16. Mass of chemical constituents. For each of the following species in the pres

Question

16. Mass of chemical constituents. For each of the following species in the present day Earth atmosphere, calculate i) the mass fraction, and the total mass in the Earth's atmosphere. (a) (1 pt) O2, mole fraction ra, = 0.21 (b) (1 pt ) CO2. Icol = 400 Pprnv (c) (1 pt) 03, assuming a constant mean volume fraction 0a = 10-6 (d) (1 pt) CFC-12, assuming a constat concentration of [CFC-12] = 2.5 * 10-10 in the atmospheric layer between the surface and 100 hPa i.e. approximately 90% of the total mass of the atmosphere), and [CFC-12]-0 everywhere above the 100 hPa pressure surface

Explanation / Answer

Mass of atmosphere = 5.15×1018 kg

Molar mass of earth atmosphere = 29 g/mol

a)

Molar mass of O2 = 32 g/mol

Mass fraction of O2 = 0.21 * 32 g/mol / 29 g/mol

= 0.23

Mass of O2 in earth’s atmosphere = 0.23 * 5.15 x 1018 kg

= 1.19 x 1018 kg

b)

[CO2] = 400 ppmv

Mole fraction of CO2 = 400/1 x 106 = 0.0004

Molar mass of CO2 = 44 g/mol

Mass fraction of CO2 = 0.0004 * 44 g/mol / 29 g/mol

= 0.000607

Mass of CO2 in earth’s atmosphere = 0.000607 * 5.15 x 1018 kg

= 3.13 x 1015 kg

c)

Volume fraction of O3 = Mole fraction of O3 = 10-6

Molar mass of O3 = 48 g/mol

Mass fraction of O3 = 10-6 * 48 g/mol / 29 g/mol

= 1.66 x 10-6

Mass of O3 in earth’s atmosphere = 1.66 x 10-6 * 5.15 x 1018 kg

= 8.52 x 1012 kg

d)

Mole fraction of CFC-12 = 2.5 x 10-10

Molar mass of CFC-12 = 120.91 g/mol

Mass fraction of CFC-12 = 2.25 x 10-10 * 120.91 g/mol / 29 g/mol

= 9.38 x 10-10

Mass of CFC-12 in earth’s atmosphere = 90 % of (9.38 x 10-10 * 5.15 x 1018 kg)

= 4.83 x 109 kg * 0.9

= 4.35 x 109 kg

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