Chapter 3 ( Stoichiometry )
Stoichiometry- Is a set of quantitative relationships among reactants and products in a
chemical reaction in terms of mass or mole, etc..
Chemical Reaction- process by which one or more substances are changed into one or
more different substances.
Chemical Equation- A shorthand expression of a chemical reaction. It shows the
reactants, products, their relative amounts and usually phases of the substances.
Ex. 2H2(g) + O2(g) → 2H2O(l)
2x2g
32g → 2x18g = 36g (Law of Conservation of Mass)
Equilibrium arrow, Electrolysis arrow, add heat arrow, Phases (aq), (s), (l), (g),
The arrow sign means ‘yields’ or ‘produces’. The left side of arrow is the reactant(s) side;
the right side of the arrow is the product(s) side. The chemical equation above is
balanced. This means:
1. the number of atoms of each element is the same on both sides of the arrow.
2. the sum of the masses on both sides of the arrow are the same.
3. the net charge on both sides of the arrow are the same.
Balancing Chemical Equations by Inspection
1. Always balance by adjusting coefficients; never adjust the subscripts.
2. Balance elements that are by themselves last( i.e. Na, H2, O2, Cl2)
3. Treat polyatomic ions as one unit(i.e. NO3-, SO42-, NH4+)
KClO3 → KCl + O2
coefficients should be (2, 2, 3)
Na + H2O → NaOH + H2 coefficients should be (2, 2, 2, 1)
C3H8 + O2 → CO2 + H2O coefficients should be (1, 5, 3, 4)
Periodic Table Group Reactivity
Elements in the same periodic table group have same charge state and react similarly, but
at different rates. Groups IA and IIA (Ca, Sr, Ba) react with water to form the hydroxide
base and H2 gas; and they react with oxygen gas to form the metal oxide.
2K + 2H2O → 2KOH + H2;
4K + O2 → 2K2O;
Ca + 2H2O → Ca(OH)2 + H2
2Ca + O2 → 2CaO
Four Basic Types of Chemical Reactions
Combination Reaction- Two or more substances react to make a new compound.
2Na + Cl2 → 2NaCl;
CaO + CO2 → CaCO3
SO2 + H2O → H2SO3; Na2O + H2O → 2NaOH
Decomposition Reaction- One substance decomposes into simpler substances. Heat,
light or electric current are usually required for these types of reaction to occur.
Metal carbonates decompose into the metal oxide and carbon dioxide:
CaCO3 → CaO + CO2
Metal sulfites decompose into the metal oxide and sulfur dioxide:
ZnSO3 → ZnO + SO2
Also: 2NaOH → Na2O + H2O;
H2SO4 → SO3 + H2O
Single Displacement Reaction- One element displaces another less reactive element in a
compound. The activity series reactions illustrate this.
____A = electropositive (metal) → AC + B
A + BC ─┤
───A = electronegative (non-metal) → BA + C
Activity of halogens in decreasing order: F > Cl > Br > I
Activity of metals in decreasing order:
Li, K, Ba, Ca, Na, Mg, Al, Mn, Zn, Cr, Fe, Co, Ni, Sn, Pb, H, Cu, Hg, Ag, Pt, Au
2Cr + 6H+ → 2Cr+3 + 3H2
Zn + H2SO4 → ZnSO4 + H2
Ca + MgSO4 → CaSO4 + Mg
F2 + 2NaCl → 2NaF + Cl2
Cl2 + 2KI → 2KCl + I2
Ag + HCl → No Reaction
Br2 + 2NaCl → No Reaction
Double Displacement Reaction- Two compounds react by exchanging cation/anion.
AB + CD → CB + AD
H2SO4 + 2KOH → K2SO4 + 2H2O (acid-base reactions)
AgNO3 + NaCl → NaNO3 + AgCl(s)
2KI + Pb(NO3)2
→ 2KNO3 + PbI2
Molar Quantities
Just like a dozen, score , gross, the mole is just a number, a huge number(6.022 x 1023.)
One hydrogen atom weighs 1a.m.u.(1.66 x 10-24g/amu)
6.022 x 1023 hydrogen atoms weigh 1g (6.022 x 1023 x 1.66 x 10-24g = 1g)
Atomic mass can be in a.m.u. if you are referring to one atom or one molecule.
