You should be reading Chapter 3 and practicing nomenclature from Chapter 2 (lots to know).

Announcements You should be reading Chapter 3 and practicing nomenclature from Chapter 2 (lots to know). Chapter 2: Please expect 1 or perhaps 2 to be collected as a Quiz grade on Tuesday. 1,2,4,13,14,16,18,20,24,27,34,38,40,42,45,50,53,54, 56,58,59,70,73,76,83,84,88,89,90,94,96,100,103,10 5,113,118,134,151,152

Key Chapter 3 Topics For Exam Chapter 3 Problem Assignment 1,2,3,7,11,13,16,18,20,25,27,29,33,35,37,42,44,4 6,50,51,58,61,64,70,71,76,79,82,84,93,96,97,99, 104,119 Exam 1 July 15 --50 minutes long --modeled after Exams in Blog (no bonus, fewer choices, --Please see me now if you having trouble. ---Don t wait and cram it won t work. Work the problems!

Key Chapter 3 Topics For Exam 1. Interconverting Moles, Mass and Atoms 2. Mass Percent From A Chemical Formula 3. Definition of Empirical and Molecular Formula 4. Determing the Empirical and Molecular Formula from combustion analysis data 5. Writing and Balancing Chemical Equations 6. Stoichiometry: Calculating Amounts of Reactants and Products 7. Limiting Reagent Problem 8. Theoretical and Actual Yields

Polyatomic Ions To Memorize Cations NH + 4 H 3 O + ammonium hydronium -1 Anions MnO 4 NO 2 permanganate nitrite -1 Anions NO 3 nitrate CH 3 CO 2 CN ClO ClO 2 ClO 3 ClO 4 H 2 PO 4 HCO 3 HSO 3 HSO 4 OH acetate cyanide hypochlorite chlorite chlorate perchlorate dihydrogen phosphate hydrogen carbonate hydrogen sulfite (bisulfite) hydrogen sulfate (bisulfate) hydroxyl -2 Anions CO 2 3 CrO 2 4 Cr 2 O 2 7 O 2 2 HPO 2 4 SO 2 3 SO 2 4 S 2 O 2 3-3 Anions PO 3 4 carbonate chromate dichromate peroxide hydrogen phosphate sulfite sulfate thiosulfate phosphate

Ionic Compound Nomenclature Practice calcium + nitrogen Potassium + carbonate anion Sulfur + cesium Na + fluorine Tin II fluoride Iron and oxygen (possible) cobalt + sulfur Copper II nitrite potassium permanganate Ni chloride (possible) Ca + oxygen Carbon disulfide Ca3N2 K2CO3 Cs2S sodium fluoride SnF2 Fe2O3 or FeO Cobalt Sulfide Cu(NO2)2 KMnO4 NiCl2 or NiCl3 CaO CS2

Chapter 3 Mass Relationships, Stoichiometry and Chemical Formulas

A known chemical formula provides lots of information: the elements in the compound, the ratios elements combined, the mass of one formula or molecular unit, and the mass in grams of 1 mole of compound in grams. Al2(SO4)3 2 atoms of Al and 3 molecules of (SO4) 2-2 moles of Al and 3 moles of (SO4) 2-1 formula unit Al2(SO4)3 = 342.17 amu Al2(SO4)3 1 mole Al2(SO4)3 = 342.17 g Al2(SO4)3 1 mole Al2(SO4)3 = 6.022 x 10 23 formula units Al2(SO4)3 1 mole Al2(SO4)3 = 2 mol Al 3+ ions 1 mole Al2(SO4)3 = 3 mol (SO4) 2- ions 1 mole Al2(SO4)3 = 12 mol O atoms = 12 x A.N. O atoms 1 mole Al2(SO4)3 = 3 mol S atoms = 3 x A.N. S atoms 1 mole Al2(SO4)3 = 2 mol Al atoms =2 x A.N. Al atoms

Molecular formulas indicates the type and the ratios of combining atoms in a covalent compound. H2O(l) covalent compound 1 molecule H2O = 1 atom O and 2 atoms of H Molecular mass H2O = (2 x 1.008)+15.99 = 18.00 amu Molar mass = 18.00 g/mol H2O 1 mole H2O = 6.022 x 10 23 molecules H2O 1 mole H2O = 2 mol H atoms = 2 x 6.02 x 10 23 H atoms 1 mole H2O = 1 mol O atoms = 6.02 x 10 23 O atoms

We must learn how to recognize conversion factors within a chemical formula, and how to convert grams to moles to molecules to atoms and vis versa. MASS(g) of compound Molar mass (g/mol) MOLES of compound MOLECULES or FORMULA UNITS Chemical Formula Avogadro s Number MOLES of elements in compound ATOMS in a compound

What s In A Chemical Formula? 2.0 g sodium carbonate --how many moles of sodium carbonate? --how many moles of Na +? --how many Na + atoms? 5 moles of magnesium phosphate --how many grams of atomic oxygen -- how many atoms of Mg 2+

What s In A Chemical Formula? Urea, (NH2)2CO, is a nitrogen containing compound used as a fertilizer around the globe? Calculate the number of moles of urea, # of molecules of urea, # hydrogen atoms present in 25.6 g of urea.

