Chemical Formulas and Chemical Compounds

Chemical Formulas and Chemical Compounds
Chapter 7 Chemical Formulas and Chemical Compounds

Ch 7, Sect 1 Chemical Names & Formulas (pp. 207-219)
Formulas tell the number and kinds of atoms in a compound In ionic compounds Relative number of atoms of each kind in a chemical compound In molecular compounds Number of atoms of each element contained in a single molecule Hydrocarbons Molecular compounds composed only of carbon and hydrogens

Formulas tell the number and kinds of atoms in a compound Ionic compound Consists of a lattice of positive and negative ions held together by mutual attraction Represents one formula unit Simplest ratio of the compound’s cations and anions Example: aluminum sulfate (at right ) Subscript to the right of Al refers to 2 atoms of Al Subscript to the right of O refers to 4 atoms of O in a sulfate ion Subscript to the right of the parentheses refers to everything inside the parentheses 3 sulfate ions (Total of 3 sulfur & 12 oxygen)

Monatomic ions are made of only one type of atom. Many atoms form a noble gas configuration by gaining or losing electrons Group 1 metals lose 1 electron to make 1+ cations Group 2 metals lose 2 electrons to make 2+ cations Nonmetals in groups 15, 16, and 17 tend to gain electrons to form anions

Monatomic ions are made of only one type of atom. Not all main group elements easily form ions Carbon and silicon tend to form covalent bonds and share electrons with other atoms Lead and tin are group 14 metals Difficult for them to lose 4 electrons to get a noble gas configuration Tend to lose 2 p electrons and keep their s electrons Can also form molecular compounds and share all four valence electrons Oxidative state is the same as the charge on the ion The charge on an ion is sometimes called an oxidation number

Monatomic ions are made of only one type of atom. d-block elements may have more than one oxidative state Most form cations with charges between 1+ and 4+ Copper cations can be Cu1+ or Cu2+ Iron cations can be Fe2+ or Fe3+ See chart to right 

Naming Monatomic Ions Monatomic cations are identified by the element name No change in the name Monatomic anions are identified by: dropping the ending of the element name Adding the ending –ide to the root name See page 209, Figure 1.1 Organized by charge Some elements include Roman numerals Part of the Stock system

Binary compounds contain atoms of two elements In an ionic binary compound, the positive and negative charges must be equal The formula can be written given the identities of the ions Example: magnesium and bromine Magnesium forms Mg2+ Bromine forms Br1- when combined with a metal The formula should include one Mg2+ and two Br1- to make the charge on the compound neutral Formula: MgBr2 The 2 is written as a subscript in a chemical formula

Binary compounds contain atoms of two elements “crossing over” or “criss-cross” method Method of balancing charges between ions in an ionic compound Write the symbols for the ions side by side Write the cation first: Al3+ O2- Cross over the charges by using the absolute value of each ion’s charge as the subscript for the other ion Check the subscripts and divide them by their largest common factor to give the smallest possible whole-number ratio of ions. Then write the formula Multiplying the charge by the subscript shows that the charge on Two Al3+ cations = 6+ Three O2- anions = 6- Largest common factor of subscripts is one Formula: Al2O3

Naming Binary Ionic Compounds Nomenclature Naming system Involves combining names of the compound’s positive and negative ions Cation is given first Anion is given second For most simple ionic compounds, Ratio of ions is not given in the name of the compound Ratio is understood based on relative charges of the compound’s ions Does not include ions with more than one possible charge

Ch 7, Sect 1 Chemical Names & Formulas (p. 211)
Sample Problem A: Write the formulas for the binary ionic compounds formed between the following elements: a. zinc and iodine b. zinc and sulfur Write the symbols for the ions side by side. Write the cation first. Zn2+ I- Zn2+ S2- Cross over the charges to give subscripts. Zn I − 2 Zn S 2− 2

Sample Problem A: Write the formulas for the binary ionic compounds formed between the following elements: a. zinc and iodine b. zinc and sulfur Check the subscripts, and divide them by their largest common factor to give the smallest possible whole-number ratio of ions. Then write the formula. The subscripts are mathematically correct because they give equal total charges of 1 x 2+ = 2+ and 2 x 1- = 2-. The smallest whole-number ratio of ions in the compound is 1:2 The subscript 1 is not written, so the formula is ZnI2 The subscripts are mathematically correct because they give equal total charges of 2 x 2+ = 4+ and 2 x 2- = 4-. The largest common factor of the subscripts is 2 The smallest whole number ratio of ions in the compound is 1:1 The subscript 1 is not written, so the formula is ZnS.

