# Chapter 4 - UCF Chemistry

```Chapter 4
Chemical Equations and
Stoichiometry
Chapter goals
• Balance equations for simple chemical
reactions.
• Perform stoichiometry calculations
using balanced chemical equations.
• Understand the meaning of a limiting
reagent.
• Calculate the theoretical and percent
yields of a chemical reaction.
• Use stoichiometry to analyze a mixture
of compounds or to determine the
formula of a compound.
Chemical Equations
• shorthand notation for chemical
change
• based on the law of conservation of
matter (A. Lavoisier, 18th century): valid
for ordinary reactions. For extraordinary
reactions, such as nuclear reactions, the
more general law of conservation of
energy holds
Symbolism
• starting materials (reactants) on left
• products on right
• reactants separated from products by 
or 

• reactants separated from each other with
+, same for products
  heat
• h light
• balanced by both mass and charge by
using coefficients
• use correct molecular formulas
Symbolism (contd…)
• trailing subscripts
(g) gas
(l) liquid
(s) solid
(aq) aqueous solution
• catalysts, special solvents, special
conditions written above and/or
below arrow,  heat, h light
Balancing Chemical Equations
• done by inspection
• usually best to begin with largest
molecule
• for reactions of organic compounds,
such as combustion reactions, balance C
atoms first. Secondly balance H atoms.
Then, balance O atoms
• convert any fractional coefficient to
whole number by multiplying all equation
times the denominator of fraction
Example, Balance
Al(s) + O2(g)  Al2O3(s)
Balancing Al
2 Al(s) + O2(g)  Al2O3(s)
Balancing O
2 Al(s) + 1.5 O2(g)  Al2O3(s)
fractional coefficients not allowed
then, multiply  2 (2  1.5 = 3)
4 Al(s) + 3 O2(g)  2 Al2O3(s)
Example, Balance
NH3(g) + O2(g)  NO2(g) + H2O(l)
Balancing H
2 NH3(g) + O2(g)  NO2(g) + 3 H2O(l)
Balancing N
2 NH3(g) + O2(g)  2 NO2(g) + 3 H2O(l)
3.5  2 = 7 O  (4 + 3) O
Balancing O
2 NH3(g) + 3.5 O2(g)  2 NO2(g) + 3 H2O(l)
4 NH3(g) + 7 O2(g)  4 NO2(g) + 6 H2O(l)
Types of Reactions
• combination
A + B  C
• decomposition
A  B + C
• combustion
A + O2  products
Combustion Reactions
• reaction with molecular oxygen
• of hydrocarbons produces
 carbon dioxide and water
e.g.. CH4(g) + O2(g)  CO2(g) + H2O(l)
(methane)
CH4(g) + 2 O2(g)  CO2(g) + 2H2O(l)
Also: 2 H2(g) + O2(g)  2 H2O (l)
Combustion Reactions
C5H12(g) + O2(g)  CO2(g) + H2O(l)
Balance C atoms
C5H12(g) + O2(g)  5CO2(g) + H2O(l)
Balance H atoms
C5H12(g) + O2(g)  5CO2(g) + 6H2O(l)
Balance O atoms,
5x2 + 6 = 16 O
C5H12(g) + 8O2(g)  5CO2(g) + 6H2O(l)
Combustion Reactions
C6H14(g) + O2(g) 
Balance C atoms
C6H14(g) + O2(g) 
Balance H atoms
C6H14(g) + O2(g) 
Balance O atoms,
C6H14(g) + 9.5O2(g)
CO2(g) + H2O(l)
6CO2(g) + H2O(l)
6CO2(g) + 7H2O(l)
6x2 + 7 = 19 O
 6CO2(g) + 7H2O(l)
Multiply both sides of equation by 2
2C6H14(g) + 19O2(g)  12CO2(g) + 14H2O(l)
Combustion Reactions
NH3(g) + O2(g)  NO(g) + H2O(g)
N atoms are balanced
Balance H atoms
2NH3(g) + O2(g)  NO(g) + 3H2O(g)
N atoms are not balanced now
2NH3(g) + O2(g)  2NO(g) + 3H2O(g)
Balance O atoms,
2 +3=5O
2NH3(g) + 2.5O2(g)  2NO(g) + 3H2O(g)
Multiply both sides of equation by 2
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)
Stoichiometry
• calculations involving chemical
equations
Steps
• write balanced chemical equation
• convert given quantities to moles
• determine limiting reagent if necessary
• calculate number of moles of desired
substance
• convert to desired units
Stoichiometry
Calculations are based on relationships similar
to those in the following example
CH4(g)(methane) + 2 O2(g)  CO2(g) + 2H2O(l)
1 molecule of CH4 reacts with 2 molecules of O2 to
produce 1 molecule of CO2 + 2 molecules of H2O
1 mole of CH4 reacts with 2 moles of O2 to
produce 1 mole of CO2 + 2 moles of H2O
16.04 g of CH4 reacts with 2x32.00 of O2 to
produce 44.01 g of CO2 + 2x18.02 g of H2O
Example: Iron reacts with steam to form
Fe3O4 and H2. Calculate mol H2 produced
by reaction of 10.0 g iron with excess
steam.
