Stoichiometry1

Stoichiometry 1

The Mole

•

Chemical reactions involve atoms and molecules.

•

The ratios with which elements combine depend

on the number of atoms not on their mass.

•

The masses of atoms or molecules depend on

the substance.

•

Individual atoms and molecules are extremely

small. Hence a larger unit is appropriate for

measuring quantities of matter.

•

A mole is equal to exactly the number of atoms in

exactly 12.0000 grams of carbon 12.

Definitions of the Mole

•

_{1 mole of a substance has a mass equal to the}

formula mass in grams

.

•

_Examples

• _{1 mole H2O is the number of molecules in 18.015 g H2O} • _{1 mole H2 is the number of molecules in 2.016 g H2.}

• 1 mole of atoms has a mass equal to the atomic weight in grams.

• 1 mole of particles = 6.02214 x 1023 particles for any

substance!

• The

Molar mass

is the mass of one mole of a

substance

The Formula Mass

•

The

formula mass

is the sum of atomic

masses in a formula.

Gram Formula Mass and Molar Mass

•

If the

formula mass

is expressed in

grams it is called a gram formula mass.

•

The gram formula mass is also known

as the

Molar Mass.

•

_{The molar mass is the number of grams}

necessary to make 1 mole of a

substance.

Formula Mass and the

Mole

•

The atomic mass of Carbon 12 is exactly

12.00000.

•

1 atomic mass unit = 1/12 of the atomic mass

of carbon 12.

•

The periodic table gives the average atomic

mass for an element relative to Carbon 12.

•

1 mole of a substance is 6.022 x 10

particles.

Gram Formula Mass

•

The

formula mass

is the sum of the

atomic masses in a formula.

•

A

gram formula mass

is the same number

expressed in grams.

•

It is also equal to Avogadro’s Number of

particles

• _{Example: H2O}

From the Periodic Table - Atomic Masses: H =1.00797, O = 15.999 The formula mass = 2(1.00797)+15.999 = 18.015

The Mole

•

The mole is connects the macro world

that we can measure with the micro

world of atoms and molecules.

•

_{A Mole is also equal to}

–

1 gram formula mass.

–

22.4 dm

of any gas measured at 0

C and

Example 1: Calculating the

Molar Mass of a Compound

•

Calculate the gram formula mass or

Molar Mass

of

Na

PO

.

Atom

#

Atomic Mass

Total

Na

3

X 23.0

=

69.0

P

1

X 31.0

=

31.0

O

4

X 16.0

=

64.0

Total

=

164.0

Example 2:

Find the mass of 2.50 moles of Ca(OH)

Find the molar mass of Calcium hydroxide and

multiply by 2.50 mol

The molar mass of

Ca(OH)

is

1 Ca 1 x 40.08 = 40.08

2 O 2 x 16.00 = 32.00

2 H 2 x 1.01 = 2.02

Calculating Moles

• The number of moles in a given mass of a

substance can be determined by dividing the

mass by the molar mass

Moles =

Mass

Example 3:

Find the number of moles in 44.46 grams of

Ca(OH)

Find the molar mass of and divide it into the

given mass

From the previous example the molar mass of

calcium hydroxide is 74.10 gmol

.

44.46 g

Ca(OH)

74.10 g mol

Ca(OH)

Example 4: Calculating

Moles

• Calculate the number of moles in 20.5 grams of Na₃PO₄

Moles =

Mass

Molar Mass

Moles =

20.5 g

164.0 g mol

Calculating Mass From

Moles

• The mass of a quantity of a substance can

be found by multiplying the number of moles

by the molar mass

Example 5 Calculating Mass

from Moles

• Calculate the mass of 2.50 moles of

Na

PO

Mass =

Moles X Molar Mass

=

=

2.50 mol x 164.0 g mol

Percentage Composition

•

According to the law of definite proportions,

compounds, contain definite proportions of each

element by mass.

•

The sum of all of the atomic masses of elements in

a formula is called the

formula mass

.

