Chemical Calculations by Sidney W. Benson: An Introduction to the Use of Mathematics in Chemistry
Chemical calculations are an essential skill for any chemistry student or professional. They allow us to quantify the properties and reactions of substances, and to predict the outcomes of experiments and processes. However, many students find chemical
calculations challenging and intimidating, especially when they involve complex formulas and equations.
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That's why Sidney W. Benson, a renowned chemist and educator, wrote Chemical Calculations: An Introduction to the Use of Mathematics in Chemistry. This book is designed to help students master the basic concepts and techniques of chemical calculations,
using simple and clear explanations, examples, and exercises. The book covers topics such as:
The mole concept and stoichiometry
Gas laws and kinetic molecular theory
Thermochemistry and calorimetry
Chemical equilibrium and Le Chatelier's principle
Acid-base reactions and pH
Oxidation-reduction reactions and electrochemistry
Nuclear chemistry and radioactivity
The book also includes appendices on mathematical review, logarithms, exponential functions, scientific notation, significant figures, units and conversions, and constants and formulas. The book is suitable for high school and college students who are
taking introductory or general chemistry courses, as well as for anyone who wants to refresh their knowledge of chemical calculations.
Chemical Calculations by Sidney W. Benson is available as a PDF file that can be downloaded from this link. The file is in Spanish, but it can be easily translated using online tools. The book is also available in print from various online retailers.
In this article, we will review some of the main concepts and techniques of chemical calculations, using examples from the book by Sidney W. Benson. We will also provide some tips and tricks to make chemical calculations easier and more accurate.
The Mole Concept and Stoichiometry
The mole is a unit of measurement that is used to count the number of atoms, molecules, ions, or other particles in a substance. One mole of any substance contains exactly 6.022 x 10^23 particles, which is known as Avogadro's number. The mole allows us
to relate the mass of a substance to the number of particles it contains, using the molar mass. The molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol).
For example, the molar mass of water (H2O) is 18.02 g/mol, which means that one mole of water has a mass of 18.02 grams and contains 6.022 x 10^23 molecules of water. The molar mass can be calculated by adding up the atomic masses of the elements that
make up the substance, using the periodic table. For example, the molar mass of sodium chloride (NaCl) is 58.44 g/mol, which is obtained by adding the atomic masses of sodium (22.99 g/mol) and chlorine (35.45 g/mol).
Stoichiometry is the study of the quantitative relationships between the reactants and products in a chemical reaction. It allows us to calculate how much of each substance is consumed or produced in a reaction, using the balanced chemical equation and
the mole concept. The balanced chemical equation shows the ratio of moles of each substance involved in the reaction, which is called the stoichiometric coefficient.
For example, the balanced chemical equation for the combustion of methane (CH4) with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O) is:
CH4 + 2 O2 -> CO2 + 2 H2O
This means that for every mole of methane that reacts, two moles of oxygen are consumed and one mole of carbon dioxide and two moles of water are produced. To calculate how much of each substance is involved in a given reaction, we can use the following
steps:
Identify the given and unknown quantities in terms of mass or moles.
Convert the given quantity to moles using the molar mass.
Use the stoichiometric coefficients to find the moles of the unknown quantity. Convert the moles of the unknown quantity to mass using the molar mass.
For example, if we want to find out how much oxygen is needed to completely burn 16 grams of methane, we can use these steps:
The given quantity is 16 grams of methane and the unknown quantity is oxygen. We can convert 16 grams of methane to moles using its molar mass: 16 g / 16.04 g/mol = 0.998 mol
We can use the stoichiometric coefficient to find out how many moles of oxygen are needed: 0.998 mol x 2 mol O2 / 1 mol CH4 = 1.996 mol O2
We can convert 1.996 moles of oxygen to grams using its molar mass: 1.996 mol x 32 g/mol = 63.87 g O2
Therefore, we need 63.87 grams of oxygen to completely burn 16 grams of methane.
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