What happens if I mix this with that?
What happens if I mix this baking soda with vinegar?
What happens if I let this spilled gasoline sit on the painted garage floor?
What happens if I pour bleach directly onto my clothes?
Predicting chemical reactions does not take place only in the lab; it is actually a part of everyday life! However, in the lab, we can be more specific and better isolate the this and the that.
So… I told you this would be long! Maybe another soda or cup of coffee?
Recall: the goal is to end up with neutral molecules. The goal is to determine the correct product formula so that the overall charge is zero.
Because of the vast scope of chemistry, there will be variations that this system will not cover. And there are always exceptions!
What do you have to start with?
This step identifies what kind of reaction you could be looking at. Is it synthesis? Decomposition? Single replacement? Double replacement? Combustion?
What does it become?
- Will it even react?
- What combination of atoms or ions is needed to form the correct product formula? For a neutral ionic compound, what subscripts are needed so that the overall charge is zero? (AKA what are the subscripts?)
- What is the FINAL balanced equation?
So, here is a "decision tree" for what to do to predict the products of a chemical reaction.
First off, will they react?
For two elements to react in a synthesis reaction, they must be able to form a stable compound together. Depending on the elements involved, this may occur through ionic bonding or covalent bonding.
They need to have opposite valences. At an introductory level, we can think of one element as tending to give up or share electrons and the other as tending to gain or share electrons so that both reach more stable valence-electron arrangements.
As a simple starting point, atoms with only a few valence electrons often tend to lose or share electrons, while atoms with nearly full valence shells often tend to gain or share electrons.
Plot twist—some elements can have more than one valence. For example, sulfur is often described at an introductory level as having valences of 2, 4, and 6.
Secondly, if they will react, what is the correct formula of the compound formed in the product?
The subscripts show the ratio of atoms in the product formula. At an introductory level, the valences of the elements can often help predict that ratio. For ionic compounds, ion charges can be criss-crossed to determine the subscripts. For simple covalent compounds, the same ratio idea can be used with typical valences as a shortcut, but the valences should not be mistaken for actual ionic charges.
For this introductory system, think of valence as an element's typical combining capacity. The numbers can help us predict ratios, but the + and − signs we sometimes attach to them do not always represent actual ionic charges.
For instance, take carbon and oxygen.
Since both are nonmetals, they form a covalent compound rather than an ionic compound.
Oxygen has a valence of 2. Since it NEEDS 2 more electrons to complete its valence shell of eight, as an introductory shortcut, you can conceptualize its combining tendency as −2.
Carbon has a valence of 4. (Since we're conceptualizing oxygen's combining tendency as −2, we'll represent carbon's as +4 for this introductory shortcut.)
If we are combining carbon and oxygen, we start off with something like this:C + O₂ → ??
So, the product has to be C?O?
As an introductory shortcut, criss-cross the valences: C₂O₄
Reduce the subscripts to the lowest whole-number ratio: C₂O₄ becomes CO₂
C + O₂ → CO₂
Thirdly, balance the equation using coefficients so that the same number of each type of atom is present on both sides:
C + O2 --> CO2 (already in balance)
Likewise:
- Magnesium has a valence of 2—it has 2 valence electrons and typically loses both when forming an ionic compound.
- Chlorine has a valence of 1—it has 7 valence electrons, so it NEEDS one more to complete its valence shell.
- Therefore, combining them gives the formula MgCl₂
The balanced equation would be:Mg + Cl₂ → MgCl₂
First off, will it decompose?
Not everything will break apart easily. Some things only break apart at high temperatures.
It seems fair to presume that, if given a predicting-products exercise, the compound will, by some means, undergo a decomposition reaction and form simpler substances.
Secondly, if… Well, this one is pretty easy. Whatever you start with breaks apart. But… into how many pieces! (Probably two.)
Usually, it will look like this:
AB → A + B
The products must be valid substances with the correct chemical formulas. They may be elements or compounds. And don't forget about those diatomic elements! For example, if decomposition produces elemental oxygen, students should write O₂, not O.
“Decompose into simpler substances” can suggest that a compound must separate into its original elements. That is not always true. A decomposition reaction produces two or more simpler substances, which may themselves be compounds.
For example:
CaCO₃ → CaO + CO₂
Neither product is simply an individual element.
Thirdly, balance the equation using coefficients so that the same number of each type of atom is present on both sides:v
2H2O → 2 H2 + O2
It could be tricky, though!
First off, will they react?For one thing to replace another… Let's say it like this… For A to replace B,A + BC --> ?? + ????
If it WILL react, we get the standard single displacement pattern:
A + BC → AC + B
For the reaction to occur, A has to be more highly reactive than B.
How would anyone know that? There is a chart—the activity series!So, sodium won't replace potassium in a compound because potassium is above sodium on the metal activity series. Etc.!Also, either B or C could represent a polyatomic ion! Usually, the polyatomic ion stays together while the element in the compound is the part being replaced.
That makes it harder to figure out what is being replaced. Look into the BC part and match one of them to the A with regard to location on the periodic table. There's a good chance that A will be in a family/group that is near the family or group of B or C. (You'll have to be open-minded about this claim when dealing with transition elements.)
If A is a metal, it will usually attempt to replace another metal or hydrogen. If A is a halogen, it will attempt to replace another halogen. Then use the appropriate activity series to see whether the replacement can occur.
