Moles and Formula Mass - Edl€™s Number 6.022 x 1023 is called “Avogadro’s Number” in honor...

Moles and Formula Mass

The Mole

1 dozen =

1 gross =

1 ream =

1 mole =

12

144

500

6.022 x 1023

There are exactly 12 grams of carbon-12 in one mole of carbon-12.

Avogadro’s Number6.022 x 1023 is called “Avogadro’s Number” in honor of the Italian chemist Amadeo Avogadro (1776-1855).

Amadeo Avogadro

I didn’t discover it. Its just named after me!

Calculations with Moles:Converting moles to grams

How many grams of lithium are in 3.50 moles of lithium?

3.50 mol Li= g Li

1 mol Li

6.94 g Li 45.1

Calculations with Moles:Converting grams to moles

How many moles of lithium are in 18.2 grams of lithium?

18.2 g Li= mol Li

6.94 g Li

1 mol Li2.62

Calculations with Moles:Using Avogadro’s Number

How many atoms of lithium are in 3.50 moles of lithium?

3.50 mol= atoms

1 mol

6.02 x 1023 atoms2.07 x 1024

Calculations with Moles:Using Avogadro’s Number

How many atoms of lithium are in 18.2 g of lithium?

18.2 g Li

= atoms Li

1 mol Li 6.022 x 1023 atoms Li

1.58 x 1024

6.94 g Li 1 mol Li

(18.2)(6.022 x 1023)/6.94

Calculating Formula MassCalculate the formula mass of magnesium carbonate, MgCO3.

24.31 g + 12.01 g + 3(16.00 g) = 84.32 g

Calculating Percentage Composition

Calculate the percentage composition of magnesium carbonate, MgCO3.

From previous slide:

24.31 g + 12.01 g + 3(16.00 g) = 84.32 g

100.00

24.31100 28.83%

84.32Mg

12.01100 14.24%

84.32C

48.00100 56.93%

84.32O

Formulas

molecular formula = (empirical formula)n [n = integer]

molecular formula = C6H6 = (CH)6 empirical formula = CH

Empirical formula: the lowest whole number ratio of atoms in a compound.Molecular formula: the true number of atoms of each element in the formula of a compound.

Formulas (continued)

Formulas for ionic compounds are ALWAYSempirical (lowest whole number ratio).

Examples:

NaCl MgCl2 Al2(SO4)3 K2CO3

Formulas (continued)

Formulas for molecular compounds MIGHTbe empirical (lowest whole number ratio).

Molecular:

H2O

C6H12O6 C12H22O11

Empirical:

H2O

CH2O C12H22O11

Empirical Formula Determination

1. Base calculation on 100 grams of compound.

2. Determine moles of each element in 100 grams of compound.

3. Divide each value of moles by the smallest of the values.

4. Multiply each number by an integer to obtain all whole numbers.

Empirical Formula Determination

Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid?

49.32g C 1 mol C=4.107 mol C

12.01 g C

6.85 16.78

1.01

g H mol Hmol H

g H

43.84 12.74

16.00

g O mol Omol O

g O

Empirical Formula Determination(part 2)

Divide each value of moles by the smallest of the values.

Carbon:

Hydrogen:

Oxygen:

4.1071.50

2.74

mol C

mol O

6.782.47

2.74

mol H

mol O

2.741.00

2.74

mol O

mol O

Empirical Formula Determination(part 3)

Multiply each number by an integer to obtain all whole numbers.

Carbon: 1.50 Hydrogen: 2.50 Oxygen: 1.00x 2 x 2 x 2

3 5 2

Empirical formula: C3H5O2

Finding the Molecular FormulaThe empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid?

1. Find the formula mass of C3H5O2

3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g


3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

2. Divide the molecular mass by the mass given by the emipirical formula.

1462

73


3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

3. Multiply the empirical formula by this number to get the molecular formula.

(C3H5O2) x 2 = C6H10O4

1462

73

Chemical Analysis

• Law of Definite Proportions:– Different samples of the same compound always contain its constituent elements in the same proportion by mass.

– This means that the ratio of atoms in a compound must be constant.

– Example: We can analyze CO2 from different sources and find that each sample has the same ratio by mass of carbon to oxygen.

Chemical Analysis• Law of Multiple Proportions:

– If two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in ratios of small numbers

– Different compounds made up of the same elements differ in the number of atoms of each kind that combine.

