0.1 Atomic masses and molecular formulas  (Page 10/10)

This is easiest to see with an example. Some of the most common chemical reactions are those in which compounds of hydrogen and carbon, called hydrocarbons, are burned in oxygen gas to form carbon dioxide and water. The simplest hydrocarbon is methane, and using the methods of this study, we can find that methane has the molecular formula, CH ₄ . The chemical equation which represents the burning of methane is:

1 CH ₄ molecule + 2 O ₂ molecules → 1 CO ₂ molecule + 2 H ₂ O molecules

It is important to note that the number of atoms of each type is conserved during the chemical reaction. The reactants and products both contain 1 carbon atom, 4 hydrogen atoms, and 2 oxygen atoms. This is called a “balanced” chemical equation, and it expresses the postulate of the Atomic Molecular Theory that the numbers of atoms of each element does not change during a chemical reaction.

In chemical algebra, we can ask and answer questions such as, “If we burn 1.0 kg of methane, what is the mass of carbon dioxide which is produced?” Such a question would clearly be of importance in understanding the production of greenhouse gases like CO ₂ . The chemical equation above expresses a relationship between the number of molecules of methane which are burned and the number of molecules of CO ₂ produced. From the equation, each molecule of CH ₄ produces one molecule of CO ₂ . Therefore, if we knew how many molecules of CH ₄ we have in a sample, we know how many molecules of CO ₂ we will produce.

The chemical equation works for any number of molecules. If we burn N molecules of CH ₄ , we produce N molecules of CO ₂ . This will work no matter what N is. Therefore, we can say that 1 mole of CH ₄ molecules will produce 1 mole of CO ₂ molecules. The chemical equation works just as well for moles as it does for molecules, since 1 mole is just a fixed number of molecules. And we know how to calculate the number of moles from a measurement of the mass of the sample.

Recall that we are interested in what happens when we burn 1.00 kg = 1000 g of methane. We just need to know the mass of a mole of methane. Since one molecule of methane has relative mass 16.0, then one mole of methane has mass 16.0 g/mol. Then the number of moles in 1000 g of methane can be calculated by dividing by the mass of 1.0 mole of methane:

$n_{{\text{CH}}_4}\mathrm{=}\frac{\text{1000}g}{\text{16}\text{.}\mathrm{0g}\mathrm{/}\text{mol}}\mathrm{=}\text{62}\text{.}5\text{moles}$

This means we have counted the number of particles of CH ₄ in our sample. And we know that the number of particles of CO ₂ produced must be the same as this, because the chemical equation shows us the 1:1 ratio of CH ₄ to CO ₂ . So, 62.5 moles of CO ₂ are produced by this reaction and this 1.0 kg sample.

We are usually more interested in the mass of the product, and we can calculate this, too. The mass of one mole of CO ₂ is found from the mass of one mole of C and two moles of O, and is therefore 44.0 g. This is the mass for one mole. The mass for 62.5 moles will be

$m_{{\text{CO}}_2}\mathrm{=}{n}_{{\text{CO}}_2}{M}_{{\text{CO}}_2}\left(\text{62}\text{.}5\text{moles}\right)\left(\text{44}\text{.}\mathrm{0g}\mathrm{/}\text{mol}\right)\mathrm{=}\text{2750}g\mathrm{=}2\text{.}\text{75}\text{kg}$

Therefore, for every 1 kg of methane burned, we produce 2.75 kg of CO ₂ .

The important conclusion from this example of chemical algebra is that it is possible to calculate masses of products from masses of reactants. We do so by using a balanced chemical equation and by understanding that the equation gives us the ratio of moles of reacting materials just as it gives us the ratio of molecules of reacting materials. This is because numbers of moles and numbers of molecules are simply different ways of counting the number of particles.

Chemical algebra is usually referred to as “stoichiometry,” a somewhat intimidating term that makes the calculations seem harder and more abstract than they are. We really only need to remember two things. First, from the Atomic Molecular Theory, a chemical reaction can be represented by a balanced chemical equation which conserves the numbers of atoms of each element. Second, the balanced equation provides the ratio of the number of product molecules to the number of reactant molecules, either in numbers of molecules or numbers of moles. Thus, we can solve problems efficiently by calculating the number of moles.

A final interesting note about Avogadro’s number is helpful in understanding what 1 mole is. A question often asked is, where did the number 6.022×10 ²³ come from? If we wanted to pick a very large number for the number of particles in a mole, why didn’t we pick something easier to remember, like 6×10 ²³ , or even 1×10 ²³ ? The value of Avogadro’s number comes from the fact that we chose 1 mole to be the number of carbon atoms in 12.01g of carbon. Since 1 carbon atom has mass 12.01 amu, then the mass of N _A carbon atoms is N _A ×12.01 amu. But 1 mole of carbon atoms has mass 12.01 g, so 12.01 g must equal N _A ×12.01 amu:

12.01 g = N _A ×12.01 amu

This means that

1 g = N _A amu

This shows that Avogadro’s number is just the conversion factor for mass between grams and amu. We didn’t randomly pick Avogadro’s number. Rather, we picked the unit of mass amu, and it turns out that there are Avogadro’s number of amu in one gram.

Review and discussion questions

1. State the Law of Combining Volumes and provide an example of your own construction which demonstrates this law.
2. Explain how the Law of Combining Volumes, combined with the Atomic Molecular Theory, leads directly to Avogadro's hypothesis that equal volumes of gas at equal temperatures and pressure contain equal numbers of particles.
3. Use Avogadro's hypothesis to demonstrate that oxygen gas molecules cannot be monatomic.
4. The density of water vapor at room temperature and atmospheric pressure is 0.737 g/L. Compound A is 80.0% carbon by mass, and 20.0% hydrogen. Compound B is 83.3% carbon by mass and 16.7% hydrogen. The density of gaseous Compound A is 1.227 g/L, and the density of Compound B is 2.948 g/L. Show how these data can be used to determine the molar masses of Compounds A and B, assuming that water has molecular mass 18.
5. From the results in Problem 4, determine the mass of carbon in a molecule of Compound A and in a molecule of Compound B. Explain how these results indicate that a carbon atom has atomic mass 12.
6. Explain the utility of calculating the number of moles in a sample of a substance.
7. Explain how we can conclude that 28g of nitrogen gas (N ₂ ) contains exactly as many molecules as 32g of oxygen gas (O ₂ ), even though we cannot possibly count this number.

By John S. Hutchinson, Rice University, 2011