Water Methanol AFR Calculation

[Flr Power]

Can anybody double check the math I have done here please?
I am not sure if this is the right way to calculate the ratio of AFR change when injecting 25 % by weight of water methanol. What would be the real AFR change?
I am injecting 510cc/min of W50/M50 by weight for 2040cc/min of petrol at peak power.

510/2040=0.25; 25%/2 = 12.5% or 255cc/min of injected methanol.
14.7 petrol stoich and 6.4 methanol stoich = 8.3 total AFR change; So 8.3x.125=1.04 AFR change?

Is the math good? Will injecting 12.5% methanol combine with petrol would give me an AFR change of 1.04?

So it looks like I would need to trim 7% petrol from my fuel map when I fully inject right?