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

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years