(Figure 1) Part A What is the molecular mass of estradiol? Express your answer w
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Question
(Figure 1)
Part A
What is the molecular mass of estradiol?
Express your answer with the appropriate units.
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Part B
What is a probable formula?
Express your answer as a chemical formula.
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The steroid hormone estradiol contains only C, H, and O; combustion analysis of a 3.47 mg sample yields 10.10 mgCO2 and 2.76 mgH2O. On dissolving 7.55 mg of estradiol in 0.500 g of camphor, the melting point of camphor is depressed by 2.10 C. [For camphor, Kf=37.7 (Ckg)/mol.](Figure 1)
Part A
What is the molecular mass of estradiol?
Express your answer with the appropriate units.
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Part B
What is a probable formula?
Express your answer as a chemical formula.
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Explanation / Answer
PART (A)
Formula for,
Depression in freezing point = 1000 * Kf * w / (W * m) ------------- (1)
Where,
depression in freezing point = 2.10 0C
Kf of camphor = 37.7 0C.kg/mol
w = weight of estradiol = 7.55 mg = 0.00755 g.
W = Weight of camphor = 0.500 g.
m = molar mass of estradiol = ?
From (1)
m = 1000 * 37.7 * 0.00755 / (0.5 * 2.10)
m = molar mass of estradiol = 271. g/mol
PART (B)
% by mass of C = 12 * 10.10 * 100 / (44 * 3.47) = 79.4 %
% by mass of H = 2 * 2.76 * 100 / (18 * 3.47) = 8.84 %
Now % by mass of O = 100 - 79.4 - 8.84 = 11.76 %
Element mass number of moles simple whole number ratio
C 79.4 79.4/12 = 6.62 6.62 / 0.735 = 9
H 8.84 8.84/1 = 8.84 8.84 / 0.735 = 12
O 11.76 11.76 / 16 = 0.735 0.735 / 0.735 = 1
Emperical formula = C9H12O
Emperical formula mass = 9(12)+12(1)+1(16) = 136
n = Molecular mass / emperical formula mass = 271 / 136 = 2.00
Therefore, Molecular formula = (Emperical formula)n = (C9H12O)2 = C18H24O2
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