5 dollars/100 dollars = 5%

I would much rather pay a 5 ppm sales tax:

5 dollars/10

^{6}dollars = 5 ppm

We do not usually refer to money in ppm, but we could. Ppm is more often found as a concentration, for example, ppm by mass or ppm by volume (sometimes referred to as ppmv). In nuclear magnetic resonance spectroscopy ppm can be used to describe the amount of chemical shift in frequency (Hz/MHz). This post focuses on the use of ppm as a measure of concentration.

**Parts-Per-Million By Volume (ppmv)**

Parts-per-million by volume is a common way of expressing a concentration in the gas phase. Gases are miscible and, in general, once allowed to come to equilibrium a gas is homogeneous, in other words its constituents are equally mixed. The SI unit of volume is the cubic meter (m

^{3}). Suppose a constituent of a gaseous mixture is at a concentration of 1 ppm. If one were to separate the mixture into its components, and measure the volume of each component, there would be 1 m

^{3}of the gas of interest for every 10

^{6 }m

^{3}of the mixture.

1 m

^{3}constituent/10

^{6 }m

^{3}of the mixture = 1 ppm

**The Ideal Gas Law**

To fully appreciate the value of such a ratio, one needs to consider what happens when the pressure or temperature changes. The ideal gas law (which is extremely accurate at atmospheric and lower pressures) states:

PV = nRT

in which p is the pressure, V is the volume, n is the number of moles (proportional to the number of molecules), R is a constant called the universal gas constant, and T is the temperature. All quantities are in SI units.

Alternatively:

V = nRT/P

Now consider a trace gas in a mixture. Let V

_{1}be the volume of the trace gas and V

_{T }be the total volume of the mixture. If we measure the concentration in ppmv we are expressing the concentration as a ratio of volumes.

V

_{1}/V

_{T}= n

_{1}RT

_{1}P

_{T}/n

_{T}RT

_{T}P

_{1}

Consider now that the gas mixture is at the same temperature and pressure. Notice that temperature and pressure divide out. If we express a concentration in ppmv, we can change the pressure or the temperature of the mixture and not change the value of the ratio.

V

_{1}/V

_{T}= n

_{1}/n

_{T}

This feature can be very useful in the atmosphere, where pressure and temperature change. Sometimes gas phase concentration is measure in milligrams per cubic meter( mg/m

^{3}). Although such units can also be useful, the expression is not insensitive to pressure and temperature. If I have a sample of gas, and I reduce the pressure of the gas, its volume will expand. The mass of the gas sample, however, is independent of pressure and does not change. Therefore the same gas mixture at a lower pressure will have a lower amount of a trace gas expressed in mg/m

^{3}, whereas its concentration measured in ppm remains unchanged.

**Converting Between mg/m**

^{3}and ppmSometimes it is important to be able to convert between the units

**mg/m**

^{3}and ppmv. To make such a conversion it is important to know some things about the situation; so it is useful to use a concrete example. Suppose the concentration of carbon dioxide in air is 380 ppmv, and we want to express this number in mg/m

^{3}at a pressure of 1 atmosphere (101,325 Pa) and a temperature of 298.15 K. The starting place is the definition of ppmv:

380 ppmv CO

_{2}= 380 x 10

^{-6}m

^{3}CO

_{2}/ 1 m

^{3}air.

The volume of air is already in the correct units; so one only needs to convert the volume of CO

_{2}to mg and the answer presents itself. If one knows how many moles of CO

_{2 }are present, it is easy to convert moles to mass. Again, I rearrange the ideal gas law:

n = PV/RT

Making sure that everything is in SI units:

n = (101,325)(380 x 10

^{-6})/(8.31441)(298.15)

Also consider:

(Mass of CO

_{2 }in grams) = n * (molecular mass of CO

_{2 }in grams)

The molecular mass is simply the sum of the atomic masses (12.01 + 16.00 + 16.00 = 44.01 grams)

Mass of CO

_{2 }in grams = (44.01)(101,325)(380 x 10

^{-6})/(8.31441)(298.15) = 0.684 g

So, at 1 atm and 298.15 K,

380 ppm CO

_{2 }= 684 mg/m

^{3}CO

_{2 }

***** Exercise for the Reader: Convert 21.2 mg/m**

^{3}of ammonia gas (NH_{3}) at a temperature of 288K and a pressure of 100,001 Pa to ppm ***If you have difficulties, state them in the comments, and I will help.

**Parts-Per-Million By Mass**

In liquids, it is common to use ppm by mass as a concentration. In general the calculations are a lot easier than in the gas phase.

