Chapter Three - Wikispaces3+-+Stoichiometry... · General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three Stoichiometry: Chemical Calculations Chapter Three .

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Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

Chapter Three

Stoichiometry: Chemical Calculations

Chapter Three

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Chapter Three

• Molecular mass: sum of the masses of the atoms

represented in a molecular formula.

• Simply put: the mass of a molecule.

• Molecular mass is specifically for molecules.

• Ionic compounds don’t exist as molecules; for

them we use …

• Formula mass: sum of the masses of the atoms or

ions present in a formula unit.

Molecular Masses and

Formula Masses

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Chapter Three

Example 3.1 Calculate the molecular mass of glycerol (1,2,3-

propanetriol).

Example 3.2 Calculate the formula mass of ammonium

sulfate, a fertilizer commonly used by home

gardeners.

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Chapter Three

Determining the Formula Mass

of Ammonium Sulfate

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Chapter Three

• Mole (mol): amount of substance that contains as

many elementary entities as there are atoms in

exactly 12 g of the carbon-12 isotope.

• Atoms are small, so this is a BIG number …

• Avogadro’s number (NA) = 6.022 × 1023 mol–1

• 1 mol = 6.022 × 1023 “things” (atoms, molecules,

ions, formula units, oranges, etc.)

– A mole of oranges would weigh about as much as the

earth!

• Mole is NOT abbreviated as either M or m.

The Mole & Avogadro’s Number

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Chapter Three

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Chapter Three

One Mole of

Four Elements

One mole each of helium,

sulfur, copper, and mercury.

How many atoms of helium

are present? Of sulfur? Of

copper? Of mercury?

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Chapter Three

Example 3.3 Determine (a) the mass of a 0.0750-mol sample

of Na, (b) the number of moles of Na in a 62.5-

g sample, (c) the mass of a sample of Na

containing 1.00 × 1025 Na atoms, and (d) the

mass of a single Na atom.

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Chapter Three

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

• Molar mass is numerically equal to atomic mass, molecular mass, or formula mass. However …

• … the units of molar mass are grams (g/mol).

• Examples:

1 atom Na = 22.99 u 1 mol Na = 22.99 g

1 formula unit KCl = 74.56 u

The Mole and Molar Mass

1 mol CO2 = 44.01 g 1 molecule CO2 = 44.01 u

1 mol KCl = 74.56 g

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Chapter Three

We can use these equalities to construct conversion factors,

such as:

Note: preliminary and follow-up calculations may be needed.

1 mol Na ––––––––– 22.99 g Na

Conversions involving Mass, Moles,

and Number of Atoms/Molecules

1 mol Na = 6.022 × 1023 Na atoms = 22.99 g Na

22.99 g Na ––––––––– 1 mol Na

1 mol Na –––––––––––––––––– 6.022 × 1023 Na atoms

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Chapter Three

We can read formulas in terms of

moles of atoms or ions.

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Chapter Three

Example 3.4 Determine (a) the number of NH4

+ ions in a 145-g

sample of (NH4)2SO4 and (b) the volume of 1,2,3-

propanetriol (glycerol, d = 1.261 g/mL) that contains

1.00 mol O atoms.

Example 3.5 An Estimation Example Which of the following is a reasonable value for the

number of atoms in 1.00 g of helium?

(a) 4.1 × 10–23 (c) 1.5 × 1023

(b) 4.0 (d) 1.5 × 1024

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Chapter Three

The mass percent composition of a compound refers

to the proportion of the constituent elements,

expressed as the number of grams of each element

per 100 grams of the compound. In other words …

Mass Percent Composition

from Chemical Formulas

X g element X % element = –––––––––––––– OR … 100 g compound

g element % element = ––––––––––– × 100 g compound

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Chapter Three

Percentage Composition of Butane

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Chapter Three

Example 3.6 Calculate, to four significant figures, the mass percent of each element in ammonium nitrate.

Example 3.7 How many grams of nitrogen are present in 46.34 g ammonium nitrate?

Example 3.8 An Estimation Example Without doing detailed calculations, determine which of these compounds contains the greatest mass of sulfur per gram of compound: barium sulfate, lithium sulfate, sodium sulfate, or lead sulfate.

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Chapter Three

• We can “reverse” the process of finding

percentage composition.

• First we use the percentage or mass of each

element to find moles of each element.

