Fumaric acid is an organic compound composed of 41.39% , 3.47% , and the rest oxygen. If 0.288 mol of fumaric acid has a mass of 33.4 g, what are the empirical and molecular fomulas of fumaric acid

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

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

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