1 ppm solute = 10

^{-6}kg solute / 1 kg solution = 1 mg solute / 1 kg solution

There is a very convenient way to do this calculation in dilute solutions of water. The density of water is 1 kg/liter (there is a small variation with temperature, but I neglect that here. Moreover, in dilute aqueous solutions most of the mass is from water:

1 ppm solute = 1 mg solute / 1 kg solution ≈ 1 mg solute/1 kg water = 1 mg solute/1 liter water

It is possible to be more accurate by correcting by the known or measured density of the solution.

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## 47 comments:

Hi, I have difficulties understanding something. You explained how to obtain ppmv to mg/m3. But I need to know how you convert say 50 ppm concentration of CO2 captured in a static gas chamber of volume 0.0029 m3 into another unit say g/day, if suppose that CO2 evolved after 5 minutes? Thanks in advance!

BTW, I see that the conversion factor as 1.96 from quick google search but plugging in the pv = nrt values won't give me that number 1.96. How's that factor 1.96 calculated for CO2?

gatoratanu@gmail.com

It is not 100% clear to me what you are trying to calculate, for example, the temperature of your gas cell matters. Also, what was the initial concentration of CO2? On the other hand, it sounds suspiciously like a homework problem. If it is a homework problem, you will have to calculate it yourself.

Can you please give me the answer for that Exercise for the reader. I'm trying to calculate it, but i don't know if i'm getting it right. Thanks.

Take it piece by piece. What would be the volume of 21.2 mg of NH3 at a temperature of 288K and a pressure of 100,001?

PV=nRT; so V=nRT/P. You know R, T, and P; so you have to convert mass to moles.

What is the molecular mass of NH3?

30ppmv

Excellent! Thank you very much for this explanation! Helped me alot

Those with experience cranking out calculations of this nature will be able to identify which R is being used. Most of us do not fall into that category, however. I am a visual learner and a big fan of unit factoring; not only to ensure my answers are in the correct units, but also to make sense of what I am calculating.

It is good practice to state which R is being used (some people may not even know there are several different Rs which can be applied to the universal gas law), and clearly state the units.

As stated in the article, "All quantities are in SI units." You are right that not everyone may understand what that means, but I have to start somewhere. I guess I will change that sentence to a link.

Nice Article Rich - well said... I was coming at this from a different angle and a large source of my confusion was recognizing I was dealing with a unit-less measure. Though not ALWAYS the case with ppm it was in this instance. Thanks!

The simpler approach is here..

To find concentration at state 2 given parameters of state 1 and state 2 and concentration at state 1, apply the formula.

C2 = C1 (P2 x T1)/(P1 x T2)

Concentration at a state 2 = concentration at state 1 corrected by pressures and temperature as shown. This can be worked from the basic gas laws

P1V1/T1 = P2V2/T2

The second equation is --

conc (mg/m3) = conc(ppm)x gas density x ((21-O2 refer)/(21-O2))

the correction of oxygen reference is required as only measurements at same oxygen values must be compared.

If the measurements and conversions are at same O2 levels then, this equation boils down to,

Conc (mg/m3) = conc(ppm) x gas density

for example,

100 (mg/m3) of carbon monoxide =

1.15 (gas density) X conc (ppm)

which gives

Conc(ppm) = 86.96 ppm.

A 12.0 liter sample of waste air from a smelter process was collected at 25 degree C and 1.00 atm pressure, and sulfur dioxide was removed. After SO2 removal, the volume of the air sample was 11.50 liters. What was the percentage by weight of SO2 in the original sample?

A 12.0 liter sample of waste air from a smelter process was collected at 25 degree C and 1.00 atm pressure, and sulfur dioxide was removed. After SO2 removal, the volume of the air sample was 11.50 liters. What was the percentage by weight of SO2 in the original sample?

How much is 30 ppm of CH4 in water expressed in ppmv?

I have been trying to understand the differences between ppm and ppmv, as far I can see ppmv has something to do with volume but how do you convert between the two?

Anonymous #1,

Looks like a homework problem. You will have to do your own homework.

Anonymous #2,

Same.

Anonymous #3,

Normally gases and vapor are measured by ppm by volume (ppmv), and liquids by mass. For a gas or a vapor it is possible to convert to mass just by converting volume to mass. For a liquid to convert to ppmv, it's a tricky issue, and I'm not sure why anyone would report a liquid contaminant in ppmv.

Can you tell me what 3 mg/cubic meter would be in ppm? this is a contaminant in the air.

Thanks. Not a math whiz.

