PO 3−4

-3 Anionsphosphate

CO 2−3CrO 2−4Cr2O

2−7

O 2−2HPO 2−4SO 2−3SO 2−4S2O

2−3

-2 Anions

carbonatechromatedichromate

peroxidehydrogen phosphate

sulfitesulfatethiosulfate

MnO−4NO−2NO−3

permanganatenitritenitrate

-1 Anions

OH−

NH+4H3O

+

Cations

-1 AnionsCH3CO

−2

CN−

ClO−

ClO−2ClO−3ClO−4H2PO

−4

HCO−3HSO−3

acetatecyanidehypochloritechloritechlorateperchloratedihydrogen phosphatehydrogen carbonatehydrogen sulfite

(bisulfite)

hydroxyl

HSO−4 hydrogen sulfate (bisulfate)

ammonium

hydronium

Polyatomic Ions Contain Multiple Covalently Bonded

Announcements

-Exam I: Wednesday July 22Note sure about time likely 1.5 hour40 multiple choice question @ 2 pts and 2 longer problems worth 10 each.

-Bring a calculator

-Chapter 3 is critical

NON-METALS

Metals: Charge of Cation = Group Number (1A, 2A, 3A)

Non-metals: Charge of Anion = Group Number - 8

Ni2+

Charge of Transition metals is not so simple--must be learned.

The group number in the periodic table can tell us the charge of the metal cation and the charge of the non-metal anion that will be formed in an ionic reaction.

Transition-Metal Cations and their NamesStock System Ion Older Name

Hydrates are ionic compounds that have “bound” water molecules attached to each formula unit.

CaSO4 . 4H2O CaSO4anhydrous calcium sulfatehydrated calcium sulfate

CaSO4 · 4H2O → 2 CaSO4 · H2O + 3 H2O(g)When naming hydrates, the word hydrate together with a greek prefix and a “.” is used in the formula.

Mono- 1Di-2Tri-3Tetra-4Penta-5Hexa-6Hepta-7Octa-8Nonyl-9Deca-10

Lithium chloride monohydrate LiCl.H2O

MgSO4.5H2O magnesium sulfate pentahydrate

BaCl2.2H2O barium chloride dihydrate

Indicate where in the periodic table the following elements are found:(a) Generic name of the Group VII elements(b) Generic names of Group IA and Group IIA elements(c) Elements that commonly form cations(d) Elements that commonly form covalent bonds(e) Elements that commonly form ionic bonds(f) Elements that commonly have multiple valency(g) Elements that commonly form anions(h) Elements that are inert (unreactive)(i) Elements that are found as molecular compounds

Make the periodic table your friend

More Nomenclature Drills

CATION ANION NAME FORMULAsodium bicarbonate

NH4OH

Na2+ S2O32- sodium thiosulfate

aluminum phosphate

KClO3

Na+ CH3COO-

copper (II) oxide

carbon disulfide

barium hydridesodium dihydrogen phosphate

KH2PO4 Potassium dihydrogen phosphate

Potassium hydrogen phosphate K2HPO4HBr (g) hydrogen bromide (gas)

HBr (aq) hydrobromic acid (aqueous)

Lithium carbonate Li2CO3K2Cr2O7 Potassium dichromate

NH4NO2 ammonium nitrite

disulfur hexafluoride S2F6NaH sodium hydride

P4O6 tetraphosphorous hexaoxide

Iron III Sulfate Fe2(SO4)3HClO2 Hypochlorous acid

Ionic Compound Nomenclature Practice

Borax is the common name of a mineral sodium tetraborate, an industrial cleaning adjunct, Na2B4O7. Suppose you are given 20.0 g of borax

(a) what is the formula mass of Na2B4O7(b) how many moles of borax is 20.0 g?(b) how many moles of boron are present in 20.0 g Na2B4O7?(c) how many grams of boron are present in 20.0 g Na2B4O7? ?(d) how many atoms of boron are present? (e) how many grams of atomic oxygen are present?