Fe(s) + H2O(g)  Fe3O4(s) + H2(g)
3 Fe(s) + 4 H2O(g)  Fe3O4(s) + 4 H2(g)
First, we convert g of Fe into moles Fe
1 mol Fe
10.0 g Fe x —————— = 0.179 mol Fe
55.85 g Fe
Second, we use Fe moles to calculate H2’s.
Water is not used because it is excess
4 mol H2
0.179 mol Fe x ————— = 0.239 mol
3 mol Fe
Example: Tin(IV) oxide reacts with carbon at
high temperature to form elemental tin and
carbon monoxide. How many kg of carbon
are needed to convert 50.0 kg of tin(IV)
oxide to elemental tin?
• Sn4+
• O2–
• SnO2
• SnO2 + C  Sn + CO
To balance the equation we need 2 O on
the right and so on.
SnO2 + 2 C  Sn + 2 CO
1x 103 g
1 mol SnO2
50.0 kg SnO2 x ———— x ——————
1 kg
150.7 g SnO2
2 moles C
12.01 g C 1 kg C
x —————— x ———— x ————
1 mol SnO2
1 mol C 1x 103 g C
7.97 kg C (we may use kmol too)
(332 moles SnO2
664 moles C)
The air in a closed container consists of
1.40 mol O2, 7.00 mol N2, with traces of other
gases. A small lamp fueled by methanol,
CH3OH, is lighted in the container. How
many mL of methanol (d=0.791 g/mL) will be
consumed when the lamp goes out?
• CH3OH(l) + O2(g) 
• CH3OH(l) + O2(g)  CO2(g) + H2O(l)
• 2 CH3OH(l) + 3 O2(g)  2 CO2(g) + 4 H2O(l)
I recommend you to practice this balancing
2 CH3OH(l) + 3 O2(g)  2 CO2(g) + 4 H2O(l)
We can determine the amount of CH3OH
out of O2’s
2 mol CH3OH
1.40 mol O2 x —————— = 0.933 mol CH3OH
3 mol O2
32.04 g CH3OH 1 mL CH3OH
0.933 mol x —————— x ——————
1 mol CH3OH
0.791 g CH3OH
= 37.8 mL CH3OH
Limiting Reagent (reactant)
• the reactant that is present in quantity
smaller to completely react other reactant; is
consumed completely during the reaction;
determines amount of product yielded
It can be also seen as the reagent that
theoretically produces the smallest amount of
product(s).
• excess reagent (reactant): the reactant
present in quantity greater than needed for
the reaction;
part of it remains after the reaction is
completed
Example: Copper(II) oxide reacts with
ammonia to form copper, water, and
nitrogen. If 236.1 g copper(II) oxide are
treated with 64.38 g ammonia, how much
copper (grams) is produced? How many g
of excess reagent remain?
•
•
•
•
CuO
NH3
CuO(s) + NH3(g)  Cu(s) + H2O(l) + N2(g)
3CuO + 2NH3(g)  3Cu(s) + 3H2O(l) + N2(g)
3CuO + 2NH3(g)  3Cu(s) + 3H2O(l) + N2(g)
1 mol CuO
236.1 g CuO x —————— = 2.968 mol CuO
79.55 g CuO
1 mol NH3
64.38 g NH3 x —————— = 3.780 mol NH3
17.03 g NH3
Now, let’s see which one of the two, CuO or NH3,
theoretically produces the smallest amount of a
product, Cu in this case.
