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
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)