Transcript 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
Balancing Chemical Equations – A Primer
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?
Balancing Chemical Equations – A Primer
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...
Balancing Chemical Equations – A Primer
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 11electrons 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-.
Balancing Chemical Equations – A Primer
Let’s add the ionic charges so we can see them.
[Na]1+ + [F ]1In 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.
Balancing Chemical Equations – A Primer
So...these two elements will bond to form a new compound.
[Na]1+ + [F ]1In words,
Sodium + Fluorine
NaF
Sodium fluoride
Balancing Chemical Equations – A Primer
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.
Balancing Chemical Equations – A Primer
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
Balancing Chemical Equations – A Primer
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
Balancing Chemical Equations – A Primer
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 (3charges), 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+
[Al]3+
[O ]2- 3
[O ]2-
2
Al2O3
In words,
Aluminum + Oxygen
Aluminum oxide
Balancing Chemical Equations – A Primer
Practice writing chemical formulas
Ca + Cl
K+P
Ca + S
Li + S
Li + Cl
Balancing Chemical Equations – A Primer
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-
Balancing Chemical Equations – A Primer
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
Balancing Chemical Equations – A Primer
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
Balancing Chemical Equations – A Primer
Got this far....well done...you are figuring it out
Balancing Chemical Equations – A Primer
Let’s try a Single Displacement Reaction.
What? A reaction involving a compound and an
element.
K + HCl
Balancing Chemical Equations – A Primer
Identify the ionic charge of each ion...
[K]1+ + [H]1+[Cl]1What 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-
Balancing Chemical Equations – A Primer
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
Balancing Chemical Equations – A Primer
Balance the chemical formula...
The Law of Conservation of Mass states “Mass of Products must equal
Mass of Reactants”
[K]1+ + [H]1+[Cl]1KCl + 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
Balancing Chemical Equations – A Primer
A little fast....okay, let’s do it step-by-step, only slower and
more methodical.
Na2O + Mg
Balancing Chemical Equations – A Primer
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+
Balancing Chemical Equations – A Primer
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
Balancing Chemical Equations – A Primer
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, O2means 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
Balancing Chemical Equations – A Primer
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.
Balancing Chemical Equations – A Primer
Practice
CaCl2 + Mg
Li3N + Ca
Na2S + Mg
H 2O + K
Balancing Chemical Equations – A Primer
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+
Balancing Chemical Equations – A Primer
Rearrange the formula, and remember, DO NOT bring the subscript charges to the
products
[Ca]2+[Cl2]1- + [Mg]2+
[Li3]1+[N]3- + [Ca]2+
[Na2]1+[S]2- + [Mg]2+
[H2]1+[O]2- + [K]1+
[Mg]2+[Cl]1- + [Ca]2+
[Ca]2+ [N]3- + [Li]1+
[Mg]2+ [S]2- + [Na]1+
[K]1+ [O]2- + [H2]1+
Balancing Chemical Equations – A Primer
Make sure the right number of atoms are in the products using the crisscross rule
CaCl2+ Mg
Li3N + Ca
MgCl2+ Ca
Ca3N2 + Li
Na2S + Mg
MgS + Na
H2O + K
K2O + H2
Balancing Chemical Equations – A Primer
Finally, balance (# Reactants = # Products)
CaCl2+ Mg
2Li3N + 3Ca
Na2S + Mg
H2O + 2K
MgCl2+ Ca
Ca3N2 + 6Li
MgS + 2Na
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.
Balancing Chemical Equations – A Primer
By the way, the names of the new compounds
CaCl2+ Mg
2Li3N + 3Ca
Na2S + Mg
H2O + 2K
MgCl2+ Ca --- Magnesium chloride
Ca3N2 + 6Li --- Calcium nitride
MgS + 2Na --- Magnesium sulfide
K2O + H2 --- Potassium oxide
The new compounds all end with the suffix ide
Balancing Chemical Equations – A Primer
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]32. 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]13. 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
Balancing Chemical Equations – A Primer
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]22. 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]13. 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?
Balancing Chemical Equations – A Primer
Your turn....
K2S + MgCl2
KF + Na2O
CaBr2 + Al2S3
K3P + Mg3N2
H2O + Ca3P2
Balancing Chemical Equations – A Primer
The answers....
K2S + MgCl2
2KCl + MgS
2KF + Na2O
K2O + 2NaF
3(CaBr2) + Al2S3
2(K3P) + Mg3N2
3(H2O) + Ca3P2
3(CaS) + 2(AlBr3)
2(K3N) + Mg3P2
3(CaO) + H3P
If you got them all write...congrats. Some wrong...try again. Resilience
and practice is all it is...
Balancing Chemical Equations – A Primer
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:
1.
The polyatomic ion will have a negative charge
2.
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
Balancing Chemical Equations – A Primer
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.
Balancing Chemical Equations – A Primer
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.
H2SO4 + Ca
H2 + CaSO4
You should note that the polyatomic SO4 stayed together throughout the reaction.
Balancing Chemical Equations – A Primer
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+
Cs2SO3 + Mg3(PO4)2
3(Cs2SO3) + Mg3(PO4)2
[Cs]1+ [(PO4)2]3+ + [Mg]2+[SO3]2Cs3PO4 + MgSO3
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
Balancing Chemical Equations – A Primer
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
Balancing Chemical Equations – A Primer
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
Balancing Chemical Equations – A Primer
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
Balancing Chemical Equations – A Primer
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)