Determine the empirical formula for a compound that is 70.79% carbon

CH₅

Explanation:

This question may be lacking some information and in order to solve it, I will assume that the compound is a hydrocarbon, and therefore the percentages of carbon and hydrogen are:

C = 70.79 %                        H = 100 - 70.79 = 29.21 %

In 100 grams of compound wil have

70.79 g C

20.21 g H

Dividing by the atomic weights to get the mol of atoms:

C= 70.71 g/ 12 g/mol = 5.90 mol

H= 29.21 g/1 g/mol = 29.21 mol

Dividing by the 5.90 to get the proportions ,

C = 5.90/5.90 = 1

H= 29.21/5.90 = 4.95 rounding to 5,

the empirical formula is 1 atom of carbon per 5 of hydrogen

CH₅

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.

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