Elemental analysis of caffeine yields the following results: 49.49% c, 5.15% h, 28.87% n, and the remainder is o. what is the molecular formula of caffeine?

The answer to your question is C₄H₅N₂O

Explanation:

Process

1.- Calculate the percent of oxygen in the sample

    Percent of oxygen = 100 - 49.49 - 5.15 - 28.87

    Percent of oxygen = 16.49 %

2.- Write the percents as grams

C = 49.49 g

H = 5.15 g

N = 28.87 g

O = 16.47 g

3.- Convert the grams to moles

C          12 g 1 mol

          49.49 g  x

             x = (49.49 x 1) 12

             x = 4.12 moles

H        1 g   1 mol

         5.15 g  x

            x = (5.15 x 1)/ 1

            x = 5.15 moles

N          14 g   1 mol

            28.87 g  x

              x = (28.87 x 1) / 14

              x = 2 mol

O           16 g 1 mol

              16.49 g x

               x = (16.49 x 1) / 16

               x = 1.03 moles

4.- Divide by the lowest number of moles

C     4.12 /  1.03 = 4

H      5.15 / 1.03 = 5

N       2 / 1.03 = 1.9 ≈ 2

O      1.03 / 1.03 = 1

5.- Write the empirical formula

                                 C₄H₅N₂O

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.