Molar relationships in chemical reactions Stoichiometry Molar relationships in chemical reactions

3.1 Chemical Equations: A Review Law of conservation of mass Relationship between reactant and products produces a balanced chemical equation Reactants  Products A + B  C + D Balancing chemical eqns

Balancing Chemical Equations Must have correct chemical formulas Change coefficients as needed to balance equation Do not change subscripts of chemical formulas Indicate physical states by writing in parentheses (s), (l), (g), (aq) Begin with an element that appears in only one reactant and product If possible, do not begin with O or H

Balancing Chemical Equations ____ N2 + ____ H2  ____ NH3 ____ KClO3  ____ KCl + ____ O2 ____ NaCl + ____ F2  ____ NaF + ____ Cl2 ____ H2 + ____ O2  ____ H2O ____ Pb(OH)2 + ____ HCl  ____ H2O + ____ PbCl2 ____ AlBr3 + ____ K2SO4  ____ KBr + ____ Al2(SO4)3 ____ CH4 + ____ O2  ____ CO2 + ____ H2O ____ C3H8 + ____ O2  ____ CO2 + ____ H2O ____ C8H18 + ____ O2  ____ CO2 + ____ H2O 10) ____ FeCl3 + ____ NaOH  ____ Fe(OH)3 + ____NaCl

Balancing Chemical Equations ____ P + ____O2  ____P2O5 ____ Na + ____ H2O  ____ NaOH + ____H2 ____ Ag2O  ____ Ag + ____O2 ____ S8 + ____O2  ____ SO3 ____ CO2 + ____ H2O  ____ C6H12O6 + ____O2 ____ K + ____ MgBr  ____ KBr + ____ Mg ____ HCl + ____ CaCO3  ____ CaCl2 + ____H2O + ____ CO2 ____ HNO3 + ____ NaHCO3  ____ NaNO3 + ____ H2O + ____ CO2 ____ H2O + ____ O2  ____ H2O2 20) ____ NaBr + ____ CaF2  ____ NaF + ____ CaBr2

3.2 Patterns of Chemical Reactivity Because groups of elements have similar chemical behavior, the periodic table can be used to predict reactions involving elements Example: Na (s) + H2O (l)  NaOH (aq) + H2 (g) Alkalai metals + water  metal hydroxides + H2

Types of Chemical Reactions Synthesis (Combination) reactions Two or more substances combine to form a new substance Often involve elements combining to form a compound A + B  C 4Na (s) + O2 (g)  2Na2O (s) sodium oxide

Types of Chemical Reactions Decomposition reactions A single substance breaks down into two or more smaller substances A  B + C H2CO3 (aq)  CO2 (g) + H2O (l) carbonic acid

Types of Chemical Reactions Combustion reactions Hydrocarbons + O2  CO2 + H2O Cpds containing C, H, O  CO2 + H2O Involve… O2 as reactant Release of heat & light Often H2O is a product

3.3 Formula Weights Atomic mass unit (amu) A unit invented to describe extremely small masses Defined as 1/12 the mass of an atom of 12C 1 amu = 1.66054 x 10-24 g

Average Atomic Masses Atomic masses given on the PT are weighted averages of all the isotopes of that element Equals the sum of the products of the (mass) x (frequency) of the isotopes mavg = Σ (m1f1) + (m2f2) + (m3f3) + …..

Average atomic mass mavg = Σ (m1f1) + (m2f2) + (m3f3) + ….. Example Chlorine occurs as two isotopes: Isotope amu freq (m x f) 35Cl 34.964 75.53% 26.41 37Cl 36.966 24.47% +9.05 Atomic mass of chlorine = 35.46

Formula & Molecular Weights Formula mass refers to ionic cpds Molecular mass refers to covalent cpds Is the sum of atomic masses of the molecule Calculate mol mass of Ca (NO3)2

Percent Composition of Formulas Percentage by mass contributed by each element in the formula

Percent Composition Determine the % composition of all elements in sucrose, C12H22O11 %C = (12 x 12) / 342 = 42.0% %H = (1 x 22) / 342 = 6.4% %O = (16 x 11) / 342 = 51.6%

3.4 The Mole Is a counting unit Avagadro’s number = 6.022 x 1023 Interconvert between moles & particles Molar mass = the mass of one mole of a substance Molar mass of elements = atomic mass in grams

Converting between particles, moles, and grams Particles and grams cannot be interconverted directly. The mole is the bridge between particles and grams.

