BONDINGBONDING

•Atoms are generally found in nature in combination held together by chemical bonds.–A chemical bond is a mutual electrical attraction between the nuclei and valence electrons of different atoms that binds the atoms together.

•There are two types of chemical bonds: ionic, and covalent.

•Atoms are generally found in nature in combination held together by chemical bonds.–A chemical bond is a mutual electrical attraction between the nuclei and valence electrons of different atoms that binds the atoms together.

•There are two types of chemical bonds: ionic, and covalent.

Introduction to BondingIntroduction to BondingIntroduction to BondingIntroduction to Bonding

•What determines the type of bond that forms?–The valence electrons of the two atoms involved are redistributed to the most stable arrangement.

–The interaction and rearrangement of the valence electrons determines which type of bond that forms.

•Before bonding the atoms are at their highest possible potential energy

•What determines the type of bond that forms?–The valence electrons of the two atoms involved are redistributed to the most stable arrangement.

–The interaction and rearrangement of the valence electrons determines which type of bond that forms.

•Before bonding the atoms are at their highest possible potential energy

Introduction to BondingIntroduction to BondingIntroduction to BondingIntroduction to Bonding

•There are understandings of bond electron interaction–One understanding of the formation of a chemical bond deals with balancing the opposing forces of repulsion and attraction•Repulsion occurs between the

negative e- clouds of each atom•Attraction occurs between the

positive nuclei and the negative electron clouds

•There are understandings of bond electron interaction–One understanding of the formation of a chemical bond deals with balancing the opposing forces of repulsion and attraction•Repulsion occurs between the

negative e- clouds of each atom•Attraction occurs between the

positive nuclei and the negative electron clouds

Introduction to BondingIntroduction to BondingIntroduction to BondingIntroduction to Bonding

•When two atoms approach each other closely enough for their electron clouds to begin to overlap–The electrons of one atom begin to repel the electrons of the other atom

–And repulsion occurs between the nuclei of the two atoms

•When two atoms approach each other closely enough for their electron clouds to begin to overlap–The electrons of one atom begin to repel the electrons of the other atom

–And repulsion occurs between the nuclei of the two atoms

Introduction to BondingIntroduction to BondingIntroduction to BondingIntroduction to Bonding

•As the optimum distance is achieved that balances these forces, there is a release of potential energy–The atoms vibrate within the window of maximum attraction/minimum repulsion

•The more energy released the stronger the connecting bond between the atoms

•As the optimum distance is achieved that balances these forces, there is a release of potential energy–The atoms vibrate within the window of maximum attraction/minimum repulsion

•The more energy released the stronger the connecting bond between the atoms

Introduction to BondingIntroduction to BondingIntroduction to BondingIntroduction to Bonding

•Another understanding of the form-ation of a chemical bond between two atoms centers on achieving the most stable arrangement of the atoms’ valence electrons–By rearranging the electrons so that each atom achieves a noble gas-like arrangement of its electrons creates a pair of stable atoms (only occurs when bonded)

•Another understanding of the form-ation of a chemical bond between two atoms centers on achieving the most stable arrangement of the atoms’ valence electrons–By rearranging the electrons so that each atom achieves a noble gas-like arrangement of its electrons creates a pair of stable atoms (only occurs when bonded)

Introduction to BondingIntroduction to BondingIntroduction to BondingIntroduction to Bonding

•Sometimes to establish this arrange-ment one or more valence electrons are transferred between two atoms–Basis for ionic bonding

•Sometimes valence electrons are shared between two atoms–Basis for covalent bonding

•Sometimes to establish this arrange-ment one or more valence electrons are transferred between two atoms–Basis for ionic bonding

•Sometimes valence electrons are shared between two atoms–Basis for covalent bonding

Introduction to BondingIntroduction to BondingIntroduction to BondingIntroduction to Bonding

•A good predictor for which type of bonding will develop between a set of atoms is the difference in their electronegativities.–Remember, electronegativity is a measure of the attraction an atom has for e-s after developing a bond

•The more extreme the difference between the two atoms, the less equal the exchange of electrons

•A good predictor for which type of bonding will develop between a set of atoms is the difference in their electronegativities.–Remember, electronegativity is a measure of the attraction an atom has for e-s after developing a bond

•The more extreme the difference between the two atoms, the less equal the exchange of electrons

Introduction to BondingIntroduction to BondingIntroduction to BondingIntroduction to Bonding

•Let’s consider the compound Cesium Fluoride, CsF.–The electronegativity value (EV) for Cs is .70; the EV for F is 4.00.•The difference between the two is

3.30, which falls within the scale of ionic character.

