Calculate the percentage of each element in acetic acid, hc2h3o2, and glucose, c6h12o6.b. these two substances have the same empirical formula. what would you expect the percentage composition of the empirical formula to b

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.

the answers would be

the first one and the second one

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer