Dears ,
Please I need your urgent support to explain how can I calculate the Al2O3 weight required to prepare Al % standard in total weight 8.8 g ?
For example : if I need to prepare 0.2 % Al from Al2O3 how many grams I have to take it from Al2O3 ?
First step:
In your example you need to have 0,2% Al from 8,8g total weight.
So the total Al weight in the 8,8g will be 0,2/100*8,8g= 0,0176g of Al.
Second step:
Your Al2O3 has a molecular weight of
2*26,98g (=Al2)+ 3*16,0 (=O3) = 53,96g + 48g= 101,96g per mol of Al2O3.
From that equation you see that you have got 53,96g of Al in 101,96g of Al2O3
or when scaling down to 1g of Al:
1g of Al in (101,96/53,96)g of Al2O3 ( just applying rule of proportion).
Third step:
For your goal of 0,0176g of Al just multiply the above result by 0,0176:
So you need 0,0176*(101,96/53,96)g = 0,033g of Al2O3
Fourth step:
general equation:
(Al%/100) * (101,96/53,96) * 8,8g of Al2O3
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