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# Answer to Question #10304 in Organic Chemistry 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.
1
2012-05-31T09:23:25-0400
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

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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