The mole, like grams or meters, is a unit of measurement that is part of the International System of Units (SI System). Essentially, the mole is used to define a specific number of something, similar to using the word “dozen” to define 12 things. In this case, a mole is { 6.022\times 10 }^{ 23 } of something. A mole can be used to describe the number of anything (pens, planets, pennies), but is almost always used to quantify the number of molecules or atoms of a compound or element.
Background
The concept of the mole was developed in 1865 by Josef Loschmidt in Austria. He used the kinetic molecular theory to estimate how many particles were contained in one cubic centimeter of gas. His research was expanded upon by French physicist Jean Baptiste in 1909. He used probability theory to estimate the number of particles in a sample of liquid or gas.
However, the best estimate was determined by Robert Millikan, the scientist who discovered the charge per electron. Through a series of experiments, he determined that a mole of electrons consisted of { 6.022\times 10 }^{ 23 } electrons. This number has come to be called Avogadro’s Number, named after Italian physicist Amadeo Avogadro.
Formal Definition
The official definition of the mole was established by the National Institute of Standards and Technology in 1971. It states that a mole is the amount of a substance that contains the same number of particles as 0.012 kilograms of carbon-12 (that is, carbon that has 6 protons, neutrons and electrons). This means that one mole of pure carbon-12 weighs exactly 12 grams.
Knowledge of a mole allows us to define the molar mass of any element or molecule. We can find the grams per mole of any substance just by looking at the periodic table. The molecular weight of any element or compound is equivalent to its molar mass. So, the molar mass of hydrogen is 1.01 grams per mole. We can use this information to do various calculations in a branch of chemistry called stoichiometry.
The Mole and Volume
The concept of the mole leads to an interesting property of gases. If we consider gases to act ideally, they conform to what is known as the gas law. The gas law is defined by the following equation:
PV=nRT
Interpreted this is:
(Pressure)(Volume) = (number of moles)(0.0821 Liters*atm/moles*K)(Temperature)
If we assume one mole of a gas at standard temperature and pressure (0°C and 1.0 atm), then the volume will always be 22.4 liters. This property applies for almost all gases and works particularly well for pure oxygen or hydrogen.
Practice
The concept of the mole can be used to convert between mass and number of particles. For example, if we are told that we have a 5-gram sample of pure hydrogen gas, how many particles are there? We can calculate this by setting up a simple dimensional analysis equation. Also, we will need the molar mass of hydrogen, which we can easily determine from a periodic table.
5\enspace grams\times \dfrac { 1\enspace mole }{ 1.01\enspace grams } \times \dfrac { { 6.022 }^{ 23 }\enspace molecules }{ 1\enspace mole } ={ 2.98\times 10 }^{ 24 }\enspace molecules
Notice that if we wanted the number of moles, we would only need the first conversion factor (1 mole per 1.01 grams).
As mentioned above, the mole is a vital concept for performing calculations in stoichiometry. These calculations try to relate the amounts of different reactants and products in chemical reactions. Often, the amount of reactants is given in grams and we are asked to solve the amount of a particular product. However, to do this, we need to convert to the number of molecules, because it is the actual molecules that are reacting. For example, water is formed from one molecule of oxygen and two molecules of hydrogen, not one gram of oxygen and two grams of hydrogen.
Let’s go over a simple stoichiometry calculation with this problem.
Determine the grams of water formed when we start with 15 grams of oxygen gas and unlimited hydrogen. The balanced reaction is { 2H }_{ 2 }+{ O }_{ 2 }\rightarrow { 2H }_{ 2 }O.
First, notice that the problem pointed out that there is unlimited hydrogen. This makes the calculation easier for us. If an amount of hydrogen were specified, we would have to calculate the amount of water formed from the specified amount of hydrogen, then another calculation for the specified amount of oxygen. Then we would have to use the smallest number as our final answer. This is because one of the two reactants would be a limiting reactant. It would be used up completely before the other reactant was and thus limit the amount of product that could be made. This is similar to wanting to bake cakes. If you only have 5 eggs and each cake requires 2 eggs, you can only make 2 complete cakes, no matter how much flour and milk you have.
