Chemistry-140 Lecture 6

Chemistry-140 Lecture 6

Chapter 3: Molecules & Compounds

Chapter Highlights

define molecules

definitions of molecular formula & structural formula understand formation of cations & anions learn names & formula for polyatomic ions molar mass of compounds percent composition empirical formula

Molecules

Molecules are tightly bound assemblies of two or more atoms. This "package" behaves as a single unit.

Some elements exist as discrete molecules;

H2, C, O2 (O3), N2, F2, Cl2, Br2, I2, P4, S8

Remember: Although He, Ne, Ar, Kr, Xe, Rn are gases they are not really molecules since they are monatomic


Formulas

Chemical formula: a collection of elemental symbols with subscripts that indicate the relative number of

atoms of each element in the substance

Molecular formula: the chemical formula of a molecular compound

Example:the molecular formula for sucrose is C12H22O11


Molecular Compounds

Compounds are pure substances that can be decomposed into one or more different pure substances

Example:

1 molecule of sucrose

12 atoms of C + 11 molecules of water

side 4: frames 05413-05903


Formulas Structural formula: emphasizes how atoms are connected and

shows any chemically active groups (functional group)

Example:the molecular formula for ethanol is C2H6O

the structural formula for ethanol is CH3CH2OH

OH (alcohol functional group) is an important chemically active group

C CH

H

H

O

H

H

H


Molecular Models

Ball & stick model for ethanol:gray = carbon white = hydrogenred = oxygen

CH

H

H

HModels of methane CH4

ball &stick space filling perspective drawing


Ions Atoms of almost all elements can gain or lose

electrons to form ions (charged species)

Compounds composed of ions are known as ionic compounds

Cations are positively charged ions Anions are negatively charged ions

Example:sodium chloride, NaCl is composed of sodium cations = Na+ and chloride anions = Cl-



Monatomic ions are comprised of single atoms, while polyatomic ions are comprised of several atoms

Examples:

monatomic: Na+ Ca2+ Fe3+ S2- Cl- N3-

polyatomic: NH4+ SO4

2- ClO3- PO4

3-

Monatomic and Polyatomic IonsChemistry-140 Lecture 6

The charge of an ion can be predicted from the elements position in the periodic table.

Monatomic ions: atoms either gain or lose electrons until they have the same number of electrons as the nearest noble gas.

metals lose electrons to form cationsnonmetals gain electrons to form anions

For example:Ar (18 electrons), K (19 electrons) and Cl (17

electrons).K K+

Cl Cl-

in order to have 18 electrons

Predicting the Charges on IonsChemistry-140 Lecture 6

The Monatomic Anionshydride

fluoride

chloride

bromide

iodide

carbide

nitride

phosphide

oxide

sulfideselenide

telluride


Ionic compounds are those compounds formed from the combination of ions.

Ionic Compounds

Please Remember!!

total cationic charge = total anionic charge

overall the material is neutralthe TOTAL charge on the compound = ZERO


Question:

What ionic compound would you expect from the combination of Mg and N?


Answer:

Mg Mg2+(magnesium ion)

N N3- (nitride ion)

In order to obtain overall neutrality 3 Mg2+ combine

with 2 N3- to yield Mg3N2 (magnesium nitride)


Positive ions.

Name plus ion, for example, aluminum ion.

Specify the charge on the ion, for example cobalt (II) and cobalt (III) ions

Ammonium, carbonium and oxonium ions all refer to different types of positive ions of ammonia, carbon and oxygen.

Naming CompoundsChemistry-140 Lecture 6

Negative ions

Simple anions end in –ide, for example chloride ion.

Polyatomic anions (often oxoanions) are not systematic and must be learned (see table 3.1):

perchlorate, chlorate, chlorite, hypochloritehydrogen phosphate, dihydrogen phosphate

carbonate, bicarbonate

ClO4- ClO3

- ClO2- ClO- HPO4

2- H2PO4-

Naming CompoundsChemistry-140 Lecture 6

The Mole

A convenient unit for matter that containsa known number of particles

Definition: the amount of substance that contains as many particles as their are atoms in exactly 12 g

of the carbon-12 isotope

1 mole = 6.022136736 x 1023

Avogadro’s Number (N)


The Mole

How big is this number???

