Empirical and Molecular Formulas Module Post-Assement

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Tamera Scott 1G
Posts: 65
Joined: Fri Sep 28, 2018 12:27 am

Empirical and Molecular Formulas Module Post-Assement

Postby Tamera Scott 1G » Mon Oct 01, 2018 8:42 pm

I was completing the post assessment for the Empirical and Molecular Formulas module, and I am having problems with this question: L-Dopa, a drug used for a treatment of Parkinson's disease, is 54.82% C, 5.62% H, 7.10% N, and 32.46% O, and has a molar mass of 197.19 g/mol. What is the molecular formula of L-dopa?

What step do I take first? Are those percentages the mass percentages, or do I have to find 54.82%, 5.62%, etc. from the 197.19 g/mol?

jguiman4H
Posts: 31
Joined: Fri Sep 28, 2018 12:29 am

Re: Empirical and Molecular Formulas Module Post-Assement

Postby jguiman4H » Mon Oct 01, 2018 9:08 pm

In this problem, we are given several mass percentages as well as the molar mass of the compound, L-Dopa, tasked to find its molecular formula.

First, we will assume that we have 100 g of the substance because our mass percentages total to 100%. So, we have 54.82g C, 5.62g H, 7.10g N, and 32.46g O.

Next, we should convert each of these elements into moles, as this will help us determine the empirical formula first. Thus:

Grams given / molar mass = given moles
54.82g C/12.01 g C = 4.565 mol C
5.62g H/1.008 g H = 5.575 mol H
7.10g N/ 14.01 g N = .5068 mol N
32.46g O/ 16.00 g O = 2.029 mol O

Our next step will be to divide by the lowest mole value achieved in the prior step. In this case, it would be N at .5068 mol. So:

4.565 mol C / .5068 mol = 9.007 C
5.575 mol H /.5068 mol = 11.00 H
.5068 mol N / .5068 mol = 1 N
2.029 mol O / .5068 mol = 4.003 O

Because we are determining the empirical formula first, we will round each of these values to the nearest whole number, as they are exactly or are extremely close to such. This also helps to check your work, as your answers should be very near whole numbers. Thus, we have:

9 C
11 H
1 N
4 O

The empirical formula is thus: C9H11NO4

We must then calculate the molar mass of the empirical formula, so (12.01 g C x 9) + (1.008 g H x 11) + (14.01 g N x 1) + (16.00 g O x 4) = 194.164 g/mol.

Our given molar mass is 197.19 g/mol. We will divide this by our empirical formula to determine if we need to multiply our empirical formula by a factor. 197.19g/194.164g = 1.016. This is close enough to 1, so we see that in this case, our empirical formula is the same as the molecular formula. If we divided the given molar mass by the empirical formula molar mass and determined the answer to be 2 or 3, we would then need to multiply the atoms in our empirical formula by that value. However, because our answer is essentially 1 when doing (given molar mass/ empirical formula molar mass), we know our empirical formula is the same as the molecular formula.

Thus, our molecular formula is also: C9H11NO4


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