Question

A given sample has 30% ethyl
alcohol by mass, the rest being water. Determine the percentage of C, H and O in the given sample​

Answer 1

Answer:

52.2% of Carbon, 13.0% of Hydrogen, and 34.8% of Oxygen

Answer 2

Answer:

Let the mass of sample be  = m gm

The mass of ethyl alcohol (C2H5OH) in it = 0.30m  g

Mass of water (H2O) = 0.70m g

No. of moles of ethyl alcohol = Mass / Molar mass = 0.30m /46  moles of Ethyl alcohol

No. of moles of carbon = 2 x (0.30m /46 ) moles

Mass of Carbon in sample = No. of moles x Molar mass = 2 x (0.30m /46 ) x 12

% of C in sample =mass of Element/total mass of sample×100

=0.1565m/m×100                                                

= 15.65 % of C in sample

No. of moles of O in ethyl alcohol = 0.30m /46  moles

No, of moles of O in water = 0.70m/18

Total no. of moles of O in sample = 0.30m/46+0.70m/18

Mass of O in sample = (0.30m/46+0.70m/18)×16 =0.7265m g

% of O in sample = 0.7265mm×100=72.65

Total no. of moles of H in sample  = 6(0.30m/46)+2(0.70m/18) =0.1169m  moles of H

Mass of H in sample = 1 x 0.1169m = 0.1169m g

% of H in sample = 0.1169m/m×100 =11.69 % of H