By, Rachel Smith and Matt Rose

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