Answer:
59.8%
Explanation:
First find the Mr of manganese (III) nitrate.
Mr of Mn(NO₃)₃ = 54.9 + (14 × 3) + (16 × 3 × 3) = 240.9
Since we have to find the percentage composition of oxygen, we need to find the Mr of oxygen in the compound, which is:
Mr of (O₃)₃ = (16 × 3) × 3 = 144
Now we can find percentage composition / percentage by mass of oxygen.
% composition = [tex]\frac{Mr\ of\ oxygen\ in\ compound}{Mr\ of\ compound}[/tex] × 100
% composition = [tex]\frac{144}{240.9}[/tex] × 100 = 59.776%
∴ % compostion of oxygen in maganese(III)nitrate is 59.8% (to 3 significant figures).
