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Lecture 4: (Lec 3B) Stoichiometry
Sections (Zumdahl 6th Edition) 3.6-3.7
Outline:
•Mass Composition of Molecules
•Conservation of Atoms
• (Modern version of the law of definite proportions)
•Balancing Chemical Reactions; Meaning
Ch 3 Problems: 3.53-3.56. (at the end of Ch 3 in Text)
(What is different about 3.56b and 3.56g?)
Discussion Problem: 3.9
Why do we need to go between moles and grams?
• Molecules go together as atoms (or moles)
• Molecules can only be weighed to
determine amount.
• We need to convert back and forth because
reactions preserve atoms (or moles of
atoms)
• Molecules represent the rearrangement and
joining together of numbers of atoms of
different types (or moles of atoms).
Review: The RYP “molecule”
Atomic Masses given on each atom
What is the molecular Formula: R3 Y3 P2Red,
Yellow, Purple
From the molecular formula and Atomic Masses, we
can compute the mass fraction of each element in the
molecule.
Our Goal: Find the Molecular Formula
•Determine the Molecular Formula from the mass of the
elements that make up the molecule.
•Elemental Analysis: The method of finding the mass
percent of each element in a molecule.
•Determine the Empirical Formula from the mass percents.
•Use the Mass Spectroscopy data,
which gives the total molecular mass,
to determine the molecular formula.
Empirical and Molecular Formulas
Empirical Formula - The simplest formula for a compound
that agrees with the elemental analysis (the process
Used to get the mass percents of each element)! The
smallest set of whole numbers of atoms.
Molecular Formula - The formula of the compound as it
exists, it may be a multiple of the Empirical
formula.
Some Examples of Compounds with the same
Elemental Ratio’s
Empirical Formula
Molecular Formula
CH2(unsaturated Hydrocarbons)
C2H4 , C3H6 , C4H8
OH or HO
H2O2
S
S8
P
P4
Cl
Cl2
CH2O (carbohydrates)
C6H12O6
Empirical and Molecular Formulas
Name
water
hydrogen
Molecular
H2O
H2O2
peroxide
ethane
C2H6
sulfur
S8
acetic acid
CH3COOH
Empirical
Empirical and Molecular Formulas
Name
water
Molecular
H2O
Empirical
H2O
hydrogen
peroxide
H2O2
HO
ethane
C2H6
CH3
sulfur
acetic acid
S8
CH3COOH
S
COH2
FIGURE 3.6: Molecules with Different
Empirical and Molecular Formulas
Ball-and-Stick Model: P4O10
Various Names
Molecular
Empirical
Empirical Weight
Molecular Weight
Ken O'Donoghue
Space Filling: Caffeine Molecule
Name
Formula
Molar Mass
C=black; N=blue; O=red; H=green
Steps to Determine Empirical Formulas from masses
Mass (g) of Element in sample
Or a relative mass
÷ AW (g/mol ) for that element
Moles of Element
Use no. of moles as subscripts.
Preliminary Formula
Change to integer subscripts:
÷ smallest, conv. to whole #.
Empirical Formula
Determining Empirical Formulas from
Measured Masses of Elements - I
Problem: The elemental analysis of a sample compound gave the
following results: 5.677g Na, 6.420 g Cr, and 7.902 g O. What is the
empirical formula and name of the compound?
Plan: First we have to convert mass of the elements to moles of the
elements using the atomic masses. Then we construct a preliminary
formula and name of the compound.
Solution: Finding the moles of the elements:
n(Na) =
n(Cr) =
n(O) =
N.B.: The masses are equivalent to the mass percents or mass fractions.
Determining Empirical Formulas from
Measured Masses of Elements - I
Problem: The elemental analysis of a sample compound gave the
following results: 5.677g Na, 6.420 g Cr, and 7.902 g O. What is the
empirical formula and name of the compound?
Plan: First we have to convert mass of the elements to moles of the
elements using the atomic masses. Then we construct a preliminary
formula and name of the compound.
