- #1
sarbot
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Appreciate some backup, checking my calculations to see where I messed up - - it's been 30years since college chemistry:
Let's say I have a compound X which has a molecular weight of 320, so, 320 grams = 1 mole.
Next, I have a 1% stock solution A of that compound X (1 gram dissolved in 100 grams water).
We know that 1 ml = 1 gram for water, and we know that approx. 20 drops are in one ml.
1000 mg. = 1 gram.
20 drops X 100 ml = 2000 drops, or, 1000 milligrams divided by 2000 drops =.5 mg of the compound X in every drop of 1% stock solution A.
Now, put that 1 drop of 1% stock solution A in 240 ml of water to create stock solution B.
That is rather close to a 1/4800th dilution.
Then use 1 drop from this stock dilution B to dissolve into 1000 ml of water which yields .000104 mg of compound X in a whole liter of water (final stock solution C).
We know that 1 mg. per liter is 1 ppm, so 0.000104 mg. would be 0.000104 ppm, close to 0.104 ppb, or 104 nanomolar.
Now, a mole = 6.022 X 10 ^23 molecules of any compound.
A nanomole = 10^-12, essentially that is 1 trillionth, so we can calculate that 10^23 minus 10^12 yields 6.022 X 10^11 molecules in a nanomole.
and 100 nanomoles is 6.022 X 10^13 molecules.
Therefore, in that one liter of water there are better than 60 trillion molecules of compound X floating around in the final stock solution C - - even though the dilution is rather incredible and ultimately results in about only one ten thousandth of a milligram of compound X in the entire liter of final stock solution C.
Let's say I have a compound X which has a molecular weight of 320, so, 320 grams = 1 mole.
Next, I have a 1% stock solution A of that compound X (1 gram dissolved in 100 grams water).
We know that 1 ml = 1 gram for water, and we know that approx. 20 drops are in one ml.
1000 mg. = 1 gram.
20 drops X 100 ml = 2000 drops, or, 1000 milligrams divided by 2000 drops =.5 mg of the compound X in every drop of 1% stock solution A.
Now, put that 1 drop of 1% stock solution A in 240 ml of water to create stock solution B.
That is rather close to a 1/4800th dilution.
Then use 1 drop from this stock dilution B to dissolve into 1000 ml of water which yields .000104 mg of compound X in a whole liter of water (final stock solution C).
We know that 1 mg. per liter is 1 ppm, so 0.000104 mg. would be 0.000104 ppm, close to 0.104 ppb, or 104 nanomolar.
Now, a mole = 6.022 X 10 ^23 molecules of any compound.
A nanomole = 10^-12, essentially that is 1 trillionth, so we can calculate that 10^23 minus 10^12 yields 6.022 X 10^11 molecules in a nanomole.
and 100 nanomoles is 6.022 X 10^13 molecules.
Therefore, in that one liter of water there are better than 60 trillion molecules of compound X floating around in the final stock solution C - - even though the dilution is rather incredible and ultimately results in about only one ten thousandth of a milligram of compound X in the entire liter of final stock solution C.