Appreciate some backup, checking my calculations to see where I messed up - - it's been 30years since college chemistry:(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# Homework Help: Help with nanomolar calculations

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