Chemical Reactions: Mole and Mass Relationships Chapter 6 Lecture Fundamentals of General, Organic, and Biological Chemistry 7th Edition McMurry, Ballantine, Hoeger, Peterson Chapter Six Chemical Reactions: Mole and Mass Relationships Julie Klare Gwinnett Technical College © 2013 Pearson Education, Inc.

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 carbon C-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 11.8668 0.1443 12.01 amu

fyi: Disambiguation For molecules, these terms are identical molar weight molar mass molecular weight molecular mass For atoms, these terms are identical atomic weight atomic mass

Molecular Weight (Mass): Mass in grams of one mole of the substance: Molar Mass of N = 14.01 g/mol Molar 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) Copyright © Cengage Learning. All rights reserved

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

Meaning that this reaction is balanced – in amu units H2(g) + I2(g) → 2HI(g) 2.016 amu 253.80 amu 1 : 126 Copyright © Cengage Learning. All rights reserved

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 Copyright © Cengage Learning. All rights reserved

And again by The Law of Identical Ratios, this must also be balanced H2(g) + I2(g) → 2HI(g) 1.00 g 126 g Copyright © Cengage Learning. All rights reserved

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

The mole (NA) has now been measured 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?

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 Worked Example 6.3 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 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?

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

Mole to mole This reaction represents the reaction of A2 (red) with B2 (blue) Chapter 6, Unnumbered Figure 2, Page 174 a) Write a balanced equation for the reaction b) How many moles of product can be made from 1.0 mole of A2? From 1.0 mole of B2?

How many moles of NH3 can be produced from 5 How many moles of NH3 can be produced from 5.00 moles of H2 according to the following equation? N2 + 3 H2  2 NH3 0.67 mol 2.00 mol 3.33 mol 7.50 mol

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 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) 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

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

10.0 grams of HI requires how many grams of H2 I2 = 253.80 g/mol H2 = 2.016 g/mol HI = 127.90 g/mol

How many grams of oxygen are needed to react with 25 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) 10.2 g O2 2.56 g O2 8.66 g O2 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

limiting reactant (reagent) 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.

limiting reactant (reagent) 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

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) Copyright © Cengage Learning. All rights reserved

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) Copyright © Cengage Learning. All rights reserved

A limiting reactant mixture on the other hand, contains an excess of one of the reactants The other reactant is thus limiting N2(g) + 3H2(g)  2NH3(g) Copyright © Cengage Learning. All rights reserved

A limiting reactant mixture on the other hand, contains an excess of one of the reactants The other reactant is thus limiting N2(g) + 3H2(g)  2NH3(g) Which reactant is in excess in this reaction mixture? Which reactant is limiting?

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 Copyright © Cengage Learning. All rights reserved

But the limiting reactant must be identified before the correct amount of product can be calculated 2H2 + O2  H2O Copyright © Cengage Learning. All rights reserved

10.0 g of Chemical A are reacted with 10.0 g of B 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 Copyright © Cengage Learning. All rights reserved

10.0 g of Chemical A are reacted with 10.0 g of B 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? A + 3B  2C molar masses: A is 10.0 g/mol B is 20.0 g/mol C is 25.0 g/mol?

10.0 g of Chemical A are reacted with 10.0 g of B Example 10.0 g of Chemical A are reacted with 10.0 g of B A + 3B  2C 1) Convert known masses of reactants to moles

10.0 g of Chemical A are reacted with 10.0 g of B Example 10.0 g of Chemical A are reacted with 10.0 g of B A + 3B  2C 2) Convert to the number of moles of product So B is limiting reactant

10.0 g of Chemical A are reacted with 10.0 g of B Example 10.0 g of Chemical A are reacted with 10.0 g of B A + 3B  2C Choose the least number of moles of product formed as limiting reactant: B is limiting

10.0 g of Chemical A are reacted with 10.0 g of B Example 10.0 g of Chemical A are reacted with 10.0 g of B 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 Calculate the number of moles of each reactant divide the number of moles of each reactant by its coefficient from the balanced equation 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) K2O = 94.2g/mol 67g K2O; K is the limiting reactant. 270 g K2O; K is the limiting reactant. 270 g K2O; O2 is the limiting reactant. 67g K2O; O2 is the limiting reactant.

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)

FYI 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

Theoretical Yield Actual 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:

WORKED EXAMPLE 6.8 Percent Yield 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 (finished)

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

If 28. 56 g of K2O is produced when 25 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) 87.54% 95% 94.82% 88%

Chapter Summary

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.

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.

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 H2(g) + I2(g) → 2HI(g) 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 Copyright © Cengage Learning. All rights reserved