Molecular Equations - Online Tutor, Practice Problems & Exam Prep

A molecular equation represents intact compounds in reactions, such as the neutralization of an acid and a base, producing water and an ionic salt. Different types of molecular equations include neutralization, gas-evolution, and precipitation reactions. In neutralization, an aqueous acid reacts with an aqueous base to form water and an ionic compound. Gas-evolution reactions produce at least one gas, while precipitation reactions yield a solid precipitate. Understanding these concepts is essential for grasping acid-base reactions and their applications in chemistry.

Molecular Equations

A molecular equation shows the intact compounds instead of their dissociated ionic forms. When we talk about a molecular equation, we're reacting aqueous reactant 1 with aqueous reactant 2 to produce product 1 plus product 2. A good example of this is we have 2 moles of perchloric acid in their aqueous form, plus 1 mole of barium hydroxide, creating or producing 1 mole of barium perchlorate plus 2 moles of water as liquid.

Now, when it comes to molecular equations, we can define different types of molecular equations based on the products they form. In a neutralization equation, we have an aqueous acid reacting with an aqueous base, just like in the example we provided, and what we tend to form as products are some type of ionic compound plus water. As we progress to later chapters discussing more advanced acid and base reactions, we'll see that sometimes we don't form water. For now, just realize that in a basic acid-base neutralization equation, we'll produce water and ionic salt as products.

Now, in a gas evolution equation, from the name we know that a gas is involved; here we still have aqueous reactant 1 and 2 reacting, and at least one of the products produced will be a gas. So, we'll have a gas plus product 2 — product 2 itself could also be a gas, but that's not always guaranteed.

Finally, we have a precipitation equation, and from the name 'precipitation', we know that a precipitate is involved. If you've watched my solubility rules videos, you would know that a precipitate represents a solid. So, in a precipitation equation, if at least one of the products formed is a solid ionic compound. So just remember, we have our molecular equation, which tends to have aqueous reactant 1 and aqueous reactant 2 reacting to create products 1 and products 2. Based on the identities of those products, we can have different types of molecular equations.

Molecular Equations Example 1

In this example question, it asks, which of the following is a precipitation reaction? So remember, in a precipitation reaction, we have aqueous reactant 1 reacting with aqueous reactant 2 to produce products. And at least one of those products has to be a solid ionic compound. So if we take a look here, in the first one, we have 2 aqueous reactants. That's cool. But here we're creating water as a liquid and an ionic compound. Here, HCl is an acid, KOH is a base. So this is actually an acid-base reaction or equation. For the next one, we have a solid reactant reacting with an aqueous reactant to produce a solid product and an aqueous product. Now remember, for it to be a precipitation reaction, both of my reactants have to be in the aqueous form. One of them is a solid, so this will not do. For the last one, we have 2 aqueous reactants and we created at least 1 solid product. Because of this, option C would have to represent our precipitation reaction.

Molecular Equations Example 2

Here we can say that a molecular equation can be written when given reactant 1 and reactant 2. If we take a look at this example question, it says, predict whether a chemical reaction occurs and write the balanced molecular equation. Alright. So, write the molecular equation. We're going to follow the given steps. So step 1 says we have to break up reactant 1 and reactant 2 into their ionic forms. If we look at reactant 1, it's composed of lithium and hydroxide. Remember, lithium is in group 1A, so it's +1. Hydroxide is a polyatomic ion, so it's -1. Magnesium sulfate, magnesium is in group 2A, so it's 2+ for its charge. Sulfate is 2-. Now if you don't quite remember how I'm coming up with these charges, make sure you go back and take a look at my videos on the periodic table and charges, and then as well as my polyatomic ions, concept videos. So, we've done that first part.

Step 2, we swap ionic partners by remembering that opposite charges attract. Apply the rules for combining ions based on the numerical values of their charges. Alright. So opposite charges attract. This positive will be attracted to this new negative. This negative will be attracted to this new positive. So they're switching partners. Now remember, if the numbers within the charges are different, they don't cancel out. They're going to criss-cross. So the 1 from here would come here, the 2 from here would come here. That would give me Li2SO4. Plus, then we're going to have the 1 from here come over here, the 2 from here come over here. That would give me Mg(OH)2. This you can find when we talked about naming ionic compounds and writing ionic compounds. So again, everything we've learned about ionic compounds up to this point is necessary to effectively write a molecular equation. Alright. So, we've written out what the products are.