Borax is the common name of a mineral sodium tetraborate, an industrial cleaning adjunct, Na2B4O7. Suppose you are given 20.0 g of borax

Solution: (a) The formula weight of Na2B4O7 is (2 × 23.0) + (4 × 10.8) + (7 × 16.0) = 201.2.b) # mol borax = (20.0 g borax) / (201.2 g mol–1) = 0.10 mol of borax, c) # mol B = 20.0 g borax)/(201.2 g mol–1) X 4 mol B/ 1 mol borax = 0.40 mol of B.d) # g B = (0.40 mol) × (10.8 g B/ mol B) = 4.3 g Be) # atoms B = 0.40 mol B x 6.02 X 1023 atoms B/1 mol B = 2.41 X 1023 atoms Bf) # g O = 20.0 g borax)/(201.2 g mol–1) X 7 mol O/ 1 mol borax x 15.99 g O/1 mol O = 11.1 g O

Empirical and Molecular Formulas

Empirical Formula

Molecular Formula

The simplest formula for a compound that gives rise to the smallest set of whole numbers of atoms.

The formula of a compound as it actually exists according to experimental data. It is a multiple of the empirical formula.

CH2O

CH2O C2H4O2 C3H6O3 C4H8O4 C5H10O530.02 60.05 90.08 120.10 150.13

A chemical formula determines the % mass of each element in a pure compound.

n x molar mass of elementmolar mass of compound x 100%

n is the number of moles of the element in 1 mole of the compound

C2H6O

%C = 2 x (12.01 g)46.07 g x 100% = 52.14%

%H = 6 x (1.008 g)46.07 g x 100% = 13.13%

%O = 1 x (16.00 g)46.07 g x 100% = 34.73%

52.14% + 13.13% + 34.73% = 100.0%

Compounds that have the same % mass of its elements have the same empirical formula!

NameMolecular Formula

Whole-Number Multiple

M(g/mol) Use or Function

formaldehydeacetic acidlactic aciderythroseriboseglucose

CH2OC2H4O2C3H6O3C4H8O4C5H10O5C6H12O6

123456

30.0360.0590.09

120.10150.13180.16

disinfectant; biological preservative

acetate polymers; vinegar(5% soln)

part of sugar metabolism

sour milk; forms in exercising muscle

component of nucleic acids and B2

major energy source of the cell

All compounds below have the same % by mass:

40.0% C6.71% H53.3% O.

A laboratory technique called “elemental analysis” can determine the % by mass of a compound and we can use this to derive a empirical and molecular formula.

%C = 52.14%%H = 13.13%%O = 34.73%

C2H6O

Mass Percent

ratio of moles

%C = a%%H = b%%O = c%

CxHyOzMolar Mass

Molar Mass

Empirical Formula

Determining the Empirical Formula from Masses of Elements

Elemental analysis of a sample of an ionic compound gave the following results: 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. Determine the empirical formula and the name of this compound?

Determining the Empirical Formula from Masses of Elements

SOLUTION: 2.82 g Na x = 0.123 mol Na

4.35 g Cl x = 0.123 mol Cl

7.83 g O x

mol Na22.99 g Na

mol Cl35.45 g Clmol O

16.00 g O = 0.489 mol O

Na1 Cl1 O3.98 NaClO4NaClO4 is sodium perchlorate.

Elemental analysis of a sample of an ionic compound gave the following results: 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. Determine the empirical formula and the name of this compound?

Problem Solving Method• Convert % to grams (Assume 100.0 g)• Convert each element in grams to moles

• Divide Moles by the smallest number of moles

• Simplify the Mole RatioMay be necessary to multiply (or divide) all moles by a common term to convert to the entire formula to whole numbers.

• Use Molar Mass to Determine Molecular Formula (molar mass must be given).

Molecularformula

Determining the Empirical Formula From Mass % Data

Dibutyl succinate is an insect repellent used against household ants and roaches. Elemental analysis or analysis indicates that % mass of the composition is 62.58% C, 9.63% H and 27.79% O. Its experimentally determined molecular mass is 230 u. What are the empirical and molecular formulas of dibutyl succinate?

Determining a Molecular Formula from Elemental Analysis and Molar Mass

Step 2: Convert masses to amounts in moles.

