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)