•When the electronegativity difference between two atoms is greater than 2.1 the bond is mostly ionic.

•Let’s consider the compound Cesium Fluoride, CsF.–The electronegativity value (EV) for Cs is .70; the EV for F is 4.00.•The difference between the two is

3.30, which falls within the scale of ionic character.

•When the electronegativity difference between two atoms is greater than 2.1 the bond is mostly ionic.

Introduction to BondingIntroduction to BondingIntroduction to BondingIntroduction to Bonding

•The take home lesson on electro-negativity and bonding is this:–The closer together the atoms are on the P.T., the more evenly their e- interact, and are therefore more likely to form a covalent bond

–The farther apart they are on the P.T., the less evenly their e- interact, and are therefore more likely to form an ionic bond.

•The take home lesson on electro-negativity and bonding is this:–The closer together the atoms are on the P.T., the more evenly their e- interact, and are therefore more likely to form a covalent bond

–The farther apart they are on the P.T., the less evenly their e- interact, and are therefore more likely to form an ionic bond.

Introduction to BondingIntroduction to BondingIntroduction to BondingIntroduction to Bonding

metal w/nonmetal = ionicmetal w/nonmetal = ionicnonmetal w/nonmetal = covalentnonmetal w/nonmetal = covalent

•In a co-valent bond:–The electronegativity difference between the atoms involved is not extreme•So the interaction between the

involved electrons is more like a sharing relationship

– It may not be an equal sharing relationship, but at least the electrons are being “shared”.

•In a co-valent bond:–The electronegativity difference between the atoms involved is not extreme•So the interaction between the

involved electrons is more like a sharing relationship

– It may not be an equal sharing relationship, but at least the electrons are being “shared”.

Introduction to Covalent Introduction to Covalent BondingBondingIntroduction to Covalent Introduction to Covalent BondingBonding

Lets look at the molecule Cl2

Lets look at the molecule Cl2

Covalent BondsCovalent Bonds

ClCl

SharedElectronsShared

Electrons

ClCl ClClClCl

Shared electrons are counted with

both atoms

Shared electrons are counted with

both atoms

Cl ClNotice 8 eNotice 8 e--

in each in each valence valence shell!!!shell!!!

ClCl H H ClCl H H

Covalent BondsCovalent Bonds

How about the molecule HCl?

How about the molecule HCl?

(Polar Covalent) shared, but not

evenly

(Polar Covalent) shared, but not

evenly

2.12.1 3.03.0

To be stable the two atoms involved in the covalent

bond share their electrons in order to achieve the

arrangement of a noble gas.

To be stable the two atoms involved in the covalent

bond share their electrons in order to achieve the

arrangement of a noble gas.

So what’s the bottom line?

• In an ion - ic bond:–The electronegativity difference is extreme, •So the atom with the stronger pull

doesn’t really share the electron– Instead the electron is essentially transferred from the atom with the least attraction to the atom with the most attraction

• In an ion - ic bond:–The electronegativity difference is extreme, •So the atom with the stronger pull

doesn’t really share the electron– Instead the electron is essentially transferred from the atom with the least attraction to the atom with the most attraction

Introduction to Ionic BondingIntroduction to Ionic BondingIntroduction to Ionic BondingIntroduction to Ionic Bonding

-- -

---

- -

- --

+

- - -

---

-- -

---

- -

- -

+

-

An electron is transferred from the sodium atom to the chlorine

atom

-- -

---

- -

- -

++ -- -

---

- -

- --

++

- - -

---

-

Notice 8 eNotice 8 e-- in each in each valence valence shell!!!shell!!!

Both atoms are happy, they both achieve the electron arrangement

of a noble gas.

Both atoms are happy, they both achieve the electron arrangement

of a noble gas.

Very Strong Very Strong Electrostatic attraction Electrostatic attraction

established…established…

IONIC BONDSIONIC BONDS

To be stable the two atoms involved in the ionic bond will

either lose or gain their valence electrons in order to

achieve a stable arrangement of electrons.

To be stable the two atoms involved in the ionic bond will

either lose or gain their valence electrons in order to

achieve a stable arrangement of electrons.

So what’s the bottom line?

•As we’ve learned so far ionic com-pounds are formed by the transfer of electrons from a metal to a nonmetal

•The ionic compound is held together by the strong electrostatic attraction between oppositely charged ions.

