Stoichiometry: Chemical Calculations NaCl, salt Buckyball, C 60 Ethanol, C 2 H 6 O

Stoichiometry: Chemical Calculations

Stoichiometry: Chemical Calculations

NaCl, saltNaCl, saltNaCl, saltNaCl, salt

Buckyball, CBuckyball, C6060

Ethanol, CEthanol, C22HH66OO

Just to Review…Please convert 12.4 fluid ounces to cm3

1 fl. oz. = 0.0295735 L

1 cm3 = 1 mL

Convert 4.5 centuries to seconds.

The Carbon-12 ScaleThe Carbon-12 Scale• The atomic mass of an element indicates

how heavy, on average, an atom of an element is when compared to an atom of another element– Unit is the atomic mass unit (amu)

• The standard scaled based on the carbon-12 isotope

• Mass of one 12C atom = 12 amu (exactly)• Note that 12C and C-12 mean the same thing

Figure 3.1 – Mass Spectrometer

• A mass spectrometer is used to determine atomic masses

Figure 3.1 – Mass Spectrometer

Isotopic Abundance

• Most elements exist in nature as a mixture of two or more isotopes.

• To determine the mass of an element, we must know the mass of each isotope and the atom percent of the isotopes (isotopic abundance)

• The mass spectrometer can determine the isotopic abundance

Avogadro’s NumberAvogadro’s Number• A sample of any element with a mass equal

to its atomic mass contains the same number of atoms, NA, regardless of the identity of the element.

– NA = 6.022 X 1023 Avogadro’s #

• It represents the number of atoms of an element in a sample whose mass in grams is numerically equal to the atomic mass of the element.

Amadeo Avogadro & Avogadro’s Number

6.02214199 x 1023

(6.022 x 1023 to 4 s.f.)

Avogadro’s NumberAvogadro’s Number

There is Avogadro’s number of particles in a mole of any substance.There is Avogadro’s number of particles in a mole of any substance.

Amadeo AvogadroAmadeo Avogadro1776-18561776-1856

Examples:

• 6.022 x 1023 H atoms in 1.008 g atomic mass H = 1.008 amu

• 6.022 x 1023 S atoms in 32.07 g atomic mass S = 32.07 amu

Mole: A Perspective on SizeThe Green Pea Analogy:

A Dramatic Reading

Examples of Mole Quantities

EOS

1 mole of stars in the universe = 6.022 x 1023 stars1 mole of pennies = 6.022 x 1023 pennies

(beats the lottery!)

1 mole of glucose molecules = 6.022 x 1023 molecules1 mole of helium atoms = 6.022 x 1023 atoms1 mole of potassium ions (K+) = 6.022 x 1023 ions

Example:water - H2O

Molecular MassMolecular mass: the sum of the atomic masses of all atoms in a molecular formula.

- (units are amu or u)

2 Hydrogen atoms2(1.01 amu)

1 Oxygen atom

+ 16.00 amu

= 18.02 amu

Sample Problem: Glucose

Glucose - C6H12O6

= 6(12.01 u) + 12(1.01 u) + 6(16.00 u)= 180.18 u

EOS

Formula Mass

Formula mass is the sum of the masses of the atoms or ions present in a formula unit – the unit for an ionic formula

Na+ Cl-

Na+

Na+Cl-

Cl-

Cl-Na+

Crystal ofsodium chloride

One Na+ and one Cl– make a formula unit for sodium chloride

The mass of one formula unit is:= 22.99 amu + 35.45 amu= 58.44 amu

Sample Problem

Example 3.1 Determine the formula or molecular mass for each of the following:

CaI2 (NH4)2S

Al(NO3)3 C6H12O6

Mole--DefinitionChemistry is a quantitative

science—we need a “counting unit” aka the

1 mole is the amount of 1 mole is the amount of substance that contains as substance that contains as many particles (atoms, many particles (atoms, molecules) as there are in molecules) as there are in 12.0 g of 12.0 g of 1212C.C.

