Question

Following hurricane Katrina, many people were housed in trailers provided by the Federal Emergency Management Agency (FEMA). A potential health concern for individuals living in the trailers is exposure to formaldehyde (due to its use in adhesives). From Sept. 19 to Oct. 7, 2006, air samples were collected from 96 new, unused travel trailers from, at a staging area in Baton Rouge, LA. Only previously unoccupied trailers were tested in order to eliminate any effects from human activities that might cause formaldehyde levels to rise. The units tested had been closed for approximately six weeks before the sampling. The air sampling data was analyzed for FEMA by the Department of Health and Human Services' Agency for Toxic Substances and Disease Registry (ATSDR) in Atlanta, GA. The baseline concentration of formaldehyde in the units averaged 1.2 ppmv (parts per million by volume) at the beginning of the test. What was the average formaldehyde concentration within the trailers, expressed in units of g/m3? Assume that the temperature inside the units was 25oC and P = 1 atm. Molecular weight of formaldehyde is 30 g/mole.

Answer 1

Ans. Step 1: 1 ppmv = 1 parts per million by volume = 1 ug/ L                     

[Note: 1 L = 10⁶ uL ; so, 1 uL is one millionth part of 1 L].

Given,

            [Formaldehyde] = 1.2 ppmv = 1.2 uL formaldehyde / 1.0 L air

Now,

Volume of formaldehyde in 1m³ air = [Formaldehyde] x Volume of air in liters

                                                = (1.2 uL formaldehyde / 1.0 L air) x 1000 L

                                                = 1200 uL                                          ; [1 m³ = 1000 L]

                                                = 1.2 mL                                            l [1 L = 1000 mL]

                                                = 0.0012 L

Step 2: Calculate the mass of formaldehyde in 1m³ air

So far, we have-

            Volume of formaldehyde = 0.0012 L

            Temperature = 25.0⁰C         , Pressure = 1 atm

It’s assumed that formaldehyde behave as ideal gas in atmosphere.

Ideal gas equation: PV = nRT      - equation 1

            Where, P = pressure in atm

            V = volume in L                   

            n = number of moles

            R = universal gas constant= 0.0821 atm L mol^-1K^-1

            T = absolute temperature (in K) = (⁰C + 273.15) K

Putting the values in above equation-     

            1.0 atm = 0.0012 L = n x (0.0821 atm L mol^-1K^-1) x 298.15 K

            Or, n = 0.0012 atm L / (24.478 atm L mol^-1) = 0.000049 mol

Now,

            Mass of formaldehyde = Moles x molar mass

                                                = 0.000049 mol x (30.031 g/mol)

                                                = 0.00147 g

Therefore,

            [Formaldehyde] = mass of formaldehyde in grams / Volume of air in m³

                                    = 0.00147 g / 1 m³

                                    = 0.00147 g/ m³