Solution To Urea Problem 1. Calculate the molar mass (MM) of urea, (NH2)2CO MM(NH2)2CO = 2 x MN + 4 x MH + MC + MO MM(NH2)2CO = (2 x 14.07g) + 4 (1.007g) + 12.01g + 15.99g MM(NH2)2CO = 60.06 g 2. # moles of (NH2)2CO in 25.6 g Mol (NH 2 ) 2 CO = 25.6 g (NH 2 ) 2 CO 1 mol (NH 2 ) 2 CO 60.06 g (NH 2 ) 2 CO = 0.426 mol (NH 2 ) 2 CO 3. # molecules of (NH2)2CO in 25.6 g Molec urea = 0.426 mol urea 6.02 1023 molecu urea 1 mol urea = 2.57 10 23 molecu urea 4. # H atoms in 25.6 g (NH2)2CO #H atoms = 0.426 mol urea 4 mol Hatoms 1 mol urea 6.02 1023 H atoms 1 mol urea = 1.03 10 24 atoms

Borax is the common name of a mineral sodium tetraborate, an industrial cleaning adjunct, Na2B4O7. In 20.0 g of borax (a) what is the formula mass of Na2B4O7 (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?

In 20.0 g of borax (a) what is the formula mass of Na2B4O7 (b) how many moles of borax and boron are present? (c) how many grams of atomic oxygen are present? Solution: The formula weight of Na2B4O7: (a) (2 22.99)+(4 10.81)+(7 15.99) = 201.15 (b) mol Borax = (20.0 g) / (201.15 g mol 1 ) = 0.0994 mol borax, mol B = 0.0994 x 4 mol B/1 mol borax = 0.3976 mol B. (c) g O = 0.0994 mol borax x 7 mol O/1mol borax x 6.022 x 10 23 O atoms /1 mol O = 4.19 x 10 23 O atom

Empirical and Molecular Formulas Empirical Formula The simplest formula for a compound that gives rise to the smallest set of whole numbers of atoms. CH 2 O Molecular Formula The formula of a compound as it actually exists according to experimental data. It may be a multiple of the empirical formula. CH 2 O C 2 H 4 O 2 C 3 H 6 O 3 C 4 H 8 O 4 C 5 H 10 O 5 30.02 60.05 90.08 120.10 150.13

The percent by mass of an element in a compound is a constant for any pure compound. Knowledge of a formula we can determine % mass and visversa. n x molar mass of element molar mass of compound x 100% n is the number of moles of the element in 1 mole of the compound C 2 H 6 O %C = 2 x (12.01 g) 46.07 g 6 x (1.008 g) %H = 46.07 g 1 x (16.00 g) %O = 46.07 g x 100% = 52.14% x 100% = 13.13% x 100% = 34.73% 52.14% + 13.13% + 34.73% = 100.0%

Compounds that have the same fractional mass (% mass) of its elements have the same empirical formula! All compounds below have the same % by mass: Name Molecular Formula 40.0% C 6.71% H 53.3% O. Whole-Number Multiple M (g/mol) Use or Function formaldehyde acetic acid lactic acid erythrose ribose glucose CH 2 O C 2 H 4 O 2 C 3 H 6 O 3 C 4 H 8 O 4 C 5 H 10 O 5 C 6 H 12 O 6 1 2 3 4 5 6 30.03 60.05 90.09 120.10 150.13 180.16 disinfectant; biological preservative acetate polymers; vinegar(5% soln) sour milk; forms in exercising muscle part of sugar metabolism component of nucleic acids and B 2 major energy source of the cell

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. Mass Percent %C = a% %H = b% %O = c% Empirical Formula C x H y Oz ratio of moles %C = 52.14% %H = 13.13% %O = 34.73% C 2 H 6 O

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. What are the empirical formula and name of the compound?

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. What are the empirical formula and name of the compound? SOLUTION: 2.82 g Na x mol Na = 0.123 mol Na 22.99 g Na 4.35 g Cl x mol Cl = 0.123 mol Cl 35.45 g Cl 7.83 g O x mol O 16.00 g O = 0.489 mol O Na 1 Cl 1 O 3.98 NaClO 4 NaClO 4 is sodium perchlorate.

Determining the Empirical Formula From Mass % Data Molecular formula 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 Ratio May 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).

Determining the Empirical Formula From Mass % Data

Determining An Empirical Formula 5 Step approach: 1. Choose an arbitrary sample size (100g). 2. Convert masses to amounts in moles. 3. Write a general formula. 4. Convert formula to small whole numbers. 5. Multiply all subscripts by a small whole number to make the subscripts integral.

Determining a Molecular Formula from Elemental Analysis and Molar Mass Dibutyl succinate is an insect repellent used against household ants and roaches. Its 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?

Step 1: Determine the mass of each element in a 100g sample. Assuming a 100 gram sample we have: C 62.58 g H 9.63 g O 27.79 g Step 2: Convert masses to amounts in moles. Step 3: Write a tentative empirical formula. C 5.21 H 9.55 O 1.74 Step 4: Convert to small whole numbers. Divide by smallest number of moles C 2.99 H 5.49 O

Step 5: Convert to a small whole number ratio. Multiply by 2 to get C 5.98 H 10.98 O 2 The empirical formula is C 6 H 11 O 2 Step 6: Now using the formula mass and molecular mass together determine the molecular formula. Empirical formula mass is 115 u. Molecular formula mass is 230 u. n = 230 amu/115 amu = 2 The molecular formula is C 12 H 22 O 4