### What does "ppm" mean?

The expression "1 ppm" means a given solute exists at a concentration of one part per million parts of the solution. These are two common ways to think about what the concentration "1 ppm" means:

a) it is one-millionth of a gram per gram of sample solution.

b) it is one gram of solute per million grams of sample solution.

Notice that the more general word 'part' is used above, but 'gram' is used in (a) and (b) just above. This is because 'gram' is used almost exclusively when parts per million is used.

The best way to explain this is by doing some examples.

Example #1: Sea water contains 3.90 x 10¯6 ppm of dissolved gold. What volume (in liters) of this sea water would contain 1.00 g of gold?

Solution:

1) 3.90 x 10¯6 ppm means this:

3.90 x 10¯6 g of Au per 1,000,000 gram of seawater

2) We use a ratio and proportion:

 3.90 x 10¯6 g Au 1 g Au ––––––––––––––––– = ––––––– 1,000,000 g seawater x

x = 3.91 x 1011 g of sea water contains 1.00 g of gold

3) Using ppm means very dilute solutions, in which case the density is assumed to be 1.00 g/mL. Convert the mass just above to volume:

3.91 x 1011 g / 1.00 g/mL = 3.91 x 1011 mL

changing to liters, we have:

1.00 g of Au per 3.91 x 108 L

Example #2: Pollutants in air and water are frequently measured in parts per million (ppm) or parts per billion (ppb). One part per million would mean that there is one gram of the pollutant in every one million grams of air. At ordinary temperature and pressure, air has a density of 0.00012 gram per cubic centimeter. What volume of air would contain one gram of sulfur dioxide, a pollutant that causes acid rain, if the sulfur dioxide concentration is 2 ppm.

Solution:

1) How many grams of air contain one gram of SO2?

2 ppm means 2 grams SO2 per million grams of air

therefore,

one gram of SO2 will be found in 500,000 g of air

2) What volume of air weighs 500,000 g? We use the density (notice I set it up in a ratio and proportion style:

0.00012 g over 1 cm3 = 500,000 g over x

x = 4.17 x 109 cm3

Example #3: A sample of oil (density = 0.89 g/mL) was found to have dioxin contamination of 2 ppm. How many mL of the oil would contain 0.01 gram of dioxin?

Solution:

1) Determine grams of oil holding 0.01 g of dioxin. Use a ratio and proportion:

(2 g dioxin over 1,000,000 g solution) = (0.01 g over x)

x = 5000 g

2) Determine the volume of oil:

0.89 g/mL = 5000 g divided by x

x = 5618 mL

Example #4: Fish need about 5 ppm O2 dissolved in water to survive. Will water with 7 mg O2 per liter sustain fish?

Solution:

We need to convert 7 mg/L into ppm:

(0.007 g / 1000 g) = (x / 1,000,00 g)

x = 7 ppm

Notice the conversion of one liter of water (with 7 mg dissolved oxygen) into 1000 g. I used the density of 1.00 g/mL for the conversion. This is acceptable because solutions in the ppm range are so dilute that the solute (in this case, the O2, has no effect on the solution density.

Look again at the mg/L value. Another way to explain getting from mg/L to ppm is to multiply both numerator and denominator by 1000:

(0.007 g / 1000 g) x (1000 / 1000) = 7 g / 1,000,000 g = 7 ppm

You may very well face problems where you get the data in mg/L format and are then asked something about ppm. Remember!

Example #5: Show why a mass of 0.145 g of KMnO4 in a 500 mL volumetric flask corresponds to 100 ppm in Mn.

Solution:

1) Using a gravimetric factor, determine grams of Mn in 0.145 g of KMnO4:

0.145 g times (54.938 g/mol over 158.032 g/mol) = 0.0504076 g

2) Calculate ppm of Mn. Use a ratio and proportion:

(0.050 g over 500 g) = (x over 1,000,000)

x = 100 ppm

Example #6: Symptoms of lead poisoning become apparent after a person has accumulated more than 20 mg in the body. Express this amount as parts per million for an 80 kg person.

Solution:

1) Convert mg and kg to grams:

20 mg = 20 x 10-3 g
80 kg = 80 x 103 g

2) Determine grams Pb per g of bodyweight:

20 x 10-3 g / 80 x 103 g = 0.25 x 10-6 per g of bodyweight

3) Refer to first description of ppm given at beginning of file:

one ppm "is one-millionth of a gram per gram of sample solution."

In this case, the "solution" is the bodyweight.

Example #7: If you eat a 6 oz. can (180g) of tuna with 0.20 ppm Hg, how much mercury do you ingest?

Solution:

1) The meaning of 0.20 ppm Hg:

Remember, one ppm "is one-millionth of a gram per gram of sample solution."

