Question

Oleum or fuming sulfuric acid available commercially in concentrations ranging from 20 to 99.9% sulfur trioxide.

What is the percent by mass of sulfur trioxide (SO3) if oleum is expressed as 3H2SO4 . 5SO3?
3H2SO4 * 5SO3

3(98.1g) + 5(80.1g) = 294.3 + 400.5 = 694.8

400.5 (SO3) * 100 = 40050

40050 / 694.8 = 57.64% (SO3)
Why was the 400.5 multiplied by 100, then divided by the 694.8???
to find percentage

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