chem 16 (unit 2 lecture)

General Chemistry 1Chemistry 16 Lecture Unit 2

MOLE CONCEPT

Mole: Latin word for pileEquivalent to 6.02 x 1023 particles or the Avogadro’s number

Molar mass is equivalent to the mass of one mole of a substance.Ex.: 1 Mole H = 1.01 grams/mole

1 Mole C = 16.00 grams/mole

Example 1. How many grams of Zinc are in 2.45 moles?

mass∈gramsof Zn=2.45moles Zn× 65.37g /molof Zn1mole

mass∈gramsof Zn=160 grams

or simply,gramsof Zn=2.45moles×65.37g /molmass∈gramsof Zn=160 grams

mass∈grams=no .of moles×Molar Mass

Example 2. How many moles of Fe/ Iron are in 150.0 g?

n=mass∈gramsMolar Mass

= 150.0 g55.85 g/mol

=2.686moles of Fe


GRAMS MOLES NO. OF PARTICLES

Example 3. How many atoms are present in 25.2 grams of Barium?


= 25.2 grams137.27 g/mol

=0.183moles Ba

no .of atoms=no .of moles × Avogadr o' snumberno .of atoms=nNno .of atoms=0.183moles (6.02×1023atoms)no .of atoms=1.10×1023atoms of Ba

no .of atoms=no .of moles × Avogadr o' snumber

SOLVE:What is the mass in grams of 2.305 x 1020 atoms of Iodine?

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MOLAR MASSES OF MOLECULAR MASS

O2 O + O AtomsO2 +

O2

O + O + O + O Atoms

1 mole of O2 = 2(6.02 x 1023) atoms/ mol= 12.04 x 1023 atoms/ mol

Mass of O2 = 2(16.00) grams/mol= 32 g/mol

1 mole of P4 of Phosphorus = 4(6.02 x 1023) atoms/ mol or 2.408 x 1024 atoms= 4 moles of P atoms= 4(30.37) mass in grams or 121.48 grams

Example 4. What is the molar mass of ethyl alcohol, CH3CH2OH?C : 2 x 12.01

amu= 24.02 amu

H : 6 x 6.06 amu

= 6.06 amu

O : 1 x 16.00 amu

= 16.00 amu

46.08 amu Molecular Mass

46.08 g/mol

Molar Mass

Example 5. What is the mass in grams of a molecule of CH3CH2OH?

mass∈grams= 46.08 grams

6.02×1023molecules(Equivalent¿1mole)=7.6454×10−23 grams

Example 6. For a sample of CO2, 24.6 moles of O2 was found to combine with carbon. How many moles of carbon is combined with this amount of oxygen?

x 24.6 moles

C + O2 CO2

1 mole

2 moles

Let x be the number of C moles combined with O2

1moleC2moles of O

= x24.6molesof O

x=24.6moles of O(1moleC )

2moles of O=12.3moleC

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Example 7. In commercial fertilizers, phosphorus is present as P2O5, phosphorus pentaoxide. What is the mass in grams of O2 combined with 21.2 grams of P?

Grams of P

Moles of P

Moles of O

Grams of O

Molesof P= 21.2 g30.974 g /mol

=0.684molesof P

Using ratio and proportion, let x be equal to the number of moles of O2

5molesO2

2moles P= x0.684 g /mol

x=1.71molesO2

Mass of Oxygen=Molar Mass×MoleMass of Oxygen=16.00×1.71Mass of Oxygen=27.36 g

SOLVE:Find the molar mass of Table Salt, NaCl and Calcium hydrogen phosphate, Ca (HPO4)2

Example 8. How many atoms of H are there in 6.08 grams of (NH4)2SO4?

Grams of (NH4)2SO4

Moles of (NH4)2SO4

Moles of H

No. of Atoms of H

Solve for the Molar Mass: N : 2 x 14.01

g= 28.02 g

H : 8 x 1.01 g

= 8.08 g

S : 1 x 32.06 g

= 32.06 g

O : 4 x 16.00 g

= 64.00 g

132.16 g/mol

Solve for the number of moles of (N H 4)2SO4


= 6.08 g132.16 g/mol

=0.0460moles of (N H 4)2SO4

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Ratio of (NH4)2SO4 to H = 1:8Let x be the no. of moles of H

x0.046moles (N H 4)2SO4

= 8moles H1mole (N H 4)2SO4

x=0.046moles(N H 4)2SO4×8moles H

1mole (N H 4 )2SO4

x=0.368moles of H

Find the number of atoms of Hno .of atoms=no .of moles × Avogadr o' snumberno .of atoms=0.368moles × (6.02×1023moles /atomof H )no .of atoms=2.22×1023 atoms of H

PERCENTAGE COMPOSITIONS

%Element= Mass of ElementTotalMass of Compound

×100%

Example 9. What is the percentage composition of Na2CO3 (Sodium Carbonate)?

