg  1 mg. They also found good results at higher 1 mg. But they said concentration? Is this right term or do they have say quantity/amount.

If you are saying 200 microliters has more moles than in 50 microlitre (Is this not a concentration?). Then in my calculation described above  3.97 mol in 1 Litre. So 1 litre = 1000 ml. So in 100microlitres it is 0.0397 M . So 100 microlitres has less moles?

Thank you and Sorry

Concentration = a certain amount per volume. Meaning: X mg per X ml (or grams per liter , or nano grams per microliter or mg per liter...) or moles/ liter (this is than the molar concentration.

Amount = total number of a certain component...; for example: 10 grams or 1 kilogram or .... (or could also be expressed as a volume).

for you its like this:

200 microliter of a certain concentration added to a sample

100 microliter ....

50 microliter...

The concentration of those 3 volumes is the same!

The amount however is different! 200 microliter holds more of the component than the 50 microliter.

But what did you do ? I still do not understand it.

You added to fungal (spores) to the 200 microliter?

About that paper: the amounts they describe are (I think) the total amount of that drug: meaning they added 1 ng , 1 microgram or 1 mg in total to a certain fixed volume... but I do not know this without the correct information.

Another oversimplified example:

component Y has a molar concentration of 8 moles/liter.

If I take 1 liter I have 8 moles.

If I take 0,5 liter I have 4 moles.

The concentration of the 1 liter and 0,5 liter is however the same:

1 liter = 8 moles, 8moles/1liter = 8moles per liter

0,5 liter = 4 moles , 4moles/0,4liter = 8 moles per liter

The total amount of moles is however different: 1 liter contains 8 moles , 0,5 liter is only 4 moles...

Thank you. I understan

If you are saying 200 microliters has more moles than in 50 microlitre (Is this not a concentration?). Then in my calculation described above  3.97 mol in 1 Litre. So 1 litre = 1000 ml. So in 100microlitres it is 0.0397 M . So 100 microlitres has less moles?

Thank you and Sorry

Yes, 200 microliter has more moles than 50, but the concentration is the same!

3,97 moles in 1 liter , 1 liter = 1000 ml , so 1000 ml will have 3,97 moles.

1 ml will have 3,97/1000 moles!

100 microliter will have 3,97/10000 moles!

However the molar concentration (M) is the same for all!

moles =/= M !

M = moles per liter! Its a concentration.

Thank you. I put that chemical on filter paper and exposed to fungi (with out direct contact). I understood you point now. The molar concentration is same for all 3 volumes (The molar concentration will only decrease when you diluted). Here in my experiment the volume concentration has an effect on fungi, which means there is more compound in 200 microlitres than in 100 microlitres? Am I right now

### #169831Calculating concentration

Posted by on 04 August 2014 - 03:35 AM in Chemistry

Thank you. phage434 and Pito. I simply took 200, 100 and 50 microlitres and used directly for my experiment (3 different volumes of same pure compound, but I would like to express as concentration). I did not diluted it. Coming to my calculation if it is 3.97 mol/L, then can it be a 0.397 M in 100 microlitres? I am not sure, if I calculated it right. I calculated based on : 3.97 mol in 1 Litre. So 1 litre = 1000 ml. So in 100microlitres it is 0.0397 M . Am I right now?

But the concentration is the same than!

If you did not dilute it you would just have added the same concentration to your sample...

The sample itself of course would have an effect on the end concentration.

I will give you a simple example:

if you have 10 moles per liter for a certain solution than this concentration does not change whether you have taken 100 microliter or 50 microliter.

The concentration is the same!

The amount of moles is different of course: for 100 microliter you have twice the amount as for 50 microliter.

1 liter = 10 moles

100 microliter = 0,1 ml = 0,1ml x 10^-3 liter , this meaning 10/10 000 moles!

50 microliter is than the half of 10/10 000 moles..

Now the end concentration of course depends on to what you add this 100 microliter or 50 microliter.

Thank you. The result of my experiment is like this, as the volume increases from 50 to 200, the results were better. I did not get good results when I used 50 microlitre. Does this mean concentration has an effect on my sample? So you are saying, even though the volume increases, concentration remain same? Am I right? Do I need say as the volume increases the results were better, (instead of saying as concentration increases the results were better)?

But what did you do with the "volumes" ?

Did you add the 200 microliter to something else?

The concentration stays the same! In the 200 microliter , 50 microliter and 100 microliter the concentration is the same!

The total amount of your substance is different of course..... because 200 microliter has more moles in it than 50 microliter!

Sorry I am totally confused with the term concentration and amount. I did not diluted 200 microlitres. I know, you understood my point. But I am confused. I am working on fungal inhibition with a pure chemical compounds described above. I took 3 volumes and tested on fungal inhibition. When I used 200 microlitre the inhibition was higher, but could not see inhibition at 50 microlitre. So If I understood correctly, I used only one concentration (is conc expressed in %?) or  (I think concentration is expressed in moles, millimoles, micromoles - if I dilute from 1 mole to 1milli mole, the concentration usually decreases because of dilution?). So in my experiment, is the quantity/amount or concentration (do not know the right term here to be used) have an effect on fungal growth. Also in one paper, authors used 1 ng  1 μg  1 mg. They also found good results at higher 1 mg. But they said concentration? Is this right term or do they have say quantity/amount.

