Question

Determine the mass of oxygen in a 7.2g sample of Al2(SO4)3. Express to two sig fig.

Answer 1

Molar mass of Al2(SO4)3,

MM = 2*MM(Al) + 3*MM(S) + 12*MM(O)

= 2*26.98 + 3*32.07 + 12*16.0

= 342.17 g/mol

mass(Al2(SO4)3)= 7.2 g

number of mol of Al2(SO4)3,

n = mass of Al2(SO4)3/molar mass of Al2(SO4)3

=(7.2 g)/(342.17 g/mol)

= 2.104*10^-2 mol

Molar mass of O = 16 g/mol

1 mol of Al2(SO4)3 has 12 moles of O

So,

moles of O = 12 moles of Al2(SO4)3

= 12 * 2.104*10^-2 mol

= 0.2525 mol

mass of O,

m = number of mol * molar mass

= 0.2525 mol * 16 g/mol

= 4.04 g

Answer: 4.0 g