**Calculating the Median Mass of One Molecule When calculating the average mass of one molecule, do the following:Calculate the molar mass of the substanceDivide it by Avogadro"s NumberBy the method, the approach 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 keep using the word "average." Due to the fact that each element in a compound has a number of isotopes, a mole of a compound (say H2O) is written of molecules of slightly different weights. For instance, hydrogen has actually two stable isotopes while oxygen has actually 3. This leads to nine different feasible combinations of isotopes.Because tbelow is no useful method to separate out all the various weights, what we wind up measuring is the average weight of one molecule, which suggests 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.

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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 including the outcome to the weight for one oxygen (15.999).Please remember that you need the molar mass first once 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 g6.022 x 1023 mol¯1**

3) Note that the final answer has actually been rounded to four substantial numbers (from 2.9915 - note usage of rounding through 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 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 way, an older name for the molar mass of an aspect is gram-atomic weight. Um, er, it"s the one the nlinux.org learned way earlier as soon as 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 need to uncover the gram-atomic weight for that one isotope (regularly dubbed the isotopic mass), not the molar mass (likewise referred to as the average atomic weight) for gold (the value we used in example #3). Wikipedia has actually a table listing the masses for all the isotopes of gold.The worth for gold-198 is 197.968 g/mol. The trouble set up is:197.968 g/mol–––––––––––––––=3.287 x 10¯23 g6.022 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 solitary atom of silver:

**Silver has 2 stable isotopes, so its molar mass is a weighted average of those two values. That implies 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 disregard it completely.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) Note that the mass of a details isotope is forced. For that, we go to the Web and look it up:

**235.0439 g/mol2) There is an additional unexplained element to this problem:Only one steady isotope of fluorine is present in nature. In other words, 100% of all fluorine atoms in nature weigh the very same amount:18.9984 g/molBy the way, the longest-lived unsecure isotope of fluorine has a half-life that is a little bit much less than 110 minutes. In a chemical sense, it ain"t existing 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 only one isotope in nature and uranium (which has actually two isotopes in nature) is limited to just one specific isotope.Example #8:**Can you swim in a billion billion (1.00 x 1018) molecules of water?

**Solution:**

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

No, you can not swim in that amount of water.I decided to look right into just how much water vapor is in an average breath. Assume 500 mL for a breath. Assume 5% water vapor by volume. That implies 25 mL of water vapor. Assume room temperature and press. 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 be required to develop 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 an additional approach, put up in dimensional analysis style:1 mol6.022 x 1023 molecules0.010 gx–––––––x––––––––––––––––––––=3.3 x 1020 molecules (to two sig figs)18.015 g1 mol