Chapter 7 Lesson Starter CCl 4 MgCl 2 What kind of information can you discern from the formulas? ...

Chapter 7 Lesson Starter

CCl4 MgCl2What kind of information can you discern from the formulas?

Which of the compounds represented is molecular and which is ionic?

Guess the name of each of the above compounds based on the formulas written.

Chemical formulas form the basis of the language of chemistry and reveal much information about the substances they represent.

Chapter 7Chemical formula- shows the number and kinds of atoms in the smallest representative unit of the substance.

Molecular formula- chemical formula written for a molecule (covalent compounds)Ex. CO2 O2 or H2O2

The formula for this compound would be C5H12O

Formula unit- chemical formula written for an ionic compound. Is the lowest whole-number ratio of ions in the compoundThe formula for this compound would be NaCl

Chapter 7Which of the following is molecular and which is ionic?

NaBr CO2 O3 NO KCl AlBr3

Ionic

Ionic

Ionic

Molecular

Molecular

Molecular

Molecular

Ionic

•Two or more nonmetals•Low melting point•Low boiling point•Example CO2

•Metal and nonmetals•High melting point•High boiling point•Solid at room temperature•Electrically neutral•Example NaCl

Chapter 7

A chemical formula indicates the relative number of atoms of each kind in a chemical compound.

For a molecular compound, the chemical formula reveals the number of atoms of each element contained in a single molecule of the compound. example: octane — C8H18

Significance of a Chemical FormulaMolecular Formulas:

The subscript after the C indicates that there are 8 carbon atoms in the molecule.

The subscript after the H indicates that there are 18 hydrogen atoms in the molecule.

Chapter 7Significance of a Chemical FormulaFormula Units:

• Note also that there is no subscript for sulfur: when there is no subscript next to an atom, the subscript is understood to be 1.

Chapter 7Names and Formulas We’re going to learn how to write names and

formulas for Binary ionic compounds ionic compounds (stock system) Polyatomic ions / compounds Molecular compounds Acids and Salts

Take out 3 sheets of paper and fold them over to make a flip chart like the picture above. Label our flip chart “Chemical Names and Formulas”

Chemical Names and Formulas

Chapter 7Chapter 7Formulas for Binary Ionic Compounds Binary ionic compound: ionic compound

composed of two elements The formula for a binary ionic compound can be

written using the identities of the compound’s ions and balancing the total numbers of positive and negative charges to be equal .Example: magnesium bromide Ions combined: Mg2+, Br–, Br– Chemical formula: MgBr2

“Crossing Over”1. Write the symbols for the ions2. Cross over the charges using the absolute value

of the charge as the subscript for the other ion3. Reduce and drop the charges

Chapter 7ExamplesWrite the formula for the binary ionic compounds formed between the following elements1.Zinc and iodine2.Zinc and sulfur3.Potassium and iodine4.Magnesium and chlorine5.Sodium and sulfur6.Aluminum and sulfur7.Aluminum and nitrogen

Chapter 7Chapter 7Naming Binary Ionic Compounds The nomenclature, or naming system, or binary

ionic compounds involves combining the names of the compound’s positive and negative ions.

The name of the cation is first, followed by the anion, the anions ending changes to “ide”: example: Al2O3 — aluminum oxide

For most simple ionic compounds, the ratio of the ions is not given in the compound’s name, because it is understood based on the relative charges of the compound’s ions.

Chapter 7ExamplesName the binary ionic compounds indicated by the following formulas1.AgCl2.ZnO

3.CaBr2

4.SrF2

5.BaO

6.CaCl27.Cs2S

Chapter 7Chapter 7Binary Ionic Compounds (Stock System) Some elements such as iron, form two or more

cations with different charges. To distinguish the ions formed by such elements,

scientists use the Stock system of nomenclature. The system uses a Roman numeral to indicate an

ion’s charge. examples: Fe2+ iron(II)

Fe3+ iron(III) Use the charge and follow the “cross over”

technique to write the formula

Chapter 7ExamplesWrite the formula and give the name for the compounds formed between the following ions: 1.Cu (II) and Br2.Fe (II) and O3.Fe (III) and O4.Pb (II) and Cl5.Hg (II) and S6.Hg (IV) and S7.Sn (II) and F

Chapter 7Binary Ionic Compounds (Stock System)

Naming ionic compounds using the stock system nomenclature:

