The empirical formula of a given compound is C6H9ON5.
Explanation:
Step 1: Obtain the mass of each element present in grams
Element % = mass in g = m
Carbon = 83.884 grams, Hydrogen = 10.486 grams, Oxygen = 18.640 grams, Nitrogen = 86.99 grams.
Step 2: Determine the number of moles of each type of atom present
m/atomic mass = Molar amount (M)
Molar amount of carbon = (83.884 1 mol ) / 12 g = 6.99
Molar amount of hydrogen = (10.486 1 mol) / 1 g = 10.49
Molar amount of oxygen = (18.64 1 mol) / 16 g = 1.17
Molar amount of nitrogen = (86.99 1 mol) / 14 g = 6.21
Step 3: Divide the number of moles of each element by the smallest number of moles
M / least M value = Atomic Ratio (R)
Atomic radius of carbon = 6.99 / 1.17 = 5.9 = 6
Atomic radius of hydrogen = 10.49 / 1.17 = 8.9 = 9
Atomic radius of oxygen = 1.17 / 1.17 = 1
Atomic radius of nitrogen = 6.21 / 1.17 = 5
Step 4: Convert numbers to whole numbers. This set of whole numbers are the subscripts in the empirical formula.
R * whole number = Empirical Formula
The empirical formula of a given compound is C6H9ON5.