Now, step 3 is important. We're going to say a reaction only occurs if a solid, gas, or liquid water is formed as a product. If both of the products formed are aqueous, meaning they're soluble, then no reaction has occurred and we stop. We just say no reaction. Here we have to use the solubility rules to determine if the products formed will be soluble or insoluble. So let's come back up here. One product that we made is lithium sulfate. Remember, GHS and COPS. Lithium is a group 1A element. So that's part of GHS. Right? So anything connected to a group 1A ion is automatically soluble, so this would be aqueous. Magnesium hydroxide. Hydroxide is part of COPS. We said that hydroxides are insoluble unless they're connected to CBS. Right? Calcium, barium, or strontium. Also, if they're connected to things from GHS, they could also be soluble. But here, hydroxide is not connected to calcium, barium, or strontium, and magnesium is not part of GHS in any way. So here, this would be solid. So we just created a solid so we know a reaction has occurred. Because remember, you have to create a solid, gas, or liquid water.

So, step 4, if necessary, balance your molecular equation by placing the correct coefficients in front of each molecule. So we're going to do that now. We're going to place the right coefficient in front of each molecule. So, coming back up, if we look at our balanced equation, or unbalanced equation, we have 2 lithiums here but only 1 here. So we're going to put a 2 here. Now we have 2 lithiums but it also gives us 2 hydroxides. That's okay because look, we have 2 hydroxides here. Then what do we see? We see that we have 1 magnesium, 1 magnesium, 1 sulfate, 1 sulfate. So, our equation is balanced. So if we write all the coefficients in front, it would be 2, 1, 1, and 1. So this would represent our balanced molecular equation.

Predict whether a chemical reaction occurs and write the balanced molecular equation.

Ag₂SO₄ (aq) + KCl (aq) →

Here in this practice question it says, predict whether a chemical reaction occurs and write the balanced molecular equation. So here we're given silver sulfate and potassium chloride. So step 1 tells us that we need to break it up into its ions. So we have silver and sulfate. Don't worry about the fact that there are two silvers there, just break it up into its ions. Silver is plus 1. It's a transition metal but its charge is always plus 1. Sulfate is a polyatomic ion, so 2 minus. Potassium is in group 1A, so it's plus 1. And chlorine is in group 7A, so it's minus 1. Now remember, we're going to swap ionic partners, opposite charges attract. This plus 1 and this minus 1 are attracted to each other. So their numbers are exactly the same in terms of their charges. When they're the same, they just cancel out. So here we'll have AgCl plus here, the numbers and the charges are not the same. When they're not the same, they don't cancel out; instead, they crisscross. So 2 comes here, 1 comes here. So it's going to give me K2SO4. Now, we have to determine if a reaction has occurred. And it only occurs if we create a solid, a gas, or liquid water. If we think back on our solubility rules, so silver, there is no rule for silver from Guyana Cash or Ca'ns, but there is a rule for chlorine. Chlorine is a halogen. Right? We know that from Guyana Cash, halogens do have exceptions. Right? So their exceptions are happy. If they're connected to mercury, silver, or lead, they will not be soluble, they will be insoluble and make a precipitate. Here, chlorine my halogen is connected to silver, so it's not going to be soluble, it's going to be a solid. So we know a reaction has occurred because we've created a solid. Now let's look at the other one. We have potassium sulfate. So potassium is group 1A. Group 1A elements if they're connected to anybody, it doesn't matter, they're always going to be soluble. So we don't even need to look at the solubility rule for sulfate, because by default the whole thing is soluble because of the presence of potassium. So here goes our equation, now we have to balance it. Here we have two silvers, but here we only have one, so I'm going to put a 2 here. Next, we have one sulfate, one sulfate. Then we have, if we look, two gets distributed to chlorine, so we have two chlorines, and then we have two potassiums here. So that means I have to put a 2 right here. Now every element within this structure is balanced. So now we're going to say my coefficients would be 1, 2, 2, and 1. So this represents my balanced molecular equation.

