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Old 03-24-2008, 08:38 AM
SBS
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Smile Convert nanomoles/L to milligrams/L for sTfR

I have a nice medical article relating to sTfR (soluble transferrin receptor) and the ability to use this laboratory test in conjunction with serum ferritin to diagnose iron deficiency anemia. I have turned this article into an Excel program. The only problem is that the routine measurement of sTfR is in nM/l (nanomoles per liter) and the medical article and software program measures sTfR in mg/L (milligrams per liter).

I could use help in converting nM/L of sTfR to mg/L. Thanks.
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Old 03-24-2008, 10:12 AM
JohnS JohnS is offline
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Default Re: Convert nanomoles/L to milligrams/L for sTfR

You need to know the molecular weight, which is also the molar mass in g/mol.

I tried to search. It seems to be a large, complex molecular with a high but somewhat uncertain molecular weight. I didn't find a good value, but you can't convert without it.
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Old 03-24-2008, 03:02 PM
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Default Re: Convert nanomoles/L to milligrams/L for sTfR

I found this on the net Hope this helps.
This question involves a series of conversions to get the answer.

First, you need to know how many "moles" of NH3 you have. To find this, you may know that there are 6.023 X10^23 molecules of ANYTHING in 1 mole. In other words, there are 6.023 X 10^23 molecules of NH3 in 1 mole of NH3. Therefore, take: 4.12 X 10^24 and divide by 6.023 X10^23 and you should get 6.84 moles of NH3.

2nd, now that you know that 4.12E24 molecules of NH3 = 6.84 moles of NH3, you need to convert to mass. (Since, I don't know if you need it in lbs or grms, I'll just assume grams.) From the periodic table of elements, you can find that there is 14.006 grams of N in every mole of N. Also, you'll find that there is 1.008 grams of H in every mole of H. Therefore, since there are 3 H's, this equals 3.024 g. And when tackin on the N, the total molecular weight of NH3 equals 17.03 g/mole of NH3. Now you can take the 6.84moles of NH3 and multiply by the 17.03 MW to get approx 116.48 g of NH3

To sum up everything just said:

4.12 E24 molecules NH3 * (1 mole NH3 / 6.023E23 molecules NH3) * (17.03 grams NH3 / 1 mole NH3) = 116.48 grams NH3.
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Old 03-24-2008, 11:42 PM
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Default Re: Convert nanomoles/L to milligrams/L for sTfR

Quote:
Originally Posted by joe
I found this on the net Hope this helps.
This question involves a series of conversions to get the answer.

First, you need to know how many "moles" of NH3 you have. To find this, you may know that there are 6.023 X10^23 molecules of ANYTHING in 1 mole. In other words, there are 6.023 X 10^23 molecules of NH3 in 1 mole of NH3. Therefore, take: 4.12 X 10^24 and divide by 6.023 X10^23 and you should get 6.84 moles of NH3.

2nd, now that you know that 4.12E24 molecules of NH3 = 6.84 moles of NH3, you need to convert to mass. (Since, I don't know if you need it in lbs or grms, I'll just assume grams.) From the periodic table of elements, you can find that there is 14.006 grams of N in every mole of N. Also, you'll find that there is 1.008 grams of H in every mole of H. Therefore, since there are 3 H's, this equals 3.024 g. And when tackin on the N, the total molecular weight of NH3 equals 17.03 g/mole of NH3. Now you can take the 6.84moles of NH3 and multiply by the 17.03 MW to get approx 116.48 g of NH3

To sum up everything just said:

4.12 E24 molecules NH3 * (1 mole NH3 / 6.023E23 molecules NH3) * (17.03 grams NH3 / 1 mole NH3) = 116.48 grams NH3.


Hi Joe
SBS isn't asking about ammonia (NH3), they are asking about are large metal "complex", which will have a very large molecular weight. As John has said, this will be over a range, and they have to decide what they will use as the average.
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