Atomic mass will be in grams if one is referring to moles of substances.
1 mole of H atoms = 1g => 6.022 x 1023 H atoms
1 mole of H2 molecule = 2g => 6.022 x 1023 H2 molecules
1 mole of K+ ions = 39g => 6.022 x 1023 K+ ions
Molecular Formula for glucose is C6H12O6 ( empirical formula is CH2O)
Formula weight (molecular weight) of glucose is:
(6 x 12amu) + (12 x 1amu) + (6 x 16amu) = 180 a.m.u. for one molecule
and 180g for 6.022 x 1023 molecules.
The empirical formula weight of glucose is (1/6)180 = 30a.m.u. OR 30g
Moles of substance = grams of substance divided by molecular weight of substance
(g/M.W.).
There are 0.5 mole in 90g of glucose: (g/M.W.) = (90g/180g/mol) = 0.5 mole.
How many molecules in 90g of glucose?
(90g)(1 mole/180g)(6.022 x 1023 molecules/1mole) = 3.011 x 1023 molecules
How many C atoms in 90g glucose?
(90g)(1 mole/180g)(6.022 x 1023 molecules/1mole)(6C atoms/ molecule) = 1.807 x 1024
carbon atoms.
Percent Composition- is the relative percent by mass of an element in a compound.
Percent composition = (total mass of atom of element divided by molecular weight of
compound) x 100
% composition of C in C6H12O6 is:
{(6 x 12g)/180g} x 100 = (72g/180g) x 100 = 40%
Hydrates- Are crystalline compounds which have water in their structures in a fixed
ratio.
CuSO4∙5H2O
FeSO4∙7H2O
Copper(II) sulfate pentahydrate
Iron(II) sulfate heptahydrate
% H2O composition in hydrate = (total mass of water/total mass of hydrate) x 100
For CuSO4∙5H2O: mass of CuSO4 is 159.623g; total mass water is 5 x 18g = 90g; total
mass of hydrate is 159.623g + 90g = 249.623g
Therefore % H2O composition in CuSO4∙5H2O is: (90g/249.623g) x 100 = 36%
Determination of Empirical and Molecular Formula
Atomic % composition → change to grams → change to moles → make moles whole
numbers
Ex. Determine empirical & molecular formula of compound that is 40% C, 6.72% H,
53.3% O. Molecular weight is determined by colligative properties to be 180g. CXHYOZ
= empirical formula?
element
%composition
C
H
O
40%
6.72%
53.3%
Change % to
gram
40g
6.72g
53.3g
Grams to moles
40/12 = 3.33
6.72/1 = 6.72
53.3/16 = 3.33
Moles whole
#’s
3.33/3.33 = 1
6.72/3.33 = 2
3.33/3.33 = 1
Therefore x = 1, y = 2, z = 1
Empirical formula is CH2O (empirical weight is 30g); molecular weight is 180g
Molecular formula ? => 180/30 = 6; So it is C6XH6YO6Z = C6H12O6
Combustion Analysis is a way of obtaining % compositions of hydrocarbon (CXHY) and
oxy-hydrocarbon (CXHYOZ) compounds. A sample gram of the compound is
burned(reacted with oxygen gas) to produce a stoichiometric amount of CO2 and H2O.
All the C goes to make CO2; all the H goes to make H2O.
αCXHY + ∞O2 → γCO2 + δH2O
αCXHYOZ + ∞O2 → γCO2 + δH2O
The following shortcut method is useful in getting the mass of each element in the gram
sample compound of a combustion analysis problem.
M = sample mass, MC = C mass, MH = H mass, MO = O mass, m(H2O) = mass H2O,
m(CO2) = mass CO2
MH = (1/9) m(H2O),
MC = (3/11) m(CO2),
MO = M – (MC + MH)
Ex. 1.51 x 10-2g organic compound containing only C, H & O is combusted to give
0.0267g CO2 and 0.0176g H2O. What is the empirical formula?