Secondly, if they will react, how many of each are needed to get the correct formula of the new compound in the product?You know what you are starting with, so the reactant side is done. Let's do aluminum and HCl as an example…Al + HCl --> ?? + ??Since Al is above H on the activity series, Al can replace H. The product side will be:
Al + HCl --> Al?Cl? + H?Product side subscript time:This should be fun!1. H is diatomic, so it will be H22. The aluminum chloride formula has to have an overall charge of zero.The subscripts show the simplest whole-number ratio of ions needed to give the compound an overall charge of zero. You can "criss-cross" the charges of two elements (then reduce mathematically to the lowest whole-number ratio) to find the right numbers.For instance, take aluminum and chlorine…
Al has a valence of 3 and commonly forms Al³⁺.
Cl has 7 valence electrons and a typical valence of 1, meaning it needs 1 more electron to complete its valence shell and commonly forms Cl⁻.
The correct formula will be AlCl3
Thirdly, balance the equation using coefficients so that the same number of each type of atom is present on both sides:
2Al + 6HCl --> 2AlCl3 + 3H2
First off, will they react?
For one thing to replace another… Let's say it like this…
AB + CD --> ???? + ????
In a double-displacement reaction, one element does not simply replace another. Instead, the positive and negative parts of two compounds exchange partners.
If it will react, then we arrive at the general form of the double replacement reaction:
AB + CD → AD + CB
Deciding what is A, B, C, and D can be hard when polyatomic ions are involved. Really, the only way to get good at this is to do it a lot. Practice makes identifying the parts much easier.Look at A and C first. Are they both the positive ions (cations)? Then look at B and D. Are they both the negative ions (anions)? (There is a chart!)?
If A and C are positive and B and D are negative, you can identify the potential "swaps."
At this point, you have to answer the question!
Will they react?In real reactions, determining whether the ion exchange actually occurs can be more complicated.
Answering this question depends on whether the ion exchange produces something that drives the reaction forward.
So…
1. Does it form water?
2. Does it form a gas?
If the ion swap produces water, a gas, or a precipitate, the reaction can occur.
Secondly, if they will react, what are the correct formulas of the new compounds in the products?The work done in the first step should have resulted in you knowing what the A, B, C, and D parts are. The potential products will be AD and CB.You know what you are starting with, so the reactant side is done. Let's do the following as an example…Fe2(SO4)3 + KOH →So, there's some AB and CD up there? That looks like a sentence in a foreign language!So this…
AB + CD → AD + CBFe2(SO4)3 + KOH →(As a temporary step, it can help to put polyatomic ions in parentheses while you figure out the new subscripts.)
Fe2(SO4)3 + K(OH)1 →
Fe2(SO4)3 + K(OH)1 → Fe?(OH)? + K?(SO4)?
Product side subscript time:This should be… never mind!1. Do that thing with the charges to get product formulas with an overall charge of zero. For instance…K (from periodic table) commonly forms K⁺, with a charge of +1.
SO₄²⁻ has a charge of −2.
Criss-cross the charges to get K2(SO4)1
Chemistry "grammar" says we don't write a subscript of 1, and we don't use parentheses around a polyatomic ion unless more than one of that ion is needed.Thus, we get:
K2SO4
Using the example from above, working through the process of getting both products to a neutral charge, we get the unbalanced (but each part is a neutrally charged molecule) equation:?Fe2(SO4)3 + ?KOH → ?K2SO4 + ?Fe(OH)3Thirdly, balance the equation using coefficients so that the same number of each type of atom is present on both sides.So, for the above example…Fe2(SO4)3 + 6KOH → 3K2SO4 + 2Fe(OH)3
In this example, the reaction occurs because Fe(OH)₃ is insoluble and forms a solid precipitate.
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Combustion Reaction (background)
First off, what do you have to start with?
In the combustion reactions we will use, look for a substance containing carbon and hydrogen reacting with oxygen (O₂).
A compound made only of carbon and hydrogen is called a hydrocarbon.
The basic pattern is:
Hydrocarbon + O₂ → ?? + ??
For these introductory problems, assume that the reaction WILL occur and that there is enough oxygen for complete combustion.
Secondly, what does it become?
This part is actually pretty predictable.
In the complete combustion of a hydrocarbon:
Carbon ends up in CO₂.
Hydrogen ends up in H₂O.
So:
Hydrocarbon + O₂ → CO₂ + H₂O
Yep. That's pretty much the product-prediction part.
For example, start with methane:
CH₄ + O₂ → ?? + ??
Methane contains carbon and hydrogen and is reacting with oxygen, so the products are:
CH₄ + O₂ → CO₂ + H₂O
Notice that we did not criss-cross charges or change subscripts to invent the products. For the complete combustion of a hydrocarbon, the expected products are CO₂ and H₂O.
Thirdly, balance the equation using coefficients so that the same number of each type of atom is present on both sides.
Start by balancing carbon and hydrogen. Save oxygen for last because oxygen appears in both products.
CH₄ + 2O₂ → CO₂ + 2H₂O
Check it:
Carbon: 1 on each side
Hydrogen: 4 on each side
Oxygen: 4 on each side
Balanced!
The Big Combustion Shortcut
If you see:
hydrocarbon + O₂
think:
CO₂ + H₂O
Then balance the equation using coefficients.
One small catch: If there is not enough oxygen, incomplete combustion can occur and products such as carbon monoxide (CO) or carbon may form. For our introductory predicting-products problems, however, assume complete combustion unless told otherwise.
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