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 equation as needed to balance Do not change subscripts formulas of chemical 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

1) ____ N 2 + ____ H 2  ____ NH 3 2) ____ KClO 3  ____ KCl + ____ O 2 3) ____ NaCl + ____ F 2  ____ NaF + ____ Cl 2 4) ____ H 2 + ____ O 2  ____ H 2 O 5) ____ Pb(OH) 2 + ____ HCl  ____ H 2 O + ____ PbCl 2 6) ____ AlBr 3 + ____ K 2 SO 4  ____ KBr + ____ Al 2 (SO 4 ) 3 7) ____ CH 4 + ____ O 2  ____ CO 2 + ____ H 2 O 8) ____ C 3 H 8 + ____ O 2  ____ CO 2 + ____ H 2 O 9) ____ C 8 H 18 + ____ O 2  ____ CO 2 + ____ H 2 O 10) ____ FeCl 3 + ____ NaOH  ____ Fe(OH) 3 + ____NaCl

Balancing Chemical Equations

11) ____ P + ____O 2  ____P 2 O 5 12) ____ Na + ____ H 2 O  ____ NaOH + ____H 2 13) ____ Ag 2 O  ____ Ag + ____O 2 14) ____ S 8 + ____O 2  ____ SO 3 15) ____ CO 2 + ____ H 2 O  ____ C 6 H 12 O 6 + ____O 2 16) ____ K + ____ MgBr  ____ KBr + ____ Mg 17) ____ HCl + ____ CaCO 3  ____ CaCl 2 + ____H 2 O + ____ CO 2 18) ____ HNO 3 + ____ NaHCO 3  ____ NaNO3 + ____ H 2 O + ____ CO 2 19) ____ H 2 O + ____ O 2  ____ H 2 O 2 20) ____ NaBr + ____ CaF 2  ____ NaF + ____ CaBr 2

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

) + H 2 O (

l

)  NaOH (

aq

) • Alkalai metals + water  + H 2 (

g

) metal hydroxides + H 2

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

) + O 2 (

g

)  2Na 2 O (

s

)

sodium oxide

Types of Chemical Reactions

•

Decomposition reactions

• A single substance breaks down into two or more smaller substances • A  B + C H 2 CO 3 (

aq

)

carbonic acid

 CO 2 (

g

) + H 2 O (

l

)

Types of Chemical Reactions

•

Combustion reactions

• Hydrocarbons + O 2  CO 2 • Cpds containing C, H, O  + H CO 2 2 O + H 2 O • Involve… – O 2 as reactant – Release of heat & light – Often H 2 O 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 12 C • 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 • m avg = Σ (

m 1 f 1

) + (

m 2 f 2

) + (

m 3 f 3

) + …..

Average atomic mass

• m avg = Σ (

m 1 f 1

) + (

m 2 f 2

) + (

m 3 f 3

) + …..

• Example • • • • Chlorine occurs as two isotopes: • Isotope amu freq (m x f) 35 Cl 37 Cl 34.964 36.966 75.53% 24.47%

Atomic mass of chlorine =

26.41

+9.05

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 (NO 3 ) 2

Percent Composition of Formulas • Percentage by mass contributed by each element in the formula % mass  mass of molar element mass of in the cpd the cpd Determine % O   % oxygen 16g  63g 3   0.76

in nitric acid, HNO 3 % O  76%

Percent Composition

• Determine the % composition of all elements in sucrose, C 12 H 22 O 11 • %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 10 23 • 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 and grams.

bridge

between particles

particles Avagadro s

   

moles molar mass

   

molar mass 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 CO 2

C

x

H

y

O

z

+ O 2 

C

O 2 and 19.65 g H +

H

2 O 2 O 18.4 g 41.7g 19.65 g • All carbon in CO 2 came from the CHO • All hydrogen in H 2 O came from CHO

Combustion Analysis

1. Determine amt of

C

in

C

HO by determining amt of

C

in the

C

O 2 produced.

%C in CO 2 x g CO 2 = g C in CHO .2729 x 41.17 = 11.23 g C = 0.935 mol C

Combustion Analysis

2. Determine amt of

H

in C

H

O by determining amt of H in the

H

2 O produced.

%H in H 2 O x g H 2 O = g H in CHO .1119 x 19.65 = 2.20 g H = 2.18 mol H

Combustion Analysis

• Determine amt of

O

in C

H O

by determining amt of H in the H 2

O

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 g

O

= .311 mol

O Mole ratio C 0.935

H 2.18

O 0.311

Empirical formula C 3 H 7 O

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

Apple Snapple Oatmeal

(www.MrBreakfast.com)

Apple Snapple Oatmeal

(4 servings) 1 medium apple or pear 3 cups apple juice or cider 1 and 1/3 cups regular rolled oats 1/4 cup raisins or chopped pitted dates 1/4 teaspoon ground cinnamon milk (if you like) brown sugar (if you like) 1 A  3 J  1.33

O  0.25

R  0.25

C    4 S

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 • C 6 H 12 O 6 • C 6 H 12 O 6 + __ +

6

O 2  O 2  __ CO 2

6

CO 2 + __ H 2 O +

6

H 2 O

Solving Stoichiometry Problems

Sample problem

• Determine mass of water produced from combustion of 1.00 g glucose • C 6 H 12 O 6 + 6 O 2  6 CO 2 • Determine moles glucose + 6 H 2 O • Determine moles water • Determine grams water

3.7 Limiting Reactants

• • • Stoichiometrically least abundant reactant

Completely consumed

in the reaction

Determines theoretical yield

reaction of the • 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 H 2 ).

Limiting Reactants

In the example below, the O 2 excess reagent .

would be the

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 C 6 H 12 + 5 O 2  2 H 2 C 6 H 8 O 4 + 2 H 2 O • 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 C 6 H 12 + 5 O 2  2 H 2 C 6 H 8 O 4 + 2 H 2 O • 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%