MC = (3/11) m(CO2) = (3/11)(0.0267g) = 0.00728g C
MH = (1/9) m(H2O) = (1/9)(0.0176g) = 0.00196g H
MO = M – (MC + MH) = 1.51 x 10-2g – (0.00728g + 0.00196g) = 0.00586g O
C=> 0.00728/12 = 0.000606 mole; H=> 0.00196/1 = 0.00196mole; O=> 0.00586/16 =
0.000366mole
C=> 0.000606/0.000366 = 1.66 =>
H=> 0.00196/0.000366 = 5.35 =>
5
16
O=> 0.000366/0.000366 = 1 =>
Multiply all by 3 to get about whole numbers
3
So empirical formula is C5H16O3
Stoichiometric Calculations: Grams to Moles to Stoichiometric Mole Ratio to Grams
Given: C6H12O6 + 6O2 → 6CO2 + 6H2O
What mass of oxygen gas is required to react with 3.6g C6H12O6 ?
3.6g =
X
OR
180g
6 x 32 g
6 x 32g = X
180g
3.6g
;
X = 3.84g
OR
3.6g glu │ 1 mol glu │6 mol O2 │ 32g O2
= 3.84g O2
│180g glu │ 1 mol glu │ 1 mol O2
How many grams of water is produced? Do
Percent Yield = actual yield
x 100
Theoretical yield
Ex. Given 2KOH + H2SO4 → K2SO4 + 2H2O
13.6 gram H2SO4 reacts with excess KOH to produce 3.6g H2O. What is % yield? M.W.
H2SO4 is 98g; M.W. H2O is 18g; XT is theoretical yield.
2x18g =
98.0g
XT
13.6g
;
XT = 5.00g
So % yield = 3.6g x 100 = 72%
5.00g
Given: CaCO3 → CaO + CO2
How many grams of CaCO3 are required to produce 13.6 g CO2 if the % yield is 82%?
Do; Ans. = 37.7g CaCO3 .
M.W. CaCO3 is 100g; M.W. CO2 is 44g
Limiting Reagent- is reactant completely consumed in a reaction. It is the reactant that
gives the smallest amount of products because once it is used up, no further reaction
proceeds. Stoichiometric calculations are based on the limiting reagent.
Given: CH4 + 2O2 → CO2 + 2H2O
8.0g of CH4 ( 0.5mole) and 24.0g of O2 ( 0.75 mole) => react to form CO2 and H2O
a) limiting reagent? b) moles of CO2 produced c) mass of unreacted excess reagent.
a) 0.5mole CH4 => 0.5mole CO2 produced;
0.75mole O2 => 0.75/2 = 0.375 mole CO2 produced.
So O2 is limiting reagent.
b) 0.375 mole CO2 produced.
c) CH4 is excess reactant. 0.75 moles O2 => 0.375 mole CH4 reacted(16g x 0.375 = 6g).
So unreacted CH4 is 8g – 6g = 2g
Solutions for Some Chapter 3/ practice problems(CHEM 1411)
1. Balance the following equations:
a) 2C5H10O2 + 13O2  10CO2 + 10H2O
b) PCl5 + H2O  H3PO4 + HCl
c) Al(OH)3 + H2SO4  ____________+ ____________
d) Na + ________  NaOH + H2
1.
