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Chapter 3Stoichiometry:
Calculations with Chemical Formulas and Equations
Stoichiometry
is one where the substance retains its identity.
Examples of physical reactions:
◦ melting / freezing◦ boiling / condensing ◦ subliming◦ subdivision◦ dissolving (solvation)
Physical Change
Stoichiometry
the atoms of one or more substances are rearranged to produce new substances with new properties.
the atoms remain the same, but their arrangement changes:
H2O H2 + O2
water is broken down into two substances, oxygen
gas and hydrogen gas. The two new substances bear no resemblance to
the water of which they are made.
Chemical Change
Stoichiometry
a) a colour changeb) gas productionc) an energy changed) precipitate formation
How to Recognize a Chemical Change
Stoichiometry
refers to balancing chemical equations and the associated mathematics.
is possible because of the law of conservation of mass.
Stoichiometry
Stoichiometry
Concise representations of chemical reactions
Chemical Equations
Stoichiometry
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Anatomy of a Chemical Equation
Stoichiometry
Reactants appear on the left side of the
equation.
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Anatomy of a Chemical Equation
Stoichiometry
Products appear on the right side of the equation.
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Anatomy of a Chemical Equation
Stoichiometry
The states of the reactants and products are written in parentheses to the right of each compound.
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Anatomy of a Chemical Equation
Stoichiometry
Coefficients are inserted to balance the equation.
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Anatomy of a Chemical Equation
Stoichiometry
Subscripts and Coefficients Give Different Information Subscripts tell the number of atoms of each
element in a molecule
Stoichiometry
Subscripts and Coefficients Give Different Information Coefficients tell the number of molecules
Stoichiometry
Matter cannot be lost in any chemical reaction.◦all the atoms must balance in the
chemical equation. same number and type on each side of the
equation◦can only change the stoichiometric
coefficients in front of chemical formulas.◦Subscripts in a formula are never
changed when balancing an equation.
Balancing Chemical Equations
Stoichiometry
Balancing Chemical Equations Balance the following equation:
Na(s) + H2O(l) H2 (g) + NaOH(aq)
count the atoms of each kind on both sides of the arrow: ◦ The Na and O atoms are balanced (one Na and one
O on each side)◦ there are two H atoms on the left and three H
atoms on the right.
Stoichiometry
Balancing Chemical Equations Balance the following equation:
Na(s) + 2 H2O(l) H2 (g) + NaOH(aq)
the 2 increases the number of H atoms among the reactants.
O is now unbalanced. have 2 H2O on the left; can balance H by
putting a coefficient 2 in front of NaOH on the right:
Stoichiometry
Balancing Chemical Equations Balance the following equation:
Na(s) + 2 H2O(l) H2 (g) + 2 NaOH(aq) H is balanced
O is balanced Na is now unbalanced, with one on the left
but two on the right.◦ put a coefficient 2 in front of the reactant:
Stoichiometry
Balancing Chemical Equations Balance the following equation:
2 Na(s) + 2 H2O(l) H2 (g) + 2 NaOH(aq)
Check our work:◦ 2 Na on each side◦ 4 H on each side◦ 2 O on each side
Balanced !!
Stoichiometry
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
CH4 (g) + O2 (g) CO2 (g) + H2O
(g)
Stoichiometry
Pb(NO3)2 (aq) + 2 NaI (aq)
PbI2 (aq) + 2 NaNO3 (aq)
Pb(NO3)2(aq) + NaI(aq) PbI2(aq) + NaNO3(aq)
Stoichiometry
Remember:◦ when changing the coefficient of a compound, the
amounts of all the atoms are changed.◦ do a recount of all the atoms every time you
change a coefficient.◦ do a final check at the end.◦ fractions are not allowed. Multiply to reach whole
numbers.
