Step-by-Step Math Tools in Wolfram|Alpha Help Your Chemistry Course Prep

The following is an excerpt of a post originally published on the Wolfram Blog. The original post can be viewed in full here.

Math is one of the main things that deters students from wanting to learn more about chemistry. Being a chemical engineering student, I understand this, especially for students who just have to get chemistry out of the way as a general education requirement. Essentially, step-by-step solutions are like your own on-demand math tutor: in addition to calculating the answer, Wolfram|Alpha shows you how it got there. Here are [three] important math skills that you will definitely use on a regular basis in your chemistry class and how they relate to different chemistry concepts.

1. Basic Algebra

A lot of chemistry skills involve the application of basic algebra. However, most of the algebra is hidden behind fancy chemistry words like “reactants,” “products,” “moles,” “molarity” and so on. If you read these problems closely enough, you will see that most of the work is really just an algebra problem.

First off, let’s look at an expression to brush up on algebra skills that may appear in equilibrium constant, K𝒸, problems. Simplify the following expression: (x + 1) — 2(x² + 10 * x). This is done by distributing the negative coefficient to all parts in the parentheses, combining like terms and ordering terms from the highest exponent to the lowest exponent. In Wolfram|Alpha, if you type “simplify (x + 1) — 2 (x^2 + 10 * x)”, you can find the step-by-step solution for this:

Fractions are another thing that you will see a lot in chemistry, particularly in stoichiometry problems. Let’s take a look at a basic example of multiplying and dividing fractions. What is the solution for 4/3 * 2/3? Wolfram|Alpha provides a comprehensive step-by-step solution as a refresher on how to multiply fractions:

Next, let’s look at a chemistry problem that uses fractions. This may be something that doesn’t look familiar to you now, but it soon will! Consider the chemical reaction AlCl3 + NaOH -> Al(OH)3 + NaCl. Assuming you have two moles of AlCl3, how many moles of NaCl can you make by running the reaction to completion? Well, you can type “2 mole AlCl3 + NaOH -> Al(OH)3 + NaCl” into Wolfram|Alpha:

Looking at the stoichiometry step-by-step results, we can see that the answer would be six moles of NaCl. Looking at step 4, your equation dwindles down to 2 moles AlCl3/1 * 3 moles NaCl/1 mole AlCl3 = 6 moles of NaCl. Not as bad as it looks at first, right?

2. Unit Conversions

Every chemistry student’s worst nightmare: unit conversions, such as converting from centimeters to meters, milliliters to liters and so on. But if you think about it, they’re just fractions, which we covered in the previous section. It is one of the most common things for students to get wrong on their first chemistry exam, usually by multiplying instead of dividing or vice versa. On top of that, most units in chemistry are in metric, also known as SI, units, which may be unfamiliar to those living in the US.

Let’s start with a more familiar unit: gallons. Let’s convert one gallon of water to liters. By typing “convert 1 gallon of water to liters”, we see that it is equal to 3.785 liters. Using this knowledge, if you had two gallons, you could multiply 3.785 by 2 to get 7.571 liters:

Let’s look at a common chemistry conversion: converting mass to moles. This is one of many dimensional analysis problems you will encounter in chemistry. You’ve just measured out 10 g of sucrose on the laboratory balance but need to know how many moles of sucrose you have. You can do this conversion by entering “10 grams of sucrose to moles” into Wolfram|Alpha:

3. Finding the Slope and Intercept of a Line

Chemists are especially fond of linear relationship, with a very popular one being the basis of Beer–Lambert law experiments. Using data collected from a spectrophotometer for samples of known concentration, a line of best fit can be formed, which is also called a linear regression. The form that this equation takes is y = mx + b, where m is the slope and b is the y intercept.

By taking a simple equation like y = 3x + 1, identifying what the slope and intercept are will be helpful as these values can be plugged into other equations or interpreted to determine specific trends or correlations. For example, if the slope is negative, that means there is an inverse correlation, versus a positive slope having a direct correlation.

Given a list of points {{1, 2}, {2, 3}, {3, 5}, {4, 6}}, Wolfram|Alpha can determine the best-fitting linear equation: “linear equation {{1, 2}, {2, 3}, {3, 5}, {4, 6}}”:

Additionally, Wolfram|Alpha can be used to rewrite an equation that is not in y = mx + b form to turn it into this form. For example, given the equation 2x + 7y — 5 = 0, what is the slope and intercept? Just enter “solve 2x + 7y — 5 = 0 for y”:

The slope would be m = –2/7, and the y intercept is b = 5/7. You can also type in “slope and y intercept of 2x + 7y — 5 = 0” and receive the answers directly, but it’s helpful to know how to identify each part in the slope-intercept form of the equation. Following this, if someone were to ask what the value of y is given that x = 5, by evaluating –2/7x + 5/7 at x = 5, the answer would be –5/7, which can be calculated using the basic algebra skills seen previously:

Being able to identify what the slope and y intercept are, and also plugging in values to determine what y is, is a very important skill for topics in chemistry, such as kinetics.

In chemistry, we use Celsius to measure temperature instead of Fahrenheit. Given a couple known temperatures in Celsius and Fahrenheit, how can we convert other temperatures without using a digital assistant? Well, let’s take five points, where the pairings will be listed as {Fahrenheit, Celsius}: {{40, 4.4}, {50, 10}, {60, 15}, {70, 21}, {80, 26}}. We can find the information for this line with “linear regression {{40, 4.4}, {50, 10}, {60, 15}, {70, 21}, {80, 26}}”:

The equation of this line is y = 0.542x - 17.24, with x being Fahrenheit and y being Celsius. So if you have a value in Celsius or Fahrenheit, you can convert to either one by plugging the value into this equation and solving for the other variable. So if you’re trying to convert from Fahrenheit to Celsius, of course just typing in “75 F to C” will do the trick, but being able to see this linear correlation can help you determine this conversion even if you don’t have access to the internet.

Check out the original blog post for even more great ideas on how to use Wolfram|Alpha, like calculating logarithms and solving for x!