3.5 Empirical Formula Analysis To determine empirical formula from percent composition data Empirical formula gives simplest ratio of atoms in a compound Therefore, always convert to moles when calculating empirical formulas Sample exercise 3.13

Combustion Analysis Determines empirical formula based upon analysis of products of combustion Example: 18.4 g of a compound of CHO produced 41.7 g CO2 and 19.65 g H2O CxHyOz + O2  CO2 + H2O 18.4 g 41.7g 19.65 g All carbon in CO2 came from the CHO All hydrogen in H2O came from CHO

Combustion Analysis %C in CO2 x g CO2 = g C in CHO Determine amt of C in CHO by determining amt of C in the CO2 produced. %C in CO2 x g CO2 = g C in CHO .2729 x 41.17 = 11.23 g C = 0.935 mol C

Combustion Analysis 2. Determine amt of H in CHO by determining amt of H in the H2O produced. %H in H2O x g H2O = g H in CHO .1119 x 19.65 = 2.20 g H = 2.18 mol H

Combustion Analysis Determine amt of O in CHO by determining amt of H in the H2O produced By law of conservation of mass, grams of CHO = gC + gH + gO So grams O = grams CHO – gC – gH 18.40 – 11.23 – 2.20 = 4.97 gO = .311 mol O Mole ratio C 0.935 H 2.18 O 0.311 Empirical formula C3H7O

3.6 Quantitative Information from Balanced Equations Balanced chemical equations are like specific recipes Balanced equations tell us …. the relative amounts of reactants (ingredients) the relative amounts of products other relevant conditions needed (e.g. heat) The unit of measure in a balanced equation is the mole Indicated by coefficients

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Stoichiometric Calculations The coefficients in the balanced equation give the ratio of moles of reactants and products.

3.6 Quantitative Information from Balanced Equations Mole ratio used to calculate actual molar amounts in a balanced eqn Sample Problem 3.16 Determine mass of water produced from combustion of 1.00 g glucose C6H12O6 + __ O2  __ CO2 +__ H2O C6H12O6 + 6 O2  6 CO2 + 6 H2O

Solving Stoichiometry Problems

Sample problem Determine mass of water produced from combustion of 1.00 g glucose C6H12O6 + 6 O2  6 CO2 + 6 H2O Determine moles glucose Determine moles water Determine grams water

3.7 Limiting Reactants Stoichiometrically least abundant reactant Completely consumed in the reaction Determines theoretical yield of the reaction When all reactants are present in stoichiometric amounts, all are limiting reagents

Limiting Reactants The limiting reactant is the reactant present in the smallest stoichiometric amount. In other words, it’s the reactant you’ll run out of first (in this case, the H2).

Limiting Reactants In the example below, the O2 would be the excess reagent.

Limiting Reagents How many moles of ammonia can be produced when 3.0 mol nitrogen and 6.0 mol of hydrogen are permitted to react? Write & balance the equation Determine the LR to determine yield Reactant  least yield is the LR What is the excess reagent? How much is left over?

Theoretical & Actual Yields Theoretical Yield Quantity of product produced when 100% of limiting reagent reacts Determined from balanced equation Actual Yield Quantity of product actually produced No reaction is 100% efficient, therefore…. Theoretical Yield does not equal Actual Yield

Sample Problem 3.20 2 C6H12 + 5 O2  2 H2C6H8O4 + 2 H2O Given 25.0 g cyclohexane and excess oxygen, determine theoretical yield of adipic acid

Percent Yield Expresses the efficiency of a reaction Is ratio of actual yield to theoretical yield Percent Yield = Actual Y/ Theroretical Y %Y = AY / TY

Sample Problem 3.20 (cont) 2 C6H12 + 5 O2  2 H2C6H8O4 + 2 H2O Theoretical yield = 43.5 g adipic acid (from first part) If actual yield of adipic acid was 33.5 g, determine percent yield of reaction %Y = Actual yield / Theoretical yield %Y = 33.5g/43.5g = .77 = 77%