04/22/23 1Kull Chem 105 Chapter 2

Chapter 3Topics• Molecular Formulas • Ionic Compounds

– Ions and charges– Naming ionic compounds

• Molecular Compounds• The Mole

– Mole to mass; mass to mole• Describing Compound Formulas

– Mass percent – percent composition– Empirical and Molecular Formulas

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Molecular Formulas

• Molecule: smallest identifiable unit of pure substance; still maintains composition and chemical properties– Formula (molecular): description of the

composition C3H6O2

– Condensed formulaCH3COCH2OH CH3OCH2CHO

– Structural formula

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Ionic Compounds

• Ions and charges

• Naming ionic compounds

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Predicting Charges on Monatomic IonsPredicting Charges on Monatomic Ions

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3.4 Molecular Compounds• Nonmetal to nonmetal• Use prefixes – mono, di, tri, etc.• Second component add –ide• Only one element in cation spot, mono not

required• Phosporus triiodide• N2F4 • Dioxygen difluoride• P4O10

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3.5 The MoleMole to mass; mass to mole

• Citric acid C6H8O7

• MgCO3

• MgSO4· 7H2O• Formula mass• Mass percent• Moles to grams• Grams to moles to molecules• % composition

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3.6 Describing compound formulas: Using percent composition

Empirical and molecular formulas

A pure compound always consists of the same A pure compound always consists of the same elements combined in the same proportions by elements combined in the same proportions by weight.weight.

Therefore, we can express molecular composition Therefore, we can express molecular composition as as PERCENT BY WEIGHTPERCENT BY WEIGHT

Ethanol, CEthanol, C22HH66OO52.13% C52.13% C13.15% H 13.15% H 34.72% O34.72% O

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Percent CompositionPercent CompositionConsider NOConsider NO22, Molar mass = ?, Molar mass = ?

What is the weight percent of N and of O?What is the weight percent of N and of O?

Wt. % O 2 (16 .0 g O per mole )46 .0 g

x 100 % 69 .6%

Wt. % N = 14.0 g N46.0 g NO2

• 100% = 30.4 %

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Calculating a formula

In In chemical analysischemical analysis we determine the % by we determine the % by weight of each element in a given amount of pure weight of each element in a given amount of pure compound and derive the compound and derive the EMPIRICALEMPIRICAL or or SIMPLESTSIMPLEST formula.formula.

PROBLEMPROBLEM: A compound of B and H : A compound of B and H is 81.10% B. What is its empirical is 81.10% B. What is its empirical formula?formula?

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• Because it contains only B and H, it Because it contains only B and H, it must contain 18.90% H.must contain 18.90% H.

• In 100.0 g of the compound there In 100.0 g of the compound there are 81.10 g of B and 18.90 g of H.are 81.10 g of B and 18.90 g of H.

• Calculate the number of moles of Calculate the number of moles of each constituent.each constituent.

A compound of B and H is 81.10% A compound of B and H is 81.10% B. What is its empirical formula?B. What is its empirical formula?

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Calculate the number of moles of Calculate the number of moles of each element in 100.0 g of sample.each element in 100.0 g of sample.

81.10 g B • 1 mol10.81 g

= 7.502 mol B

18.90 g H • 1 mol1.008 g

= 18.75 mol H

A compound of B and H is 81.10% A compound of B and H is 81.10% B. What is its empirical formula?B. What is its empirical formula?

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Now, recognize that Now, recognize that atoms combine in the atoms combine in the ratio of small whole numbers.ratio of small whole numbers.

1 atom B + 3 atoms H --> 1 molecule BH1 atom B + 3 atoms H --> 1 molecule BH33

oror1 mol B atoms + 3 mol H atoms ---> 1 mol B atoms + 3 mol H atoms ---> 1 mol BH1 mol BH33

moleculesmolecules

Find the ratio of moles of elements Find the ratio of moles of elements in the compound.in the compound.

A compound of B and H is 81.10% A compound of B and H is 81.10% B. What is its empirical formula?B. What is its empirical formula?

