Chemical Reactions in Aqueous
Solutions
1


How do we know whether a reaction occurs?
What observations indicate a reaction has
occurred?
In your groups, make a list of changes that
indicate a chemical reaction has occurred.
2
Ag+
Cl-
+
AgCl
Pb2+ + CrO42 PbCrO4
Cr3+ + 3OH Cr(OH)3
return
3
Cesium in a bathtub
Ba(OH)2.8H2O
+ NH4Cl
Brainiac Thermite movie
Thermite: Al + Fe2O3
return 4
2Al + 3Br2  2AlBr3
P4 + 5O2  P4O10
2Na + Cl2  2NaCl
return
5

(NH4)2Cr2O7(s)  Cr2O3(s) +
4H2O(g) + N2(g)
 CuSO4.5H2O
 2NI3
 CuSO4 + 5H2O(g)
 N2 + 3I2
Heat CuSO4
NI3
return
6
Na + 2H2O 
NaOH + H2(g)
Cu + 2H+ 
Cu2+ + H2(g)
Mg + 2HCl 
MgCl2 + H2
return
7
Cu + 2Ag+  Cu2+ + 2Ag
Zn + Sn2+ 
Zn2+ + Sn
return 8
Cesium in a bathtub
Exploding Whale
Building Demolition
return
9




Substances behave differently
when they are placed in water,
specifically ionic versus covalent compounds.
One breaks apart in water, the
other does not.
Which one is more likely to be pulled apart
by water molecules?
Electrolytes are ionic and strong acid
solutions (e.g., GatoradeTM); Nonelectrolytes
are covalent compounds (e.g., sugar); weak
electrolytes are in between.
Strong/Weak
Electrolytes



Strong electrolyte: substance that, when
dissolved in water, results in a solution that
can conduct electricity (NaCl)
Weak electrolyte: substance that is a poor
conductor of electricity when dissolved in
water (CH3COOH – vinegar)
Nonelectrolyte: substance that doesn’t
conduct electricity when dissolved in water
(CH3OH – methanol)
Strong/Weak Electrolytes

3 substances: A2X, A2Y, and A2Z. Which one
is the strongest electrolyte? The weakest?
(Water has been omitted for clarity.)


Most reactions in general chemistry take
place in an aqueous environment. What does
that mean?
Terms:
◦ Solution: homogeneous mixture of two or more
substances
◦ Solute: substance present in smaller amount
◦ Solvent: substance present in greater amount
◦ Aqueous solution: solvent is water

Three general categories:
◦ Precipitation: insoluble (solid) product is formed
from aqueous solutions
◦ Acid-base neutralization: acid and base react to
form water and a salt (ionic compound)
◦ Oxidation-Reduction: electrons are transferred
between atoms in reaction
 Combination
 Decomposition
 Single-replacement (metal or hydrogen)



Precipitation reactions always begin with two
ionic compounds.
Example: NaCl (aq) + AgNO3 (aq)  ?
Draw these compounds in two separate
aqueous environments. What are the
possible products when they are combined?
+

NO3Na+
Ag+
Ag+
NO3-
Na+
Cl-
NO3-
Ag+
Cl-
Na+
Cl-
Write formulas of products (based on
charges), predict phases (Solubility Rules),
and balance the equation.
If not covered by the rules, it is probably insoluble.
19
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

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



Determine if the following ionic compounds
will be soluble (aq) or insoluble (s) in water:
K2CO3
BaSO4
PbI2
NaClO4
Ag2S
(NH4)3PO4
Cu(OH)2








Determine if the following ionic compounds
will be soluble (aq) or insoluble (s) in water:
K2CO3
soluble (aq)
BaSO4
insoluble (s)
PbI2
insoluble (s)
NaClO4
soluble (aq)
Ag2S
insoluble (s)
(NH4)3PO4
soluble (aq)
Cu(OH)2
insoluble (s)

There are 3 ways to represent ppt reactions:
◦ As whole compounds (molecular equation)
◦ As ionic species (ionic equation) – more accurate
◦ As participants in reaction (net ionic equation)


Any aqueous ionic substance is written as a
compound (e.g., AgNO3), but this isn’t
accurate. What does this look like in water?
Solids, liquids, and gases remain as
compounds.
22

Molecular equation:

Ionic equation (write separate ions for soluble
(aq) compounds):
NaCl(aq) + AgNO3(aq)  AgCl(s) + NaNO3(aq)
Na+(aq) + Cl-(aq) + Ag+(aq) +NO3-(aq)  AgCl(s) +
Na+(aq) + NO3-(aq)


Net ionic equation (cancel any identical ion
on both sides of the equation, called
spectator ions): Ag+(aq) + Cl-(aq)  AgCl(s)
Note: s, l, and g stay together!!!!!
Ag+/Cl- ions
Chemistry
humor, ha ha!