The Stock System of Nomenclature Some elements form two or more cations with different charges To distinguish the ions formed by such elements, use the Stock system of nomenclature The system used Roman numerals to indicate the ion’s charge The numeral in parentheses is placed immediately after the metal name Some d-block elements, Sn, Pb, etc. Names of metals that commonly form only one cation do NOT include a Roman numeral No element forms more than one monatomic anion

Sample Problem B: Write the formula and give the name for the compound formed by the ions Cr3+ and F-. Write the symbols for the ions side by side. Write the cation first. Cr3+ F- Cross over the charges to give the subscripts. Cr F −1 3 Check the subscripts and write the formula. The subscripts are correct because they give charges of 1 x 3+ = 3+ and 3 x 1- = 3- The largest common factor of the subscripts is 1, so the smallest whole-number ratio of the ions is 1:3. The formula is CrF3 Since chromium forms more than one ion, so the name must be followed by a Roman numeral indicating its charge. The compound’s name is chromium (III) fluoride.

Compounds Containing Polyatomic Ions Most are negatively charged Most are oxyanions Polyatomic ions that contain oxygen Some elements can combine with oxygen to form more than one type of oxyanion Name depends on number of oxygens in the polyatomic ion: Greater number of oxygens: -ate Smaller number of oxygens: -ite Sometimes more than two types of oxyanions: Hypochlorite, chlorite, chlorate, perchlorate (see chart at right )

Compounds Containing Polyatomic Ions Named in the same way as binary ionic compounds Cation named first Anion named second Example: Silver nitrite: AgNO2 Silver nitrate: AgNO3 When multiples of a polyatomic ion are present in a compound, the formula for the polyatomic ion is enclosed in parentheses Calcium hydroxide: Ca(OH)2 Phosphorus sulfate: P2(SO4)3

Sample Problem C: Write the formula for tin (IV) sulfate. Write the symbols for the ions side by side. Write the cations first. Sn SO42- Cross over the charges to give subscripts. Add parentheses around the polyatomic ion if necessary. Sn (SO4) 2− 4 Check the subscripts and write the formula. The total positive charge is 2 x 4+ = 8+. The total negative charge is 4 x 2- = 8- The charges are equal. The largest common factor of the subscripts is 2, so the smallest whole-number ratio of ions in the compound is 1:2. The correct formula is Sn(SO4)2.

Naming Binary Molecular Compounds Molecular compounds are composed of individual covalently bonded units, or molecules Not simplest ratio as in ionic compounds There are 2 nomenclature systems to name binary molecules: Newer system: Stock system Requires knowledge of oxidation numbers Section 2 of chapter 7 Older system: Based on use of prefixes See prefix list to the right  Either naming system is acceptable Unless specified otherwise

Rules for Naming Binary Molecular Compounds The element that has the smaller group number is usually given first. If both elements are in the same group, the element whose period number is greater is given first. The element is given a prefix only if it contributes more than one atom to a molecule of the compound. Example: Carbon monoxide (CO) Carbon dioxide (CO2) Dinitrogen trioxide (N2O3) Pentaphosphorus decoxide (P5O10)

Rules for Naming Binary Molecular Compounds The second element is named by combining A prefix indicating the number of atoms contributed by the element The root of the name of the element The ending –ide With few exceptions, the ending –ide indicates that a compound contains only two elements

Rules for Naming Binary Molecular Compounds The o or a at the end of a prefix is usually dropped when the word following the prefix begins with another vowel. Examples: Monoxide instead of mono-oxide Pentoxide instead of penta-oxide In general, the order of nonmetals in binary compound names and formulas is: C, P, N, H, S, I, Br, Cl, O, and F.

