You have to figure out a way to determine the (average) mass of your gold nanoparticles, which depends on their (average) size. So perhaps you should determine their size first. 

Yes Kåre surely it depends on their average size. The calculation formula is:

Microgram/ml =2 x 10^11 (particles/ml) * gold density (gm/cm3) * (1/6) (22/7) * (average diameter of particles nm)3  *10^-21 (cm3/nm3) * 10^6 (ugm/gm)

If you take the gold nanoparticles true density: 19.32 g/cm3

Microgram/ml = 20.24^ -4 *(average diameter of particles in nanometer)3

Best regards

 The Size of AuNPs is 30 nm. 

Then you have everything required to do the remaining math yourself. Don't forget to thank Mohammed for his kind help. You should expect some uncertainty in the results, due to size variation around 30 nm, and possible deviations from perfect spherical shape. The provider ought to have information about that, if you should need it.

Yes Monique

Microgram/ml = 20.24^ -4 *(average diameter of particles in nanometer)3

Microgram/ml = 20.24^ -4 *(30nm)3

                         = 54.64 

Best regards

@Mohammed. You should encourage Monique to do some work on her own; how can she become a serious scientist otherwise? That said, I also enjoy doing the computations myself :-).

 @Mohammed - Thank you so much for your advice and guiding - this will surely help me. I appreciate your effort!

@Kåre - Please don't assume that I did not do work on my own. 

Many thanks. 

Dear Dr. Mohammed

Is this equation applied for silver nano particles as well?

Thank you very much in advance

Best regards

Yes, Hanan it is applied for silver nano particles as well. Please use the silver nanoparticles true density: 10.5 g/cm3 instead of gold true density.

Best regards.

Mohammed sir,

I want to know how this formulae for converting the concentration(particles/ml) of nanoparticles into microgram/ml came. Is there any reference or journal paper for this calculation.

As, I am working on the mix nanoparticles I want to know is it possible that using this formulae I can also calculate the unknown concentration of nanoparticles in the suspension.

@Mohammed - This information is so helpful, thank you for sharing. In your original equation what did the numbers, 10^-21 (cm3/nm3) * 10^6 (ugm/gm), represent. I am trying to do the same calculation with a 500 nm particles.

Mohammed Hussein j. H. Al'Atia

Please let me know if we can apply this formula for silica gold core shell nanoparticles

Thanks

Yes dear Gagandeep Kaur you can apply the formula for silica gold core shell nanoparticles providing that you have good idea about the average shell thickness and the average diameter of particles.

Best regards

Mohammed Hussein j. H. Al'Atia

Thank you so much sir.

Dr. Mohammed

In your original equation below, what does (1/6) represent?

Microgram/ml =2 x 10^11 (particles/ml) * gold density (gm/cm3) * (1/6) (22/7) * (average diameter of particles nm)3  *10^-21 (cm3/nm3) * 10^6 (ugm/gm)

Thank you for your help

Dear Analicia M. Carcano

Thank you for your feedback

The volume of spherical particle = 4 π r3 /3 Where r is the radius ,

Now in terms of particle diameter D:

r=D/2 then r3=D3/8

Volume of spherical particle = 4 π D3 /3*8 =(1/6)π D3

Best regards

Dr. Mohammed

In your original equation below, what does (22/7) represent?

Microgram/ml =2 x 10^11 (particles/ml) * gold density (gm/cm3) * (1/6) (22/7) * (average diameter of particles nm)3  *10^-21 (cm3/nm3) * 10^6 (ugm/gm)

Thank you for your help

Dear Mayra Beltran

Thank you for your feedback

Volume of spherical particle = 4 π D3 /3*8 =(1/6)π D3

Where π is the constant ratio = 22/7

Best regards

Thank you very much Dr Mohammed for your answer, finally you know of any bibliographic reference where this equation is found is to respect copyright and be able to correctly cite the source of information. Thanks again.

Dear Mayra Beltran

Thanks again for your feedback. In this respect, units' conversion factors are not required to cite by references as it is not equation, it is only a common practice steps.