• Then we can obtain the empirical formula by

finding the smallest whole-number ratio of moles.

– Find the whole-number ratio by dividing each number

of moles by the smallest number of moles.

Chemical Formulas from Mass

Percent Composition

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Chapter Three

Example 3.9 Phenol, a general disinfectant, has the composition

76.57% C, 6.43% H, and 17.00% O by mass.

Determine its empirical formula.

Example 3.10 Diethylene glycol, used in antifreeze, as a softening

agent for textile fibers and some leathers, and as a

moistening agent for glues and paper, has the

composition 45.27% C, 9.50% H, and 45.23% O by

mass. Determine its empirical formula.

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Chapter Three

• A molecular formula is a simple integer multiple of the empirical formula.

• That is, an empirical formula of CH2 means that the molecular formula is CH2, or C2H4, or C3H6, or C4H8, etc.

• So: we find the molecular formula by:

= integer (nearly) molecular formula mass

empirical formula mass

We then multiply each subscript in the empirical formula by the

integer.

Relating Molecular Formulas

to Empirical Formulas

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Chapter Three

Example 3.11 The empirical formula of hydroquinone, a

chemical used in photography, is C3H3O, and

its molecular mass is 110 u. What is its

molecular formula?

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Chapter Three

• … is one method of determining empirical

formulas in the laboratory.

• This method is used primarily for simple organic

compounds (that contain carbon, hydrogen,

oxygen).

– The organic compound is burned in oxygen.

– The products of combustion (usually CO2 and H2O)

are weighed.

– The amount of each element is determined from the

mass of products.

Elemental Analysis …

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Chapter Three

Elemental Analysis (cont’d)

The sample is

burned in a

stream of oxygen

gas, producing …

… H2O, which is absorbed

by MgClO4, and …

… CO2, which is

absorbed by NaOH.

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Chapter Three

Elemental Analysis (cont’d)

If our sample

were CH3OH,

every two

molecules of

CH3OH …

… would give two

molecules of CO2 …

… and four

molecules of H2O.

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Chapter Three

Example 3.12 Burning a 0.1000-g sample of a carbon–

hydrogen–oxygen compound in oxygen yields

0.1953 g CO2 and 0.1000 g H2O. A separate

experiment shows that the molecular mass of

the compound is 90 u. Determine (a) the mass

percent composition, (b) the empirical formula,

and (c) the molecular formula of the compound.

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Chapter Three

• A chemical equation is a shorthand description of a

chemical reaction, using symbols and formulas to represent

the elements and compounds involved.

Writing Chemical Equations

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Chapter Three

• Sometimes

additional

information

about the

reaction is

conveyed in

the equation.

Writing Chemical Equations

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Chapter Three

Balancing Equations Illustrated

The equation is

balanced by changing

the coefficients …

How can we tell

that the

equation is not

balanced?

… not by changing

the equation …

… and not by

changing the

formulas.

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Chapter Three

• If an element is present in just one compound on each

side of the equation, try balancing that element first.

• Balance any reactants or products that exist as the

free element last.

• In some reactions, certain groupings of atoms (such

as polyatomic ions) remain unchanged. In such cases,

treat these groupings as a unit.

• At times, an equation can be balanced by first using a

fractional coefficient(s). The fraction is then cleared

by multiplying each coefficient by a common factor.

Guidelines for Balancing

Chemical Equations

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Chapter Three

Example 3.13 Balance the equation

Fe + O2 Fe2O3 (not balanced)


C2H6 + O2 CO2 + H2O


H3PO4 + NaCN HCN + Na3PO4

Example 3.16 A Conceptual Example Write a plausible chemical equation for the reaction between water and a liquid molecular chloride of phosphorus to form an aqueous solution of hydrochloric acid and phosphorus acid. The phosphorus-chlorine compound is 77.45% Cl by mass.

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Chapter Three

• A stoichiometric factor or mole ratio is a

conversion factor obtained from the stoichiometric

coefficients in a chemical equation.

• In the equation: CO(g) + 2 H2(g) CH3OH(l)

– 1 mol CO is chemically equivalent to 2 mol H2

– 1 mol CO is chemically equivalent to 1 mol CH3OH

– 2 mol H2 is chemically equivalent to 1 mol CH3OH

Stoichiometric Equivalence

and Reaction Stoichiometry

1 mol CO ––––––––– 2 mol H2

1 mol CO ––––––––––––– 1 mol CH3OH

2 mol H2 ––––––––––––– 1 mol CH3OH

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Chapter Three

One car may be equivalent

to either 25 feet or 10 feet,

depending on the method of

parking.

