2.2: Counting Atoms

Atoms are unimaginably small, so small that, as mentioned earlier, their existence to us is almost purely abstract. Yet chemists discuss them and the changes they undergo with a level of detail that implies deep familiarity. Where does that insight come from? While modern instrumentation has certainly given many valuable tools to examine molecular structures, a common starting point in gathering information about a given material is simply weighing it. It was through careful mass measurements of reactants and products that Antoine and Marie Lavoisier discovered the Conservation of Mass, and simply weighing matter remains a fundamental step – indeed, it is a ubiquitous staring point – with which chemists investigate matter. 

Figure 2-3. A heaping teaspoon of table sugar. Detail from a photo by by Mali Maeder (Public domain via Creative Commons)

Consider the sample of common table sugar shown at left (Figure 2-3). It corresponds to what a cooking recipe might refer to a “heaping teaspoon”. Note that table sugar is not glucose, the product of photosynthesis, but sucrose, C₁₂H₂₂O₁₁. How many sugar molecules are there in the sample? We can determine this with just two pieces of information: the total mass of the sample and the mass of each individual sugar molecule. Both numbers are easy to get. In the lab, an analytical balance (Figure 2-4) is used to measure the mass of samples; let's say that doing so in this case indicates that our heaping teaspoon has a mass of 7.382 g. 

As for the second piece of information we need, the mass of a sucrose molecule, you can calculate this easily enough and there will be many occasions where you will need to perform similar calculations, so we explain this process in some detail below. The mass of a single molecule of sucrose, or its molecular weight, is simply the sum of the atomic weights of each of its component atoms. Most Periodic Tables include the atomic weight of each element in units of amu, or atomic mass units. In this case, we find that the atomic weight of carbon, hydrogen and oxygen are 12.011, 1.008 and 15.999 amu, respectively. Because there are 12 carbon atoms, 22 hydrogen atoms and 11 oxygen atoms, a single molecule of sucrose has a mass of 342.297 amu (see box below for a more detailed explanation).

 

Figure 2-4. An example of a modern analytical balance. It is capable of measuring mass to a tenth of a milligram (0.0001 g). Photo by XericX is licensed under CC BY 2.0.

 

Calculating molecular weights.

To calculate the mass of a single molecule you need to have the molecular formula. For sucrose, it is C₁₂H₂₂O₁₁. You then need to find the masses of all the atoms present by multiplying the number of each atom type by its respective mass. In this case:

     12 carbon atoms, each with a mass of 12.011 amu:     (12 × 12.011 amu) = 144.132 amu

 +  22 hydrogen atoms, each with a mass of 1.008 amu:   (22 × 1.008 amu)   = 22.176 amu

 +  11 oxygen atoms, each with a mass of 15.999 amu:    (11 × 15.999 amu)   = 175.989 amu

                                                                                                                 = 342.279 amu

The process is the same for any molecule. The molecular weight is equal to the sum of the products of the atomic weight of each element and the number of atoms of that type in the molecule. Using sigma notation (expressing the summation of similar terms), the following defines the operation. 

\[ molecular\ weight = \sum (m_E)(n_E) \]

Here, the E subscript represents each element that appears in the molecular formula, m_E is the atomic mass of that element, and n_E is the subscript for element E in the molecular formula. Using this notation for sucrose, C₁₂H₂₂O₁₁, we would write:

\[ molecular\ weight\ (sucrose) = \sum (m_E)(n_E)  = 12(12.011\ amu)\ +\ 22(1.008\ amu)\ +\ 11\ (15.999\ amu) = 349.279 amu \nonumber \]

 

Getting back to the problem at hand - how many molecules of sucrose are in the heaping teaspoon - we now have two terms, the mass of the sucrose and the molecular weight of individual sucrose molecules.  You should note that these two quantities employ different mass units  – the total mass of the sugar is in units of grams, but that for a single molecule of sucrose is in amu. We must therefore know how to convert between these units to find how many sucrose molecules are in the sample. This is analogous to trying to add distances on the legs of a trip when some values are in miles and others are in kilometers. They can’t be added together directly so a conversion factor between the two units is needed. In the case of trip distances, 0.61 mile = 1 km, whereas to solve our sucrose problem we can use the following conversion:

\[1\ amu = 1.66054 × 10^{-24} g \nonumber \] 

We can now express the mass of a single sucrose molecule in grams as shown below.

