Balancing Chemical Equations – A Primer

Balancing Chemical Equations – A Primer
Balancing seems hard...and it can be. So, we need a process or methodology to help us do it correctly. Let’s start with a reaction that bonds two chemicals together... Sodium and Fluorine Na + F

Na + F So what? Where do I start. One idea is the Periodic Table... Okay...but this one does not tell me much....or does it?

Recall, the elements in the “far right” column (Family 18) have a stable electron configuration (i.e., a full outer or valence shell of electrons). All other elements seek to have this stable configuration. In some reactions, elements will give or take electrons from other elements to achieve stability (Ionic Reaction). An element that is only one column away from Column 18 needs to add or subtract on electron (one negative charge). Likewise, an element that lies two columns away must add or subtract two electrons (two negative charges). When electrons are given or taken, the element becomes an ION. Since electrons have a negative charge, adding one electron gives the ion a charge of 1-. giving away one electron makes the ion charge 1+, giving two electrons makes the charge 2+, and so on...

Ouch...okay, how does it work? Na + F Sodium (Na) is found in Column #1. Na is element #11. This tells you that Na has 11+ protons and 11- electrons with an overall charge of zero. In this column, elements have one electron in their valence shell. Na wants to get rid of that one electron. If it does, Na has 11+ charges and 10- charges for an overall charge of 1+ Fluorine (F) is found in Column #17. It is element #9. This means F has 9+ protons and 9- electrons for 0 charge. F has seven electrons in its valence shell. To be stable, F will add one electron. Now, the F ion has 9+ protons and 10- electrons. The net results is an ionic charge of 1-.

Let’s add the ionic charges so we can see them. [Na]1+ + [F ]1- In this format, it seems that Na would give F the one electron it wants to get rid of...and F would take one electron from Na. In both cases, the valence shell of Na and F becomes full and stable.

So...these two elements will bond to form a new compound. [Na]1+ + [F ]1- NaF In words, Sodium + Fluorine Sodium fluoride

Now, what happens if magnesium (Mg) reacts with fluorine (F)? Mg is element #12. It resides in Column #2, and to become stable, Mg must give away two electrons. As an ion with 12+ protons and 10- electrons, the Mg ion has a charge of 2+. Writing the chemical formula [Mg]2+ + [F ]1- MgF2 In words, Magnesium + Fluorine Magnesium fluoride But why F2? One atom of Mg has two electrons to give away, but one atom of F can only take one electron. That means, I need two F atoms to take up the two available electrons. F2 tells me that there are two atoms of F in the compound MgF2.

Keep going... Let’s add sodium (Na) and oxygen (O). Na has one electron to give away to get a full valence shell of electrons. O is element #8. Element O has six electrons in its valence shell. To be stable, O wants to add two electrons Writing the chemical formula [Na]1+ + [O ]2- Na2O This tells us that you need two atoms of Na (...each giving away one electron...) and one atom of O (....one atom taking two electrons...) for a reaction to occur to form a new compound. In words, Sodium + Oxygen Sodium oxide

Add magnesium (Mg) and oxygen (O). Mg has two electron to give away to get a full valence shell of electrons. As you know, O wants to add two electrons to get a full valence shell. Writing the chemical formula [Mg]2+ + [O ]2- MgO Wait, it is MgO. Why is it not Mg2O2? If one atom of Mg has two electrons to give away AND one atom of O takes two electrons, I only need one Mg atom and one O atom for the ionic reaction to occur and a new compound to form. Thus, it is MgO In words, Magnesium + Oxygen Magnesium oxide

React aluminum (Al – element #13) and oxygen (O). Al has three electrons is its outer valence shell. If Al gives those three electrons away, its valence shell will be full. By giving three electrons away (3- charges), the ion Al has a charge of 3+. As you know, O wants to add two electrons to get a full valence shell. The ion O has a 2- charge. Writing the chemical formula [Al]3+ + [O ]2- Al2O3 Al2O3 tells me that the compound has two atoms of Al and three atoms of O. The fast way to get the correct chemical formula is crisscross the charges down [Al]3+ [O ]2- 3 [Al] [O ]2- Al2O3 In words, Aluminum + Oxygen Aluminum oxide

Practice writing chemical formulas Ca + Cl K + P Ca + S Li + S Li + Cl

Step #1 – Identify the ionic charge for each... [Ca]2+ + [Cl]1- [K]1+ + [P]3- [Ca]2+ + [S]2- [Li]1+ + [S]2- [Al]3+ + [Cl]1-

