In this article, you will learn how stoichiometry is used in reactions, more importantly how Avogadro's constant is exploited for different means. Stoichiometry is crucial to all parts of chemistry, as without specific calculations, different experiments would not be possible. Stoichiometry underpins the whole of chemistry.
- In this article, you will learn what is stoichiometry and its meaning
- Avogadro and his constant (the mole concept)
- Calculations involving the mole concept using gases, volumes, and masses
- Example calculations using mole formulae
Meaning of Stoichiometry in Reactions
What is Stoichiometry?
Stoichiometry: the relationship between the reacting quantities of compounds.
Here we will explore stoichiometry in the context of the mole concept and Avogadro's constant, the ideal gas equation, and mass spectra. This will allow us to make balanced equations, and determine empirical and molecular formulae.
The Mole and Avogadro's Constant
Where do we start thinking about "how much" of a substance is in a particular weighted amount? This was a question puzzling scientists for centuries, one being Amedeo Avogadro.
Avogadro and the Constant
Avogadro began to look at the relationship between the atomic number of atoms and the amount of atoms present. Atomic number defines the amount of protons present in the nucleus of the atom in question.
You can find out the atomic number of any compound by looking at its positioning in the periodic table. Elements in the periodic table are arranged by increasing atomic number.
Suppose we consider carbon with atomic number of 12. How many atoms are there in 12 grams of carbon-12? Avogadro found the number to be 6.02214179 x 1023.
Avogadro's constant (or Avogadro's number) is 6.02214179 x 1023
This constant is often simplified to 6.022 x 1023 for making calculations simpler. The constant is unitless as it refers to a specific amount, so you can say 6.02214179 x 1023 atoms if needed.
Please note when using the rounded form of Avogadro's number (6.022 x 1023) you need to be mindful how this will affect your future calculation and manipulation of errors.
The Mole
The mole is a concept extrapolated from Avogadro's constant. The number Avogadro discovered (6.02214179 x 1023) is often referred to as a mole.
1 mole = 6.02214179 x 1023 atoms
This is a simple analogy you use in daily life, such as a "dozen of ..." or a "pair of ...". Here you can use it exchangeably in chemical contexts. Such as "one mole of carbon-12" will refer to 6.02214179 x 1023 of carbon-12 atoms.
Stoichiometry Used in Reaction Equations
Using the Mole: Calculations and Conversions
Here we will explore how to use the defined concept of the mole for calculations and conversions involving mass and concentrations.
The Mole and Relative Atomic Mass
How do we know how many moles is in a particular weighted amount? We need to know precisely this to measure out the precise mass, and thus the exact amount of moles we need for balanced reactions to take place accordingly.
\[ \mbox{moles (n)} = \frac{ \mbox{mass in grams}}{ \mbox{atomic relative mass in grams per mole}} \]
or simply \( n= \frac{m}{A_r} \)
Can you see how the highlighted units in the equation above cancel out to give a number of moles?
In fact, the relative atomic mass of elements is related to the 1/12 of the grams of one mole of carbon-12 modules. Hence, the units of relative atomic mass are grams per mole.
An infographic to help you understand how the concept of the mole, mass, and molar mass come together. With this you can know the formula to any unknown value you are trying to calculate. Source: Klimczyk, Vaia Original
In fact, you can use the formula above to determine the amount of moles of any substance or compound, by replacing the relative atomic mass to the molar mass of the compound.
\[ \mbox{moles (n)} = \frac{ \mbox{mass in grams}}{ \mbox{molar mass in grams per mole}} \]
or simply \( n= \frac{m}{M_r} \)
With the formula above, you can find the amount of moles of any compound to use in reactions and experiments.
The Mole and Mass of Compounds
Alternatively, you can also manipulate the formula to determine the mass of a compound for a specific reaction. This is especially useful if you have a specific number of moles you need for an experiment and you need to know how much to weight out a specific element or substance.
\[ \mbox{mass in grams} = \mbox{moles (n)} \times \mbox{molar mass in grams per mole} \]
\[ \mbox{molar mass in grams per mole} = \mbox{mass in grams)} \times \mbox{moles (n)} \]
The Mole and Concentrations
You can use the concept of a mole to calculate the concentrations of solutions. Instead of using the formulae above with the molar mass or relative atomic mass, we can use a different formula that is based on volumes and concentrations.
