1Chemical Formula Relationships

Chapter 7

2Outline of material covered in Chem23/25

Introduction / Key Concepts: Matter and energy Measurement and chemical calculations Atoms Atomic theory: The nuclear model of the atom Nuclear chemistry The quantum model of the atom Chemical bonding Molecules, the language and math of chemical

structures and reactions Chemical nomenclature, writing formulas Describing chemical reactions Structure and shape

Molecular Formula (recap)

Formula of a chemical or a molecule List of elements it contains, with amounts of each

Glucose (a sugar)Molecular Formula = C6H12O6

6 atoms carbon, C 12 atoms hydrogen, H 6 atoms oxygen, O

3Atomic mass

The average mass of the atoms of an element as theyoccur in nature

Atomic mass units, amu

1 amu = 1/12th the mass of one carbon-12 atom

The mass of an electron is very small so more than 99.9% ofthe mass of the carbon-12 atom are the 6 protons and 6neutrons (about the same mass)

1 amu the weight of one subatomic particle (neutron orproton)

1 amu = 1.66 x 1024 g (a very small number!)

126c

41 copper atom1 sulfur atom

4 oxygen atoms

eg copper(II) sulfate

copper(II) is Cu2+sulfate is SO42-

We learned how to figure out the molecular formula inthe last lecture

Molecular Formula

copper(II) sulfate is CuSO4

1 copper atom1 sulfur atom

4 oxygen atoms

Formula masscopper(II) sulfate is CuSO4

1 molecule of copper(II) sulfate weighs

(1 x 63.546) + (1 x 32.066) + (4 x 15.9994) = 159.6096 amu

x 1.66 x 1024 g/amu

2.65 x 10-22 g

5Calcium ion: Ca2+ Phosphate ion: PO43

Calcium phosphate: Ca3(PO4)2

3 x 40.08 amu = 120.24 amuCa

2 x 30.97 amu = 61.94 amuP

O 8 x 16.00 amu = 128.00 amu

310.18 amuCa3(PO4)2

Example

What is the formula mass of calcium phosphate?

How not to do chemistry

1 g Cu + 1 g H2SO4 2 g CuSO4 ? No!

In the first lecture we defined chemistry as the study of matter atthe particulate level - atoms and molecules

To understand a chemical change, a chemical reaction, or achemical property, we need to know how many atoms andmolecules are involved

1 molecule of Cu reacts with 1 molecule of H2SO4, but themolecules have different weights, so they dont combine insimple gram-for-gram ratios

6Remember the mole..?One mole is the amount of

substance which contains as many"elemental entities" (eg, atoms,molecules, ions, electrons) as thereare atoms in 12 g of carbon-12

Avogadros number 6.022 x 1023

1 mole of pennies is $ 6.022 x 1021

$890 billion for every person on theplanet!

1 mole carbon atoms (12.0 g)

Avogadros Number 6.022 x 1023

6.022 x 1023 atoms of carbon (12C) weighs 12.0 g

6.022 x 1023 atoms of hydrogen (1H) weighs 1.0 g

Do you notice a pattern here..?

7These are the Atomic Weights in grams!

6.022 x 1023 atoms of carbon (12C) weighs 12.0 g (the weight 12.011 is an average that includes 13C and 14C)6.022 x 1023 atoms of hydrogen (1H) weighs 1.0 g (the weight 1.0079 is an average that includes 2H and 3H)

How heavy is a mole of

sulfur

32.065 g

196.97 g

28.086 g

gold

silicon

81 mole of...C

Zn

Ni

CaI2

S

Cu

Hg

Br2

Thats all well and good for atomsBut what about molecules?

Simply calculate the molecular mass and then change the unitsfrom amu to grams (g).This gives the weight of one mole of molecules: the molar mass

9Molar mass

6.022 x 1023 molecules (1 mole) of glucose weighs 180.2 g

6.022 x 1023 molecules (1 mole) of CO2 weighs 44.01g

Molar mass for non-molecular compounds

Mass of one mole of the formula unit

Example: Sodium chloride exists as an orderly, repeatingpattern of sodium and chlorine atoms, but there is nosodium chloride molecule--its hypothetical particle, theformula unit, has the formula NaCl:

One sodium atom and one chlorine atom The molar mass of sodium chloride is

22.99 + 35.45 = 58.44g

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1 mole of ...Hg2I2

CaBr2

CuBr2CuS

ZnS

NiCO3

CuSO4

How many carbon dioxide molecules arein 2.0 moles of carbon dioxide?

2.0 mol CO2 x6.02 x 1023 molecules CO2

mol CO2=

1.2 x 1024 molecules CO2

Working with moles

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Moles to grams? You want to make copper(II) sulfate from the reaction of copper

with H2SO4. You want 0.0640 moles of copper(II) sulfate: howmany grams is this?

1. Determine the formula of copper(II) sulfate: CuSO42. Calculate the molar mass:

63.546 g/mol Cu + 32.066 g/mol S + (4 x 15.9995) g/mol O= 159.61 g/mol CuSO4

Given: 0.0640 mol Wanted: g

159.61 gmol

0.0640 mol x = 10.2 g

Grams to moles? You set up the reaction and make the copper(II) sulfate. You weigh

the product and it is 8.60 g. How many moles is this?