See page 217, Figure 1.5 Notice how the binary molecular compounds are written and named

Sample Problem D: a. Give the name for As2O5 b. Write the formula for oxygen difluoride A molecule of the compound contains two arsenic atoms, so the first word in the name is: diarsenic The five oxygen atoms are indicated by adding the prefix pent- to the word oxide. The complete name is: diarsenic pentoxide The first symbol in the formula is that for oxygen. Oxygen is first in the name because it is less electronegative than fluorine Since there is no prefix, there must be only one oxygen atom The prefix di- in difluoride shows that there are two fluorine atoms in the molecule The formula is OF2

Some covalent compounds are a network with no single molecules. Each atom is joined to all its neighbors in a covalently bonded three-dimensional network There are no distinct units in these compounds Similar to a crystal in an ionic compound The subscripts indicate the smallest whole-number ratio of the atoms in the compound Naming such compounds is similar to naming molecular compounds. Examples: SiC: silicon carbide SiO2: silicon dioxide Si3N4: trisilicon tetranitride

Acids are solutions of water and a special type of compound. An acid is a distinct type of molecular compound Classified as: Binary acids: acids that consist of two elements Usually hydrogen and one of the halogens Fluorine, chlorine, bromine, or iodine Oxyacids: acids that contain hydrogen, oxygen, and a third element Usually a nonmetal The nonmetal and oxygen are usually a polyatomic ion First recognized as a specific class of compounds based on their properties in solutions of water In chemical nomenclature, acid usually refers to a solution in water of one of these special compounds rather than to the compound itself Example: HCl refers to a water solution of hydrogen monochloride

Acids are solutions of water and a special type of compound. Many polyatomic ions are produced by the loss of hydrogen ions from oxyacids A few examples are seen to the right 

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 Calcium sulfate (CaSO4) is a salt containing an anion from sulfuric acid Some salts contain anions in which one or more hydrogen atoms from the acid are retained Such anions are named by adding the word hydrogen or the prefix bi- to the anion The best known such anion comes from carbonic acid: H2CO3 HCO3- is called hydrogen carbonate ion or bicarbonate ion

Ch 7, Sect 2 Oxidation Numbers (pp. 220-223)
Oxidation Numbers (Oxidation states) Charges on ions making up ionic compounds Reflect general electron distribution of compound among bonded atoms in a molecular compound or polyatomic ion Do not have an exact physical meaning Useful in: Naming compounds Writing formulas Balancing chemical equations Helpful in studying certain reactions Assigned to the atoms composing the compound or ion

Specific Rules Used to Assign Oxidation Numbers (Oxidation states) As a general rule in assigning oxidation numbers shared electrons assumed to belong to more electronegative atom in each bond Atoms in a pure elements have oxidation number of zero Example: pure sodium, oxygen, phosphorus, and sulfur have oxidation number = zero More electronegative element in a binary molecular compound assigned number equal to negative charge it would have as an anion. Less electronegative element assigned number equal to positive charge it would have as a cation Fluorine has an oxidation number of 1- in all of its compounds Most electronegative element

Specific Rules Used to Assign Oxidation Numbers (Oxidation states) Oxygen has an oxidation number of 2- in almost all compounds Exceptions: When oxygen forms peroxides (H2O2), oxidation number = 1- When oxygen forms compounds with fluorine (OF2), oxidation number = 2+ Hydrogen has oxidation number of 1+ in all compounds containing elements more electronegative than hydrogen Hydrogen has oxidation number of 1- in compounds with metals The algebraic sum of the oxidation numbers of all atoms in a neutral compound equals zero The algebraic sum of the oxidation numbers of all atoms in a polyatomic ion equals the charge of the ion

Specific Rules Used to Assign Oxidation Numbers (Oxidation states) Although rules 1-7 apply to covalently bonded atoms, oxidation numbers can also be assigned to atoms in ionic compounds A monatomic ion has an oxidation number equal to the charge of the ion For example; Na+ has a charge of 1+ Ca2+ has a charge of 2+ Cl- has a charge of 1-

