The elementary analysis showed that an organic compound contained c, h, n and o as its only elementary constituents. a 1.279 g sample was burned completely as a result of which 1.60 g of carbon dioxide and 0.77 g of water were obtained. a separately weighed 1.625 g sample contained 0.216 g of nitrogen. what is the empirical formula of the compound?

The answer to your question is:   C₃H₇NO₃

Explanation:

Data

                                  CxHyNzOa

sample = 1.279 g

CO₂ = 1.60 g

H₂O = 0.77 g

sample 2 = 1.625 g      0.216 g of N

Process

1.- Calculate the mass and moles of carbon (C)

Molecular mass of C0₂ = 44 g

Atomic mass of C = 12

                                    44 g of CO₂  12 g of C

                                    1.6 g of CO₂  x

                                    x = (1.6 x 12) / 44

                                    x = 0.44 g of C

                                    12 g of C   1 mol of C

                                    0.44 g        x

                                    x = (0.44 x 1) / 12

                                    x = 0.037 moles of C

2.- Calculate the mass and moles of hydrogen (H)

                                    18 g of H₂O  2 g of H

                                    0.77 g           x

                                    x = (0.77 x 2) / 18

                                    x = 0.086 g of H

                                    1 g of H  1 mol of H

                                   0.086 g    x

                                    x = (0.086 x 1) / 1

                                    x = 0.086 moles of H

3.- Calculate the percent of Nitrogen in sample 2

                                    1.625 g  100%

                                     0.216 g   x

                                     x = (0.216 x 100) / 1.625

                                     x = 13.29 %        

                            1.279 g of sample  100 %

                            x                              13.29%

                            x = (13.29 x 1.279) / 100

                           x = 0.17 g of Nitrogen

                           14 g of N 1 mol

                           0.17 g of N  x

                           x = (0.17 x 1) / 14

                           x = 0.012 moles of N

4.- Calculate the mass of Oxygen and the moles      

Mass of Oxygen = 1.279 - 0.44 - 0.086 - 0.17

                           = 0.583 g

Moles of oxygen

                               16 g of O  1 mol

                               0.583 g   x

                               x = (0.583 x 1) / 16

                               x = 0.036 moles of O

5.- Divide moles by the lowest number of moles

Carbon          0.037 / 0.012  = 3.08 ≈ 3

Hydrogen     0.086 / 0.012 = 7.1 ≈ 7

Nitrogen        0.012 / 0.012 = 1

Oxygen          0.036 / 0.012 = 3              

6.- Write the empirical formula

                                                  C₃H₇NO₃

4d^55s^1 ( third choice)