Balancing Chemical Equations
Stoichiometry
(s) - solid (l) - liquid (g) - gas (aq) - aqueous (meaning that the compound
is dissolved in water.) (ppt) - precipitate (meaning that the reaction
produces a solid which falls out of solution.) Reaction conditions occasionally appear above or
below the reaction arrow◦ e.g., "Δ" is often used to indicate the addition of heat).
States of Reactants and Products
Stoichiometry
Balancing Chemical Equations
Complete 3.11 to 3.14
Reaction Types
Stoichiometry
Examples:N2 (g) + 3 H2 (g) 2 NH3 (g)
C3H6 (g) + Br2 (l) C3H6Br2 (l)
2 Mg (s) + O2 (g) 2 MgO (s)
Two or more substances react to form one product
Combination Reactions
Stoichiometry
2 Mg (s) + O2 (g) 2 MgO (s)
Stoichiometry
Examples:CaCO3 (s) CaO (s) + CO2 (g)
2 KClO3 (s) 2 KCl (s) + O2 (g)
2 NaN3 (s) 2 Na (s) + 3 N2 (g)
One substance breaks down into two or more substances
Decomposition Reactions
Stoichiometry
Examples:CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g)
Rapid reactions that produce a flame
Most often involve hydrocarbons reacting with oxygen in the air
Combustion Reactions
Stoichiometry
Complete Problems 3.15 to 3.20
Balancing Chemical Reactions
Formula Weights
Stoichiometry
Sum of the atomic weights for the atoms in a chemical formula
So, the formula weight of calcium chloride, CaCl2, would be
Ca: 1(40.08 amu) + Cl: 2(35.45 amu)
110.98 amu These are generally reported for ionic
compounds
Formula Weight (FW)
Stoichiometry
Sum of the atomic weights of the atoms in a molecule
For the molecule ethane, C2H6, the molecular weight would be
Molecular Weight (MW)
C: 2(12.01 g/mol)+ H: 6(1.01 g/mol)
30.08 g/mol
Stoichiometry
formula and molecular weights are often called molar mass.
The formula weight (in amu) is numerically equal to the molar mass (in g/mol).
Always report molar mass in g/mol. Always round the atomic weights to 2
decimal places.
Molecular & Formula Weights
Stoichiometry
Complete questions 3.21 and 3.22
Formula & Molecular Weights
Stoichiometry
Percentage composition is obtained by dividing the mass contributed by each element (number of atoms times AW) by the formula weight of the compound and multiplying by 100.
Percent Composition
% element =(number of atoms)(atomic weight)
(molar mass of the compound)x 100
Stoichiometry
So the percentage of carbon in ethane (C2H6) is…
Percent Composition
%C = (2)(12.01 g/mol)(2(12.01) + 6(1.01))g/mol
24.02 g/mol30.08 g/mol
= x 100
= 79.85%
Stoichiometry
• the formula weight is 98.09 amu (do calculation)
• % H = 2(1.01 amu) x 100 = 2.06 % H98.09 amu
• % S = 32.07 amu x 100 = 32.69 % S 98.09 amu
• % O = 4(16.00 amu) x 100 = 65.25 % O 98.09 amu
the sum of the percent compositions will always equal 100 %
% Composition of H2SO4
Stoichiometry
Complete questions 3.24 and 3.26 Calculate the % composition of each
element in each compound.
Percent Composition
Moles
Stoichiometry
Avogadro’s Number
6.022 x 1023
1 mole of 12C has a mass of 12 g
Stoichiometry
By definition, these are the mass of 1 mol of a substance (i.e., g/mol)◦ is the atomic weight on the periodic table◦ The formula weight (in amu’s) will be the same
number as the molar mass (in g/mol)
Molar Mass
Stoichiometry
Using Moles
Moles provide a bridge from the molecular scale to the real-world scale
Stoichiometry
Molar Calculations Using the mass and molar mass to calculate
the number of MOLES of a compound or element.◦ m Mass grams
g◦ n amount of matter moles mol◦ M Molar mass grams per mole
g/mol
number of = mass moles molar mass
n = mM
Stoichiometry
Sample Calculation
If you have 20.2 grams of Potash, how many moles do you have?