Best regards

Thanks Dr Mohammed just a last question the numbers 10E-21 y 10E6 in the equation what that it's mean?. Thanks again.

Dear Mayra Beltran

Thank you for your feedback

10^-21 is the conversion factors from cm3 to nm3

10^6 is the conversion factors from ugm to gm

Best regards

Dear Dr Muhammad,

I am working with 3 sizes of gold nano particles 5nm, 20nm and 40nm with 4.92E+13 - 6.01E+13, 5.89E+11-7.19E+11 and 6.44E+10-7.87E+10 concentrations particles/ml respectively. Could you please help me to covert these concentrations to ug/ml?

Best regards

Mehwish

Mehwish Fida

Your question has been answered separately: https://www.researchgate.net/post/How_to_calculate_the_concentration_of_gold_Nanoparticles

Mohammed Hussein j. H. Al'Atia

You state 'Where π is the constant ratio = 22/7'.

Pi is not a ratio - it is a completely irrational number that cannot be expressed as a fraction. If so, why not use 355/113 which is closer?

Dear Alan F Rawle ,

you are right:

a) Pi is not a ratio, and

b) 355/113 is closer to Pi than 22/7.

But, nearly everywhere there is a 'but'.

The term '22/7' is well known to me as a simple ratio, which is very close to Pi.

I remember, I have learned that '22/7' as a numerical substitute*) for Pi in a very early course of theoretical physics (last millenium).

One might also use your '355/113', but dividing by 7 is much easier done by head (mental arithmetic) or by pencil, at least for me, than by 113. You know, we had times without any electronic calculator. Incredible nowadays...

*) According to the 'Background' section of

https://en.wikipedia.org/wiki/Proof_that_22/7_exceeds_%CF%80

such a substitute by a ratio is called 'Diophantine approximation'

(I must admit, that I have never heard that term up to now)

https://en.wikipedia.org/wiki/Diophantine_approximation

Best regards

G.M.

Gerhard Martens

Yes, I think that all scientists will know that 22/7 is an approximation for pi. I was being a little facetious in suggesting 355/113 (I could have used the even better 52163/16604). I collect old text books (among other things) and a couple of the more interesting ones are:

• The 'proof' that pi is 3 1/6 or 19/6 (The Measure of the Circle Perfected in January 1845 John Davis Providence 1854). There's no way that Mr Davis is convinced that he is wrong in the 156 pages of the book. Laborious calculations. Sad... I attach a Google Books download
• Squaring the circle. Numerous examples

Yes, many times approximations are useful (we still teach Dalton's Atomic Theory in schools - all atoms are indivisible - even though it's patently untrue. It has uses.). Reminds me of Rothchild's Law: 'For every phenomenon, however complex, someone will eventually come up with a simple and elegant theory. This theory will be wrong'. Maybe in your field of XRD then the Scherrer equation for crystallite size by line broadening could be an example - sure, small size increases the broadening but there's also strain and instrumental broadening to consider. We can then work with Williamson-Hall or Rietveld...

may I know the reference or Paper that uses the formula (Concentration Particle / ml into mg/ml) like your discussion in this link (https://www.researchgate.net/post/How-to-convert-particles-mL-to- microgram-mL) Prof Mohammed Hussein j. H. Al'Atia and Monique Engelbrecht

Ridhwan Haliq You need to open a separate question (as you have done) rather than confusing this thread. It is likely that you'll not get a reply here as people will not look at old questions generally.

Yes Kåre surely it depends on their average size. The calculation formula is:

Microgram/ml =2 x 10^11 (particles/ml) * gold density (gm/cm3) * (1/6) (22/7) * (average diameter of particles nm)3  *10^-21 (cm3/nm3) * 10^6 (ugm/gm)

If you take the gold nanoparticles true density: 19.32 g/cm3

Microgram/ml = 20.24^ -4 *(average diameter of particles in nanometer)3.

Is there any formula like the above for calculating microplastic particle conversion from particles/mL to microgram/mL?