Question #cb91b

Dec 15, 2017

${C}_{4} {H}_{4} {O}_{4}$

Explanation:

First we need to calculate the mass of each element in our compound. Since we have not been given a mass for our compound, I would just use 100 g for ease of calculation.

0.4239 x 100 g = 42.39 g
0.0347 x 100 g = 3.47 g
0.5514 x 100 g = 55.14 g

Now we need to convert these to moles in order to find the empirical formula. It is necessary to do this because a mole weighs a different amount for each element.

We can convert grams to moles using the following equation:
Number of moles (mol) = mass (g) / Mr (g/mol)

Mr carbon = 12
Mr oxygen = 16
Mr hydrogen = 1

Usually we would use molar mass to a higher degree of accuracy e.g. 3 decimal places, but since we are only using this calculation to determine ratios, and not exact values, we are just going to use simple Mr for ease of calculation.

42.39 / 12 = 3.53
3.47 / 1 = 3.47
55.14 / 16 = 3.45

The elements are in a 1 : 1 : 1 ratio, so we can say our empirical formula is:

$C H O$

Now we need to convert this to our molecular formula. The total molar mass of the compound in our empirical formula is:

(1 x 12) + (1 x 16) + (1 x 1) = 29 g/mol

We know the molecular mass of our compound is 116 g/mol, so to determine the number of times our empirical formula goes into our molecular formula, we divide it by 29:
116 / 29 = 4

The empirical formula goes into the molecular formula 4 times, so the molecular formula is:

${C}_{4} {H}_{4} {O}_{4}$

This is the molecular formula of Fumaric Acid.

We can check this is correct by calculating the molecular formula mass:

(4 x 12) + (4 x 16) + (4 x 1) = 116 g/mol