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

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.