Question

Calculate the mass of aluminum in 500 g of Al(C2H3O2)3 ?

Hey guys! Working on a really hard chemistry assignment. If you could explain how to solve a problem like this to me, it would be awesome :)

Answer 1

`"70 g"`.

Explanation:

First, we need to find the number of moles of `Al(C_2H_3O_2)_3` in `"500 g"`. To do that, we need to find the molar mass (mass of `1` mole) of `Al(C_2H_3O_2)_3` by adding up the individual molar masses of its constituent elements:

`Al(C_2H_3O_2)_3 = Al + (3xx2xxC) + (3xx3xxH) + (3xx2xxO)`

These individual molar masses can be found on the periodic table. For example, here's the molar mass of `Al`:

After we've found all of the individual molar masses on the periodic table, we can plug them into our first equation:

`Al + (3xx2xxC) + (3xx3xxH) + (3xx2xxO)`
`= 26.98 + (3xx2xx12.01) + (3xx3xx1.008) + (3xx2xx16.00)`
`"= 204.11 g/mol"`

Now, we can divide `"500 g"` by `"204.11 g/mol"` to find how many moles of `Al(C_2H_3O_2)_3` are in `"500 g"`:

`"moles" = "mass"/"molar mass" = "500 g"/"204.11 g/mol"`
`"= 2.45 mol"`

In `1` mole of `Al(C_2H_3O_2)_3`, there `1` mole of aluminum.
So, in `2.45` moles of `Al(C_2H_3O_2)_3`, there would be `2.45` moles of aluminum.

So, to find the mass of `2.45` moles of aluminum, we'd just need to multiply the molar mass (mass of `1` mole) of aluminum by `2.45` moles:

`"mass" = "molar mass" xx "moles" = "26.98 g/mol" xx "2.45 mol"`
`"= 66.1 g"`

Finally, since there is only `1` significant figure in the amount provided to us in the question (`"500 g"`), there should only be `1` significant figure in our answer.
So, we'd have to round `"66.1 g"` to `"70 g"`. :)