So, to solve this problem, we need to determine how many moles of oxygen gas we have. In the equation, we see that for every one mole of oxygen gas we have, we can produce two moles of water. This is determined by looking at the coefficients in the equation. Once we know how many moles of oxygen there are, we can use this 1-to-2 relationship to determine moles of water. Finally, using the molar mass of water, we can convert the number of moles of water to grams of water and we will be complete!
Step 1: Grams of oxygen to moles of oxygen.
Molar mass of { O }_{ 2 }:
2\enspace atoms\enspace \times \dfrac { 16\enspace grams }{ mole\enspace of\enspace oxygen } =\dfrac { 36\enspace grams }{ mole }
15\enspace grams\times \dfrac { 1\enspace mole }{ 36\enspace grams } =0.44\enspace moles\enspace { O }_{ 2 }
Step 2: Moles of of oxygen to moles of water
0.44\enspace moles\enspace { O }_{ 2 }\times \dfrac { 2\enspace moles\enspace { H }_{ 2 }O }{ 1\enspace mole\enspace { O }_{ 2 } } =0.89\enspace moles\enspace { H }_{ 2 }O
Step 3: Moles of water to grams of water
Molar mass of water: 2\left( H \right) +\left( O \right) =2\left( 1.01 \right) +\left( 16 \right) =18.02\dfrac { grams }{ mole }
0.89\enspace moles\enspace { H }_{ 2 }O\times \dfrac { 18.02\enspace grams }{ 1\enspace mole } =16.02\enspace grams\enspace { H }_{ 2 }O
Thus, if we start with 15 grams of oxygen gas, we can form 16.02 grams of water.
Problems
Try the following stoichiometry problems. Be sure to pay careful attention to all the steps and look at the previous example to see how to solve them. Full solutions are given below.
- How many particles are in 24 grams of pure hydrogen gas? (Hint: be careful with how you write hydrogen gas. Is there only one atom per molecule?)
- How many grams of { HBO }_{ 2 } are formed with 18 grams of { B }_{ 2 }{ H }_{ 6 }?\enspace { B }_{ 2 }{ H }_{ 6 }?+{ 3O }_{ 2 }\rightarrow { 2HBO }_{ 2 }+{ 2H }_{ 2 }O
- How many liters of { CO }_{ 2 } gas are formed when 2.5 moles of { O }_{ 2 } react? { C }_{ 3 }{ H }_{ 5 }+{ 5O }_{ 2 }\rightarrow { 3CO }_{ 2 }+{ 4H }_{ 2 }O
- How many moles of water will react with 8 grams of { Li }_{ 2 }O?\enspace { Li }_{ 2 }O+{ H }_{ 2 }O\rightarrow 2LiOH
- Challenge Problem: How many grams of carbon dioxide are formed if 45 grams of { C }_{ 2 }{ H }_{ 5 }OH are burned in the presence of 60 grams of oxygen? { C }_{ 2 }{ H }_{ 5 }OH+{ 3O }_{ 2 }\rightarrow { 2CO }_{ 2 }+{ 3H }_{ 2 }O (Hint: remember the concept of limiting reactants mentioned above? That concept will need to be applied here.)
Solutions
1. How many particles are in 24 grams of pure hydrogen gas?
Hydrogen is one of the diatomic molecules, so pure hydrogen actually exists as { H }_{ 2 }. Therefore, the molar mass is 2.02\dfrac { grams }{ mole }.
24\enspace grams\times \dfrac { 1\enspace mole }{ 2.02\enspace grams } \times \dfrac { { 6.022\times 10 }^{ 23 }\enspace particles }{ 1\enspace mole } ={ 7.15\times 10 }^{ 24 }\enspace particles\enspace of\enspace hydrogen\enspace gas
2. How many grams of { HBO }_{ 2 } are formed with 18 grams of { B }_{ 2 }{ H }_{ 6 }? { B }_{ 2 }{ H }_{ 6 }+{ 3O }_{ 2 }\rightarrow { 2HBO }_{ 2 }+{ 2H }_{ 2 }O
This is a straightforward problem, similar to the example given above.
