Question

A compound analyzes as 79.08% carbon 5.54% hydrogen and 15.38 percent nitrogen. What is the empirical and molecular formula if the molar mass is 273.36 g/ Mol?

Answer 1

Answer:

emperical-C6H5N,Molecular formula-C18H15N3

Explanation:

Lets take 100 g of that compound,

:masses of

carbon-79.08g

hydrogen-5.54g

nitrogen-15.38g

Now find moles of each

therefore,no of moles of:

carbon-79.08÷12=6.58 mol

hydrogen-5.54÷1.008=5.49mol

nitrogen-15.38÷14=1.09 mol

Now as moles of N are smallest among all there fore divide it with moles of other elements tomget simplest ratio

C : H : N = 6 :5 : 1

Therefore ,emperical formula is C6H5N

where as molecular formula is emperical formula × n

where n =molarmass÷emerical formula mass

n= molar mass (273.36)÷Emperical formula mass(molar mass of c6h5n)

n= 273.36÷91=3

so molecular formula will be C6H5N×3=C18H15N3