I am doing some exam review and do not know how to do these...Calculate the number of Nitrate ions in 3.99 g or aluminum nitrate.
? number of NO3 = 3.99g AlNO3 x(1mol AlNO3/212.996 g AlNO3)x(1mol NO3/1mol AlNO3)x(6.02*10^23 number of NO3/1mol NO3)=11*10^21 number of NO3
OK - first you need the ENTIRE mass of the aluminum nitrate (gram formula mass). You need the correct formula for aluminum nitrate, then add up the atomic masses from the periodic chart. Then, you need to convert 3.99 g of aluminum nitrate to moles of aluminum nitrate. (grams/formula mass = moles) Then, you need to take into account how many moles of nitrate ions are in a mole (look at the correct formula for aluminum nitrate). Then multiply by the number of moles you calculated (moles of ions per molecule x #moles). This would be moles of ions. You need individual ions you would the multiply by 6.02 x 10^23 ions/mole. SO: grams of sample divided by formula mass x #nitrate ions per molecule x (6.023 x 10^23 ions per mole) = # ions