1. Reverse cross the subscripts 2. Double check that the nonmetals charge is

accurate3. Add a positive for the metal and negative for

the nonmetalExample: Cr1F3 Cr3F1 Cr3+F1-

Chapter 7ExamplesName the binary ionic compounds indicated by the following formulas using stock system nomenclature

1.SnI42.CoF3

3.CuO

4.CoI25.HgI26.FeS

7.PbS2

Chapter 7

The name of the ion with the greater number of oxygen atoms ends in -ate. = nitrate

The name of the ion with the smaller number of oxygen atoms ends in -ite. = nitrite

Naming Binary Ionic Compounds Polyatomic Ions

2NO

3NO

Chapter 7Naming Binary Ionic Compounds Containing Polyatomic Ions,

Some elements can form more than two types of oxyanions. example: chlorine can form

ClO2ClO

3ClO4ClO

• In this case, an anion that has one fewer oxygen atom than the -ite anion has is given the prefix hypo-.

• An anion that has one more oxygen atom than the -ate anion has is given the prefix per-.

hypochlorite chlorite chlorate perchlorate

Chapter 7Polyatomic Ions

Chapter 7Understanding Formulas for Polyatomic Ionic Compounds

Chapter 7ExamplesName the binary ionic compounds containing polyatomic ions as indicated by the following formulas

1.Ca(OH)2

2.KClO3

3.NH4OH

4.Fe2(CrO4)3

5.KClO

6.Fe(ClO)2

7.Ba(OH)S2

Chapter 7Chapter 7Formulas for Binary Ionic Compounds Containing Polyatomic Ions

Chapter 7ExamplesWrite the formulas for the binary ionic compounds containing polyatomic ions 1.Lithium nitrate2.Copper (II) sulfate3.Sodium carbonate4.Calcium nitrite5.Potassium perchlorate6.Barium hydroxide7.Iron (II) hypochlorate

Chapter 7Chapter 7Naming Binary Molecular Compounds Unlike ionic compounds, molecular compounds

are composed of individual covalently bonded units, or molecules.

As with ionic compounds, there is also a Stock system for naming molecular compounds.

The old system of naming molecular compounds is based on the use of prefixes. examples: CCl4 — carbon tetrachloride (tetra- = 4)

CO — carbon monoxide (mon- = 1)CO2 — carbon dioxide (di- = 2)

Chapter 7Prefixes for Naming Covalent Compounds

Chapter 7ExamplesWrite the names for the binary ionic compounds containing polyatomic ions 1.Carbon tetraiodide2.Phosphorus trichloride3.Dinitrogen trioxide4.Carbon dioxide5.Dinitrogen pentoxide6.Sulfur hexafluoride7.Iodine trichloride

Chapter 7Chapter 7Formulas for Binary Molecular Compounds Use the stock system for naming molecular

compounds The ending of the name of the last atom is

changed to “ide” Note that if the first element is only one atom you

do not need to write the “mono” examples: carbon tetrachloride — CCl4

carbon monoxide — COcarbon dioxide — CO2

Chapter 7ExamplesWrite the names for the binary covalent compounds

1.SO3

2.ICl53.PBr5

4.XeF4

5.N2O5

6.SF6

7.N2O3

Chapter 7Chapter 7Acids and Salts An acid is a certain type of molecular compound.

Most acids used in the laboratory are either binary acids or oxyacids. Binary acids are acids that consist of two elements,

usually hydrogen and a halogen. Oxyacids are acids that contain hydrogen, oxygen,

and a third element (usually a nonmetal).

In the laboratory, the term acid usually refers to a solution in water of an acid compound rather than the acid itself. example: hydrochloric acid refers to a water solution of

the molecular compound hydrogen chloride, HCl

Chapter 7Acids and Salts

Many polyatomic ions are produced by the loss of hydrogen ions from oxyacids. examples:

An ionic compound composed of a cation and the anion from an acid is often referred to as a salt.

Table salt, NaCl, contains the anion from hydrochloric acid, HCl. Calcium sulfate, CaSO4, is a salt containing the anion from sulfuric acid, H2SO4.