•

If it is expressed in grams, then it is called a gram

formula mass or

molar mass

.

•

If it represents the sum of all of the masses of all

of the elements in a molecule then it is called a

molecular mass

.

•

To find the percentage of each element in a

Percentage Composition

•

The percent by mass of each element in a

compound is equal to the percentage that

its atomic mass is of the formula mass.

•

_{Example: Calculate the percentage of}

oxygen in potassium chlorate, KClO

Atomic masses: K = 39.09, Cl = 35.45 and O = 16.00. Formula mass = 39.09 + 35.45+ 3(16.00) = 122.54

Example 2

• _{Calculate the percentage by mass of each element in}

potassium carbonate, K₂CO₃

First calculate the formula mass for K₂CO₃ . Find the atomic

mass of each element from the periodic table. Multiply it by the number of times it appears in the formula and add up the total

2 Potassium atoms K 2 x 39.10 = 78.20 1 carbon atom C 1 x 12.01 = 12.01 3 Oxygen atoms O 3 x 16.00 = 48.00 Total = 138.21

To find the percent of each element divide the part of the formula mass that pertains to that element with the total formula mass

Percent of Potassium K = 78.20 X 100 =56.58 % 138.21

Percent of Carbon C = 12.01 X 100 = 8.69 % 138.21

Empirical Formula Determination

The empirical formula is the simplest ratio of

the numbers of atoms of each element that

make a compound.

To find the empirical formula of a compound:

1. Divide the amount of each element (either in mass or percentage) by its atomic mass. This calculation gives you moles of atoms for each element that

appears in the formula

Example 1

Analysis of a certain compound showed that 32.356 grams of compound contained 0.883 grams of

hydrogen, 10.497 grams of Carbon, and 27.968

grams of Oxygen. Calculate the empirical formula of the compound.

First divide the amount by the atomic mass to get the number of moles of each kind of atom in the formula

Hydrogen H = 0.883 g = 0.874 mol 1.01 g mol-1

Carbon C = 10.497 g = 0.874 mol 12.01 g mol-1

Oxygen O = 27.968 g = 1.748 mol 16.00 g mol-1

Molecular Formula

• To calculate the molecular formula from the empirical formula it is necessary to know the molecular (molar) mass.

• Add up the atomic masses in the empirical formula to get the factor

• Divide this number into the molecular formula mass.

• If the number does not divide evenly you probably have a mistake in the empirical formula or its

formula mass

Molecular Formula Example

compound in the previous example, HCO₂ is 90.0. Calculate the molecular formula.

• The empirical formula mass of is 1 H 1.0 x 1 = 1.0

1 C 12.0 x 1 = 12.0 2 O 16.0 x 2 = 32.0 Total 45.0

• Note that 45 is exactly half of the molecular mass of 90. • So the formula mass of HCO₂ is exactly half of the

Part 2: Stoichiometry Problems

Stoichiometry Problems

•

Stoichiometry

problems involve the

calculation of amounts of materials in a

chemical reaction from known quantities in

the same reaction

•

The substance whose amount is known is the

given substance

Mass to Mass Problems

• Goal:

To calculate the mass of a substance that

appears in a chemical reaction from the mass of

a given substance in the same reaction.

• The

given substance

is the substance whose

mass is known.

Steps in a Mass to Mass

Problem

1.

Find the gram formula masses for the given

and the required substances

2.

Convert the given mass to moles by

dividing it by the molar mass

3.

Multiple the given moles by the mole ratio

to get the moles of the required substance

4.

Multiple the number of moles of the

Example 1 Mass-Mass

Problem

• Glucose burns in oxygen to form CO

and

H

O according to this equation:

C

H

O

+ 6 O



6 CO

+ 6 H

O

How many

grams of CO

are produced from

Example 1 Mass-Mass

Problem

Glucose burns in oxygen to form CO₂ and H₂O according to this equation:

C₆H₁₂O₆ + 6 O₂  6 CO₂ + 6 H₂O

How many grams of CO₂ are produced from burning 45.0 g of glucose?

