Chemistry 3211 Chapter 1

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Chemistry 3211 Chapter 1
Elemental Analysis -
Mass H2O increase in mass of drying tube
Mass CO2 increase in mass of CO2 absorption tube
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Problem 2 (pg 11) The combustion of an 8.23 mg
sample of unknown substance gave 9.62 mg of CO2
and 3.94 mg of H2O. Another sample, weighing
5.32 mg, gave 13.49 mg of AgCl in a halogen
analysis. Determine the precentage composition
and the empirical formula for this organic
compound.
C (2.62 mg C / 8.23 mg) 100 31.8 C
H (0.44 mg H / 8.23 mg) 100 5.3 H
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Determination of chlorine Another sample,
weighing 5.32 mg, gave 13.49 mg of AgCl
CXHYClZ ? Cl-(aq)
Ag(aq) ? AgCl(s)
Cl (3.34 mg Cl / 5.32 mg) 100 62.8 Cl
31.8 C 5.3 H 62.8 Cl 99.9 total
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Empirical formula
1.50
3.0
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Empirical formula C3H6Cl2
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Molecular formula
C3H6Cl2 113 g/mol C6H12Cl4 226
g/mol C9H12Cl6 339 g/mol
Selection of the correct molecular formula
requires a knowledge of the molecular weight.
Classical methods for determining molecular
weights include
1) Vapor density methods
2) Colligative properties freezing pt
depression, boiling pt elevation, osmotic pressure
3) Titration methods
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4. A compound had the empirical formula C9H8O4.
When a mixture of 5.02 mg of the unknown and
50.37 mg of camphor was prepared, the melting
point of a portion of this mixture was determined
to be 156oC. What is the molecular mass of this
substance?
For camphor Tf 180oC and Kf 40oC per molal
of moles of solute molality of Kg of
solvent
mmoles solute 0.6 .05037 .03 mmoles cmpd
MW 5.02 mg/ .03 mmoles 167 mg / mmol 167
g/mol
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Home work assignment 1 Chapter 1 (page
11) 1, 3, 6, 8 and 11.