Be sure to answer all parts Determine the percent compositio

Be sure to answer all parts. Determine the percent composition by mass of the artificial sweetener aspartame (C4HigN20). %C- %N =

molar mass of C₁₄H₁₈N₂O₅ = 294.3 g/mol

% of C

molar mass of C = 12.0107 g/mol then molar mass of 14 C = 14 X 12.0107 = 168.1498 gm

294.3 gm aspartame = 100 % then 168.1498 gm C = 168.1498 X 100 / 294.3 = 57.14 %

% of C = 57.14 %

% of H

molar mass of H = 1.00794 g/mol then molar mass of 18 H = 18 X 1.00794 = 18.14292 gm

294.3 gm aspartame = 100 % then 18.14292 gm H = 18.14292 X 100 / 294.3 = 6.16 %

% of H = 6.16 %

% of N

molar mass of N = 14.0067 g/mol then molar mass of 2 N = 2 X 14.0067 = 28.0134 gm

294.3 gm aspartame = 100 % then 28.0134 gm N = 28.0134 X 100 / 294.3 = 9.52 %

% of N = 9.52 %

% of O

molar mass of O = 15.999 g/mol then molar mass of 5 O = 5 X 15.999 = 79.995 gm

294.3 gm aspartame = 100 % then 168.1498 gm O = 79.995 X 100 / 294.3 = 27.18 %

% of O = 27.18 %