Ch 7, Sect 2 Oxidation Numbers (p. 221)
Sample Problem E: Assign oxidation numbers to each atom in the following compounds or ions: UF6 Start by placing known oxidation numbers above the appropriate elements Rule 3: Fluorine always has an oxidation number of 1- Multiply known oxidation numbers by the appropriate number of atoms Place the totals underneath corresponding elements There are 6 fluorine atoms: 6 x 1- = 6- The compound UF6 is molecular. According to guidelines, the sum of the oxidation numbers = 0 Total of positive oxidation numbers = 6+ Divide total calculated oxidation numbers by appropriate number of atoms Only one uranium atom in molecule, oxidation number = 6+

b. H2SO4 Oxygen and sulfur are each more electronegative than hydrogen Rule 5: Hydrogen has and oxidation number of 1+ in all compounds containing elements that are more electronegative than it. Oxygen is not combined with a halogen H2SO4 is not a peroxide Oxygen’s oxidation number = 2- Place these known oxidation numbers above the appropriate symbols Place the total of the oxidation numbers underneath The sum of all the numbers must = zero and there is only one sulfur atom Because (2+) + (8-) = 6-, the oxidation number of sulfur must be 6+

ClO3- To assign oxidation numbers to the elements ClO3-, proceed as in parts a and b. Remember, however, that the total of the oxidation numbers should equal the overall charge of the anion, 1- Rule 7: The algebraic sum of the oxidation numbers of all atoms in a polyatomic ion is equal to the charge on the ion The oxidation number of a single oxygen atom in the ion is 2- The total oxidation number due to all three atoms = 1- For the chlorate ion to have a 1- charge, chlorine must be assigned an oxidation number of +5

Many nonmetals have multiple oxidation numbers See page 223, Figure 2.1 These numbers can be used in the same way as ionic charges to determine formulas Formula for binary compound between sulfur and oxygen: Sulfur commonly has a 4+ or 6+ charge Oxygen has a 2- charge You could have compounds: SO2 SO3

Many nonmetals have multiple oxidation numbers Stock system: Uses Roman numerals to denote oxidation numbers Can be used as an alternative to the prefix system for naming binary molecules SO2 can be called: Prefixes: Sulfur dioxide Stock: Sulfur (IV) oxide SO3 can be called: Prefixes: Sulfur trioxide Stock: Sulfur (VI) oxide You do need to figure out oxidation numbers when using Stock system

Many nonmetals have multiple oxidation numbers The international body that governs nomenclature has endorsed the Stock system More practical for complex compounds Prefix-based names and Stock system names are used interchangeably for many simpler compounds

Ch 7, Sect 3 Using Chemical Formulas (pp. 225-232)
Formula mass is the sum of the average atomic masses of a compound’s elements. Like individual atoms have characteristic average masses, so do: Molecules Formula units Ions We know water (dihydrogen monoxide) is made of 2 hydrogen atoms and 1 oxygen atom. To find the mass of water, add the masses of the three atoms that make up water 2 H atoms x 1.01 u/1 H atom = u 1 O atom x u/1 O atom = u Average mass of H2O molecule = u The mass of a water molecule is referred to as a molecular mass

Formula mass is the sum of the average atomic masses of a compound’s elements. The mass of a NaCl formula unit is not a molecular mass NaCl is ionic, not molecular Formula mass Sum of the average atomic masses of all atoms represented by its formula Same procedure used for any molecule, formula unit, or ion We will round all average atomic masses to two decimal points.

Ch 7, Sect 3 Using Chemical Formulas (p. 226)
Sample Problem F: Find the formula mass of potassium chlorate, KClO3. The mass of a formula unit of KClO3 is found by summing the masses of one K atom, one Cl atom, and three O atoms. The required atomic masses can be found on the periodic table. In the calculation, each atomic mass has been rounded to two decimal points. 1 K atom x u/1 K atom = u 1 Cl atom x u/1 Cl atom = u 3 O atoms x u/1 O atom = u Formula mass of KClO = u

The molar mass of a compound is numerically equal to its formula mass. Molar mass = mass in grams of one mole of a substance 1 mole = x 1023 particles (atoms, ions, molecules, etc.) Determined in the same way as formula mass Units for formula mass: u Units for molar mass: g/mol A compound’s molar mass is numerically equal to its formula mass Sample Problem F: mass of KClO3 = u Molar mass of KClO3 = g/mol

