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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. Molecular Formulas.
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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 Kull Chem 105 Chapter 2
Molecular Formulas • Molecule: smallest identifiable unit of pure substance; still maintains composition and chemical properties • Formula (molecular): description of the composition C3H6O2 • Condensed formula CH3COCH2OH CH3OCH2CHO • Structural formula
Ionic Compounds • Ions and charges • Naming ionic compounds
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
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
3.6 Describing compound formulas: Using percent compositionEmpirical and molecular formulas A pure compound always consists of the same elements combined in the same proportions by weight. Therefore, we can express molecular composition as PERCENT BY WEIGHT Ethanol, C2H6O 52.13% C 13.15% H 34.72% O
Percent Composition Consider NO2, Molar mass = ? What is the weight percent of N and of O?
Calculating a formula In chemical analysis we determine the % by weight of each element in a given amount of pure compound and derive the EMPIRICALor SIMPLESTformula. PROBLEM: A compound of B and H is 81.10% B. What is its empirical formula?
A compound of B and H is 81.10% B. What is its empirical formula? • Because it contains only B and H, it must contain 18.90% H. • In 100.0 g of the compound there are 81.10 g of B and 18.90 g of H. • Calculate the number of moles of each constituent.
A compound of B and H is 81.10% B. What is its empirical formula? Calculate the number of moles of each element in 100.0 g of sample.
A compound of B and H is 81.10% B. What is its empirical formula? Now, recognize that atoms combine in the ratio of small whole numbers. 1 atom B + 3 atoms H --> 1 molecule BH3 or 1 mol B atoms + 3 mol H atoms ---> 1 mol BH3 molecules Find the ratio of moles of elements in the compound.
A compound of B and H is 81.10% B. What is its empirical formula? Take the ratio of moles of B and H. Always divide by the smaller number. But we need a whole number ratio. 2.5 mol H/1.0 mol B = 5 mol H to 2 mol B EMPIRICAL FORMULA = B2H5
B2H6 B2H6 is one example of this class of compounds. A compound of B and H is 81.10% B. Its empirical formula is B2H5. What is its molecular formula? Is the molecular formula B2H5, B4H10, B6H15, B8H20, etc.?
A compound of B and H is 81.10% B. Its empirical formula is B2H5. What is its molecular formula? We need to do an EXPERIMENTto find the MOLAR MASS. Here experiment gives 53.3 g/mol Compare with the mass of B2H5 = 26.66 g/unit Find the ratio of these masses. Molecular formula = B4H10
Data to Determine the formula of a Sn—I Compound • Reaction of Sn and I2 is done using excess Sn. • Mass of Sn in the beginning = 1.056 g • Mass of Sn remaining (recovered) = 0.601 g • Mass of iodine (I2) used = 1.947 g (See p. 125) Convert these masses to moles
Tin and Iodine Compound • Reaction of Sn and I2 is done using excess Sn. Mass of iodine (I2) used = 1.947 g Mass of Sn initially = 1.056 g Mass of Sn recovered = 0.601 g Mass of Sn used = 0.455 g Find the mass of Sn that combined with 1.947 g I2. Find moles of Sn used:
Tin and Iodine Compound Now find the number of moles of I2 that combined with 3.83 x 10-3 mol Sn. Mass of I2 used was 1.947g. How many mol of iodine atoms? = 1.534 x 10-2 mol I atoms
Tin and Iodine Compound Now find the ratio of number of moles of moles of I and Sn that combined. Empirical formula is SnI4
Naming Ionic and Covalent Compounds 3.86. 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) K3PO4 3.86 (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 trifluoride 3.87 (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 FormulaNH4+ 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
Molecules, Compounds, & the Mole 3.102. 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 • True. · 100% = 84.1% C • True. • False. 57.1 g C8H18 · = 9.07 g H
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
Empirical and Molecular Formulas • 3.56Nicotine, 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?
Determining Formulas from Mass Data 3.61Zinc 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 Zn 9.70 g I · = 0.0764 mol I The empirical formula is ZnI2
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