Reaction of lead
(II) nitrate and
potassium iodide.
What is the
precipitate?
Write the molecular,
ionic, and net ionic
equations.

Acid: substance that breaks apart in water to
form H+ (e.g., HCl, HNO3, CH3COOH, lemon, lime,
vitamin C).
◦ HA(aq)  H+(aq) + A-(aq)

Base: substance that breaks apart in water to
form OH- (e.g., NH3, DranoTM, Milk of
MagnesiaTM)
◦ MOH(aq)  M+(aq) + OH-(aq)
acid
strong acid
base
weak acid
26



You need to KNOW these!!!
Strong acids: HCl, HBr, HI, HClO4, HClO3, H2SO4, HNO3
Strong bases: LiOH, KOH, NaOH, RbOH, CsOH,
Ca(OH)2, Sr(OH)2, Ba(OH)2
27


What does “neutralization” mean?
How might these two solutions react when
combined?
Na+
OHNa+
+
Cl-
H+
OH-
H+
Cl-




Neutralization reaction: reaction between
acid and base; products are salt (ionic
compound) and water
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
Acid + base  salt + water
What are the ionic and net ionic equations
for these reactions?


HF(aq) + NaOH(aq)

KOH (aq) + H2SO4 (aq)

HCl
Acid + Base → H2O + Salt
Mg(OH)2
Water is created from H+
and OH-
2HCl(aq) + Mg(OH)2(s) → 2H2O(l) + MgCl2(aq)
The salt is created from
spectator ions


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
Complete and balance these equations. Write
ionic and net ionic equations, if applicable.
Na2S(aq) + CuCl2(aq) 
MgSO4(aq) + BaCl2(aq) 
KNO3(aq) + CaCl2(aq) 
CuSO4(aq) + NaOH(aq) 
HF(aq) + NaOH(aq) 
KOH(aq) + H2SO4(aq) 
31
If not covered by the rules, it is probably insoluble.
32

Na2S(aq) + CuCl2(aq)  2 NaCl(aq) + CuS(s)
◦ Cu2+(aq) + S2-(aq)  CuS(s)

MgSO4(aq) + BaCl2(aq)  MgCl2(aq) + BaSO4(s)
◦ Ba2+(aq) + SO42-(aq)  BaSO4(s)

2 KNO3(aq) + CaCl2(aq)  2 KCl(aq) +
Ca(NO3)2(aq)
◦ No reaction

CuSO4(aq) + 2 NaOH(aq) Cu(OH)2(s) +
Na2SO4(aq)
◦ Cu2+(aq) + 2 OH-(aq)  Cu(OH)2(s)

HF(aq) + NaOH(aq) H2O (l) + NaF (aq)
◦ H+ (aq) + OH- (aq)  H2O (l)

2KOH(aq) + H2SO4(aq)  2H2O (l) + K2SO4 (aq)
◦ H+ (aq) + OH- (aq)  H2O (l)
33

Determine the products of the reaction.
Identify the phase of each compound, and
balance the equation. Also write the ionic
and net ionic equations.
Molecular: Na2S + Cr(NO3)3 

Complete Ionic:

Net Ionic:

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

Oxidation-Reduction (redox)
reactions: electron-transfer reactions
When iron rusts, it loses electrons to form a
cation, oxygen gain electrons to form an
anion: 4 Fe(s) + 3 O2(g)  2 Fe2O3(s)
Use oxidation number rules to determine gain
and loss of electrons.
Oxidation numbers are assigned as if
elements in compounds completely
transferred electrons (like in ionic
compounds).