Step 3: Write a tentative empirical formula. C5.21H9.55O1.74

Step 4: Convert to small whole numbers.C2.99H5.49ODivide by smallest number of moles

Step 1: Determine the mass of each element. Assume a 100 gram sample and use the given data we have:

C 62.58 g H 9.63 g O 27.79 g

Step 5: Convert to a small whole number ratio.Multiply by 2 to get C5.98H10.98O2The empirical formula is C6H11O2

Step 6: Now using the empirical formula mass and molecular mass together determine the molecular formula. Empirical formula mass is 115 u.

Molecular formula mass is 230 u.

The molecular formula is C12H22O4

n = Molecular mass/empirical mass

Empirical formula mass = 6(12.01) + 1.008(11) + 2(16.00) = 115 u.

= 230 amu/115 amu = 2

Determining a Molecular Formula from Elemental Analysis and Molar Mass

During physical activity, lactic acid (M = 90.08 g/mol) forms in muscle tissue and is responsible for muscle soreness at fatigue. Elemental analysis shows that this compound contains 40.0 mass% C, 6.71 mass% H, and 53.3 mass% O.

(a) Determine the empirical formula of lactic acid.(b) Determine the molecular formula.

MassPercent

EmpiricalFormula

MolecularFormula

Molecular Mass

Understand what is asked: What is the formula CxHyOz

Determining a Molecular Formula from Elemental Analysis and Molar Mass

1. Assume there are 100. g of lactic acid then use % mass:

40.0 g C 6.71 g H 53.3 g O1 mol C12.01g C

1 mol H1.008 g H

1 mol O16.00 g O

3.33 mol C 6.66 mol H 3.33 mol O

C3.33 H6.66 O3.333.33 3.33 3.33

CH2O empirical formula

mass of CH2Omolar mass of lactate 90.08 g

30.03 g3

C3H6O3 is the molecular formula

3. The molecular formula will be a whole number multiple of the empirical formula determined BY THE MOLAR MASS ratio

2. The red numbers are the number of moles of atoms in lactic acid. This is what we use in the formula C3.33 H6.66 O3.33

Combustion analysis is a chemical technique where an unknown hydrocarbon (containing elements C,H,O) is burned forming carbon dioxide and water. All of the C and H in the hydrocarbon appears in the CO2 and H2O. We use this data to determine the empirical and molecular formula of a compounds containing C, H, and O. (Nitrogen can also be detected with an NO2 absorber.)

Unknown sample CxHyOz

All carbon is in CO2 and all H is in the H2O!

CxHyOz CO2 + H2Oexcess O2

Key Riff: All C and H in the unknown sample end up as CO2 and H2O in products of combustion. O you get by difference from beginning and given sample mass.

Combustion analysis gives CO2 and H2O as products. We will use conservation of mass to determine the amounts of C, H and O in a compound.

(Nitrogen can also be detected with an NO2 absorber.)

Determining a Molecular Formula from Combustion Analysis

Vitamin C (M=176.12 g/mol) is a hydrocarbon composed of C,H, and O and is found in many citrus fruits. When a 1.000-g sample of vitamin C is placed in a combustion chamber and burned, the following masses are recorded:

mass of CO2 absorber after combustion =85.35gmass of CO2 absorber before combustion =83.85gmass of H2O absorber after combustion =37.96gmass of H2O absorber before combustion =37.55g

What is the empirical and molecular formula, CxHYOz of vitamin C?

Solving a Combustion Analysis Problem

g CO2 mol CO2 mol C g C

g H2O mol H2O mol H g H

g of O = g of sample – (g of C + g of H)

1. We are trying to find out the empirical formula and the molecular formula. They may be the same---they may not: Both will have a: CxHyOz formula.