1.Make sure that the equations is balanced

2.Divide the mass of the given by its molar mass

45.0 g C

H

O

180.0 g mol

C

Example 1 Mass-Mass

Problem

_{Glucose burns in oxygen to form CO}

2

and H

O

according to this equation:

C

H

O

+ 6 O



6 CO

+ 6 H

O

How many

grams of CO

are produced from

burning

45.0 g of glucose

?

1. Make sure that the equations is balanced

2. Divide the mass of the given by its molar

3.

45.0 g C

H

O

6 mol CO

180.0 g mol

C

6

H

O

1 mol C

H

O

Example 1 Mass-Mass

Problem

Glucose burns in oxygen to form CO₂ and H₂O according to

this equation:

C₆H₁₂O₆ + 6 O₂  6 CO₂ + 6 H₂O

How many grams of CO₂ are produced from burning 45 g

of glucose?

1.Make sure that the equations is balanced 2.Divide the mass of the given by its molar 3.Multiply by the mole ratio

4.Multiply by the molar mass of the required

45.0 g C

H

O

x

6 mol CO

x

44.0 g mol

CO

180.0 g mol

C

H

O

1 mol C

H

O

Example 2 Mass-Mass

Problem

What mass of Barium chloride is required to react

with 48.6 grams of sodium phosphate according to

the following reaction:

Example 2

What mass of Barium chloride is required to

react with 48.6 grams of sodium phosphate

according to the following reaction

2 Na

PO

+ 3BaCl



Ba

(PO

)

+ 6 NaCl

48.6g Na

PO

x

3 mol BaCl

x

208.3 g mol

BaCl

164.0 g mol

Na

PO

2 mol Na

PO

Molar Masses: Na

PO

= 3(23.0)+31.0+4(16.0) =164 g mol

Example 3

What mass of carbon dioxide is produced

from burning 100 grams of ethanol in oxygen

according to the following reaction :

Example 3

What mass of carbon dioxide is produced

from burning 100 grams of ethanol in oxygen

according to the following reaction :

C

H

OH + 3 O



2 CO

+ 3 H

O

Molar Masses:

C

C

H

H

OH = 2(12) +6(1)+ 16 = 46

OH

CO

= 12 + 2(16) = 44.0

100.0 g C

H

OH

x

2 mol CO

44.0 g mol

-1

CO

46.0 g mol

1 mol C

2

H

OH

Mass to Volume Problems

• Goal:

To calculate the volume of a gas that

appears in a chemical reaction from the mass of

a given substance in the same reaction.

• The

given substance

is the substance whose

mass is known.

• The

required substance

is the gas whose

volume is to be determined.

• Remember 1 mole of any gas at STP is equal to

22.4 dm

. STP is defined as 0

C and 1

Steps in a Mass to Volume

Problem

1.

Find the gram formula masses for the given

substance.

2.

Convert the given mass to moles by dividing it by

the molar mass

3.

Multiple the given moles by the mole ratio to get

the moles of the required substance

4.

Multiple the number of moles of the required

substance by the molar volume, 22.4 dm

mol

,

to get the volume of the required substance.

5.

This procedure is only valid if the required

Example 1 Mass-Volume

Problem

• Sucrose burns in oxygen to form CO

and H

O

according to this equation:

C

H

O

+ 12 O



12 CO

+ 11 H

O

What volume of CO

measured at STP is

Example 1 Mass-Volume

Problem

_{Sucrose burns in oxygen to form CO}

2

and H

O

according to this equation:

C

H

O

+ 12 O



12 CO

+ 11 H

O

What volume of CO

measured at STP is

produced from burning 100 g of sucrose?

