Chapter 3 stoichiometry

Stoichiometry

Chapter 3Stoichiometry:

Calculations with Chemical Formulas and Equations

Chemistry, The Central Science, 10th editionTheodore L. Brown, H. Eugene LeMay, Jr.,

and Bruce E. Bursten

Stoichiometry

Law of Conservation of Mass

• States that the total mass of a substance present before and after the chemical reaction is the same

2HCl + BaO → H2O + BaCl2

Stoichiometry

Law of Definite Proportions(Joseph Proust, 1799)

• States that all samples of a compound have the same composition, i.e. the same proportion of mass by the constituent elements

• Elemental composition of a pure compound is always the same regardless of the source

Stoichiometry

Law of Multiple Proportions(John Dalton)

• States that the mass of one element that combines with a fixed mass of the other element are in ratios of small whole numbers if two elements are combined to form more than one compound.

Example: CO and CO2

Stoichiometry

Chemical Equations

Concise representations of chemical reactions

Stoichiometry

Anatomy of a Chemical Equation

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Stoichiometry

Anatomy of a Chemical Equation

Reactants appear on the left side of the

equation.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Stoichiometry

Anatomy of a Chemical Equation

Products appear on the right side of the

equation.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Stoichiometry

Anatomy of a Chemical Equation

The states of the reactants and products are written in parentheses to the bottom right of each compound.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Stoichiometry

Anatomy of a Chemical Equation

Coefficients are inserted to balance

the equation.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

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

• Subscripts tell the number of atoms of each element in a molecule

• Coefficients tell the number of molecules

Stoichiometry

Types of Chemical Reaction

Stoichiometry

Combination Reactions

• 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

Stoichiometry

2 Mg (s) + O2 (g) 2 MgO (s)

Stoichiometry

Decomposition Reactions

• 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

Stoichiometry

Combustion Reactions

• 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

Stoichiometry

The Atomic and Molecular Mass

Stoichiometry

Atomic Mass

• The mass of atoms of elements expressed in atomic mass units (amu)

• The amu is defined by assigning a mass of exactly 12 amu to the mass of the 12C isotope of carbon.

Stoichiometry

Isotopes

• Isotopes are two or more atoms (nuclides) having the same atomic number (Z) but different mass numbers.

Stoichiometry

Average Atomic Mass/ Atomic Weight

• The average of the isotopic masses, weighed according to the naturally occuring abundances of the isotopes of an element.

• Unit: amu or u; g/mol

Stoichiometry

Sample Problems

Ex1] Calculate the atomic mass of Ne given the percent natural abundances: 90.48%, 0.27%, and 9.26% for the three isotopes respectively (ans : 20.19 amu)

Ex2] Two naturally occuring isotopes of boron, boron-10 and boron-11, have the masses of 10.012937 u and 11. 009305 u, respectively. Calculate the percent natural abundances of each isotope given the atomic mass of boron as 10.81 u (ans: B-10:20% and B-11:80%)

Stoichiometry

Formula Weight (FW)• Sum of the atomic weights for the atoms

in a chemical formula• So, the formula weight of calcium

chloride, CaCl2, would be Ca: 1(40.1 amu) + Cl: 2(35.5 amu)

111.1 amu

• These are generally reported for ionic compounds

Stoichiometry

Molecular Weight (MW)

• Sum of the atomic weights of the atoms in a molecule

• For the molecule ethane, C2H6, the molecular weight would be

C: 2(12.0 amu)+ H: 6(1.0 amu)

30.0 amu

Stoichiometry

Percent Composition

The percent composition of a component in a compound is the percent of the total mass of the compound that is due to that component.

% A =(number of A atoms)(atomic weight of A)

(FW of the compound)x 100

where A = any element in the periodic table

Stoichiometry

Percent Composition

So the percentage of carbon in ethane, C2H6, is…

%C =(2)(12.0 amu)

(30.0 amu)

24.0 amu

30.0 amu= x 100

= 80.0%

Stoichiometry

Sample Problems

Ex3] Calculate the percentage composition of each element in CCl3Br. (MW=198.26)

Stoichiometry

Moles

Stoichiometry

Avogadro’s Number

• defined as the ratio of the number of constituent particles in a sample to the amount of substance.