24SO

3NO

34PO

sulfuric acid H2SO4 sulfate

nitric acid HNO3 nitrate

phosphoric acid H3PO4 phosphate

Chapter 7

Acid Formulas Binary acids: Add hydro and change the anions

ending to “ic” Example: HF hydroflouric acid

Oxyacids: For the polyatomic ion with the greater amount of oxygen's change the anions ending from “ate” to “ic”. For the one with fewer oxygens change the anions ending from “ite” to “ous” Examples

HNO3 nitric acid HNO2 nitrous acid

Chapter 7ExamplesWrite the names for the binary covalent compounds 1.HCl2.HBr3.HI

4.H3PO4

5.H2SO3

6.H2SO4

7.CH3COOH

Chapter 7

Section 2

Chapter 7Chapter 7Oxidation Numbers

In an ionic compound: the charges of the ions reflect the electron distribution of the compound.

oxidation numbers are assigned to the atoms in the compound or ion. In order to indicate the general distribution of

electrons among the bonded atoms in a compound or a polyatomic ion,

Unlike ionic charges, oxidation numbers do not have an exact physical meaning: rather, they serve as useful “bookkeeping” devices to help keep track of electrons.

Chapter 7Chapter 7Assigning Oxidation Numbers1. The atoms in a pure element have an oxidation

number of zero. examples: Na, O2, P4, and, S8, have oxidation # = 02. Fluorine has an oxidation number of –1 in all of its

compounds because it is the most electronegative element.

examples: HF the F = -13. Oxygen usually has an oxidation number of –2.

Exceptions: In peroxides, such as H2O2, O = –1. With fluorine, such as OF2, O = +2.

Chapter 7Chapter 7Assigning Oxidation Numbers4. Hydrogen has an oxidation number of +1; it has

an oxidation number of –1 when with metals. Examples:

HF H = +1

5. The more-electronegative element (right, top) is assigned a negative number equal to its charge

examples: NaCl: Cl = -1

Chapter 7Chapter 7

Assigning Oxidation Numbers6. The algebraic sum of the oxidation numbers of

all atoms in an neutral compound is equal to zero.

7. The algebraic sum of the oxidation numbers of all atoms in a polyatomic ion is equal to the charge of the ion.

Chapter 7Chapter 7Assigning Oxidation NumbersSample Problem: Assign oxidation numbers to each atom in the following compounds or ions:a. UF6

b. H2SO4

c. ClO3-

d. CI4e. Na2O2

f. NO2-

Chapter 7Chapter 7Using Oxidation Numbers for Formulas and Names

As shown in the table in the next slide, many nonmetals can have more than one oxidation number.

These numbers can sometimes be used in the same manner as ionic charges to determine formulas. example: What is the formula of a binary compound

formed between sulfur and oxygen?From the common +4 and +6 oxidation states of sulfur, you could predict that sulfur might form SO2 or SO3. Both are known compounds.

Chapter 7Common Oxidation States of Nonmetals

Chapter 7Chapter 7Using Oxidation Numbers for Formulas and NamesUsing oxidation numbers, the Stock system, introduced in the previous section for naming ionic compounds, can be used as an alternative to the prefix system for naming binary molecular compounds.

Prefix system Stock systemPCl3 phosphorus trichloride phosphorus(III) chloride

PCl5 phosphorus pentachloride phosphorus(V) chloride

N2O dinitrogen monoxide nitrogen(I) oxide

NO nitrogen monoxide nitrogen(II) oxideMo2O3 dimolybdenum trioxide molybdenum(III) oxide

Chapter 7

Section 3

Chapter 7Chapter 7

A chemical formula indicates: the elements present in a compound the relative number of atoms or ions of each element

present in a compoundChemical formulas also allow chemists to calculate a

number of other characteristic values for a compound:

formula mass molar mass percentage composition

Section 3 Using Chemical Formulas

Chapter 7Chapter 7

Formula Masses The formula mass is the sum of the average atomic masses of all atoms represented in its formula. example: formula mass of water, H2O

average atomic mass of H: 1.01 amuaverage atomic mass of O: 16.00 amu


1.01 amu2 H atoms 2.02 amuH atom

16.00 amu1 O atom 16.00 amuO atom

average mass of H2O molecule: 18.02 amu

Chapter 7Chapter 7

Formula Masses The mass of a water molecule can be

referred to as a molecular mass. The mass of one formula unit of an ionic

compound, such as NaCl, is not a molecular mass.

The mass of any unit represented by a chemical formula (H2O, NaCl) can be referred to as the formula mass.


Chapter 7Chapter 7

Formula Masses, continuedSample Problem FFind the formula mass of potassium chlorate, KClO3.