Popcorn kernels covering the continental US $$-Dollars-$$ a national debt ($ 3.6 trillion) computer counting at 10 million particles/second

1 mole = 6.022136736 x 1023

Avogadro’s Number (N)


The Mole

What is the Chemical Significance?

A mole of any element (or compound) always contains the same number of atoms (or molecules)

Since each type of atom has a different atomic mass, a mole of atoms of one element has a different mass from the mass of a mole of a different element

Example: 1 mole of 16O has mass = 16.0 g while 1 mole of 19F has mass = 19.0 g


Molar Mass The mass in grams of 1 mole of atoms of any element is the

molar mass of that element

Molar mass is conventionally shown as M and expressed in grams/mole (g/mol)

For any element, the molar mass in grams is equal to the atomic mass in atomic mass units (amu).

Example:Molar mass of Na = mass of 1 mol of Na atoms

= 22.98 g/mol = mass of 6.022 x 1023 Na atoms


Mass Moles Conversion

The ability to convert from moles to mass and mass to moles is absolutely essential

MASS MOLES CONVERSION

Moles to Mass Mass to Moles

(Moles) = gramsgrams1 mole

(Grams) = moles

1 molegrams

molar mass 1/molar mass


Question:

How many moles are in 454 g of silicon?

A Question of Conversion


A Question of Conversion

Answer:

The molar mass of silicon is 28.09 g/mol (from the periodic table!).

Convert the mass of silicon to its equivalent in moles

(454 g Si) = 16.2 mol Si1 mole Si28.09 g Si


Question:

What is the mass of 2.50 moles of lead (Pb)?

Another Question of Conversion


Another Question of Conversion

Answer:

The molar mass of lead is 207.2 g/mol (where else but from the periodic table!).

Convert the moles of lead to its equivalent mass

(2.50 mol Pb) = 518 g Pb207.2 g Pb1 mol Pb


Question:

A graduated cylinder contains 25.4 mL of mercury (Hg). If the density of mercury is 13.534 g/mL, how many moles of mercury are in the cylinder? How many atoms of Hg are in the cylinder?

A Real Calculation


A Real Calculation

Method:

Requires certain conversion!

Volume, mL Mass, g Moles Atoms

density molar mass

Avogadro’s Number

x g/mL x atoms/molx mol/g


A Real Calculation

Answer:

(25.4 mL Hg) = 344 g Hg

(344 g Hg) = 1.71 mol Hg

(1.71 mol Hg) = 1.03 x 1024 atoms Hg

13.534 g Hg1 mL Hg

1 mol Hg200.6 g Hg

6.022 x 10 atoms Hg1 mol Hg

23




Chapter Highlights

define molecules

definitions of molecular formula & structural formula understand formation of cations & anions learn names & formula for polyatomic ions molar mass of compounds percent composition empirical formula

Molecules, Compounds & the Mole

Molar Mass M, is the mass of a mole of molecules of a particluar substance = Molecular Weight

Example:

The molecular weight of PBr3 = the atomic weight of P plus 3 x the atomic weight of Br

MW (PBr3) = AW (P) + 3[AW (Br)]= (30.97 amu) + 3(79.90 amu)= 270.7 amu

Thus: 1 mole of PBr3 has a mass of 270.7 g


A Mole of ………….


NaCl

CoCl2.6H2ONiCl2.6H2O K2Cr2O7

CuSO4.5H2O

Question (compare to example 3.9):

You have 23.2 g of ethanol, C2H6O.

How many moles are contained in this mass of ethanol?

How many molecules of ethanol are contained in 23.2 g?

How many atoms of carbon are contained in 23.2 g of ethanol?

What is the average mass of one molecule of ethanol?


Method:

Conversions using molar mass & Avogadro’s number!

Mass, g Moles Molecules Number of C atomsmolar

massAvogadro’s

Number

x C atoms/moleculex molecules/molx mol/g


Answer:Step 1: Calculate the molar mass of ethanol, C2H6O.

2 mol of C per mole of ethanol = (2 mol C)

= 24.02 g C

6 mol of H per mole of ethanol = (6 mol H)

= 6.048 g H

1 mol of O per mole of ethanol = (1 mol O)

= 16.00 g O

Molar mass of ethanol = (24.02 + 6.048 + 16.00) g =

12.01 g C1 mol C

1.008 g H1 mol H

16.00 g O1 mol O

46.07 g/mol


Answer:

Step 2: Calculate the number of moles of C2H6O.