•There is a tremendous amount of energy stored in the bonds formed in an ionic compound.

•As we’ve learned so far ionic com-pounds are formed by the transfer of electrons from a metal to a nonmetal

•The ionic compound is held together by the strong electrostatic attraction between oppositely charged ions.

•There is a tremendous amount of energy stored in the bonds formed in an ionic compound.

Bond Energies and BondingBond Energies and BondingBond Energies and BondingBond Energies and Bonding

• It takes a lot of energy (A.K.A. bond energy) to pull the two ions apart once they have established their stable arrangement through bonding

• Energy can be released or absorbed when ions form

—Removing electrons from atoms requires an input of energy

—Remember from last chapter this energy is called ionization energy

• It takes a lot of energy (A.K.A. bond energy) to pull the two ions apart once they have established their stable arrangement through bonding

• Energy can be released or absorbed when ions form

—Removing electrons from atoms requires an input of energy

—Remember from last chapter this energy is called ionization energy

Bond Energies and BondingBond Energies and BondingBond Energies and BondingBond Energies and Bonding

Energy and Ionic BondingEnergy and Ionic BondingEnergy and Ionic BondingEnergy and Ionic Bonding

• On the other hand adding electrons to atoms releases energy into the environment—Remember this has to do with

the atoms affinity for electrons—Sometimes this energy is used

to help remove the electron from another atom

• The ionization energy to remove 1 e- from each atom in a mole of Na atoms is 495.8 kJ

• On the other hand adding electrons to atoms releases energy into the environment—Remember this has to do with

the atoms affinity for electrons—Sometimes this energy is used

to help remove the electron from another atom

• The ionization energy to remove 1 e- from each atom in a mole of Na atoms is 495.8 kJ

•A mol of Cl atoms releases 348.6 kJ when an e- is added to the atom—Notice that it takes more energy

to remove Na’s e- than the amount released from the Cl atoms.

•Forming an ionic bond is a multi-step process—The final step releases

a substantial amount of energy (a.k.a. the driving force)

•A mol of Cl atoms releases 348.6 kJ when an e- is added to the atom—Notice that it takes more energy

to remove Na’s e- than the amount released from the Cl atoms.

•Forming an ionic bond is a multi-step process—The final step releases

a substantial amount of energy (a.k.a. the driving force)

Energy and Ionic BondingEnergy and Ionic BondingEnergy and Ionic BondingEnergy and Ionic Bonding

At the beginning

there is solid sodium and chlorine gas. Na(s) & Cl2(g)

At the beginning

there is solid sodium and chlorine gas. Na(s) & Cl2(g)

Crystal FormationCrystal FormationCrystal FormationCrystal Formation

ENERGY INENERGY INENERGY INENERGY IN

A mol of sodium is converted from a solid to a gas: Na(s) + energy Na(g)

A mol of sodium is converted from a solid to a gas: Na(s) + energy Na(g)

One electron is then removed from each sodium atom of form a sodium cationNa(g) + energy Na+(g) + e-

One electron is then removed from each sodium atom of form a sodium cationNa(g) + energy Na+(g) + e-

ENERGY INENERGY INENERGY INENERGY IN

Energy is required to break the bond holding 0.5mol of Cl2 molecules together to form a mole of chlorine atoms Cl2(g) + energy 2Cl(g)

Energy is required to break the bond holding 0.5mol of Cl2 molecules together to form a mole of chlorine atoms Cl2(g) + energy 2Cl(g)

ENERGY INENERGY INENERGY INENERGY IN

The next step involves adding an electron to each chlorine atom to form a chloride anion: Cl(g) + e- Cl-(g) + energy

The next step involves adding an electron to each chlorine atom to form a chloride anion: Cl(g) + e- Cl-(g) + energy

ENERGY OUTENERGY OUTENERGY OUTENERGY OUT

The final step provides the driving force for the reaction. Na+(g) + Cl-(g) NaCl(s) + energy

The final step provides the driving force for the reaction. Na+(g) + Cl-(g) NaCl(s) + energy

ENERGY OUTENERGY OUTENERGY OUTENERGY OUT

•Energy released in the final step is called the lattice energy

—Energy released when the crystal lattice of an ionic solid is formed

•For NaCl, the lattice energy is 787.5 kJ/mol, which is greater than the input of energy in the previous steps

—The lattice energy provides enough energy to allow for the formation of the sodium ion

•Energy released in the final step is called the lattice energy

—Energy released when the crystal lattice of an ionic solid is formed

•For NaCl, the lattice energy is 787.5 kJ/mol, which is greater than the input of energy in the previous steps

—The lattice energy provides enough energy to allow for the formation of the sodium ion

Crystal FormationCrystal FormationCrystal FormationCrystal Formation

• We can use the lattice energy as a method for measuring the strength of the bond in ionic compounds.—The amount of energy

necessary to break a bond is called bond energy.