Solution: Finding the moles of the elements:
5.678 g ( Na ) 5.678 g ( Na ) ⋅1mol
n ( Na ) =
=
= 0.2469mol ( Na )
Aw ( Na )
22.99 g
1 mol Cr
= 0.12347 mol Cr
52.00 g Cr
1 mol O
n(O) = 7.902 g O x
= 0.4939 mol O
16.00 g O
n(Cr) = 6.420 g Cr x
Determining Empirical Formulas from
Masses of Elements - II
Constructing the preliminary formula:
Converting to integer subscripts (dividing all by smallest subscript):
Rounding off to whole numbers:
Determining Empirical Formulas from
Masses of Elements - II
Constructing the preliminary formula:
Na0.2469 Cr0.1235 O0.4939
Converting to integer subscripts (dividing all by smallest subscript):
Na1.99 Cr1.00 O4.02
Rounding off to whole numbers:
Na2CrO4
Sodium Chromate
Determining the Molecular Formula from
Elemental Composition and Molar Mass - I
Problem: The sugar burned for energy in cells of the body is Glucose
(M = 180.16 g/mol), elemental analysis shows that it contains
40.00 mass % C, 6.719 mass % H, and 53.27 mass % O.
(a) Determine the empirical formula of glucose.
(b) Determine the Molecular formula.
Plan: We are only given mass %, and no weight of the compound so we
will assume 100g of the compound, and % becomes grams, and
we can do as done previously with masses of the elements.
Solution:
Mass Carbon =
Mass Hydrogen =
Mass Oxygen =
Determining the Molecular Formula from
Elemental Composition and Molar Mass - I
Problem: The sugar burned for energy in cells of the body is Glucose
(M = 180.16 g/mol), elemental analysis shows that it contains
40.00 mass % C, 6.719 mass % H, and 53.27 mass % O.
(a) Determine the empirical formula of glucose.
(b) Determine the Molecular formula.
Plan: We are only given mass %, and no weight of the compound so we
will assume 100g of the compound, and % becomes grams, and
we can do as done previously with masses of the elements.
Solution:
Mass Carbon = 40.00% x 100g/100% = 40.00 g C
Mass Hydrogen = 6.719% x 100g/100% = 6.719g H
Mass Oxygen = 53.27% x 100g/100% = 53.27 g O
99.989 g Cpd
Determining the Molecular Formula from
Elemental Composition and Molar Mass - II
Converting from Grams of Elements to moles:
Moles of C =
Moles of H =
Moles of O =
Constructing the preliminary formula:
Converting to integer subscripts, ÷ all subscripts by the smallest:
Determining the Molecular Formula from
Elemental Composition and Molar Mass - II
Converting from Grams of Elements to moles:
Moles of C = Mass of C x 1 mole C = 3.3306 moles C
12.01 g C
1 mol H
Moles of H = Mass of H x
= 6.6657 moles H
1.008 g H
Moles of O = Mass of O x 1 mol O = 3.3294 moles O
16.00 g O
Constructing the preliminary formula C 3.33 H 6.67 O 3.33
Converting to integer subscripts, ÷ all subscripts by the smallest:
C 3.33/3.33 H 6.667 / 3.33 O3.33 / 3.33 = C1H2O1 = CH2O
Alternative to finding Empirical Formula
Begin with relative masses:
40.00 mass % C, 6.719 mass % H, and 53.27 mass % O.
Preliminary-Empirical-formula is
C 40.00 H 6.719 O53.27 = C3.33 H 6.67O3.33
12.01
1.008
Empirical
16
Formula
C 3.33 H 6.67 O3.33 = C1 H 2O1 or CH 2O
3.33
3.33
3.33
Obtaining the empirical formula requires dividing the preliminary
one by the smallest valued subscript; but this is not necessarily
the complete answer. There is a bit of an art to it.
Compounds with empirical formula CH2O
Determining the Molecular Formula from
Elemental Composition and Molar Mass - III
(b) Determining the Molecular Formula:
The formula weight of the empirical formula is:
Whole-number multiple =
M compound
empirical formula mass
Therefore the Molecular Formula is:
=
Determining the Molecular Formula from
Elemental Composition and Molar Mass - III
(b) Determining the Molecular Formula:
The formula mass of the empirical formula is:
1 x C + 2 x H + 1 x O = 1 x 12.01 + 2 x 1.008 + 1 x 16.00 = 30.03 g/mol
s= Whole-number multiple =
s =
MW of the compound =
empirical formula mass
180.16 g (Mf)
30.03 g (Ef)
Therefore the Molecular Formula is:
C1x6H2x6O1x6 =
C6H12O6
= 6.00 = 6
Mf
Ef
Obtain the Molecular Formula with no guessing
• Eg. A hydrocarbon sample is analyzed, to contain 0.2778
g C, 0.0116g H, and 0.3699 g O. The Molecular Mass is
180.16 g (from Mass Spec). What is the molecular
formula?