1) An atom (or molecule) in its elemental
state has an oxidation number of 0.
2) An atom in a monatomic ion (Na+, Cl-) has
an oxidation number identical to its charge.
3a) Hydrogen has an oxidation number of +1,
unless it is combined with a metal, in which
case it has an oxidation number of –1.
3b) Oxygen usually has an oxidation number
of -2. Oxygen in peroxides (O22-) has an
oxidation number of -1.
36


3c) Halogens usually have an oxidation
number of -1 (except when bonded to
oxygen or in polyatomic ions).
4) The sum of oxidation numbers is 0 for a
neutral compound and is equal to the net
charge for a polyatomic ion.
(Example: NaCl = 0, SO42- = -2)
◦ 4a) For binary ionic compounds, the position of the
element in the periodic table may be useful:
◦ Group IA: +1; Group IIA: +2; Group VIIA: –1;
Group VIA: –2; Group VA: –3
37

H2SO4
 H = +1; O = –2
 S is unknown, so leave this for last.
◦ The overall charge on this compound is 0.
◦ Use algebra to solve for S:
◦ 2(+1) + 1(x) + 4(-2) = 0

Solve for each element: MgCr2O7
38

Determine values of the oxidation number of
each element in these compounds or ions:
H2O
SO2
CCl4
H2O2
Fe3(PO4)2
MnO4NaNO3
KClO4
39

Determine values of the oxidation number
of each element in these compounds or
ions:
H2O
H: +1, O: -2
SO2
S: +4, O: -2
CCl4
C: +4, Cl: -1
H2O2
H: +1, O: -1
Fe3(PO4)2
Fe: +2, P: +5, O: -2
MnO4- Mn: +7, O: -2
NaNO3 Na:+1, N:+5, O: -2
KClO4 K: +1, Cl: +7, O: -2
40



Oxidized: atom, molecule, or ion becomes
more positively charged
◦ Loss of electrons is oxidation (LEO)
Reduced: atom, molecule, or ion becomes
less positively charged (reduced charge)
◦ Gain of electrons is reduction (GER)
Or: OIL RIG (oxidation is loss;
reduction is gain)
Redox 1
Redox 2



The substance oxidized causes the other
substance to be reduced and is called the
reducing agent.
The substance reduced causes the other
substance to be oxidized and is called the
oxidizing agent.
4 Fe(s) + 3O2(g)  2Fe2O3(s)


Identify the element or ion oxidized/reduced.
Also identify the oxidizing agent and the
reducing agent.
Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)
Reduced
Oxidized





Identify the oxidation number of each
element in the compounds or ions below:
Ba(ClO3)2
SO32For the reaction below, identify what has
been oxidized and reduced; identify the
oxidizing agent and the reducing agent.
Cu(s) + 2AgNO3(aq)  2Ag(s) + Cu(NO3)2(aq)

Combination (1 product)
◦ Na(s) + Cl2(g) 

Decomposition (1 reactant) – usually give off
gases
◦ CuCO3(s) 

Single Replacement (or Displacement) (start
and end with an element and a compound)
◦ Zn(s) + HCl(aq) 
46

Metals tend to react with water to form bases:
◦ 2Na (s) + 2H2O  2NaOH + H2
◦ MgO (s) + H2O  Mg(OH)2

metals
Nonmetals tend to react with water to Alkali
form
+ water
acids:
◦ 2F2 (g) + 2H2O  4HF + O2
◦ CO2 (g) + H2O  H2CO3
CO2 (s) + H2O (l)
 H2CO3 (aq)

element + element
◦

metal + nonmetal
◦

Na(s) + Cl2(g)
 ionic compound

nonmetal + nonmetal
compound
◦

H2(g) + O2(g) 
 compound
C(s) + O2(g)
 covalent

Why are these redox reactions?
49

Mg + O2
K + Cl2


Compound  2 elements; element +
compound; or 2 compounds
Oxides, peroxides
◦ Give off O2

Nitrates
◦ Give off NO2 , NO2-

Carbonates
◦ Give off CO2

Ammonium salts
◦ Give off NH3
51




NH4Cl (s) 

NiCO3(s) 
CuO(s) 






element + compound  compound +
element
(The more metallic/active element in the
compound is displaced.)