Therefore the tuna contains 0.20 x 10-6 g of Hg per gram of tuna.

2) Calculate Hg in 180 g of tuna:

0.20 x 10-6 g of Hg per g of tuna times 180 g of tuna = 3.6 x 10-5 g of Hg

Example #8: It is estimated that 3 x 105 tons of sulfur dioxide enters the atmosphere daily owing to the burning of coal and petroleum products. Assuming an even distribution of the sulfur dioxide throughout the earth's atmosphere (which is not the case), calculate in parts per million by weight the concentration of SO2 added daily to the atmosphere. The weight of the atmosphere is 4.5 x 1015 tons.

Solution:

1) Set up a ratio and proportion:

You are adding 3 x 105 parts per 4.5 x 1015 parts.

How many parts per 1 x 106 is this?

2) Solve it:

3 x 105 parts is to 4.5 x 1015 parts as x is to 1 x 106 parts

x = 0.000067 ppm

Example #9: A solution used to chlorinate a home swimming pool contains 7% chlorine by mass. An ideal chlorine level for the pool is one part per million chlorine. If you assume densities of 1.10 g/mL for the chlorine solution and 1.00 g/mL for the swimming pool water, what volume of the chlorine solution in liters, is required to produce a chlorine level of 1.00 ppm in an 18,000 gallon swimming pool?

Solution:

1) Convert 18,000 gallons to liters:

The conversion factor we will use is 1 gallon = 3.7854 L

18,000 gal x 3.7854 L/gal = 6.81372 x 104 L

2) Determine how many grams of pool water this is:

6.81372 x 107 mL x 1.00 g/mL = 6.81372 x 107 g

Note change from L to mL.

3) At 1 ppm, how much chlorine is required? Use a ratio and proportion:

(1 g chlorine ÷ 106 g pool water) = (x ÷ 6.81372 x 107 g of pool water)

x = 68.1372 g chlorine required

4) What amount of 7% (by mass) chlorine solution is required to deliver 68.1372 g of chlorine?

(68.1372 g ÷ 0.07) = (x/1)

x = 973.3886 g of chlorine solution required

5) What volume (in liters) is this?

973.3886 g ÷ 1.10 g/mL = 884.8987 mL

To three sig figs (which seems reasonable to the ChemTeam), the answer is 0.885 L.

Example #10: Find the number of millimoles of solute in 250. mL of a solution that contains 4.20 ppm CuSO4

Solution:

1) Among the various definitions of ppm, this one can be found:

1 mg of solute per 1 kg of solution equals 1 ppm

2) Given 4.20 ppm, that means:

4.20 mg CuSO4 per 1 kg of solution

3) At these low concentrations, it is safe to assume that the density of the solution is 1.00 g/mL. That means:

4.20 mg CuSO4 per 1 L (or 1000 mL) of solution

4) 250. mL of solution means:

4.20 mg / 4 = 1.05 mg of CuSO4 in 250. mL of solution

5) Calculate millimoles:

1.05 mg / 159.607 mg/mmole = 0.006579 millimoles

Example #11: Convert 6.30 ppm to molarity. The molecular weight is 584 g/mol.

Solution:

1) 6.30 ppm means a solution containing 6.30 parts of solute per million parts of solvent. We will let the part be identified as grams.

6.30 g per 1,000,000 grams

2) Let the density of the solution be 1.00 g/mL:

6.30 g per 1,000,000 mL

3) Divide by 1000:

0.00630 g per 1,000 mL

4) Note the silent conversion of mL to L:

0.00630 g/L divided by 584 g/mol = 1.08 x 10¯5 M

5) A very common way to express ppm is like this:

1 ppm = 1 mg/L

6) For our problem, we have this from above:

6.30 g per 1,000,000 mL

7) Divide each side by 1,000 to get:

0.00630 g per 1000 mL

8) Change grams to milligrams and mL to L:

6.30 ppm = 6.30 mg / L

9) And the solution proceeds from there.

Example #12: A solution of CO2 in water contains 0.000540 g of CO2 per 500.0 g of solution. What is the concentration of CO2 in ppb?

Solution:

1) Determine grams of CO2 per 1 gram of solution:

0.000540 g CO2 / 500.0 g solution = 1.080 x 10¯6 g CO2 / 1 g solution

2) ppb means "part per billion" by weight:

The expanded version would be grams CO2 per billion grams of solution.

3) Multiply top and bottom of the ratio obtained in step 1 by 109 (one billion).

 1.080 x 10¯6 g CO2 109 –––––––––––––––– x –––– = 1080 g CO2 per 109 g solution 1 g 109

1080 g CO2 per 109 g solution = 1080 ppb