Sample Problem G: What is the molar mass of barium nitrate, Ba(NO3)2? One mole of barium nitrate contains exactly one mole of Ba2+ ions and two moles of NO3- ions. The two moles of NO3- ions contain two moles of N atoms and six moles of O atoms Therefore, the molar mass of Ba(NO3)2 is calculated as follows: 1 mol Ba x g Ba/1 mol Ba = g Ba 2 mol N x g N/1 mol N = g N 6 mol O x g O/1 mol O = g O Molar mass of Ba(NO3) = g/mol

Molar mass is used to convert from moles to grams. Molar mass of a compound can be used as a conversion factor Relate moles to grams Amount in moles x molar mass (g/mol) = mass in grams See Figure 3.2 on page 228

Sample Problem H: What is the mass in grams of 2.50 mol of oxygen gas? Given: mol of O2 Unknown: mass of O2 in grams Plan: moles of O2 to grams of O2 To convert amount of O2 in moles to mass of O2 in grams, multiply by the molar mass of O2 Solve: Find molar mass of O2 2 mol O x 𝑔 𝑂 1 𝑚𝑜𝑙 𝑂 = g (mass of 1 mol of O2) 2.50 mol O2 x 𝑔 𝑂2 1 𝑚𝑜𝑙 𝑂2 = 80.0 g O2 Check your work: The answer is correctly given in three significant figures The answer is close to an estimated value of 75 g 2.50 mol x 30 g/mol

Ch 7, Sect 3 Using Chemical Formulas (p. 229-230)
Sample Problem I: Ibuprofen, C13H18O2, is the active ingredient in many nonprescription pain relievers. Its molar mass is 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? Given: 33 g of C13H18O2, molar mass g/mol Unknown: moles C13H18O2 Plan: convert grams to moles To convert mass of ibuprofen in grams to amount of ibuprofen in moles Solve: 33g C13H18O2 x 1 𝑚𝑜𝑙 𝐶13𝐻18𝑂 𝑔 𝐶13𝐻18𝑂2 = 0.16 mol C13H18O2 Check your work: Checking each step show that the arithmetic is correct, sig figs have been used correctly, and units have been canceled as desired.

Sample Problem I: Ibuprofen, C13H18O2, is the active ingredient in many nonprescription pain relievers. Its molar mass is g/mol b. How many molecules of ibuprofen are in the bottle? Given: 33 g of C13H18O2, molar mass g/mol Unknown: molecules C13H18O2 Plan: convert moles to molecules To find the number of molecules of ibuprofen, multiply amount of C13H18O2 in moles to Avogadro’s number Solve: 0.16 mol C13H18O2 x 𝑥 1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 1 𝑚𝑜𝑙 = 9.6 x 1022 molecules of C13H18O2 Check your work: Checking each step show that the arithmetic is correct, sig figs have been used correctly, and units have been canceled as desired.

Sample Problem I: Ibuprofen, C13H18O2, is the active ingredient in many nonprescription pain relievers. Its molar mass is g/mol c. What is the total mass in grams of carbon in 33 grams of ibuprofen? Given: 33 g of C13H18O2, molar mass g/mol Unknown: total mass of C Plan: convert mol C13H18O2 to mol C to grams C To find the mass of carbon present in the ibuprofen, the two conversion factors needed are the amount of carbon in moles per mole of C13H18O2 Solve: 0.16 mol C13H18O2 x 13 𝑚𝑜𝑙 𝐶 1 𝑚𝑜𝑙 𝐶13𝐻18𝑂2 x 𝑔 𝐶 1 𝑚𝑜𝑙 𝐶 = 25 g C Check your work: Checking each step show that the arithmetic is correct, sig figs have been used correctly, and units have been canceled as desired.

Percent composition is the number of grams in one mole of a compound Percentage composition Percentage by mass of each element in a compound To find percentage composition: 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑖𝑛 𝑠𝑎𝑚𝑝𝑙𝑒 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑎𝑚𝑝𝑙𝑒 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑 x 100 = % element in compound Mass percentage of an element in a compound is the same regardless of the size of the sample A simpler way to find percentage of an element in a compound is to determine how many grams of the element are present in one mole of the compound Divide that value by the molar mass of the compound Multiply by 100 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑖𝑛 1 𝑚𝑜𝑙𝑒 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑 x 100 = % element in compound