3 mol Cu
2.968 mol CuO ————— = 2.968 mol Cu
3 mol CuO
And, now, ammonia, NH3, …
3CuO + 2NH3(g)  3Cu(s) + 3H2O(l) + N2(g)
3 mol Cu
3.780 mol NH3————— = 5.670 mol Cu
2 mol NH3
Then, the amount of Cu we can obtain with CuO,
2.968 mol, is less than what we can get with NH3,
i.e., CuO is the LR.
Conclusion: we must use the amount of CuO
to calculate the amount of Cu yielded and for the
amount of NH3 that has reacted.
2.968 mol CuO
3.780 mol NH3
───────── = 0.989 < ───────── = 1.89
3 (coeff.)
2 (coeff.)
3CuO + 2NH3(g)  3Cu(s) + 3H2O(l) + N2(g)
How many moles of Cu? 2.968 mol Cu
63.55 g Cu
2.968 mol Cu x ————— = 188.6 g Cu
1 mol Cu
How many g of NH3 in excess?
started with 3.780 mol NH3. The reacted from CuO,
2 mol NH3
2.968 mol CuO x ————— = 1.979 mol NH3 used up
3 mol CuO
Then, by subtraction
left over
3.780 – 1.979 = 1.801 mol NH3
17.03 g
1.801 mol NH3 x ———— = 30.67 g NH3
1 mol
4 Li(s)
start (given) 4.0 mol
reacted
-4.0 mol
Finish
0 mol
+
O2(g)  2 Li2O(s)
1.0 mol
0 mol
-1.0 mol
+2.0 mol
0 mol
2.0 mol NoLR
Which one is the limit reagent, Li or O2?
2 mol Li2O
4.0 mol Li ───────= 2 mol Li2O theoret. produced
4 mol Li
2 mol Li2O
1.0 mol O2───────=2 mol Li2O theoret. produced
1 mol O2
Then, there is no L. R. Both react completely
4 Li(s)
+ O2(g) 
Start (given)4.0 mol 0.5 mol
reacted -2.0 mol
-0.5 mol
Finish
2.0 mol
0 mol
2 Li2O(s)
0 mol
+1.0 mol
1.0 mol O2 LR
Which one is the limit reagent, Li or O2?
2 mol Li2O
4.0 mol Li ───────= 2 mol Li2O theoret. produced
4 mol Li
2 mol Li2O
0.5 mol O2───────= 1 mol Li2O theort. produced
1 mol O2
Then, O2 is the L. R. 1.0 mole of Li2O formed; 2 mole of
Li excess
4 Li(s)
start
8.0 mol
reacted -8.0 mol
Finish 0 mol
Homework…
+
O2(g) 
3.0 mol
-2.0 mol
1.0 mol
2 Li2O(s)
0 mol
+4.0 mol
4.0 mol Li LR
Example: Dihydrogen sulfide and sulfur
dioxide react to form sulfur & water. How
much sulfur is formed when 5.00 g
dihydrogen sulfide are mixed with 5.00 g of
sulfur dioxide? How many g of excess
reagent remain after reaction?
Equation
H2S(g) + SO2(g)  S8(s) + H2O(l)
balancing (try it)
16 H2S(g) + 8 SO2(g)  3 S8(s) + 16H2O(l)
Firstly: calculate moles (of molecules) of H2S
and SO2. Then, determine Limiting Ractant
1 mol H2S
5.00 g H2S x ————— = 0.147 mol H2S
34.1 g H2S
1 mol SO2
5.00 g SO2 x ————— = 0.0780 mol SO2
64.1 g SO2
How much S8?
16 H2S(g) + 8 SO2(g)  3 S8(s) + 16 H2O(l)
3 mol S8
0.147 mol H2S x ————— = 0.0276 mol S8
16 mol H2S (smaller)
3 mol S8
0.0780 mol SO2 x ————— = 0.0292 mol S8
8 mol SO2
H2S is the L.R.
Theoretically, 0.147 mol of H2S produces 0.0276
mol S8, and 0.0780 mol SO2 produces 0.0292
mol S8. Then, H2S is the Limiting Reactant, and
SO2 is the excess reagent.