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But we need a whole number ratio. But we need a whole number ratio. 2.5 mol H/1.0 mol B = 5 mol H to 2 mol B2.5 mol H/1.0 mol B = 5 mol H to 2 mol BEMPIRICAL FORMULA = BEMPIRICAL FORMULA = B22HH55

Take the ratio of moles of B and H. Take the ratio of moles of B and H. AlwaysAlwaysdivide by the smaller number.divide by the smaller number.

18.75 mol H7.502 mol B

= 2.499 mol H1.000 mol B

= 2.5 mol H1.0 mol B

A compound of B and H is 81.10% A compound of B and H is 81.10% B. What is its empirical formula?B. What is its empirical formula?

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A compound of B and H is 81.10% B. Its A compound of B and H is 81.10% B. Its empirical formulaempirical formula is B is B22HH55. What is its . What is its molecular formulamolecular formula??

Is the molecular formula BIs the molecular formula B22HH55, B, B44HH1010, , BB66HH1515, B, B88HH2020, etc.? , etc.?

BB22HH66 is one example of this class of compounds. is one example of this class of compounds.

B2H6

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A compound of B and H is 81.10% B. Its A compound of B and H is 81.10% B. Its empirical formula is Bempirical formula is B22HH55. What is its molecular . What is its molecular formula?formula?

We need to do an We need to do an EXPERIMENTEXPERIMENT to find to find the MOLAR MASS.the MOLAR MASS.

Here experiment gives Here experiment gives 53.3 g/mol53.3 g/molCompare with the mass of BCompare with the mass of B22HH55 = = 26.66 g/unit26.66 g/unit Find the ratio of these masses.Find the ratio of these masses.

53.3 g/mol26.66 g/unit of B2H5

= 2 units of B2H51 mol

Molecular formula = BMolecular formula = B44HH1010

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Data to Determine the Data to Determine the formula of a Sn—I formula of a Sn—I

CompoundCompound• Reaction of Sn and IReaction of Sn and I22 is done using excess Sn. is done using excess Sn.• Mass of Sn in the beginning = 1.056 gMass of Sn in the beginning = 1.056 g• Mass of Sn remaining (recovered) = 0.601 gMass of Sn remaining (recovered) = 0.601 g• Mass of iodine (IMass of iodine (I22) used = 1.947 g) used = 1.947 g

(See p. 125)(See p. 125)Convert these masses to molesConvert these masses to moles

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• Reaction of Sn and IReaction of Sn and I22 is done using excess Sn. is done using excess Sn.

Mass of iodine (IMass of iodine (I22) used = 1.947 g) used = 1.947 g

Mass of Sn initially = 1.056 gMass of Sn initially = 1.056 gMass of Sn recovered = 0.601 gMass of Sn recovered = 0.601 gMass of Sn used = 0.455 gMass of Sn used = 0.455 g Find the mass of Sn that combined with 1.947 g IFind the mass of Sn that combined with 1.947 g I22..

Find moles of Sn used:Find moles of Sn used:0.455 g Sn • 1 mol

118.7 g = 3.83 x 10-3 mol Sn

Tin and Iodine CompoundTin and Iodine Compound

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Tin and Iodine CompoundTin and Iodine CompoundNow find the number of moles of INow find the number of moles of I22 that combined that combined

with with 3.83 x 103.83 x 10-3-3 mol Sn mol Sn. Mass of I. Mass of I22 used was 1.947g. used was 1.947g.

1.947 g I2 • 1 mol253.81 g

= 7.671 x 10-3 mol I2

How many mol of How many mol of iodine atomsiodine atoms? ?

