Learn to Calculate Molecular Weight

Here's How to Do It

Vincent  Summers
Chemical reactions often require the use of a specific amount of each reacting chemical (reactant) for best results. The correct numbers minimize losses and increase products. For instance, in making salt, sodium chloride, from sodium metal and chlorine gas,

2 Na + Cl2 → 2 NaCl

the correct weights to use are 46 grams of sodium and 71 grams of chlorine-or a fraction (the same fraction) of each. Thus 23 grams of sodium and 35.5 grams of chlorine is also perfectly acceptable, as is 4.6 grams of sodium and 7.1 grams of chlorine-or pounds, for that matter-4.6 pounds of sodium and 7.1 pounds of chlorine.

Breaking Down the Simple Example Above

Why use the numbers 46 and 71? It is because of the nature of the reaction and the use of atomic and molecular weights. Examine a periodic table of the elements (such as this one), and you will see that sodium has an atomic weight of 23.1,2 In the reaction cited above, two atoms of sodium are used, hence the number 46.

As for chlorine-two atoms of chlorine, due to their great reactivity, ordinarily exist as a diatomic ("two atoms") molecule, hence the number for chlorine we use is not an atomic weight, but a molecular weight-71.

Now the reaction proceeds readily (even dangerously so) to produce sodium chloride or table salt. Notice the number of atoms of each kind balances on both sides of the equation. Two sodium atoms and one diatomic chlorine molecule produce two sodium atoms and two chlorine atoms. How much does the sodium chloride weigh? If 46 grams of sodium and 71 grams of chlorine are combined, then 46 + 71 = 117 grams of sodium chloride are produced. The molecular weight of sodium chloride is 117.

So far we have discussed atomic weights and molecular weights, albeit of the simplest sort. Now many compounds have much more complex chemical formulas. How may their molecular weights be ascertained? A comprehensive listing is not possible here, but another example should cover by far the majority of cases.

Calculating the Molecular Weight of Sodium Ferrocyanide

Consider the example of sodium ferrocyanide, Na4Fe(CN)6. Notice the additional feature of parentheses in this example. This structure contains sodium, iron (Fe), carbon and nitrogen (C and N). There are four atoms of sodium-nothing new there. There is one atom of iron-nothing new there.3 The CN within the parentheses means the two atoms, carbon and nitrogen, are connected together. The six after the right parenthesis means that there are six of those CN units involved. Thus, the calculation for molecular weight becomes:

4 x 23 + 1 x 56 + 6 x (12 + 14) = 92 + 56 + 156 = 304.

The molecular weight of Na4Fe(CN)6 is 304. But wait! Sodium ferrocyanide ordinarily exists as the decahydrate (ten water molecules of crystallization), Na4Fe(CN)6• 10 H2O. Thus its full molecular weight actually becomes the 304 plus the weight of ten molecules of water (at 18 each) or 484.

1 Some rounding off is used to simplify the explanation. Sodium actually has an atomic weight of 22.990.
2 The weight in grams, in this instance 23, is based on Avogadro's number, which is constant and equals 6.022 x 1023. Thus 6.022 x 1023 sodium atoms weigh 23 grams.
3 When no number follows the symbol for an atom, it is understood to mean there is only one.

References and Resources:

Georgia Tech University - American Scientist - An Exact Value for Avogadro's Number

"Avogadro's number atomic and molecular weight," by Uri Lachish.

Published by Vincent Summers

My secular expertise includes 23 years of experience at the National Radio Astronomy Observatory, with a share in NASA's extended Voyager 2 effort. I formerly wrote for Demand Studios, Bukisa, Suite 101, Exa...  View profile

9 Comments

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  • Zona Zirconia9/12/2010

    Excellent article, high 5!

  • Tiffany Booth9/9/2010

    Great info- Thanks =)

  • Danielle Olivia Tefft9/2/2010

    You are a good teacher, Vincent. I don't remember molecular weight calculations being so well explained to me in high school!

  • Jeanne Baney9/1/2010

    I must say I'm glad I don't need to do this often.

  • David B. Bolick9/1/2010

    Flashback to the 70s

  • Vonda J. Sines9/1/2010

    Once again, you made it seem simple. Congrats.

  • Michele Starkey9/1/2010

    Vincent, you and I have just met and already I've decided that not only do I admire your intellect, I'm secretly wishing I understood half of what you've shared here with me! I will impress my friends with this shared knowledge though! cheers :)

  • Michael Segers9/1/2010

    Frankly, Vincent, the only weight I'm concerned about is my own.

  • Jenny Heart9/1/2010

    You are such a wizard!

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