Answer
L-DOPA, $C_{9}H_{11}NO_{4}$
%$ C = 54.82$ %
%$ H = 5.623$ %
%$ N = 7.104$ %
%$ O = 32.45$ %
Vitamin E, $C_{29}H_{50}O_{2}$
%$ C = 80.87$ %
%$ H = 11.70$ %
%$ O = 7.430$ %
Vanillin, $C_{8}H_{8}O_{3}$
%$ C = 63.15$ %
%$ H = 5.300$ %
%$ O = 31.55$ %
Work Step by Step
The mass of a single atom is equal to its molar mass multiplied by the atomic mass unit. The atomic mass unit is defined as the one-twelfth of the mass of a carbon atom with 12 nucleons.
The percent composition by mass is defined as the mass of a component divided by the mass of the entire system.
We also have to take into account all the atoms of a single element that are in a compound.
L-DOPA, $C_{9}H_{11}NO_{4}$
%$ C = \frac{m(C)}{m(C_{9}H_{11}NO_{4})}\times 100$ $=\frac{9\times12.01 amu }{197.2 amu}\times100=54.82$ %
%$ H = \frac{m(H)}{m(C_{9}H_{11}NO_{4})}\times 100$ $=\frac{11\times1.008 amu }{197.2 amu}\times100=5.623$ %
%$ N = \frac{m(N)}{m(C_{9}H_{11}NO_{4})}\times 100$ $=\frac{14.01 amu }{197.2 amu}\times100=7.104$ %
%$ O = \frac{m(O)}{m(C_{9}H_{11}NO_{4})}\times 100$ $=\frac{4\times16 amu }{197.2 amu}\times100=32.45$ %
Vitamin E, $C_{29}H_{50}O_{2}$
%$ C = \frac{m(C)}{m(C_{29}H_{50}O_{2})}\times 100$ $=\frac{29\times12.01 amu }{430.7 amu}\times100=80.87$ %
%$ H = \frac{m(H)}{m(C_{29}H_{50}O_{2})}\times 100$ $=\frac{50\times1.008 amu }{430.7 amu}\times100=11.70$ %
%$ O = \frac{m(O)}{m(C_{29}H_{50}O_{2})}\times 100$ $=\frac{2\times16 amu }{430.7amu}\times100=7.430$ %
Vanillin, $C_{8}H_{8}O_{3}$
%$ C = \frac{m(C)}{m(C_{8}H_{8}O_{3})}\times 100$ $=\frac{8\times12.01 amu }{152.14 amu}\times100=63.15$ %
%$ H = \frac{m(H)}{m(C_{8}H_{8}O_{3})}\times 100$ $=\frac{8\times1.008 amu }{152.14 amu}\times100=5.300$ %
%$ O = \frac{m(O)}{m(C_{8}H_{8}O_{3})}\times 100$ $=\frac{3\times16 amu }{152.14amu}\times100=31.55$ %
