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Converting units of a grouped expression...

Fri Aug 28, 2009 4:13 pm

Hey.

Hope someone can clear something up for me. I'm working with some dissolved nutrients - Having measures for Nitrate (NO3, µg/l), Nitrite (NO2, µg/l) and Ammonium (NH4, µg/l). Grouping these in a pool called Dissolved Inorganic Nitrogen (DIN) using µg/l seems to be a straight forward addition. Example:
NO2=0,56 µg/l NO3=12,89µg/l NH4=6,55µg/l => DIN = 20 µg/l

Converting each component into µM (µmoles/l) using the molar weights of each substance (NO2 = 46 µg/µmole, NO3 = 62µg/µmole and NH4 = 18µg/µmole) is all fine - but there is a big difference in taking the DIN µg/l and the collected µg/µmole (126 µg/µmole) and computing a DIN in µM versus just adding up the individual components... :?: How is this different?

Example:
NO2: 0,56 µg/l = 0,012 µM
NO3: 12,89 µg/l = 0,21 µM ------> added up this gives 0,582 µM DIN
NH4: 6,55 µg/l = 0,36 µM

DIN: 20 µg/l = 0,159 µM (using the collected 126 µg/µmole of all components)... So the results are 0,582 µM DIN vs. 0,159 µM :?: :?:

What is the correct if any and what is the reason this logic does not work. Is there some sort of scientific logic that rules out one of the above methods?
All answers are appreciated

/Jack

Re: Converting units of a grouped expression...

Thu Dec 10, 2009 6:38 pm

*bump

Don't anyone have an answer to this?

Re: Converting units of a grouped expression...

Fri Dec 11, 2009 2:28 am

Jack wrote:Don't anyone have an answer to this?


Jack:

I'm not a chemist and have forgotten much of my high school chemistry as it has nothing to do with my field, so I can't answer your question. However, if you post your question to the Convert and Calculate section of:http://forum.onlineconversion.com, Mrs. X or someone else there can answer your question.
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