a) C5H10O2 + O2 → CO2 + H2O
(2, 13, 10, 10)
b) PCl5 + H2O → H3PO4 + HCl
(1, 4, 1, 5)
c) Al(OH)3 + H2SO4 → Al2(SO4)3 + H2O
(2, 3, 1, 6)
d) Na + H2O → NaOH + H2
(2, 2, 2, 1)
2. Write balanced chemical equations to correspond to each of the following
descriptions:
a) Heptane, C7H16, is combusted in air
b) When sulfur trioxide gas reacts with water, a solution of sulfuric
acid forms
c) Lithium metal is added to water
d) ZnCO3(s) is decomposed upon heating
2. C7H16 + O2 → CO2 + H2O
SO3 + H2O → H2SO4
Li + H2O → LiOH + H2
ZnCO3 → ZnO + CO2
(1, 11, 7, 8)
(1, 1, 1)
(2, 2, 2, 1)
(1, 1, 1)
3. Determine the formula weights of each of the following compounds:
a) Ca(C2H3O2)2
b) CH3CH2COOH
c) (NH4)3PO4
3. Determine formula weight:
Ca(C2H3O2)2 : 40g + (4x12g) + (6x1g) + (4x16g) = 158g
b) CH3CH2COOH : (3x12g) + (6x1g) + (2x16g) = 74g
c) (NH4)3PO4 : (3x14g) + (12x1g) + 31g + (4x16g) = 149g
4. Calculate the percentage compositions in the following compounds;
a) Benzoic acid, C7H6O2
b) Urea, (NH2)2CO
c) Laughing gas, N2O
4. Percentage compositions:
a) benzoic acid, C7H6O2 mol wt. is:
(7x12) + (6x1) + (2x16) = 84 + 6 + 32 = 122g
%comp. C=> (84/122) x 100 = 68.85%
H => (6/122) x 100 = 4.92%
O => (32/122) x 100 = 26.23%
b) (NH2)2CO N=> (28/60)102 ; H => (4/60)102 ; C => (12/60)102 ; O => (16/60)102!!!
47%
6.7%
20%
26.7%
2
2
c) N2O N => (28/44)10 = 63.6% O => (16/44)10 = 36.4%
5. Gallium comes in two isotopic forms (Ga-69 => 68.926g & Ga- 71 => 70.926g)
Atomic mass of Gallium is 69.726g. Calculate percent abundance of both isotopes. Let x
be the fractional abundance of isotope 70.926-Ga. Then:
69.726 = 70.926x + 68.926(1-x)
x = 0.8000/2 = 0.4000
% abundance of isotope 70.926 is 0.4000(100) = 40%
% abundance of isotope 68.926 is 60%.
6. Complete the following conversions;
a) moles of 0.005 g water
c) grams mass of 2.50 mole O2
b) atoms in 3.50 mole NaOH
d) carbon atoms in 10.5 g C6H12O6
6. (a) 0.005g │ 1mole H2O = 2.77 x 10-4 mole = 3 x 10-4 mole
│ 18g H2O
(b) 3.5moles │6.022 x 1023 molecules│3atoms = 6.32 x 1024 atoms
│ 1 mole
│1molecule
(c) 2.5 moles │ 32g O2 = 80g O2
│ 1 mole
(d) F.W. C6H12O6 = 180g = glucose = glu
(10.5g-glu)│1mole-glu│6.02 x 1023molecules-glu│6C = 2.10 x 1023 C atoms
│180g-glu) │1 mole-glu
│1 molecule-glu)
7. Determine the empirical formulas of each of the compounds if a sample
contains the following compositions:
a) 62.1% C, 5.21% H, 12.1% N, and 20.7% O
b) 0.104 mol K, 0.052 mol C, and 0.152 mol O
c) 11.66g iron and 5.01g oxygen
7. (a) CWHXNYOZ Epirical Formula?
62.1% C, 5.21% H, 12.1% N, 20.7% O
Assume 100g sample. Then:
62.1g/12g = 5.175moles C; 5.21g/1g = 5.21moles H; 12.1g/14g = 0.86 mole N
20.7g/16g = 1.29 moles O
Divide smallest mole number into each calculated mole:
5.175/0.86 = 6;
5.21/0.86 = 6; 0.86/0.86 = 1; 1.29/0.86 = 1.5
Multiply all subscripts by 2 to get all of them as whole numbers.
Therefore:
Empirical Formula of CWHXNYOZ is C12H12N2O3
(b) KXCYOZ Empirical Formula?
Divide smallest mole number into each mole.
i.e. (0.104/.052) = 2moles K; (0.052/0.052) = 1mole C; (0.152/0.052) = 3moles O
Therefore empirical formula is : K2CO3
(c) Get equivalent mole value for each mass.
Fe 11.66g/56g = 0.21 mole; O 5.01g/16 g = 0.313 mole
Divide smallest mole into each calculated mole.
Fe  1, O  1.5
Multiply subscripts by 2
Therefore Empirical formula For FexOy is Fe2O3
8. Determine the molecular formula of each of the following compounds?
a) empirical formula HCO2, molar mass = 90.0 g/mol
b) 75.69% C, 8.80% H, and 15.51% O by mass; molar mass 206g
8. (a) First find empirical formula mass.
