1)26.4 % Carbon 3.3 % Hydrogen
70.3 % Oxygen
Molar Mass: 91.0 g/mol
Empirical Formula:
Molecular Formula:

In the given case where 26.4 % Carbon , 3.3 % Hydrogen , 70.3 % Oxygen , and Molar Mass: 91.0 g/mol the:

Given:

Carbon (C) = 26.4 %  

Hydrogen (H) = 3.3 %  

Oxygen (O) = 70.3 %  

Molar mass of compound = 91.0 g/mol  

Determination of the empirical formula of the compound:-

The ratio is CHO = 1 : 1.5 : 2

multiply with 2 to find correct and complete number ratio

C = 1 × 2 = 2

H = 1.5 × 2 = 3

O = 2 × 2 = 4

Thus,  the Empirical formula =>

Mass × n = molar mass

12 × 2 + 1 × 3 + 16 × 4 = 91

24 + 3 + 64 = 91

91 = 91

Thus moles are 1 which means

molecular formula =  ()

Thus, here in the given data:

Molecular Formula: ()

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1. Empirical formula => CH2O2

2. Molecular formula => C2H4O4

Explanation:

From the question given above, the following data were obtained:

Carbon (C) = 26.4 %

Hydrogen (H) = 3.3 %

Oxygen (O) = 70.3 %

Molar mass of compound = 91.0 g/mol

Empirical formula =..?

Molecular formula =..?

1. Determination of the empirical formula of the compound.

C = 26.4 %

H = 3.3 %

O = 70.3 %

Divide by their molar mass

C = 26.4 /12 = 2.2

H = 3.3 /1 = 3.3

O = 70.3 /16 = 4.39

Divide by the smallest

C = 2.2 /2.2 = 1

H = 3.3 /2.2 = 2

O = 4.39 /2.2 = 2

Empirical formula => CH2O2

2. Determination of the molecular formula of the compound.

Molar mass of compound = 91.0 g/mol

Empirical formula => CH2O2

Molecular formula => [CH2O2]n

We shall determine the value of n as follow:

[CH2O2]n = 91

[12 + (2×1) + (2×16)]n = 91

[12 + 2 + 32]n = 91

46n = 91

Divide both side by 46

n = 91/46

n = 2

Molecular formula => [CH2O2]n

Molecular formula => C2H4O4

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8