To determine the amount of sulfur we use H2S:
16 H2S(g) + 8 SO2(g)  3 S8(s) + 16 H2O(l)
3 mol S8
0.147 mol H2S x ————— = 0.0276 mol S8
16 mol H2S
MW (S8) = 8 x 32.066 = 256.6 g/mol
256.5 g S8
0.0276 mol S8 x ————— = 7.08 g S8
1 mol S8
How many g of excess reagent remain after
reaction? That is SO2
We need to determine the amount of SO2 that reacts
with 0.147 mol of H2S (the LR):
16 H2S(g) + 8 SO2(g)  3 S8(s) + 16 H2O(l)
8 mol SO2
0.147 mol H2S x ————— = 0.0735 mol SO2
16 mol H2S
0.0780 mol SO2 initial – 0.0735 mol reacted = 0.0045 mol
remaining after reaction
MW (SO2) = 32.066 + 2 x 16.00 = 64.07 g/mol
64.07 g SO2
0.0045 mol SO2 x —————— = 0.29 g SO2 (excess)
1 mol SO2
Percent Yield
• Theoretical Yield: the maximum amount of
product that can be obtained from a chemical
reaction. It is the one we calculate from the
chemical equations
• Actual Yield: the amount of product that is
experimentally obtained from a reaction—it is
less than the theoretical yield (due to …)
actual yield
% yield = ———————— x 100
theoretical yield
% yield  100%
Example: Benzene, C6H6, reacts with
bromine to form bromobenzene, C6H5Br, and
hydrogen bromide. Reaction of 8.00 g
benzene with excess bromine yielded 12.85
g bromobenzene. Calculate % yield of
bromobenzene.
• actual yield = 12.85 g C6H5Br
• C6H6 + Br2  C6H5Br + HBr
• To calculate %yield we need to know the
theoretical yield. We will use 8.0 g C6H6 (LR)
because bromine is the excess reagent
C6H6 + Br2  C6H5Br + HBr
MW (C6H6) = 6 x 12.011 + 6 x 1.008 = 78.11 g/mol
How many moles of C6H6?
1mol C6H6
8.00 g C6H6 x —————— = 0.102 mol C6H6
78. 11 g C6H6
How many moles of C6H5Br are produced?
1mol C6H5Br
0.102 mol C6H6 x —————— = 0.102 mol C6H5Br
1 mol C6H6
Converting moles of C6H5Br to grams… The theoretic.
yield
157.0 g C6H5Br
0.102 mol C6H5Br x ——————— = 16.0 g C6H5Br
1 mol C6H5Br
the percent yield of C6H5Br
actual yield
% yield = ———————— x 100
theoretical yield
12.85 g C6H5Br
% yield = ——————— x 100 = 80.3%
16.0 g C6H5Br
64.0 g of methanol, CH3OH, were expected to be
produced through the reaction
CO(g) + 2H2(g)  CH3OH(l)
One student got 56.0 g of methanol for that reaction
in the laboratory. What was the %yield of methanol?
64.0 g of methanol is the theoretical yield
(expected)
56.0 g is the actual yield (in the laboratory)
56.0 g
% Yield = ————  100 = 87.5 %
64.0 g
Analysis of mineral sample
The mineral cerussite is mostly PbCO3, but
other substances are present. To analyze for
the PbCO3 content, a sample of mineral is
first treated with nitric acid to dissolve PbCO3
PbCO3(s) + 2 HNO3(aq)  Pb(NO3)2(aq) + H2O(l) + CO2(g)
On adding sulfuric acid, lead(II) sulfate precipitates.
Pb(NO3)2(aq) + H2SO4(aq)  PbSO4(s) + 2 HNO3(aq)
Solid PbSO4(s) is isolated and weighed. Suppose a
0.583 g sample of mineral produced 0.628 g PbSO4.
What is the mass % of PbCO3 in the mineral sample?
