MHS Chemistry
Moles & Mole Ratios

There is a lot of talk in every chemistry class about “moles.” In chemistry these are not furry little animals or big freckles, but a name for a counting number, similar to “dozen.” In chemistry, a mole is the number of protons or neutrons needed to make up one gram. This is a huge huge number.

1 mole = 602 000 000 000 000 000 000 000 = 6.02x1023

You can see that it’s easier to say “one mole” than either of the numbers above.  Also, it doesn't matter if it's atoms or donuts, a mole is the same number of things.

Since pretty much all the mass of any atom comes from it’s protons and neutrons, we can extend a concept from the fall. We learned that the atomic weight of an element was the average of the masses of it’s isotopes, and is given for each element of the periodic table. For hydrogen, this value is 1.00794 atomic mass units, for chlorine its 35.453 amu, etc. These are “atomic mass units per atom,” which is impractical (way too tiny!). But since a mole of amu’s is a gram, we can use these exact same numbers for atomic weights in “grams per mole.” Hydrogen weighs 1.00794 grams per mole of atoms; chlorine weighs 35.453 grams per mole of atoms, etc.

This is good because almost no one ever deals with just one or two atoms. We deal with enough to see, which means trillions and trillions of them – measurable numbers or fractions of moles.

The other really good thing about moles is that they allow chemical equations to become useful tools for chemical calculations. In the reaction

1N2 + 3H2 ® 2NH3

it appears that one molecule of nitrogen gas combines with three molecules of hydrogen gas to form two molecules of ammonia gas. This isn’t exactly true (the mechanism is more complicated than that), but the idea is true, and better if we generalize it. We should actually say that “one portion of nitrogen gas combines with three portions of hydrogen gas to form two portions of ammonia gas.” A portion could be a molecule, a dozen molecules, or (most likely in chemistry:) a mole of molecules.

So according to the reaction written above, nitrogen and hydrogen combine in a one to three ratio to form ammonia. And moles are the way we measure chemical portions. Since we have a balanced reaction, we can use the proportions built into it to know that two portions of nitrogen gas will need six portions of hydrogen gas, or that one portion of hydrogen gas will be enough to produce two-thirds of a portion of ammonia gas.

Generally, to find a portion (usually in moles) of a chemical from a known portion of another, we use a mole ratio. For instance, if we want to find out how much nitrogen would be needed to react completely with 9 moles of hydrogen according to the reaction above, we would set up a ratio:

 ratio from equation = ratio from question 1 N2 = x 3 H2 9 H2

Solving for x reveals that three moles of nitrogen would be needed to use up nine moles of hydrogen.

There’s a problem with moles, however: We don’t have any tool that can measure them directly, and the number is way way way too big to count practically. We are saved by the idea of atomic weight however. Let’s say we want to know what mass of nitrogen would be needed to make 6 moles of ammonia. We already balanced the reaction

1N2 + 3H2 ® 2NH3

so we can go directly to the ratio calculation to find out how many moles of nitrogen would be needed.
 1 N2 = x 2 NH3 6 NH3

x = 3 moles of nitrogen

But how much is 3 moles of nitrogen? Well, we know that nitrogen is N2, so we can find out the molecular weight. Add up all the atomic weights from the formula: 14.0 + 14.0 = 28.0 grams per mole. Since this includes the word “per” we know it’s actually a fraction, so we can set up another ratio:

 28.0 g N2 = x 1 mole 3 moles

x = 84.0 grams (from algebra!).

Facts to keep in mind:

1. balance the reaction
2. convert grams to moles
3. use mole ratios from the equation
4. convert back to grams

Molecular weight = atomic weights added up  (  carbon dioxide = CO2 = 12.0 + 2x16.0 = 44.0 g/mole  )

one mole = 6.02x1023 of anything (but usually atoms or molecules)

at 0 °C and 1 atmosphere of pressure (standard conditions), 1 mole of any gas = 22.4 L