Assignment 1
1. The element mercury has seven naturally occurring isotopes. Based on the isotopic
masses and abundances shown below, calculate the average atomic mass of mercury.
Isotope Mass (amu) Abundance (%)
196
Hg 195.9658 0.146
198
Hg 197.9668 10.02
199
Hg 198.9683 16.84
200
Hg 199.9683 23.13
201
Hg 200.9703 13.22
202
Hg 201.9706 29.80
204
Hg 203.9735 6.85
Answer: Atomic mass = Fraction x mass (amu)
1) Hg 196 = 0.146/100 = 0.00146
2) Hg 198 = 10.02/100 = 0.1002
3) Hg 199 = 16.84/100 = 0.1684
4) Hg 200 = 23.13/100 = 0.2313
5) Hg 201 = 13.22/100 = 0.1322
6) Hg 202 = 29.80/100 = 0.2980
7) Hg 204 = 6.85/100 = 0.0685
0.00146(195.9658) + 0.1002(197.9668) +
0.1684(198.9683) + 0.2313(199.9683) +
0.1322(200.9703) + 0.2980(201.9706) +
0.0685(203.9735) = 200.6091 = 200.61
, 2. Dieldrin is an insecticide composed of carbon, hydrogen, oxygen, and chlorine. A
2.416 g sample of dieldrin is subjected to combustion analysis, yielding 3.350 g
CO2and 0.458 g H2O. Mass spectrometry gives a molecular weight of 381 g mol -1.
From a separate experiment, it is found that there are 2:1 ratio of C:Cl atoms.
Based on this information, what is the molecular formula of dieldrin?
Answer:
C, H, O, Cl + O2 CO2 + H2O
2.416 g 3.350g 0.458g
Number of moles of CO2 = 3.350g/44.01g x 1 mol =
0.7612 mol
Number of moles of H2O = 0.458g/18.016g x 1 mol =
0.02542 mol
Number of moles of H = 0.02542 mol x 2 = 0.0508 mol
Number of moles of Cl = 0.07612 mol / 2 = 0.03806 mol
Mass of C = 12.01g/mol x 0.07612 mol = 0.9142g
Mass of H = 1.008g/mol x 0.0508 mol = 0.0512g
Mass of Cl = 35.45g/mol x 0.03806 mol = 1.349g
Mass of O = 2.146g – (0.9142g + 0.0512g + 1.349g) =
0.1016g
Number of moles C = 0.9142g/12.01gmol = 0.07612 mol /
0.00635 = 11.987 = 12 moles
Number of moles H = 0.0512g/1.008gmol = 0.0508
mol/0.00635 = 7.99 = 8 moles
Number of moles Cl = 1.349g/35.45g/mol = 6 moles
Number of moles of O = 0.1016g/16.00gmol =
0.00635/0.00635 = 1 mol
= C12H8Cl6O
1. The element mercury has seven naturally occurring isotopes. Based on the isotopic
masses and abundances shown below, calculate the average atomic mass of mercury.
Isotope Mass (amu) Abundance (%)
196
Hg 195.9658 0.146
198
Hg 197.9668 10.02
199
Hg 198.9683 16.84
200
Hg 199.9683 23.13
201
Hg 200.9703 13.22
202
Hg 201.9706 29.80
204
Hg 203.9735 6.85
Answer: Atomic mass = Fraction x mass (amu)
1) Hg 196 = 0.146/100 = 0.00146
2) Hg 198 = 10.02/100 = 0.1002
3) Hg 199 = 16.84/100 = 0.1684
4) Hg 200 = 23.13/100 = 0.2313
5) Hg 201 = 13.22/100 = 0.1322
6) Hg 202 = 29.80/100 = 0.2980
7) Hg 204 = 6.85/100 = 0.0685
0.00146(195.9658) + 0.1002(197.9668) +
0.1684(198.9683) + 0.2313(199.9683) +
0.1322(200.9703) + 0.2980(201.9706) +
0.0685(203.9735) = 200.6091 = 200.61
, 2. Dieldrin is an insecticide composed of carbon, hydrogen, oxygen, and chlorine. A
2.416 g sample of dieldrin is subjected to combustion analysis, yielding 3.350 g
CO2and 0.458 g H2O. Mass spectrometry gives a molecular weight of 381 g mol -1.
From a separate experiment, it is found that there are 2:1 ratio of C:Cl atoms.
Based on this information, what is the molecular formula of dieldrin?
Answer:
C, H, O, Cl + O2 CO2 + H2O
2.416 g 3.350g 0.458g
Number of moles of CO2 = 3.350g/44.01g x 1 mol =
0.7612 mol
Number of moles of H2O = 0.458g/18.016g x 1 mol =
0.02542 mol
Number of moles of H = 0.02542 mol x 2 = 0.0508 mol
Number of moles of Cl = 0.07612 mol / 2 = 0.03806 mol
Mass of C = 12.01g/mol x 0.07612 mol = 0.9142g
Mass of H = 1.008g/mol x 0.0508 mol = 0.0512g
Mass of Cl = 35.45g/mol x 0.03806 mol = 1.349g
Mass of O = 2.146g – (0.9142g + 0.0512g + 1.349g) =
0.1016g
Number of moles C = 0.9142g/12.01gmol = 0.07612 mol /
0.00635 = 11.987 = 12 moles
Number of moles H = 0.0512g/1.008gmol = 0.0508
mol/0.00635 = 7.99 = 8 moles
Number of moles Cl = 1.349g/35.45g/mol = 6 moles
Number of moles of O = 0.1016g/16.00gmol =
0.00635/0.00635 = 1 mol
= C12H8Cl6O
