CHAPTER 3: Stoichiometry I: Equations, the Mole, and Chemical Formulas

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1 CHAPTER 3: Stoichiometry I: Equations, the Mole, and Chemical Formulas 3.1 Chemical Equations C 5 H 1 + O H O + CO reactants products Use subscripts (s), (l), (g), (aq) to indicate physical states Balancing Chemical Equations Al + H SO 4 Al (SO 4 ) 3 + H Be C + H O Be(OH) + CH 4 CH 4 + Br CBr 4 + HBr Cl + NaI NaCl + I CO + O CO CO + KOH K CO 3 + H O Cu(NO 3 ) CuO + NO + O Fe O 3 + CO Fe + CO Fe O 3 + CO Fe 3 O 4 + CO H + Br HBr H O H O + O HCl + CaCO 3 CaCl + H O + CO K + H O KOH + H KClO 3 KCl + O KClO 3 KClO 4 + KCl KNO 3 KNO + O KO + H O + CO KHCO 3 + O Mg + O MgO N + H NH 3 (NH 4 ) Cr O 7 Cr O 3 + H O + N N H 4 + N O 4 N + H O NaHCO 3 Na CO 3 + H O + CO NH 3 + CuO Cu + N + H O NH 3 + O N + H O NH 3 + O NO + H O NH 4 NO N + H O NH 4 NO 3 N O + H O O 3 O P 4 O 10 + H O H 3 PO 4 PbS + O PbO + SO S + HNO 3 H SO 4 + NO + H O S 8 + O SO XeF + H O Xe + HF + O Zn + AgCl ZnCl + Ag Types of Chemical Reactions Combustion: reaction with oxygen C + O CO CH 4 + O CO + H O Fe + O Fe O 3 C 1 H O 11 + O CO + H O Neutralization: reaction of acid and base to yield water and salt KOH + H 3 PO 4 K 3 PO 4 + H O NaOH + H SO 4 Na SO 4 + H O

2 3. The Mole Atomic Masses Atomic masses originally defined as relative masses based on H. Changed to O to improve accuracy, and to 1 C (1961). Masses can be determined accurately with a mass spectrometer. For atoms which exist as isotopes, the atomic mass given is the weighted average of the isotopoc masses. 1 C = 1 (by definition), 98.89% 13 C = , 1.11% To calculate weighted average, multiply each isotopic mass by its relative abundance. C = ( )(0.9889) + ( )(0.0111) = = however, relative abundance is only accurate to 4 significant figures, so the atomic mass of C is given as 1.01 The Mole 6 Li = , 7.4% 7 Li = , 9.58% Li = (6.0151)(0.074) + ( )(0.958) = 6.94 A mole is a way to count atoms 1 mole = large number of atoms since atoms are too small to count, we must count them indirectly by weighing then, so we define the mole in terms of mass 1 mole = number of atoms in 1 g of 1 C Because 1 is the atomic mass of 1 C, and the atomic mass scale is a relative scale, any time you have a mass in grams equal to the atomic mass of an element, you have one mole of the element. 1 mole = x 10 3 atoms = atomic mass in grams for Fe, 1 mole Fe = 6.0 x 10 3 atoms Fe = 1 mole Fe g Fe x = mol Fe mol Fe x 1 mol Fe = g Fe g Fe x x 103 atoms Fe =1.078 x 10 atoms Fe

3 Molecular Weight molecular mass = sum of masses of all atoms in molecule for C H 6, molecular mass = (1.01 amu) + 6(1.008 amu) = amu 1 mol C H 6 = 6.0 x 10 3 molecules C H 6 = g C H 6 molar mass - mass in grams of one mole of a substance g C H 6 x 1 mol C H g C H 6 = mol C H g C H 6 x 6.0 x 103 molecules C H g C H =.00 x 10 3 molecules C H x H atoms molecules C H 6 x 1 molecule C H = 1.01 x 10 4 H atoms Empirical Formulas C H 4 O C (1.01) = 4.0 H 4(1.008) = 4.03 O (16.00) = 3.00 %C = 4.0 x 100% = 40.00% %H = 4.03 x 100% = 6.714% %O = 3.00 x 100% = 53.9% Determining the Formula of a Compound 93.70% C x 1 mol C 1.01 g C = 7.80 mol C 6.30% H x 1 mol H g H = 6.4 mol H reduce to a simple ratio 7.80 mol C 1.5 mol C = 6.4 mol H 1 mol H = 5 mol C 4 mol H empirical formula is C 5 H 4 empirical formula mass = 5(1.01) + 4(1.008) = molecular mass (determined by experiment) = molecular mass formula mass = = so molecular formula = (empirical formula) = C 10H 8

