Please show all work and steps. I need help, and this answer didn’t work. Thank

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Please show all work and steps. I need help, and this answer didn’t work. Thank you!! PartA A2.52-g sample of a compound containing only carbon hydrogen, nitrogen, oxygen, and sulfur was burned in excess oxygen to yield 4.23 g of CO2 and 1.01 g of H2O. Another sample of the same compound, of mass 4.14 g. yielded 2.11 g of SO3. A third sample, of mass 5.66 g. was burned under different conditions to yield 227 g of HNOs as the only ntrogen containing product Calculate the empirical formula of the compourd Express your answer as a chemical formula. Enter the elements in the order: C, H, S, N,O. xX e2 C,H,O SN Prevlous Answers Request Answer X Incorrect; Try Again; 5 attempts remaining

Explanation / Answer

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ANS:

From the first sample ,

% C = [( molar mass of C / Molar mass of CO2 ) x ( mass of CO2 / mass of sample ) ]x100

        = [(12 / 44 )x ( 4.23 / 2.52 ) ] x 100

        = 45.8

% H = [( molar mass of H / Molar mass of H2O ) x ( mass of H2O / mass of sample ) ]x100

        = [(2 / 18 )x ( 1.01 / 2.52 ) ] x 100

        = 4.45

% O is = 100 -( % C + % H )

            = 100 -( 45.8 +4.45)

           = 49.75

No.of moles of O is , = mass/molar mass

                                 = 49.75 / 16

                                 = 3.1

No.of moles of C is = mass / molar mass

                              = 45.8 / 12

                               = 3.82 moles

No.of moles of H is = mass / molar mass

                              = 4.45 / 1

                               = 4.45 moles

From the second sample,

So % S =  [( molar mass of S / Molar mass of SO3 ) x ( mass of SO3 / mass of sample ) ]x100

             =  [(32 / 80 )x ( 2.11 / 4.14 ) ] x 100

             = 20.4

No.of moles of S is = mass / molar mass

                              = 20.4 / 32

                               = 0.64 moles

From the third sample,

So % N =  [( molar mass of N / Molar mass of HNO3 ) x ( mass of HNO3 / mass of sample ) ]x100

             =  [(14 / 63 )x ( 2.27 / 5.66 ) ] x 100

             = 8.91

No.of moles of N is = mass / molar mass

                              = 8.91 / 14

                               = 0.64 moles

Ratio of the no.of moles of C : H : O : S : N   is = 3.82 :4.45:3.1:0.64 : 0.64

in simple ineger ratio is : (3.82/0.64) :(4.45/0.64):(3.1/0.64):(0.64/0.64) : (0.64/0.64)

                                          5.9 : 6.9 : 4.9 : 1 : 1

                                       ~ 6 : 7 : 5 : 1 : 1

So emperical formula is : C6H7O5SN

THANK YOU

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