**Calculating the Median Mass of One Molecule When calculating the average mass of one molecule, execute the following:Calculate the molar mass of the substanceDivide it by Avogadro"s NumberBy the way, the technique to calculate the average mass of one atom of an aspect is precisely the same as for calculating the average mass of one molecule of a compound.Also, note that I save using the word "average." Since each aspect in a compound has actually several isotopes, a mole of a compound (say H2O) is written of molecules of slightly various weights. For example, hydrogen has actually 2 steady isotopes while oxygen has 3. This leads to nine different feasible combicountries of isotopes.Because there is no helpful means 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 actually the weight calculated. (That particular truth often gets tested.)Example #1:**What is the average mass of one molecule of H2O?1) Calculate the molar mass.

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**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 also adding the result to the weight for one oxygen (15.999).Please remember that you need the molar mass initially once trying to discover 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 g6.022 x 1023 mol¯1**

3) Note that the last answer has been rounded to four considerable figures (from 2.9915 - note usage of rounding with 5 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 gAvogadro"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 g6.022 x 1023 mol¯1**

By the means, an older name for the molar mass of an facet is gram-atomic weight. Um, er, it"s the one the steustatiushistory.org learned way ago when he was just a sprout.Example #4:Determine the mass (in grams) of an atom of gold-198.

**Keep in mind that this question asks about one particular isotope. For that, we have to uncover the gram-atomic weight for that one isotope (regularly called the isotopic mass), not the molar mass (also called the average atomic weight) for gold (the worth we supplied in example #3). Wikipedia has actually 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 g6.022 x 1023 mol¯1**

Keep in mind that this is not an average. It is the actual mass of each gold-198 atom.Example #5:Calculate the mass of a solitary atom of silver:

**Silver has two secure 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 conscious of this bereason you may have actually an instructor that emphasizes the average element of the calculation while others may neglect it entirely.Example #6:**Determine the mass of 125 atoms of palladium.

**Solution:**

**Done in dimensional analysis style.106.42 g1 mol125 atoms–––––––x–––––––––––––––x––––––––=2.21 x 10¯20 g1 mol6.022 x 1023 atoms1**

Example #7:Determine the mass of one molecule of U235F6.

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**Solution:**1) Keep in mind that the mass of a specific isotope is required. For that, we go to the Net and also look it up:

**235.0439 g/mol2) There is one more inexplicable element to this problem:Only one stable isotope of fluorine is existing in nature. In other words, 100% of all fluorine atoms in nature weigh the exact same amount:18.9984 g/molBy the method, the longest-lived unsteady isotope of fluorine has a half-life that is a little much less than 110 minutes. In a chemical sense, it ain"t current in nature!3) Calculate the molar mass of U235F6:349.0343 g/mol4) 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 fairly than average mass? This is because fluorine has just one isotope in nature and uranium (which has 2 isotopes in nature) is limited to just one particular isotope.Example #8:**Can you swim in a billion billion (1.00 x 1018) molecules of water?

**Solution:**

**The ideal means to identify if you deserve to swim in this amount of water is to recognize the mass of water existing. I"ll do it dimensional analysis style.1 mol18.015 g1.00 x 1018 moleculesx–––––––––––––––––––x––––––––=0.0000299 g6.022 x 1023 molecules1 mol**

No, you can"t swim in that amount of water.I decided to look into just how much water vapor is in an average breath. Assume 500 mL for a breath. Assume 5% water vapor by volume. That suggests 25 mL of water vapor. Assume room temperature and push. 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 gExample #9:How many type of water molecules would certainly be forced to produce one drop (0.010 g)?

**Solution:**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/molecule2) 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 one more method, set up in dimensional analysis style:1 mol6.022 x 1023 molecules0.010 gx–––––––x––––––––––––––––––––=3.3 x 1020 molecules (to 2 sig figs)18.015 g1 mol