• This energy is equal to the lattice energy, but —Bond energy moves into the

system—Lattice energy moves out of

the system

• We can use the lattice energy as a method for measuring the strength of the bond in ionic compounds.—The amount of energy

necessary to break a bond is called bond energy.

• This energy is equal to the lattice energy, but —Bond energy moves into the

system—Lattice energy moves out of

the system

Crystal FormationCrystal FormationCrystal FormationCrystal Formation

+3760.2+3760.2-3760.2-3760.2MgOMgO+2634.7+2634.7-2634.7-2634.7CaF2CaF2

+700.1+700.1-700.1-700.1NaINaI

+751.4+751.4-751.4-751.4NaBrNaBr+787.5+787.5-787.5-787.5NaClNaCl+759.0+759.0-759.0-759.0LiILiI+817.9+817.9-817.9-817.9LiBrLiBr+861.3+861.3-861.3-861.3LiClLiCl

kJ/mol (in)kJ/mol (in)kJ/mol (out)kJ/mol (out)Bond energyBond energyLattice energyLattice energyCompoundCompound

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• In the construction of a crystal lattice, depending on the ions involved there can be small “pores” develop between ions in the ionic crystal.—Some ionic compnds have

enough space between the ions that water molecules can get trapped in between the ions

• Ionic compounds that absorb water into their pores form a special type of ionic compound called a hydrate.

• In the construction of a crystal lattice, depending on the ions involved there can be small “pores” develop between ions in the ionic crystal.—Some ionic compnds have

enough space between the ions that water molecules can get trapped in between the ions

• Ionic compounds that absorb water into their pores form a special type of ionic compound called a hydrate.

Hydrate FormationHydrate FormationHydrate FormationHydrate Formation

• Hydrates typically have different properties than their dry versions - A.K.A. anhydrides—Anhydrous CuSO4 is nearly colorless

—CuSO4•5 H2O is a bright blue color

• When Copper (II) Sulfate is fully hydrated there are 5 water molecules present for every Copper ion.—The hydrated name would be Copper (II) Sulfate Pentahydrate

• Hydrates typically have different properties than their dry versions - A.K.A. anhydrides—Anhydrous CuSO4 is nearly colorless

—CuSO4•5 H2O is a bright blue color

• When Copper (II) Sulfate is fully hydrated there are 5 water molecules present for every Copper ion.—The hydrated name would be Copper (II) Sulfate Pentahydrate

Hydrate FormationHydrate FormationHydrate FormationHydrate Formation

• Have you ever bought a new purse or camera and found a small packet of crystals labeled – do not eat?—These crystals are there to

absorb water that might lead to mildew or mold

• The formula of a hydrate is XAYB • Z H2O (Z is a coefficient indicating how many waters are present per formula unit)

• Have you ever bought a new purse or camera and found a small packet of crystals labeled – do not eat?—These crystals are there to

absorb water that might lead to mildew or mold

• The formula of a hydrate is XAYB • Z H2O (Z is a coefficient indicating how many waters are present per formula unit)

Hydrate FormationHydrate FormationHydrate FormationHydrate Formation

• The FDA requires manufacturers to provide nutritional information on the labels of processed food products.—Dietary guidelines are

based on the percent of calories that the average person should consume from fats, carbohydrates, proteins, etc.

• The FDA requires manufacturers to provide nutritional information on the labels of processed food products.—Dietary guidelines are

based on the percent of calories that the average person should consume from fats, carbohydrates, proteins, etc.

Percent CompositionPercent CompositionPercent CompositionPercent Composition

• percent composition in a compnd can be determined in 2 ways—The 1st is by calculating the

percent composition by mass from a chemical formula.

—The 2nd is a lab scenario where an unknown compound is chemically broken up into its individual components and percent compo-sition is determined by analyzing the results.

• percent composition in a compnd can be determined in 2 ways—The 1st is by calculating the

percent composition by mass from a chemical formula.

—The 2nd is a lab scenario where an unknown compound is chemically broken up into its individual components and percent compo-sition is determined by analyzing the results.

