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Let us takeone more example. What is thepercentage of carbon, hydrogen and oxygen

Answer»

There are 2 C atoms, 6 H atoms and 1 O2atom in 1 molecule of ethanol.So,mass of 1 molecule = 2 X mass of C atom + 6 X mass of H atom + 1 X massof O2atom

= (2 X 12) + (6 X 1) + (1 X 16) {atomic wt. of C atom is 12,H atom is 1 and O2 atom is16}

= 24 + 6 +16 = 46.

Now, % composition ofcarbonatom is = (2 X mass of C atom)/mass of ethanol molecule= 2 X 12 / 46=24 / 46= 0.5217

To convert it into %;-

=0.5217 * 100%= 52.17 %

% composition ofhydrogen= (6 X mass of H atom)/mass of ethanol mol.= 6 X 1 / 46=0.1304=13.04%

% composition ofoxygen= (1 X mass of oxygen atom)/mass of ethanol mol.= 1 X 16 / 46= 16 / 46

To convert it into %;-

=0.3478 * 100%

=34.78%

thus, % composition of ethanol is52.17% C, 13.14% H and 34.78% Oby weight


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