Predict whether a chemical reaction occurs and write the balanced molecular equation.

MgBr₂ (aq) +  NaC₂H₃O₂ (aq) →

Here in this practice question, it says predict whether a chemical reaction occurs and write the balanced molecular equation. So step 1, we're going to break up these ionic compounds into their different ionic forms. So we have magnesium bromide. Magnesium is in group 2a, so it's 2+. Bromine is in group 7a, so it's -1. Don't worry about the fact that there are 2 bromines here. We're just breaking them up into their ions. Here we have sodium acetate. Sodium is in group 1a, so it's +1. Acetate is a polyatomic ion, so C2H3O2-. Step 2 is opposite charges attract, we're going to swap ionic partners. So we're going to say here that +1 and -1, they're opposite charges so they're attracted to one another. If the numbers in the charges are exactly the same, when they combine they just cancel out. Remember, when you write an ionic compound, we traditionally write the positive ion first, followed by the negative ion. Our next product, when the numbers in the charges are different, they don't cancel out. Instead, they crisscross. So 2 comes over here and 1 comes over here. This gives us Mg2(C2H3O2)2. Let's think about our solubility rules because now we have to determine if a reaction has occurred. A reaction has occurred. Remember, it only occurs if you create a solid, liquid, water, or a gas. If we think about our solubility rules, Na is a group 1a element. Anything connected to group 1a is soluble, so this is aqueous. And then we have magnesium acetate. Remember acetate, anything connected to an acetate ion will also be soluble. We just created 2 aqueous products. And remember, we said that if we create 2 aqueous products, then we would say no reaction has occurred. So here we just stop and we say no reaction has happened. Later on when we talk about net ionic equations, we'll see exactly why that is. But for now, just realize when writing out your molecular equation, if you get 2 aqueous products, no reaction has occurred.

Determine the balanced equation for the neutralization equation

Ca(OH)₂ (aq) + HCN (aq) →

Here in this practice question, it says determine the balanced equation for the neutralization equation. Now before we actually do it, realize that this is a neutralization equation because it possesses an acid and a base. The acid is in the form of hydrocyanic acid, and the base is calcium hydroxide. Alright. Now step 1, we have to break it up into its ionic forms, both ionic compounds. So calcium hydroxide possesses calcium which is in group 2a, so its charge is 2+. Hydroxide is a polyatomic ion and its charge is minus 1. Here we have our acid. Now the acid will break up into H+, and then CN-, which is your cyanide ion. Now remember at this point, opposites attract. Opposite charges will attract one another. When the numbers in the charges are different, though, they don't just cancel out, they crisscross. So 2 comes here and then one comes here. So that's going to give me Ca2(CN)2. Remember, we write the positive ion first, and then the 2 is for the entire CN ion, so put it in parentheses and then a little 2. Alright. Plus, then we have here, we have plus 1 and minus 1, the numbers are the same, so when they combine they just cancel out. Now, H+ and OH-, think about it. When they combine, it gives us HOH, which can be written as water. And remember, water in these molecular equations will exist as a liquid. So just remember, if you form HOH, that's just water. So we know a reaction has definitely occurred because if you create a solid, a gas, or liquid water, we know a reaction has occurred. Alright. So we know water is a liquid, but what about calcium cyanide? Here, calcium doesn't exactly have its own solubility rule. It's just part of exceptions. The cyanide ion isn't even talked about in terms of solubility rules. And because it's not stated within solubility rules, we can assume that it's going to be a solid. Now we have our equation, but we have to still balance it. If we look, we have 1 calcium, 1 calcium. We have 2 hydroxides here, but we don't see hydroxide exactly within this product side. That's because it's part of water. So let's just not worry about hydroxide for right now. For CN, we have 1 CN here but 2 over here. Right? 2 of them over here. So let's put a 2 here. Alright. Now let's go back. If we look we have 2 hydroxides, right, being distributed here, so that means we have 2 oxygens there and we have 2 hydrogens. This 2 is getting distributed to this H, so that also means we have another 2 hydrogens. So that's 4 hydrogens total on the product side. So if we go to the, on the reactant side, if we go to the product side, we would have to place a 2 right here because that 2 would get distributed to the H's to give us 4 hydrogens, and to the oxygens to give us 2 oxygens. And we would see that both sides would match. So here, this would be 1-2-1-2 for the coefficients of our balanced molecular equation. Now if you want more in-depth steps in terms of balancing chemical equations, make sure you go back and take a look at my topic videos on balancing chemical equations. At this point, we've done enough of these that I feel like we've mastered it, so we're not going through every single step with it. And also, these questions are pretty basic or simple in terms of balancing these chemical equations, but if you're still having trouble, make sure you go back and take a look at those earlier videos to keep on practicing. So at this point, we know how to balance molecular equations. Just remember to rely on the rules to guide you to your answer.

