A 4.50 g sample is analyzed and found to contain 3.22 g of carbon, 0.300 g of hydrogen, 0.221 g of nitrogen, and the rest is oxygen. The molar mass of the actual compound is found to be 285 g/mol. Find the empirical formula and then the molecular formula.

The empirical formula and the molecular formula are the same : C₁₇H₁₉NO₃

Given

3.22 g of carbon, 0.300 g of hydrogen, 0.221 g of nitrogen, and the rest is oxygen.

Required

The empirical formula and then the molecular formula.

Solution

Mass of Oxygen = 4.5 - (3.22+0.3+0.221)=0.759 g

Mol ratio of elements C : H : N : O =

3.22/12 : 0.3/1 : 0.221/14 : 0.759/16 = 0.268 : 0.3 : 0.016 : 0.047

Divide by 0.016 :

16.75 : 18.75 : 1 : 2.9375 = 17 : 19 : 1 : 3

(C₁₇H₁₉NO₃)n = 285

(12.17+19+14+3.16)n = 285

(285)n = 285

n = 1

Then the empirical formula and the molecular formula are the same

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be