Can anyone help me to understand the calculations in the attached file?? I want to understand the amount of HAuCl4.4H2O (in mmoles) to obtain 2 wt.% of Au in 2 g of Tetra butyl titanium. Thanks
hi,
2 wt% means 2 gm in 100 mL.
Ti(OC4H9)4 was taken 2 mL, it means the solute (HAuCl4, 4H2O) is present 0.04 g (2 wt%).
The molecular weight of solute is 411.84 g mol-1. so 0.04 g equivalent to 0.0973 mmol of HAuCl4, 4H2O. Only 47.8 % Au present in HAuCl4,4H2O.
So to make 2 wt% Au, you need 0.0465 mmol HAuCl4, 4H2O dissolve in 2 mL of Ti(OC4H9)4 solution.
Thank you so much. but still I could get the last point (So to make 2 wt% Au, you need 0.0465 mmol HAuCl4, 4H2O dissolve in 2 mL of Ti(OC4H9)4 solution.). As you said " 0.04 g equivalent to 0.0973 mmol of HAuCl4, 4H2O" and 47.8% of 0.04 g is 0.0188g=18.8mg. Could you please explain it to me again?
Thanks
411.84 mg equivalent to 1 mmol, hence 18.8 mg equivalent to 0.0465 mmol of HAuCl4, 4H2O.
Hope you will understand this.
Yeah I understand this but I couldn't get this point that how this amount of HAuCl4, 4H2O (0.0486mmol=0.02 g 0r 20 mg) contain 2 wt. % of Au as they write in the paper that theoretically in this amount 2 wt. % is Au or is it according to TiO2 ratio? Could you please explain it.
Thanks
I calculated again and it seems that 2 wt.% Au ratio is with TiO2, not of HAuCl4, 4H2O.
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