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?

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

i believe you are right for all except the 3rd ( look at sqdancefan his answer is much better)

it should have an internal chamber and two thermometers sticking out of the top. the internal chamber will also be closed off from the outside.