How many milligrams of aluminum nitrate contains 3.5 x 1018 oxygen atoms?

Question

Mass of aluminum

Answer 1

Hi Matt,

 

The formula for aluminum nitrate = Al(NO3)3.

 

The molar mass of Al(NO3)3 = 213 g/mol.  That means 213 g Al(NO3)3 = 1 mol Al(NO3)3 = 6.02 x 1023 Al(N03)3.

 

you can see that there are 9 x 6.02 x 1023 oxygen atoms in each mol.

 

 

3.5 x 1018 Oxygen atoms x (1 mol Al(NO3)/ 9 x 6.02 x 10 23 Oxygen atoms) x (213 g Al(NO3)3/ 1mol Al(NO3)3 )   

 

= 1.4 x 10-4 g Al(N03)3 = 0.14 mg Al(N03)3.