Chapter 7Chapter 7

Formula Masses, continuedSample Problem F Solution

The mass of a formula unit of KClO3 is found by adding the masses of one K atom, one Cl atom, and three O atoms.Atomic masses can be found in the periodic table in the back of your book. In your calculations, round each atomic mass to two decimal places.


Chapter 7

Formula Masses, continuedSample Problem F Solution, continued

Chapter 7

39.10 amu1 K atom 39.10 amuK atom

35.45 amu1 Cl atom 35.45 amuCl atom

16.00 amu3 O atoms 48.00 amuO atom


formula mass of KClO3 = 122.55 amu

Chapter 7Chapter 7

Molar Masses The molar mass of a substance is equal to the

mass in grams of one mole, or approximately 6.022 × 1023 particles, of the substance. example: the molar mass of pure calcium, Ca, is

40.08 g/mol because one mole of calcium atoms has a mass of 40.08 g.

The molar mass of a compound is calculated by adding the masses of the elements present in a mole of the molecules or formula units that make up the compound.


Chapter 7Chapter 7

Molar Masses, continued One mole of water molecules contains

exactly two moles of H atoms and one mole of O atoms. The molar mass of water is calculated as follows.


1.01 g H2 mol H 2.02 g Hmol H

16.00 g O1 mol O 16.00 g Omol O

molar mass of H2O molecule: 18.02 g/mol

• A compound’s molar mass is numerically equal to its formula mass.

Chapter 7

Calculating Molar Masses for Ionic Compounds

Chapter 7 Section 3 Using Chemical Formulas

Chapter 7Chapter 7

Molar Masses, continuedSample Problem GWhat is the molar mass of barium nitrate, Ba(NO3)2?


Chapter 7

Molar Masses, continuedSample Problem G SolutionOne mole of barium nitrate, contains one mole of Ba, two moles of N (1 × 2), and six moles of O (3 × 2).

Chapter 7

137.33 g H1 mol Ba 137.33 g Bamol Ba

14.01 g2 mol N 28.02 g Nmol N

16.00 g O6 mol O 96.00 g Omol O


molar mass of Ba(NO3)2 = 261.35 g/mol

Chapter 7Chapter 7

Molar Mass as a Conversion Factor The molar mass of a compound can be used

as a conversion factor to relate an amount in moles to a mass in grams for a given substance.

To convert moles to grams, multiply the amount in moles by the molar mass:

Amount in moles × molar mass (g/mol) = mass in grams


Chapter 7

Mole-Mass Calculations


Chapter 7Chapter 7Molar Mass as a Conversion Factor, continuedSample Problem HWhat is the mass in grams of 2.50 mol of oxygen gas?


Chapter 7Chapter 7Molar Mass as a Conversion Factor, continued

Sample Problem H SolutionGiven: 2.50 mol O2

Unknown: mass of O2 in gramsSolution:

moles O2 grams O2

amount of O2 (mol) × molar mass of O2 (g/mol) = mass of O2 (g)


Chapter 7Chapter 7Molar Mass as a Conversion Factor, continuedSample Problem H Solution, continued

Calculate the molar mass of O2.

16.00 g O2 mol O 32.00 gmol O

22

22

32.00 g O2.50 m 80.0ol O

mol Og O


Use the molar mass of O2 to convert moles to mass.

Chapter 7Converting Between Amount in Moles and Number of Particles


Chapter 7Chapter 7Molar Mass as a Conversion Factor, continuedSample Problem IIbuprofen, C13H18O2, is the active ingredient in many nonprescription pain relievers. Its molar mass is 206.31 g/mol.a. If the tablets in a bottle contain a total of 33 g of

ibuprofen, how many moles of ibuprofen are in the bottle?

b. How many molecules of ibuprofen are in the bottle?

c. What is the total mass in grams of carbon in 33 g of ibuprofen?