1. Find the molar mass of the given substance

Molar mass of

C

C

H

H

O

O

=

12 (12.0) +22 (1.0) +

11 (16.0)

Example 1: Mass-Volume

Problem

Sucrose burns in oxygen to form CO

and H

O

according to this equation:

C

H

O

+ 12 O



12 CO

+ 11 H

O

What volume of CO

measured at STP is

produced from burning 100 g of sucrose?

2. Find moles of the given:

100 g C

H

O

=

0.292 moles

342 g mol

C

12

H

O

Example 1: Mass-Volume

Problem

_{Sucrose burns in oxygen to form CO}

2

and H

O

according to this equation:

C

H

O

+ 12 O



12 CO

+ 11 H

O

What volume of CO

measured at STP is

produced from burning 100 g of sucrose?

3. Multiply by the mole ratio:

100.0 g C

H

O

12 moles CO

342.0 g mol

C

12

H

O

1 mole C

H

O

Example 1: Mass-Volume

Problem

_{Sucrose burns in oxygen to form CO}

_{and H}

_O

according to this equation:

C

H

O

+ 12 O



12 CO

+ 11 H

O

What volume of CO

measured at STP is

produced from burning 100 g of sucrose?

4. Multiply by the molar volume, 22.4 dm

mol

.

100.0 g C

H

O

x

12 moles CO

22.4 dm

mol

CO

342.0 g mol

C

H

O

Example 2 Mass-Volume

Problem

What volume of carbon dioxide gas would be

produced by reacting 25.0 g of sodium

carbonate with hydrochloric acid according

to the following reaction:

Na

CO

+ 2 HCl



2 NaCl

+ CO

+

H

O

Example 2 Mass-Volume

Problem

25.0 g Na

CO

x

1 mole CO

x

22.4 dm

mol

CO

106.0 g mol

Na

CO

1 moles

Na

CO

= 5.28 dm

of CO

Molar Mass:

Na

Na

CO

=2(23.0)+ 12.0 +3(16.0) =106.0

• What volume of carbon dioxide gas would be produced by reacting 25.0 g of Sodium carbonate with

Solutions and

Stoichiometry

• Many times the reactants and/or products

of chemical reactions are water solutions.

• In these cases the concentration of the

solution must be determined in order to

determine amounts of reactants or

products

• The concentration of a solution is a

Molarity

• The most common concentration unit is Molarity

Molarity Calculations

How many grams of NaOH are required to

prepare 250 cm

of 0.500 M solution?

– Molar Mass of NaOH = 23+16+1 = 40.0 g/mol

– 250 cm

= 0.250 dm

(0.500 mol) x

(40.0 g) x (0.250 dm

)

₌

_{5.00 g}

Molarity Calculations

Calculate the concentration of a NaCl

solution that contains 24.5 g of NaCl in

250 cm

of solution

.

– Molar mass of NaCl = 23.0 + 35.5 = 58.5

(24.5 g NaCl)

X

1

=

1.67 M

Stoichiometry Calculations

Involving Solutions 1

Copper metal reacts with nitric acid according to the following reaction:

8 HNO₃ (aq) + 3 Cu  3 Cu(NO₃)₂ (aq) + 4 H₂O (l) + 2 NO (g)

What volume of 8.00 M HNO₃ would be required to

consume a copper penny whose mass is 3.08 grams?

(3.08 g Cu ) (8 mol HNO₃) (1 dm3) ( 1000 cm3)

(63.55 g mol-1 Cu ) (3 mol Cu) (8 mol HNO

3) (1 dm3)

Stoichiometry Calculations

Involving Solutions 2

3 solution is required to

precipitate the sodium chloride in 10 cm3 of a salt

solution. What is the concentration of the solution?