Molar Mass KCl (calculated): 74.55 g/mol Mass KCl: 20.2 g
20.2 g = 0.271 moles KCl 74.55 g/mol
if the units work out, you are correct.
Stoichiometry
How many moles are present in 0.750 g of ammonium phosphate?
Ammonium phosphate: (NH4)3PO4
Molar mass: (3(14.01) + 12(1.01) + 30.98 + 4(16.00)) g/mol = 149.13 g/mol
Number of moles = 0.750 g = 5.03 x 10-3 mol 149.13 g/mol (NH4)3PO4
Example
Stoichiometry
Example What is the mass of 12.31 mol of NaOH ?
◦ Step 1. Calculate molar mass 1 x 22.00 g/mol 1 x 16.00 g/mol 1 x 1.01 g/mol 40.00 g/mol
◦ Step 2. Calculate mass (12.31 mol)(40.00 g/mol) = 492.4 g NaOH
Stoichiometry
What is the mass of 2.1 x 10-4 mol of diarsenic trioxide?
Diarsenic trioxide: As2O3
Molar mass: (2(74.92) + 3(16.00)) g/mol = 197.84
g/mol Mass = (2.1 x 10-4 mol)(197.84 g/mol)
= 0.042 g As2O3
Example
Stoichiometry
ExampleHow many moles of caffeine (C8H10N4O2 ) is required to reach the 12.0 g lethal dose in a 140 pound female? Step 1. Molar Mass
8 x 12.01 g/mol 10 x 1.01 g/mol 4 x 14.01 g/mol 2 x 16.00 g/mol 194.22 g/mol
Step 2. Calculate Moles 12.00 g = 0.06179 mol caffeine 194.22 g/mol
Stoichiometry
Calculating Atoms…
We can also find the number of atoms of a substance using Avogadro’s Number:
6.022 x 1023 atoms/mol Example: Calculate the number of atoms in 0.500 moles
of silver?0.500mol Ag x 6.022x1023 atoms = 3.01 x
1023 atoms mol of Ag
Stoichiometry
The Same Goes…
For molecules! Avogadro’s number represents the number of
particles, whether atoms or molecules.
Eg. Calculate the number of molecules in 4.99mol of methane. 4.99 mol x 6.022 x 1023 molecules = 3.00 x 1024 molecules
mol of CH4
Stoichiometry
How many molecules are present in 0.045 mol of H2O?
Number of atoms = (0.045 mol)(6.022 x 1023
particles/mol) = 2.7 x 1022 molecules H2O
Example
Stoichiometry
How many moles are present in 8.75 x 1026 molecules of Vitamin C?
Moles = 8.75 x 1026 molecules = 1450 mol 6.022 x 1023 particles/molVitamin C
Example
Stoichiometry
What is the mass of 2.1 x 1022 molecules of N2O?
Step 1. Convert molecules to moles:◦ Moles = 2.1 x 1022 molecules = 0.0349
mol N2O 6.022 x 1023 particles/mol
Step 2. Convert moles to mass:◦ Mass = (0.0349 mol)((2(14.01) + 16.00)
g/mol) = 1.54 g N2O
Example
Stoichiometry
How many molecules are present in 4.65 t of CO2?
Step 1: Convert mass to moles◦ Moles = (4.65 t)(1000 kg/t)(1000 g/kg) = 1.06 x
105 (12.01 + 2(16.00)) g/mol mol CO2
Step 2: Convert moles to molecules◦ Molecules = (1.06 x 105 mol)(6.022 x 1023
particles/mol) = 6.36 x 1028 molecules CO2
Example
Stoichiometry
Complete questions 3.32 to 3.42, even (leave 3.32 for last)
The Mole
Finding Empirical Formulas
Stoichiometry
Calculating Empirical Formulas
One can calculate the empirical formula from the percent composition
Stoichiometry
1. Assume we start with 100 g of sample.2. The mass percent then translates as the
number of grams of each element in 100 g of sample.
