Chapter 2 Mole & Formulas

7/30/2019 Chapter 2 Mole & Formulas

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Copyright (c) 1999 by Harcourt Brace & Company

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Formulas, Compounds,

and The Mole

Calculations


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Counting Atoms

Mg burns in air (O2) toproduce whitemagnesium oxide, MgO.

How can we figure out

how much oxide isproduced from a givenmass of Mg?

Copyright 1999 by Harcourt Brace & Company

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Counting Atoms

Chemistry is a quantitative sciencewe need a counting unit.

The MOLE 1 mole is the amount of substance that

contains as many particles (atoms,molecules) as there are in 12.0 g of12C.


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Particles in a Mole

6.0221367 x 1023

Avogadros Number

Amedeo Avogadro

1776-1856


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Particles in a Mole

6.0221367 x 1023

Avogadros Number

Amedeo Avogadro1776-1856

There is Avogadros number of

particles in a mole of any substance.


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Molar Mass1 mol of12C

= 12.00 g of C= 6.022 x 1023 atoms

of C

12.00 g of12C is its

MOLAR MASS

Taking into account allof the isotopes of C,

the molar mass of C is

12.011 g/mol


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Molar Mass

1 mol of12C = 12.00 g of C= 6.022 x 1023 atoms of C

12.00 g of12C is its MOLAR MASS

Taking into account all of the isotopes ofC, the molar mass of C is 12.011 g/mol

13

Al

26.9815

atomic number

symbol

atomic weight

Find molar

mass fromperiodic

table


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PROBLEM: How many molesare represented by 0.200 g of

Mg?


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Mg?

Mg has a molar mass of 24.3050 g/mol.


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Mg?


0.200 g 1 mol

24.31 g= 8.23 x 10-3 mol


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Mg?


How many atoms in this piece of Mg?

0.200 g 1 mol

24.31 g= 8.23 x 10-3 mol


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Mg?



0.200 g 1 mol

24.31 g= 8.23 x 10-3 mol

8.23 x 10-3 mol 6.022 x 1023 atoms1 mol


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Mg?



= 4.95 x 1021 atoms Mg

0.200 g 1 mol

24.31 g= 8.23 x 10-3 mol

8.23 x 10-3 mol 6.022 x 1023 atoms1 mol


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Device for Remembering

How to Do Mass/MoleCalculations

mass

# moles MW

mass

MW# moles# moles =

mass # moles MW=

MW =mass


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MOLECULAR WEIGHT

AND MOLAR MASSMolecular weight is the sum of the

atomic weights of all atoms in the

molecule.

Molar mass = molecular weight in

grams



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What is the molar massof ethanol, C2H6O?

1 mol contains

2 mol C (12.01 g C/1 mol) = 24.02 g C

6 mol H (1.01 g H/1 mol) = 6.06 g H

1 mol O (16.00 g O/1 mol) = 16.00 g O

TOTAL = molar mass = 46.08 g/mol


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Formula = C8H9NO2

Molar mass = 151.2 g/mol

Tylenol


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How many moles of alcohol arethere in a standard can of beer if

there are 21.3 g of C2H6O?

(a) Molar mass of C2H6O = 46.08 g/mol

(b) Calc. moles of alcohol


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How many moles of alcohol arethere in a standard can of beer if

there are 21.3 g of C2H6O?

(a) Molar mass of C2H6O = 46.08 g/mol

(b) Calc. moles of alcohol

21.3 g 1 mol

46.08 g= 0.462 mol


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How many moleculesof alcoholare there in a standard can of

beer if there are 21.3 g of C2H6O?

We know there are 0.462 mol of C2H6O.


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How many moleculesof alcoholare there in a standard can of

beer if there are 21.3 g of C2H6O?

= 2.78 x 1023 molecules

We know there are 0.462 mol of C2H6O.

0.462 mol 6.022 x 1023 molecules

1 mol


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How many atoms of C are there ina standard can of beer if there are

21.3 g of C2H6O?

We know there are 2.78 x 1023 molecules.

Each molecule contains 2 C atoms.

Therefore, the number of C atoms is


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How many atoms of C are there ina standard can of beer if there are

21.3 g of C2H6O?

= 5.57 x 1023 C atoms

We know there are 2.78 x 1023 molecules.

Each molecule contains 2 C atoms.

Therefore, the number of C atoms is

2.78 x 1023 molecules 2 C atoms

1 molecule


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Empirical and Molecular

FormulasA pure compound always consists of the

same elements combined in the sameproportions by weight.

Therefore, we can express molecular

composition as PERCENT BYWEIGHT



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Department, Harcourt Brace & Company, 6277 SeaHarbor Drive, Orlando, Florida 32887-6777


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Empirical and Molecular

FormulasA pure compound always consists of the

same elements combined in the sameproportions by weight.

Therefore, we can express molecularcomposition as PERCENT BYWEIGHT

Ethanol, C2H6O

52.13% C, 13.15% H,

34.72% O


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Percent CompositionConsider some of the family of nitrogen-

oxygen compounds:

NO2, nitrogen dioxide and closelyrelated, NO, nitrogen monoxide (or

nitric oxide)

Structure of NO2

Chemistry of NO,

nitrogen monoxide


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Percent Composition

Consider NO2, Molar mass = ?