Metal Displacement

Hydrogen Displacement
◦ Zn(s) + Cu(NO3)2(aq)  Cu(s) + Zn(NO3)2(aq)
◦ Mg(s) + HCl(aq)  MgCl2(aq) + H2(g)
Alkali Metals
Zn+SnCl2
53
54


The higher
the metal on
the activity
series, the
more active
that metal.
Translation:
higher
metals on
the chart will
form ions as
products.
56
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




Na + H2O
Fe + H2O
Fe + Cr(NO3)2
Ni + Pb2+
Ag + Mg2+
Zn + Co2+
React w/ cold
water
React w/ steam
React w/ acid
57
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



Four metals (A, B, C, and D) are tested and
the following observations are recorded.
A + B+  no reaction
C + B+  reaction occurs
D + A+  reaction occurs
D + B+  no reaction




Burning hydrocarbons
CxHyOz + O2 (g)  CO2 (g) + H2O (g)
Methanol, CH3OH
_CH3OH (l) + _O2 (g)  _CO2 (g) + _H2O (g)






Precipitation: use Solubility Rules
(AB + CD  AD + CB)
Acid-Base Neutralization: acid + base 
salt + water
(AB + CD  AD + H2O)
Combination: start with elements (A + B 
AB)
Decomposition: often produces gas
(AB  A + B)
Single Displacement: use Activity Series to
predict if a reaction occurs (A + BC  B +
AC)
Combustion:
◦ Hydrocarbon + O2(g)  CO2(g) + H2O(g)
+
Na Na Na
Na Na Na
Cl Cl
+
ClH+
ClH+
ClH+
Cl-
H+
Ca Ca Ca
Ca Ca Ca Ca
+
OHCl-
H+
ClH+
K+
OH-
K+
1.
2.
3.
4.
5.
6.
7.
Co(s) + AgNO3(aq) 
Fe(s) + HCl(aq) 

Na2CO3(s) 
Ca(s) + H2O(l) 
CaCO3(s) 


HClO4(aq) + KOH (aq) 
BaCl2(aq) + Na2SO4(aq) 
64
8. HgO(s)



9. LiOH(aq) +
H2SO4(aq) 
10. Na2CrO4(aq) + Ni(NO3)2(aq) 
11. Li(s) + O2(g) 
12. Mg(OH)2(aq) + 2HCl(aq) 
13. NH3(g) + HCl(g) 
14. NiCO3(s)
15. Ca(s) + F2(g) 


Single disp. 1. Co(s) + 2AgNO3(aq)  Co(NO3)2(aq) + 2Ag(s)
Single disp. 2. Fe(s) + 2HCl(aq)  FeCl2(aq) + H2(g)
Decomp.
3. Na2CO3(s)  Na2O(s) + CO2(g)
Single disp. 4. Ca(s) + 2H2O(l)  Ca(OH)2(aq) + H2(g)
Decomp.
5. CaCO3(s) + heat  CaO (s) + CO2(g)
Acid-base neut. 6. HClO4(aq) + KOH(aq)  KClO4(aq) + H2O(l)
Precip.
7. BaCl2(aq) + Na2SO4(aq)  BaSO4(s) + 2NaCl(aq)
Decomp. 8. 2HgO(s)  2Hg(l) + O2(g)
Acid-base neut. 9. 2LiOH(aq) + H2SO4(aq)  Li2SO4(aq) + 2H2O(l)
Precip. 10. Na2CrO4(aq) + Ni(NO3)2(aq)  2NaNO3(aq) + NiCrO4(s)
Combo. 11. 4Li(s) + O2(g)  2Li2O(s)
Acid-base neut. 12. Mg(OH)2(aq) + 2HCl(aq)  2MgCl2(aq) + H2O(l)
Combo. 13. NH3(g) + HCl(g)  NH4Cl(s)
Decomp. 14. NiCO3(s)  NiO(s) + CO2(g)
Combo. 15. Ca(s) + F2(g)  CaF2(g)





Predict products of the following reactions.
Write correct phases for the products and
balance each equation:
___ Al (s) + ___ NaNO3 (aq) 
___ Na (s) + ___ O2 (g) 
___ Na2SO4 (aq) + ___ Pb(NO3)2 (aq) 

Sections 8.4-8.5


We use the balanced
chemical equation to
determine relative
amounts of reactants
and products.
How can we predict
the amount of
product we should
get?
69




Mass to mass conversions
Given mass of reactant, find mass of precipitate:
MgCl2 + 2 AgNO3 2 AgCl + Mg(NO3)2
1.2482 g MgCl2 and excess AgNO3
◦ Convert grams of MgCl2 to moles using molar mass
◦ Convert moles of MgCl2 to moles of AgCl using mol-mol
ratio
◦ Convert moles of AgCl to grams using molar mass




NaOH
+
10.54 g
Al(NO3)3  Al(OH)3 + NaNO3
xs
?
How many grams of aluminum hydroxide can be
made by starting with 10.54 grams of sodium
hydroxide? (Hint: Is the equation balanced?)
The mass calculated is the theoretical yield (the
most that can theoretically be made in lab).



