chapter six chemical reactions: mole and mass relationships fundamentals of general, organic, and...

Chapter SixChemical Reactions:

Mole and Mass Relationships

Fundamentals of General, Organic, and Biological

Chemistry7th Edition

Chapter 6 Lecture

© 2013 Pearson Education, Inc.

Julie KlareGwinnett Technical College

McMurry, Ballantine, Hoeger, Peterson

• 6.1 The Mole and Avogadro’s Number• 6.2 Gram–Mole Conversions• 6.3 Mole Relationships and Chemical Equations• 6.4 Mass Relationships and Chemical Equations• 6.5 Limiting Reagent and Percent Yield

Goals

1. What is the mole, and why is it useful in chemistry? Be able to explain the meaning and uses of the mole and Avogadro’s

number.

2. How are molar quantities and mass quantities related? Be able to convert between molar and mass quantities of an element or

compound.

3. What are the limiting reagent, theoretical yield, and percent yield of a reaction? Be able to take the amount of product actually formed in a reaction,

calculate the amount that could form theoretically, and express the results as a percent yield.

6.1 The Mole and Avogadro’s Number

How to balance reactions by counting molecules

Counting molecules: Conceptually

• The reactants in a chemical reaction must be balanced

• One molecule of hydrogen (H2) reacts with one molecule of iodine (I2), creating two hydrogen iodide molecules (HI)

• Of course, we can’t visually count molecules to achieve this one-to-one ratio

• In fact, the only tool available is an analytical mass balance (using grams!)

Molecular Weight

• But how do we relate a sample’s mass in grams to the number of molecules it contains?

• We begin with the concept of ‘molecular weight’ (molecular mass)

• Let’s start by utilizing the Ch 2 concept of mass by ‘amu’ – Note: We will be focusing on molecules for a while

• Recall, atomic weight is the average mass of an element’s isotope atoms (review next slide)

• By analogy, molecular weight (MW) is the mass of a compound’s molecules

– A molecule’s molecular weight is the sum of the atomic weights for all the atoms in the molecule

– A salt’s formula weight is the sum of the atomic weights for all the atoms in the formula unit

fyi: Atomic Weight Calculation

C-12 is 98.89% of all natural carbonC-13 is 1.11% of all natural carbon

The mass of C-12 is: 12 amu (by definition)The mass of C-13 is: 13.003355 amu

Atomic weight = [(% isotope abundance) × (isotope mass)]

[98.89% x 12 amu] + [1.11% x 13.0034 amu] =

(0.9889)(12 amu) + (0.0111)(13.0034 amu) = 12.0111

12.01 amu

11.8668 0.1443

fyi: Disambiguation

• For molecules, these terms are identical– molar weight – molar mass– molecular weight – molecular mass

• For atoms, these terms are identical– molar weight – molar mass – atomic weight– atomic mass

Copyright © Cengage Learning. All rights reserved

12

Molar Mass of N = 14.01 g/molMolar Mass of N2 = 28.02 g/mol

Molar Mass of H2O = 18.02 g/mol

(2 × 1.008 g) + 16.00 g

Molar Mass of Ba(NO3)2 = 261.35 g/mol

137.33 g + (2 × 14.01 g) + (6 × 16.00 g)

Molecular Weight (Mass): Mass in grams of one mole of the substance:

Molecular weight• So if the mass (atomic weight) of one atom of

hydrogen (H) is 1.008 amu – then the mass (molecular weight) of one hydrogen

molecule (H2) is 2.016 amu

• If the atomic weight of one iodine atom (I) is 126.90 amu– then the molecular weight of one iodine molecule (I2) is

253.80 amu

• Finally, the molecular weight of each HI molecule must be: – 1.008 amu + 126.90 amu = 127.91 amu

Counting molecules: by mass

• Here is where we left our reaction making two HI molecules from H2 and I2

• We learned in chapter 2 that the iodine atom (I) is approximately 126 times heavier than the hydrogen atom (H)– H = 1.008 amu – I = 126.90 amu

• So mathematically, it is necessary that the iodine molecule (I2) is approximately 126 times heavier than the hydrogen molecule (H2) – H2 = 2.016 amu