1 mole is the amount of 1 mole is the amount of substance that contains as substance that contains as many particles (atoms, many particles (atoms, molecules) as there are in molecules) as there are in 12.0 g of 12.0 g of 1212C.C.

MOLE!MOLE!

518 g of Pb, 2.50 mol

One-mole Amounts

AnalogiesWe can group items and count by grouping:

12 eggs = 1 dozen eggs

We can also group items and

count by weighing:

Grass seed and nails —Purchased by the POUND, not by the item.

Molar Mass

The molar mass, MM, in grams/mole, is numerically equal to the sum of the masses (in amu) of the atoms in the formula

Molar mass is the mass of one mole of a particular substance.

Molar Masses of Some Substances

Equivalencies

1 mole of any substance contains Avogadro’s number of particles and the mass on the periodic table expressed in grams.

1 mol of C = 12.01 g of C= 6.022 x 1023 atoms of C

1 mol of O2 = 32.00 g of O2= 6.022 x 1023 molecules

1 mol of NaCl = 58.44 g of NaCl= 6.022 x 1023 formula units of NaCl

The Mole and Reactions

Example: consider the formation of carbon dioxide

EOS

Answer: one doesn’t!

At the molecular level ...

Problem: how does one mass out a single carbon atom? Note that the mass in grams is ~2.00 x 10–23 g!

The Solution ...

EOS

For carbon, mass out:

2.0 × 10–23 g atom–1 × 6.0 × 1023 atoms mol–1

= 12 g C

Use a measurable amount – molar quantities

MOLECULAR WEIGHT VS. MOLAR MASS

MOLECULAR WEIGHT VS. MOLAR MASS

Molecular weight = sum of the atomic

weights of all atoms in the molecule (in

amu)

Molar mass = molecular weight in grams

(in g/mol)

We will use molar mass in all problems in

this chapter!!!!

Equivalencies

same number different entities macroscopic amounts EOS

Must Memorize Elements THAT EXIST AS DIATOMIC MOLECULES Must Memorize Elements THAT

EXIST AS DIATOMIC MOLECULES

Remember Mr. BrINClHOFRemember Mr. BrINClHOFOROR

HON17 !!!HON17 !!!

Practice problems1. Calcium carbonate, CaCO3, is the

principal mineral found in marble and limestone. How many moles are in 188.0 g of CaCO3?

Practice problems

2. What is the mass, in grams, of 0.329 mol of spearmint oil, C10H14O?

Practice problems3. Find the mass of a single lead atom.

Practice problems

4. How many individual lead atoms are in a 1.000 g sample of this metal?

Practice problems5. a. Calculate the number of moles of aluminum in

a solid cube that measures 3.40 cm on a side. (d=2.70 g/cm3).

b. How many atoms of aluminum are in the same sample?

Practice problems6. a. How many molecules of oxygen, O2,

are in 0.00100 grams of this gas?

b. How many atoms?

Molar Mass

Definition: Describes the proportion of elements in a compound using a percent

Equal to the mass of each element present in 100 g of compound!

Percent Composition by Mass

Mass of elementPercent composition of an element = 100

Mass of compound

Sample ProblemExample: Sodium carbonate is a compound used in the manufacture of soap and glass. Determine the percent composition by mass of each element in this compound.

Sample ProblemExample : Determine the percent by

mass of water in Al2(SO4)3∙18H2O.

Sample ProblemExample: Magnetite, Fe2O3, is one of the principal iron containing ores. How much elemental iron can be obtained from a metric ton (103 kg) of this ore, assuming 100 % recovery? (hint: first find % iron in Fe2O3)

Determining FormulasDetermining Formulas

In chemical analysis we determine the % by weight of each element in a given amount of pure compound and derive the EMPIRICAL or SIMPLEST formula.

PROBLEM: A compound of B and H is 81.10% B. What is its empirical formula?

