LECTURE 7 CHM 151 ©slg

1. The Mole: Molecules and Compounds

2. % Composition from Formulas

3. Empirical Formula from %

Topics:

Mercury is a liquid metal with a density of 13.534 g/cm3. If you measured out 75.0 mL of Hg into a graduated cylinder, how many atoms of Hg would be in the sample?

75.0 mL Hg= ? atoms Hg

Question: 75.0 mL Hg = ? Atoms Hg

Relationships: 1 mL or cm3 Hg = 13.534 g Hg 200.59 g Hg = 1 mol Hg 1 mol Hg = 6.022 x 1023 atoms HgSetup and solve: mL---> g ---> mol --->atoms

Warm-up:

75.0 mL Hg

1 mL Hg

13.534 g Hg

200.59 g Hg

1mol Hg

1 mol Hg

6.02 x1023 atoms Hg

= 30.46 x 1023 atoms = 3.05 X 1024 atoms Hg

mL g mol atoms

Molecules, Compounds, and the Mole

Let us now extend the use of molar mass, M, to include all particles chemists need to measure: not just atoms but also especially ions and molecules....

The basic principle is this: whenever you weigh out the“formula weight” of any substance or species in grams, you have A’s number of particles of that species, and the molar mass of that species...

Molar Mass of Molecules

The formula of any molecule describes the number of atoms making up one unit of that molecule:

Br2 The diatomic bromine molecule, as bromine is found in nature: the formula tells us that 2 atoms of bromine are contained in every molecule.

By extension, 2 moles of bromine atoms are contained in every 1 mole of bromine molecules. The calculation of the molar mass of molecular bromine then looks like this:

The atomic weight of Br, from the PT, is 79.904 amu’s.

Therefore:

2 moles of Br = 2 X 79.904 g = 159.808 g

And the molar mass, M, of Br2 is 159.808 g/mol

Now let’s try the molar mass of CH3CH2OH, ethylalcohol:

Molar Mass, M, CH3CH2OH

Element # of atoms M, g/mol total

C 2 12.01 24.02

H 6 1.008 6.048

O 1 16.00 16.00

Total 46.068

MM, CH, CH33CHCH22OH, 46.07 g/molOH, 46.07 g/mol

Molar Mass of Ionic Compounds

The formula of an ionic compound indicates the simplest ratio of ions present in any sample of the compound. It is this “formula unit” that we use for calculating the molar mass.

Actually, we needn’t ask what kind of compound we are getting the M for; we simply calculate for all atoms found in the formula of any species!

Cl2 (2 Cl) Fe(CN)2 (1Fe 2C 2N) (NH4)2CO3 (2N 8H 1C 3O)

M, CH3CH2OH, 46.07 g/mol, use in problems:

Given a Given a massmass, or , or volumevolume and and densitydensity, , solve for:solve for:

a) moles of compound or individual atoms

b) grams of individual atoms

c) number of molecules or atoms

M, CH3CH2OH, 46.07 g/mol, use in problems:

Given a Given a massmass, or , or volumevolume and and densitydensity, , solve for:solve for:

a) moles of compound or individual atoms

b) grams of individual atoms

c) number of molecules or atoms

How many moles of ethyl alcohol are contained in a sample that weighs 33.95 g? (CH3CH2OH, 46.07 g/mol).

Question: 33.95 g CH3CH2OH = ? mol CH3CH2OH

Relationship: 46.07 g CH3CH2OH = 1 mol

Setup and Solve: ( g ---> mol)

33.95 g CH3CH2OH= ? mol CH3CH2OH

g ----------> mol

33.95 g CH3CH2OH = ? mol CH3CH2OH

46.07 g

1 mol

= .7369 mol CH3CH2OH

1 molecule CH3CH2OH,

2 C atoms

contains:

6 H atoms

1 O atom

C C

H

H

H H

H

O

H

1 mole CH3CH2OH,

2 mole C atoms

contains:

6 moles H atoms 1 mole O atoms

6.022x1023 molecules of CH3CH2OH

2 X 6.022x1023

Carbon atoms

6 X 6.022x1023

Hydrogen atoms

1 X 6.022x1023

Oxygen atoms

How many moles of hydrogen are contained in33.95 g CH3CH2OH?

Question: 33.95 g CH3CH2OH = ? mol H

Relationship: 46.07 g CH3CH2OH = 1 mol CH3CH2OH 1 mol CH3CH2OH = 6 mol H

Setup and Solve: ( g ---> mol CH3CH2OH ---> mol H)

33.95 g CH3CH2OH= ? mol H

33.95 g CH3CH2OH = ? mol H

46.07 g CH 3CH 2OH

CH 3CH 2OH

CH 3CH 2OH

1 mol 6 mol

1 mol

H

= 4.422 mol H

g mol alcohol mol H

How many grams of hydrogen are contained in33.95 g CH3CH2OH?

