Answer:
[tex]\%\ Composition\ of\ sulfur=40.1\ \%[/tex]
Explanation:
Percent composition is percentage by the mass of element present in the compound.
Given , Mass of sulfur= 32.1 amu
Mass of oxygen = 16.0 amu
Mass of sulfur trioxide [tex]SO_3[/tex] = 32.1 amu + 3*16.0 amu = 80.1 amu
[tex]\%\ Composition\ of\ sulfur=\frac{Mass_{sulfur}}{Mass_{SO_3}}\times 100[/tex]
[tex]\%\ Composition\ of\ sulfur=\frac{32.1\ amu}{80.1\ amu}\times 100=40.1\ \%[/tex]