Balancing Chemical Equations (Simplified) - Online Tutor, Practice Problems & Exam Prep

Balancing chemical equations requires ensuring that the type and number of atoms on both sides of the equation are equal. Coefficients, such as 2 for hydrogen and 1 for oxygen, are distributed to calculate the total number of each atom. For example, 2 H₂O yields 4 hydrogens and 2 oxygens on both sides. This process confirms that the law of conservation of mass is upheld, as the total mass and number of atoms remain constant throughout the reaction.

Balancing Chemical Equations (Simplified) Concept 1

In this series of videos, we're going to take a look at balancing chemical equations. When balancing, always make sure the type and number of atoms on both sides of the arrow are equal. We're going to say in a balanced equation, the numbers that are in red are referred to as the coefficients. So we have 2, 1, and 2 as the coefficients for this balanced equation. And when we're talking about type and number of atoms, we'd say that these coefficients get distributed. So this 2 would get distributed to this hydrogen. There's already 2 of them, so 2 times 2 gives me 4 hydrogens. This is just a 1, so 1 distributed is 1 times 2 which is 2 oxygens. On the other side, the 2 gets distributed, so it'd be 2 times the 2 hydrogens would give me 4 hydrogens. But there's also 1 oxygen here, so 2 times 1 gives me 2 oxygens. The types of atoms are the same on both sides. I have hydrogen and hydrogen, oxygen and oxygen, and then the numbers of each are equal on both sides. 4 hydrogens, 4 hydrogens, 2 oxygens, 2 oxygens. This is how we ensure a chemical equation is balanced. The types and numbers of each atom are the same on both sides of the arrow. Now click on the next video, and let's take a look at an example question where we go more refined at our approach to balancing a chemical equation.

Balancing Chemical Equations (Simplified) Example 1

Here we're told to write the balanced equation for the following by inserting the correct coefficients in the blanks. Alright. So we have here C₄H₁₀ gas reacting with O₂ gas to produce and give us H₂O liquid and CO₂ gas. Right. So step 1, we have to set up a list for the elements that are reactants, and another list for the elements that are products. Right. So on the reactant side we have 4 carbons, 10 hydrogens, and 2 oxygens. On the product side we have 1 carbon, 2 hydrogens, and we have 3 oxygens. We can see the lists don't match and that's because the equation is imbalanced.

So step 2, we're going to start from the top and go down both lists to determine how many of each element is present. Now here, this is important. If a polyatomic ion is present on both sides, then treat it as a single unit. So for example, if one side had 1 phosphate ion, and the other side had 3 phosphate ions, you make sure that this side here also has 3 phosphate ions. In this particular equation, we don't have to worry about polyatomic ions, there are none.

And then here, step 3. Starting from the top and going down both lists, begin balancing each element to ensure they match. Sometimes you may have a decimal or fraction as a coefficient, and so multiply the equation by 2. Alright. So let's start doing this. We have 4 carbons here but only 1 here, so I'm going to put a 4 right here. That 4 gets distributed to the carbon, but it also gets distributed to the oxygen. So it's going to be 4 times 2, which is 8, plus one more is 9 oxygens now. Now continue down the list. The carbons are balanced because they're both the same. Keep going down the list. Hydrogens. Here we have 10 and here we have 2. We need this side to also have 10, so I'm going to put a coefficient of 5 here. The 5 gets distributed to the hydrogen, so 5 times 2 gives me 10 hydrogens, but then it also gets distributed to the oxygen. So 5 times 1 is 5 oxygens here, and remember we still have another 8 over here. So that's 5 oxygens plus 8 oxygen gives us 13 total oxygens on the product side. Finally, we have to look at the oxygens. Here we have 2 and here we have 13. So you have to think of a number that I can multiply by 2, which will give me 13. If I put a 6 here, that's 6 times 2, which is 12. It's not enough. If I put a 7 here, that's 7 times 2, which is 14. That's too much. So we gotta find something in between 6 and 7, and that would be 6.5. 6.5 times 2 gives me 13. Both lists match because now my equation is balanced. But remember step 3 says that we cannot have fractions or decimals as coefficients. When that happens, I multiply the entire equation by 2. Okay? So this whole thing will get multiplied by 2. So this coefficient of 1 gets multiplied by 2, this coefficient of 6.5 gets multiplied by 2, this coefficient of 5 gets multiplied by 2, and then this coefficient of 4 gets multiplied by 2. So these are the new coefficients that we're going to have for each of these compounds. So we're going to have 2 C₄H₁₀ gas + 13 O₂ gas, gives me or produces 10 H₂O liquid + 8 CO₂ gas. So here this would represent the coefficients for each of the compounds and all we have to do is insert the correct coefficients in the blanks provided. K. So those would be the values that you put, 2, 13, 10, and 8.

Write the balanced equation for the following by inserting the correct coefficients in the blanks.

