Determine the empirical and molecular formulas of a compound that contains 56.36% o and 43.64% p and has a molar mass of 284 g/mo

Step 1;  calculate  the number of moles   of  each  element  (moles  =%  composition /  molar mass)

that  is  P =  43.64 / 30.974  =  1.41  moles
            O =  56.36 /   15.999=  3.52  moles

then   find  the  mole  ratio  by   diving  both  mole  with  smallest number  of mole

that  is  P =  1.41/1.41  =  1  mole
             O= 3.52 /1.41 =  2.5  moles
multiply  by  both  mole  ratio  by  2  to  remove   the  decimal
p=  1 x2  =  2 moles
O=  2.5  x2 =  5 moles
therefore  the  empirical   formula  =  P2O5

molecular  formula  is  calculated  as  follows

(P2O5)  n =  284  g/mol
{(30.974  x2)  +   (15.999x5)}n=  284  g/mol
141.943n=  284  g/mol
divide  both    side by  141.943
n=2

therefore  the  molecular  formula =   (P2O5)2  =  P4O10

  the volume of  0.220 m k2s    solution  required  to  completely  react with 175 ml  of 0.115m co(no3)2  calculate the moles  of co(no3)2 =  molarity x volume/1000175 x0.115/1000= 0.02  moleswrite  the reacting  equationk2s(aq)  +  co(no3)2  = 2 kno3(aq)  + cos(s)by  use  of mole ratio between k2s to co(no3)2 which  is  1: 1 the moles of k2s is therefore = 175  x0.115/1000= 0.02  molesvolume = moles/molarity=0.02/0.22=0.09  ml  of k2s