HCO2  1 + 12 + 2(16) = 45g
If molecular formula mass is 90g
90g/45g = 2
(HCO2)2 = H2C2O4
(b) % composition given, Molar mass given  206g for CxHyOz
Moles of C is 75.69g/12g = 6.31moles
Moles of H is 8.8g/1g = 8.8moles
Moles of O is 15.51g/16g = 0.97mole
C  6.31/0.97 = 6.5; H  8.8/0.97 = 9.07; O  0.97/0.97 = 1.00
Multiply by 2. Then C  13; H  18; O  2
Then CxHyOz is C13H18O2 = empirical and molecular formula. OR
mole of C = (206g/12g)( 0.7569) = 12.99 mole  13 moles = x
mole of H = (206g/1g)( 0.088) = 18.12 mole  18 moles = y
mole of O = (206g/16g)( 0.1551) = 1.99 moles  2 moles = z
 C13H18O2 = molecular formula = empirical formula
9. Combustion analysis of toluene, a common organic solvent, produces
5.86 mg of CO2 and 1.37 mg of H2O. If the compound contains only
carbon and hydrogen, what is its molecular formula? Its molecular
mass is 92 g/ mol. Assume CO2 & H2O masses in grams.
9
(3/11)5.86g = 1.60g C; (1/9)1.37g = 0.15g H  toluene sample mass is 1.75g.
% mass C is 1.60g/1.75g x 100 =91.43%
% mass H is 0.15/1.75 x 100 = 8.6%
So moles C in 92g/mol is (0.9143 x 92g)/12g = 7.01 moles  7.00 moles = x
So moles H in 92g/mol is (0.086 x 92g)/1g = 7.9 moles  8.00 moles = y
So for CxHy molecular formula is C7H8
10. A 0.1005 g sample of organic solvent is combusted, producing 0.2829
g of CO2 and 0.1159 g of water. If the compound has a molar mass
of 156 g/mol, what is its molecular formula?
(Remember compound could contain oxygen!!)
10 (3/11)(0.2829g) = 0.07715g C; (1/9)(0.1159g) = 0.01288g H.
Mass of C & H in sample is (0.07715 + 0.01288) = 0.09003g
Therefore sample contains O of mass (0.1005- 0.09003) = 0.01047g O
So the compound is CxHyOz
Moles of C in molecular mass is (0.07715g x 156g)/(0.1005 x 12g) = 9.97  10
Moles of H in molar mass is (0.01288g x 156g)/(0.1005g x 1g) = 19.99  20
Moles of O in molar mass is (0.01047g x 156g)/(0.1005g x 16g) = 1.01  1.00
Then x = 10; y = 20; z = 1
Therefore molecular formula of CxHyOz is C10H20O
11. The complete combustion of octane, C8H18, a component of gasoline,
proceeds as follows:
2 C8H18 + 25 O2  16 CO2 + 18 H2O
a) how many moles of CO2 are produced when 1.50 mol octane
reacted?
b) How many grams water are produced in this reaction?
c) How many moles oxygen gas are required to form 90.0 g water?
11 (a) 16 CO2/2 C8H18 = X /1.5; X = 12.0 moles CO2
(b) 18H2O/2 C8H18 = X/1.5; X = 13.5 moles H2O  13.5x18g/mol = 243g H2O
(c) (90g water)/18g/mol = 5 moles; (25mol)/(18mol) = (X moles O2)/5moles
H2O; X = 6.94 moles oxygen gas.
12. In a certain experiment 2.50 g of NH3 reacts with 2.85 g of O2.
Which reactant is the limiting reactant? How many grams of NO
form? How much excess reactant remains after the liming reactant
is completely consumed?
12. 4NH3 + 5O2  4NO + 6H2O
(a) 2.50g NH3 is 0.147mole  0.147 mole NO produced
2.85g O2 is 0.089mole  (4/5)(0.089mole) = 0.071mole NO produced.
Therefore O2 is limiting reagent.
