. A compound contains only C, H, N, and O. It contains 37.0 % C and 42.5% O (both by mass), and there are 2 O atoms for every 1 N atom. Combustion [involving reaction with excess O2(g)] of 0.157 g of the compound produced 0.0310 g of H2O. In another, separate experiment, 0.103 g of the compound produced 0.0230 g of NH3(g). If the molar mass of the compound is 681g/mol, determine its molecular formula.

C₂₁H₁₅N₉O₁₈

Explanation:

Molecular formula is the ratio of atoms that are present in 1 molecule of the compound. We need to find moles of all atoms to find this ratio.

In a basis of 100, moles of C and O are:

Moles C:

37.0 * (1mol / 12g) = 3.083 moles C

Moles O:

42.5g * (1mol / 16) = 2.656 moles O

Now, to find moles of H, we need determine moles of H2O produced:

0.0310g H2O * (1mol / 18g) = 1.72x10⁻³ moles H2O * 2 = 3.44x10⁻³ moles H

These moles were produced when 0.157g of the compound react. In a basis of 100g:

3.44x10⁻³ moles H * (100g / 0.157g) = 2.194 moles H

In the same way, moles of N are:

0.0230g NH3 * (1mol / 17g) = 1.35x10⁻³ moles H2O

These moles were produced when 0.103g of the compound react. In a basis of 100g:

1.35x10⁻³ moles H * (100g / 0.103g) = 1.313 moles N

Empirical formula is (The simplest whole number ratio of atoms presents in a molecule). Dividing in the low number of moles (Moles N):

C: 3.083 moles C / 1.313 moles N = 2.3

O: 2.656 moles O / 1.313 moles N = 2.0

N: 1.313 moles N / 1.313 moles N = 1

H: 2.194 moles H / 1.313 moles N = 1.67

This ratio times 3 (To have the whole number ratio):

C: 7

O: 6

N: 3

H: 5

The empirical formula is:

C₇H₅N₃O₆

And weighs:

C: 7*(12g/mol)= 84

H: 5 * (1g/mol) = 5

N: 3 * (14g/mol) = 42

O: 6* (16g/mol) = 96

227g/mol

As the molecular mass of the compound is 681g/mol:

681 / 227 = 3

The empirical formula times 3 is the molecular formula, that is:

C₇H₅N₃O₆ × 3

the electronic configuration of an atom with eight electrons is 1s²2s²2p⁴. the orbital diagrams are shown in the figure.

as per the hunds rule:

1. electrons tend to avoid being in the same orbital as per as practicable.

2. electrons in different orbitals of the same energy will have their spin parallel.

this can be illustrated from the multiplicity of the orbitals of the atom. as there are four electrons in the p orbital and p orbital has 3 sub-shells. the electrons can be fed in different ways as shown in the figure but the correct orbital diagram will be as per the hunds rule.