Dear Aarif,

Probably, the authors should have used a concentration of 200 mg/L for each salt (to give 10 mg of metal in 1000 mg of TiO2; i.e. 1%, but always in 'nominal terms').

On the other hand, the formula of chloroplatinic acid is wrong in the reference.

Kind regards,

Javier

Francisco Javier Peñas Thank you sir for the calculations. Sir, could you please inform how to calculate 200 mg/l value that gives 10 mg of metal. Thankyou once again.

It's easy from the moment you are taking 50 mL of each salt solution (containing 200 mg/L).

Aarif Hussain Shah

Not sure why there are issues here other than the formulae are not correct in the question or the attached paper. Chloroplatinic acid is usually formulated as H2PtCl6.xH2O (x = nominally 6) and chloroauric acid as HAuCl4.yH2O. The x's and y's are important because both salts are hygroscopic and weighing out exact quantities is a problem. It's usual to measure the real metal loading afterwards by AAS or ICP - one usually finds that the actual loading is less than that aimed for because of (metal salt) wash out and the underestimation of 'x' and 'y'. The catalyst preparation isn't conventional either - normally one would just saturate the pores of the support (and this requires a knowledge or measurement of the pore volume). The calculations are not correct either: to make a 1 (wt)% loading of metal (final total 100 g catalyst) on support requires 1 g of metal and 99 g (not 100 g) of support. So, I suspect that chemists or catalyst engineers were not the authors of the paper.

Let's take one example. A desired loading of 1% Pt on 1 g of TiO2 requires 0.01 g of Pt and 0.99 g of TiO2. In order to get 0.01g of Pt (atomic weight 195.084) requires [517.90/195.084]*0.01 g of the chloroplatinic salt precursor (the MW of H2PtCl6.6H2O is 517.90) or ~ 0.0265 g of the salt. It's freely soluble in water so you'd dissolve the approximate amount in water and use this solution with the 0.99 g of the TiO2. Now the key here is the pore volume of the TiO2 (and, of course, that x is not equal to 6 but practically greater). It wouldn't be normal in catalyst engineering to just dissolve 0.0265 g of the precursor in any amount of water. One would want to just saturate the pores. Imagine a TiO2 with a known pore volume of 1 mL/g then the ideal situation would be to dissolve 0.0265 g of the salt precursor in 0.99*1 mL = 0.99 mL of water and add this to the dried (0.99 g) TiO2 support and then carry out the reduction step. This is not really practical and that's the reason why most people would make 25 or 100 g of final catalyst.

Alan F Rawle Thanks a lot Sir. I really appreciate this. You explained this in a lucid way. Thanks once again

Alan F Rawle Sir, could you please explain how will the calculations/formula change if we want to load more than 1 metal on support. For example in above problem, if we want to make 1% Pt + 1 % Au on 1 gram of TiO2, how is the amount of salt precursors to be calculated?

Aarif Hussain Shah

Identical to shown before (making a 2% metal loading catalyst). However, you need to be careful by what you mean by "%" - do you mean weight percent or molar percent? Now, do you want to end up with an alloy of Pt and Au or a mix of Pt/TiO2 plus Au/TiO2? You need to define what method you'll be using for the reduction of the metal salts to metal.

Alan F Rawle

Sir, it is 1 wt % as mentioned in your earlier answer. Sir, let's suppose we want to make (1 wt %) Pt, Au codoped TiO2 nanocomposite by sol-gel or hydrothermal method. So please throw some light in this case. Thanking you in advance.

Aarif Hussain Shah

Your answers are still not clear:

• In your post above you state making 1% Pt, Au co-doped but earlier state 1% Pt and 1% Au making 2% total. Which is it?
• You still do not tell me whether you wish to have an alloy or a mix of components? The latter would be accomplished by separate reductions and then mixing the reduced components.
• You still do not tell me whether you require 1 wt% Pt plus 1% wt% Au (making a total of 2 wt%) or a 2% loading consisting of equal molar proportions of Au and Pt
• How will you carry out the reduction from metal salt to metal? This is important as the chemistry may negate certain routes.

Without the above answers I can only speculate and I want to answer your question correctly and precisely just once without going backwards and forwards as we are doing now. I see the next question as being 'what are the masses if we use Pd instead of Pt and we use 12.7 g of support and introduce 0.1% Cu doping into the system etc etc'?

I'm trying to show you how to fish not give you the fish...

Alan F Rawle Sir, I'm sorry for not making my point clear at first place.

• Let's suppose, I wish to make a nanocomposite (Pt, Au Codoped TiO2) having 1wt % of Au and 1 Wt % of Pt.
• Precursors will be HAuCl4 or HPtCl6 in deionized water (as mentioned in the reference paper attached with the queation)

PS: Thankyou sir for making us learn how to fish rather than giving a fish. I really appreciate your efforts. Please help us in learning the tricks of this doping/co-doping calculations.