– Example: Carbon can form two different compounds with oxygen, CO and CO2. The ration of oxygen in each compound is 1:2

Chemical Analysis

• Because of these two laws AND the Law of conservation of mass…..

• You can determine the empirical formula from percent composition.

• We can analyze percent composition of a sample by various methods:– Mass Spectrometer

– Burning in a combustion train

Pure O2 inCO2 is absorbed

H2O is absorbed

Sample is burned

completely to

form CO2 and

H2O

Burning a 0.1000-g sample of a carbon–hydrogen–oxygen compound in oxygen yields 0.1953 g CO2 and 0.1000 g H2O.

Determine the percent composition of the compound.

The amount of CO2 tells us how much carbon was in the sample.

Carbon:

Cg105932.5

Cmol1

Cg01.12

COmole1

Cmol1

COg01.44

COmole1COg1953.0

2

22

22

Burning a 0.1000-g sample of a carbon–hydrogen–oxygen compound in oxygen yields 0.1953 g CO2 and 0.1000 g H2O.

Determine the percent composition of the compound.

The amount of H2O tells us how much hydrogen was in the sample.

Hydrogen:

Hg107811.1

Hmol1

Hg008.1

OHmole1

Hmol2

OHg02.18

OHmole1OHg1000.0

3

22

22

Burning a 0.1000-g sample of a carbon–hydrogen–oxygen compound in oxygen yields 0.1953 g CO2

and 0.1000 g H2O.

Calculate the percent composition of the sample.

Carbon:

%30.53100Sampleg1000.0

Cg105932.5C%

2

Hydrogen:

%19.11100Sampleg1000.0

Cg101187.1H%

2

Burning a 0.1000-g sample of a carbon–hydrogen–oxygen compound in oxygen yields 0.1953 g CO2

and 0.1000 g H2O.

Calculate the percent composition of the sample.

Oxygen:

%51.35%19.11%30.53%100O%

So our sample has a percent composition of: 53.30% C, 11.19% H and 35.51% O.

Now we can calculate empirical formula!!!!! YIPEE!!!!

Hydrates

• Compounds that have a specific number of water molecules attached to them.

• Copper sulfate in it’s normal state has the five water molecules associated with it.

• The formula is CuSO4 ‧ 5H2O

• The name is copper (II) sulfate pentahydrate

Hydrates

• When hydrates are heated the water molecules are evaporated off

• What is left is CuSO4.

• We call this anhydrous copper (II) sulfate

• Anhydrous means the water is gone!

Hydrates and Composition• Cupric chloride, CuCl2, when heated to 100C is dehydrated. If 0.235 g of CuCl2 · x H2O gives 0.185 g of CuCl2on heating, what is the value of x?

OHg050.0

CuClg185.0OHCuClg235.0

2

222

First calculate how much H2O has been heated off:


23

2

22 CuClmole10x38.1

CuClg45.134

CuClmole1CuClg185.0

Convert CuCl2 and H2O to moles

OHmole10x77.2OHg02.18

OHmole1OHg050.0 2

3

2

22


3

3

3

3

10x38.1

10x77.2:

10x38.1

10x38.1

Now find the mole ratio of CuCl2:H2O

1:2.01

CuCl2‧2H2O

Mass Spectrometry

Courtesy www.lab-initio.com

Mass Spectometry

Learning Objective for EK 1.D.2:LO 1.14 The student is able to use data from mass spectrometry to identify theelements and the masses of individual atoms of a specific element.

Purpose of Mass Spectrometry

Produces spectra of masses from elements in a sample of material. The isotopic composition of sample of an element The relative atomic mass of an element

Produces spectra of masses from molecules in a sample of material, and fragments of the molecules.

Purpose of Mass Spectrometry

Used to determine the elemental composition of a sample the masses of particles and of molecules potential chemical structures of molecules by

analyzing the fragments the identity of unknown compounds by determining

mass and matching to known spectra the isotopic composition of elements in a molecule

Mass Spectometry

Mass Spectometry

• Sample is vaporized

• Passed through a beam of

electrons (creates cations)

• Accelerated through a

magnetic field which changes

their path depending on their

mass

• Lightest ions have greater

deflection

• Position on detector plate

gives accurate values of their

masses

Basics of Mass Spectrometry

Sample of element X

X-20X-18

20𝑋 → 20𝑋+ + 𝑒−

Shot with an electron

Detected by the spectrometer

Mass of 20𝑋 = 20𝑋+

Appears on spectra as mass:charge (M/Z)

Relative isotopic mass of X-20 = 20 M/Z of X-20 = 20/1 = 20


Sample of element X

X-20X-18

10 15 18 20

Rela

tive

abun

dan

ce

M/Z

Spectra of element X

8

4

How does the height of X+ -20 compare to X +-18?