Here’s what students ask on this topic:

What is a molecular equation in chemistry?

A molecular equation in chemistry shows the complete chemical formulas of the reactants and products involved in a reaction, without indicating their ionic forms. For example, the reaction between perchloric acid (HClO₄) and barium hydroxide (Ba(OH)₂) can be written as:

$2{\mathrm{HClO}}_4\left(\mathrm{aq}\right)+{\mathrm{Ba}}_{(OH)}\left(\mathrm{aq}\right)\to {\mathrm{Ba}}_{(ClO})+2H_{2}O(l)$

This equation shows the intact compounds rather than their dissociated ions, providing a clear picture of the substances involved in the reaction.

What are the different types of molecular equations?

There are three main types of molecular equations: neutralization, gas-evolution, and precipitation reactions. In a neutralization reaction, an aqueous acid reacts with an aqueous base to form water and an ionic salt. For example:

$2{\mathrm{HClO}}_4\left(\mathrm{aq}\right)+{\mathrm{Ba}}_{(OH)}\left(\mathrm{aq}\right)\to {\mathrm{Ba}}_{(ClO})+2H_{2}O(l)$

In a gas-evolution reaction, at least one of the products is a gas. In a precipitation reaction, at least one of the products is a solid precipitate, indicating the formation of an insoluble ionic compound.

How do you identify a precipitation reaction in a molecular equation?

A precipitation reaction in a molecular equation can be identified by the formation of a solid product, known as a precipitate. This occurs when two aqueous solutions react to form an insoluble ionic compound. For example, when aqueous solutions of silver nitrate (AgNO₃) and sodium chloride (NaCl) are mixed, a white precipitate of silver chloride (AgCl) forms:

${\mathrm{AgNO}}_3\left(\mathrm{aq}\right)+{\mathrm{NaCl}}_{(aq)}\to {\mathrm{AgCl}}_{(s)}+{\mathrm{NaNO}}_3\left(\mathrm{aq}\right)$

The solid AgCl is the precipitate, indicating a precipitation reaction.

What is a gas-evolution reaction in a molecular equation?

A gas-evolution reaction in a molecular equation involves the formation of at least one gaseous product when two aqueous reactants react. For example, when hydrochloric acid (HCl) reacts with sodium bicarbonate (NaHCO₃), carbon dioxide gas (CO₂) is produced:

${\mathrm{HCl}}_{(aq)}+{\mathrm{NaHCO}}_3\left(\mathrm{aq}\right)\to {\mathrm{NaCl}}_{(aq)}+H_{2}O\left(l\right)+{\mathrm{CO}}_2\left(g\right)$

The CO₂ gas is the evolved gas, indicating a gas-evolution reaction.

What is a neutralization reaction in a molecular equation?

A neutralization reaction in a molecular equation occurs when an aqueous acid reacts with an aqueous base to produce water and an ionic salt. For example, the reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) can be written as:

${\mathrm{HCl}}_{(aq)}+{\mathrm{NaOH}}_{(aq)}\to {\mathrm{NaCl}}_{(aq)}+H_{2}O\left(l\right)$

In this reaction, HCl and NaOH neutralize each other to form water (H₂O) and sodium chloride (NaCl), an ionic salt.