Icp-aes analysis of nanoparticles

This is essentially your answer now as 2552 ppm. This is the amount of Fe in the solution of digested particles. This was made by diluting 0.5 mL of the original solution into 9.5 mL concentrated nitric acid, which is the same as diluting by a factor of 20. To calculate how much analyte was in the original batch that was synthesized we multiply the previous value by 20 again, i.e., [link] . This is the final amount of Fe concentration of the original batch when it was synthesized and made soluble in hexanes.

Calculating stoichiometric ratio

Moving from calculating the concentration of individual elements now we can concentrate on the calculation of stoichiometric ratios in the bi-metallic nanoparticles.

Consider the case when we have the iron and the copper elements in the nanoparticle. The amounts determined by ICP are:

• Iron = 1.429 mg/L.
• Copper = 1.837 mg/L.

We must account for the molecular weights of each element by dividing the ICP obtained value, by the molecular weight for that particular element. For iron this is calculated by [link] , and thus this is molar ratio of iron. On the other hand the ICP returns a value for copper that is given by [link] .

To determine the percentage iron we use [link] , which gives a percentage value of 42.15% Fe.

To work out the copper percentage we calculate this amount using [link] , which leads to an answer of 57.85% Cu.

In this way the percentage iron in the nanoparticle can be determined as function of the reagent concentration prior to the synthesis ( [link] ).

Determining concentration of nanoparticles in solution

The previous examples have shown how to calculate both the concentration of one analyte and the effective shared concentration of metals in the solution. These figures pertain to the concentration of elemental atoms present in solution. To use this to determine the concentration of nanoparticles we must first consider how many atoms that are being detected are in a nanoparticle. Let us consider that the Fe ₃ O ₄ nanoparticles are of 7 nm diameter. In a 7 nm particle we expect to find 20,000 atoms. However in this analysis we have only detected Fe atoms, so we must still account for the number of oxygen atoms that form the crystal lattice also.

For every 3 Fe atoms, there are 4 O atoms. But as iron is slightly larger than oxygen, it will make up for the fact there is one less Fe atom. This is an over simplification but at this time it serves the purpose to make the reader aware of the steps that are required to take when judging nanoparticles concentration. Let us consider that half of the nanoparticle size is attributed to iron atoms, and the other half of the size is attributed to oxygen atoms.

As there are 20,000 atoms total in a 7 nm particle, and then when considering the effect of the oxide state we will say that for every 10,000 atoms of Fe you will have a 7 nm particle. So now we must find out how many Fe atoms are present in the sample so we can divide by 10,000 to determine how many nanoparticles are present.

In the case from above, we found the solution when synthesized had a concentration 51,040 ppm Fe atoms in solution. To determine how how many atoms this equates to we will use the fact that 1 mole of material has the Avogadro number of atoms present, [link] .

1 mole of iron weighs 55.847 g. To determine how many moles we now have, we divide the values like this:

The number of atoms is found by multiplying this by Avogadro’s number (6.022x10 ²³ ):

For every 10,000 atoms we have a nanoparticle (NP) of 7 nm diameter, assuming all the particles are equivalent in size we can then divide the values, [link] . This is the concentration of nanoparticles per liter of solution as synthesized.

Combined surface area

One very interesting thing about nanotechnology that nanoparticles can be used for is their incredible ratio between the surface areas compared with the volume. As the particles get smaller and smaller the surface area becomes more prominent. And as much of the chemistry is done on surfaces, nanoparticles are good contenders for future use where high aspect ratios are required.

In the example above we considered the particles to be of 7 nm diameters. The surface area of such a particle is 1.539 x10 ^-16 m ² . So the combined surface area of all the particles is found by multiplying each particle by their individual surface areas.

To put this into context, an American football field is approximately 5321 m ² . So a liter of this nanoparticle solution would have the same surface area of approximately 1.5 football fields. That is allot of area in one liter of solution when you consider how much material it would take to line the football field with thin layer of metallic iron. Remember there is only about 51 g/L of iron in this solution!

Bibliography

• (External Link)
• A. Scheffer, C. Engelhard, M. Sperling, and W. Buscher, W. Anal. Bioanal. Chem. , 2008, 390 , 249.
• H. Nakamuru, T. Shimizu, M. Uehara, Y. Yamaguchi, and H. Maeda, Mater. Res. Soc., Symp. Proc. , 2007, 1056 , 11.
• S. Sun and H. Zeng, J. Am. Chem. Soc. , 2002, 124 , 8204.
• C. A. Crouse and A. R. Barron, J. Mater. Chem. , 2008, 18 , 4146.