Chemical Reactions (chemical changes): Ch 11.
I.
Name: ___________________________________________
Chemical reactions are changes in which new substances are produced. Important vocabulary: The word “substance” refers
always to either an element or a compound”.
A.
During chemical reactions, bonds between atoms are broken:
i. covalent bonds (connecting nonmetal atoms to nonmetal atoms) are broken,
ii. metallic bonds (connecting metal atoms to metal atoms) are broken,
iii. and/or ionic bonds (connecting metal ions to nonmetal ions) are broken.
B.
During a chemical reaction, the freed-up atoms and/or ions scramble around, and then connect to different atoms or ions,
making new substances. In this unit of study, you will recognize when a new substance has been made by seeing that a
different chemical formula is being used.
C. Sometimes the new substances produced are in a different state (and sometimes not). Observing a change of state (solid to
liquid to gas, and vice versa) does not guarantee that a new substance has been produced, since, (for example) solid water,
liquid water, and water vapor are all the same substance. But sometimes, in the process of producing a new substance, that
substance will be in a different state than that of the original chemicals.
D. Atoms are not created nor destroyed during chemical reactions.
i. Since atoms have mass (although each different atom has a different mass): mass is not created nor destroyed (not gained
nor lost) during chemical reactions, thus we say that mass is conserved. Both the identity of each atom, how many of each
atom, and the mass of each atom is conserved during a chemical change. This fact is the “law of conservation of mass”.
E. The way we write about a particular chemical change is to write an “equation” with chemical formulas for the original
substances and the final substances. This is called a “chemical equation”! When 2 chemicals are combined, we place a
“+” between them; when 2 new chemicals are made, we separate them from the original chemicals by using an arrow, and
then we place a “+” between the 2 new chemicals. Ex: A + B  C + D
When we go to write about a particular chemical change for real, we won’t be using “A, B, C, or D”; instead, we must be sure
that the chemical formulas of our original substances are correct, and we must be sure that the chemical formulas of our
new substances are also correct. Additionally, we must account for every atom, making sure that an atom that wasn’t
present originally hasn’t miraculously appeared from nowhere.
For Ex: HNO3 + NaHCO3  NaNO3 + H2O + CO2
Count all the atoms to the left of the arrow, and fill in the blanks:
H = _____, N = _____, O = _____, Na = ____, C = ____
Now count all the atoms to the right of the arrow, and fill in the blanks:
H = _____, N = _____, O = _____, Na = ____, C = ____
They should be equal, because, for mass to be conserved, the number of each atom must be conserved, and the identity of each
atom must be conserved.
Chemical equations and “balancing”: A chemical equation is, in a way, a recipe telling what the ingredients (called
“reactants”) are that should be mixed together to make specific “products”. But, when we start thinking about the
amounts of ingredients (reactants) to use, and the amounts of products we hope to make, we need to include whole
numbers called “coefficients”. When we use coefficients correctly, we show our understanding of conservation of mass.
II.
i.
ii.
i.
ii.
____Al
iii.
The coefficient is a “multiplier”. It doubles, triples, quadruples (etc.) the amount of the substance it is placed in front of.
We insert coefficients in front of one or two substances at a time, NOT in front of all of the substances at once, in order
to “balance” the number of atoms of each element, making the number of atoms of each atom found in our ingredients
equivalent to the number of atoms of that element found in our products. The chemical equation is “balanced” when the
number of atoms of each element found on the reactant (left) side of the arrow is equal to the number of atoms of that
same element on the product (right) side of the arrow. Some chemical equations are balanced using no coefficients at all
(as in the example I gave you above)! But, most need coefficients. An unbalanced equation, giving the correct identity
of substances, but no amounts, is called a “skeleton equation”. Ex: Na + Cl2  NaCl
The coefficients used for an equation should be the smallest combination of whole numbers possible to achieve
balancing of the atoms. Ex: 2 Na + 1 Cl2  2 NaCl (2:1:2) is used rather than (4:2:4) or (6:3:6) etc.
B. Strategies or tips for balancing chemical equations. You should memorize these asap!
Pick an element that appears in only 1 reactant and 1 product to start with.
Determine the LCM for the element you have picked. Ex (below) – let’s pick fluorine.
The subscripts are 2 and 3; the least common multiple of 2 and 3 is 6. Remember, a coefficient is a multiplier, so,
let’s place a coefficient of 3 in front of the F2 and a coefficient of 2 in front of AlF3, so as to now have 6 atoms of F
both on the left and on the right of the arrow….Remember, it is the arrow that separates the original substances (the
reactants) from the newly produced substances (the products).
+ ____F2  ____AlF3
becomes
____Al
+
3 F2  __2_ AlF3
because (3 x 2 = 6 and 2 x 3 = 6)
After balancing one element (such as fluorine), pick a second element to work with that appears in the same compound as
the already-balanced element. In this example, the compound with an already-balanced element is AlF3, and the next
element to work with would therefore be Al. So, with “2 AlF 3” already on the right, put a 2 in front of the Al on the left.
___2__Al
+_
3 F2 
__2_ AlF3
Atom inventory:
reactants
2Al
6F
products
2Al
6F
iv. Balance these two atoms last in a combustion reaction: hydrogen and oxygen. I usually do H before O. The reason is, oxygen
atoms occur in 2 products. You will need to add oxygen atoms as well as multiply in order to correctly balance the oxygens.
Ex:
_____ CH4 + _____ O2  _____ CO2 + _____H2O
Balance the “C” first:
__1__ CH4 + _____ O2  __1__ CO2 + _____H2O
Balance the “H” next:
__1__ CH4 + _____ O2  __1__ CO2 + __2__H2O
Notice, to balance the “O”, you must now add the 2 oxygens from CO2 to the 2 oxygens from “2 H2O”.
2 + 2 = 4 oxygen atoms are on the right side. Last, put a “2” in front of O2 on the left side, in order to balance the 4 oxygens:
__1__ CH4 + __2__ O2  __1__ CO2 + __2__H2O
v.
If you are having difficulty with a combustion problem, double a coefficient and re-work the problem based on the assumption that
the new double coefficient is correct. This corrects problems with needing a “half” coefficient, which occur during combustion
reactions in which there are an even number of carbon atoms. Ex: 2 ½ can’t be a coefficient; but, 5 can be.
vi. As long as you have a polyatomic ion both to the left and to the right of the arrow, keep the ion together (don’t work with it as
individual separate atoms). Ex: in Al(NO3)3, work it as 3 of (NO3)
vii.
Re-write H2O as HOH whenever (OH) is present on the opposite side of the arrow.