KEY: 1 mol PbCO3  1 mol Pb(NO3)2  1 mol PbSO4
Calculating moles and grams
First we calculate moles of PbSO4 (FW = 303.3)
1 mol PbSO4
0.628 g PbSO4x —————— = 0.00207 mol PbSO4
303.3 g PbSO4
Now, we determine moles of PbCO3
1 mol Pb(NO3)2 1 mol PbCO3
0.00207 mol PbSO4x——————— x ———————
1 mol PbSO4
1 mol Pb(NO3)2
= 0.00207 mol PbCO3
that must be converted to g
grams and % of PbCO3 (FW = 267.2 g/mol)
g PbCO3 out of mol PbCO3
267.2 g PbCO3
0.00207 mol PbCO3 x —————— = 0.553 g PbCO3
1 mol PbCO3
Now the % PbCO3 in the mineral sample (0.583 g)
0.553 g PbCO3
Mass percent PbCO3= ——————— x 100 = 94.9%
0.583 g sample
Combustion Analysis
• sample is burned in oxygen
• C  CO2
• H  H2O
Example:
• compound containing only C, H, & O
• combustion of 28.64 mg sample of
compound 
88.02 mg CO2
27.03 mg H2O
• MW = 286.5 (it is a given data)
• determine molecular formula
Example:
• CxHyOz + O2  CO2
•
•
•
•
•
•
+ H2O
C  CO2, H  H2O, O  CO2 & H2O
det. mass of C and of H
det. mass of O
calc. mol C, H, & O
det. empirical formula
det. molecular formula
Example:
1 mmol CO2
1 mmol C
88.02 mg CO2 x —————— x —————— x
44.01 mg CO2 1 mmol CO2
12.01 mg C
x —————— = 24.02 mg C = 2.000 mmol of C
1 mmol C
… now H
Example:
1 mmol H2O
2 mmol H
27.03 mg H2O x —————— x —————— x
18.02 mg H2O 1 mmol H2O
1.008 mg H
x —————— = 3.02 mg H = 3.000 mmol of H
1 mmol H
… now mg of oxygen
mg sample = mg C + mg H + mg O
28.64 mg = 24.02 mg C + 3.024 mg H + mg O
mg O = 28.64 – 24.02 – 3.024 = 1.60 mg O
Example:
1 mmol C
24.02 mg C x ————— = 2.000 mmol of C
12.01 mg
1 mmol H
3.024 mg H x ————— = 3.000 mmol of H
1.008 mg H
1 mmol O
1.60 mg O x ————— = 0.100 mmol of O
16.00 mg O
Now, we will divide all by the smallest…
Example:
C2.0H3.0O0.1
0.1
0.1
0.1
C20H30O is the empirical formula
weight of E.F. = 2012.01 + 301.008 + 16.00
= 286.4
MW
286.5
n = —————— = ————— = 1
W of EF
286.4
molecular formula = C20H30O (same as emp)
Problem: Methane, CH4, burns in oxygen.
(a) What are the products of the reaction? CO2 + H2O
(b) Write the balance equation for the reaction.
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
(c) What mass (g) of O2 is required for complete
combustion of 25.5 g of methane?
Firstly, we need to calculate moles of methane
1 mol CH4
25.5 g CH4 x ─────── = 1.59 mol CH4
Then, use
16.04 g CH4
coeffients
and molar mass O2
2 mol O2 32.00 g O2
1.59 mol CH4 —————  ————— = 102 g O2
1 mol CH4 1 mol O2
Problem: Hexane, C6H14, burns in oxygen to give
CO2 and H2O.
(a) Write a balance equation for the reaction.
2 C6H14(l) + 19 O2(g)  12 CO2(g) + 14 H2O(l)
(b) If 215 g C6H14 are mixed with 215 g of O2, what
masses of CO2 and H2O are produced?
Firstly, we need to calculate moles of C6H14 and O2
1 mol C6H14
2.50
215 g C6H14x ─────── = 2.50 mol C6H14 ───= 1.25
86.17 g C6H14
2
1 mol O2
215 g O2 ————— = 6.72 mol O2
32.00 g O2
Now, the Limiting Reagent
6.72
─── = 0.354
19
(It seems to be O2)
2 C6H14(l) + 19 O2(g)
 12 CO2(g) + 14 H2O(l)
12 mol CO2
2.50 mol C6H14 x ──────── = 15.0 mol CO2
2 mol C6H14
12 mol CO2
6.72 mol O2 x ─────── = 4.24 mol CO2
19 mol O6
O2 is the L.R.