18\enspace grams\enspace { B }_{ 2 }{ H }_{ 6 }\times \dfrac { 1\enspace mole }{ 27.67\enspace grams } \times \dfrac { 2\enspace moles\enspace { HB }O_{ 2 } }{ 1\enspace mole\enspace { B }_{ 2 }{ H }_{ 6 } } \times \dfrac { 43.82\enspace grams }{ 1\enspace mole } =57.01\enspace grams\enspace { HBO }_{ 2 }
3. How many liters of { CO }_{ 2 } gas are formed when 2.5 moles of { O }_{ 2 } react? { C }_{ 3 }{ H }_{ 5 }+{ 5O }_{ 2 }\rightarrow { 3CO }_{ 2 }+{ 4H }_{ 2 }O
Remember, 1 mole of a standard gas has a volume of 22.4 liters.
2.5\enspace mol\enspace { O }_{ 2 }\times \dfrac { 3\enspace mole\enspace { CO }_{ 2 } }{ 5\enspace mole\enspace { O }_{ 2 } } \times \dfrac { 22.5L }{ 1\enspace mole } =33.6\enspace liters\enspace { CO }_{ 2 }
4. How many moles of water will react with 8 grams of { Li }_{ 2 }O? { Li }_{ 2 }O+{ H }_{ 2 }O\rightarrow 2LiOH.
Even though both molecules we are dealing with are on the same side of the equation, the approach is still exactly the same.
8\enspace grams\enspace { Li }_{ 2 }O\times \dfrac { 1\enspace mole }{ 29.88\enspace grams } \times \dfrac { 1\enspace mole\enspace { H }_{ 2 }O }{ 1\enspace mole\enspace { Li }_{ 2 } } =0.27\enspace mole\enspace { H }_{ 2 }O
5. Challenge Problem: How many grams of carbon dioxide are formed if 45 grams of { C }_{ 2 }{ H }_{ 5 }OH are burned in the presence of 60 grams of oxygen? { C }_{ 2 }{ H }_{ 5 }OH+{ 3O }_{ 2 }\rightarrow { 2CO }_{ 2 }+{ 3H }_{ 2 }O.
In this problem, we are given specific amounts of both reactants and asked how many grams of a product are formed. To solve, we use the same approach we have been practicing so far. However, we must also consider which reactant will get used up first. This is what is known as a limiting reactant problem because one of the reactants will be completely used up first, limiting the amount of product that can be formed. To solve this, we must solve for how much of the product each amount of reactant can form, then choose the smallest number as our final answer.
45\enspace grams\enspace { C }_{ 2 }{ H }_{ 5 }OH\times \dfrac { 1\enspace mole }{ 46.07\enspace grams } \times \dfrac { 2\enspace mole\enspace { CO }_{ 2 } }{ 1\enspace mole\enspace { C }_{ 2 }{ H }_{ 5 }OH } \times \dfrac { 44.01\enspace grams }{ 1\enspace mole } =86.03\enspace grams\enspace { CO }_{ 2 }
Or, if we use { O }_{ 2 }:
60\enspace grams\enspace { O }_{ 2 }\times \dfrac { 1\enspace mole }{ 16\enspace grams } \times \dfrac { 2\enspace mole\enspace { CO }_{ 2 } }{ 3\enspace mole\enspace O_{ 2 } } \times \dfrac { 44.01\enspace grams }{ 1\enspace mole } =110\enspace grams\enspace { CO }_{ 2 }
Comparing the two results, we see that the amount of { CO }_{ 2 } formed must be 86.03 grams and that { C }_{ 2 }{ H }_{ 5 }OH is our limiting reactant.
By now, the concept of the mole hopefully makes a lot more sense for you! Remember to practice carefully and deliberately and soon enough, applying the mole will become second nature.
If you have any tips and tricks for using the mole in stoichiometry calculations, be sure to leave a comment!
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