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 :)

1 Answer
Apr 29, 2018

"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:

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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". :)