Question: 33.95 g CH3CH2OH = ? g H

Relationship: 46.07 g CH3CH2OH = 1 mol CH3CH2OH

1 mol CH3CH2OH = 6 mol H

1 mol H = 1.008 g H

Setup and Solve: ( g CH3CH2OH ---> mol CH3CH2OH ---> mol H -----> g H)

33.95 g CH3CH2OH = ? g H

g alcohol mol alcohol mol H

33.95 g CH3CH2OH

= ? g H

46.07 g CH 3CH 2OH

CH 3CH 2OH

CH 3CH 2OH

1 mol 6 mol

1 mol

H

= 4.457 g H

g H

1 mol H

1.008 g H

How many grams of carbon are contained in 33.95 g CH3CH2OH?

Question: 33.95 g CH3CH2OH = ? g C

Relationship: 46.07 g CH3CH2OH = 1 mol CH3CH2OH 1 mol CH3CH2OH = 2 mol C 1 mol C = 12.01 g C

Setup and Solve: ( g CH3CH2OH ---> mol CH3CH2OH ---> mol C -----> g C)

Group Work

g alcohol mol alcohol mol C

33.95 g CH3CH2OH

= ? g C

46.07 g CH 3CH 2OH

CH 3CH 2OH

CH 3CH 2OH

1 mol 2 mol

1 mol

C

= 17.70 g C

g C

1 mol C

12.01 g C

Solution

How many atoms of carbon are contained in 33.95 g CH3CH2OH?

Question: 33.95 g CH3CH2OH = ? atoms C

Relationship: 46.07 g CH3CH2OH = 1 mol CH3CH2OH 1 mol CH3CH2OH = 2 mol C 1 mol C = 6.02 x 1023 atoms C

Setup and Solve: ( g CH3CH2OH ---> mol CH3CH2OH ---> mol C -----> atoms C)

33.95 g CH3CH2OH = ? atoms C

g alcohol mol alcohol mol C atoms C

g alcohol mol alcohol mol C

33.95 g CH3CH2OH

= ? atoms C

46.07 g CH 3CH 2OH

CH 3CH 2OH

CH 3CH 2OH

1 mol 2 mol

1 mol

C

= 8.873 x 1023 atoms C

atoms C

1 mol C

6.02x1023 atoms C

What mass of ethyl alcohol, CH3CH2OH, would contain2.06 X 1024 atoms of carbon?

Question: 2.06 X 1024 atoms C = ? g CH3CH2OH

Relationship: 6.02 x 1023 atoms C = 1 mol C 2 mol C = 1 mol CH3CH2OH 1 mol CH3CH2OH = 46.07 g CH3CH2OH

Setup and Solve: (atoms C-----> mol C ---> mol CH3CH2OH ---> g CH3CH2OH )

Group Work

SOLUTION

2.06 X 1024 atoms C

= ? g CH3CH2OH = 78.8 g CH3CH2OH

6.022 x 1023

atoms C

1 mol C

2 mol C

1 molCH 3CH 2OH

CH 3CH 2OH1 mol

46.07 g CH 3CH 2OH

atoms C mol C mol CH3CH2OH g CH3CH2OH

New TOPIC:% Composition from Formula of the Compound

The formula of a compound can be used to determinethe mass % of each element by using the molar relationships we have learned...

Consider the following:

What is the % by mass of Carbon, Hydrogen and Oxygenin a sample of ethyl alcohol, CH3CH2OH?

Mass % of Elements in a Compound

This is done using the formula weight calculations andthe approach:

% by mass = total mass of one element (the part) X 100 % total mass of compound (the whole)

based on one mole of the compound

% Composition of Ethyl Alcohol, CH3CH2OH

Molar Mass Calculation:2 C = 2 X 12.011 = 24.022 g C6 H = 6 X 1.008 = 6.048 g H 1 O = 1 X 15.999 = 15.999 g O 46.069 g/mol CH3CH2OH

%C = 24.022 g C X 100 = 52.144% C 46.069 g CH3CH2OH

%H = 6.048 g H X 100 = 13.13% H 46.069 g CH3CH2OH

%O = 15.999 g O X 100 = 34.728% O 46.069 g CH3CH2OH

CHECKING.....

52.144% C 13.13 % H 34.728% O 100.002 % CH3CH2OH

= 100.00% (2 digits allowed after decimal)

Formula from % Composition

If one can go from formula to % composition,one should be able to go from % composition to the formula of the compound.

This is quite true (almost):

We can take % composition back to the “empirical formula”, which describes the simplest mole ratio ofatoms in the formula... That is not always the same thing...

Empirical Vs Molecular Formulas

Name MolecularFormula

EmpiricalFormula

“n”

Acetylene C2H2 (CH)n 2

Benzene C6H6 (CH)n 4

VinylAcetylene

C4H4 (CH)n 6

For “organic” or “molecular carbon containing compounds”, there exist a list of compounds which share almost every conceivable empirical formula. To determine which compound one has fromanalytical data, one needs the molar mass aswell, as we will see...

For most ionic compounds and simple molecularcompounds, the empirical and molecular formulasare identical.

