# Answer on Organic Chemistry Question for tori

Question #10304

Ascorbic acid is used to improve the nutrient of crops. it is composed of 36.20% carbon, 4.90% hydrogen, and 48.90% oxygen. The molar mass of ascorbic acid is 176.0 grams per mole. Determine the empirical and molecular formulas for ascorbic acid.

Expert's answer

This is not ascorbic acid because it is composed of 40.92% carbon, 4.58%

hydrogen and 54.5% oxygen.

This is unknown substance. Let's determine it's

molecular formula.

The first step will be to assume exactly 100 g of this

substance. This means in 100 g of this compound, 36.20 g will be due to carbon,

4.90 g will be due to hydrogen, and 48.90 g will be due to oxygen. We will need

to compare these elements to each other stoichiometrically. In order to compare

these quantities, they must be expressed in terms of moles. So the next task

will be to convert each of these masses to moles, using their respective atomic

weights:

(36.20 g C) x (1 mol C) / (12.01 g C) = 3.014 mol C

(4.90 g

H) x (1 mol H) / (1.008 g H) = 4.861 mol H

(48.90 g O) x (1 mol O) / (16.00 g

O) = 3.056 mol O

Take notice that since the composition data was given to

four significant figures, the atomic weights used in the calculation were to at

least four significant figures. Using fewer significant figures may actually

lead to an erroneous formula.

Now that the moles of each element are

known, a stoichiometric comparison between the elements can be made to determine

the empirical formula. This is achieved by dividing through each of the mole

quantities by which ever mole quantity is the smallest number of moles. In this

example, the smallest mole quantity is the moles of carbon (3.014

mol):

(3.014 mol C) / (3.014 mol) = 1.000 C = 1 C

(4.861 mol H) /

(3.014 mol) = 1.613 H = 1.6 H

(3.056 mol O) / (3.014 mol) = 1.014 O = 1

O

The ratio of C:H:O has been found to be 1:1.6:1, thus the empirical

formula is: CH1.6O. Again, as a reminder, this is the simplest formula for the

compound, and not necessarily the molecular formula. Suppose we know that the

molecular weight of this compound is 176 g/mol. With this information, the

molecular formula may be determined. The formula weight of the empirical formula

is 29.6 g/mol. Divide the molecular weight by the empirical formula weight to

find a multiple:

(176.0 g/mol) / (29.6 g/mol) = 5.95 = 6

The

molecular formula is a multiple of 6 times the empirical formula:

C(1 x

6) H(1.6 x 6) O(1 x 6) which becomes C6H10O6

hydrogen and 54.5% oxygen.

This is unknown substance. Let's determine it's

molecular formula.

The first step will be to assume exactly 100 g of this

substance. This means in 100 g of this compound, 36.20 g will be due to carbon,

4.90 g will be due to hydrogen, and 48.90 g will be due to oxygen. We will need

to compare these elements to each other stoichiometrically. In order to compare

these quantities, they must be expressed in terms of moles. So the next task

will be to convert each of these masses to moles, using their respective atomic

weights:

(36.20 g C) x (1 mol C) / (12.01 g C) = 3.014 mol C

(4.90 g

H) x (1 mol H) / (1.008 g H) = 4.861 mol H

(48.90 g O) x (1 mol O) / (16.00 g

O) = 3.056 mol O

Take notice that since the composition data was given to

four significant figures, the atomic weights used in the calculation were to at

least four significant figures. Using fewer significant figures may actually

lead to an erroneous formula.

Now that the moles of each element are

known, a stoichiometric comparison between the elements can be made to determine

the empirical formula. This is achieved by dividing through each of the mole

quantities by which ever mole quantity is the smallest number of moles. In this

example, the smallest mole quantity is the moles of carbon (3.014

mol):

(3.014 mol C) / (3.014 mol) = 1.000 C = 1 C

(4.861 mol H) /

(3.014 mol) = 1.613 H = 1.6 H

(3.056 mol O) / (3.014 mol) = 1.014 O = 1

O

The ratio of C:H:O has been found to be 1:1.6:1, thus the empirical

formula is: CH1.6O. Again, as a reminder, this is the simplest formula for the

compound, and not necessarily the molecular formula. Suppose we know that the

molecular weight of this compound is 176 g/mol. With this information, the

molecular formula may be determined. The formula weight of the empirical formula

is 29.6 g/mol. Divide the molecular weight by the empirical formula weight to

find a multiple:

(176.0 g/mol) / (29.6 g/mol) = 5.95 = 6

The

molecular formula is a multiple of 6 times the empirical formula:

C(1 x

6) H(1.6 x 6) O(1 x 6) which becomes C6H10O6

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