© 2009, Prentice-Hall, Inc. Chapter 3: Stoichiometry Formula Weight A formula weight is the sum of the atomic weights for the atoms in a chemical formula

© 2009, Prentice-Hall, Inc.

Chapter 3: StoichiometryChapter 3: StoichiometryFormula Weight•A formula weight is the sum of the atomic weights for the atoms in a chemical formula.•So, the formula weight of calcium chloride, CaCl2, would be

Ca: 1(40.1 amu) + Cl: 2(35.5 amu) 111.1 amu

•Formula weights are generally reported for ionic compounds.Calculate the formula weight for:•Aluminum hydroxide•Calcium nitrate


Molecular Weight (MW)

• A molecular weight is the sum of the atomic weights of the atoms in a molecule.

• For the molecule ethane, C2H6, the molecular weight would be

C: 2(12.0 amu)

30.0 amu+ H: 6(1.0 amu)


Percent Composition

One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation:

% element =(number of atoms)(atomic weight)

(FW of the compound)x 100


Percent Composition

• Calculate the percentage of carbon in ethane (C2H6)

• Calculate the percentage of oxygen in water


The MoleThe Mole

• 6.02 x 1023 things (atoms, particles, molecules…)• 1 mole of 12C has a mass of 12 g.• How many carbon atoms in .350 mol of glucose?

Avogadro’s Number


Molar Mass

• By definition, a molar mass is the mass of 1 mol of a substance (i.e., g/mol).– The molar mass of an element is the mass number

for the element that we find on the periodic table.– The formula weight (in amu’s) will be the same

number as the molar mass (in g/mol).• What is the mass of 5 moles of water?


Using Moles

Moles provide a bridge from the molecular scale to the real-world scale.

Number of particles


Mole Relationships

• One mole of atoms, ions, or molecules contains Avogadro’s number of those particles.

• One mole of molecules or formula units contains Avogadro’s number times the number of atoms or ions of each element in the compound.

Practice• Calculate the number of moles of sulfur

dioxide in 130 g.

• Calculate the mass in grams of .6 moles of calcium nitrate.


Empirical Formulas from AnalysesEmpirical Formulas from Analyses

One can calculate the empirical formula from the percent composition.

Helpful Rhyme: % to mass, mass to mole, divide by small, times ’til whole.


Calculating Empirical FormulasThe compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA.

remember: % to mass, mass to mole, divide by small, times ’til whole.

% to mass: assume 100 g

mass to mole:

divide by small:

times ’til whole


para-aminobenzoic acidC7H7NO2

Practice

Ascorbic acid, vitamin C, contains 40.92% C, 4.58% H, and 54.50% O. What is the empirical formula for ascorbic acid?

Molecular from Empirical


Molecular formulas are a whole number multiple of the empirical formula.

The empirical formula for ascorbic acid was C3H4O3. It’s molecular weight is 176. What is the molecular formula?


Combustion Analysis

• Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this.– C is determined from the mass of CO2 produced.

– H is determined from the mass of H2O produced.

– O is determined by difference after the C and H have been determined.

1. Calculate the moles of CO2 produced (same as mols of C)

2. Calculate the moles of H2O produced Multiply this number by two to calculate the number of moles of H atoms present in the original compound.

3. Calculate the mass of C and H

4. If there is another element present (typically O) in the combusted substance then calculate its mass by subtracting the mass of C and H from the total mass of the combusted sample.

5. Once you have determined the mass of each piece of the original compound the problem can be solved just like an empirical formula problem.

Steps for Combustion Analysis Problems

EXAMPLE 1:

A 0.255g sample of isopropyl alcohol was burned in a combustion train and 0.561g CO2 and 0.306g H2O were formed. Determine the empirical formula of ascorbic acid.

Steps to Follow:

(1) Calculate the moles of Carbon and Hydrogen in the compound.

(2) Calculate the moles of Oxygen in the compound.

(3) The mole ratios are the subscripts of the empirical formula. Use the “Empirical Formula Rhyme” to complete the rest of the problem.

Combustion Analysis

• A sample of a common alcohol with a mass of 4.599 g, containing C, H and O, was combusted in excess oxygen to yield 8.802 g of CO2 and 5.394 g of H2O.

• The alcohol contains only carbon, hydrogen, and oxygen. What is the empirical formula of the alcohol?

EXAMPLE 2:

The combustion of 62.63 grams of a compound which contains only C,H and O yields 134.43 grams of CO2 and 13.75 grams of H2O. What is the empirical formula of the compound?

– If the compound’s molecular weight is 156 g/mol, determine the molecular formula.

EXAMPLE 3:

The combustion of 40.10 g of a compound which contains only C, H, Cl and O yields 58.57 g of CO2 and 14.98 g of H2O. Another sample of the compound with a mass of 75.00 g is found to contain 22.06 g of Cl. What is the empirical formula of the compound?

– (Hint: Use the mass data to calculate the % composition of each element in the compound.)


Reactions

Chemical equations are concise representations of chemical reactions.