– I2 = 253.80 amu

Copyright © Cengage Learning. All rights reserved 17

• Meaning that this reaction is balanced – in amu units

H2(g) + I2(g) → 2HI(g)

2.016 amu 253.80 amu 1 : 126


• So by the law of identical ratios, this reaction must also be balanced – in gram units

H2(g) + I2(g) → 2HI(g)

2.016 grams 253.80 grams 1 : 126

Now notice we have gone

from amu to grams


• And again by The Law of Identical Ratios, this must also be balanced

H2(g) + I2(g) → 2HI(g)

1.00 g 126 g

Mass / Number relationship for I2 & H2

• So in order control the number ratio of I2 to H2 molecules at 1:1 we can measure out any of the following – so long we keep the mass ratio at 1:126

H2(g) + I2(g) → 2HI(g)

2.016 amu 253.80 amu 2.016 g 253.80 g

1.00 g 126 g

0.0079 g 1.0000 g

But how many molecules is that?

• We know that: – 2.016 amu of H2 contains one molecule, and

– 253.80 amu of I2 contains one molecule

• Our final question is, how many molecules are contained in: – 2.016 g of H2

– 253.80 g of I2

That amount we call The Mole

• A mole is the amount of substance whose mass in grams is numerically equal to its molecular weight – 2.016 g of H2 contains one mole of H2 molecules

– 253.80 g of I2 contains one mole of I2 molecules

• The mole (NA) has now been measured – One mole of any molecule contains 6.022 × 1023

molecules – One mole of any salt contains 6.022 × 1023 formula

units

Getting from amu to grams• The number ‘126.904,’

typically placed as shown, functions BOTH: – as the mass in amu per

atom– and the mass in grams per

mole of atoms • So I has a mass of

126.904 grams per mole of I atoms– I = 126.904 g/mol

The technical definition of a mole


How many silicon atoms are in 0.532 moles of silicon?

a. 8.83 × 10−25 atoms

b. 3.81 × 1023 atoms

c. 2.6 × 1023 atoms

d. 3.8 × 1023 atoms

6.2 Gram–Mole Conversions

fyi: Synonyms

• The terms Molecular Weight (section 6.1) and Molar Mass (this section – 6.2) are identical – Also, the term Molecular Mass is allowed – Also, the term Molar Weight is allowed

• When we defined Molecular Weight in Section 6.1, we had not yet define the Mole – so you couldn’t have understood the more

common term “Molar” Mass

• The molar mass of water is 18.02 g/mol• So how many moles of water are there in 27

grams?

• Most importantly, molar mass serves as a conversion factor between numbers of moles and mass in grams for use in dimensional analysis

Molar Mass: Mole to Gram Conversion

Ibuprofen is a pain reliever used in Advil. Its molecular weight is 206.3 g/mol.

If a bottle of Advil contain 0.082 mole of ibuprofen, how many grams of ibuprofen does it contain?

Worked Example 6.3

WORKED EXAMPLE 6.3 Molar Mass: Mole to Gram Conversion (Continued)

We now have the tools to convert:

• grams → moles → number of atoms • grams → moles → number of molecules • grams → moles → number of formula units

• And we can go backwards


What is the mass in grams of 3.2 × 1022 molecules of water?

a. 0.0029 g

b. 339 g

c. 0.90 g

d. 0.96 g

6.3 Mole Relationships and Chemical Equations

Molar coefficients: using Mole ratios as

conversion factors

• Coefficients in a balanced chemical equation tell us the necessary ratio of moles of reactants – and how many moles of each product are

formed

• Especially useful, coefficients can be put in the form of mole ratios – which act as conversion factors when

setting up dimensional analysis calculations

This reaction represents the reaction of A2 (red) with B2 (blue)

a) Write a balanced equation for the reactionb) How many moles of product can be made from 1.0 mole of A2? From 1.0 mole of B2?