(23.2 g C2H6O) =

Number of moles of ethanol in 23.2 g is 0.504 mol.

1 mol C H O46.07 g C H O

2 6

2 6

0.504 mol C2H6O


Answer:Step 3: Calculate the number of carbon atoms.

(0.504 mol C2H6O)

= 3.04 x 10 23 molecules of C2H6O

3.04 x 10 23 molecules of C2H6O

=

Number of carbon atoms in 23.2 g of ethanol is 6.07 x 10 23 .

6.022 x 10 molecules C H O1 mol C H O

232 6

2 6

6.07 x 10 23 atoms of C

2 C atoms1 molecule C H O2 6


Answer:Step 4: Calculate the mass of 1 molecule of C2H6O.

=

The mass of 1 molecule of ethanol, C2H6O, is 7.650 x 10 -23 g.

46.07 g C H O1 mol C H O

2 6

2 6

7.650 x 10 -23 g/molecule of C2H6O

1 mol6.022 x 10 molecules23


Percent Composition Composition can be given by the mass of each element

relative to the total mass of the compound = Mass Percentage

Mass percentage N in NH3 =

= x 100 % = Mass percentage H in NH3 =

= x 100 % =

mass of H in 1 mol of NHmass of 1 mole of NH

3

3

14.01 g N17.030 g NH3

mass of N in 1 mol of NHmass of 1 mole of NH

3

3

3.024 g H17.030 g NH3

82.27 %

17.76 %


Empirical & Molecular Formulas

Percentage composition can be used to determine a simplest or empirical formula

Empirical or simplest formulas show the simplest ratio of the numbers of atoms of each element in a substance.

Example, C6H6 is the molecular formula showing the numbers of C and H atoms in the molecule benzene. CH is the empirical formula showing the simplest ratio of atoms.

Therefore, to convert an empirical formula to a molecular formula we need a molar mass!


Example 3.10:

Eugenol has a molar mass of 164.2 g/mol and is73.14 % C and 7.37 % H with the remainder O. What are the molecular and empirical formulas for eugenol?


findmole ratio

Method:

% B y mol B

AxBy

% A x mol Ax mol Ay mol B}

ratios givesformulaconvert weight

percentage to moles


Answer:

Step 1: Find the number of moles of C and H in a 100 g sample of eugenol.

(73.14 g C) =

(7.37 g H) =

1 mol C12.011 g C

1 mol H1.008 g H

6.089 mol C

7.31 mol H


Answer:

Step 2: Find the number of moles of O in a 100 g sample of vanillin by difference.

100.00 g = (73.14 g C + 7.37 g H) + mass of O

mass of O = 100.00 g - (73.14 g C + 7.37 g H)

= 19.49 g O

19.49 g O =

1 mol O15.999 g O

1.218 mol O


Answer:Step 3: Calculate the ratio of moles = empirical formula.

= = 4.999

= = 6.00

= = 1.000

C4.999H6.00O

Mole CMole O

Mole HMole O

Mole OMole O

O mol 1.218C mol 6.089

O mol 1.218H mol 7.31

O mol 1.218O mol 1.218

C5H6Oempirical formula


Answer:Step 4: Determine the molecular formula from the

empirical formula and the molar mass.

M(empirical formula) = [5(MC) + 6(MH) + (MO)]

= [5(12.011) + 6(1.008) + (15.999]

= [60 + 6 + 16]

= 82 g/mol

Determined molar mass of Eugenol is 164 g/mol.

C10H12O2

empirical formula

C5H6Omolecular formula


X 2



Chapter Highlights

define molecules

definitions of molecular formula & structural formula definition of allotrope understand formation of cations & anions learn names & formula for polyatomic ions molar mass of compounds percent composition empirical formula

Determining & Using Formulas Empirical Formula can be determined by a

number of different experiments

Sn + I2 SnxIy


Example 3.11:

1.056 g of tin metal and 1.947 g of solid iodine are allowed to react in 100 mL of ethylacetate. After the reaction is complete (all the iodine has reacted), 0.601 g of tin is recovered.

What is the empirical formula of the product formed from the reaction between Sn & I2 in this experiment ?

Determining & Using Formulas


Answer:

Step 1: Calculate the mass of Sn that reacted with the I2.

