A2.52 g sample of a compound containing only carbon, hydrogen, nitrogen, oxygen, and sulfur was burned in excess o2 to yield 3.52 g of co2 and 0.839 g of h2o. another sample of the same compound, of mass 4.14 g, yielded 1.75 g of so3. a reaction to determine the nitrogen content was carried out on a third sample with a mass of 5.66 g and yielded 1.88 hno3 (assume that all of the nitrogen in hno3 came from the compound).
calculate the empirical formula of the compound. express your answer as a chemical formula. enter the elements in the order: c, h, s, n, o.

The answer to your question is:  C₄H₄NO₂S

Explanation:

Data

2.53 g CvHwNxOySv       ⇒   3.52g CO2  + 0.839g H2O

4.14 g  CvHwNxOySv       ⇒  1.75 g SO3

5.66g  CvHwNxOySv      ⇒    1.88 g HNO3

Empirical formula

For Carbon

MW CO2 = 44g

                         44g of CO2 12 g of C

                          3.52 of CO2   x

                            x = (3.52 x 12)/44

                            x = 0.96 g of C

                         12 g   1 mol

                         0.96g   x

                        x = (0.96 x 1) / 12

                         x = 0.08 mol of C

Percent of Carbon in the sample

                        2.53 g of sample    100%

                        0.96 g of Carbon       x

                          x = (0.96 x 100) / 2.53

                          x = 37.94 % of Carbon in the sample

For Hydrogen

MW H2O = 18 g

                        18g of H2O  2 g of H2

                        0.839 g       -    x

                         x = 0.093 g of H2

                        1 mol of H2  1 g of H2

                        x                   0.093 g of H2

                        x = 0.093 mol of H2

Percent of H2 in the sample

                        2.53 g of sample    100%

                       0.093                       x

                        x = 3.68 % of Hydrogen in the sample

For Sulfur

MW of SO3 = 80 g

                       80g of SO3  32 g

                       1.75 g             x

                        x = 0.7 g of S

                       32g S   1 mol

                       0.7g   x

                       x = 0.022 mol of S

Percent of S in the sample

                       4.14 g  100%

                        0.7 g   x

                       x = 16.9%

For Nitrogen

MW of HNO3   = 63g

                          63g of HNO3 14 g of N

                           1.88 g             x

                          x = 0.42 g of N

                          14 g   1 mol of N

                          0,42 g of N     x

                          x = 0.03 mol of N

Percent of N in the sample

                           5.66 g     100%

                           0.42 g   x

                           x = 7.4%

Percent of Oxygen = 100 - (37.94 + 3.68 + 16.9 + 7.4)

Percent of oxygen = 34.08

Grams of Oxygen                2.53 g of sample 100%

                                                 x                         34.08

                                              x = 0.86 g of Oxygen

                                         16 g of O   1 mol

                                         0.86 g of O  x

                                         x = 0.054 mol of Oxygen

Divide by the lowest # of mol

C =  0.08/ 0.022 = 3.6 ≈ 4

H = 0.093/ 0.022 = 4.2  ≈ 4

N = 0.022/ 0.022 = 1

O = 0.053/ 0.022 = 2.4  ≈ 2

S = 0.03/ 0.022 = 1.4  ≈  1

Empirical formula

                                  C₄H₄NO₂S

answer: mole of the gas

answer: mole of the gasexplanation: