Quantities of Reactants and Products

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Quantities of Reactants and Quantities of Reactants and ProductsProducts

Chapter 4Chapter 4

Antoine Lavoisier1743-1794. “Fatherof modern chemistry.”Recognized true elements.Used quantitative measurementsin chemical reactions.

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• Lavoisier: mass is conserved in a chemical reaction.• Chemical equations: descriptions of chemical

reactions.• Two parts to an equation: reactants and products:

Chemical EquationsChemical Equations

2H2 + O2 2H2O

Reactants Product

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H2 O2H2O

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• Stoichiometric coefficients: numbers in front of the chemical formulas give numbers of molecules or atoms reacting (and numbers being produced).

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CH4 + O2 CO2 + H2O

Count atoms:

Reactants: Products: 1 C 1 C

4 H 2 H2 O 3 O

is not balanced. (Why?)

Law of Conservation of Mass:All reactions must be balanced

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Balance reactions only by changing coefficients, not by altering chemical formula

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Combustion is the burning of a substance in oxygen:

C3H8(g) + 5O2(g) 3CO2 (g) + 4H2O(l)

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Which is correct? (Blue=A; Red=B)a) A2 + B A2Bb) A2 + 4B 2 AB2

c) 2A + B4 2 AB2

d) A + B2 AB2

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Percentage Composition from FormulasPercentage Composition from Formulas

Percent composition is the atomic weight for each element divided by the formula weight of the compound multiplied by 100:

Atomic and Molecular WeightsAtomic and Molecular Weights

100

Compound of FWAWElement of Atoms

Element %

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Percentage Composition from FormulasPercentage Composition from Formulas

FW= (2x1) + (1x32)+ (4x16)= 98 amu

FW of O in H2SO4 = 4 x 16 = 64 amu

What is % O in H2SO4 (by mass)?

%O = 64 x 100 = 65.3% 98

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The Mole*The Mole*•The “amu” is an “atomic mass unit.”•O has a mass of 16 amu – but we can’t weigh out anything in amu•If we want to keep the number “16” for the mass of oxygen in some real units (like grams) then we are dealing with a whole bunch of atoms (in 16 g of oxygen).•That bunch of atoms is called a mole.•Experimentally, 1 mole = 6.02 x 1023 things (atoms)•This number is called Avogadro’s number.

*MSJ Ch 3 pp 100-104

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This amount of P or Ca each contains Avogadro’s number (6.02 x 1023) of atoms of P and Ca.

The mole is defined so that The mole is defined so that oneone molemole of a substance has a mass of a substance has a massequal to its AW or MW equal to its AW or MW in gramsin grams

Basically, you are replacing amu with grams, e.g.The mass of a P atom is 31 amu.The molar mass of P is 31 grams.

The mass of a Ca atom is 40 amuThe molar mass of Ca is 40 grams

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The MoleThe Mole

Experimentally, 1 mole of 12C has a mass of exactly 12 g. (recall from Ch. 2)

MolarMolar Mass MassMolar mass: mass in grams of 1 mole of substance Units: g/mol or g.mol-1.Mass of 1 mole of 12C = 12 g exactly

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This photograph shows

one mole of :

solid NaCl (58.5 g),

liquid H2O (18 g), and

gaseous N2 (28 g).

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Molar MassMolar Mass

Molar mass: sum of the molar masses of the atoms:

The MoleThe Mole

What is Molar Mass of H2SO4?

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Interconverting Masses, Moles, and Numbers of Interconverting Masses, Moles, and Numbers of ParticlesParticles

The MoleThe Mole

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Example 1: 5.00 g of P(a) contains mol of P(b) contains atoms of P

Example 2: 5.00 x 1024 atoms of C(a) equals mol of C(b) has mass equal to grams.

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Example 3: 3.5 mol CO2:(a) has what mass?

(b) contains how many molecules of CO2?

(c) contains how many atoms of O?

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Start with mass % of elements (i.e. empirical data) and calculate a formula.

Empirical Formulas from AnalysesEmpirical Formulas from Analyses

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Empirical Formulas from Analyses*Empirical Formulas from Analyses*

N: 25.9 g x mol = 1.85 mol 14 g

Example: compound of N and OGiven analysis: N: 25.9%; O: 74.1%

Assume 100g; N: 25.9 g; O: 74.1 gChange to mol:

Preliminary emp. Formula: N1.85O4.63

Clean it up: divide both by 1.85:Get N1O2.5; get rid of fractions, multiply both by 2:Get N2O5 which is the empirical (simplest) formula

O: 74.1 g x mol = 4.63 mol 16 g

What are some possible molecular formulas?

*MSJ Ch 3 pp 104-108

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Molecular Formula from Empirical FormulaMolecular Formula from Empirical FormulaOnce we know the empirical formula, we need the MW to find the molecular formula.Subscripts in the molecular formula are always whole-number multiples of subscripts in the empirical formula.

Empirical Formulas from AnalysesEmpirical Formulas from Analyses

Example: suppose compound of C and H has empirical formula of C3H8 and a MW = 176 g/mol. What is the molecular formula?

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Balanced chemical equation gives number of molecules (or moles) that react to form products.

Quantitative Information from Quantitative Information from Balanced EquationsBalanced Equations

Interpretation: balanced equation gives us the ratio of number of moles of reactant to product (or v.v.).These ratios are called stoichiometric ratios.

Example: 2 H2 + O2 2 H2O

Molecules: 2 1 2

Moles: 2 1 2

Ratio of O2:H2O = 1:2 (either molecules or moles)

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The ratio of grams of reactant cannot be directly related to the grams of product.

Quantitative Information from Quantitative Information from Balanced EquationsBalanced Equations

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StoichiometryStoichiometry Problem:aluminum sulfide + water aluminum hydroxide + hydrogen sulfide

(a) Write balanced reaction:

(b) How many g aluminum hydroxide obtained from 10.5 g of aluminum sulfide?

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StoichiometryStoichiometry

Problem: 2 NaN3(s) 2Na(s) + 3 N2(g)

(a) how many mol N2 produced from 2.50 mol NaN3?

(b) how many g NaN3 needed to form 6.00 g N2

(c) how many g NaN3 needed to produce 10.0 ft3 of N2?

(1.00 ft3 = 28.3 L; density of N2 = 1.25 g/L)

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If the reactants are not present in stoichiometric amounts, at end of reaction some reactants are still present (in excess).Limiting Reactant: one reactant that is consumed.

Limiting ReactantsLimiting Reactants

O2 INXS

O2

H2

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RECIPE: 3 cups flour + 4 eggs + 2 cups sugar cake --------------------------------------------------------

How many cakes can be made from these amounts?What is the “limiting reactant” (LR)What’s left over and how much of it is left over?

On Hand: cups flour eggs cups sugar 20 20 20

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RECIPE: 3 cups flour + 4 eggs + 2 cups sugar cake --------------------------------------------------------

How many cakes can be made from these amounts?What is the “limiting reactant” (LR)What’s left over and how much of it is left over?

On Hand: cups flour eggs cups sugar 24 28 10

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Limiting ReagentLimiting Reagent(a)Assume a reactant (any one) is LR. Calculate stoichiometric amount of product (any product) formed.

(b) Pick another reactant and make it the LR.Calculate the stoichiometric amount of same product formed.

(c) Whichever reactant gives the smaller amount of product is the Limiting Reagent

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Limiting ReagentLimiting Reagent

The LR is always used up in a chemical reaction. Everything else is in excess (INXS).

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Limiting ReagentLimiting Reagent

(a) which is LR?(b) how many grams NO formed?(c) how much of excess reactant remains?

Problem:

4 NH3 + 5 O2 4 NO + 6 H2O

2.25 g NH3 mixed with 3.75 g O2 and allowed to react

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Theoretical YieldsTheoretical Yields

Limiting ReactantsLimiting Reactants

100yield lTheoretica

yield ActualYield %

The amount of product predicted from stoichiometry taking into account limiting reagents is called the theoretical yield.

The percent yield relates the actual yield (amount of material recovered in the laboratory) to the theoretical yield:

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Theoretical YieldTheoretical YieldProblem:

C6H6 + Br2 C6H5Br + HBr

(a) theoretical yield of C6H5Br when 30.0 g of C6H6 reacts with 65.0 g of Br2?

(b) if actual yield of C6H5Br is 56.7 g, calculate %yield