Percent CompositionPercent CompositionPercent CompositionPercent Composition

What is the percent composition of

Hydrogen & Oxygen in Water (H2O)?

What is the percent composition of

Hydrogen & Oxygen in Water (H2O)?

(2•1) + (1•16)=(2•1) + (1•16)=18 g H2O18 g H2O

1st Assume you have a mole of the compound in question, and calculate its molar mass

1st Assume you have a mole of the compound in question, and calculate its molar mass

2nd Use the MM of each component and the MM of the compound to calculate the percent by mass of each component

2nd Use the MM of each component and the MM of the compound to calculate the percent by mass of each component(2•1) = 2 g/mol(2•1) = 2 g/molH:H:

18 g/mol18 g/molx 100 = x 100 = 11.1%11.1%

O:O:100% – 11.1% = 88.9%100% – 11.1% = 88.9%

• In this method, the mass of the sample is measured, then the sample is decomposed or separated into the component elements

• The masses of the component ele-ments are then determined and the percent composition is calculated as before—divide the mass of each element

by the total mass of the sample and multiply by 100.

• In this method, the mass of the sample is measured, then the sample is decomposed or separated into the component elements

• The masses of the component ele-ments are then determined and the percent composition is calculated as before—divide the mass of each element

by the total mass of the sample and multiply by 100.

Calculating PC Using Analysis DataCalculating PC Using Analysis Data

Find the percent composition of a compound that contains 1.94g of carbon, 0.48g of Hydrogen, and

2.58g of Sulfur in a 5.0g sample of the compound.

Find the percent composition of a compound that contains 1.94g of carbon, 0.48g of Hydrogen, and

2.58g of Sulfur in a 5.0g sample of the compound.• Calculate the percents for each

component by the equation: (Component Mass/Total Sample Mass) x 100

• Calculate the percents for each component by the equation: (Component Mass/Total Sample Mass) x 100C: 1.94g/5.0g x 100 = 38.8%C: 1.94g/5.0g x 100 = 38.8%H: 0.48g/5.0g x 100 = 9.6%H: 0.48g/5.0g x 100 = 9.6%

S: 2.58g/5.0g x 100 = 51.6%S: 2.58g/5.0g x 100 = 51.6%

Empirical FormulasEmpirical Formulas

• Percent compositions can be used to calculate the a simple chemical formula of a compound, called an empirical formula—Empirical formula is the

simplest ratio of the atoms in a compound

—Ionic compounds are always written as empirical formulas

• Percent compositions can be used to calculate the a simple chemical formula of a compound, called an empirical formula—Empirical formula is the

simplest ratio of the atoms in a compound

—Ionic compounds are always written as empirical formulas

Empirical FormulasEmpirical Formulas

• Procedure for calculating Empirical Formula—convert the percent compositions into moles

—compare the mols of each compo-nent to calculate the simplest whole number ratio

o divide each amount in moles by the smallest of the mole amounts

o This sets up a simple ratio

• Procedure for calculating Empirical Formula—convert the percent compositions into moles

—compare the mols of each compo-nent to calculate the simplest whole number ratio

o divide each amount in moles by the smallest of the mole amounts

o This sets up a simple ratio

Calculate the empirical formula of a compound that is 80.0% Carbon and 20.0% Hydrogen by

mass

Calculate the empirical formula of a compound that is 80.0% Carbon and 20.0% Hydrogen by

massSince we don’t know the original mass of the sample, we can assume a 100 g sample:

Since we don’t know the original mass of the sample, we can assume a 100 g sample:• We have 80 grams of Carbon and 20

grams of Hydrogen• We need to calculate the number of

moles of each element that we have.

• We have 80 grams of Carbon and 20 grams of Hydrogen

• We need to calculate the number of moles of each element that we have.

80.0g C80.0g C1 mole C1 mole C

12.01 g C12.01 g C== 6.66 mol C6.66 mol C

Calculating Empirical FormulasCalculating Empirical Formulas

20.0g H20.0g H 1 mole H1 mole H

1.008 g H1.008 g H== 19.8 mol H19.8 mol H

• Now we need to calculate the smallest whole number ratio in order to find the empirical formula.

• Divide each component by the smallest number in moles

• Now we need to calculate the smallest whole number ratio in order to find the empirical formula.

• Divide each component by the smallest number in moles

6.66 mol6.66 mol

6.66 mol6.66 mol

= 1= 1

= 2.97= 2.97

CH3CH3

Determine the empirical formula of a compound

containing 2.644g of Au and 0.476g of Cl.