Chapter 7Chapter 7Molar Mass as a Conversion Factor, continuedSample Problem I SolutionGiven: 33 g of C13H18O2

molar mass 206.31 g/molUnknown: a. moles C13H18O2

b. molecules C13H18O2

c. total mass of CSolution: a. grams moles

13 18 213 18 2 13 18 2

13 18 2

1 mol C H Og C H O mol C H O

206.31 g C H O


Chapter 7Chapter 7Molar Mass as a Conversion Factor, continuedSample Problem I Solution, continued

b. moles molecules

23

13 18 2 13 18 26.022 10 moleculesmol C H O molecules C H O

mol

13 18 2

13 18 2

13 mol C 12.01 g Cmol C H O g Cmol C H O mol C


c. moles C13H18O2 moles C grams C

Chapter 7Chapter 7Molar Mass as a Conversion Factor, continuedSample Problem I Solution, continued

23

2

13 18

2

2

13 18 2

6.022 10 molecules0.16mol C

9.6 10 molecules C

H Omol

H O

13 18 213 18 2

13 mol C 12.01 g C0.16 mol C H Omol C H O mol C

25 g C

13 18 213 18 2

13 18 213 18 2

1 mol C H O33 g C H O

206.31 g C H O0.16 mol C H O


a.

b.

c.

Chapter 7Chapter 7

Percentage Composition It is often useful to know the percentage by mass of a particular element in a chemical compound.

To find the mass percentage of an element in a compound, the following equation can be used.

mass of element in sample of compound 100mass of sample of compound

% element in compound


• The mass percentage of an element in a compound is the same regardless of the sample’s size.

Chapter 7Chapter 7

Percentage Composition, continued The percentage of an element in a compound can be calculated by determining how many grams of the element are present in one mole of the compound.

mass of element in 1 mol of compound 100molar mass of compound

% element in compound


• The percentage by mass of each element in a compound is known as the percentage composition of the compound.

Chapter 7

Percentage Composition of Iron Oxides


Chapter 7

Click below to watch the Visual Concept.

Visual Concept


Percentage Composition

Chapter 7

Percentage Composition Calculations


Chapter 7Chapter 7

Percentage Composition, continuedSample Problem J

Find the percentage composition of copper(I) sulfide, Cu2S.


Chapter 7Chapter 7

Percentage Composition, continuedSample Problem J SolutionGiven: formula, Cu2SUnknown: percentage composition of Cu2SSolution:

formula molar mass mass percentage

of each element


Chapter 7Chapter 7

Percentage Composition, continuedSample Problem J Solution, continued

63.55 g Cu2 mol Cu 127.1 g Cumol Cu

32.07 g S1 mol S 32.07 g Smol S


Molar mass of Cu2S = 159.2 g

Chapter 7Chapter 7

Percentage Composition, continuedSample Problem J Solution, continued

2

127.1 g Cu 100159.2 g C

79.8 uS

5% Cu

2

32.07 g S 100159.2 g C

20. Su S

15%


Chapter 7

Preview

• Lesson Starter• Objectives• Calculation of Empirical Formulas• Calculation of Molecular Formulas

Section 4 Determining Chemical Formulas

Chapter 7Chapter 7Section 4 Determining Chemical Formulas

Lesson Starter Compare and contrast models of the

molecules NO2 and N2O4. The numbers of atoms in the molecules

differ, but the ratio of N atoms to O atoms for each molecule is the same.


Objectives Define empirical formula, and explain how the

term applies to ionic and molecular compounds. Determine an empirical formula from either a

percentage or a mass composition. Explain the relationship between the empirical

formula and the molecular formula of a given compound.

Determine a molecular formula from an empirical formula.

Chapter 7

Empirical and Actual Formulas

Chapter 7Section 4 Determining Chemical Formulas


An empirical formula consists of the symbols for the elements combined in a compound, with subscripts showing the smallest whole-number mole ratio of the different atoms in the compound.

For an ionic compound, the formula unit is usually the compound’s empirical formula.

For a molecular compound, however, the empirical formula does not necessarily indicate the actual numbers of atoms present in each molecule. example: the empirical formula of the gas diborane is BH3,

but the molecular formula is B2H6.


Calculation of Empirical Formulas To determine a compound’s empirical formula from its percentage composition, begin by converting percentage composition to a mass composition.

• Assume that you have a 100.0 g sample of the compound.

• Then calculate the amount of each element in the sample.

• example: diborane• The percentage composition is 78.1% B and 21.9% H.• Therefore, 100.0 g of diborane contains 78.1 g of B and

21.9 g of H.


Calculation of Empirical Formulas, continued Next, the mass composition of each

element is converted to a composition in moles by dividing by the appropriate molar mass.

1 mol B78.1 g B 7.22 mol B10.81 g B

1 mol H21.9 g H 21.7 mol H1.01 g H

• These values give a mole ratio of 7.22 mol B to 21.7 mol H.