@Dotty LeMieux: without identifying the contaminant in question, it's a nonsensical question. Read through the blog post, and you should be able to understand why.

Very nice information.

I have a little problem here. I am working with 1MCP gas (Used for delaying fruit ripening). I have a stock solution of 1MCP having 110ppm solution in 510 ml glass jar. I want to make another working solution from stock solution having concentration 3ppm in 200ml glass jar. how should I prepare this from stock solution?

i want to convert sulphur dioxide microgram per cubic meter to kg per day.how to convert it

Hello

I have a CO2 sensor that gives me the indication in PPM.

I want to use this sensor to measure the CO2 produced in a wine fermentation tank.

in this case how do we convert the PPM indicated to g / m3, knowing that the tank has a capacity of 1000l and has 800l of liquid in fermentation.

Thanks

If you just want the total CO2 that is in the empty 200 liters. Convert from ppm to g/m^3 as explained above, then realize that 1 m^3 = 1000 liters, so the gas phase part of your tank is 1/5 of a cubic meter. If you are interested in total CO2, however, you need to worry about what's dissolved in your liquid.

That's not a unit conversion problem as much as a Henry's Law problem. Henry's Law will only work if the gas phase CO2 is in equilibrium with CO2 in solution. You'd probably be better off measuring pH before and after fermentation. Page 53 of my expert report http://www.holocaust-history.org/irving-david/rudolf/affweb.pdf will give you a hint of how to do it. I don't have time to give a detailed explanation. Do some searching on Henry's Law, Carbonic acid, and pH.

I still can not get how you convert from mg/m3 to PPMV, its not clear to me. Could you more clear brief it. I have sample where gas spec is 5.7 mg/Sm3 in LNG. lets say LNG density is 454 kg/m3. could you help how to convert this 5.7 mg/Sm3 to ppmv.

Taking the ppmv definition you highligthed above (1ppmv= 1 cu. m constituent/10^6 cu. m of the mixture), my naive question is: does this mean that there is the same amount of "things" within these volumes? That is, is there the same amount of constituent "atoms" (it can be moles also, I guess) in the 1 cu. m of volume as there are mixture "atoms" in the 10^6 cu m of volume?

Thanks in advance...

Hello Rich,

In my environmental engineering class, we have a similar homework problem:

The total mercury concentration in the San Francisco Bay Area is reported to be 2.1 mg/m3 in air. Report this concentration in ppmv. Assume the air temperature is 20ºC.

My solution:

T = 20ºC + 273.15 = 293.15 K

AWHg = Atomic Weight of Mercury = 200.59 g/mol

Assumptions:

p = 1 atm

n = 1 mol

V_(20ºC)=nRT/p= (1 mol ∙0.082056 (L∙atm)/(K∙mol) ∙293.15 K)/(1 atm)=24.055 L

2.1 mg/m^3 =(〖ppm〗_v ∙200.59 g/mol)/(24.055 L) ∙ (273.15 K)/(293.15 K) ∙ (1 atm)/(1 atm)

〖ppm〗_v=(2.1 mg/m^3 ∙ 24.055 L)/(200.59 g/mol) ∙ (293.15 K)/(273.15 K) ∙ (1 atm)/(1 atm)

〖ppm〗_v=0.27

Is it correct for me to assume p = 1 atm and n = 1 mol or am I assuming erroneously? I can see pressure cancelling out, but I don't think the number of moles cancel out. What do you think?

Thank you so much,

- Denise

hey I have a question we are trying to find the weight needed for a substance but the thing is we don't the molar masses of lets call it A and B. for sure we need 10 of B in grams and we need 1600ppm of A would it be .016 grams or 1.6 grams since we get .0016% for every 10g

Hey there, I'm trying to determine if a particular chemical will kill bedbugs in it's gaseous form, rather than it's liquid form. I've read the studies that claim this chemical will do so at 362 ppm at 1029 ppm-hours. I know that ppm-hours is the exposure time but I don't really know how long that is. The claim is the bed bugs are dead after 6 hours. Also, the chemical weight is 25g. In gas/vapor form, what is the ppm? Not a chemist, just a consumer trying to do her homework.

Anon,

A measurement in ppm-hours is known as a Ct value, i.e., concentration x time. 1029 ppm-hours is the dosage you'd get from 1029 ppm in one hour, the dosage you get from 2058 ppm in a half hour, the dosage you get from 514.5 ppm in 2 hours. If you had a concentration of 362 ppm, you would have to wait 1029/362 = 2.84 hours or 2 hours and 50 minutes.