= 1.534 x 10-2 mol I atoms

7.671 x 10-3 mol I2 2 mol I atoms1 mol I2

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Tin and Iodine CompoundTin and Iodine CompoundNow find the ratio of number of moles of Now find the ratio of number of moles of

moles of I and Sn that combined.moles of I and Sn that combined.1.534 x 10-2 mol I3.83 x 10-3 mol Sn

= 4.01 mol I1.00 mol Sn

Empirical formula is Empirical formula is SnISnI44

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Naming Ionic and Covalent Compounds3.86. Name each of the following compounds, and tell which ones are best described as ionic:(c) SrSO4 (h) Al2S3(d) Ca(NO3)2 (i) PCl3(e) XeF4 ( j) K3PO43.86 (a) chlorine trifluoride (f) oxygen difluoride

(b) nitrogen trichloride (g) potassium iodide, ionic(c) strontium sulfate, ionic (h) aluminum sulfide, ionic(d) calcium nitrate, ionic (i) phosphorus trichloride(e) xenon tetrafluoride (j) potassium phosphate, ionic

3.87. Write the formula for each of the following compounds, and tell which ones are best described as ionic:(b) boron triiodide(c) aluminum perchlorate(e) potassium permanganate(f ) ammonium sulfite(g) potassium dihydrogen phosphate(h) disulfur dichloride(i) chlorine trifluoride3.87 (a) NaOCl, ionic (f) (NH4)2SO3, ionic

(b) BI3 (g) KH2PO4, ionic(c) Al(ClO4)3, ionic (h) S2Cl2 (d) Ca(CH3CO2)2, ionic (i) ClF3(e) KMnO4, ionic (j) PF3

3.88. Complete the table by placing symbols, formulas, and names in the blanks.Cation Anion Name Formula______ ______ ammonium bromide ______Ba2 ______ __________________ BaS______ Cl iron(II) chloride ____________ F __________________ PbF2

Al3 CO3 __________________ ______3.88 Cation Anion Name Formula

NH4+ Br– ammonium bromide NH4BrBa2+ S2– barium sulfide BaSFe2+ Cl– iron(II) chloride FeCl2Pb2+ F– lead(II) fluoride PbF2Al3+ CO32– aluminum carbonate Al2(CO3)3Fe3+ O2– iron(III) oxide Fe2O3

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Molecules, Compounds, & the Mole3.102. Which of the following statements about 57.1 g of octane,

C8H18, is (are) not true?(a) 57.1 g is 0.500 mol of octane.(b) The compound is 84.1% C by weight.(c) The empirical formula of the compound is C4H9.(d) 57.1 g of octane contains 28.0 g of hydrogen atoms.3.102 (a) True. 0.500 mol C8H18 · = 57.1 g

C8H18

(b) True. · 100% = 84.1% C(c) True.(d) False. 57.1 g C8H18 · = 9.07 g H

114.2 g C8H18

1 mol C8H18

(8)(12.01) g C114.2 g C8H18

(18)(1.008) g H114.2 g C8H18

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Percent Composition

3.104 A metal M forms a compound with the formula MCl4. If the compound is 74.75% chlorine, what is the identity of M?

3.104 Molar mass MCl4 = 189.7 g

Atomic weight M = 189.7 g MCl4 – (4)(35.453) g Cl = 47.9 g M M is Ti, titanium

74.75 g Cl100.00 g MCl4

= (4)(35.453) g Cl

molar mass MCl4

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Empirical and Molecular Formulas

• 3.56 Nicotine, a poisonous compound found in tobacco leaves, is 74.0% C, 8.65% H, and 17.35% N. Its molar mass is 162 g/mol. What are the empirical and molecular formulas of nicotine?

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Determining Formulas from Mass Data

3.61 Zinc metal (2.50 g) combines with 9.70 g of iodine to produce zinc iodide, ZnxIy. What is the formula of this ionic compound?

3.61 2.50 g Zn · = 0.0382 mol Zn9.70 g I · = 0.0764 mol I

The empirical formula is ZnI2

1 mol Zn65.39 g Zn

1 mol I126.9 g I

0.0764 mol I0.0382 mol Zn

= 2 mol I

1 mol Zn

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3.63. Write formulas for all of the compounds that can be made by combining the cations NH4 and Ni2 with the anions CO3 and SO4.

3.63 (NH4)2CO3

(NH4)2SO4

NiCO3

NiSO4