Chapter 3

Stoichiometry

Atomic Mass

• Carbon-12 is assigned a mass of exactly 12.00 atomic mass units (amu)

• Masses of other elements are compared to Carbon-12

• Comparisons are made using a mass spectrometer

– Compares ratios of isotopes present

Mass Spectrometer

Average Atomic Mass

• An average of the atomic masses of all naturally occurring isotopes of an element

• Average atomic mass is the mass value on the periodic table

• No atom has the exact mass listed on the table

• Carbon = 12.01 amu– This means the majority of C is Carbon-12

• Used to weigh large numbers of atoms

Example• Lead occurs in 4 main isotopes 1.40% has

a mass of 203.973amu. 24.10% has a mass of 205.975amu. 22.10% has a mass of 206.976, and 52.40% has a mass of 207.977amu. What is average atomic mass of Lead?

The Mole

• 1 mole = 6.022x1023 particles

• Avogadro’s Number– He did not determine the number!

• The average atomic mass of an element expressed in grams contains 1 mole of atoms of that element.

• 22.99 g of Sodium is one mole of sodium

• 40.08 g of Calcium is one mole of calcium

Example

• How many moles of sodium atoms are in 2.83 grams of sodium?

Example

• How many atoms of sodium are there in 2.83 grams?

Molar Mass

• The mass in grams of one mole of an element or compound– Units of grams/mole (g/mol)

• Molar mass of sodium = 22.99 g/mol

• Molar mass of compound is the sum of all atoms present

• Molar Mass of Water (H2O)

• 2(1.01) + 16.00 = 18.02 g/mol

Example

• Determine the molar mass of

• Mg3N2

• C2H5O2N

• Ca(NO3)2

Example

• 18.1g Calcium Nitrate = ? moles

Example

• How many nitrate ions are present in 18.1 grams of Calcium Nitrate?

Percent Composition

• The percent by mass each element contributes to the compound

(# of atoms of element)(atomic mass) X 100

(molar mass of compound)

Example

• What is the percent composition of Calcium Nitrate?

Homework

• P. 123 #’s 21,34,36,40,44,49,52,53

Objectives

• Compare Empirical and Molecular Formulas

• Calculate Empirical and Molecular Formulas

• Balance and Interpret chemical equations

• Perform stoichiometry calculations

Empirical and Molecular Formulas

• Empirical Formula

– A chemical formula with the simplest whole number ratio

• CH2O (30.03 g/mol)

• Molecular Formula

– The exact chemical formula for a compound

• C3H6O3 (90.09 g/mol)

Calculating Empirical Formulas

• Assume 100 g of compound – (Percents become grams)

• Or use given masses

• Covert grams to moles

• Divide each by the smallest # of moles

• If you don’t have whole #’s multiply by an integer to get whole numbers– 3.33 * 3 4.5 * 2 8.25 *4

Example

A compound contains 49.31%C, 43.79%O and the rest H. The molar mass is 146 g/mol. What is the empirical and molecular formula?

Combustion Analysis

• Compounds containing C,H, and O are often burned to determine the chemical makeup.

– The products of combustion are CO2 and H2O

• These are collected and massed– Carbon and Hydrogen come from

compound– Oxygen comes from compound and air

Combustion Analysis

Example

The combustion of 17.27 grams of a compound which contains only C,H and O yields 44.06 grams of CO2 and 11.27 grams of H2O. What is the empirical formula of the compound?

Balanced Chemical Equations

• Gives the recipe for a reaction• Reactants on left / Products on the right• Coefficients tell us the atomic/mole ratios• State symbols tell what state of matter • (s) solid, (l) liquid, (g) gas, (aq) solution

2H2(g) + O2(g) 2H2O(l)

Means – 2 mol of gaseous Hydrogen reacts with 1 mol of gaseous Oxygen to form 2 mol of liquid Water

Balancing Equations

• Use coefficients to get the same number of atoms on both sides of the equation

– Due to Law of Conservation of Mass

– Coefficient – number in front of a formula

• Balance the equation

Fe2S3 + HCl FeCl3 + H2S

Cont.

• Combustion Equations– When there are an even number of Carbons

in the reactant you may have to double the reaction.

C2H6 + O2 CO2 + H2O

Stoichiometry

• A process for comparing amounts in reactions

• Compares reactant to reactant or reactant to product

• Start with a balanced equation

• Process

Grams A Moles A Moles B Grams BMolar Mass A Molar Ratio Molar Mass B

Cont.

• Could Include– Density– Percent

• Could start or end with moles

• Read the problem and determine the best way to proceed

Example

6.00 grams of hydrochloric acid is reacted with barium hydroxide to form barium chloride and water. How many grams of barium chloride are formed?

Example

4.82 grams of propane (C3H8) is burned with excess oxygen to form carbon dioxide and water. What volume of water vapor is formed if the density is .8044 g/L?

Homework

• P. 125 #’s 66,74,79,86

Objectives

• Describe limiting and excess reactants

• Be able to calculate the limiting reactant

• Compare theoretical and actual yield

• Describe and calculate percent yield

Limiting Reactant

• The reactant that runs out in a chemical reaction.

• When one reactant is used up the reaction stops

• Based on mole ratio NOT on mass

Example

What is the limiting reactant when 16.00g of Hydrogen gas reacts with 64.00g of oxygen gas to produce water

Example

What mass of carbon dioxide gas is produced when 4.00g of methane (CH4) reacts with 32.00g of oxygen gas

Theoretical and Actual Yield

• Theoretical yield is the amount of product you should get based on calculations

• Actual yield is the amount of product you get when the reaction is run– Actual yield cannot be more than the

theoretical yield. (Conservation of mass)– Actual yield is usually less than theoretical b/c

of product loss

Percent Yield

• Measures how much product a reaction produces

• High percent yields are good

Percent Yield = Actual X 100

Theoretical

Example

1000. grams of calcium phosphate reacts with 1000. grams of 98% sulfuric acid to produce calcium sulfate and phosphoric acid. When the reaction is run 1100. g of calcium sulfate was produced. What is the percent yield of the reaction?

Homework

• Stoichiometry Worksheet