• Molar Mass NaCl = 23.0 + 35.5 = 58.5 g/mol 0.500 mol

AgNO_{3 X} 0.0150 _dm3 x 1 mol NaCl X 58.5 g mol-1 NaCl

dm3 1 mol AgNO

3

= 0.439 g of NaCl

Cookie Recipe

• 1 cube butter • 1 cup canola oil • 2 cups white sugar • 1 egg

• 1 teaspoon vanilla extract • 1/2 teaspoon salt

• 1 teaspoon baking soda • 4 1/2 cups all-purpose flour • 1 cup oatmeal

• 1 (12 ounce) package chocolate chips

In my cupboard I have:

• 5 cubes butter • 8 cups canola oil • 8 cups white sugar • 12 eggs

• 20 teaspoons vanilla extract • 1 pound salt

• 40 teaspoons baking soda • 45 cups all-purpose flour • 30 cups oatmeal

• 5 (12 ounce) packages chocolate chips

Limiting Reagent

• _{Although we have been basing our calculations thus}

far on only one of the reactants in a chemical

reaction, the reaction will only occur if we have all of the reactants

• The mole ratio determines how much of each reactant we need for the reaction

• Often we have an excess of one of the reactantsThen not all of that reactant will be used up. There will be some left over.

• It is known as the excess reagent.

• The other reactant will be used up and it will determine the amount of product we can form.

Limiting Reagent

To determine which of the reagents is the limiting reagent

1. Calculate the number of moles of each reactant

2. Multiply first reactant by the appropriate mole ratio to get the number of moles of the second reactant that you need.

3. Compare the amount of the second reactant you have to the amount you need .

4. If you have more than you need it is in excess and the first reactant is the limiting reagent

5. If you have less of the second reactant than you need it becomes the limiting reagent

Limiting Reagent Example 1

Barium chloride reacts with potassium phosphate as follows: 3 BaCl₂ (aq) + 2 K₃PO₄(aq)  6 KCl (aq) + Ba₃(PO₄)₂ (s)

Calculate the mass of barium phosphate that could be formed when a solution containing 10.00 g of potassium phosphate is added to a solution containing 12.00 g of barium chloride.

Molar mass potassium phosphate = 3(39.10) + (30.97) + 4(16.00) = 212.27 g mol-1

Molar mass barium chloride = (137.34) + 2(35.45) = 208.24 g mol-1

Molar mass barium phosphate = 3(137.34)+ 2(30.97)+(8)(16.00) = 601.96 g mol-1

Moles barium chloride = 12.00g / 208.24 g mol-1 _{= 0.05762 mol}

Moles potassium phosphate = 10.00g / 212.27 g mol-1 _{= 0.04711 mol}

The mole ratio is 3 mol BaCl₂ to 2 mol K₃PO₄. While there are more moles of BaCl₂ than K₃PO₄, It is not 1.5 times greater. Therefore BaCl₂ is the limiting reagent and all other calculations will be based on barium chloride.

(0.05762 mol BaCl₂) (1mol Ba₃(PO₄)₂) (601.96 g mol-1 _Ba

3(PO4)2 )

Percent Yield

Stoichiometry allows us to calculate the amounts of

reactants required or the amounts of products generated from a chemical reaction.

Chemical reactions frequently do not proceed to completion.

Hence the amount of product recovered is often less than would be predicted from stoichiometric calculations.

In these situations it is helpful to calculate a percent yield.

Percent Yield

The

Theoretical Yield

is defined as the amount of

product(s) calculated using Stoichiometry

calculations

The

Actual Yield

is the amount of product that can

actually be recovered when the reaction is done in

a lab.

The

Percent Yield

is calculated as follows

Percent Yield

Iron reacts with copper sulfate in a single replacement reaction as follows

Fe (s) + CuSO₄ (aq)  FeSO₄ (aq) + Cu (s)

30.00 grams of iron metal were added to excess were added to excess copper sulfate dissolved in a water solution. 22.50 grams of copper were recovered. Calculate the theoretical yield of copper in this experiment .

1. First calculate the theoretical yield

22.50 g Cu

(30.00 g Fe) (1 mol Cu ) (63.55 g mol-1 Cu)

(55.85 g mol-1 Fe) (1 mol Fe )

2. Divide the actual yield by the theoretical yield and multiply by 100