One mole of CO may be

equivalent to one mole of

CH3OH, or to one mole of CO2,

or to two moles of CH3OH,

depending on the reaction(s).

Concept of Stoichiometric Equivalence

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Chapter Three

Outline of Simple Reaction Stoichiometry

Note: preliminary and/or follow-up calculations may be needed.

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Chapter Three

Example 3.17 When 0.105 mol propane is burned in an

excess of oxygen, how many moles of oxygen

are consumed? The reaction is

C3H8 + 5 O2 3 CO2 + 4 H2O

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Chapter Three

Outline of Stoichiometry Involving Mass

To our simple

stoichiometry

scheme …

… we’ve added a

conversion from

mass at the

beginning …

… and a

conversion to

mass at the end.

Substances A and B

may be two reactants,

two products, or

reactant and product.

Think: If we are given

moles of substance A

initially, do we need

to convert A to

grams?

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Chapter Three

Example 3.18 The final step in the production of nitric acid involves

the reaction of nitrogen dioxide with water; nitrogen

monoxide is also produced. How many grams of nitric

acid are produced for every 100.0 g of nitrogen dioxide

that reacts?

Example 3.19 Ammonium sulfate, a common fertilizer used by

gardeners, is produced commercially by passing

gaseous ammonia into an aqueous solution that is 65%

H2SO4 by mass and has a density of 1.55 g/mL. How

many milliliters of this sulfuric acid solution are required

to convert 1.00 kg NH3 to (NH4)2SO4?

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Chapter Three

• Many reactions are carried out with a limited

amount of one reactant and a plentiful amount of

the other(s).

• The reactant that is completely consumed in the

reaction limits the amounts of products and is

called the limiting reactant, or limiting reagent.

• The limiting reactant is not necessarily the one

present in smallest amount.

Limiting Reactants

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Chapter Three

Limiting Reactant Analogy

If we have 10

sandwiches, 18

cookies, and 12

oranges …

… how many

packaged meals

can we make?

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Chapter Three

When 28 g (1.0

mol) ethylene

reacts with … … 128 g (0.80 mol)

bromine, we get …

… 150 g of 1,2-

dibromoethane, and

leftover ethylene!

1. Why is ethylene left

over, when we started

with more bromine

than ethylene? (Hint:

count the molecules.)

2. What mass of ethylene

is left over after

reaction is complete?

(Hint: it’s an easy

calculation; why?)

Molecular View of the Limiting Reactant Concept

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Chapter Three

• We recognize limiting reactant problems by the

fact that amounts of two (or more) reactants are

given.

• One way to solve them is to perform a normal

stoichiometric calculation of the amount of

product obtained, starting with each reactant.

• The reactant that produces the smallest amount

of product is the limiting reactant.

Recognizing and Solving

Limiting Reactant Problems

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Chapter Three

Example 3.20 Magnesium nitride can be formed by the

reaction of magnesium metal with nitrogen gas.

(a) How many grams of magnesium nitride can

be made in the reaction of 35.00 g of

magnesium and 15.00 g of nitrogen? (b) How

many grams of the excess reactant remain after

the reaction?

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Chapter Three

• The theoretical yield of a chemical reaction is the

calculated quantity of product in the reaction.

• The actual yield is the amount you actually get

when you carry out the reaction.

• Actual yield will be less than the theoretical yield,

for many reasons … can you name some?

Yields of Chemical Reactions

actual yield Percent yield = ––––––––––––– × 100 theoretical yield

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Chapter Three

Actual Yield of ZnS Is Less than

the Theoretical Yield

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Chapter Three

Example 3.21 Ethyl acetate is a solvent used as fingernail polish remover.

What mass of acetic acid is needed to prepare 252 g ethyl

acetate if the expected percent yield is 85.0%? Assume that

the other reactant, ethanol, is present in excess. The

equation for the reaction, carried out in the presence of

H2SO4, is

CH3COOH + HOCH2CH3 CH3COOCH2CH3 + H2O

Acetic acid Ethanol Ethyl acetate

Example 3.22 A Conceptual Example What is the maximum yield of CO(g) obtainable from 725 g

of C6H14(l), regardless of the reaction(s) used, assuming no

other carbon-containing reactant or product?