\[ mass\ of\ one\ sucrose\ molecule = (342.30\ \cancel{amu}) \dfrac {(1.66054 × 10^{-24} g)}{(1\ \cancel{amu})} = 5.68 × 10^{-22} g \nonumber \]

Knowing the total mass of all of the sucrose molecules and that of a single sucrose molecule reduces the problem to one of simple ratios that can be solved algebraically.

\[ \dfrac {number\ of\ sucrose\ molecules}{total\ mass} = \dfrac{1\ molecule}{mass\ of\ one\ molecule} \nonumber \]

\[ number\ of\ sucrose\ molecules = \dfrac{1\ molecule}{mass\ of\ one\ molecule} × total\ mass \nonumber \]

\[ number\ of\ sucrose\ molecules = \dfrac{1\ molecule}{(1.66054 × 10^{-24} \cancel{g})} × 7.382\ \cancel{g} = 1.30 × 10^{22}\ molecules \nonumber \]

So that modest spoonful of table sugar has a staggering 13 sixtillion molecules. That’s 13,000,000,000,000,000,000,000 individual particles of sugar. But a question one could ask at this point concerns the value of the above information: why would knowing the number of sucrose molecules in a given sample be useful? To address this reasonable concern, let’s return to the balanced equation for photosynthesis:

\[ \ce{6 CO2 + 6 H2O    →     C6H12O6 + 6 O2} \nonumber \]

The quantitative relationships specified by the respective coefficients of the compounds are not mass ratios, but ratios of the numbers of molecules involved. In other words, the equation is not saying that six grams of carbon dioxide react with six grams of water to produce one gram of glucose and six grams of oxygen (that would have the Lavoisiers spinning in their graves!). Rather, it is specifying the ratios of the number of molecules: six molecules of carbon dioxide react with six molecules of water to produce one molecule of glucose and six molecules of oxygen gas. While we can’t easily count the number of molecules of anything, at least not directly, we can calculate that number after weighing the material. From that point we can then determine how much of a given product can be made from the available reactants, or how much of a reactant is needed to produce a desired amount of product. 

Because each individual molecule is so small, any measurable mass will have a very large number of them, as the above example illustrates, making for some cumbersome math. So for convenience, and convenience alone, chemists developed a unit of quantity that streamlines these sorts of calculations. That unit is the mole (absurdly abbreviated as “mol”) and it has a value of 6.02 × 10²³ “particles”. It is important to keep in mind that this is simply a number, an incomprehensibly large number to be sure, but still just a number. Specifically, it is the number of atoms of carbon-12 corresponding to a total mass of exactly 12 grams. It is often referred to as Avogadro’s number, or Avogadro’s constant, N_A, in honor of the Italian chemist whose work on gasses led to the first reliable atomic mass tabulation and the concept of the mole. 

The utility of the unit can perhaps be most effectively explained with the following graphic. Here, the exact number signified by “X” is the same on the left and the right. 

Simply put: the mole is the amount of any given pure material that makes its mass – in grams (this unit is critical) – have exactly the numerical value as the mass of a single particle of that substance in amu. That may seem convoluted, but the idea is simple in practice. For example, a carbon-12 atom weighs exactly 12 amu and one mole of such atoms, that is 6.02 ×10²³ carbon-12 atoms, weighs exactly 12 grams. Likewise, we calculated above that a single molecule of sucrose has a mass of 342.30 amu, so one mole of sucrose will therefore have a mass of 342.30 grams. Another way of expressing this is that sucrose has a molar mass of 342.30 grams. To reiterate, the mole is just a number; just as the word “dozen” corresponds to twelve objects (it doesn’t matter what the objects are), the word “mole” corresponds to 6.02 × 10²³ particles, regardless of what the particles are. One can have a dozen doughnuts, or eggs, or basset hounds. Similarly one can have a mole of iron atoms, sucrose molecules, or any other chemical species. And we will be able to find the mass of those collections of 6.02 × 10²³ atoms or molecules by simply using the atomic weights of the component atoms. 

Just how big is Avogadro’s number? Over the years chemistry teachers have searched in vain for ways to effectively convey its magnitude. A variation on one of these: envision a device that could count water molecules at a rate of one million per second; it would take more than 19 billion years to reach one mole, a mere 18.02 g, of water - quite a long time, especially in light of the fact that the universe is thought to be only 13.8 billion years old. This is a reflection of just how small individual atoms and molecules are, and the mole provides a convenient means to quantify them without constantly using very large numbers. Another example: there are roughly the same number of molecules of water in a single teaspoon of water (5 mL), as there are teaspoons of water in all of the world’s oceans.

