Question

What is (i) `%O` in `"aluminum sulfate"` by mass? And (ii) given a `1.00*g` mass of this salt, how many formula units does this mass represent?

Answer 1

Well, `%O="Mass of oxygen in aluminum sulfate"/"Mass of aluminum sulfate"=56%`

Explanation:

`O%=(12xx16.00*g*mol^-1)/(342.15*g*mol^-1)xx100%=??%`

For formula units present in `1.000*g` of the salt, we work out the molar quantity.......

`"Moles of aluminum sulfate"=(1.000*g)/(342.15*g*mol^-1)=2.923xx10^-3*mol`

And since we know that `1*mol` contains `N_A="Avogadro's Number"=6.022xx10^23*"particles"*mol^-1`, we has ..............

`2.923xx10^-3*molxx6.022xx10^23*mol^-1=1.760xx10^21*"formula units"`.