Mole to mole


a. 0.67 mol

b. 2.00 mol

c. 3.33 mol

d. 7.50 mol

How many moles of NH3 can be produced from 5.00 moles of H2 according to the following equation? N2 + 3 H2 2 NH3

6.4 Mass Relationships and Chemical Equations

Putting it all together

• The actual amounts of substances used in the laboratory must be weighed out in grams

• Furthermore, customers generally want the quantity of product reported in terms of grams

We already have the necessary tools

• Mass to mole conversions • Mole to mole conversions • Mole to mass conversions

• Mole to mole conversions (section 6.3) are carried out using mole ratios as conversion factors

If you have 9.0 moles of H2, how many moles of NH3 can you make?

• 2) Mole-to-mass and mass-to-mole conversions (from section 6.2) are carried out using molar mass as a conversion factor

• 3) Mass to mass conversions cannot be carried out directly

• If you know the mass of A and need to find the mass of B

• first convert the mass of A into moles of A

• then carry out a mole to mole conversion to find moles of B

• then convert moles of B into the mass of B

• So after collecting the fundamental data, always follow this three step process for mass to mass

• 1) convert grams of A to moles via molar mass• 2) convert moles of A to B via mole ratio • 3) convert mole of B to grams via molar mass

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Let’s go back to our hydrogen / iodine reaction giving hydrogen iodide

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I2 = 253.80 g/mol H2 = 2.016 g/molHI = 127.90 g/mol

10.0 grams of HI requires how many grams of H2


How many grams of oxygen are needed to react with 25.0 g of K according to the following equation? 4 K(s) + O2(g) 2 K2O(s)

a. 10.2 g O2

b. 2.56 g O2

c. 8.66 g O2

d. 5.12 g O2

6.5 Limiting Reagent and Percent Yield

Vocabulary:Reactant = Reagent

limiting reactant (reagent)

• Reactants are not always perfectly balanced

• When running a chemical reaction, we generally ‘overcharge’ one of the reactants

As our mechanic has ‘overcharged’ tires


• The limiting reactant is the reactant that runs out first

• The reactant that never runs out is called the excess reactant

So which is ‘limiting here?’ Tires or car bodies.


• In order to make this ‘reaction’ balanced stoichiometrically, how many tires should our mechanic actually have on hand?

32 tires would perfectly ‘balance’ this ‘reaction’

• Back to chemistry


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A stoichiometric mixture of reactants

contains the relative amounts (in moles) of reactants that matches the coefficients in the balanced equation

N2(g) + 3H2(g) 2NH3(g)


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A stoichiometric mixture of reactants

• Contains the relative amounts (in moles) of reactants that matches the coefficients in the balanced equation

N2(g) + 3H2(g) 2NH3(g)


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A limiting reactant mixtureon the other hand, contains an excess of one of the reactants The other reactant is thus limiting

N2(g) + 3H2(g) 2NH3(g)

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A limiting reactant mixture

Which reactant is in excess in this reaction mixture?

Which reactant is limiting?

on the other hand, contains an excess of one of the reactants The other reactant is thus limiting

N2(g) + 3H2(g) 2NH3(g)


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Definition: Limiting Reactant

N2(g) + 3H2(g) 2NH3(g)

• The limiting reactant (reagent) is the reactant that runs out first and thus constrains (limits) the amounts of product(s) that can form H2 in the previous example is limiting


60

• But the limiting reactant must be identified before the correct amount of product can be calculated

2H2 + O2 H2O


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Example

10.0 g of Chemical A are reacted with 10.0 g of B

• What information do you need to calculate the mass of the product (C) that will be produced?

• The mole ratios between A, B, and C – ie, the balanced reaction equation

• The molar masses of A, B, and C

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Example


• What information do you need to calculate the mass of the product (C) that will be produced?

A + 3B 2C• molar masses:

– A is 10.0 g/mol– B is 20.0 g/mol– C is 25.0 g/mol?

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Example


A + 3B 2C• 1) Convert known masses of reactants to moles

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Example


A + 3B 2C• 2) Convert to the number of moles of product

So B is limiting reactant

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Example


A + 3B 2C• Choose the least number of moles of product

formed as limiting reactant: B is limiting

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Example


A + 3B 2C• 3) Convert moles of C to grams of C using the molar

mass

shortcut to limiting reactant

• How do we determine quickly which reactant is limiting – and by deduction, which is in excess?

