Final answer:
The mass of oxygen in a 7.2 g sample of Al2(SO4)3 is calculated using the mole ratio and converting to grams, resulting in 4.0 g of oxygen to two significant figures.
Step-by-step explanation:
To determine the mass of oxygen in a 7.2 g sample of Al2(SO4)3, we need to use the mole ratio to calculate how many moles of oxygen are in the compound and then convert those moles to grams.
The formula mass of Al2(SO4)3 is 342.14 amu. Oxygen's part of the formula mass comes from three sulfate groups, with each sulfate containing four oxygen atoms, which gives us 12 oxygen atoms in total. The atomic mass of one oxygen atom is approximately 16.00 amu.
Firstly, calculate the total mass of oxygen in one formula unit of Al2(SO4)3:
(12 oxygen atoms) x (16.00 amu/oxygen atom) = 192.00 amu
This means that for one mole of Al2(SO4)3, you have 192.00 grams of oxygen.
Then calculate how much oxygen is in a 7.2 g sample:
192.00 g oxygen / 342.14 g Al2(SO4)3 x 7.2 g Al2(SO4)3 = 4.03 g oxygen
The mass of oxygen in the 7.2 g sample of Al2(SO4)3 is 4.03 grams, when expressed to two significant figures: 4.0 g.