Empirical Formula from % Composition:

Let’s take our ethyl alcohol compound back to itsformula from its percent composition.The trick is to use 100 grams of sample whenever one is calculating from mass %. Then we have:

52.144% C = 52.14 g carbon 13.13 % H = 13.13 g hydrogen 34.728% O = 34.73 g Oxygen

Our procedure will be:% element ---> g element ---> mol element ---> simplest mole ratio

Method of choice: make a chart:

element grams Molarmass

#moles Simplestmole ratio

List eachseparately

Use basedon 100 gwhen %given

Use massof 1 molof atoms

#g X 1 mol #g

(calculate and insert)

Divide allmoles bysmallest #of moles

Exp dataifprovided

“atomicweightonly”

Chart out your information in this fashion to determine formula:

grams Molarmass

Moles:#g/ M

Simplestmole ratio

C 52.14 g 12.01 4.341 4.341/2.171 =2.000

H 13.13 g 1.008 13.03 13.03/2.171= 6.002

O 34.73 g 16.00 2.171 2.171/2.171= 1.000

Moles: 52.14 g C X 1 mol C = 4.341 mol C 12.01 g C

Your empirical formula is the simplest ratio of molesof the elements in the formula, for which you have determined that:

For every 1.00 mole of O atoms, you have 6.002 mole of H atoms and 2.000 mole of C atoms

Empirical formula: C2H6O

Empirical Formula using Experimental Data

A new compound weighing 0.678 g, containing xenon and fluorine, was made from a mixture of the gases in sunlight. If the xenon showed a mass of 0.526 g, what is the empirical formula of the compound?

Note:0.678 g compound - 0.526 g Xe = .152 g F2 gas

Although the gas itself is diatomic, when combined into a compound it exists as individual atoms, and the atomic weight of F. 19.0 g/mol, is used for formula calculation, not 38.0 g/mol F2.

Grams M Moles: #g/ M Simplest mole ratio

Xe .526 131.29 .00401 .00401/.00401 = 1

F .152 19.00 .00800 .00800/.00401 = 2

EMPIRICAL FORMULA: XeF2

Molecular and Empirical Formula

Nicotine, a poisonous compound found in tobaccoleaves, is 74.0% C, 8.65% H, and 17.35% N. Its molar mass is 162 g/mol.

What are the empirical and molecular formulas of this compound?

Note: When the molar mass is included in the problem, the exact molecular formula can be determined...

grams Molar mass Moles Simplest mole ratio

C 74.0 g 12.01 6.16 6.16/1.238 = 4.98 H 8.65 g 1.008 8.58 8.58/1.238 = 6.93N 17.35 g 14.01 1.238 1.238/1.238= 1.000

Empirical Formula: C5H7N

In order to obtain the molecular formula, you must divide the molar mass (mass of molecule) by the empirical formula mass (the mass of the simplest ratio of atoms...)

Empirical Formula Mass:

5C = 5 X 12.01 = 60.057H = 7 X 1.008 = 7.0561N = 1 X 14.01 = 14.01 81.12 g/mol

Molar mass = 162 = 2.00 = “n”Emp. Form. mass 81.12

(C5H7N)n = (C5H7N)2 : C10H14N2

Hydrated Compounds

Crystalline ionic solids (salts!) are frequently found innature or are produced from aqueous solutions with a specific number of water molecules associated with each set of formula ions:

CuSO4. 5H2O NiCl2.6H2O CaSO4.2H2O

Frequently the color of the salt depends on the presence of these “waters of hydration.”

The number of water molecules associated with a particular salt is characteristic but not easy to predict: therefore the value is determined experimentally:

The hydrated salt is weighed, heated carefully to drive off the water, and reweighed.

The mass of the water driven off is calculated, converted to moles and compared to moles of the parent, anhydrous salt to determine theformula of the hydrate...

Ex 3.15: Naturally occurring hydrated copper(II) chlorideis called eriochalcite. If 0.235 g of CuCl2.xH2O is heatedto drive off the water, 0.185 g residue remains. What is the value of x?

0.235 g hydrate - 0.185 g parent salt = 0.050 g H2O

1Cu= 1 X 63.55 = 63.552Cl = 2 X 35.45 =70.90 134.45 g/mol CuCl2

2H= 2 x 1.008= 2.0161O= 1 x 16.00= 16.00 18.02 g/mol H2O

grams Molar mass Moles Simplest mole ratio

CuCl2 0.185 g 134.45g/mol .00138 .00138/.00138= 1H2O 0.050 g 18.02 g/mol .00278 .00278/.00138= 2

Formula of eriochalcite: CuCl2. 2H2O

GROUP WORK: Do both problems, be sure all names are on sheet, hand in as you leave!

1. If 7.572 g Cu (63.546 g/mol) reacts with 1.910 g S(32.066 g/mol) to form a binary compound CuxSy, whatis the empirical formula of the compound formed?

2. If 1.023 g of a hydrated compound, CuSO4. xH2Oshows a mass of 0.654 g when dehydrated, what is the formula of the compound? (CuSO4, 159.6 g/mol;H2O, 18.02 g/mol).

The answers are Cu2S, CuSO4. 5 H2O; show work to prove...

End, Unit One material