To find out how many grams are in 4.65 moles of Al(NO₂)₃ Find out what the molar mass of Al(NO₂)₃ is Al = 26.98 g/mol Al N = 14 g/mol N O = 16 g/mol O Next, you have to look at the subscripts and figure out which they belong to, in this case: Al = 26.98 g/mol Al N₃ = 42 g/mol N₃ O₆ = 96 g/mol O₆ Finally, add the numbers together, so: 26.98 g/mol Al + 42 g/mol N₃ + 96 g/mol O₆ = 164.98 g/mol Al(NO₂)₃ Now, you have 4.65 mol Al(NO₂)₃ so 164.98 g/mol Al(NO₂)₃ × 4.65 mol Al(NO₂)₃ = 767.157 grams of Al(NO₂)₃
