Simple Concentration Stoic Problem?

[Lori]

Homework Statement

What mass of aluminum nitrate should be dissolved in 225 ml of water to make a solute with a total nitrate concentration of 0.866 M ?

Homework Equations

M = n/L

The Attempt at a Solution

So, i keep getting the wrong answer. THe answer is suppose to be 13.8 g but i get a different number
:

.226 L * (0.866 mols/L) = 0.19485 mols needed

Nitrate molar mass is just 16*3 + 14 + 26.98= 88.98 g/mol

0.19485 * 62 = 17.33 grams of Aluminum nitrate

 

[Charles Link]

The chemical formula for aluminum nitrate is $Al (NO_3)_3$. I think that might solve your problem.

 

[Lori]

The chemical formula for aluminum nitrate is $Al (NO_3)_3$. I think that might solve your problem.

I get that the molar mass for (NO3)3 is 186 but i still get 40 g of Al(NO3)3

 

[Charles Link]

I get that the molar mass for (NO3)3 is 186 but i still get 40 g of Al(NO3)3

The molar mass of $Al (NO_3)_3$ is 213. Meanwhile, how many moles do you need? With $(NO3)_3$ you don't need .195 moles.

 

[Lori]

isnt the nitrate concentration referring to just nitrate?

 

[Charles Link]

isnt the nitrate concentration referring to just nitrate?

But for every mole of $Al(NO_3)_3$ you get 3 moles of $(NO_3)^-$ in the solution. You do need .195 moles of $(NO_3)^-$. That part you got correct.

 

[Lori]

The molar mass of $Al (NO_3)_3$ is 213. Meanwhile, how many moles do you need? With $(NO3)_3$ you don't need .195 moles.

Oh my goodness! I keep forgetting you can just get mole-mole ratio from the molecule itself... thanks!
I spent like so long on this problem ._.