Calculation of Empirical Formulas, continued To find the smallest whole number ratio,

divide each number of moles by the smallest number in the existing ratio. 7.22 mol B 21.7 mol H: 1 mol B : 3.01 mol H

7.22 7.22

• Because of rounding or experimental error, a compound’s mole ratio sometimes consists of numbers close to whole numbers instead of exact whole numbers.

• In this case, the differences from whole numbers may be ignored and the nearest whole number taken.


Calculation of Empirical Formulas, continuedSample Problem L

Quantitative analysis shows that a compound contains 32.38% sodium, 22.65% sulfur, and 44.99% oxygen. Find the empirical formula of this compound.


Calculation of Empirical Formulas, continuedSample Problem L SolutionGiven: percentage composition: 32.38% Na, 22.65% S, and 44.99% OUnknown: empirical formulaSolution: percentage composition mass composition composition in moles smallest whole-number

mole ratio of atoms


Calculation of Empirical Formulas, continuedSample Problem L Solution, continued

1 mol Na32.38 g Na 1.408 mol Na

22.99 g Na

1 mol S22.65 g S 0.7063 mol S32.07 g S

1 mol O44.99 g O 2.812 mol O16.00 g O

Chapter 7Chapter 7

Calculation of Empirical Formulas, continuedSample Problem L Solution, continued

Smallest whole-number mole ratio of atoms: The compound contains atoms in the ratio 1.408 mol Na:0.7063 mol S:2.812 mol O.

1.408 mol Na 0.7063 mol S 2.812 mol O: :0.7063 0.7063 0.7063

1.993 mol Na :1 mol S : 3.981 mol O


Rounding yields a mole ratio of 2 mol Na:1 mol S:4 mol O.

The empirical formula of the compound is Na2SO4.


Calculation of Molecular Formulas The empirical formula contains the smallest possible whole numbers that describe the atomic ratio.

The molecular formula is the actual formula of a molecular compound.

An empirical formula may or may not be a correct molecular formula.

The relationship between a compound’s empirical formula and its molecular formula can be written as follows.

x(empirical formula) = molecular formula

Chapter 7Chapter 7Section 4 Determining Chemical FormulasCalculation of Molecular

Formulas, continued

The formula masses have a similar relationship.x(empirical formula mass) = molecular formula mass

To determine the molecular formula of a compound, you must know the compound’s formula mass. Dividing the experimental formula mass by the empirical

formula mass gives the value of x. A compound’s molecular formula mass is

numerically equal to its molar mass, so a compound’s molecular formula can also be found given the compound’s empirical formula and its molar mass.

Chapter 7Comparing Empirical and Molecular Formulas

Chapter 7Section 4 Determining Chemical Formulas

Chapter 7

Click below to watch the Visual Concept.

Visual Concept

Section 4 Determining Chemical FormulasComparing Molecular and

Empirical Formulas

Chapter 7Chapter 7Calculation of Molecular Formulas, continuedSample Problem NIn Sample Problem M, the empirical formula of a compound of phosphorus and oxygen was found to be P2O5. Experimentation shows that the molar mass of this compound is 283.89 g/mol. What is the compound’s molecular formula?


Chapter 7Chapter 7Calculation of Molecular Formulas, continued

Sample Problem N SolutionGiven: empirical formulaUnknown: molecular formulaSolution: x(empirical formula) = molecular formula x

molecular formula massempirical formula mass


Chapter 7Chapter 7Calculation of Molecular Formulas, continuedSample Problem N Solution, continuedMolecular formula mass is numerically equal to molar mass.

molecular molar mass = 283.89 g/mol molecular formula mass = 283.89 amu

empirical formula mass mass of phosphorus atom = 30.97 amumass of oxygen atom = 16.00 amuempirical formula mass of P2O5 = 2 × 30.97 amu + 5 × 16.00 amu = 141.94 amu


Chapter 7Chapter 7Calculation of Molecular Formulas, continued

Sample Problem N Solution, continuedDividing the experimental formula mass by the empirical formula mass gives the value of x.

x =

283.89 amu 2.0001141.94 amu


The compound’s molecular formula is therefore P4O10.

2 × (P2O5) = P4O10

End of Chapter 7 Show

Chapter 7 Lesson Starter CCl 4 MgCl 2 What kind of information can you discern from the formulas? ...

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