# Conversion of microMolar DOC to ppm

1. Jun 29, 2009

### Bacat

In a lot of the scientific literature, dissolved organic carbon (DOC) is given in microMolar units of carbon. Modern instruments give DOC in parts per million (ppm). I'm trying to figure out how to convert uM to ppm.

Since uM is dependent on the molecular weight, I already have a problem. DOC is actually a distribution of dissolved compounds so there is no single molecular weight. I do know that approximately 65% of the DOC in seawater is <1000 Daltons, but what is an appropriate value to use for molecular weight in this case?

Values for DOC in seawater range, but blanks are often reported around 25-30 uM C. Blanks are the measurement of DOC on a sample that has no dissolved carbon in it, so it represents the contribution from the measurement method. I'm trying to convert this value into ppm so I can compare it to a modern system blank which is reported in ppm.

Once I decide on a molecular weight to use, what is the actual calculation to convert from uM to ppm? Is this correct?

$$ppm = MW \times \mu M \;C$$

where ppm is parts per million (mass), uM C is microMolar carbon, and MW is molecular weight in grams/mole.

Any help is much appreciated.

2. Jun 30, 2009

### chemisttree

In DOC measurements, the CO2 resulting from either an oxidation or combustion is measured and reported. What molecular weight do you think is important in this case?

Generally, ppm and $$\mu mol$$ are equivalent. Remember that DOC measures CO2 obtained via either combustion or oxidation. Don't worry yourself about the MW of the particular organic species responsible for that CO2. Knowing this, can you hypothesize about the reason for a blank seawater sample having a 25-30 $$\mu M$$ value?

3. Jun 30, 2009

### Bacat

I believe uM only equals ppm if the molecular weight equals the density of the solution (water, for example, would be about 1000, or 1025 for seawater, kg/m^3). Am I missing something?

Measuring DOC:

In the literature, CO2 evolves from combustion (high temperature with a platinum catalyst) or oxidation (via persulfate at 100C). But this is really non-purgible organic carbon (NPOC). The purgable organic carbon is purged before the oxidation step, via acidification, and measured. The total organic carbon (TOC) is calculated from the total carbon (TC) by subtracting the inorganic carbon (IC).

$$TOC = TC - IC$$

If the sample is filtered to a known pore size, then the TOC is called DOC (dissolved organic carbon). This size is somewhat arbitrary, but <= 45 um particles in solution is commonly used.

Blank System Response:

The 25-30 uM C in the system blank comes from a variety of sources.

Most DI water and distilled water has a small amount of carbon in it (this is attributed to volatiles from the lab atmosphere dissolving into the water), on the order of about 10-15 uM C, or even higher (according to Peltzer and Brewer, 1993 Marine Chemistry).

The instrument itself also contributes to the system blank, though these mechanisms are less well understood. It is thought the type of catalyst of the high temperature combustion apparatus contributes to the system blank, but also issues with sample injection volumes, heating of a sample within the injection syringe, and so forth all contribute somewhat (ibid).

I'm afraid I'm still in somewhat of a quandry, though I thank you for your response.

4. Jun 30, 2009

### Bacat

I think I understand what I was missing now.

When I have a micromole of carbon, I am already defining a molecular weight of 12 grams/mole.

$$\mu M\;C = \frac{10^{-6}\;moles\;C}{1\;L\;H_{2}O} = \frac{10^{-6}\;moles\times \;12g/mol}{10^{6}g\;H_{2}O}=\frac{12g \times 10^{-6}}{10^{6}g\;H_2O}=ppm$$

Therefore,

$$ppm=\mu M \times MW \times 10^{-6}\;$$ if density of solution is 1000 kg/m^3 (water).

Is this correct?

But then I only get a value of .0003 for ppm from 25uM C...which seems much too low to me.

5. Jul 1, 2009

### chemisttree

Arrrgh! I should have said that $$\mu mol$$ per mol is ppm! Alternatively, you could use milligrams per liter as ppm.

6. Sep 2, 2010

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Bacat, I think you have everything right except, I got 1 L = 10^3 g not 10^6 g