• Strategy --- Scale the masses first to give the correct
masses for the actual molecular mass.
MW
180.16
s=
=
= 259.4589
M ( E .F . )
0.2778 + 0.0467 + 0.3699
• Therefore the new masses are 72.0719 g C, 12.1063 g H,
95.9818 g O. = s ⋅ 0.2778 gC , s ⋅ 0.0467 gH , s ⋅ 0.3699 gO
C 72.0719 H 12.1063 O95.9818 = C6.001 H12.01O5.999
12.01
1.008
16.00
How Chemical Analysis is done:
eg :
(n + 12 m)O2 + Cn H m → mH 2O + nCO2
Mass Carbon = fC*Mass CO2
Mass Hydrogen = fH*MassH2O
12
12 3
fC =
=
= = 0.273
12 + 2 ⋅16 44 11
1⋅ 2
2 1
fH =
= = = 0.111
2 + 16 18 9
Chemical Analysis and empirical formula
Know the mass of the sample
From the chemical analysis:
Obtain the mass of H2O and CO2
Mass Carbon = fC*Mass CO2
Mass Hydrogen = fH*MassH2O
12
12 3
=
= = 0.273
12 + 2 ⋅16 44 11
1⋅ 2
2 1
fH =
= = = 0.111
2 + 16 18 9
fC =
Our Goal: Determine the fraction (by mass) of each
element in the original sample
e.g. Mass Carbon = mass CO2*(3/11)
Mass Fraction Carbon = Mass Carbon / mass sample
Mass Fraction Hydrogen = Mass Hydrogen / mass sample
Mass Extra (e.g. N or O) = Mass sample – (mass C + mass H)
Ascorbic acid ( Vitamin C ) - I contains only C , H , and O
• Upon combustion in excess oxygen, a 6.49 mg
sample yielded 9.74 mg CO2 and 2.64 mg H2O.
• Calculate its Empirical formula!
• Mass C:
• Mass H:
• Mass Oxygen =
Ascorbic acid ( Vitamin C ) - I
contains only C , H , and O
• Upon combustion in excess oxygen, a 6.49 mg
sample yielded 9.74 mg CO2 and 2.64 mg H2O.
• Calculate its Empirical formula!
• Mass C: 9.74 x10-3g CO2 x(12.01 g C/44.01 g CO2)
= 2.65 x 10-3 g C
• Mass H: 2.64 x10-3g H2O x (2.016 g H/18.02 gH2O)
= 2.95 x 10-4 g H
• Mass Oxygen = 6.49 mg - 2.65 mg - 0.30 mg
= 3.54 mg O = 3.54x10-3 g O
Ascorbic acid combustion – cont’d
• Moles C =
• Moles H =
• Moles O =
• Divide each by smallest:
• Moles C =
• Moles H =
• Moles O =
Vitamin C combustion - II
• C = 2.65 x 10-3 g C / ( 12.01 g C / mol C ) =
= 2.21 x 10-4 mol C
• H = 0.295 x 10-3 g H / ( 1.008 g H / mol H ) =
= 2.92 x 10-4 mol H
• O = 3.54 x 10-3 g O / ( 16.00 g O / mol O ) =
= 2.21 x 10-4 mol O
•
•
•
•
Divide each by smallest (2.21 x 10-4 ):
C = 1.00
Multiply each by 3: C = 3.00 = 3.0
H = 1.32
(to get ~integers)
H = 3.96 = 4.0
O = 1.00
O = 3.00 = 3.0
C3H4O3
If the empirical formula of ascorbic acid is
C3H4O3, (empirical mass 88 g/mol) and the
molecular mass of ascorbic acid is 176 g/mol, what
is the molecular formula?
1. C3H4O3
2. C6H8O6
3. C9H12O9
4. C12H16O12
5. None of the above
Change: The meaning of a chemical reaction
•
•
•
•
•
A rxn represents a change.
Burn Sugar: Sugar becomes CO2 and Water.
Sugar on the left (with O2) before reaction
Products on the right after the change.
