Question

A compound contains 57.2 percent carbon, 6.1 percent hydrogen, 9.5 percent nitrogen, and 27.2 percent oxygen. What the empirical formula of the compound?

Answer 1

Answer:

So the empirical formula is C14H18N2O5

Explanation:

C = 57.2% = 12g/mol

H = 6.1% = 1g/mol

N = 9.5% = 14g/mol

O = 27.2% = 16g/mol

Empirical Formula for compound hmm

Assume

C = 57.2g

H = 6.1g

N = 9.5g

O = 27.2g

So we have

C = 57.2g/12g =  4.76  moles

H = 6.1g/1g      =  6.10   moles

N = 9.5g/14g   =  0.68  moles

O = 27.2g/16g =  1.70   moles

Divide each mole value by the smallest number of moles calculated.  Round to the nearest whole number.

C =  4.76  moles / 0.68 moles = 7

H = 6.10   moles / 0.68 moles = 9

N = 0.68  moles / 0.68 moles = 1

O = 1.70   moles / 0.68 moles = 2.5

Ok so we now have the ratios but for O it's 2.5, have to be whole numbers so we will need to double all the numbers.

C = 14

H = 18

N =  2

O =  5

So the empirical formula is C14H18N2O5