Mass Relationships in Chemical Reactions

Chapter 3

By definition: 1 atom 12C “weighs” 12 amu

On this scale

1H = 1.008 amu

16O = 16.00 amu

Atomic mass is the mass of an atom in atomic mass units (amu)

Micro Worldatoms & molecules

Macro Worldgrams

3.1

Natural lithium is:

7.42% 6Li (6.015 amu)

92.58% 7Li (7.016 amu)

(7.42% x 6.015) + (92.58% x 7.016)100

= 6.941 amu

3.1

Average atomic mass of lithium:

The mole (mol) is the amount of a substance that contains as many elementary entities as there

are atoms in exactly 12.00 grams of 12C

3.2

1 mol = NA = 6.0221367 x 1023

Avogadro’s number (NA)

Molar mass is the mass of 1 mole of in gramseggsshoes

marblesatoms

1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g

1 12C atom = 12.00 amu

1 mole 12C atoms = 12.00 g 12C

1 mole lithium atoms = 6.941 g of Li

For any element

atomic mass (amu) = molar mass (grams)

3.2

One Mole of:

C S

Cu Fe

Hg

3.2

1 g = 6.022 x 1023 amu

3.2

M = molar mass in g/mol

NA = Avogadro’s number

1 amu = 1.66 x 10-24 g

Do You Understand Molar Mass?

How many atoms are in 0.551 g of potassium (K) ?

1 mol K = 39.10 g K

1 mol K = 6.022 x 1023 atoms K

0.551 g K 1 mol K39.10 g K

x x 6.022 x 1023 atoms K1 mol K

=

8.49 x 1021 atoms K

3.2

Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule.

SO2

1S 32.07 amu

2O + 2 x 16.00 amu SO2 64.07 amu

For any molecule

molecular mass (amu) = molar mass (grams)

1 molecule SO2 = 64.07 amu

1 mole SO2 = 64.07 g SO2 3.3

Do You Understand Molecular Mass?

How many H atoms are in 72.5 g of C3H8O ?

1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O

1 mol H = 6.022 x 1023 atoms H

5.82 x 1024 atoms H

3.3

1 mol C3H8O molecules = 8 mol H atoms

72.5 g C3H8O1 mol C3H8O

60 g C3H8Ox

8 mol H atoms

1 mol C3H8Ox

6.022 x 1023 H atoms

1 mol H atomsx =

Percent composition of an element in a compound =

n x molar mass of elementmolar mass of compound

x 100%

n is the number of moles of the element in 1 mole of the compound

C2H6O

%C =2 x (12.01 g)

46.07 gx 100% = 52.14%

%H =6 x (1.008 g)

46.07 gx 100% = 13.13%

%O =1 x (16.00 g)

46.07 gx 100% = 34.73%

52.14% + 13.13% + 34.73% = 100.0%

3.5

Types of FormulasTypes of Formulas Empirical FormulaEmpirical Formula

The formula of a compound that The formula of a compound that expresses the expresses the smallest whole number ratiosmallest whole number ratio of of the atoms present.the atoms present.

Ionic formula are always empirical formulaIonic formula are always empirical formula

Molecular FormulaMolecular Formula

The formula that states the The formula that states the actualactual number number of each kind of atom found in of each kind of atom found in one moleculeone molecule of of the compound.the compound.

To obtain an To obtain an Empirical FormulaEmpirical Formula

1.1. Determine the mass in grams of each Determine the mass in grams of each element present, if necessary.element present, if necessary.

2.2. Calculate the number of Calculate the number of molesmoles of of each each element.element.

3.3. Divide each by the smallest number of moles Divide each by the smallest number of moles to obtain the to obtain the simplest whole number ratio.simplest whole number ratio.

4.4. If whole numbers are not obtainedIf whole numbers are not obtained** in step 3), in step 3), multiply through by the smallest number multiply through by the smallest number that will give all whole numbersthat will give all whole numbers

** Be careful! Do not round off numbers prematurelyBe careful! Do not round off numbers prematurely

A sample of a brown gas, a major air pollutant, A sample of a brown gas, a major air pollutant,

is found to contain 2.34 g N is found to contain 2.34 g N

and 5.34g O. Determine a formula and 5.34g O. Determine a formula

for this substance.for this substance.

require require molemole ratios so convert grams to moles ratios so convert grams to moles

moles of N = moles of N = 2.34g of N 2.34g of N = 0.167 moles of N= 0.167 moles of N

14.01 g/mole14.01 g/mole

moles of O = moles of O = 5.34 g5.34 g = 0.334 moles of O = 0.334 moles of O

16.00 g/mole16.00 g/mole

Formula:Formula:

0.334 0.167ON 0.167 0.334 2

0.167 0.167

N O NO

Calculation of the Molecular FormulaCalculation of the Molecular Formula

A compound has an empirical formula of NOA compound has an empirical formula of NO22. . The colourless liquid, used in rocket engines The colourless liquid, used in rocket engines has a molar mass of 92.0 g/mole. What is the has a molar mass of 92.0 g/mole. What is the molecular formula molecular formula of this substance?of this substance?

empirical formula mass: 14.01+2 (16.00) = 46.01 empirical formula mass: 14.01+2 (16.00) = 46.01 g/molg/moln = n = molar massmolar mass = = 92.0 g/mol92.0 g/mol emp. f. mass 46.01 g/mol emp. f. mass 46.01 g/moln = 2n = 2

NN22OO44

Empirical Formula from % CompositionEmpirical Formula from % Composition

A substance has the following composition by mass: 60.80 % Na ; 28.60 % B ; 10.60 % H What is the empirical formula of the substance?