Empirical Formula Calculations

Percent to massMass to molesDivide by Small

Multiply ‘til whole

% g moles empirical formulaDivide by smallest

mole value

Percent to mass: (always assume 100 g sample!)

A compound of B and H is 81.10% B. What is its empirical formula?A compound of B and H is 81.10% B. What is its empirical formula?

Mass to moles:

81.10 g B • 1 mol

10.81 g = 7.502 mol B81.10 g B •

1 mol10.81 g

= 7.502 mol B

18.90 g H • 1 mol

1.008 g = 18.75 mol H18.90 g H •

1 mol1.008 g

= 18.75 mol H


Now, recognize that atoms combine in the ratio of small whole numbers.

Find the ratio of moles of elements in the compound by “dividing by small”


But we need a whole number ratio!

Must multiply each ratio by the smallest integer available to obtain whole numbers (multiply ‘til whole)


Multiply ‘til Whole Hints

Common Possible Endings:

.33 x 3

.25 x 4

.67 x 3

.50 x 2

Sample ProblemExample: Bicarbonate of soda is used in products like Alka-Seltzer and generally relieves an upset stomach. Determine the empirical formula of this compound based on the following percent composition: 27.36% Na, 1.200%H, 14.30% C, 57.14% O.

Sample ProblemExample: A 25.00 gram sample of an orange compound contains 6.64 g of potassium, 8.84 g of chromium, and 9.52 g of oxygen. Find its empirical formula.

A compound of B and H is 81.10% B. Its empirical formula is B2H5. What is its molecular formula?

Is the molecular formula B2H5, B4H10,

B6H15, B8H20, etc.?

BB22HH66 is one example of this class of compounds. is one example of this class of compounds.

B2H6

We need to do an EXPERIMENT to find the MOLAR MASS.

Here experiment gives 53.3 g/molCompare with the mass of B2H5 = 26.66 g/unit

Find the ratio of these masses.53.3 g/mol

26.66 g/unit of B2H5 =

2 units of B2H51 mol

53.3 g/mol26.66 g/unit of B2H5

= 2 units of B2H5

1 mol

Multiply all of the subscripts by the ratio and obtain the molecular formula:

Molecular formula = B4H10

Sample ProblemExample : A certain compound has the empirical formula C2H4O. Its molar mass is about 90 g/mole. What is the molecular formula?

Sample ProblemExample: A hydrate of magnesium iodide has the formula MgI2 ∙ X H2O. To determine the value of X, a student heats a sample of the hydrate until all the water is gone. A 1.628 g sample of hydrate is heated to constant mass of 1.072 g. What is the value of X?

Writing and Balancing Chemical Equations

• All chemical reactions have two parts:

• ReactantsReactants - the substances you start with (on left side of arrow)

• Products Products - the substances you end up with (on right side of arrow)

• The reactants turn into the products.

• Reactants Products

In a chemical reaction…

• The way atoms are joined is changed.• Atoms aren’t created or destroyed; they just

combine together in new ways.• Can be described using sentences, symbols or

word equations:

Example:

Copper reacts with chlorine to form copper (II) chloride.

Copper + chlorine copper (II) chloride

Cu + Cl2 CuCl2

Symbols used in Equations• The arrow separates the reactants from the

products; means “reacts” or “yields”

• The plus sign = “and”

• Subscripts are used to describe the number of atoms in a FORMULA.

• Coefficients are used to describe the number of molecules or formula units in the REACTION. They are the only things changed when balancing a reaction.

States of Matter• Solid--(s) after the formula

• Precipitate -- a solid formed in a reaction

• Gas--(g) after the formula

• Liquid—(l) after the formula

• Aqueous-- (aq) after the formula - dissolved in water.

used after a product indicates a gas (same as (g))

used after a product indicates a solid or precipitate (same as (s))

Other Symbols used in Equations

• indicates a reversible reaction (More later)

• show that heat is supplied to the reaction

• is used to indicate a catalyst used or supplied, in this case, platinum.

heat ,

Pt

Chemical Equations

EOS

Example: consider the formation of water

H2(g) + O2(g) H2O(g)

Law of Conservation of Mass must be obeyed …

therefore, equations must be balanced.

