Answer:
49.92% is the mass percent of oxygen (O)in sulfur dioxide.
Explanation:
Atomic mass of oxygen = 16 g/mol
Number of oxygen atoms in sulfur dioxide molecule = 2
Molar mass of sulfur dioxide =16 g/mol+ 16 g/mol + 32.1 g/mol= 64.1 g/mol
Percentage of oxygen in sulfur dioxide:
[tex]\frac{16g/mol+16 g/mol}{64.1 g/mol}\times 100=49.92\%[/tex]
49.92% is the mass percent of oxygen (O)in sulfur dioxide.