3. From these masses, calculate the number of moles (use atomic weight from Periodic Table)
4. The lowest number of moles becomes the divisor for the others. (gives a mole ratio greater than 1)
5. Adjust mole ratios so all numbers are whole (1, 2, etc)
6. The lowest whole-number ratio of moles is the empirical formula.
Finding empirical formula from mass percent data:
Stoichiometry
Calculating Empirical Formulas
The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA.
Assumptions:1. These are % by mass; in 100g of the compound,
the percentages represent masses in grams.2. Mass divided by molar mass gives moles of each
atom.
Stoichiometry
Calculating Empirical Formulas
Assuming 100.00 g of para-aminobenzoic acid, each percentage converts into a mass:
C: 61.31 g x = 5.105 mol C
H: 5.14 g x = 5.09 mol H
N: 10.21 g x = 0.7288 mol N
O: 23.33 g x = 1.456 mol O
1 mol12.01 g
1 mol14.01 g
1 mol1.01 g
1 mol16.00 g
Stoichiometry
Calculating Empirical Formulas
Calculate the mole ratio by dividing by the smallest number of moles:
C: = 7.005 7
H: = 6.984 7
N: = 1.000
O: = 2.001 2
5.105 mol0.7288 mol
5.09 mol0.7288 mol
0.7288 mol0.7288 mol
1.458 mol0.7288 mol
Stoichiometry
Calculating Empirical Formulas
These are the subscripts for the empirical formula:
C7H7NO2
Stoichiometry
Complete questions 3.43 to 3.46
Empirical Formulas
Stoichiometry
Once we know the composition of a compound the next step is to determine the molar mass.
From these two pieces of information the actual molecular formula can be determined:◦ take the ratio of the actual molar mass divided by
the molar mass determined by the empirical formula
◦ multiply the atoms in the empirical formula by the results of the ratio.
Molecular Formulas
Stoichiometry
Complete questions 3.47 to 3.50
Stoichiometry
Combustion Analysis
Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this◦ C is determined from the mass of CO2 produced
◦ H is determined from the mass of H2O produced◦ O is determined by difference after the C and H have
been determined
Stoichiometry
Elemental Analyses
Compounds containing other elements are analyzed using methods analogous to those used for C, H and O
Stoichiometry
Complete questions 3.51 & 3.52
Stoichiometry
Stoichiometric Calculations
The coefficients in the balanced equation give the ratio of moles of reactants and products
Stoichiometry
Stoichiometric CalculationsFrom the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant)
Stoichiometry
Stoichiometric Calculations
C6H12O6 + 6 O2 6 CO2 + 6 H2O
Stoichiometry
Work on questions 3.57 to 3.66 Copy examples done in class.
Stoichiometric Calculations
Limiting Reactants
Stoichiometry
How Many Cookies Can I Make?
You can make cookies until you run out of one of the ingredients
Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat)
Stoichiometry
How Many Cookies Can I Make?
In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make
Stoichiometry
Limiting Reactants
The limiting reactant is the reactant present in the smallest stoichiometric amount
Stoichiometry
Limiting Reactants
The limiting reactant is the reactant you’ll run out of first (in this case, the H2)
Stoichiometry
Limiting Reactants
In the example below, the O2 would be the excess reagent
Stoichiometry
Complete questions 3.69 to 3.76
Stoichiometry
The theoretical yield is the amount of product that can be made◦ In other words it’s the amount of product possible
as calculated through the stoichiometry problem This is different from the actual yield, the
amount one actually produces and measures
Theoretical Yield
Stoichiometry
A comparison of the amount actually obtained to the amount it was possible to make
Percent Yield
Actual YieldTheoretical YieldPercent Yield = x 100
Stoichiometry
Complete questions 3.77 to 3.80