Consider a sample size of 100 gramsThis will contain: 60.80 grams of Na, 28.60 grams of

B, and 10.60 grams HDetermine the number of moles of eachDetermine the simplest whole number ratio

1. Write balanced chemical equation

2. Convert quantities of known substances into moles

3. Use coefficients in balanced equation to calculate the number of moles of the sought quantity

4. Convert moles of sought quantity into desired units

Mass Changes in Chemical Reactions

3.8

Other units

Molarity Moles solute / L solution

Gases 22.4 L = 1 mole of ANY GAS at STP

Methanol burns in air according to the equation

2CH3OH + 3O2 2CO2 + 4H2O

If 209 g of methanol are used up in the combustion, what mass of water is produced?

grams CH3OH moles CH3OH moles H2O grams H2O

molar massCH3OH

coefficientschemical equation

molar massH2O

209 g CH3OH1 mol CH3OH

32.0 g CH3OHx

4 mol H2O

2 mol CH3OHx

18.0 g H2O

1 mol H2Ox =

235 g H2O

3.8

Limiting Reagents

3.9

Method 1

Pick A Product Try ALL the reactants The lowest answer will be the correct answer The reactant that gives the lowest answer will

be the limiting reactant

Limiting Reactant: Method 1 10.0g of aluminum reacts with 35.0 grams of chlorine

gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced?

2 Al + 3 Cl2 2 AlCl3 Start with Al:

Now Cl2:

10.0 g Al 1 mol Al 2 mol AlCl3 133.5 g AlCl3

27.0 g Al 2 mol Al 1 mol AlCl3

= 49.4g AlCl3

35.0g Cl2 1 mol Cl2 2 mol AlCl3 133.5 g AlCl3

71.0 g Cl2 3 mol Cl2 1 mol AlCl3

= 43.9g AlCl3

Solving for Multiple Products

Once you determine the LR, you should only start with it!

A + B X + Y + Z

A X

B X Let’s say B is the LR!

To find Y and Z

B Y

B Z

There is no need to use A to find Y and Z

It will give you the wrong answer – a lot of extra work for nothing

Method 2

Convert one of the reactants to the other REACTANT

See if there is enough reactant “A” to use up the other reactants

If there is less than the GIVEN amount, it is the limiting reactant

Then, you can find the desired species

Do You Understand Limiting Reagents?

In one process, 124 g of Al are reacted with 601 g of Fe2O3

2Al + Fe2O3 Al2O3 + 2Fe

Calculate the mass of Al2O3 formed.

g Al mol Al mol Fe2O3 needed g Fe2O3 needed

ORg Fe2O3 mol Fe2O3 mol Al needed g Al needed

124 g Al1 mol Al

27.0 g Alx

1 mol Fe2O3

2 mol Alx

160. g Fe2O3

1 mol Fe2O3

x = 367 g Fe2O3

Start with 124 g Al need 367 g Fe2O3

Have more Fe2O3 (601 g) so Al is limiting reagent3.9

Use limiting reagent (Al) to calculate amount of product that can be formed.

g Al mol Al mol Al2O3 g Al2O3

124 g Al1 mol Al

27.0 g Alx

1 mol Al2O3

2 mol Alx

102. g Al2O3

1 mol Al2O3

x = 234 g Al2O3

2Al + Fe2O3 Al2O3 + 2Fe

3.9

Finding Excess Practice 10.0g of aluminum reacts with 35.0 grams of

chlorine gas 2 Al + 3 Cl2 2 AlCl3

We found that chlorine is the limiting reactant, and 43.8 g of aluminum chloride are produced.

35.0 g Cl2 1 mol Cl2 2 mol Al 27.0 g Al

71 g Cl2 3 mol Cl2 1 mol Al= 8.8 g Al USED!

10.0 g Al – 8.8 g Al = 1.2 g Al EXCESS

Given amount of excess reactant

Amount of excess reactant actually used

Note that we started with the limiting reactant! Once you determine the LR, you should only start with it!

Theoretical Yield is the amount of product that would result if all the limiting reagent reacted.

Actual Yield is the amount of product actually obtainedfrom a reaction.

% Yield = Actual Yield

Theoretical Yieldx 100

3.10