g Determine the empirical formula for a compound that contains C, H and O. It contains 40.92% C, 4.58% H, and 54.50% O by mass. Must show your work on scratch paper to receive credit.

The empirical formula for the compound is C3H4O3

Explanation:

The following data were obtained from the question:

Carbon (C) = 40.92%

Hydrogen (H) = 4.58%

Oxygen (O) = 54.50%

The empirical formula for the compound can be obtained as follow:

C = 40.92%

H = 4.58%

O = 54.50%

Divide by their molar mass

C = 40.92/12 = 3.41

H = 4.58/1 = 4.58

O = 54.50/16 = 3.41

Divide by the smallest i.e 3.41

C = 3.41/3.41 = 1

H = 4.58/3.41 = 1.3

O = 3.41/3.41 = 1

Multiply through by 3 to express in whole number

C = 1 x 3 = 3

H = 1.3 x 3 = 4

O = 1 x 3 = 3

The empirical formula for the compound is C3H4O3

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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