When calculating the average mass of one molecule, do the following:
By the way, the technique to calculate the average mass of one atom of an element is exactly the same as for calculating the average mass of one molecule of a compound.
Also, note that I keep using the word 'average.' Since each element in a compound has several isotopes, a mole of a compound (say H2O) is composed of molecules of slightly different weights. For example, hydrogen has two stable isotopes while oxygen has three. This leads to nine different possible combinations of isotopes.
Since there is no practical way to separate out all the different weights, what we wind up measuring is the average weight of one molecule, which means that no one, single molecule has the weight calculated. (That particular fact often gets tested.)
Example #1: What is the average mass of one molecule of H2O?
1) Calculate the molar mass.
The molar mass of water is 18.015 g/mol. This was calculated by multiplying the atomic weight of hydrogen (1.008) by two and adding the result to the weight for one oxygen (15.999).
Please remember that you need the molar mass first when trying to find the average mass of one molecule.
2) Divide the substance's molar mass by Avogadro's Number.
18.015 g/mol ––––––––––––––– = 2.992 x 10¯23 g 6.022 x 1023 mol¯1
3) Note that the final answer has been rounded to four significant figures (from 2.9915 - note use of rounding with five rule). Also, note that the unit of mole cancels.
Example #2: Calculate the average mass (in grams) of one molecule of CH3COOH (molar mass = 60.0516 g/mol)
molar mass ---> 60.0516 g/mol ––––––––––––––– = 9.972 x 10¯23 g Avogadro's Number ---> 6.022 x 1023 mol¯1
Example #3: Determine the average mass in grams for one atom of gold (molar mass = 196.666 g/mol).
196.666 g/mol ––––––––––––––– = 3.266 x 10¯23 g 6.022 x 1023 mol¯1
By the way, an older name for the molar mass of an element is gram-atomic weight. Um, er, it's the one the ChemTeam learned way back when he was just a sprout.
Example #4: Determine the mass (in grams) of an atom of gold-198.
Note that this question asks about one specific isotope. For that, we must find the gram-atomic weight for that one isotope (often called the isotopic mass), not the molar mass (also called the average atomic weight) for gold (the value we used in example #3). Wikipedia has a table listing the masses for all the isotopes of gold.
The value for gold-198 is 197.968 g/mol. The problem set up is:
197.968 g/mol ––––––––––––––– = 3.287 x 10¯23 g 6.022 x 1023 mol¯1
Note that this is not an average. It is the actual mass of each gold-198 atom.
Example #5: Calculate the mass of a single atom of silver:
Silver has two stable isotopes, so its molar mass is a weighted average of those two values. That means that what is calculated in the video is actually the average mass of a single atom of silver.
You should be aware of this because you may have an instructor that emphasizes the average aspect of the calculation while others may ignore it completely.
Example #6: Determine the mass of 125 atoms of palladium.
Done in dimensional analysis style.
106.42 g 1 mol 125 atoms ––––––– x ––––––––––––––– x –––––––– = 2.21 x 10¯20 g 1 mol 6.022 x 1023 atoms 1
Example #7: Determine the mass of one molecule of U235F6 .
1) Note that the mass of a specific isotope is required. For that, we go to the Internet and look it up:
2) There is one more unusual aspect to this problem:
Only one stable isotope of fluorine is present in nature. In other words, 100% of all fluorine atoms in nature weigh the same amount:18.9984 g/mol
By the way, the longest-lived unstable isotope of fluorine has a half-life that is a bit less than 110 minutes. In a chemical sense, it ain't present in nature!
3) Calculate the molar mass of U235F6:
4) Calculate the actual mass of one molecule of U235F6:
349.0343 g/mol / 6.022 x 1023 molecules/mol = 5.796 x 10¯22
5) Why would I say actual mass rather than average mass? This is because fluorine has only one isotope in nature and uranium (which has two isotopes in nature) is restricted to just one specific isotope.
Example #8: Can you swim in a billion billion (1.00 x 1018) molecules of water?
The best way to determine if you can swim in this amount of water is to determine the mass of water present. I'll do it dimensional analysis style.
1 mol 18.015 g 1.00 x 1018 molecules x ––––––––––––––––––– x –––––––– = 0.0000299 g 6.022 x 1023 molecules 1 mol
No, you can't swim in that amount of water.
I decided to look into how much water vapor is in an average breath. Assume 500 mL for a breath. Assume 5% water vapor by volume. That means 25 mL of water vapor. Assume room temperature and pressure. Use PV = nRT:
(1.00 atm) (0.025 L) = (n) (0.08206 L atm / mol K) (293 K)
n = 0.00103978 mol
(0.00103978 mol) (18.015 g/mol) = 0.0187 g
Example #9: How many water molecules would be required to produce one drop (0.010 g)?
1) Determine average mass of one molecule of water:
18.015 g/mol / 6.022 x 1023 molecules/mol = 2.99153 x 10¯23 g/molecule
2) Determine number of molecules in one drop of water:
0.010 g / 2.99153 x 10¯23 g/molecule = 3.3 x 1020 molecules (to two sig figs)
3) Here's another approach, set up in dimensional analysis style:
1 mol 6.022 x 1023 molecules 0.010 g x ––––––– x –––––––––––––––––––– = 3.3 x 1020 molecules (to two sig figs) 18.015 g 1 mol
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