I think you’re misunderstanding how molecular efficiency and overall yield are related.

So for example let’s say your starting distillate is 1000 grams of:

80% CBD

3% THC

3% CBC

2% CBG

1% CBDV

89% Total Cannabinoids

That’s 800g of CBD if you have 3% CBD loss then (800 - (800 x 0.03)) = 776g of CBD in your final product or 24 grams of CBD loss

Let’s assume the same 3% loss for CBG and CBDV

CBG: 1000 x 0.02 = 20g of CBG to start with

(20 - (20 x 0.03)) = 19.4g of CBG recovered or a 0.6g loss

CBDV: 1000 x 0.01 = 10g of CBDV to start

(10 - (10 x 0.03)) = 9.7g of CBDV recovered or a 0.3g loss

Then for CBC and THC we have

THC: 1000g x 0.03 = 30g of THC to start

CBC: 1000g x 0.03 = 30g of CBC to start

Let’s say we totally remove them… so 60g of loss

Now let’s say we also remove other impurities via Chromatography and so our total cannabinoid number increases from 89% to 95%… indicating a 6% loss in mass from impurities

1000g x 0.06 = 60g of impurities lost during purification

1000g - (24g + 0.6g + 0.3g + 60g + 60g) = 855.1g of “T-free”

So if we have 776g of CBD in our final product and it weighs 855.1g then it’s 90.75% CBD in your final product. CBG is elevated to 2.27% and CBDV is elevated to 1.13%…

So it is possible to get 95% CBD T-free from something that is say 85% CBD and 90% total cannabinoids…

Not trying to rant on anyone, just sharing knowledge