Answer:
Molecular formula → C₄H₄O₄
Empircal formula → CHO
Explanation:
The centesimal composition for the maleic acid is:
41.39 g / 100 g of maleic acid → C
3.47 g / 100 g of maleic acid → H
100 - (41.39 + 3.47) / 100 g of maleic acid → 55.14 g / 100 g of maleic acid → O
Let's determine the molar mass (g/mol) → 15 g / 0.129 mol = 116.2 g/mol
So, we must do these rules of three:
100 g of maleic acid have 41.39 g of C, 3.47 g of H and 55.14 g of O
Therefore, 116.2 g of maleic must have:
(116.2 . 41.39) / 100 = 48 g of C
(116.2 . 3.47) / 100 = 4 g of H
(116.2 . 55.14) / 100 = 64 g of O
We convert the mass to moles:
48 g . 1mol / 12 g = 4 moles of C
4 g . 1mol / 1g = 4 moles of H
64 g . 1mol / 16 g = 4 moles of O
Molecular formula → C₄H₄O₄
Empircal formula → We can divide the sub-index by 4 to reach the lowest coefficients → CHO
