By, Rachel Smith and Matt Rose Chapter 3: Stoichiometry
Dec 24, 2015
By, Rachel Smith and Matt Rose
Chapter 3: Stoichiometry
BasicsIt is used to find the quantities of materials
consumed and produced in chemical reactions
Average mass: total mass/amount
Once you have the average mass you can use it to find the estimate of how much of something you will need to meet a certain weight or amount
Example: The average weight of jelly beans is 5.0 grams so how many beans will you have to weigh to have 1000?
Atomic MassesThe most accurate method for comparing masses
of atoms involves the use of the mass spectrometer Atoms are passed into a beam of high-speed
electrons which knock electrons off the atoms being analyzed and change them into positive ions
Example: When 12C and 13C are analyzed in a mass spectrometer the ratio of their masses are mass 13C/mass 12C=1.0836129
Mass of 13C=(1.0836129)(12 amu)= 13.003355 amuThe mass of 12C is exactly 12 atomic mass units
Average Atomic MassIt is known that natural carbon is composed of
98.89% 12C atoms and 1.11% 13C atoms
Average atomic mass= 98.89% of 12 amu+ 1.11% of 13.003355 amu(0.9889)(12amu) + (.0111)(13.003355amu)=
12.01amu
12 amu is the atomic mass for carbon
The MoleAvogradro’s number- one mole of something
consists of 6.022x10^23 units of that substance
In exactly 12 grams of 12C mean that 12 grams of 12C contains 6.022x10^23 atoms
Example: How many moles are there in 43.26 grams of carbon?(43.26g)x(1mol/12.01g)= 3.60 moles of carbon in
43.26 grams
Molar MassMass of 1 mole of methane: CH4
Mass of 1 mol C= 12.01gMass of 4 mol H= 4x1.008gMass of 1 mol CH4= 16.04g
16.04g represents 1 mole of methane molecules which is the molar mass
Molar Mass- the mass in grams of one mole of the compound
Percent Composition of Compounds
Can be found by comparing the mass of each element present in 1 mole of the compound to the total mass of 1 mole compound
Example: Mass percent of carbon in ethanol(mass of C in 1 mol C2H5OH/mas of 1 mol
C2H5OH) x 100%(24.02g/46.07g) x 100% = 52.14%
Section 3.7 Determining the Formula of a Compound
Solving for Empirical Formula• Determine mass percentage, and base that on 100 grams• From the amount of each type of element in the 100 grams determine
moles of a given element by dividing by the atomic mass• Divide each amount of moles by the smallest mole number present
• If each of the resulting numbers is whole, then that is your empirical compound
• If even a single whole number is not whole, you must multiply by an integer which will make all the number whole, the resulting numbers then make you empirical formula
Section 3.7 Determining the Formula of a Compound
Determining Molecular Formula from Empirical Formula
Obtain Empirical Formula
Calculate mass of Empirical Formula
Calculate Ration of Molar Mass over Empirical Formula
Section 3.7 Determining the Formula of a Compound
Ex. A compound is found to be 38.67% Carbon, 16.22% Hydrogen and 45.11% Nitrogen. The molecular mass is found to be 72.12 g, determine Molecular formula
100g x .3876 = 38.67g C
38.67g C x (1 mole C/12.01 g C) = 3.220 mol C
3.220/3.220 = 1
100g x .1622 = 16.22g H
16.22g H x (1 mole H/1.008g H) = 16.09 mol H
16.09/3.220 = 4.997 5
100g x .4511 =45.11g N
45.11g N x (1 mole N/14.01g N) = 3.220 mol N
Empirical Formula Mass1 x 12.01 g C= 12.01 g C5 x 1.008 g H= 5.040 g H1 x 14.01 g N= 14.01 g N-----------------------------------------------------Empirical Formula Mass = 36.06 g
72.12/36.06 = 2
2 x C₁H₅N₁ Molecular Formula C₂H₁₀N₂
3.220/3.220 = 1
Empirical Formula: C₁H₅N₁
Section 3.8 Chemical EquationsChemical Equations
Chemical Equations have 2 side• Left side are the reactants, right side are the products
Important to remember that in a reaction atoms are never created or destroyed so the same amount of atoms should be found on each side of the equation
Chemical Equations tell us 2 important things• The amount of each given atom• The state of each atom
(g)-gas (l)-liquid (s)-solid (aq)-aqueous solution
Ex. 2C₂H₆(l) + 7O₂(g) 4CO₂(g) + 6H₂O(g)
4 Carbon 4 Carbon
12 Hydrogen 12 Hydrogen
14 Oxygen 14 Oxygen
Section 3.9 Chemical Balancing Equations
Balancing an Equation
Done mostly through trial and error
The main objective is to always have the same number of atoms on both side
• In a chemical equation the formulas may no be changed in any manner, only amount may be changed
Writing and Balancing Equations
Determine what type of reaction is occurring, reactants, products and physical states
Create a summarizing unbalanced equation
Balance equation by inspection and trial and error
• Determine coefficients
• Don’t change the identity of any atoms or formulas
Section 3.10 Stoichiometric EquationsChemical equations determine the number of atoms and
compounds by moles• However in application settings all amounts of materials are
determined by mass
Stoichiometry• The branch of chemistry concerned with the proportions in which
elements are combined in compounds and the quantitative relationships between reactants and products in chemical reactions¹
Solving for Stoichiometric Equations
Balance the equation
Convert masses into moles
Set up the appropriate mole rations
Convert back to grams
Section 3.10 Stoichiometric Equations
Ex. What amount of oxygen would it take to react fully with 96.1 grams of Propane?
91.6g C₃H₈(g) + 5O₂(g)3CO₂(g) + 4H₂O(g)
96.1 grams C₃H₈ · · · = 349 g
Section 3.11 The Concept of Limiting Reagent
Stoichiometric mixture-contain relative amounts of that reactants that match the numbers in the equation
When doing a stoichiometric equation you must consider the possibility that not all the reactants will be used because the amount of one may be limited
Limiting Reagent-The reactant which runs out first and thus limits the amount of product that can be formed
How to fund limiting Reagent
• Find moles
• Find mole ratios
• Find the reactants which does not satisfy the equation
Section 3.11 The Concept of Limiting Reagent
Ex. In a reaction with between methane and oxygen, 67.02 grams of Methane and 103.45 grams of oxygen gas are present. Which is the limiting reagent
CH₄(g) + 2O₂(g) CO₂(g) + 2H₂O(g)
103.45g · = 3.233 mol O₂
3.222 mol · = 1.616 mol of CH₄ to react
67.02g · = 4.176 mol CH₄
4.176 mol CH₄ · = 8.352 mol O₂ to react
O₂ is the limiting reagent