Step #2 – Bond the ions... [Ca]2+ + [Cl]1- CaCl [K]1+ + [P]3- KP [Ca]2+ + [S]2- CaS [Li]1+ + [S]2- LiS [Al]3+ + [Cl]1- AlCl

Step #3 – Identify the correct formula using the ionic charges and the crisscross process... [Ca]2+ + [Cl]1- CaCl2 [K]1+ + [P]3- K3P [Ca]2+ + [S]2- CaS [Li]1+ + [S]2- Li2S [Al]3+ + [Cl]1- AlCl3

Got this far....well done...you are figuring it out

Let’s try a Single Displacement Reaction. What? A reaction involving a compound and an element. K + HCl

Identify the ionic charge of each ion... [K]1+ + [H]1+[Cl]1- What will be displaced? According to the Law of Electric Charges, like charges repel while opposite charges attract. Logically, K1+ will displace the similarly charged H1+ and bond with the opposite charged Cl1-

Write the chemical formula... [K]1+ + [H]1+[Cl]1- KCl + H2 Why H2? Hydrogen is a gas that cannot exist alone. Hydrogen bonds with itself to form H2

Balance the chemical formula... The Law of Conservation of Mass states “Mass of Products must equal Mass of Reactants” [K]1+ + [H]1+[Cl]1- KCl + H2 Yet, the formula shows one atom of H in the reactants but two atoms of H in the products. This violates the Law of Conservation of Mass. We must have started with two atoms of H. This is written as... K+ 2[HCl] KCl + H2 Now, 2[HCl] tells you there are two atoms of H and two atoms of Cl. According to the law, I need two atoms of Cl in the products... K + 2HCl 2[KCl] + H2 Oops, 2[KCl] suggests 2 atoms of K 2K + 2HCl 2KCl + H2

A little fast....okay, let’s do it step-by-step, only slower and more methodical. Na2O + Mg

Using the Periodic Table, identify the ionic charges of each element Na is element #11. It is found in Column 1. To realize a stable electron configuration, Na will give away one electron. This gives Na a full outer valence shell. The Na ion will have 11+ protons and 10- electrons. It has one more + than – and, thus, the Na ion has a charge of 1+. Repeat for O. O is element #8. It is found two columns to the right of stability. Thus, O will take two electrons to establish electron stability. The O ion will have 8+ protons and 10- electrons. With two additional negative electrons, the O ion has a charge of 2-. Do Mg on your own. If you get a 2+ charge for the Mg ion, you get it [Na2]1+[O]2- + [Mg]2+

Since like-electrical charges repel (Law of Electric Charges), Mg will displace Na. On the product side, the formula is now... [Na2]1+[O]2- + [Mg]2+ [Na]1+ + [Mg]2+[O]2- To keep things apparent, I carried the ionic charges to the products of the reaction. Wait. What happened to Na2? The subscript 2 told us that the compound Na2O had two atoms of Na giving one electron each and one atom of O needing two electrons. Na-O-Na or more correctly Na2O

What next? Make sure the correct number of atoms for each new product are identified using the ionic charges and the crisscross rule. Na2O + Mg Na + MgO Okay...ahhh...hmmm...How did [Mg]2+[O]2- become MgO? Mg2+ means Mg needs to give away two electrons to become stable. Likewise, O2- means O needs to take two electrons to become stable. Thus, one atom of Mg giving two electrons will bond with one atom of O taking two electrons. Only one atom of each ion is needed. Thus, MgO

Next, the chemical equation must be balanced. That is, the number of atoms of each element of products must equal the number of atoms of each element of reactants. Why? Two reasons: (1) the Law of Conservation of Mass and (2) matter cannot be created or destroyed. If I start with two atoms of Na, there must be two Na atoms in the products. Likewise if there are two atoms of an element in the products, I must have started with two atoms of that element in the reactants. Na2O + Mg 2Na + MgO In the reactions, Na2 gives two atoms of Na. As well, there is one atom of O and one atom of Mg in the reactants. Looking at the products, there are two Na, one O and one Mg. Reactants = Products. This is a correctly balanced chemical equation.