If you ran that reaction in lab and only
obtained
4.9548 g (actual yield) of Al(OH)3, what
percent of the maximum did you obtain?
Percent yield = Actual / Theoretical x 100%
Theoretical will be calculated; actual yield will
be given in a problem (or obtained in lab)




Analogy: making martinis
I have 7 martini glasses, 36 olives, 12 shots
of vodka, and 8 stirrers
Each martini requires 1 glass, 2 olives, 2
shots of vodka, and 1 stirrer
How many martinis can I make?
Limiting Reactant
73
74

How would you define limiting reagent?
75
2H2 + O2  2H2O
What is the limiting reactant?
76
2H2 + O2  2H2O
What is the limiting reactant?
2H2 + O2  2H2O
What is the limiting reactant?
78
2H2 + O2  2H2O
What is the limiting reactant?
79
2H2 + O2  2H2O
What is the limiting reactant?
80
2H2 + O2  2H2O
What is the limiting reactant?
81







Al(s) + Cl2(g)  AlCl3(s)
Begin with 1.67 g Al (s) and 1.67 g Cl2 (g).
Calculate mass of product that can be made from
each reactant?
1) Which reactant produces less product?
2) How many grams of AlCl3(s) will be produced,
theoretically?
3) What is the percent yield if you obtained 1.76 g
AlCl3 in the lab?
4) How many grams of the excess reagent will be
left over?

Hydrogen and chlorine react to yield
hydrogen chloride. How many grams of HCl
are formed from reaction of 3.56 g of
hydrogen and 8.94 grams of chlorine?

Sections 9.5-9.6

Consider the reaction:
◦ 3CaCl2(aq) + 2Na3PO4(aq)
6NaCl(aq)

 Ca3(PO4)2(s) +
If we mix 25.0 mL of 0.200 M CaCl2
solution with 50.0 mL of 0.250 M Na3PO4
solution, what mass of precipitate is
formed?
85




Complete the equation below (with phases)
and balance it.
K2S(aq) + AgNO3(g) 
If you combine 10.21 mL of 0.152 M K2S
with 1.0092 g AgNO3, how much solid
product can be formed?
If 0.2744 g of product were actually formed,
what is the percent yield?

Titration: method of determining the
concentration (or volume) of an unknown
solution by using a solution with a known
concentration (standard)
◦ moles = M * V


Equivalence point: point at which acid has
completely reacted with and been
neutralized by base (moles acid = moles
base)
End point: point at which the indicator
changes color (slight change in solution’s
pH)





Titrant in buret (usually)
Unknown in flask (usually)
Can’t use dilution calculation!
Need mol-mol ratio of acid-base
Solution stoichiometry: use molarity
to convert between volume and
moles (instead of molar mass)
◦ M = mol / L


23.78 mL of 0.2500 M NaOH neutralized 20.00
mL of HCl. What is the concentration of HCl?
Start with a balanced equation.
Worked Ex. 3.14, Problems 3.20, 3.21, 3.22

How many mL of 1.018 M H2SO4 are needed
to neutralize 20.00 mL of 0.9989 M NaOH?
◦ Write balanced equation
◦ Determine number of moles of known
◦ Calculate mole ratio to determine moles of
unknown
◦ Divide by concentration to find volume

23.48 mL of lithium hydroxide are
required to neutralized 15.39 mL of
1.20 M phosphoric acid. What is the
concentration of base?