• Three step program 1 Calculate the number of moles of each reactant2 divide the number of moles of each reactant by

its coefficient from the balanced equation3 the reactant with the smaller result is limiting


If 56 g of K is reacted with 56 g of oxygen gas according to the equation below, indicate the mass of product that can be made and identify the limiting reactant. 4 K(s) + O2(g) 2 K2O(s)

a. 67g K2O; K is the limiting reactant.

b. 270 g K2O; K is the limiting reactant.

c. 270 g K2O; O2 is the limiting reactant.

d. 67g K2O; O2 is the limiting reactant.

K2O = 94.2g/mol

Theoretical vs actual yield

• Theoretical Yield– The maximum amount of a given product that can

be formed once a limiting reactant has been completely consumed – it is a calculation

• Actual yield – The amount of product that is actually produced in

a reaction – it is a measurement– It is usually less than the maximum expected

(theoretical yield)

• Theoretical yield is found by using the amount of limiting reactant calculated in a mass-to-mass calculation

• For the chemist in the lab, the actual yield is found by weighing the amount of product obtained

FYI

• Theoretical Yield– For our auto mechanic, the theoretical yield is

8 finished cars

• Actual yield – If one car body was damaged beyond repair,

the actual yield would have been 7 finished cars

• Percent yield is the percent of the theoretical yield actually obtained from a chemical reaction

• For our car mechanic:

The combustion of acetylene gas (C2H2) produces carbon dioxide and water as indicated in the following reaction

2 C2H2 (g) + 5 O2 (g) → 4 CO2 (g) + 2 H2O (g)

When 26.0 g of acetylene is burned in sufficient oxygen for complete reaction, the theoretical yield of CO2 is 88.0 g

Calculate the percent yield for this reaction if the actual yield is only 72.4 g CO2

Analysis—The percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100

WORKED EXAMPLE 6.8 Percent Yield

WORKED EXAMPLE 6.8 Percent Yield (finished)

The element boron is produced commercially by the reaction of boric oxide with magnesium at high temperature.

B2O3 (l) + 3 Mg (s) → 2 B (s) + 3 MgO (s)

What is the theoretical yield of boron when 2350 g of boric oxide is reacted with 3580 g of magnesium? The molar masses of boric oxide and magnesium are 69.6 g/mol and 24.3 g/mol, respectively

WORKED EXAMPLE 6.9 Mass to Mole Conversions: Limiting Reagent and Theoretical Yield

WORKED EXAMPLE 6.9 Mass to Mole Conversions: Limiting Reagent and Theoretical Yield (Continued)


If 28.56 g of K2O is produced when 25.00 g K is reacted according to the following equation, what is the percent yield of the reaction? 4 K(s) + O2(g) 2 K2O(s)

a. 87.54%

b. 95%

c. 94.82%

d. 88%

Chapter Summary

1. What is the mole, and why is it useful in chemistry?

• A mole refers to Avogadro’s number 6.022 × 1023 formula units of a substance.

• One mole of any substance has a mass (a molar mass) equal to the molecular or formula weight of the substance in grams.

• Because equal numbers of moles contain equal numbers of formula units, molar masses act as conversion factors between numbers of molecules and masses in grams.

2. How are molar quantities and mass quantities related?

• The coefficients in a balanced chemical equation represent the numbers of moles of reactants and products in a reaction.

• The ratios of coefficients act as mole ratios that relate amounts of reactants and/or products.

• By using molar masses and mole ratios in factor-label calculations, unknown masses or molar amounts can be found from known masses or molar amounts.

3. What are the limiting reagent, theoretical yield, and percent yield of a reaction?

• The limiting reagent is the reactant that runs out first.

• The theoretical yield is the amount of product that would be formed based on the amount of the limiting reagent.

• The actual yield of a reaction is the amount of product obtained.

• The percent yield is the amount of product obtained divided by the amount theoretically possible and multiplied by 100%.

The official name for what we are doing is Stoichiometry


amu to grams

• Recall: in Ch 2 we calculated the masses in amu needed to balance this reaction

H2(g) + I2(g) → 2HI(g)

2.016 amu 253.80 amu

– So the mass ratio of I2 to H2 needed is actually: 125.90

chapter six chemical reactions: mole and mass relationships fundamentals of general, organic, and...

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