Determine the empirical formula of a compound

containing 2.644g of Au and 0.476g of Cl.

Calculating Empirical FormulasCalculating Empirical Formulas

2.664g Au2.664g Au1 mol Au1 mol Au

197 g Au197 g Au= .01352mol Au= .01352mol Au

.476g Cl.476g Cl1 mol Cl1 mol Cl

35.4 g Cl35.4 g Cl= .01345mol Cl= .01345mol Cl

.01345mol.01345mol

.01345mol

.01345mol

= 1= 1

= 1= 1

AuClAuCl

Molecular FormulasMolecular FormulasMolecular FormulasMolecular Formulas

• The empirical formula for a compound provides the simplest ratio of the atoms in the compound

• However, it does not tell you the actual numbers of atoms in each molecule of the compound —For instance the empirical formula for glucose is CH2O (1:2:1)

—While the molecular formula for glucose is C6H12O6

• The empirical formula for a compound provides the simplest ratio of the atoms in the compound

• However, it does not tell you the actual numbers of atoms in each molecule of the compound —For instance the empirical formula for glucose is CH2O (1:2:1)

—While the molecular formula for glucose is C6H12O6

• A molecular formula indicates the numbers of each atom involved in the the compound—The molecular formula is

always a multiple of the empirical formula

—To calculate the molecular formula you must have 2 pieces of info.• Empirical formula• Molar mass of the unknown

compound (always given)

• A molecular formula indicates the numbers of each atom involved in the the compound—The molecular formula is

always a multiple of the empirical formula

—To calculate the molecular formula you must have 2 pieces of info.• Empirical formula• Molar mass of the unknown

compound (always given)

Molecular FormulasMolecular FormulasMolecular FormulasMolecular Formulas

Calculating Molecular FormulaCalculating Molecular FormulaFind the molecular formula of a

compound that contains 56.36 g of O and 54.6 g of P. The molar mass

of the compound is 189.5 g/mol.

Find the molecular formula of a compound that contains 56.36 g of O and 54.6 g of P. The molar mass

of the compound is 189.5 g/mol.1st find the Emp. Formula:1st find the Emp. Formula:

56.36 g O56.36 g O1 mol O1 mol O

15.99 g O15.99 g O= 3.525mol O= 3.525mol O

54.6g P54.6g P1 mol P1 mol P

30.97 g P30.97 g P= 1.763mol P= 1.763mol P

1.763mol1.763mol

1.763mol1.763mol

= 1.99= 1.99

= 1= 1

PO2PO2

PO2: (1•30.97g P)+(2•15.99g O)= 62.95g

PO2: (1•30.97g P)+(2•15.99g O)= 62.95g

MM Given in the problem = 189.5 g/molMM Given in the problem = 189.5 g/mol

= 3.01= 3.01

Now determine the mass of the empirical formula:Now determine the mass of the empirical formula:

Calculating Molecular FormulaCalculating Molecular Formula

189.5 g/mol189.5 g/mol

62.95g62.95g

Molecular Formula = 3(PO2)Molecular Formula = 3(PO2) P3O6P3O6

One More: A Good 1One More: A Good 1

Methyl acetate is a solvent commonly used in some paints,

inks, and adhesives. Determine the molecular formula for methyl

acetate, which has the following chemical analysis: 48.64% C,

8.16% H, and 43.20% O. The Molar Mass of the compound in

question is reported as 74g/mol.

Methyl acetate is a solvent commonly used in some paints,

inks, and adhesives. Determine the molecular formula for methyl

acetate, which has the following chemical analysis: 48.64% C,

8.16% H, and 43.20% O. The Molar Mass of the compound in

question is reported as 74g/mol.1st determine the empirical formula2nd determine the molecular formula1st determine the empirical formula2nd determine the molecular formula

48.64g C48.64g C

8.16g H8.16g H

12.01 g C12.01 g C

1 mole C1 mole C= 4.050mol C= 4.050mol C

1.008 g H1.008 g H

1 mole H1 mole H= 8.095 mol H= 8.095 mol H

43.20g O43.20g O15.99 g O15.99 g O

1 mole O1 mole O= 2.702 mol O= 2.702 mol O

2.702 mol2.702 mol

2.702 mol2.702 mol

2.702 mol2.702 mol

= 1.50= 1.50

= 2.99= 2.99

= 1.00= 1.00

C1.5H3O1C1.5H3O122 = C3H6O2= C3H6O2= 36+6+32= 36+6+32=74g=74g74g74gC3H6O2C3H6O2