Answer
a.) Percent composition of aspartame:
%C = 57.13%
%H = 6.18%
%N = 9.52%
%O = 27.18 %
b.) Percent composition of carborundum:
%Si = 70.05%
%C = 29.95%
c.) Percent composition of aspirin:
%C = 59.99%
%H = 4.48%
%O = 35.52%
Work Step by Step
Recall that percent composition is calculated by dividing the mass of a single element in one mole of a compound by the molar mass of the entire compound.
Take note that the "mass of a single element in one mole of a compound" means you must also consider how many atoms of that element are present in the compound. So you multiply the "number of atoms" with the "atomic mass" of that element. You will get a quotient from 0 to 1 so you multiply it by 100 to get the percentage.
a.) The molar mass of $C_{14}H_{18}N_{2}O_{5}$ (aspartame) is 294.34 amu.
%C = $\frac{14 \times 12.011 amu C}{294.34 amu} \times 100 $ = 57.13% C
%H = $\frac{18 \times 1.01 amu H}{294.34 amu} \times 100 $ = 6.18% H
%N = $\frac{2\times 14.01 amu N}{294.34 amu} \times 100 $ = 9.52% N
%O = $\frac{5 \times 16.00 amu O}{294.34 amu} \times 100 $ = 27.18% O
b.) The molar mass of SiC (carborundum) is 40.097 amu.
%Si = $\frac{1 \times 28.086 amu Si}{40.097 amu} \times 100 $ = 70.05% Si
%C = $\frac{1 \times 12.011 amu C}{40.097 amu} \times 100 $ = 29.95% C
c.) The molar mass of $C_{9}H_{8}O_{4}$ (aspirin) is 180.17 amu.
% C = $\frac{9 \times 12.01 amu C}{180.17 amu} \times 100 $ = 59.99% C
% H = $\frac{8 \times 1.01 amu H}{180.17 amu} \times 100 $ = 4.48% H
% O = $\frac{4 \times 16.00 amu O}{180.17 amu} \times 100 $ = 35.52% O