33.3%

66.7%


10 15 18 20

Rela

tive

abun

dan

ce

M/Z

Spectra of element X

8

4

Isotopic composition of element X is33% X-18 and 66.7% X-20 33.3%

66.7%

4𝑥18 +(8𝑥20)

12= 19.3

or 33.3 𝑥18 + 66.7 𝑥20

100= 19.3

The relative atomic mass can be calculated :

Mass Spectrum of Neon

Using the data from the mass spectrum above, calculate the average atomic mass of neon.

Mass Spectrum of Bromine, Br2

Bromine has two isotopes: 50.69% 79Br and 49.31% 81Br

79Br+ 81Br+

[79Br81Br]+

[79Br79Br]+

[81Br81Br]+

Molecular Ion Peaks

Fragments

Mass Spectrum of CO2

Molecular ion peak[CO2]+ = 44

[CO]+ = 28 [O]+ = 16 [C]+ = 12

Fragment Peaks

Stoichiometry

Sherlock Holmes, in Sir Arthur Conan Doyle’s A Study in Scarlet

“In solving a problem of this sort, the grand thing is to be able to reason backward. This is a very useful accomplishment, and a very easy one, but people do not practice it much.”

Review: Chemical EquationsChemical change involves a reorganization of

the atoms in one or more substances.

C2H5OH + 3O2 2CO2 + 3H2O

reactants products

1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water

When the equation is balanced it has quantitative significance:

Calculating Masses of Reactants and Products

1. Balance the equation.

2. Convert mass or volume to moles, if necessary.

3. Set up mole ratios.

4. Use mole ratios to calculate moles of desired substituent.

5. Convert moles to mass or volume, if necessary.

Working a Stoichiometry Problem6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed.

1. Identify reactants and products and write the balanced equation.

Al + O2 Al2O3

b. What are the reactants?

a. Every reaction needs a yield sign!

c. What are the products?

d. What are the balanced coefficients?

4 3 2

Working a Stoichiometry Problem6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed?

4 Al + 3 O2 2Al2O3

=6.50 g Al

? g Al2O3

1 mol Al

26.98 g Al 4 mol Al

2 mol Al2O3

1 mol Al2O3

101.96 g Al2O3

6.50 x 2 x 101.96÷ 26.98 ÷ 4 = 12.3 g Al2O3

Limiting ReactantThe limiting reactant is the reactant that is consumed first, limiting the amounts of products formed.

Calculating Limiting Reactants

1. Calculate the amount of product produced using each amount of reactant. (Yes, 2 stoich problems!)

2. Whichever reactant produces the LEAST amount (moles OR mass) of product is your limiting reactant.

Working a Limiting Reactant ProblemNitrogen gas can be prepared by passing gaseous ammonia over solid copper(II) oxide at high temperatures. The other products of the reaction are solid copper and water. If a sample containing 18.1 g of NH3 is reacted with 90.4 g of CuO, which is the limiting reactant and how many grams of N2 will be formed?

1. Identify reactants and products and write the balanced equation.

NH3

+

CuO N2(g)

b. What are the reactants?a. Every reaction needs a yield sign!

c. What are the products?

d. What are the balanced coefficients?

2 3 3Cu(s)+ + H2O(g)3

Working a Limiting Reactant Problem

= 14.9 g N2

1 mol NH3

17.03 g NH3

18.1 g NH3

2 mol NH3

1 mol N2

1 mol N2

28.02 g N2

NH3

+

CuO N2(g)2 3 3Cu(s)+ + H2O(g)3

=90.4 g CuO

10.6 g N2

1 mol CuO

79.55 g CuO 3 mol CuO

1 mol N2

1 mol N2

28.02 g N2

Calculate the amount of product produced using each amount of reactant.

CuO produeces the smaller amount so CuO is the limiting reactant. 10.6 g N2 are produced.

Theoretical and Actual Yield

Theoretical yield: The calculated amount of product produced.Actual yield:The amount produced by performing the experiment.

The percent yield can be calculated using the equation below:

100)g(yieldltheoretica

)g(yieldactualyield%

Moles and Formula Mass - Edl€™s Number 6.022 x 1023 is called “Avogadro’s Number” in honor...

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