Answer to Question #163611 in Organic Chemistry for fofo

Question

. (a) A researcher analyzed an unknown solid, extracted from the bark of spruce trees, to determine its percentage composition. An 11.32-mg sample was burned in a combustion apparatus. The carbon dioxide (24.87 mg) and water (5.82 mg) were collected and weighed. From the results of this analysis, calculate the percentage composition of the unknown solid. (b) Determine the empirical formula of the unknown solid. PROBLEMS 14782_01_Ch1_p001-014.pp3.qxd 1/25/08 10:11 AM Page 13 (c) Through mass spectrometry, the molecular mass was found to be 420 g/mole. What is the molecular formula? (d) How many aromatic rings could this compound contain?

Answer 1

a. Mass of sample= 11.32mg

Mass of CO₂= 24.87mg

Mass of C in CO₂= 12/44 x 24.87 = 6.78mg

Mass of H₂O= 5.82mg

Mass of H in H₂O = 2/18 x 5.82mg = 0.65mg

% composition of C in sample= 6.78/11.32 x 100% = 59.89%

% by mass of H in sample= 0.65/11.32 x 100= 5.74%

% by mass of O in sample = 100% - (58.89% + 5.74%)= 34.37%

b. Let's find the mole ratios

C= 58.89/12 = 4.9075

H= 5.74/1 = 5.74

O= 34.37/16= 2.1481

Divide by the smallest

C= 4.9075/2.1481 = 2.28

H= 5.74/2.1481 = 2.67

O= 2.1481/2.1481 = 1.00

Multiply through by 3

E.F = C₇H₈O₃

c. Molecular mass= 420g/mol

n(E.F) = 420

n(12x7 + 8 + 48) = 420

n= 420/140= 3

M.F= 3(E.F) = 3(C₇H₈O₃)

= C₂₁H₂₄O₉

D. Three aromatic rings (suggested from n=3)