Balancing Equations

Chemical “bookkeeping” of atoms involved in the reaction:

H2(g) + O2(g) H2O(g)

H – 2 O – 2Reactants

H – 2 O – 1Products

COEFFICIENTS must be added so reactant atoms EQUAL product atoms!

Note the imbalance in oxygen atoms

Hints & Tips for Balancing Equations

• Take one element at a time, working from left to right except for H and O. Save H for next to last and O for last.

• IF EVERYTHING BALANCES EXCEPT FOR O, and there is no way to balance O with a whole number, double all the coefficients and try again. (Because O is a diatomic element)

• (Shortcut) polyatomic ions that appear on both sides of the equation should be balanced as independent units

Balancing Equations

EOS

Balancing Equations Practice

Balance the following chemical equations using the appropriate coefficients:

____ Al(s) + _____ Br2 (l) _____ Al2Br6 (s)

____ Na3PO4 + ____ Fe2O3 ____ Na2O + ____ FePO4

Types of EquationsSynthesis or Combination

Equation in Symbols: A + B ABSample Equation:

2Cu (s) + O2 (g) 2 CuO (s)

Predicting Products:

Elements Compounds OR

Compounds

More Complex Compounds

Types of EquationsDecomposition

Equation in Symbols: AB A + BSample Equation:

2 CuO (s) 2Cu (s) + O2 (g)

Predicting Products:Compounds Elements OR More Complex Compounds

Compounds

Types of EquationsSingle Replacement

Equation in Symbols: A + BC AB + C• Metal replacing metal• Nonmetal replacing nonmetal

Sample Equation:

Mg (s) + CuCl2 (aq) Cu (s) + MgCl2 (aq)

Predicting Products:Cations replace cations (can also have anions replacing anions)

Types of EquationsDouble Replacement – 2 ionic compounds

Equation in Symbols: AX + BY BX + AY

Sample Equation: 2AgNO3(aq) + CuCl2 (aq) Cu (NO3)2 (aq) + 2AgCl (s)

Predicting Products:Cations switch places; solid formed

(must be driving force)

Types of EquationsCombustion

Equation in Symbols: CxHy + O2 CO2 + H2O

Sample Equation:

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

Predicting Products:

Hydrocarbons react to form CO2 and

H2O

Stoichiometric Equivalence and Reaction Stoichiometry

CS2 + 3O2 CO2 + 2 SO2

Interpretation in terms of moles:

1 mole of CS2 + 3 moles of O2 form:

1 mole of CO2 + 2 moles of SO2

Stoichiometric Equivalence and Reaction Stoichiometry

CS2 + 3O2 CO2 + 2 SO2

Conversion factors extracted from balanced equation:

1 mole of CS2 3 moles of O2

etc.

3 moles of O2 1 mole of CO2

These ratios are called MOLE RATIOS!

Stoichiometric EquivalentsCoefficients from a balanced chemical equation show molar equivalents of reactants and products

==> form conversion factors

In the formation of water:2 mol H2 = 1 mol O2

2 mol H2 = 2 mol H2O

1 mol O2 = 1 mol H2O

2H2 + O2 2H2O

Sample Problem

Example 3.12 Using the equation below, determine:

CS2 + 3O2 CO2 + 2SO2

• the number of moles of oxygen required to react with 1.38 mol of carbon disulfide

• the number of moles of SO2 produced from 1.38 moles of carbon disulfide.

Sample ProblemExample 3.13

For 2NH3 + H2SO4 (NH4)2SO4 determine:

• the mass of product possible when 1.43 mol of NH3 are reacted with an excess of sulfuric acid.

• the mass of NH3 required to react completely with 35.00 g of sulfuric acid.

• the mass of sulfuric acid required to form 1000 grams of product.

