1) The percent by mass is calculated by atomic mass of element multiply by number of that element in molecule divided by molecular mass of molecule.
Mr(HC₂H₃O₂) = 4 · Ar(H) + 2· Ar(C) + 2 · Ar(O).
Mr(HC₂H₃O₂) = 4 · 1 + 2 · 12 + 2 · 16.
M(HC₂H₃O₂) = 60; relative moecular mass of acetic acid.
Ar(H) = 1; the relative atomic mass of hydrogen atom.
Ar(C) = 12; the relative atomic mass of carbon atom.
Ar(O) = 16; the relative atomic mass of oxygen atom.
ω(H)= 4 · Ar(H) ÷ Mr(HC₂H₃O₂) · 100%.
ω(H) = (4 · 1 ÷ 60) · 100%.
ω(H) = 6.67 %; the percentage of hydrogen in acetic acid.
ω(C)= 2 · Ar(C) ÷ Mr(HC₂H₃O₂) · 100%.
ω(C) = (2 · 12 ÷ 60) · 100%.
ω(C) = 40%; the percentage of carbon in acetic acid.
ω(O)= 2 · Ar(O) ÷ Mr(HC₂H₃O₂) · 100%.
ω(O) = (2 · 16 ÷ 60) · 100%.
ω(O) = 53.33%; the percentage of oxygen in acetic acid.
2) Mr(C₆H₁₂O₆) = 12 · Ar(H) + 6 · Ar(C) + 6 · Ar(O).
Mr(C₆H₁₂O₆) = 12 · 1 + 6 · 12 + 6 · 16.
M(C₆H₁₂O₆) = 180; relative moecular mass of glucose.
ω(H)= 12 · Ar(H) ÷ Mr(C₆H₁₂O₆) · 100%.
ω(H) = (12 · 1 ÷ 180) · 100%.
ω(H) = 6.67 %; the percentage of hydrogen in glucose.
ω(C)= 6 · Ar(C) ÷ Mr(C₆H₁₂O₆) · 100%.
ω(C) = (6 · 12 ÷ 180) · 100%.
ω(C) = 40%; the percentage of carbon in glucose.
ω(O)= 6 · Ar(O) ÷ Mr(C₆H₁₂O₆) · 100%.
ω(O) = (6 · 16 ÷ 60) · 100%.
ω(O) = 53.33%; the percentage of oxygen in glucose.
The percentages of each element composing acetic acid and glucose will the same.