Practice CaCl2 + Mg Li3N + Ca Na2S + Mg H2O + K

Use the Periodic Table to identify the Ionic Charges... [Ca]2+[Cl2]1- + [Mg]2+ [Li3]1+[N]3- + [Ca]2+ [Na2]1+[S]2- + [Mg]2+ [H2]1+[O]2- + [K]1+

Rearrange the formula, and remember, DO NOT bring the subscript charges to the products [Ca]2+[Cl2]1- + [Mg]2+ [Mg]2+[Cl]1- + [Ca]2+ [Li3]1+[N]3- + [Ca]2+ [Ca]2+ [N]3- + [Li]1+ [Na2]1+[S]2- + [Mg]2+ [Mg]2+ [S]2- + [Na]1+ [H2]1+[O]2- + [K]1+ [K]1+ [O]2- + [H2]1+

Make sure the right number of atoms are in the products using the crisscross rule CaCl2+ Mg MgCl2+ Ca Li3N + Ca Ca3N2 + Li Na2S + Mg MgS + Na H2O + K K2O + H2

Finally, balance (# Reactants = # Products) CaCl2+ Mg MgCl2+ Ca 2Li3N + 3Ca Ca3N2 + 6Li Na2S + Mg MgS + 2Na H2O + 2K K2O + H2 All looks good...except why 2Li3? 2 X 3 = 6. This makes six atoms of Li in the reactants which equals the number of Li in the products.

By the way, the names of the new compounds CaCl2+ Mg MgCl2+ Ca --- Magnesium chloride 2Li3N + 3Ca Ca3N2 + 6Li --- Calcium nitride Na2S + Mg MgS + 2Na --- Magnesium sulfide H2O + 2K K2O + H Potassium oxide The new compounds all end with the suffix ide

It doesn’t end here. What happens when two compounds react? CaCl2+ Na3P The same rules apply. 1. Use the Periodic Table to identify the ionic charge of each part of the reaction. [Ca]2+[Cl2]1- + [Na3]1+[P]3- 2. Rearrange the products so + and – charges join, bringing the charges across but NOT the subscripts. [Ca]2+[Cl2]1- + [Na3]1+[P]3- [Ca]2+[P]3- + [Na]1+[Cl]1- 3. Within the products, use the crisscross rule to identify the correct number of atoms in each new compound. CaCl2 + Na3P Ca3P2 + NaCl 4. Balance the equation so there are equal numbers of each atom on both sides of the equation (i.e., atoms of reactants = atoms of products) 3(CaCl2) + 2(Na3P) Ca3P2 + 6(NaCl) I inserted brackets around the compounds to show the PREFIX NUMBERS apply to the entire compound. 3(CaCl2) means there are three Ca atoms and 3X2 = 6 atoms of Cl. Check all compounds and atoms --- the formula is balanced

Let’s do another KCl + MgO The same rules apply. 1. Use the Periodic Table to identify the ionic charge of each part of the reaction. [K]1+[Cl]1- + [Mg]2+[O]2- 2. Rearrange the products so + and – charges join, bringing the charges across but NOT the subscripts. [K]1+[Cl]1- + [Mg]2+[O]2- [K]1+[O]2- + [Mg]2+[Cl]1- 3. Within the products, use the crisscross rule to identify the correct number of atoms in each new compound. KCl + MgO K2O + MgCl2 4. Balance the equation so there are equal numbers of each atom on both sides of the equation (i.e., atoms of reactants = atoms of products) 2(KCl) + MgO K2O + MgCl2 Too easy, eh?

Your turn.... K2S + MgCl2 KF + Na2O CaBr2 + Al2S3 K3P + Mg3N2 H2O + Ca3P2

The answers.... K2S + MgCl2 2KCl + MgS 2KF + Na2O K2O + 2NaF 3(CaBr2) + Al2S3 3(CaS) + 2(AlBr3) 2(K3P) + Mg3N2 2(K3N) + Mg3P2 3(H2O) + Ca3P2 3(CaO) + H3P If you got them all write...congrats. Some wrong...try again. Resilience and practice is all it is...