Concentration: amount of solute present in
a given amount of solution
Molarity: moles of solute in 1 L of solution
◦ moles solute / liters solution; M = mol / L
1.00 M NaCl = 1.00 mol NaCl in 1 L solution
Knowing concentration, volume, and/or
mass allows us to calculate:
◦ 1) mass of solid needed to make a solution of a
certain concentration, or
◦ 2) concentration of solution if mass and volume
are known.
92
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


1.00 M NaCl = 1.00 mol NaCl in 1 L solution
0.500 M NaCl = 0.500 mol NaCl in 1 L sol’n.
To find moles of solute: M * V = mol
How many moles in 5.67 mL of 0.500 M
NaCl?
What volume is needed to make a 0.15 M
solution using 5.67 mol of solute?
93
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

What is the concentration of Cl- ions in a
0.150 M FeCl3 solution?
1 mol FeCl3  1 mol Fe3+ + 3 mol Cl0.150 M FeCl3 x (3 mol Cl- / 1 mol FeCl3) =
0.450 M Cl-

We can calculate concentration (molarity) if
we know mass and volume.
◦ Mass solute ÷ molar mass = moles solute
◦ Moles solute ÷ volume solution = Molarity

Or we can calculate the mass of solute
needed to make a solution of certain
concentration (molarity).
◦ Molarity * volume solution = moles solute
◦ Moles solute * molar mass = mass solute needed



What is the concentration of a 1.00 L solution made
by dissolving 5.00 g NaCl in water?
5.00 grams NaCl in a 500.0 mL solution?
What mass of NaCl do we need to make 1.00 L of a
0.500 M solution?

Dilution: lowering the concentration of
solution (by adding water)
◦ number of moles of solute stays the same,
amount of solvent increases

The number of moles of solute stays the
same; volume increases (lower molarity)
◦ moles = M * V
◦ Mconc * Vconc = Mdil * Vdil

or
M1 V 1 = M 2 V 2
Example: What is the final concentration of
solution made by adding 100.0 mL of 3.00
M HCl to 300.0 mL of DI water?
97

How would you prepare 500.0 mL of 0.500 M
H2SO4 starting with concentrated (18.0 M)
solution? (Hint: What variable are you solving
for?)
◦ Remember to add acid to water.


How much water do you need to add to 25.0
mL of a 4.50 M NaOH solution to make a
1.00 M NaOH solution?
1) 13.9 mL of 18.0 M H2SO4 2) 87.5 mL of
water added to 25.0 mL of 4.50 M NaOH

Sections 5.8-5.10


Percent by mass: percent by mass of each
element in a compound
H2O2
◦
◦
◦
◦
◦

2 moles H, 2 moles O
H = 2 mol x 1.008 g/mol H = 2.016 g H
O = 2 mol x 16.00 g/mol O = 32.00 g O
H2O2 = 2.016 g H + 32.00 g O = 34.02 g
%H = 2.016 g H / 34.02 g H2O2 x 100% = 5.926%
What is %O? Solve 2 ways…..




What is the percent by mass of Cr, S, and O in
Cr2(SO4)3?
Cr: (104 / 392.21) * 100 = 26.52%
S: (96.21 / 392.21) *100 = 24.53%
O: (192 / 392.21) * 100 = 48.95%


We can reverse this calculation.
Know percentages, need to calculate
empirical formulas.

Calculate the empirical formulas:
◦ 50% S, 50% O
◦ 43.64% P, 56.36% O

Assume 100 g of the material:
◦
◦
◦
◦

50 g O x 1 mol / 16.0 g = 3.125 mol O
50 g S x 1 mol / 32.066 g = 1.559 mol S
mol O / mol S = 3.125 / 1.559 = 2.004 or 2
SO2
Assume 100 g of material:
◦ 43.64 g P  mol P (1.4089 mol P)
◦ 56.36 g O  mol O (3.5225 mol O)
◦ Ratio of two moles (2.5 O / 1 P); need whole numbers
103
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
We just found the empirical formula to be
P2O5. This compound has a molecular mass
of 141.9 g/mol.
We know the molecular formula has a molar
mass of 283.8 g/mol. How many multiples of
P2O5 do we need to reach a mass of 283.8?

A compound contains 30.4 % nitrogen and
69.6 % oxygen. The molecular mass of the
compound is 92 g/mol.
◦ What is the empirical formula of the compound?
◦ What is the molecular formula of the compound?
105
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
A compound contains 40.00% C, 6.71% H,
and the rest is O. Determine the empirical
formula of this compound.
If the molar mass of the substance is 180.16
g/mol, determine the molecular formula of
this substance.