4 Combustion Train Oxygen in Sample H O absorber [Mg(ClO 4 ) ] CO absorber [NaOH] g of a compound containing C, H, and O g CO and 3.53 g H O 1.01 g C g CO x g CO = g C.016 g H 3.53 g H O x g H O = g H g compound g C g H = g O g C x 1 mol C 1.01 g C = mol C = 1.33 mol C g H x 1 mol H g H = mol H = 1.00 mol H g O x 1 mol O g O = mol O = mol O multiply by 3 to convert fractions C 4 H 3 O empirical formula mass = from experiment, molecular mass = so molecular formula = C 8 H 6 O 4 Oxygen carbon dioxide water vapor out 3.4 Mass Relationships in Equations Ca 3 N + 6H O 3Ca(OH) + NH 3 What mass of calcium hydroxide can be produced by the reaction of g calcium nitride with excess water? 1) convert mass of calcium nitride to moles g Ca 3 N x \f(148.6 g Ca 3 N ) = mol Ca 3 N ) convert moles of calcium nitride to moles of calcium hydroxide mol Ca 3 N x 3 mol Ca(OH) 1 mol Ca 3 N =.04 mol Ca(OH) 3) convert moles of calcium hydroxide to mass of calcium hydroxide.04 mol Ca(OH) x g Ca(OH) 1 mol Ca(OH) = g Ca(OH) H O + Cl O HOCl How many grams of Cl O are required to make g HOCl? 1 mol HOCl g HOCl x 5.46 g HOCl x 1 mol Cl O mol HOCl x g Cl O 1 mol Cl O = 1.4 g Cl O

5 3.5 Limiting Reactants Ca 3 (PO 4 ) + 8C Ca 3 P + 8CO What mass of Ca 3 P can be produced from g Ca 3 (PO 4 ) and 5.00 g C? g Ca 3 (PO 4 ) x 1 mol Ca 3(PO 4 ) g Ca 3 (PO 4 ) = mol Ca 3 (PO 4 ) 5.00 g C x 1 mol C 1.01 g C =.08 mol C.08 mol C mol Ca 3 (PO 4 ) = > 8 1 soc is in excess and Ca 3(PO 4 ) will run out first 1 mol Ca 3 P mol Ca 3 (PO 4 ) x 1 mol Ca 3 (PO 4 ) x g Ca 3P 1 mol Ca 3 P = 9.37 g Ca 3 P Zn + AgNO 3 Ag + Zn(NO 3 ) What mass of Ag can be fromed from 3. g Zn and 4.35 g AgNO 3? 1 mol Zn 3. g Zn x g Zn = mol Zn 4.35 g AgNO 3 x 1 mol AgNO g AgNO = mol AgNO 3 3 Zn is in excess mol Ag g Ag mol AgNO 3 x mol AgNO x 3 1 mol Ag =.76 g Ag 4LiH + AlCl 3 LiAlH 4 + 3LiCl What mass of LiAlH 4 can be formed from g LiH and g AlCl 3? 1 mol LiH g LiH x g LiH = 5.03 mol LiH g AlCl 3 x 1 mol AlCl g AlCl = mol AlCl mol LiH = mol AlCl 3 1 so LiH is in excess mol AlCl 3 x 1 mol LiAlH 4 1 mol AlCl x g LiAlH mol LiAlH = 8.46 g LiAlH 4 4 Percent Yield percent yield = actual yield theoretical yield x 100% example: if 4.04 g LiAlH 4 is actually obtained, % yield = 4.04 g 8.46 g x 100% = 84.5%