Answer:
Molecular formula: C₄H₄O₄
Empirical formula: CHO
Explanation:
Centesimal composition of fumaric acid:
41.39% C
3.47% H
(100% - 41.39% - 3.47%) = 55.14% O
In 100 g of fumaric, we got:
41.39 g of C
3.47 g of H
55.14 g of O
If we take account that we have x mol of fumaric in x mass of the same compound (g/mol), we can determine molar mass.
33.4 g / 0.288 mol → 116 g/mol
Now, we can prepare this rules of three:
In 100 g of fumaric we have 41.39 g of C, 3.47 g of H, 55.14 g of O
Then 116 g of fumaric will have:
(116 . 41.39) / 100 = 48 g C
(116 . 3.47) / 100 = 4 g H
(116. 55.14) / 100 = 64 g O
If we convert the mass to moles, we reach the molecular formula:
48 g . 1mol /12 g = 4 moles C
4 g . 1mol /1g = 4 moles H
64 . 1mol/16 g = 4 moles O
Molecular formula: C₄H₄O₄
Empirical formula: CHO