Sample Problem J: Find the percentage composition of copper (I) sulfide, Cu2S. Given: formula Cu2S Unknown: percentage composition of Cu2S Plan: The molar mass of the compound must be found Then the mass of each element present in one mole of the compound is used to calculate the mass percentage of each element Solve: 2 mol Cu x 𝑔 𝐶𝑢 1 𝑚𝑜𝑙 𝐶𝑢 = g Cu 1 mol S x 𝑔 𝑆 1 𝑚𝑜𝑙 𝑆 = g S Molar mass of Cu2S = g Cu g S = g Cu2S 127.1 𝑔 𝐶𝑢 𝑔 𝐶𝑢2𝑆 x 100 = 79.84% Cu 32.07 𝑔 𝑆 𝑔 𝐶𝑢2𝑆 x 100 = 20.14% S

Sample Problem K: As some salts crystallize from a water solution, they bind water molecules in their crystal structure. Sodium carbonate forms such a hydrate, in which 10 water molecules are present for every formula unit of sodium carbonate. Find the mass percentage of water in sodium carbonate decahydrate, Na2CO3•10H2O, which has a molar mass of g/mol. Given: chemical formula Na2CO3•10H2O Molar mass of Na2CO3•10H2O Unknown: mass percentage of H2O Plan: The mass of water per mole of sodium carbonate decahydrate must be found This value is then divided by the mass of one mole of Na2CO3•10H2O Solve: One mole of Na2CO3•10H2O Molar mass of H2O = g/mol 10 mol H2O x 𝑔 𝐻2𝑂 1 𝑚𝑜𝑙 𝐻2𝑂 = g H2O

Sample Problem K: As some salts crystallize from a water solution, they bind water molecules in their crystal structure. Sodium carbonate forms such a hydrate, in which 10 water molecules are present for every formula unit of sodium carbonate. Find the mass percentage of water in sodium carbonate decahydrate, Na2CO3•10H2O, which has a molar mass of g/mol. Solve: Molar mass of Na2CO3•10H2O is g/mol We know that 1 mol of Na2CO3•10H2O is g. The mass percentage of 10 mol H2O in one mole of Na2CO3•10H2O can be calculated. Mass % of H2O in Na2CO3•10H2O = 𝑔 𝐻2𝑂 𝑔 𝑁𝑎2𝐶𝑂3 •10𝐻2𝑂 x 100 = 62.97% H2O Checking shows that the arithmetic is correct and that units cancel as desired.

Ch 7, Sect 4 Determining Chemical Formulas (pp. 233-237)
Empirical formula Consists of the symbols for the elements combined in a compound With subscripts showing the smallest whole-number ratio of the different atoms in the compound For ionic compounds, the formula unit is usually the empirical formula For a molecular compound, the empirical formula does not necessarily indicate the actual formula See example at right 

Empirical formulas show the whole number ratio of elements in a compound. To determine a compound’s empirical formula from its percentage composition: Begin by converting percentage composition to a mass composition Diborane: 78.1% boron and 21.9% hydrogen Assume you have g sample of the compound Calculate the amount of each element in the sample Diborane: g boron and 21.9 g hydrogen Convert the mass composition of each element to a composition in moles by dividing the appropriate molar mass 78.1 g B x 1 𝑚𝑜𝑙 𝐵 𝑔 𝐵 = 7.22 mol B 21.9 g H x 1 𝑚𝑜𝑙 𝐻 1.01.𝑔 𝐻 = 21.7 mol H This is not a ratio of the smallest whole numbers

Empirical formulas show the whole number ratio of elements in a compound. Divide each mass by the smallest number in the existing ratio 7.22 𝑚𝑜𝑙 𝐵 = 1 mol B 21.7 𝑚𝑜𝑙 𝐻 = or 3 mol H In this case, take the nearest whole number for moles of H Diborane has the ratio of 1B: 3H This is the empirical formula Sometimes mass composition is known instead of % composition In that case, convert mass composition to comparison in moles Then calculate the smallest whole-number ratio of atoms