1. Determine the formula of copper(II) sulfate: CuSO42. Calculate the molar mass:

63.546 g/mol Cu + 32.066 g/mol S + (4 x 15.9995) g/mol O= 159.61 g/mol CuSO4

Given: 8.60 g Wanted: mol

1 mol 159.61 g

8.60 g x = 0.0539 mol

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How many molecules?How many molecules in 454g (1 pound) of water?1. Determine the formula: H2O2. Calculate the molar mass:

16.00 g/mol O + 2(1.008 g/mol) H = 18.02 g/mol H2O

Given: 454g Wanted: number of molecules

Path: g H2O -------------> mol H2O ----------------> molecules H2O

PER:1 mol 18.02g

6.02 x 1023 moleculesmol

1 mol18.02 g

454 g x x 6.02 x 1023 moleculesmol

= 1.52 x 1025molecules

Percentage:

% of A = parts of A X 100total parts

The percentage composition of acompound is the percentage by mass of

each element in the compound.

Percentage composition

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Example 1: calculate the percentage offluorine in CaF2

calculate the molar mass:

40.08g Ca + (2 x 19.00)g F = 78.08g CaF2

%F = x 100

= x 100

= 48.67 %

(2 x 19.00)g F78.08g CaF2

g F g CaF2

Percentage:

% of A = mass of A X 100total mass

Example 1: calculate the percentage offluorine in CaF2

Common mistakes: 2/3 of the atoms in CaF2 are F, but the percentagecomposition of F is not 66.67% because the Ca and Fatoms have different masses

When asked for the %F this is percentage by mass ofthe element F, not the percentage by mass of themolecule F2

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Example 2: percentage composition sugar

Determine the percentage composition of sugar, C12H22O11

In one mole of C12H22O11:

(12 x 12.01)g C+ (22 x 1.008)g H+ (11 x 16.00)g O 342.30 g C12H22O11

The molar mass of sugar is 342.3g

. (12 x 12.01)g C X 100 = 42.10% C342.30 g C12H22O11

. (22 x 1.008)g H X 100 = 6.479% H342.30 g C12H22O11

. (11 x 16.00)g O X 100 = 51.42% O342.30 g C12H22O11

Check:42.10% + 6.479% + 51.42% = 100.00%

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How many grams of carbon in 150gsugar?

Sugar is 42.10% carbon, 6.479% hydrogen and 51.42% oxygen

In every 100g sugar 42.10g is carbon, 6.479g is hydrogen and51.42g is oxygen

Given: 150g sugar Wanted: g carbon

Path: g sugar -------------------> g carbonPer: 42.10 g carbon 100g sugar

150g sugar x = 63.2 g carbon 42.10 g carbon 100g sugar

Empirical formula

The lowest whole-number ratio of atoms of theelements in a compound

The empirical formula of C2H4 (ethylene) is CH2 The empirical formula of C3H6 (propylene) is CH2 !

All alkanes (compounds with the general formulaCnH2n) have the same empirical formula and the samepercentage composition

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Empirical formula

Example: Write the empirical formula of benzene, C6H6

C6H6: Both divisible by 6EF: CH

Example: Write the empirical formula of octane, C8H18

C8H18: Both divisible by 2EF: C4H9

How to Find an Empirical Formula

1. Find the masses of different elements in a sample ofthe compound (quantitative analysis)

2. Convert the masses into moles of atoms of thedifferent elements

3. Express the moles of atoms as the smallest possibleratio of integers

4. Write the empirical formula, using the number foreach atom in the integer ratio as the subscript in theformula

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Many empirical formula questionscan be solved with the following table:

Element Grams MolesMoleRatio

FormulaRatio

EmpiricalFormula

Example

What is the empirical formula of a compound thatanalyzes as 79.95% carbon, 9.40% hydrogen, and10.65% oxygen?

Element Grams Moles MoleRatio

FormulaRatio

EmpiricalFormula

C

H

O

79.95

9.40

10.65

79.95 g C

12.01 g/mol C

9.40 g H

1.008 g/mol H

10.65 g O

16.00 g O

6.657

9.33

0.6656

6.657

0.6656

9.33

0.6656

0.6656

0.6656

10.00

14.0

1.000

10

14

1

C10H14O

Smallest number: divide by this to find

mole ratios

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Example A sample of a pure compound is found to contain 1.61g of

phosphorous and 2.98g of fluorine. What is its empirical formula?

Element Grams Moles MoleRatio

FormulaRatio

EmpiricalFormula

P

S

1.61

2.98

1.61 g

30.97 g/mol

2.98 g

19.00 g/mol

0.0520

0.157

0.0520

0.0520

0.157

0.0520

1.00

3.02

1

3 PF3

Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright 2007 Brooks/Cole, a part of the Thomson Corporation.

Discovering the molecular formula

The molecular formula of a compound is found bydetermination of the number of empirical formula units in themolecule

For example, if the empirical formula of a compound is CH2,the true molecular formula could be

CH2 (one empirical formula unit) or C2H4 (two empirical formula units) or C3H6 (three empirical formula units) or C4H8 (four empirical formula units) .etc

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1. Determine the empirical formula of the compound

2. Calculate the molar mass of the empirical formula unit

3. Determine the molar mass of the compound (you wouldmeasure this, or for chem23 purposes it will be given)

4. Divide the molar mass of the compound by the molar mass ofthe empirical formula unit to get n, the number of empiricalformula units per molecule

5. Write the molecular formula

How to Find the Molecular Formula of aCompound

What is the molecular formulaof a compound with the empirical formulaC5H10O and a molar mass of 258 g/mol?

Molar mass of C5H10O:5(12.01 g/mol C) + 10(1.008 g/mol H) + 16.00 g/mol O

= 86.13 g/mol C5H10O

C15H30O3

258 g/mol molecule

86.13 g/mol EF unit= 3 EF units

Example