Moving right along, we are now ready to apply what we have learned to molecules. This is really easy if you have learned the preceding -- so if it seems difficult, back up and study more the background.
A molecule is a collection of atoms in a fixed ratio. For example, a molecule of water contains 1 oxygen atom and 2 hydrogen atoms. Now what about 14 molecules of water? Well 14 molecules of water would contain 14 oxygen atoms and 28 hydrogen atoms. Similarly 1 million molecules of water would contain 1 million oxygen atoms and 2 million hydrogen atoms. See where we are going with this? One mole of water will contain one mole of oxygen atoms and 2 moles of hydrogen atoms. So what would be the mass of one mole of water?
1 mole of oxygen atoms = 16.00 gms
2 moles of hydrogen atoms = 2.016 gms
So one mole of water = 16.00g + 2.016 g = 18.02 g.
This is called the molar mass of the molecule (or the molecular weight). Then we can get the following table (be sure that you understand how):
|compound||MW(amu)||Molar Mass(g/mol)||Mass of 1 molecule(g)|
|NH3||17.04||17.04||2.830 x 10-23|
|H2O||18.02||18.02||2.992 x 10-23|
|CH2Cl2||84.93||84.93||1.410 x 10-23|
Let's do some examples:
What is the molar mass of caffeine (C8H10N4O2)?
8 moles C * (12.011 g/(mol C)) = 96.088 g
10 moles H * (1.0079 g/(mol H)) = 10.079 g
4 moles N * (14.0067 g/(mol N)) = 56.0268 g
2 mol O * (15.9994 g/(mol O)) = 31.9988 g
And the total is 194.193 g which is the molar mass of caffeine.
A can of artificially sweetened drink may contain 70 mg of aspartame(Nutra Sweet). How many moles is this? The formula is C14H18N2O5
First let's calculate the formula weight so that we can then calculate how many moles .070 g of the substance is.
C: 14 moles * (12.011 g/mol) = 168.154 g
H: 18 moles * (1.0079 g/mol) = 18.1422 g
N: 2 moles * (14.0067 g/mol) = 28.0134 g
O: 5 moles * (15.9994 g/mol) = 79.997 g
This gives the molar mass to be 294.307 g/mol. So now we can calculate the number of moles:
.070 g/(294.307g/mol) = 2.4 x 10-4 mol.
How many molecules is this?
2.4 x 10-4 mole * (6.022 x 1023 molecules/mol) = 1.4 x 1020molecules.
Here we make use of the Law of Constant Composition: any sample of a pure compound always consists of the same elements combined in the same proportions by mass.
For example, every molecule of ammonia is NH3 and 17.03 g of ammonia always has 14.01 g of N and 3.02 g of H. Therefore we can now calculate the percent composition of each element in ammonia:
% by wt. of N in ammonia: 14.01 g n/17.03 g NH3 * 100 = 82.27%
% by wt. of H in ammonia: 100-82.27 = 17.73%
Let's work another example with our favorite molecule, caffeine. From the previous problem we know that
8 mol C = 96.088 g
10 mol H = 10.079 g
4 mol N = 56.0268 g
2 mol O = 31.9988 g
for a molar mass of 194.1926 g/mol.
Then the % by weight of each element would be:
C : 96.088/194.1926 * 100 = 49.48
H : 10.079/194.1926 * 100 = 5.19
N : 56.0268/194.1926 * 100 = 28.85
O : 31.9988/194.1926 * 100 = 16.48
and the total is, of course, 100.00
So what, you say. Well, I will show you. Let's use reverse logic.
You analyze an unknown compound and find experimentally that it consists of N at 87.42 percent and H at 12.58 percent. What is the unknown compound?
Well, what we want to find is the number of moles of N and H in the formula for the compound. so we are looking for x and y in NxHy.
Let's take 100 grams for a basis utilizing the % changed into grams. If we have 100 grams of the compound and 87.42% of the compound is N then the compound will have 87.42 g of N. Similarly there will be 12.58 g of H.
87.42g N * (1 mol N/((14.01 g N)) = 6.240 moles N
12.58 g H * (1 mol H/((1.008 g H)) = 12.48 moles H
Therefore the mole ratio of H to N is
12.48moles H/(6.24 moles N) = 2.00 moles H/1.00 moles N
So in 1 molecule of the unknown there are twice as many moles of H to N. Therefore the formula could be NH2 or N2H4 or N3H6. So which is it? I don't have enough information to tell at this point. However I can define a new quantity for you : the empirical formula is the formula with the simplest ratio of numbers. So the empirical formula for this compound is NH2. The correct formula is called the molecular formula. The extra information that I need is the experimental determination of the molar mass and this is given as 32.0 g/mol. Well the empirical formula has a molar mass of 16.0 g so I know that the molecular formula is N2H4. Q.E.D.
Here is an Excellent Tutoring Site for these topics.
Here is another site where you can practice.
|Now take a practice quiz to help you understand if you understand the basic concepts.|
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|The test will only submit when you have answers all of the questions correctly.|
|If you are not taking this course for credit please do not answer all the questions correctly for I don't want to be flooded with email answers to the tests.|
Web Author: Dr. Leon L. Combs
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