Properties of New state must equal properties of old state
plus the change between them.
• For example; There are 300 people in this room: We
count noses. Two people leave. How many in the room?
– We do not need to recount the number of people in the room to
know there are 298 people. We kept track of the change, and
do the math.
New = Initial + Change (NIC Table)
Change ≡ X ≡ ΔX = New − Initial
How to Balance Equations
• Mass Balance (or Atom Balance)- same number of
each element on each side of the equation:
(1) start with simplest element (or largest molecule)
(2) progress to other elements
(3) make all whole numbers
(4) re-check atom balance
1 CH4 (g) + __O2 (g)
1 CO2 (g) + __H2O (g)
1 CH4 (g) +__ O2 (g)
1 CO2 (g) + 2 H2O (g)
1 CH4 (g) + 2 O2 (g)
1 CO2 (g) + 2 H2O (g)
• Make charges balance. (Remove “spectator” ions.)
Ca2+ (aq) + 2 OH- (aq) + Na+
Ca(OH)2 (s) + Na+
DEMO: Methane Bubbles
Extra Information
• More Information and extra problems follow.
• When working extra problems (or problems in
the text): Be sure to cover up the solution
before you start the problem.
Calculate MW and % composition of NH4NO3.
•
•
•
2 mol N x
4 mol H x
3 mol O x
Molar mass = M =
%N =
%H =
%O =
g N x 100%
g cpd
gH
x 100%
g cpd
=
=
g O x 100% =
g cpd
Check:
100% total ?
Calculate M and % composition of NH4NO3.
•
•
•
2 mol N x 14.01 g/mol = 28.02 g N
4 mol H x 1.008 g/mol = 4.032 g H
3 mol O x 15.999 g/mol = 48.00 g O
80.05 g/mol
%N = 28.02g N x 100%
80.05g
= 35.00%
%H =
4.032g H x 100% =
80.05g
%O =
48.00g O x 100% = 59.96%
80.05g
99.997%
5.037%
Calculate the Percent Composition of
Sulfuric Acid H2SO4
Molar Mass of Sulfuric acid =
2(1.008g) + 1(32.07g) + 4(16.00g) = 98.09 g/mol
2(1.008g H)
x 100% =
98.09g
2.06% H
%S = 1(32.07g S) x 100% =
98.09g
32.69% S
%O = 4(16.00g O) x 100% =
98.09 g
65.25% O
%H =
Check =
100.00%
Calculating Mass Percentage and Masses of
Elements in a Sample of a Compound - I
Problem: Sucrose (C12H22O11) is common table sugar.
( a) What is the mass percent of each element in sucrose?
( b) How many grams of carbon are in 24.35 g of sucrose?
(a) Determining the mass percent of each element:
mass of C per mole sucrose =
mass of H / mol =
mass of O / mol =
total mass per mole =
Finding the mass fraction of C in Sucrose & % C :
mass of C per mole
Mass Fraction of C =
=
mass of 1 mole sucrose
=
To find mass % of C =
Calculating Mass Percentage and Masses of
Elements in a Sample of a Compound - I
Problem: Sucrose (C12H22O11) is common table sugar.
( a) What is the mass percent of each element in sucrose?
( b) How many grams of carbon are in 24.35 g of sucrose?
(a) Determining the mass percent of each element:
mass of C per mole sucrose = 12 x 12.01 g C/mol = 144.12 g C/mol
mass of H / mol = 22 x 1.008 g H/mol =
22.176 g H/mol
mass of O / mol = 11 x 16.00 g O/mol =
176.00 g O/mol
total mass per mole = 342.296 g/mol
Finding the mass fraction of C in Sucrose & % C :
mass of C per mole
144.12 g C/mol
Mass Fraction of C =
=
mass of 1 mole sucrose
342.30 g Cpd/mol
= 0.4210
To find mass % of C = 0.4210 x 100% =
42.10%
Calculating Mass Percents and Masses of
Elements in a Sample of Compound - II
(a) continued
Mass % of H =
mol H x M of H x 100% =
mass of 1 mol sucrose
Mass % of O =
mol O x M of O x 100% =
mass of 1 mol sucrose
(b) Determining the mass of carbon:
Mass (g) of C = mass of sucrose x ( mass fraction of C in sucrose)
Mass (g) of C =
Calculating Mass Percents and Masses of
Elements in a Sample of Compound - II
(a) continued
Mass % of H =
mol H x M of H x 100% = 22 x 1.008 g Hx 100%
mass of 1 mol sucrose
342.30 g
= 6.479% H
Mass % of O =
mol O x M of O x 100% = 11 x 16.00 g Ox 100%
mass of 1 mol sucrose
342.30 g
= 51.417% O
(b) Determining the mass of carbon:
Mass (g) of C = mass of sucrose x ( mass fraction of C in sucrose)
Mass (g) of C = 24.35 g sucrose x 0.421046 g C
1 g sucrose
= 10.25 g C
Calculating the Mass of an Element in a
Compound: Ammonium Nitrate
How much Nitrogen is in 455 kg of Ammonium Nitrate?