• A. 1, 1, 3, 1
• B. 2, 3, 3, 1
• C. 1, 3, 6, 2
• D. 2, 3, 6, 1

Determine the total sum of the coefficients after balancing the following equation.  

In this practice question, it says, determine the total sum of the coefficients after balancing the following equation. Alright. So this one's a little different from what we're accustomed to seeing. Here, they're basically asking us to find all the coefficients for each compound and then add them up together because we want the sum, the total sum of all the coefficients. Alright. So we're gonna start out with our list. We're gonna say we have carbon, hydrogen, sulfur, and oxygen. On the other side, the list matches carbon, hydrogen, sulfur, and oxygen. Now we're gonna say we have 2 carbons, 5+1, 6 hydrogens, 1 sulfur, and 2 oxygens. On the other side, we have 1 carbon, 2 hydrogens, 1 sulfur, and then we have what? 2 oxygens, 3 oxygens, 5 oxygens, 5 oxygens here. Alright. So let's balance this out. The number of carbons doesn't match. Here we have 2 and here we have 1. So we're gonna place a coefficient of 2 right here. That 2 gets distributed to not only the carbon, but also the oxygen. So we're gonna have 2 carbons and then 2 times 2 is 4 oxygens, 5 oxygens, and then 7 oxygens. Next, balance out your hydrogens. Here this is 6, but here this is only 2, so I'm going to put a 3 here. So 3 times 2 gives me 6 hydrogens, but then we have 3 oxygens here plus another 4 here, so that's 7, plus another 2 is 9 oxygens. Don’t worry, we’re just going down the list, we’re gonna eventually get to the oxygens. Sulfurs are the same, both are 1 and 1, so let's not do anything to that. Then here we have 2 oxygens and here we have 9 oxygens. So we have to think of a number to multiply 4 by. It well, I kind of said it. We have to think of a number to multiply 2 by to get to 9. We can't do 4 because 4 times 2 gives us 8. We can't do 5 because 5 times 2 gives us 10. The number that we have to place here is 4.5. 4.5 times 2 would give me 9. But remember, you're not allowed to have fractions or decimals as coefficients. They have to be whole numbers. In these situations, remember, all we do is we multiply this by 2. So all the coefficients will get multiplied by 2. So when we do that, we're gonna have 2, 9, 4, 6, and 2 as the coefficients for each one of these compounds. So then our equation would be C9H4O2 → CO2 gas, H2O liquid plus S2O gas. So our coefficients would be 2, 9, 4, 6, and 2. This is in our final answer because remember, we have to add them all up. We need the total sum of all the coefficients. So 2 + 9 gives me 11, 11 + 4 equals 15, plus 6 gives me 21, plus this final 2 gives me 23 as the total sum for all the coefficients in this balanced equation. So 23 would be our final answer.

Here’s what students ask on this topic:

What is the first step in balancing a chemical equation?

The first step in balancing a chemical equation is to write down the unbalanced equation with the correct chemical formulas for all reactants and products. This provides a clear starting point. Next, count the number of atoms of each element on both sides of the equation. This helps identify which elements need to be balanced. For example, in the equation H₂ + O₂ → H₂O, you would count 2 hydrogen atoms and 2 oxygen atoms on the reactant side, and 2 hydrogen atoms and 1 oxygen atom on the product side.

Why is it important to balance chemical equations?

Balancing chemical equations is crucial because it ensures the law of conservation of mass is upheld. This law states that matter cannot be created or destroyed in a chemical reaction. By balancing the equation, we ensure that the number of atoms for each element is the same on both sides of the equation. This means the total mass of the reactants equals the total mass of the products, reflecting the true nature of the chemical reaction.

How do coefficients help in balancing chemical equations?

Coefficients are numbers placed in front of the chemical formulas in an equation to indicate the number of units of each substance. They help balance the equation by ensuring the same number of atoms for each element on both sides. For example, in the balanced equation 2 H₂ + O₂ → 2 H₂O, the coefficient 2 in front of H₂ and H₂O ensures there are 4 hydrogen atoms and 2 oxygen atoms on both sides of the equation.

Can you provide an example of a balanced chemical equation?

Sure! Let's consider the combustion of methane (CH₄). The unbalanced equation is CH₄ + O₂ → CO₂ + H₂O. To balance it, we adjust the coefficients: CH₄ + 2 O₂ → CO₂ + 2 H₂O. Now, we have 1 carbon atom, 4 hydrogen atoms, and 4 oxygen atoms on both sides, making the equation balanced.

What are some common mistakes to avoid when balancing chemical equations?

Common mistakes include changing the subscripts in chemical formulas instead of using coefficients, which alters the substances involved. Another mistake is not properly distributing coefficients to all atoms in a compound. For example, in H₂O, a coefficient of 2 means 2 H₂O, not H₂O₂. Also, forgetting to balance polyatomic ions as a whole unit when they appear unchanged on both sides of the equation can lead to errors.