I have 68nm spherical Au NPs, which gave me a concentration of 1.22 x 10^9 particles/mL. How can I calculate the molarity without the need for TEM?
If you are certain that your concentration is 1.22e9 NPs/mL, You could simply use Avagadro's number to convert it to moles. 1 mole= 6.02e23 NPs. My calculation shows 0.2e-11 M is the concentration of your 68 nm Au NPs. You can actually calculate the number of gold atoms (assuming fcc packing, although some discrepancies due to nano nature) in a certain volume based on this concentration, and run ICP-MS to check if that makes sense or not.
Alternatively, find the amount of gold from ICPMS, and calculate the concentration from the size of NP, sample volume, crystal packing etc. Let me know if you want to discuss more.
Modelling your NPs as monodispersed (that is all have the same 68 nm diameter), and assuming that the mass density of Au in NPs is the same as bulk Gold (19,32 g/cm3), the molarity should be around 2x10^-5 M.
The calculation must be performed as follows:
1. calculate the volume Vnp of a single NP considered as a sphere, 4/3 pi r^3 (convert in cm3)
2. calculate the mass of a single NP m_np as the product of the above Vnp and the density of Gold expressed in g/cm3
3. calculate the total Au mass in 1 mL dispersion as the product of the above mass m_np and the NPs concentration you reported (1,22x10^9 NP/mL in your case), and then multiply by 1000 to get the mass per Liter.
4. Divide the above Au mass per Liter by the mass of a mole of Gold which is about 197 g. You obtain the desired molarity.
Simple, of course, but based on some (strong) assumption. First of all, we have assumed that all NPs are spherical and monodispersed, which is generally only a coarse approximation. Second, we assumed that the Au mass density in NP form is the same as in a bulk piece, which may be an approximation as much better as larger a NP is (you may carry on a literature search to find a density value for NPs of your size or similar). Third, we have assumed that the only (or dominant) Au in your dispersion is in form of NP, that is no residues are present in solurion, which is good-to-very good approximation.
And you probably saw different molarity in Giuseppe's calculation and mine. Both are correct. Its just that you need to ask yourself whether you need the molarity of Au NPs or Molarity of Au as an element. Remember, the number of gold atoms and the number of gold NPs are different in your solution- that's why the molarity is different based on what you want to calculate. What reaction further down the lane you are trying to do, or what exactly you are trying to find out- elaborate more.
For example, if you are reacting this gold NP with something else, then you need the ,molarity of gold NPs. Thanks.
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