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
Answer:
The empirical formula is CHO
The molecular formula is C4H4O4
Explanation:
Step 1: Data given
Mass of fumaric acid = 33.4 grams
Moles fumaric acid = 0.288 moles
Fumaric acid contains:
41.39 % C
3.47 % H
The rest of the compound is O
Step 2: Calculate the % O
100 - 41.39 - 3.47 = 55.14 %
Step 3: calculate mass
Mass C = 0.4139 * 33.4 grams = 13.82 grams
Mass H = 0.0347 * 33.4 grams = 1.16 grams
Mass O = 0.5514 * 33.4 grams = 18.42 grams
Step 4: Calculate moles
Moles = mass / molar mass
Moles C = 13.82 grams / 12.01 g/mol = 1.15 moles
Moles H = 1.16 grams / 1.01 g/mol = 1.15 moles
Moles O = 18.42 grams / 16.0 g/mol = 1.15
The empirical formula is CHO
The molecular weight is 29.02 g/mol
Step 5: Calculate molar mass of fumaric acid
Molar mass = mass / moles
Molar mass = 33.4 grams / 0.288 moles
Molar mass = 115.97 g/mol
Step 6: Calculate molecular formula
115.97 / 29.02 = 4
We have to multiply the empirical formula by 4
4*(CHO) = C4H4O4
The molecular formula is C4H4O4
