The MoleThe MoleChapters 10 & 11Chapters 10 & 11

What is a mole?What is a mole?

http://www.topnews.in/health/scientists-identify-protein-marks-difference-between-mole-and-melanoma-2990

What is a mole?What is a mole?

http://scienceblogs.com/neurophilosophy/2009/08/the_star_nosed_moles_amazing_appendages.php

http://animals.nationalgeographic.com/animals/mammals/naked-mole-rat/

• A unit of measurement for molemolecules• # of particles of any chemical substance• 1 mole = 6.02x1023 particles

This is known as Avagodro’s #

• 1mole of Carbon = 6.02x1023 atoms of Carbon

• 1mole of water = 6.02x1023 molecules of H2O

• 1mole of glucose = 6.02x1023 molecules of C6H12O6

• 1mol of popcorn kernels = 6.02x1023 kernels

(enough to bury the entire US under 9 feet of popcorn kernels!)

What is a mole in CHEMISTRY?What is a mole in CHEMISTRY?

http://www.holidayforeveryday.com/?p=451

Practical Application of the MolePractical Application of the Mole

= mass of 1 mole of a substance in grams

= average atomic mass given in periodic table

Substance Molar Mass

Carbon 12.0 g

Magnesium 24.3 g

Copper 63.5 g

Lead 207.2 g

= sum of the atomic masses of each element represented in formula

Practical Application of the MolePractical Application of the Mole

Compound Formula Formula Mass

Water H2O 2(1) + 1(16) = 18g

Calcium Hydroxide Ca(OH)2 1(40) + 2(16) + 2(1) = 74g

Iron (III) Chloride FeCl3 1(55.8) + 3(35.4) = 162.0g

Potassium Nitrate KNO3 1(39) + 1(14) + 3(16) = 101g

Learning CheckLearning Check

• Prozac, C17H18F3NO, is a widely used antidepressant that inhibits the uptake of serotonin by the brain. Find its molar mass.

17 x 12 = 204g

18 x 1 = 18g

3 x 19 = 57g

1 x 14 = 14g

1 x 16 = 16g

309g

Calculations with Molar MassCalculations with Molar Mass

molar mass Grams Moles

Calculations with Molar MassCalculations with Molar Mass

Practice Example #1Aluminum is often used for the structure of light-weight bicycle frames. How many grams of Al are in 3.00 moles of Al?

Given Solving For

3.00 moles of Al _____ g Al

3.00 moles of Al 27 g of Al

1 mole of Al= 81g of Al

Learning CheckLearning Check

The artificial sweetener aspartame (Nutra-

Sweet) formula C14H18N2O5 is used to

sweeten diet foods, coffee and soft drinks.

How many moles of aspartame are present

in 225 g of aspartame?

225g aspartame 1 mole aspartame

294g aspartame= 0.76 mol aspartame

molar mass molar mass Avogadro’s # Avogadro’s # Grams Moles Particles

Calculations with Molar MassCalculations with Molar Mass

Multistep Practice Problem #1 Multistep Practice Problem #1

How many atoms of Cu are present in 35.4 g of Cu?

= 3.4 X 10= 3.4 X 1023 23 atoms Cuatoms Cu

35.4 g Cu 1 mol Cu 6.02 X 1023 atoms Cu

63.5 g Cu 1 mol Cu

How many atoms of K are present in 78.4 g of K?

Multistep Practice Problem #2 Multistep Practice Problem #2

= 1.21 x 10= 1.21 x 102424 atoms K atoms K

78.4 g K 1 mol K 6.02 X 1023 atoms K

39.0 g K 1 mol K

What is the mass (in grams) of 1.20 X 1024 molecules of glucose (C6H12O6)?

Multistep Practice Problem #3 Multistep Practice Problem #3

= 358 g glucose= 358 g glucose

1.20x1024 molecules of C6H12O6 1 mol glucose 180 g glucose

6.02 X 1023 molecules 1 mole glucose

How many atoms of O are present in 78.1 g of oxygen gas?

Multistep Practice Problem #4 Multistep Practice Problem #4

78.1 g O2 1 mol O2 6.02 X 1023 molecules O2 2 atoms O

32.0 g O2 1 mol O2 1 molecule O2

= 2.94 x 10= 2.94 x 102424 atoms of O atoms of O

Gases at STP (Standard Temperature & Pressure) with the same number of particles will occupy the same volume

Calculations with Molar VolumeCalculations with Molar Volume

One mole of any gas at STP = 22.4L22.4L

Calculations with Molar VolumeCalculations with Molar Volume

Molar Volume Practice ProblemMolar Volume Practice Problem

A student fills a 1.0L flask with carbon dioxide (CO2) at standard temperature and pressure. How many molecules of gas are in the flask?