(original mass of Sn) - (mass of Sn recovered after the reaction)

= (mass of Sn consumed in the reaction)

= (1.056 - 0.601) g = 0.455 g Sn


Answer:

Step 1: Find the number of moles of Sn and I2 used to create the sample of SnxIy.

(0.455 g Sn) =

(1.947 g I2) =

BUT, remember there are 2 atoms of I in each molecule of I2,

therefore, moles of I = 2 x (7.671 x 10-3) =

1 mol Sn118.7 g Sn

1 mol I253.81 g I

2

2

3.83 x 10-3 mol Sn

7.671 x 10-3 mol I2

1.534 x 10-2 mol I


Answer:

Step 3: Calculate the ratio of moles = empirical formula.

= = 4.01

= = 1.00

Mole IMole Sn

Mole SnMole Sn

1.534 x 10 mol I3.83 x 10 mol Sn

-2

-3

3.83 x 10 mol Sn3.83 x 10 mol Sn

-3

-3

SnI4The empirical formula is therefore

What is the molecular formula ?


Example 3.12:

What mass of copper(I) sulfide, Cu2S, may be obtained

from 2.00 kg of copper ?

Using Chemical Formulas


Answer:

Step 1: Find the number of moles of Cu in 2.00 kg of copper.

(2.00 kg Cu) =

1000 g Cu1 kg Cu

31.5 mol Cu

1 mol Cu63.55 g Cu


Answer:

Step 2: Use the fact that there are 2 Cu atoms per molecule of Cu2S to determine the moles of Cu2S and then calculate the mass of Cu2S.

(31.5 mol Cu)

=

1 mol Cu S2 mol Cu

2

2510 g Cu2S

159.2 g Cu S1 mol Cu S

2

2


Question:In the laboratory, you weigh out 1.023 g of hydrated copper (II) sulfate, CuSO4 . xH2O (blue). After heating in a porcelain crucible you are left with 0.603 g of anhydrous cobalt(II) sulfate, CuSO4 (white). What is the value of x in CuSO4 . xH2O ?

Determining the Formula of a Hydrated Compound


Method:

Write out an equation to describe the reaction and assign values to known quantities and identify unknown quantities.

CuSO4 . xH2O + heat CuSO4 + xH2O

1.023 g 0.654 g + ? g


Answer:

Step 1: Find out the mass of water removed by heating.

(mass of hydrated compound) - (mass of anhydrous compound)

= mass of water

(1.023 g - 0.654 g) =0.369 g H2O


Answer:

Step 2: You want to know how many moles of H2O is associated with each mole of CuSO4; the

ratio! Simply convert the masses you have to moles!


OH g 18.05

OH mol 1 OH g 936.02

22

4

42 CuSO g 159.6

CuSO mol 1 OH g 654.0

2.05 x 10-2 mol H2O

4.10 x 10-3 mol CuSO4

Answer:

Step 3: The ratio of moles of H2O to moles of CuSO4 is x.

=

=

The formula for hydrated copper (II) sulfate is

4

2

CuSO molesOH moles

43-

2-2

CuSO mol 10 x 4.10OH mol 10 x 2.05

CoSO4 . 5H2O

4

2

CuSO mol 1OH mol 5.00


Question (Chapter 3, #105):

The weight percentage of oxygen in an oxide that has formula MO2 is 15.2 %. What is the element M ?

Some More Practice!!??!!


Answer:

Step 1: Since the formula is MO2 we know that 1 mol of the compound contains 2 mol of O = (2 x 16.00 g) = 32.00 g of O.

Since this is 15.2 % of the total…...

=

(32 x 100) = (X x 15.2)

= X =210.5 g/mol

32.00 g / mol O g / mol MO2X

15.2 %100 0%.

320015 2.


Answer:

Step 2: If the molar mass of the compound is 210.5 g/mol then …...

atomic mass of element M =

(210.5 g/mol - 32.00 g/mol) =

From the periodic table we identify the element M asHf (hafnium) with an atomic mass of 178.49 g/mol

178.5 g/mol


Textbook Questions From Chapter #3Molecular Formulas: 14, 16 Ions & Ion Charges: 20, 23, 24Ionic Compounds: 28, 32Naming Compounds: 36, 42, Molar Mass & Moles: 46, 48, 50, 52, 62Percent Composition: 64, 66 Empirical & Molecular Formulas 68, 70, 74, 79, 81Extras 91, 105, 107, 111, 120


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Chemistry-140 Lecture 6