**Calculating the median Mass the One Molecule when calculating the mean mass that one molecule, do the following:Calculate the molar massive of the substanceDivide it by Avogadro"s NumberBy the way, the an approach to calculate the average mass the one atom of an aspect is specifically the same as because that calculating the typical mass the one molecule that a compound.Also, note that I save using words "average." because each aspect in a compound has several isotopes, a mole of a compound (say H2O) is written of molecules of slightly different weights. Because that example, hydrogen has two stable isotopes when oxygen has three. This leads to nine different feasible combinations that isotopes.Since over there is no practical way to separate out all the various weights, what we wind up measuring is the average weight that one molecule, which method that no one, single molecule has actually the load calculated. (That certain fact often gets tested.)Example #1:**What is the median mass of one molecule of H2O?1) calculation the molar mass.

You are watching: Mass of one molecule of water in grams

**The molar massive of water is 18.015 g/mol. This was calculated by multiply the atomic load of hydrogen (1.008) by 2 and including the an outcome to the weight for one oxygen (15.999).Please remember that you require the molar mass very first when make the efforts to discover the typical mass the one molecule.2) division the substance"s molar fixed by Avogadro"s Number.18.015 g/mol–––––––––––––––=2.992 x 10¯23 g6.022 x 1023 mol¯1**

3) keep in mind that the last answer has been rounded come four far-ranging figures (from 2.9915 - note use of round off with 5 rule). Also, keep in mind that the unit that mole cancels.Example #2:calculation the mean mass (in grams) that one molecule the CH3COOH (molar fixed = 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:recognize the median mass in grams for one atom of yellow (molar massive = 196.666 g/mol).

**196.666 g/mol–––––––––––––––=3.266 x 10¯23 g6.022 x 1023 mol¯1**

By the way, one older name for the molar massive of an element is gram-atomic weight. Um, er, it"s the one the altoalsimce.org learned means back once he was simply a sprout.Example #4:identify the massive (in grams) of one atom of gold-198.

**Note that this question asks about one specific isotope. For that, us must uncover the gram-atomic weight for the one isotope (often dubbed the isotopic mass), no the molar mass (also called the typical atomic weight) for gold (the value we provided in instance #3). Wikipedia has actually a table listing the masses for all the isotope of gold.The value for gold-198 is 197.968 g/mol. The problem collection up is:197.968 g/mol–––––––––––––––=3.287 x 10¯23 g6.022 x 1023 mol¯1**

Note the this is not an average. That is the actual mass of each gold-198 atom.Example #5:calculation the massive of a single atom the silver:

**Silver has two secure isotopes, therefore its molar mass is a weighted average of those 2 values. That way that what is calculated in the video clip is in reality the mean mass that a single atom of silver.You have to be mindful of this since you may have actually an instructor the emphasizes the average aspect of the calculation while rather may neglect it completely.Example #6:**identify the massive of 125 atom of palladium.

**Solution:**

**Done in dimensional evaluation 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 that U235F6.

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**Solution:**1) keep in mind that the mass of a specific isotope is required. For that, us go come the Internet and also look that up:

**235.0439 g/mol2) over there is one more unusual aspect to this problem:Only one steady isotope of fluorine is current in nature. In other words, 100% of all fluorine atoms in nature sweet the same amount:18.9984 g/molBy the way, the longest-lived rough isotope that fluorine has actually a half-life the is a bit less than 110 minutes. In a chemistry sense, the ain"t present in nature!3) calculate the molar fixed of U235F6:349.0343 g/mol4) calculation the actual massive of one molecule the U235F6:349.0343 g/mol / 6.022 x 1023 molecules/mol = 5.796 x 10¯22 5) Why would I speak actual mass quite than average mass? This is since fluorine has only one isotope in nature and also uranium (which has two isotope in nature) is restricted to just one certain isotope.Example #8:**deserve to you swim in a exchange rate billion (1.00 x 1018) molecule of water?

**Solution:**

**The best way to determine if you deserve to swim in this lot of water is to recognize the massive of water present. I"ll execute it dimensional evaluation style.1 mol18.015 g1.00 x 1018 moleculesx–––––––––––––––––––x––––––––=0.0000299 g6.022 x 1023 molecules1 mol**

No, you can"t swimming in the amount the water.I determined to watch into just how much water vapor is in an median breath. I think 500 mL because that a breath. I think 5% water vapor by volume. That means 25 mL that 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 gExample #9:How many water molecules would certainly be required to develop one drop (0.010 g)?

**Solution:**1) identify average fixed 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 autumn of water:0.010 g / 2.99153 x 10¯23 g/molecule = 3.3 x 1020 molecule (to two sig figs)3) Here"s one more approach, collection up in dimensional analysis style:1 mol6.022 x 1023 molecules0.010 gx–––––––x––––––––––––––––––––=3.3 x 1020 molecule (to 2 sig figs)18.015 g1 mol