2. We must convert the data to mol C and g C and mol H and g H. The mol O will come from the sample data and difference.

g CO2 = 85.35 g - 83.85 g = 1.50 g CO2 g H2O = 37.96 g - 37.55 g = 0.41 g H2O

g O = difference in mass: 1.000 g Vit C - 0.409 g - 0.046 g = 0.545 g O

176.12 g/mol

88.06 g= 2.000 Molecular Formula C6H8O6

g C = 1.50 g CO2 ×1 mol CO2

44.01 g CO2× 1 mol C

1 mol CO2× 12.01 g C

1 mol C= 0.409 g C

g H = 0.41 g H2O×1 mol H2O

18.02 g H2O× 2 mol H

1 mol H2O× 1.008 g H

1 mol H= 0.046 g H

mol C = 0.409 g C× 1 mol C12.01 g C

= 0.0341 mol C

mol H = 0.046 g H× 1 mol H1.008 g H

= 0.0456 mol H

mol O = 0.545 g O× 1 mol O16.00 g O

= 0.0341 mol OC1H1.3O1

Empirical Formula C3H4O3

x 3

Take ratio of given molecular mass to empirical mass to get multiplication factor

C0.0341H0.0456O0.0341Divide through

by 0.0341

Must be a whole number

Copper metal is obtained from copper(I) sulfide containing ores in multistep-extractive process. After grinding the ore into fine rocks, the first step is to heat it strongly with oxygen gas to form powdered cuprous oxide and gaseous sulfur dioxide.

(0) Write a balanced chemical equation for this process(a) How many moles of oxygen are required to fully roast 10.0 mol of copper(I) sulfide?(b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(I) sulfide is roasted?(c) How many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide?

Another stoichiometry example from Silberberg

(a) How many moles of oxygen are required to roast 10.0 mol of copper(I) sulfide?

(b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(I) sulfide is roasted?

Cu2S(s) + O2(g) Cu2O(s) + SO2(g)

3 mol O22 mol Cu2S

= 15.0 mol O2mol O2 = ? = 10.0 mol Cu2S x

= 641g SO2g SO2 = 10.0 mol Cu2S x 2mol SO22mol Cu2S

64.07g SO21 mol SO2

x

(0) Write a balanced chemical equation for this process

2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g)

unbalanced

2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g)

kg O2 = 0.959

1 kg O21000 g O2

kg O2 = 20.0 mol Cu2O

20.0mol Cu2O x3mol O22mol Cu2O

32.00g O21 mol O2

(c) How many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide?

kg O2 = 2.86 kg Cu2O x103g Cu2Okg Cu2O

1 mol Cu2O143.10g Cu2O

x

xxkg O2 =

Do You Understand Limiting Reagents?

124.0 g of solid Al metal is reacted with 601.0 g of iron III oxide to produce iron metal and aluminum oxide. Calculate the mass of aluminum oxide formed.

124 g of Al are reacted with 601 g of Fe2O3

g Al mol Al mol Feg Fe2O3 mol Fe2O3 mol Fe produced

124 g Al1 mol Al27.0 g Al

x1 mol Fe2O3

2 mol Alx

160. g Fe2O31 mol Fe2O3

x = 367 g Fe2O3

Start with 124 g Al need 367 g Fe2O3

We are given 601 g Fe2O3 a large excess so Al is limiting reagent

2Al + Fe2O3 Al2O3 + 2Fe

124 g Al 234 g Al2O3

We now use the limiting reagent (Al) found previously to calculate the amount of product that can be formed.

g Al mol Al mol Al2O3 g Al2O3

1 mol Al27.0 g Al

x1 mol Al2O3

2 mol Alx

102. g Al2O31 mol Al2O3

x =

2Al + Fe2O3 Al2O3 + 2Fe

Phosphorus trichloride is a commercially important compound used in the manufacture of pesticides, gasoline additives, and a number of other products. It is made by the direct combination of phosphorus P4 and molecular chlorine. Suppose 323 g Cl2 is combined with 125 g P4. Determine the amount of phosphorous trichloride can be produced when these reactants are combined.

Do You Understand Limiting Reagents?

