Help I Need A Chemist !!

I need a chemist to work out the answers to the question of "What were the original weights of the samples?", the samples are all methylamphetamine and the difference in purity would be due to the addition of MSM (MethylSulfonylMethane), need the formula/equation showing working out of the answers.

?g at 75% purity - became - 2.50g at 4% purity - what was the original weight? ?g at 85% purity - became - 1.51g at 71% purity - what was the original weight? ?g at 85% purity - became - 1.64g at 71% purity - what was the original weight? ?g at 80% purity - became - 1.7 at 63% purity - what was the original weight?


Comments


[12 Points] RobertoEnjoysDrugs:

Im not a chemist, but if I am correct, this should just be basic math.

2.50g @ 04% = 0.13g @ 75%

1.51g @ 71% = 1.26g @ 85%

1.64g @ 71% = 1.37g @ 85%

1.70g @ 63% = 1.34g @ 80%

My process was: Multiply outcome grams by outcome percentage to recieve amount of pure substance [1.51g * (.71) = 1.07g pure substance]. Then divide amount of pure substance by original precentage to get original weight [1.07g (pure) / .85 = 1.26g]


[6 Points] LedLevee:

High school must've been a fun time for you.


[2 Points] diOpAnonMu:

Well, if you have 2.5g at 4% purity, then you have .1g of meth (multiply). If you had .1g of meth at 75% purity, then .1g/.75 = .13g

This is assuming that you're assessing purity by mass and not volume.


[1 Points] fuckintristan:

Message backdoor nobaby


[0 Points] Vender_BBMC:

Tree fiddy