Done yet...ah, no. Next, we have to consider POLYATOMIC IONS. Say what? These are two elements that SHARE electrons. This is called a COVALENT BOND. For the moment, these are elements located on the right side of the Periodic Table. Thus, two negatively charged elements bonded. How do two negatives stay together...that breaks the rule that like-charges repel? The “partners” have almost the same ability or strength to attract electrons. Since they are nearly the same, they share electrons instead. Two things to note: The polyatomic ion will have a negative charge The polyatomic ion always stays together. For example, PO4 is a polyatomic. It is PO4 on both the reactant and the product side of the chemical formula

Note the Polyatomic Ions in the following compounds. H2SO4 – Hydrogen sulphate or Sulfuric acid KOH – Potassium hydroxide Ca3(PO4)2 – Calcium phosphate Mg(NO3)2 – Magnesium nitrate Cs2CO3 – Cesium carbonate LiNO2 --Lithium nitrite LiNO3 – Lithium nitrate NaClO3 – Sodium chlorate Mg(ClO2)2 – Magnesium chlorite You may observe that the suffix endings of the compound names end in either ITE or ATE. This occurs when there are two polyatomics with a different charge or bonding configuration of the “same” compound. For example, there is NO2 nitrite and NO3 nitrate. The lesser on is ITE and the larger one is ATE.

So, bonds away.... H2CO3 + Ca Identify the charges --- [H2]1+[SO4]2- + [Ca]2+ Rearrange the formula (+ attracts -). NOTE: H2 remains H2 because H does not exist by itself. [H2]1+[SO4]2- + [Ca]2 [H2]1+ + [Ca]2+[SO4]2- Add the correct charges to each new part H2SO4 + Ca H2 + CaSO4 Balance the equation (...equal numbers both sides). It is balanced. You should note that the polyatomic SO4 stayed together throughout the reaction.

Another polyatomic bonds away.... Cs2SO3 + Mg3(PO4)2 [Cs2]1+[SO3]2- + [Mg]2+[(PO4)2]3+ [Cs2]1+[SO3]2- + [Mg]2+[(PO4)2]3+ [Cs]1+ [(PO4)2]3+ + [Mg]2+[SO3]2- Cs2SO3 + Mg3(PO4)2 Cs3PO4 + MgSO3 3(Cs2SO3) + Mg3(PO4)2 2(Cs3PO4) + 3(MgSO3) Check by counting the atoms of each part on both sides the equation Reactants Products 3X2 = 6 Cs 2X3 = 6 Cs 3 X 1 = 3 SO3 3 X1 = 3 SO3 3 Mg 3X1 = 3 Mg 2 PO4 2X1 = 2 PO4

Another change...ions with more than one ionic charge Some ions have more than one charge. For example, copper is either Cu1+ or Cu2+, lead is Pb2+ or Pb4+ and iron is Fe2+ and Fe 3+ Which one do you use? The clue is in the written name of the compound. The compound name will include a ROMAN NUMERAL between the parts. NOTE: The ionic charge always seems to be positive. There are no multiple negative charges. So... Iron II sulphide is [Fe]2+[S]2- or FeS since same-charges cancel Iron III sulphide is [Fe]3+[S]2- or Fe2S3 Copper I nitride is [Cu]1+[N]3- or Cu3N Lead IV oxide is [Pb]4+[O]2- to become PbO2 since 4/2 =2 Lead II oxide is [Pb]2+[O]2- to become PbO Nickel III sulphate is [Ni]3+[SO4]2- to be Ni2(SO4)3 Titanium IV nitrite is [Ti]4+[NO2]1- to form Ti(NO2)4 Titanium III nitrite is [Ti]3+[NO2]1- or Ti(NO2)3

Finally...PREFIXES Some compounds have names such as... carbon dioxide -- CO2 trimagnesium diphosphide -- Mg3P2 carbon monoxide – CO disodium hydrogen phosphate -- Na2HPO4 The prefix indicates the number of atoms of that part of the formula Prefix Number of Atoms mono 1 di 2 tri 3 tetra 4 penta 5

All that is left...for now...is practice. Practice, resilience, questions and practice again.... Ca + Li2O Na3P + K2SO4 H2 + NO3 Cs2S + Mg Al + K2O KF + Sr Na2O + CaF2 K3PO4 + MgCl2 NaOH + H2SO4 AlF3 + Ca3(PO4)2

ANSWERS for FINAL PRACTICE Ca + Li2O CaO + 2Li 2(Na3P) + 3(K2SO4) 3(Na2SO4) + 2(K3P) H2 + NO3 H2NO3 Cs2S + Mg MgS + 2Cs 2Al + 3K2O Al2O3 + 6K 2(KF) + Sr SrF2 +2K Na2O + CaF2 2(NaF) + CaO 2(K3PO4) + 3(MgCl2) 6KCl + Mg3(PO4)2 2(NaOH) + H2SO4 Na2SO4 + 2(H2O) 2(AlF3) + Ca3(PO4)2 2(AlPO4) + 3(CaF2)

Balancing Chemical Equations – A Primer

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