Ammonium Nitrate = NH4NO3 The Formula Mass of Cpd is:
4 x H = 4 x 1.008 = 4.032 g
2 x N = 2 X 14.01 = 28.02 g
Therefore mass fraction N:
3 x O = 3 x 16.00 = 48.00 g
28.02 g Nitrogen
80.052 g
=
80.052 g Cpd
Mass N in sample =
Calculating the Mass of an Element in a
Compound: Ammonium Nitrate
How much Nitrogen is in 455 kg of Ammonium Nitrate?
Ammonium Nitrate = NH4NO3 The Formula Mass of Cpd is:
4 x H = 4 x 1.008 = 4.032 g
2 x N = 2 X 14.01 = 28.02 g
Therefore mass fraction N:
3 x O = 3 x 16.00 = 48.00 g
28.02 g Nitrogen
80.052 g
= 0.35002249 g N / g Cpd
80.052 g Cpd
455 kg x 1000g / kg = 455,000 g NH4NO3
455,000 g Cpd x 0.35002249 g N / g Cpd = 1.59 x 105 g Nitrogen
or: 455 kg NH4NO3
X 28.02 kg Nitrogen = 159 kg Nitrogen
80.052 kg NH4NO4
Determining a Chemical Formula from
Combustion Analysis - I
Problem: Erthrose (M = 120 g/mol) is an important chemical
compound used often as a starting material in chemical
synthesis, and contains Carbon, Hydrogen, and Oxygen.
Combustion analysis of a 700.0 mg sample yielded:
1.027 g CO2 and 0.4194 g H2O.
Plan: We find the masses of Hydrogen and Carbon using the mass
fractions of H in H2O, and C in CO2. The mass of Carbon and
Hydrogen are subtracted from the sample mass to get the mass
of Oxygen. We then calculate moles, and construct the empirical
formula, and from the given molar mass we can calculate the
molecular formula.
Determining a Chemical Formula from
Combustion Analysis - II
Calculating the mass fractions of the elements:
Mass fraction of C in CO2 =
Mass fraction of H in H2O =
Calculating masses of C and H:
Mass of Element = mass of compound x mass fraction of element
Determining a Chemical Formula from
Combustion Analysis - II
Calculating the mass fractions of the elements:
mol C x M of C
Mass fraction of C in CO2 =
=
mass of 1 mol CO2
= 1 mol C x 12.01 g C/ 1 mol C = 0.2729 g C / 1 g CO2
44.01 g CO2
mol H x M of H
=
mass of 1 mol H2O
2 mol H x 1.008 g H / 1 mol H
=
= 0.1119 g H / 1 g H2O
18.02 g H2O
Mass fraction of H in H2O =
Calculating masses of C and H
Mass of Element = mass of compound x mass fraction of element
Determining a Chemical Formula from
Combustion Analysis - III
Mass (g) of C =
Mass (g) of H =
Calculating the mass of O:
Calculating moles of each element:
C=
H=
O=
Determining a Chemical Formula from
Combustion Analysis - III
Mass (g) of C = 1.027 g CO2 x 0.2729 g C = 0.2803 g C
1 g CO2
0.1119 g H
Mass (g) of H = 0.4194 g H2O x
= 0.04693 g H
1 g H2 O
Calculating the mass of O:
Mass (g) of O = Sample mass -( mass of C + mass of H )
= 0.700 g - 0.2803 g C - 0.04693 g H = 0.37277 g O
Calculating moles of each element:
C = 0.2803 g C / 12.01 g C/ mol C = 0.02334 mol C
H = 0.04693 g H / 1.008 g H / mol H = 0.04656 mol H
O = 0.37277 g O / 16.00 g O / mol O = 0.02330 mol O
C0.02334H0.04656O0.02330 = CH2O formula weight = 30 g / formula
120 g /mol / 30 g / formula = 4 formula units / cpd => C4H8O4
Chemical Equations
Qualitative Information:
Reactants
Products
States of Matter: (s) solid
(l) liquid
(g) gaseous
(aq) aqueous
2 H2 (g) + O2 (g)
But also Quantitative Information!