• 6.022 x 1023 particles / 1 mole of entity

Particles= atoms/molecules/ions

Stoichiometry

Avogadro’s Number

• 6.022 x 1023

• 1 mole of 12C has a mass of 12 g

Stoichiometry

Molar Mass

• By definition, these are the mass of 1 mol of a substance (i.e., g/mol)– The molar mass of an element is the mass

number for the element that we find on the periodic table

– The formula weight (in amu’s) will be the same number as the molar mass (in g/mol)

Stoichiometry

Using Moles

Moles provide a bridge from the molecular scale to the real-world scale

Stoichiometry

Mole Relationships

• One mole of atoms, ions, or molecules contains Avogadro’s number of those particles

• One mole of molecules or formula units contains Avogadro’s number times the number of atoms or ions of each element in the compound

Stoichiometry

Sample Problems

Ex4]

a) How many molecules are there in

3.10 moles of H2O? (1.87x1024)

b) How many grams are there in 3.8

moles in H2O? (68 g)

Stoichiometry

Finding Empirical & Molecular Formulas

Stoichiometry

• Empirical Formula– Simplest whole-number ratio of atoms.

• Molecular Formula– Represents the molecule of a substance

and shows the actual number of atoms in a molecule.

Stoichiometry

EMPIRICAL

CH2

CH

CO2

CH2O

MOLECULAR

C2H4

C6H6

CO2

C5H10O5

n2615

Stoichiometry

Calculating Empirical Formulas

Stoichiometry

Guidelines in Determining Mole Ratio

• If the mole ratio has the following value:

Mole Ratio Multiply by Factor

a.50 2

a.33 or a.66 3

a.25 or a.75 4

a.20, a.40, a.60, a.80 5

Where a = 0, 1, 2, 3...

Stoichiometry

Calculating Molecular Formula from Empirical Formula

• Molecular formula (MF) can be determined from empirical formula (EF) using the following guidelines:

MF = (EF) n

where n = Molecular Weight, MWEmpirical Formula Weight, EFW

Stoichiometry

Sample Problem

Dibutyl succinate is an insect repellent used against roaches. Its composition is 62.58% C, 9.63%H and 27.79%O.Its experimentally determined molecular mass is 230 amu. Calculate the empirical formula and the molecular formula.

Stoichiometry

Sample Problems

Ex6]

Sorbitol, used as a sweetener in some sugar-free foods, has a molecular mass of 182 u and a mass percent composition: 39.56% C, 7.74% H, and 52.7% O. What are the empirical formula and the molecular formula of sorbitol?

Stoichiometry

Calculating Empirical FormulasEx 5]

Ascorbic acid, commonly known as Vitamin C, is essential for supporting different metabollic reactions in our body. It contains 40.92 %C, 4.58 %H, 54.50 %O by mass. What is the empirical formula of vitamin C?

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

CxHyOz + O2 → CO2 + H2O

Stoichiometry

Sample ProblemsEx7]

Caproic acid, the substance responsible for the aroma of dirty gym socks and running shoes, contain C, H, and O only. On combustion, 0.450g of sample of caproic acid gives 0.418g of H2O and 1.023g CO2. What is the EF of caproic acid? If the molecular weight of caproic acid is 116.2 g/mol, what is its MF?

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

Sample ProblemsEx8] Given the unbalanced equation:

Al(s) + O2(g) Al2O3

a.How many mole of O2 is needed to produce 10 mole of Al2O3 (15 mol)

b.How many moles of Al are needed to completely react with 52.00 g O2? (2.167 mol)

c.How many grams of O2 are needed to produce 158.00 g Al2O3 (74.382 g)

Stoichiometry

Limiting Reactants

Stoichiometry

• Limiting Reactant – The reactant in a chemical reaction that

limits the amount of product that can be formed.

• Excess Reactant– The reactant in a chemical reaction that

remains when the limiting reactant is completely consumed.

Stoichiometry

Analogy: Cheese Sandwich

Supposing: 10 slices loaf bread + 10 slices cheese = X sandwiches

Questions: 1. How many sandwich can we make?2. What is the limiting ingredient?3. What is the excess reactant?

Stoichiometry

Limiting Reactants

• The limiting reactant is the reactant present in the smallest stoichiometric amount– In other words, it’s the reactant you’ll run out of first (in

this case, the H2)

Stoichiometry

Theoretical Yield• The theoretical yield is the amount of

product that can be produced based on calculation.– 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.

Stoichiometry

Percent Yield

A comparison of the amount actually obtained to the amount predicted.

Actual Yield, (AY)Theoretical Yield, (TY)Percent Yield = x 100

Stoichiometry

Sample ProblemsEx9]

How many grams of NO(g) can be produced in the reaction of 1.00 mol of NH3 and 1.00 mol O2?

NH3 (g) + O2 (g) NO(g) + H2O(l)