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
Answers
Answer:
52.2% of Carbon, 13.0% of Hydrogen, and 34.8% of Oxygen
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