Cl2 is the limiting reagent and determines PCl3

P4 (s) + Cl2 (g) → PCl3 (l)Step 1. Converts words to formulas by knowing the nomenclature

Step 2. Balance, balance.......P4 (s) + 6Cl2 (g) → 4PCl3 (l)

mol PCl3 = 125 g P4 ×1 mol P4

123.88 g P4× 4 mol PCl3

1 mol P4= 4.04mol PCl3

mol PCl3 = 323 g Cl2 ×1 mol Cl2

70.91 g Cl2× 4 mol PCl3

6 mol Cl2= 3.04mol PCl3

g PCl3 = 125 g P4 ×1 mol P4

123.88 g P4× 4 mol PCl3

6 molCl2× 137.32 g PCl3

1 mol PCl3= 417.5 g PCl3

Hydrazine(N2H4) and dinitrogen tetraoxide(N2O4), ignite on contact to form nitrogen gas and water vapor. How many grams of nitrogen gas form when 1.00 x 102 g of N2H4 and 2.00 x 102 g of N2O4 are mixed?

Hydrazine(N2H4) and dinitrogen tetraoxide(N2O4), ignite on contact to form nitrogen gas and water vapor. How many grams of nitrogen gas form when 1.00 x 102 g of N2H4 and 2.00 x 102 g of N2O4 are mixed?

SOLUTION: N2H4(l) + N2O4(l) N2(g) + H2O(l)

1.00x102g N2H4 = 3.12mol N2H4mol N2H4

32.05g N2H4

3.12mol N2H4 = 4.68mol N23 mol N2

2mol N2H4

2.00x102g N2O4 = 2.17mol N2O4mol N2O492.02g N2O4

2.17mol N2O4 = 6.51mol N23 mol N2

mol N2O4

N2H4 is the limiting reactant because it produces less product, N2, than does N2O4.

4.68mol N2mol N2

28.02g N2 = 131g N2

2 43

• Actual yield = results from real lab reaction

• Theoretical yield is based on limiting reagent calculation.

• If actual = theoretical the reaction is said to be quantitative.

A + B(reactants)

C(main product)

D (side products)

When we do chemical reactions in the laboratory side-reactions occur and reduce the amount of “actual product” obtained vs the “theoretical amount” obtained in a calculation

Theoretical Yield is the amount of product that wouldresult if all the limiting reagent reacted.

Actual Yield is the amount of product actually obtainedfrom a reaction in the laboratory. It will be given in a word problem.

% Yield = Actual Yield

Theoretical Yieldx 100

The % Yield of a chemical reaction is the ratio of product mass obtained in the lab over the theoretical (calculated on paper value) X 100.

Learning Check: Calculating Percent YieldSilicon carbide (SiC) is made by allowing sand (silicon dioxide) to react with powdered carbon under high temperatures. Carbon monoxide is formed as a by-product. Suppose100.0 kg of sand is processed, 51.4 kg of SiC is recovered. What is the percent yield of SiC from this process?

write a balanced equation

use stoichiometry in balanced equation

find g product predicted

percent yield actual yield/theoretical yield x 100

Learning Check: Calculating Percent Yield

SiO2(s) + 3C(s) SiC(s) + 2CO(g)

100.0 kg SiO2 mol SiO2 60.09 g SiO2

103 g SiO2 kg SiO2

= 1664 mol SiO2

mol SiO2 = mol SiC = 1664

40.10 g SiC1 mol SiC

kg1000 g

= 66.73 kg SiCxkg SiC = 1664 mol SiC x

x 100 = 77.0%51.4 kg66.73 kg% Yield =kg theoretical

kg actual x 100 =

A student reacts 30.0 g benzene, C6H6, with 65.0 g bromine, Br2, to prepare bromobenzene, C6H5Br. (MW C6H6 = 78.102, C6H5Br 156.99, Br2 = 159.808). After the reaction was complete, the student recovered 56.7 g C6H5Br. Determine the limiting reagent, the theoretical yield, and the overall % yield?

Learning Check: Calculating Percent Yield

1. Write a balanced equation

2. Use stoichiometry in balanced equation

3. Find g product predicted limiting reagent

4. Calculate percent yield actual yield/theoretical yield x 100

A student reacts 30.0 g benzene, C6H6, with 65.0 g bromine, Br2, to prepare bromobenzene, C6H5Br. (MW C6H6 = 78.102, C6H5Br 156.99, Br2 = 159.808). After the reaction was complete, the student recovered 56.7 g C6H5Br. Determine the limiting reagent, the theoretical yield, and the overall % yield?

C6H6 + Br2 C6H5Br + HBr

Learning Check: Calculating Percent Yield