2 H2O (g)
Using A Table to Balance
Molecular Reaction: aCH 4 + bO2 → cCO2 + dH 2O
Molecular
Species
CH4
O2
CO2
H2O
Coefficients
a=1
b=2
c=1
d=2
C
1
0
1
0
H
4
0
0
2
O
0
2
2
1
Equations to balance: (Three equations and 4 unknowns)
a = c =1
From the first equation:
Second equation: d = 2a = 2Third equation:
3 Equations to Solve
1⋅ a + 0 ⋅ b = 1⋅ c + 0 ⋅ d
4⋅ a + 0⋅b = 0⋅c + 2⋅ d
0 ⋅ a + 2 ⋅ b = 2 ⋅ c + 1⋅ d
b = c + 12 d = 2
Verify the Solution
1 ⋅1 + 0 ⋅ 2 = 1 ⋅1 + 0 ⋅ 2
4 ⋅1 + 0 ⋅ 2 = 0 ⋅1 + 2 ⋅ 2
0 ⋅1 + 2 ⋅ 2 = 2 ⋅1 + 1 ⋅ 2
Mass Percent Composition of Na2SO4
Na2SO4 = 2 atoms of Sodium + 1 atom of Sulfur + 4 atoms of Oxygen
Elemental masses (g)
2 x Na = 2 x 22.99 = 45.98
1xS=
4xO=
Check
% Na + % S + % O = 100%
Percent of each Element
% Na = Mass Na / Total mass x 100%
% Na =
% S = Mass S / Total mass x 100%
%S=
%O=
%O=
Mass Percent Composition of Na2SO4
Na2SO4 = 2 atoms of Sodium + 1 atom of Sulfur + 4 atoms of Oxygen
Elemental masses (g)
2 x Na = 2 x 22.99 = 45.98
1 x S = 1 x 32.07 = 32.07
4 x O = 4 x 16.00 = 64.00
142.05
Check
Percent of each Element
% Na = Mass Na / Total mass x 100%
% Na = (45.98 / 142.05) x 100% =32.37%
% S = Mass S / Total mass x 100%
% S = (32.07 / 142.05) x 100% = 22.58%
% O = Mass O / Total mass x 100%
% O = (64.00 / 142.05) x 100% = 45.05%
% Na + % S + % O = 100%
32.37% + 22.58% + 45.05% = 100.00%
Flow Chart of Mass Percentage Calculation
Moles of X in one
mole of Compound
Multiply by MW (X in g / mol)
Mass (g) of X in one
mole of compound
Divide by mass (g) MW of one mo
of compound
Mass fraction of X
Multiply by 100 %
Mass % of X
Adrenaline is a very Important
Compound in the Body - I
• Analysis gives :
•
C = 56.8 %
•
H = 6.50 %
•
O = 28.4 %
•
N = 8.28 %
• Calculate the
Empirical Formula !
Adrenaline - II
• Assume 100g!
• C=
• H=
• O=
• N=
Divide by smallest
•
•
•
•
C=
H=
O=
N=
=>
Adrenaline - II
•
•
•
•
•
•
•
•
•
•
Assume 100g!
C = 56.8 g C/(12.01 g C/ mol C) = 4.73 mol C
H = 6.50 g H/( 1.008 g H / mol H) = 6.45 mol H
O = 28.4 g O/(16.00 g O/ mol O) = 1.78 mol O
N = 8.28 g N/(14.01 g N/ mol N) = 0.591 mol N
Divide by smallest (0.591) =>
C = 8.00 mol C = 8.0 mol C
or
H = 10.9 mol H = 11.0 mol H
O = 3.01 mol O = 3.0 mol O C8H11O3N
N = 1.00 mol N = 1.0 mol N