One skill that will be necessary to use routinely in laboratory work will be converting masses of compounds or elements into the corresponding number of moles and vice versa. The illustrative examples below should be helpful in developing your ability to perform this type of calculation. The two examples below demonstrate that molar masses can be used to convert the mass of a material to moles and vice versa, depending on the context. It is critical that you get comfortable manipulating molar masses in this way, especially for experimental work. 

 

Example 

Problem 2.2: A teaspoon is a unit of volume often used in cooking. It is equal to 5 mL. A teaspoon of salt (sodium chloride, NaCl) has a mass of 5.9 grams. How many moles does this correspond to?

Solution

Use the atomic weights of the component elements to find the molar mass of sodium chloride and use it as a conversion factor to find the corresponding number of moles.

 \[ (5.9\ \cancel{g\  \ce{NaCl}})(\dfrac{1\ mol\ \ce{NaCl}}{58.44\ \cancel{g\ \ce{NaCl}}}) = 0.10 mol\  \ce{NaCl} \nonumber \]

Note that while molar masses usually are expressed in units of grams per mole, the inverse form, moles per gram, is used in the above calculation so that the units of mass arithmetically cancel, yielding a product with units of moles. A useful approach to any conversion problem such as this, and we will see many of them, will be to make sure the units cancel appropriately to give the necessary dimensions in the product - if the units do not cancel properly you probably set up the equation improperly. Pay attention to units and you will avoid many errors in your work.

 

Example 

Problem 2.3: Calculate the mass of 0.155 mol of sucrose, C₁₂H₂₂O₁₁.

Solution

Use the molar mass of sucrose (calculated above) as a conversion factor as shown below.

 \[ mass_{sucrose}\ = (0.155\ \cancel{mol\ \ce{C12H22O11}})(\dfrac{342.3\ g\ \ce{C12H22O11}}{1\ \cancel{mol\ \ce{C12H22O11}}} ) = 53.1\ g\  \ce{C12H22O11} \nonumber \]

 

 

Example 

Problem 2.4: Calculate the number of iron atoms in 256.2 g of Fe₂O₃.

Solution. This problem has several distinct steps and it can be helpful to map out the logic involved before jumping into calculations: multi-step problems are sort of like road-trips - to prevent going down dead-ends or taking unnecessarily circuitous paths, it is best to map your path in its entirety before starting out. In this case, we can map it out as illustrated below.

In the above scheme, the units expressed in each box can be converted to those of the next box using the conversion factors shown above each arrow. You first convert the the mass of Fe₂O₃ to moles using the molar mass, then using the molecular formula you find the moles of iron contained within the sample. Finally, use Avogadro’s number to find the quantity of individual atoms. This can be written in a single string of factors as shown below. Note that the only unit left uncancelled is “Fe atoms”.

\[ number_{Fe\ atoms}\ = (256.2 \cancel{g \ce{Fe2O3}})(\dfrac{1 \cancel{mol\ \ce{Fe2O3}}}{159.7 \cancel{g\ \ce{Fe2O3}}})(\dfrac{2 \cancel{mol\ \ce{Fe}}}{1\ \cancel{mol\ \ce{Fe2O3}}})(\dfrac{6.02×10^{23}\ Fe\ atoms}{1 \cancel{mol\ \ce{Fe}}}) = 1.93×10^{24}\  \ce{Fe}\ atoms \nonumber \]

While each individual step can be done in isolation, it is helpful to combine them as shown above to explicitly show the logical connections between the steps and to make sure the units properly cancel. 

 

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Additional Problems.

 

Problem 2.5: Calculate the mass of the following:

a) 0.12 mol butane 

b) 56.1 mol butyric acid (C₄H₇O₂H)

c) 4.59 mol zinc oxide (ZnO) 

d) 1050 mol 2,2,4-trimethyloctane

 

Problem 2.6 How many moles are in the following samples of pure materials?

a) 1.000 kg gold (1 kg = 1000 g)

b) 5.32 g aluminum oxide (Al₂O₃)

c) 75 g triacontane (C₃₀H₆₂)

d) 102 g propan-2-ol

 

Problem 2.7: How many carbon atoms are in 500 mg of vitamin C? Vitamin C has the molecular formula C₆H₈O₆. (1 g = 1000 mg)