Even More Step-by-Step Math Tools in Wolfram|Alpha For Chemistry Course Prep
In a previous cross-post, we explored different math concepts found in chemistry class and how Wolfram|Alpha can help walk you through them step by step. Here is a continuation of that post, offering a look at even more ways to improve your studies.
You can view the original post in full on the Wolfram Blog by clicking here.
4. Solving for x
Another important concept needed to succeed in your first chemistry class is how to solve for x using inverse operations. Rearranging numbers and variables around to isolate x and determining the unknown value are so important in solving for different values, such as energy, mass, volume and so on.
Let’s take an example using multiplication and division to solve for x. Given the equation 8 * x = 3 * 0.08206 * 298, what is the value of x? This is actually an equation taken straight from the ideal gas law with numeric values plugged in. Ideal gas law problems allow us to solve for pressure, volume, amount of substance and temperature. In all cases, the effort distills down to an expression with x in it to be solved for. After combining terms, Wolfram|Alpha solves this via “solve 8 * x = 3 * 0.08206 * 298 for x”:
We could include results from the ideal gas law calculator for an example problem. Determine the volume of 1 mole of methane gas at 150 Kelvin and 1 atm:
Here is another problem that uses exponent rules: “solve for x when 10^2 = 200/sqrt(x)”. After simplifying 10² to 100, the rest of the steps are shown, which gives a good refresher for using exponents to isolate variables:
Let’s take a look at a more complicated example. This one involves solving for x in a quadratic function, which appears in equilibrium and buffer problems. Given that 0.0125 = (x)*(x + 0.1) / (3 – x), what is x? This one is much more difficult to do by hand, so let’s look at what Wolfram|Alpha has to say for “solve 0.0125 = (x)*(x + 0.1) / (3 – x)”:
There is a step-by-step solution available for this problem as well, but the two answers after clicking the Approximate form button are x = –0.25790 and x = 0.14540. This is very useful in chemistry, especially when solving equilibrium reactions. However, you will encounter solving for x anywhere in chemistry that involves an equation.
5. Logarithms
For the most part, logarithms are computed using calculators when solving acid-base problems, but it is helpful to know how to use them. In chemistry, the main types of logarithms that are used are base 10 and base e, or the natural log (the ln button on your calculator). Unless the logarithm is equal to a whole number, you would probably have to use a calculator to solve these. An important concept to know about logarithms is how to switch back and forth from the exponential form to the logarithmic form. For example, what is the exponential form of log10(100) = 2? It would be 10² = 100. Given this example, you should be able to determine the value of a variable given a logarithm problem. For example, what is the value of x in log10(x) = 4? Wolfram|Alpha has a great step-by-step solution for this:
Logarithms are mainly used in pH calculations, since pH is measured on a base-10 scale, meaning that a pH of 3 is 10 times “stronger” than a pH of 4. That being said, many times you would end up with a pH that is not a whole number. You can also use Wolfram|Alpha to calculate logarithms that would be harder to compute by hand, such as the value of x in the equation log10(x) = 3.5. “Solve log10(x) = 3.5” generates an answer of 3162.3, as seen here after clicking the Decimal form button:
The process for solving this problem is similar to the previous one, but it is very hard to calculate a decimal exponent, so using a computer or calculator to determine this answer is necessary. It also is worth mentioning that log(x) in Wolfram|Alpha is not interpreted as base 10, but as the natural log. If you need base 10, make sure to type log10(x).
6. Finding the Roots of a Quadratic Equation
The last important concept to know for your first chemistry course is finding roots of quadratic equations. A quadratic equation has the form a * x2 + b * x + c = 0 with the values of x given by –b ± Sqrt[b² – 4ac]/2a, which is known as the quadratic formula. Of course, solving for roots like this often trips people up, but using the computer to solve it for you is much quicker and exact! In the world of chemistry, solving for the roots of a quadratic function is useful when solving equilibrium problems, particularly Ksp equations, where there is no denominator like there usually is.
Consider the Ksp problem where you need to solve for the roots for the equation x * (0.001 + x) = 1.5 * 10–5. This would be a case where the terms have to be rearranged first so that the equation is equal to 0 before using the quadratic formula. There is a detailed step-by-step solution for this problem:
Sometimes equations containing cubic terms, like x³, and higher may appear in your homework problems. An example of this is x³ – x² + 9 * x – 9 = 0. What are the values of x in this case? Wolfram|Alpha will kindly determine the roots with the input “solve x^3– x^2 + 9 * x – 9 = 0”:
Keep in mind that there are more ways to solve for the roots of an equation than the quadratic formula. Wolfram|Alpha provides different types of step-by-step solutions that can teach you to solve the same equations by hand, whether it be factor method, quadratic formula or completing the square.
Math and Chemistry Go Hand in Hand
As you learn more about chemistry concepts, you can see how intertwined the math is with chemistry. It really is beautiful how it all comes together in the end, but it’s important to be comfortable with the algebra behind it in order to be able to see the connectedness. I recommend using Wolfram|Alpha Pro for solving both math and chemistry problems, since many of the problems will give you the step-by-step solutions for solving them, which can help you have much success in your chemistry classes!
For more chemistry-focused step-by-step solutions, see these posts from the Wolfram|Alpha Chemistry Team:
