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

many thanks I appreciate your effort