Aarif Hussain Shah

Again, I’m not sure why these calculations are so difficult… Isn't there someone local that can assist here? OK, the important data for calculation are:

Atomic weights of Pt and Au respectively are: 195.085 and 196.97 respectively

Molecular masses of the (hypothetical) HAuCl4.6H2O and H2PtCl6.6H2O are 447.8782 g/mol and 517.9 g/mol respectively

The part for the Pt is the same as before: A desired loading of 1% Pt on 1 g of TiO2 requires 0.01 g of Pt and 0.99 g of TiO2. Likewise, for Au… Making a total metal mass of 0.02 g metal and 0.98 g of support

0.01 g of Pt requires 517.90/195.084]*0.01 g of the chloroplatinic salt precursor (the MW of H2PtCl6.6H2O is 517.90) or ~ 0.0265 g of the salt

0.01 g of Au requires 447.8782/196.97)*0.01 g of the chloric acid precursor or ~ 0.0227 g of the precursor

Thus, 0.0265 g of the Pt salt precursor and 0.0227 g of the Au salt precursor could be dissolved in 0.98 mL of water (if the pore volume of the TiO2 is 1 mL/g) and stirred with 0.98 g of the TiO2. This would be a co-impregnation resulting in a nominal 2% total metal loading with 1% each of the metals after reduction. As I stated earlier, this (0.98 mL H2O) is somewhat impractical and would probably be scaled to 25 or 100 g of the final material.

For 50% molar (not weight) proportions of the metals, you are fortunate (or unfortunate) that the atomic masses of Pt and Au are very similar and thus the weights for equal masses and molar quantities would effectively be very close. However, to be pedantic, if we required equal (50%) molar proportions of each of the metals then 1 mole of Pt is 195.085 g and 1 mole of Au is 196.97 g. 0.005 moles of each are ~ 0.975 and 0.985 g of each of the metals. 0.975 g Pt and 0.985 g Au give a total metal mass of 1.96 g. We’d scale appropriately for 0.02 g (for a 2% total mass loading on 0.98 g support).

Thus, for 0.02 g of total metal on 0.98 g of support where the Pt and Au are in equimolar proportions would be (0.02/1.96)*0.975 g Pt and (0.02/1.96)*0.985 g Au or ~ 0.01 g (Pt) and 0.01g (Au) – virtually the same as I said before. The mass of metal salts would be calculated as before…

Now here is an exercise for you to show you’ve understood what I have written. Please indicate your calculations in RG, ask for help from anyone, but not me, preferably in your institute. Your homework problem is:

Consider a 2 wt% total metal loading of a PdAu system on a support where they would be both in equal weights (1% each) and where they are in equimolar proportions. You are given the atomic mass of Pd as 106.42 with your Pd precursor being H2PdCl6.6H2O (molecular mass 321.154 + [6*18.01528] or ~ 429.25 g). The gold precursor as before... What weights of the Pd and Au precursor are required for 25 g of support?

How could you reduce to metals in such a system when you’ve co-impregnated the (TiO2) support with these metal salts?

Aarif Hussain Shah

Your homework answer - did you get this correct?

What weights of the Pd and Au precursor are required for 25 g of support?... For equal masses and equimolar proportions..

Atomic mass Pd = 106.42; Au = 196.97

Precursors: H2PdCl6.6H2O = 429.25; HAuCl4.6H2O = 447.88

2% PdAu 25 g total catalyst means..

0.5 g metal (0.25 g Pd & 0.25 g Au) and 24.5 g support

Equal weights:

0.25 g Pd requires [429.25/106.42]*0.25 g of the H2PdCl6.6H2O precursor or ~ 1.008 g

0.25 g Au requires [447.88/196.97]*0.25 g of the HAuCl4.6H2O precursor or ~ 0.568 g

Equimolar proportions (50% molar mass of each Pd and Au and 2% final loading):

Same total metal content (2%) at 0.5 g metal

Pd + Au = 303.39

Equimolar = 106.4/303.39 [Pd] and 196.97/303.39 [Au]

For 0.5 g total metal this is (106.4/303.39)*0.5 Pd and (196.97/303.39)*0.5 Au

Or 0.175 g Pd and 0.325 g Au

This will come from 0.175*(429.25/106.42) g of H2PdCl6.6H2O ~ 0.706 g

And 0.325*(447.88/196.97) g of HAuCl4.6H2O ~ 0.739 g

Check my math!

Alan F Rawle

Sir, I calculated for 2 Wt%.

Thank-you once again.

Aarif Hussain Shah

I understand you require 3 wt% WO3 with (97 wt%) TiO2.

• So why is the atomic weight of Ti relevant other than as a component of titanium isopropoxide and TiO2?
• What is the molecular weight of TiO2? (this question is easy)
• Are you also starting with WO3?

Alan F Rawle Sir, should the values be like this for 10 gms of 3 wt% WO3-TiO2 nanocomposite?

(284.22/79.866)*9.7=34.51 gms

Volume of titanium isopropoxide = 34.51/0.96 = 35.95 ml

• mol. wt of TiO2 = 79.866 g/mol
• Yes sir, starting with WO3 , so for 0.3 gms Tungsten loading, we need ( 231.84/183.84)*0.3 = 0.37 gms WO3 ( W= 183.84, WO3 = 231.84 )

Aarif Hussain Shah

Now you are totally confusing me (again!).

You state you want 3 wt% WO3 on TiO2. Then in your second bullet point above you state '0.3 g Tungsten loading'. Which is it? Why is the atomic weight of W relevant other than as a component of WO3? How could or would you make W from WO3?

Alan F Rawle

Sir, I wish to make total 10 gms nanocomposite... so it will have 0.3 gms WO3 & 9.7 gms TiO2. Hope i got it clear now