It is often approximated that toxic effects are linear in CT, but that's not always valid.

You used the wrong R value. The R value in PV=nRT is 0.08026. 8.314 is used for energy only.

It's simply a question of units. If you keep all of your quantities in compatible units you will be okay. the units should cancel out and give you the correct units. I always use SI units; so pressure in Pascals, volume in cubic meters, T in Kelvin.

If you use liters and atmospheres, you will use a different R than I do, but I recommend using SI units.

Thank you for your blog Rich.

Why is there a difference in result between your calculations and if I treat 380 ppmv as I treat liquids:

44,6 moles/m3

44 gCO2/mole

1964 gCO2/m3 x 380 ppm = 740 mg/m3

Thank you.

I get n/V = P/RT =101325/(8.31441*298.15)=

40.87 moles/m3

44 g CO2/mole

1798 g CO2/m3 x 380 ppm = 680 mg/m3

The real difference is that I used 298.15 K, and you assumed 273.15 K as the temperature.

Thank you. That also opened my eyes to what R stands for.

All the best.

Hi Rich

how to convert and express ppm of CO2 to mg/g when 1 g of material is burned ?

thanks

i'm not sure if you'll be able to help or not but I have to do a chemistry project and use the pH value of sulfurous acid (H2SO3) which is the acid rain that is produced from SO2 in the air. I need to use the pH value of sulfurous acid to determine the concentration of SO2 in the air in ppm.

Hi, thanks for a great site.

You state the following which I have some trouble understanding

"Mass of CO2 in grams = (44.01)(101,325)(380 x 10-6 )/(8.31441)(298.15) = 0.684 g

So, at 1 atm and 298.15 K,

380 ppm CO2 = 684 mg/m3 CO2"

Why is it 0.684 g and not 0.684 g/m3?

The result is 684 mg CO2 per m3 air, right?

Anonymous, that's a subtle question. I was calculating the mass in a cubic meter but the left side of the equation simply stated mass of CO2.

I could have stated

Mass of CO2 (in grams) per cubic meter of air =0.684 g/m3

The left side of the equation must equal the right side of the equation. You are following the math and the substance correctly.

I just calculated the numerator and denominator separately. Since the denominator is 1 at this point, I could have done it either way.

I have a billion cubic feet of natural gas with 60 PPM of H2S how much sulfur can i generate

RAMA, Sounds like a homework problem. All the tools you need to solve it are in my post.

Two places in your article define ppmv as volume of constituent divided by volume of total (constituent plus air).

"1 m3 constituent/10^6 m3 of the mixture = 1 ppm"

"Let V1 be the volume of the trace gas and VT be the total volume of the mixture. If we measure the concentration in ppmv we are expressing the concentration as a ratio of volumes. V1/VT = n1RT1PT/nTRTTP1"

While working on the CO2 problem, you state "The starting place is the definition of ppmv: 380 ppmv CO2 = 380 x 10^-6 m3 CO2/1 m3 air."

From earlier explanations, this should mean 380 m3 CO2 for every 10^6 m3 of the

mixture. Therefore:380 ppmv CO2 = 380 m3 CO2/(10^-6 m3 CO2 plus air)

How did the mixture change to nothing but air in the example, compared to earlier statements? Is this an approximation?

Anonymous,

CO2 is a constituent of air. If we were dealing with a pollutant not normally present in air, we could use the same calculation: it would be an approximation, but a very good one.

Great article. I am trying to find a way to convert atmospheric ammonia concentration from ppm to percent. Specifically, I want to know what the lower explosive level (LEL) of 15% would be in ppm, as well as the upper explosive level (UEL) of 28 % as ppm. All I am finding online is a simple conversion (15% = 150,000 ppm), but I am not confident that is correct.

Thank you!

Percent is part-per-hundred (1/100). Ppm is part-per-million (1/1,000,000).

1/100 = 10,000/1,000,000

in regards to and are generally have to be groups you would like to , specifically Us involving in which option.

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Hi Rich How do i convert 10 PPM of H2 gas into moles

Thanks,

NG

If I have 10,000 ppm H2S and a Total S mass % of 0.795 in a crude oil sample then what will the resulting S mass % be once all of the H2S has been removed?

Sorry if I'm missing something obv here but I think there's a slight miscalc in the post:

'' Mass of CO2 in grams = (44.01)(101,325)(380 x 10-6 )/(8.31441)(298.15) = 0.684 g ''

The answer should be mass CO2 = 0.000684 g

looks like you accidentally used .380 instead of 380x10^-6

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