Ch 7, Sect 4 Determining Chemical Formulas (p. 234)
Sample 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. Given: % composition (32.38% Na, 22.65% S, 44.99% O) Unknown: empirical formula Plan: change % composition  mass composition  molar composition  smallest whole-number mole ratio of atoms Solve: mass composition: g Na, g S, g O Composition in moles: 32.38 g Na x 1 𝑚𝑜𝑙 𝑁𝑎 𝑔 𝑁𝑎 = mol Na 22.65 g S x 1 𝑚𝑜𝑙 𝑆 𝑔 𝑆 = mol S 44.99 g O x 1 𝑚𝑜𝑙 𝑂 𝑔 𝑂 = mol O

Sample 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. Find the smallest whole-number mole ratio of atoms: 1.408 𝑚𝑜𝑙 𝑁𝑎 = or 2 𝑚𝑜𝑙 𝑆 = or 1 2.812 𝑚𝑜𝑙 𝑂 = or 4 Rounding each number in the ratio to the nearest whole-number yields a mole ratio of 2 mol Na : 1 mol S : 4 mol O The empirical formula of the compound is Na2SO4 Check your work: calculating % composition of the compound reveals values that agree reasonably well with the given percentage composition.

Sample Problem M: Analysis of a g sample of a compound known to contain only phosphorus and oxygen indicates a phosphorus content of g. What is the empirical formula of this compound? Given: sample mass = g Phosphorus mass = g Unknown: empirical formula Plan: mass composition  composition in moles  empirical formula Solve: The mass of oxygen is found by subtracting the phosphorus mass rom the sample mass g sample – g phosphorus = g oxygen Composition in moles: 4.455 g P x 1 𝑚𝑜𝑙 𝑃 𝑔 𝑃 = mol P

Sample Problem M: Analysis of a g sample of a compound known to contain only phosphorus and oxygen indicates a phosphorus content of g. What is the empirical formula of this compound? Composition in moles: 5.717 g O x 1 𝑚𝑜𝑙 𝑂 𝑔 𝑂 = mol O Smallest whole-number ratio: 𝑚𝑜𝑙 𝑃 = 1 mol P 𝑚𝑜𝑙 𝑂 0/1431 = mol O The number of oxygen atoms is not close to a whole number If you multiply each number in the ratio by 2, the number of oxygen atoms becomes mol, close to 5 mol. Simplest whole number ratio of P to O atoms = 2:5 Empirical formula: P2O5

Molecular formulas give the types and numbers of atoms in a compound. Empirical formula Gives smallest possible whole numbers that describe the atomic ratio Molecular formula Actual formula of a molecular compound Example: Empirical formula for diborane: BH3 Molecular formula for diborane: B2H6 Relationship between a compound’s empirical formula and its molecular formula can be written as: x(empirical formula) = molecular formula The number represented by x is a whole-number multiple indicating the factor by which subscripts are multiplied to get the molecular formula

Molecular formulas give the types and numbers of atoms in a compound. Formula masses will have a similar relationship x(empirical formula mass) = molecular formula mass To determine the molecular formula, you must know: Empirical formula Molecular formula mass Remember: a compound’s molecular formula mass is equal numerically to its molar mass A compound’s molecular formula can be found given the compound’s empirical formula and its molar mass.

Sample Problem N: In 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 g/mol. What is the compound’s molecular formula? Given: empirical formula (P2O5) Unknown: molecular formula Plan: x(empirical formula) = molecular formula x = 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑚𝑎𝑠𝑠 𝑒𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑚𝑎𝑠𝑠 Solve: Molecular formula mass is numerically equal to molar mass. Changing g/mol of the compound’s molar mass to u yields the compound’s molecular formula mass Molecular molar mass = g/mol Molecular formula mass = u

Sample Problem N: In 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 g/mol. What is the compound’s molecular formula? Solve: Empirical formula mass if found by adding the masses of each of the atoms indicated in the empirical formula Mass of phosphorus atom = u Mass of oxygen atom = u Empirical formula mass of P2O5 = (2 x u) + (5 x u) = u Dividing experimental formula mass by the empirical formula mass gives the value of x. The formula mass is numerically equal to the molar mass

Sample Problem N: In 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 g/mol. What is the compound’s molecular formula? Solve: x = 𝑢 𝑢 = The compound’s molecular formula is therefore P4O10. 2 x (P2O5) = P4O10 Check Your Work: Checking the arithmetic shows that it is correct.

Chemical Formulas and Chemical Compounds

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Chemical Formulas and Chemical Compounds

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