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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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mlhickey mlhickey
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2 years ago
An unknown compound is composed of 65.44% C, 29.07% O, and 5.49% H. A sample weighing 5.34 g, when dissolved in 60.00 g H2O, lowers the freezing point to -0.600 °C. What is the molecular formula of the compound?  (Kf for water = 1.86 °C/m; C = 12.0, O = 16.0, H = 1.0 g/mol).

▸ C3H3O

▸ CHO

▸ C14H11O6

▸ C5H5O5

▸ C7H5O3
Textbook 
General Chemistry: Principles and Modern Applications

General Chemistry: Principles and Modern Applications


Edition: 11th
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sgreersgreer
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Anonymous
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Anonymous
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Quote from: priyan (A year ago)
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ΔT = i * Kf * m
Since the unknown compound is organic compound (because it has more C%), it will not dissociate.
So i = 1
0.6 = m * 1.86
molality, m =0.6/1.86 = 0.32258 mol solute /Kg water
gm of solute / Kg water = 5.34 / (60/1000) = 89 gm of solute / Kg water
For both the above calculations basis is 1 Kg of water so,
0.32258 mol = 89gm / molar mass in gm/mol
molar mass of solute = 275.9 gm/mol
Assume that you have 100 gm of unknown, the composition will therefore be
C - 65.44 gm ; O - 29.07 gm ; H - 5.49 gm
moles of C = 65.44 gm / 12 gm/mol = 5.4533 mol
moles of H = 5.49 gm / 1.0079 g/mol = 5.4469 mol
moles of O = 29.07 gm / 16 gm/mol = 1.8168 mol
C : H : O = 5.4533 : 5.4469 : 1.8168
divide with 1.8168
C:H:O = 3.0015 : 2.9980 : 1
the ratio can be approximated to 3 : 3 : 1
So the empirical formula is C3H3O whose molar mass is = 3*12+3*1.0079+16 = 55.0237gm/mol
275.9 gm/mol / 55.0237 gm/mol = 5.0142 ~5
so the molecular formula will b 5 times of empirical formula
C15H15O5 will be the molecular formula
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