Welcome to Mole Island

1 mole = 22.4 L @ STP

1 mol = molar mass

1 mol = 6.02 x 1023 particles

Stoichiometry Island Diagram

Mass

Particles

Volume Mole Mole

Mass

Known UnknownSubstance A Substance B


Volume

Particles

M

V

P

Mass Mountain

Liter Lagoon

Particle Place

Mole Island


Mass

Particles

Volume Mole Mole

Mass

Volume

Particles

Known Unknown

Substance A Substance B

1 mole = molar mass (g)Use coefficientsfrom balanced

chemical equation1 mole = 22.4 L @ STP

1 mole =

6.022 x

1023 partic

les

(atoms o

r molecu

les)

1 mole = 22.4 L @ STP

1 mole = 6.022 x 10 23 particles

(atoms or molecules)

1 mole =

molar m

ass (g

)

(gases) (gases)

Volume(liquids)

USE

DENSITY

Volume (liquids)

US

E

DE

NSI

TY

Sample Problem

Example : How many milliliters of liquid water can be produced by the combustion of 775 mL of octane with oxygen? Assume that the volumes of the octane and the water are measured at 20oC where the densities are 0.7025 g/mL for octane and 0.9982 g/mL for water.

C8H18(l) + O2(g) CO2(g) + H2O(l)

Reaction Yields

Theoretical yieldTheoretical yield – predicted amount of product formed from the limiting reagent, based only on the stoichiometry of the reaction

If all worked perfectly ...

2H2(g) + O2(g) 2 H2O(g)Example: 1 mol H2 will produce 1 mol of water

Actual yield – amount of product produced

In practice, actual < theoretical: errors, poor technique, etc. ...

Limiting Reactants (Reagents)

EOS

Definition: Chemical reactant that is completely consumed in a reaction and therefore limits the quantity of product formed.

**Depends on stoichiometry of reaction

Limiting Reagents (cont.)

EOS

Limiting Reagents (cont.)How many meals can be made from 105 sandwiches, 202 cookies, and 107 oranges?

1105 105

1

1202 101

2

1107 107

1

mealsandwiches x meals

sandwich

mealcookies x meals

cookies

mealoranges x meals

orange

Cookies limit the total number of whole meals with excess sandwiches and oranges

Excess Reactant = Reactant left over when limiting reactant is used up

Steps for Determining Limiting Reactant

1. Write a balanced equation.2. Convert to moles!3. Take first reactant, calculate theoretical

yield of any product.4. Repeat #3 for second reactant.5. Compare results. Whichever reactant

gives the LEAST amount of the product is the limiting reactant and determines the theoretical yield. The other is in excess.

Sample Problem

CS2(l) + 3 O2(g) CO2(g) + 2 SO2(g)

Determine the theoretical yield of product if one starts with 1.20 mol of CS2 and 3.83 mol O2

Sample ProblemExample: For the following reaction, determine

the theoretical yield of product if one starts with 105 g of CS2 and 145 g of O2

CS2(l) + 3O2(g) CO2(g) + 2SO2(g)

Percent Yield

EOS

100actua

theorpercent y

eticie d

all x

l

Reaction yields are expressed as a ratio in the form of a percentage:

ExampleFor the following reaction, determine the theoretical yield if one starts with 1.20 g of antimony and 2.40 g of iodine.

2 Sb(s) + 3 I2(s) 2 SbI3(s)

Determine theoretical yield first:

ExampleFor the following reaction, determine the theoretical yield if one starts with 1.20 g of antimony and 2.40 g of iodine.

2 Sb(s) + 3 I2(s) 2 SbI3(s)Determine theoretical yield first:

If 3.00 g of product are actually formed, what is the percent yield?

Acknowledgements

• Thomson/Brooks Cole (Textbook Publishers)

• Mark P. Heitz, State University of New York at Brockport, Prentice Hall Publishers (EOS Slides)

Documents

Stoichiometry: Chemical Calculations NaCl, salt Buckyball, C 60 Ethanol, C 2 H 6 O