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Chapter Three

• Solute: the substance being dissolved.

• Solvent: the substance doing the dissolving.

• Concentration of a solution: the quantity of a

solute in a given quantity of solution (or solvent).

– A concentrated solution contains a relatively large

amount of solute vs. the solvent (or solution).

– A dilute solution contains a relatively small

concentration of solute vs. the solvent (or solution).

– “Concentrated” and “dilute” aren’t very quantitative …

Solutions and

Solution Stoichiometry

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Chapter Three

Molarity (M), or molar concentration, is the amount

of solute, in moles, per liter of solution:

• A solution that is 0.35 M sucrose contains 0.35

moles of sucrose in each liter of solution.

• Keep in mind that molarity signifies moles of

solute per liter of solution, not liters of solvent.

Molar Concentration

moles of solute Molarity = –––––––––––––– liters of solution

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Chapter Three

Preparing 0.01000 M KMnO4

Weigh 0.01000

mol (1.580 g)

KMnO4.

Dissolve in water. How much

water? Doesn’t matter, as long as

we don’t go over a liter.

Add more water

to reach the

1.000 liter mark.

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Chapter Three

Example 3.23

What is the molarity of a solution in which 333 g potassium hydrogen carbonate is dissolved in enough water to make 10.0 L of solution?

Example 3.24

We want to prepare a 6.68 molar solution of NaOH (6.68 M NaOH).

(a) How many moles of NaOH are required to prepare 0.500 L of 6.68 M NaOH?

(b) How many liters of 6.68 M NaOH can we prepare with 2.35 kg NaOH?

Example 3.25

The label of a stock bottle of aqueous ammonia indicates that the solution is 28.0% NH3 by mass and has a density of 0.898 g/mL. Calculate the molarity of the solution.

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Chapter Three

• Dilution is the process of preparing a more dilute

solution by adding solvent to a more concentrated

one.

• Addition of solvent does not change the amount of

solute in a solution but does change the solution

concentration.

• It is very common to prepare a concentrated stock

solution of a solute, then dilute it to other

concentrations as needed.

Dilution of Solutions

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Chapter Three

Visualizing the Dilution of a Solution

We start and

end with the

same amount of

solute.

Addition of

solvent has

decreased the

concentration.

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Chapter Three

Dilution Calculations …

• … couldn’t be easier.

• Moles of solute does not change on dilution.

• Moles of solute = M × V

• Therefore …

Mconc × Vconc = Mdil × Vdil

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Chapter Three

Example 3.26

How many milliliters of a 2.00 M CuSO4 stock solution are needed to prepare 0.250 L of 0.400 M CuSO4?

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Chapter Three

• Molarity provides an additional tool in

stoichiometric calculations based on

chemical equations.

• Molarity provides factors for converting

between moles of solute (either reactant or

product) and liters of solution.

Solutions in Chemical

Reactions

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Chapter Three

If substance A is a

solution of known

concentration …

If substance B is in

solution, then … … we can start with

molarity of A times

volume (liters) of the

solution of A to get

here. … we can go from

moles of substance B

to either volume of B

or molarity of B.

How?

Adding to the previous stoichiometry scheme …

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Chapter Three

Example 3.27 A chemical reaction familiar to geologists is that used to identify limestone. The reaction of hydrochloric acid with limestone, which is largely calcium carbonate, is seen through an effervescence—a bubbling due to the liberation of gaseous carbon dioxide:

CaCO3(s) + 2 HCl(aq) CaCl2(aq) + H2O(l) + CO2(g)

How many grams of CaCO3(s) are consumed in a reaction with 225 mL of 3.25 M HCl?

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Chapter Three

Cumulative Example The combustion in oxygen of 1.5250 g of an

alkane-derived compound composed of

carbon, hydrogen, and oxygen yields 3.047 g

CO2 and 1.247 g H2O. The molecular mass of

this compound is 88.1 u. Draw a plausible

structural formula for this compound. Is there

more than one possibility? Explain.

Chapter Three - Wikispaces3+-+Stoichiometry... · General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three Stoichiometry: Chemical Calculations Chapter Three .

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