Hello all . in this thread I would like to present the basics of calculating proportions on the example of mephedrone synthesis calculations.
at the very beginning I will start with the fact that proportion counting is one of the main steps before starting work when carrying out any laboratory activities.
Here we go: in order to count how many grams or how many ml of a given product we have to add during the reaction, we need to know the molar masses of the products used.
an example would be the calculation of the proportion to perform methylamination in mephedrone synthesis :
So this is how: we need to know the molar masses of the following reagents: 2 bromo 4 methylpyrophenone, 40% methylamine aqueous solution, the amount of solvent to dissolve the bromoketone and the amount of aqueous hydrochloric acid that will be needed in the acidification phase.
So we assume that we will work on 4kg bk4:
how much do we need ml of 40% methylamine solution?
per 1mol of bk4 there is 2mol of methylamine + appropriate excess, depending on whether the synthesis is cold or warm. the excess range fluctuates from 1-2.5. for hot synthesis, an excess of 2.5 is recommended.
therefore: for 1 mole of bk we need 5 mole of methylamine.
for 17.62 mol of bk4 we need 88.1 mol of methylamine.
so: for 4 kg of bk4 we need 12l of solvent and almost 10l of a 40% solution of methylamine in water.
tip: when counting the amount of methylamine in water, it is worth assuming that there is less than 40% there because it is a strongly oxidizing compound ... it is worth assuming that it has 36% or even 33% as long as it is standing
the last thing to count for this synthesis is the amount of hydrochloric acid needed for acidification. generally the ratio is 2: 1, i.e. 2g bk4: 1ml of hcl acid ... but sometimes it is better to count so that later it does not turn out that we gave too much of it.
at the very beginning I will start with the fact that proportion counting is one of the main steps before starting work when carrying out any laboratory activities.
Here we go: in order to count how many grams or how many ml of a given product we have to add during the reaction, we need to know the molar masses of the products used.
an example would be the calculation of the proportion to perform methylamination in mephedrone synthesis :
So this is how: we need to know the molar masses of the following reagents: 2 bromo 4 methylpyrophenone, 40% methylamine aqueous solution, the amount of solvent to dissolve the bromoketone and the amount of aqueous hydrochloric acid that will be needed in the acidification phase.
1mol bk4 = 227g
1mol methylamine = 31,06g
1mol hcl = 36,46g
amount of solvent needed to dissolve bromoketone "1g bk = 1ml of solvent + x ml of solvent needed for complete dissolution" we will assume that for 1g we will need 3ml1mol methylamine = 31,06g
1mol hcl = 36,46g
So we assume that we will work on 4kg bk4:
- 4kg bk4 = 4000g bk4 mol=?
- 4000/227=17,62mol
how much do we need ml of 40% methylamine solution?
per 1mol of bk4 there is 2mol of methylamine + appropriate excess, depending on whether the synthesis is cold or warm. the excess range fluctuates from 1-2.5. for hot synthesis, an excess of 2.5 is recommended.
therefore: for 1 mole of bk we need 5 mole of methylamine.
for 17.62 mol of bk4 we need 88.1 mol of methylamine.
88,1 x 31,06 = 2736,39g methylamine hcl
to get the amount needed for a 40% solution, we have to divide our result by 40%
2736,39 / 0,4 =6840,98g ( 40% methylamine in water )
to convert g to ml we have to divide it by the density of methylamine = 0.69 g / cm3
6840,98 / 0,69 =9914,46 ml
to get the amount needed for a 40% solution, we have to divide our result by 40%
2736,39 / 0,4 =6840,98g ( 40% methylamine in water )
to convert g to ml we have to divide it by the density of methylamine = 0.69 g / cm3
6840,98 / 0,69 =9914,46 ml
so: for 4 kg of bk4 we need 12l of solvent and almost 10l of a 40% solution of methylamine in water.
tip: when counting the amount of methylamine in water, it is worth assuming that there is less than 40% there because it is a strongly oxidizing compound ... it is worth assuming that it has 36% or even 33% as long as it is standing
the last thing to count for this synthesis is the amount of hydrochloric acid needed for acidification. generally the ratio is 2: 1, i.e. 2g bk4: 1ml of hcl acid ... but sometimes it is better to count so that later it does not turn out that we gave too much of it.
So, after carrying out the entire reaction correctly in the acidification phase, we will have no more than 80% of what we gave for the start. so it was 17.62 mol and now the maximum bd 14.09 mol.
3198,43g bk4 in 12000ml toluen
14,09mol hcl =513,72g Hcl
36% hcl in water = 513,72 / 0,36 = 1427g or 1230 ml . + 10%excess= 1353ml
that is, as you can see from the calculations. for acidification you really need a little less ml of acid than if you assume that for 2g bk4 you need 1ml of acid ... of course it is worth adding acid in portions because it cannot always be sure that the entire reaction was 100% correct ... in my opinion, thanks to the theoretical calculation of the amount of acid needed we can minimize the risk of the pH dropping below 5. thanks to this, we will prevent the formation of dirty impurities that need to be cleaned later.
14,09mol hcl =513,72g Hcl
36% hcl in water = 513,72 / 0,36 = 1427g or 1230 ml . + 10%excess= 1353ml
that is, as you can see from the calculations. for acidification you really need a little less ml of acid than if you assume that for 2g bk4 you need 1ml of acid ... of course it is worth adding acid in portions because it cannot always be sure that the entire reaction was 100% correct ... in my opinion, thanks to the theoretical calculation of the amount of acid needed we can minimize the risk of the pH dropping below 5. thanks to this, we will prevent the formation of dirty impurities that need to be cleaned later.
