The answer is: the percent composition of chromium is 20.52%.
Ar(Cr) = 52; atomic mass of chromium.
Mr(BaCrO₄) = 253.3; molecular mass of barium chromate.
ω(Cr) = Ar(Cr) ÷ Mr(BaCrO₄) · 100%.
ω(Cr) = 52 ÷ 253.3 ·100%.
ω(Cr) = 20.52%; the percent composition of chromium in barium chromate.
