Question

For the following question, please explain specifically what it is you're doing for each step/where your number is coming from (ex.: 1. calculating molar mass... 2. Diving by moles...etc)... This will help me better understand how to solve the problem... thank you!

1a) Adipic Acid contains 49.84% O, and 6.85% H by mass. What is the empircal formula?

1b) The chemical process of rust is described by the following equation:

4Fe (s) + 3O2 (g) --> 2Fe2O3 (s)

The maximum of rust that can be produced from 56 g Fe and 32 g O2 is?

Answer 1

1a) Assume that you have 100 gms of compound
O - 49.84 gms
   H - 6.85 gms
   C - (100-49.84+6.85) = 43.31 gms
  converting masses into moles, by dividing gms by individual molecular weights
O = 49.84 gm / 16 g/mol = 3.115 moles
C = 43.31 gm / 12.0107 g/mol = 3.60595136 moles
H = 6.85 gm/1.00794 g/mol = 6.796039 moles
To simplify the ratio which we need to find empirical formula , ratios for every 3 carbons is taken so that
you can get everything simplified
3.60595136/ 3= 1.201983
Now divide H and O with 1.201983 (just consider the whole numbers)
H - 6.796039 / 1.201983 = 5
O - 3.115 / 1.201983 = 2.595 = 3
C₃H₅O₃ is the empirical formula.

1b) from the reaction stoichiometry it is clear that 4 moles of Fe react with 3 moles of O₂
moles of iron, Fe = 56 gm / 56 gm/mol = 1 mol
   moles of O₂ = 32 gm / 32gm/mol = 1 mol
If 1 mole of Fe and O₂ react you get only 0.5 mole of rust or Fe₂O₃
So to get 1 mole of Fe₂O₃ 2 moles of Fe + 3 moles of O₂ must react (Fe is limiting agent)
which will make = 2*56 gm + 3*16 gm = 112+48 = 160 gms to make 1 mole of Fe₂O₃
But from 1 mole of Fe you can make 0.5 mole of Fe₂O₃ which is half of 160
Therefore maximum rust that can be produced = 80 gms