To start off, how do we define concentration? It is based on the amount of substance in a particular volume of liquid. For a solution, it would be a precise amount of solute in an exact volume of solvent. To put moles into this concept, we can write out the equation:
\[ \mbox{ concentration in mol dm} ^{-3} = \frac{ \mbox{moles (n)}}{ \mbox{volume in dm}^3} \]
or simply \( C= \frac{n}{V} \)
The unit dm3 refers to 1 liter. Therefore 1 dm3 = 1 L = 1000 mL
Alternatively, you can use the above formula to determine the amount of moles in a particular solution from a known concentration.
\[ \mbox{moles (n)} = \mbox{concentration in mol dm} ^{-3} \times \mbox{volume in dm}^3 \]
A convenient triangle for you to understand how the three concepts of mole, concentration, and volume fit together. From this triangle you can make equations. Source: Klimczyk, Vaia original
This concept is well described with the infographic on the side. You can use it to know the formula of the value you are looking to calculate, be it moles or concentration or volume.
Ideal Gas Equation
So how does the mole concept fit into calculations that are not based on mass or concentrations, but rather gases? Here we will explore the ideal gas law and how it applies to the mole concept.
Think about gases, what does the volume they occupy depend on? That's the pressure and temperature. An ideal gas' volume will proportionally and linearly increase with increasing temperature or decreasing pressure.
Considering the behaviours of ideal gases based on temperature and pressure, a formula can be made (which we call the ideal gas law or general gas law):
\[ pV=nRT \]
In the formula above p = pressure in pascals (Pa), V = volume in m3, n = number of moles, R = is the gas constant (8.31 JK-1 mol-1), and T = temperature in kelvin (K).
The gas constant (R = 8.31 JK-1 mol-1) never changes for any calculation.
You can use the above equation to calculate some physical constant regarding a specific gaseous atom or compound, be it the volute from the number of moles known under standard conditions, or the other way round.
Stoichiometry in Reactions: Examples and Calculations in chemical reactions
Here we will explore examples of stoichiometry in reactions. Stoichiometry can be used in different types of reactions and below are some common problems and stoichiometry calculations in chemical reactions you will come across.
Examples of Calculations involving mass and weight
We will explore how to make calculations relating to converting mass and weighted amounts into moles, and vice versa.
How many moles are in 20 grams of NaOH?
The molar mass of NaOH is 39.997 g/mol, which you can round to 40 g/mol.
Now divide the weighted amount by the molar mass:
\( \frac{20g}{40gmol^{-1}} = 0.5mol \)
Thus there are 0.5 moles in 20 grams of NaOH.
Now consider if you need to weigh out a specific amount for a reaction. Take the example below as a reference.
You need 0.75 moles of magnesium (Mg) for a reaction. How many grams should you weigh out?
The relative atomic mass of Mg is 24.305 g/mol, which you can round off to 24.
You can use the first infographic to determine which formula you need to use.
Now multiply the number of moles by the relative atomic mass.
\( 0.75mol \times 24gmol^{-1} = 18g \)
Thus, you need to weigh out 18 grams of Mg for the reaction.
Great! Now you can use the same principle to calculate concentrations and volumes for solutions.
Examples of Calculations involving solutions, volumes, and molar concentrations.
Here we will explore how to calculate concentrations from a given amount of moles, or vice versa.
What is the concentration of a solution consisting of 0.5 moles of NaCl in 0.25 litres?
Use the second infographic to determine which formula you can use.
Divide the amount of moles by the volume.
\( \frac{0.5mol}{0.25L} = 2molL^{-1} \)
Thus the concentration is 2 mol dm-3
Examples of Calculations involving gases.
Explore how you can use the ideal gas law to calculate the volume of any gas if the conditions are provided or mentioned.
If a gas takes up 10 cubic meters of volume, how many moles does it have? Assume conditions of 101 kPa and 300K.
Use the equation \( pV=nRT \) , to rearrange it into \( n= \frac{pV}{RT} \)
Plug in the numbers: \( n= \frac{101 \times 10}{8.314 \times 300} \)
n = 0.405 moles
Thus you can manipulate the basic general gas formula to get any desired unknown, be it the volume, moles, or even the conditions the gas in under.
Stoichiometry in Reactions - Key takeaways
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