Question

A compound has a chemical composition of 97.6 g carbon, 4.9 g hydrogen, 52 g oxygen, 45.5 g nitrogen. What is the empirical formula?
a. C3H3O2N2
b. C10H6O4N4
c. C5H3O2N2
d. CHON

Answer 1

First  step  is  to  calculate  the  mole  of each   element
that  is;
carbon  97.6/12=8.13moles
hydrogen= 4.9/1=4.9  moles
oxygen 52/16=3.25  moles
nitrogen=45.5/14=3.25 moles

step  two  is  to  calculate  the mole ratio  by   dividing with  the  smallest  number of  moles
that  is  divide  each  mole  with  3.25moles
 carbon=8.13/3.25 =5/2
hydrogen=4.9/3.25= 3/2
oxygen=3.25/3.25=1
nitrogen=3.25/3.25=1
step  3;  multiply  all  the  mole  ratio by  2  to  remove  the  fraction
carbon=5/2  x2 =5
hydrogen=3/2 x2=3
oxygen=1 x2=2
nitrogen  =1x2=2
therefore  the  empirical  formula  is C5H3O2N2(answer c)