Answer:
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:
<em>Moles C:</em>
37.0 * (1mol / 12g) = 3.083 moles C
<em>Moles O: </em>
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
<h3>C₂₁H₁₅N₉O₁₈</h3>