Anatomy of a Chemical Equation

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)



Reactants appear on the left side of the equation.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)



Products appear on the right side of the equation.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)



The states of the reactants and products are written in parentheses to the right of each compound.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)


Subscripts and Coefficients Give Different Information

• Subscripts tell the number of atoms of each element in a molecule.


Subscripts and Coefficients Give Different Information

• Subscripts tell the number of atoms of each element in a molecule

• Coefficients tell the number of molecules.



Coefficients are inserted to balance the equation.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)


Combustion Reactions

• Examples:– CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

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

• These are generally rapid reactions that produce a flame.

• Most often involve hydrocarbons reacting with oxygen in the air.

• Excess oxygen CO2

• Insufficient oxygen CO


Combination Reactions

• Examples:– 2 Mg (s) + O2 (g) 2 MgO (s)

– N2 (g) + 3 H2 (g) 2 NH3 (g)

– C3H6 (g) + Br2 (l) C3H6Br2 (l)

• In this type of reaction two or more substances react to form one product.


• In a decomposition one substance breaks down into two or more substances.

Decomposition Reactions

• Examples:– CaCO3 (s) CaO (s) + CO2 (g)

– 2 KClO3 (s) 2 KCl (s) + O2 (g)

– 2 NaN3 (s) 2 Na (s) + 3 N2 (g)

Categories of Decomposition (and Composition ) Reactions

a) carbonates metallic oxide + CO2 CaCO3 _____ + _____

b) chlorates metallic chloride + O2 NaClO3 _____ + _____

c) hydroxides metallic oxide + H2O Mg(OH)2 _____ + _____

d) oxy acids nonmetal oxide + H2O H2SO4 _____ + _____

e) binary compounds 2 elements NaCl _____ + _____

• Notice that in carbonates, hydroxides, and oxy acids, the oxygen combines with both the metal AND the nonmetal!!

• Every time you try to write the formula for a new ionic compound, you must look up the ___________ of the ions and ___________ them if they are different!!

CaO CO2

NaCl O2

MgO H2O

SO3

Na Cl2

charges cross

H2O


Quantitative Information from Quantitative Information from Balanced EquationsBalanced Equations

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

You can only change the coefficients to balance (never subscripts)

You can only use whole numbers to balance

Balancing Hints:–Balance the metals first.–Balance the ion groups next.–Balance the other atoms.–Save the non ion group oxygen and

hydrogen until the end.

• Balance the following equations;

__Ca + __H2O → __H2 + __ Ca(OH)2

__ Cu + __ O2 → __ CuO

__ Na + __ O2 → __ Na2O

__ Fe + __ HCl → __ FeCl2 + __ H2

__ Fe + __ Br2 → __ FeBr3

__ C4H8 + __ O2 → __ CO2 + __ H2O

Practice writing reactionsWrite the balanced chemical equation for:•The combination reaction the occurs between lithium metal and fluorine gas.

•The reaction that occurs when solid sodium azide (NaN3) decomposes rapidly into it’s constituent elements as an airbag is inflated.

•Si2H6 burns when exposed to air


Stoichiometric Calculations

Starting with the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant).

• Balance this equation showing the burning of methane below, to answer the following questions.

• __ CH4 + __ O2 → __ CO2 + __ H2O

How many grams of oxygen are required to produce 2.23 g of carbon dioxide?

How many grams of water will be produced when 34.0 g of methane is burned?


Stoichiometric Calculations

• How many grams of water are produced in the combustion of 1 g of glucose?


3.7 Limiting Reactants3.7 Limiting Reactants

• You can make cookies until you run out of one of the ingredients.

• Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat).

How Many Cookies Can I Make?


How Many Cookies Can I Make?

• In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make.


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.

Let’s Try This Problem Together• H2(g) + O2(g) → H2O(l)

a. Calculate the stoichiometric ratio of moles H2 to moles O2

b. Calculate the moles of water created when 1.50 mol H2 is mixed with 1.00 mol O2

c. Calculate the limiting reactant (H2 or O2) for the mixture in (b)

Let’s Try This Problem Together

• For the reaction: Na2O + H2O → NaOH

What weight of NaOH could be made from 12.4 g of Na2O and 42.1 g of H2O?

What would be the limiting reagent if 100 g each of Na2O and H2O were allowed to react?

Practice

• Part of the SO2 that is introduced into the atmosphere by combustion of sulfur containing compounds ends up being converted to sulfuric acid according to the following reaction:

SO2 (g) + O2 (g) + H2O(l) H2SO4 (aq)

How much H2SO4 can be formed from 5.0 moles of SO2, 2.0 moles of O2, and an unlimited supply of H2O?


Theoretical Yield

• The theoretical yield is the maximum amount of product that can be made.– In other words it’s the amount of product possible

as calculated through the stoichiometry problem.

• This is different from the actual yield, which is the amount one actually produces and measures.


Percent Yield

One finds the percent yield by comparing the amount actually obtained (actual yield) to the amount it was possible to make (theoretical yield).

Actual YieldTheoretical YieldPercent Yield = x

100

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© 2009, Prentice-Hall, Inc. Chapter 3: Stoichiometry Formula Weight A formula weight is the sum of the atomic weights for the atoms in a chemical formula