1.0 L CO2 1 mol CO2 6.02 x 1023 molecules of CO2

22.4 L CO2 1 mol CO2

= 2.69 x 10= 2.69 x 102222 molecules CO molecules CO22

Empirical & Molecular FormulasEmpirical & Molecular Formulas

• The mass of each element in a compound compared to the total mass

Percent Hydrogen of 1 mole of H2O

% H = 2.0 g H

= 11%18 g H2O

% Composition Using Experimental Data

Knowns Unknowns

8.2 g = 60.3% MgMass Mg = 8.2 g % Mg = 13.6 g

Mass O = 5.4 g

% O = 5.4 g

= 39.7% OTotal Mass = 13.6 g

13.6 g

What is the empirical formula of a compound containing 25.9% N and 74.1% O?

Step 1: Assume 100 g of the compound (only necessary if given percents to start)

25.9% N = 25.9 g N74.1% O = 74.1 g O

Step 2: Convert grams to moles 25.9 g N 1 mol N= 1.85 mol N

14.0 g N

74.1 g O 1 mol O= 4.63 mol O

16.0 g O

Step 3: Divide both mole amounts by the smaller of the 2 numbers

1.85 mol N= 1 mol N

1.85

4.63 mol O = 2.50 mol O

1.85

Step 4: Multiply both to obtain whole numbers

1 mol N x 2 = 2 mol N2.5 mol O x 2=5 mol O N2O5

Calculating Empirical FormulasCalculating Empirical Formulas

Calculating Molecular FormulasCalculating Molecular Formulas

Step 1:Step 1: Calculate the empirical formula

N2O5 (answer from previous problem)

Step 2:Step 2: Based on the empirical formula, calculate the formula mass

2x14 = 285x16 = 80

108g

Step 3:Step 3: Divide the given molar mass by the calculated formula mass

If the molar mass of the compound is 324g, what is the molecular formula?

324g/108g = 3

Step 4: Multiply the subscripts in the empirical formula by the answer found in #3

N6O15

Stoichiometry: Mole-MoleStoichiometry: Mole-Mole

A balanced formula indicates the number of moles of each compound.

Ex: 2HCl + Zn ZnCl2 + H2

2 moles HCl+1 mole Zinc yield 1 mole Zinc Chloride +1 mole hydrogen gas

Based on this formula, how many moles of HCl are needed to react with 2.3 moles of Zn?

2.3 mol Zn 2 mol HCl

= 4.6 mol HCl1 mol Zn

Stoichiometry: Mass-MassStoichiometry: Mass-Mass

In a thermite reaction, powdered aluminum reacts with iron (III) oxide to produce aluminum oxide and molten iron. What mass of aluminum oxide is produced when 2.3g of aluminum reacts with iron (III) oxide?

Balanced Equation

Stoichiometry

Al + Fe2O3 Al2O3 + Fe2 2

2.3g Al 1 mol Al 1 mol Al2O3 102g Al2O3

27g Al 2 mol Al 1 mol Al2O3

= 4.3g Al2O3

Stoichiometry: Mass-VolumeStoichiometry: Mass-Volume

Sodium Bicarbonate (NaHCO3) can be used to extinguish a fire. When heated, it decomposes into sodium carbonate, carbon dioxide and water. If a sample contains 4.0g NaHCO3, what volume of CO2 gas is produced?

Balanced Equation

Stoichiometry

NaHCO3 Na2CO3 + H2O + CO2 2

4.0g NaHCO3 1 mol NaHCO3 1 mol CO2 22.4L CO2

84g NaHCO3 2 mol NaHCO3 1 mol CO2

= .53L CO2

Limiting ReactantsLimiting Reactants

The limiting reactant limiting reactant in any equation • would produce a smaller amount of product • is completely used up first • determines the amount of product made

Which reactant in an equation is the limiting reactant?Identify the limiting reactant when 1.7 g of sodium reacts with 2.6L of chlorine gas at STP to produce sodium chloride.

Step 1 Write the balanced equation

Step 2 Determine the mass of product that would be produced from each given reactant quantity.

2 Na + Cl2 2 NaCl

1.7g Na 1mol Na 2 mol NaCl 58g NaCl= 4.3g NaCl

23g Na 2 mol Na 1 mol NaCl

2.6L Cl2 1 mol Cl2 2 mol NaCl 58g NaCl= 13g NaCl

22.4L Cl2 1 mol Cl2 1 mol NaCl

Percent YieldPercent Yield

Percent Yield = Actual yield

X 100Expected yield

Expected yield is determined from a stoichiometry problem.

Actual yield is the amount of product measured or recovered.

A piece of copper with a mass of 5.00g is placed in a solution of silver (I) nitrate containing excess AgNO3. The silver metal produced has a measured mass of 15.2g. What is the percent yield for this reaction?

Step 1 Write balanced equation

Step 2 Find expected mass of product

Cu + 2 AgNO3 2 Ag + Cu(NO3)2

5.00g Cu 1mol Cu 2 mol Ag 107.9g Ag= 17.0g Ag

63.5g Cu 1 mol Cu 1 mol Ag

Percent Yield= Actual yield = 15.2g Ag X 100 = 89.4%

Expected yield 17.0g Ag