44.01 g CO2
4.24 mol CO2  —————— = 187 g CO2
1 mol CO2
14 mol H2O 18.02 g H2O
6.72 mol O2————— —————— = 89.2 g H2O
19 mol O2
1 mol H2O
Prob…
2 C6H14(l) + 19 O2(g)
 12 CO2(g) + 14 H2O(l)
What mass of excess reagent remains after reaction? C6H14
The 6.72 moles of O2 react with C6H14
2 mol C6H14
6.72 mol O2—————— = 0.707 mol C6H14
19 mol O2
initial amount of C6H14 was 2.50 mol − 0.707 mol reacted =
1.79 mol of C6H14 remaining
86.17 g C6H14
1.79 mol C6H14 ——————— = 154 g C6H14 remaining
1 mol C6H14
Problem
A mixture of CuSO4 and CuSO4•5H2O has a mass of 1.245 g.
After heating to drive off all the water, the mass is only 0.832
g. What is the mass percent of CuSO4•5H2O in the mixture?
FW CuSO4•5H2O = 159.61 + 518.02 = 249.69 g/mol
CuSO4
H2O
initial mass − final mass = 1.245 g − 0.832 g = 0.413 g H2O
The amount of CuSO4•5H2O in the mixture can be calculated:
249.69 g CuSO4•5H2O
0.413 g H2O —————————— = 1.14 g CuSO4•5H2O
518.02 g H2O
1.14 g CuSO4•5H2O
% CuSO4•5H2O = —————————  100 = 91.6%
1.245 g mixture
Problem
Nickel forms a compound with CO, Ni(CO)x. To determine its
formula, you carefully heat a 0.0973-g sample in air to convert
the Ni in 0.0426 g NiO and the CO in 0.100 g of CO2. What is
the empirical formula of Ni(CO)x?
From moles of NiO and CO2 we can calculate moles of Ni and CO:
molar mass of NiO = 74.71 g/mol
of CO2 = 44.01 g/mol
1 mol NiO 1 mol Ni
0.0426 g NiOx ────── x ────── = 0.000570 mol Ni
74.71 g NiO 1 mol NiO
1 mol CO2 1 mol CO
0.100 g CO2 x ─────── x ─────── = 0.00227 mol CO
44.01 g CO2 1 mol CO2
0.00227 mol CO
────────── = 4
0.000570 mol Ni
Emp. Form. = Ni(CO)4
Aluminum chloride is made by treating scrap
aluminum with chlorine.
2 Al(s) + 3 Cl2(g)  2 AlCl3(s)
If you begin with 2.70 g of Al and 4.05 g of Cl2,
a) Which reactant is limiting?
1 mol Al
2.70 g Al=0.100 mol Al 4.05 g Cl2 = 0.0571 mol
27.0 g Al
2 mol AlCl3
0.0571 mol Cl2   = 0.0381 mol AlCl3
3 mol Cl2
Cl2 is the L.R.
2 mol AlCl3
0.100 mol Al = 0.100 mol AlCl3
2 mol Al
Contd…
2 Al(s)
+ 3 Cl2(g)  2 AlCl3(s)
b) What mass of AlCl3 can be produced?
We must use Cl2, the LR, to calculate that
0.0381 mol AlCl3 is produced from the 0.0571 molCl2
133.35 g AlCl3
0.0381 mol AlCl3   = 5.08 g AlCl3
1 mol AlCl3
Contd…
2 Al(s)
+ 3 Cl2(g)  2 AlCl3(s)
b) What mass of the excess reactant remains when the
reaction is finished?
That is Al.
The remaining Al will be the initial minus the reacted.
For calculating the reacted we must use Cl2, the LR.
2 mol Al
0.0571 mol Cl2   = 0.0381 mol Al (reacted)
3 mol Cl2
0.100 mol (initial) − 0.0381 mol (reacted) = 0.0619 mol
remaining
27.0 g Al
0.0619 mol Al   = 1.67 g Al
1 mol Al
Styrene consists of only C and H. If 0.438 g styrene is
burned in excess oxygen and produces 1.481 g CO2
and 0.303 g H2O, what is the empirical formula of
styrene?
(styrene) CxHy(l) + O2(g)  CO2(g) + H2O(l)
mol of C in CO2 = mol of C in styrene
mol of H in H2O = mol of H in styrene
1 mol CO2 1 mol C
1.481 g CO2  = 0.03365 mol C
44.01 g CO2 1 mol CO2
1 mol H2O
2 mol H
0.303 g H2O   = 0.0336 mol H
18.02 g H2O 1 mol H2O
Cntd…
To determine the empirical formula we must
divide the two number of moles by the
smallest
C0.03365 H0.0336
0.0336
0.0336
The empirical formula is
CH
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