(b) Grams NO formed is (0.071mole x 30g/mol) = 2.13g NO
(c) NH3 is excess reactant. How much? 0.147mole – (4/5)(0.089mole)= 0.147 – 0.071=
0.076 mole = 17g(0.076mol) = 1.29 grams excess NH3.
13. Given 30.0 g benzene and 65.0 g bromine in the following reaction:
C6H6 + Br2  C6H5Br + HBr
If the actual yield of C6H5Br, bromobenzene, is 56.7 g, what is the
percentage yield?
78g
160g
157g
13. C6H6 + Br2 → C6H5Br + HBr (30g benzene, 65g bromine gas)
30g
65g
56.7g (yield of product)
C6H6 mol wt. is 78g; Br2 mol wt. is 160g; C6H5Br mol wt. is 157g
All stoichiometric relationships are 1 to 1. Limiting reagent?
For Benzene (157g/78g) = (X/30g); X = 60.38g bromobemzene produced.
For Bromine gas (157g/160g) = (X/65g); X = 63.78g bromobenzene produced.
Therefore benzene is limiting reagent. Then theoretically, 30g benzene should
produce 60.38g of bromobenzene for a 100% yield. But since only 56.7g was
produced, percent yield is (56.7g/60.38g) x 100 = 94% yield.
Chapter 3/ practice problems(CHEM 1411)
1. Balance the following equations:
a) C5H10O2 + O2  CO2 + H2O
b) PCl5 + H2O  H3PO4 + HCl
c) Al(OH)3 + H2SO4  ____________+
d) Na + ________  NaOH + H2
____________
2. Write balanced chemical equations to correspond to each of the following
descriptions:
a) Heptane, C7H16, is combusted in air
b) When sulfur trioxide gas reacts with water, a solution of sulfuric
acid forms
c) Lithium metal is added to water
d) ZnCO3(s) is decomposed upon heating
3. Determine the formula weights of each of the following compounds:
a) Ca(C2H3O2)2
b) CH3CH2COOH
c) (NH4)3PO4
4. Calculate the percentage compositions in the following compounds;
a) Benzoic acid, C7H6O2
b) Urea, (NH2)2CO
c) Laughing gas, N2O
5. Gallium comes in two isotopic forms (Ga-69 => 68.926g & Ga- 71 => 70.926g)
Atomic mass of Gallium is 69.726g. Calculate percent abundance of both isotopes.
6. Complete the following conversions;
a) moles of 0.005 g water
c) grams mass of 2.50 mole O2
b) atoms in 3.50 mole NaOH
d) carbon atoms in 10.5 g C6H12O6
7. Determine the empirical formulas of each of the compounds if a sample
contains the following compositions:
a) 62.1% C, 5.21% H, 12.1% N, and 20.7% O
b) 0.104 mol K, 0.052 mol C, and 0.152 mol O
d) 11.66g iron and 5.01g oxygen
8. Determine the molecular formula of each of the following compounds?
a) empirical formula HCO2, molar mass = 90.0 g/mol
b) 75.69% C, 8.80% H, and 15.51% O by mass; molar mass 206g
9. Combustion analysis of toluene, a common organic solvent, produces 5.86
mg of CO2 and 1.37 mg of H2O. If the compound contains only carbon and
hydrogen, what is its molecular formula? Its molecular mass is 92 g/ mol.
10. A 0.1005 g sample of organic solvent is combusted, producing 0.2829 g
of CO2 and 0.1159 g of water. If the compound has a molar mass of 156
g/mol, what is its molecular formula?
11. The complete combustion of octane, C8H18, a component of gasoline,
proceeds as follows:
2 C8H18 + 25 O2  16 CO2 + 18 H2O
a) how many moles of CO2 are produced when 1.50 mol octane reacted?
b) How many grams water are produced in this reaction?
c) How many moles oxygen gas are required to form 90.0 g water?
12. In a certain experiment 2.50 g of NH3 reacts with 2.85 g of O2. Which
reactant is the limiting reactant? How many grams of NO form? How much
excess reactant remains after the liming reactant is completely consumed?
13. Given 30.0 g benzene and 65.0 g bromine in the following reaction:
C6H6 + Br2  C6H5Br + HBr
If the actual yield of C6H5Br, bromobenzene, is 56.7 g, what is the
percentage yield?