if 88.9g of scandium and 88.9g of oxygen react to give scandium oxide and a left over of oxygen 4.19g ,what percentage by mass of scandium oxide is the oxygen
I've tried to make sense of this problem and I've failed. I assume you want to know percent O in Sc2O3 but the figures don't come out anywhere even close. Is this a paper problem or is it a lab experiment. According to my calculations, if all of the Sc were used, you should have had about half of the oxygen left over. Perhaps I just don't understand what you want as an answer.
To find the percentage by mass of oxygen in scandium oxide, we need to determine the masses of oxygen and scandium oxide.
First, let's calculate the mass of scandium oxide:
Mass of scandium oxide = total mass of oxygen and scandium - mass of leftover oxygen
= 88.9g + 88.9g - 4.19g
= 173.7g - 4.19g
= 169.51g
Next, we can find the mass of oxygen in scandium oxide:
Mass of oxygen in scandium oxide = mass of leftover oxygen = 4.19g
Finally, we can calculate the percentage by mass of oxygen in scandium oxide:
Percentage by mass = (mass of oxygen in scandium oxide / mass of scandium oxide) * 100
= (4.19g / 169.51g) * 100
= 2.47%
Therefore, the percentage by mass of oxygen in scandium oxide is approximately 2.47%.