What is the weight percent of N and of O?


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Percent Composition



Wt. % N =14.0 g N

46.0 g NO2 100% = 30.4 %


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Percent Composition



Wt. % O 2 (16 .0 g O per mole )46 .0 g

x 100% 69.6%

Wt. % N =14.0 g N

46.0 g NO2 100% = 30.4 %


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Percent Composition



Wt. % O 2 (16 .0 g O per mole )46 .0 g x 100% 69.6%

Wt. % N =14.0 g N

46.0 g NO2 100% = 30.4 %

What are the weight percentages of

N and O in NO?


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Determining Formulas

In chemical analysis we determine the % byweight of each element in a given amount ofpure compound and derive the

EMPIRICALorSIMPLESTformula.

PROBLEM: A compound of B and H is81.10% B. What is its empirical

formula?


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

Because it contains only B and H, it

must contain 18.90% H.

In 100.0 g of the compound there are

81.10 g of B and 18.90 g of H.

Calculate the number of moles of each

constitutent.


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element in 100.0 g of sample.

81.10 g B

1 mol

10.81 g = 7.502 mol


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element in 100.0 g of sample.

81.10 g B

1 mol

10.81 g = 7.502 mol

18.90 g H 1 mol

1.008 g

= 18.75 mol


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

1 atom B + 3 atoms H --> 1 molecule BH3

or

1 mol B atoms + 3 mol H atoms --->

1 mol BH3 molecules

Find the ratio of moles of elements in the

compound.


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Take the ratio of moles of B and H. Always

divide by the smaller number.


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18.75 mol H

7.502 mol B =

2.499 mol H

1.000 mol B =

2.5 mol

1.0 mol B


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But we need a whole number ratio.

2.5 mol H/1.0 mol B = 5 mol H to 2 mol B

EMPIRICAL FORMULA = B2H5



18.75 mol H

7.502 mol B =

2.499 mol H

1.000 mol B =

2.5 mol

1.0 mol B


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A compound of B and H is 81.10% B.

Its empirical formula is B2H5. What is

itsmolecular formula?Is the molecular formula B2H5, B4H10,

B6H15, B8H20, etc.?

B2H6

B2H6 is one example of this class of compounds.

A d f B d H i 81 10% B


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its molecular formula?

We need to do an EXPERIMENT to findthe MOLAR MASS.

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

= 26.66 g/unit

Find the ratio of these masses.

A d f B d H i 81 10% B


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We need to do an EXPERIMENT to find

the MOLAR MASS.

Here experiment gives 53.3 g/mol

The mass of B2H5 = 26.66 g/unit


53.3 g/mol

26.66 g/unit of B2H5=

2 units of B2H

51 mol

A compound of B and H is 81 10% B


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Molecular formula = B4H10

We need to do an EXPERIMENT to find

the MOLAR MASS.

Here experiment gives 53.3 g/mol

The mass of B2H5 = 26.66 g/unit


53.3 g/mol

26.66 g/unit of B2H5=

2 units of B2H

51 mol


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Determine the formula of acompound of Sn and I using the

following data.

Reaction of Sn and I2 is done using excessSn.

Mass of Sn in the beginning = 1.056 g Mass of iodine (I2) used

= 1.947 g

Mass of Sn remaining= 0.601 g

See p. 139


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Find the mass of Sn that combined with1.947 g I2.

Mass of Sn initially = 1.056 g

Mass of Sn recovered = 0.601 gMass of Sn used = 0.455 g

Find moles of Sn used:

Tin and Iodine Compound


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Find the mass of Sn that combined with1.947 g I2.

Mass of Sn initially = 1.056 g

Mass of Sn recovered = 0.601 gMass of Sn used = 0.455 g

Find moles of Sn used:

0.455 g Sn 1 mol118.7 g

= 3.83 x 10-3 mol Sn



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Now find the number of moles of I2 thatcombined with 3.83 x 10-3 mol Sn. Massof I2 used was 1.947 g.


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1.947 g I2 1 mol253.81 g

= 7.671 x 10-3 mol I2


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This is equivalent to 2 x 7.671 x 10-3 or

1.534 x 10-2 mol iodine atoms

1.947 g I2 1 mol253.81 g

= 7.671 x 10-3 mol I2


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Now find the ratio of number of moles ofmoles of I and Sn that combined.


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1.534 x 10-2 mol I

3.83 x 10-3 mol Sn =

4.01 mol I

1.00 mol Sn


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Empirical formula is SnI4

1.534 x 10-2 mol I

3.83 x 10-3 mol Sn =

4.01 mol I

1.00 mol Sn


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Hydrated Compounds

Compounds which watermolecules are associated with the

ions of the compound Usually determine experimentally

Weigh the hydrated product dryby heat

weigh the anhydrousproduct convert g to moles

find mole ratio

Chapter 2 Mole & Formulas

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