If you are saying 200 microliters has more moles than in 50 microlitre (Is this not a concentration?). Then in my calculation described above  3.97 mol in 1 Litre. So 1 litre = 1000 ml. So in 100microlitres it is 0.0397 M . So 100 microlitres has less moles?

Thank you and Sorry

### #169824Calculating concentration

Posted by on 04 August 2014 - 02:53 AM in Chemistry

Thank you. phage434 and Pito. I simply took 200, 100 and 50 microlitres and used directly for my experiment (3 different volumes of same pure compound, but I would like to express as concentration). I did not diluted it. Coming to my calculation if it is 3.97 mol/L, then can it be a 0.397 M in 100 microlitres? I am not sure, if I calculated it right. I calculated based on : 3.97 mol in 1 Litre. So 1 litre = 1000 ml. So in 100microlitres it is 0.0397 M . Am I right now?

But the concentration is the same than!

If you did not dilute it you would just have added the same concentration to your sample...

The sample itself of course would have an effect on the end concentration.

I will give you a simple example:

if you have 10 moles per liter for a certain solution than this concentration does not change whether you have taken 100 microliter or 50 microliter.

The concentration is the same!

The amount of moles is different of course: for 100 microliter you have twice the amount as for 50 microliter.

1 liter = 10 moles

100 microliter = 0,1 ml = 0,1ml x 10^-3 liter , this meaning 10/10 000 moles!

50 microliter is than the half of 10/10 000 moles..

Now the end concentration of course depends on to what you add this 100 microliter or 50 microliter.

Thank you. The result of my experiment is like this, as the volume increases from 50 to 200, the results were better. I did not get good results when I used 50 microlitre. Does this mean concentration has an effect on my sample? So you are saying, even though the volume increases, concentration remain same? Am I right? Do I need say as the volume increases the results were better, (instead of saying as concentration increases the results were better)?

### #169822Calculating concentration

Posted by on 04 August 2014 - 02:33 AM in Chemistry

Thank you. phage434 and Pito. I simply took 200, 100 and 50 microlitres and used directly for my experiment (3 different volumes of same pure compound, but I would like to express as concentration). I did not diluted it. Coming to my calculation if it is 3.97 mol/L, then can it be a 0.397 M in 100 microlitres? I am not sure, if I calculated it right. I calculated based on : 3.97 mol in 1 Litre. So 1 litre = 1000 ml. So in 100microlitres it is 0.0397 M . Am I right now?

### #169814Calculating concentration

Posted by on 03 August 2014 - 10:07 AM in Chemistry

Thank you. The density is 0.87 g/mL. The final volume I took for my experiment is 200, 100 and 50 microlitres (directly from the pure synthetic bottle for my experiment). However, I need to write it as final concentration, not as volume.

Is this correct way I am doing: 0.87g/ml = 870g/L = 870*98/100 = 852.6 g = 852.6/214.34 = 3.97 mol/L. So in 100 microlitres it is 0.397 and in 50 microlitres it is 0.19 M? Am I right? Thank you

### #169811Calculating concentration

Posted by on 03 August 2014 - 07:03 AM in Chemistry

Hi
Could any one help with this query. I would like to calculate the concentration that I used from a pure chemical Methyl laurate. The purity is 98%. MW is 214.34. I took directly 200, 100 and 50 microlitres for my experiment. Just wondering, how do I express this in concentration? such as in Molarity or in Micrograms/ml

### #158584Protein quantification

Posted by on 05 August 2013 - 05:44 AM in Protein and Proteomics

Here it is:

Sample 1= 0.28, sample 2 = 0.14, sample 3= 0.37

Thank you

### #158560What is the reason of doing proteomics work?

Posted by on 04 August 2013 - 03:02 PM in Protein and Proteomics

Hello all,

As a newbie, I am interested to know the rationale of doing proteomics work. As many of us know that "proteome analysis is required to determine which proteins have been conditionally expressed, how strongly, and whether any posttranslational modifications are affected". (between health and diseased tissue i.e., human or plant tissue)

What is the next step, If I know a protein that is expressed in response to biotic and abiotic factors? Is it only useful for writing a papers? or How can this results be useful?

Thank you

### #158557Protein quantification

Posted by on 04 August 2013 - 11:38 AM in Protein and Proteomics

thank you. Is it not correct?  could you please explain it in detail?

### #158553Protein quantification

Posted by on 04 August 2013 - 08:42 AM in Protein and Proteomics

Hello

I just performed bradford assay. My BSA stock is 1mg/ml. I did standard curve using 10, 20, 30, 40 and 50 ug (1mg/ml BSA stock). These are the values:

ug      0.D

10 = 0.21

20=  0.42

30= 0.69

40= 0.89

50 =1.05

Y = 0.0215x+0.007 and R2 = 0.9926

I extracted my protein and dissolved in 500 microlitres of IEF buffer: After I did my bradford assay, I got the following values:

ug/10 ul =  sample 1) 12.69767, sample 2) 6.186047, sample 3) 16.88372

ug/ul = sample 1)1.26, sample 2) 0.61, sample 3) 1.68

Does it mean my 1st sample has 1.26 mg/ml? I have doubt  because I dissolved in 500 microlitres of IEF buffer to dissolve my pellet (from this I took 10ul) and to store at -20C

I also need same amount (600ug) I can add upto 300-350 ul for  for my IPG strips. But I got different values, Is there any formulae to calculate