Na : 2 x 23.00 g = 46.00 gC : 1 x 12.01 g = 12.01 gO : 3 x 16.00 g = 48.00 g

Molar Mass

=

106.01 g/mol

%Na= 46.00 g106.01 grams

×100%=43.39%

%C= 12.01g106.01grams

×100%=11.33%

%O= 48.00g106.01 grams

×100%=45.28%

SOLVE:A sample of a liquid with a mass of 8.657 g, was decomposed into its element and gave 5.217 g C, 0.9620 g H and 2.478 g O. What is the percent composition of the compound?

CLASSIFICATION OF CHEMICAL FORMULA

1. Empirical Formula (EF): or the simplest formula, gives a relative number of atoms in each element of a compound.

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Ex.: HO, CH2, CH

2. Molecular Formula (MF): gives the actual number of atoms of each element in a molecule of a compound.Ex.: H2O2 EF = HO

C2H4 EF = CH2

3. Structural Formula: shows how atoms are linked together in a molecule of a compound.Ex.: CH3COOH

Determining the Empirical Formula

Example 10. An unknown sample was found to be with 3.60 g C, 0.61 g H and 38.07 g I. Find the empirical formula.

moles of C=mass∈gramsMolar Mass

= 3.60g12.01g /mol

=0.300moles C

moles of H=mass∈gramsMolar Mass

= 0.61 g1.01 g/mol

=0.604moles H

moles of I=mass∈gramsMolar Mass

= 38.07 g126.90 g/mol

=0.300moles I

C 0.3000.300

H 0.6040.300

I 0.3000.300

=C H 2 I , empirical formula

SOLVE:A compound was found to have 73.9% of C, 4.14% of H and 21.89% of O. What is the

empirical formula of the compound? (Hint: Let the total percentage be equivalent to a 100 grams)

Determining the Molecular Formula

Example 11. For a compound with an empirical formula of CH2I, determine the molecular formula if the molecular mass is 281.86 grams.

Find the EF Mass:

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Molecular FormulaMassEmpirical Mass

=281.86 g140.93 g

=2

Conclusion: Molecular Mass is twice the Empirical Mass

MF = 2 (EF)MF = 2 (CH2I)MF = C2H4I2

Example 12. In the analysis or organic compounds, a common technique used is to burn the compound and collect the CO2 and H2O formed.

A Compound contains C, H and O. Combustion of 21.36 mg yields 32.02 mg of CO2

and 8.74 mg of H2O. The molecular mass of the compound is 176.1 amu. What are the EF and MF?

%C∈CO2=12.01g

44.01 grams×100%=27.29%

%H∈H 2O= 2.02g18.02grams

×100%=11.21%

Mass of C∈ product=32.02mg× 27.29%mgC100mgCO2

=8.74mgof C

Mass ofH∈ product=8.74mg× 11.21%mgC100mgH 2O

=0.98mgof H

Mass of O=21.36mgcompound−(8.74mgC+0.98mg H )Mass of O=11.64mg

millimoles of C=mass∈gramsMolar Mass

= 8.74 g12.01 g/mol

=0.728millimolesC

millimoles of H=mass∈gramsMolar Mass

= 0.980 g1.01g /mol

=0.970millimoles H

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C : 1 x 12.01 g

= 12.01 g

H : 2 x 1.01 g

= 2.02 g

I : 1 x 126.9 g

= 126.9 g

EF Mass

=

140.93 g


millimoles of O=mass∈gramsMolar Mass

= 11.64 g16.00g /mol

=0.728millimoles H

C 0.7280.728

H 0.9700.728

O 0.7280.728

=3 (C¿¿1H 1.33O1)3=C3H 4O3 , empirical formula¿

Find the EF Mass:

Molecular FormulaMassEmpirical FormulaMass

=176.1 g88.07 g

=1.99∨2

MF = 2 (EF)MF = 2 (C3H4I3)MF = C6H8I6

GASESFrom the Flemish word, “chaos”, dubbed by Van Helmonth

Properties of Gases:1. Large volumes (low density)2. Volume changes to the volume of the container3. Shape changes to the shape of container4. Easily compressed5. Rapidly mix6. Exert pressure7. Expands when heated

Four variables that will describe a gas:1. Pressure, P2. Volume, V3. Temperature, T4. Number of moles, n

PRESSURE is a unique property of gas compressibility or expandability is due to the collision of walls of the container

Pressure = Force per unit area or P = F/A

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C : 3 x 12.01 g

= 36.03 g

H : 4 x 1.01 g

= 4.04 g

O : 3 x 16.00 g

= 48.00 g

EF Mass

=

88.07 g


Atmospheric Pressure: In 1642, Torricelli discovered that there is pressure exerted in a Mercury or Hg in disk. Under normal conditions at sea level, it has a height of 760 mm of Hg or 1 atm. Height of Hg varies with altitude.

1 standard atmosphere = 760 mm H at sea level1 standard atmosphere = 760 torr

SI Unit for pressure: Pascal or Pa1 Pa = 1 newton/meter2

1 atm = 101, 325 Pa1 atm = 101.325 KPa

DIFFERENT GAS LAWS

I. BOYLE’S LAW: by British Robert BoyleAt constant temperature and moles, as pressure increases, volume decreases. There is an inverse relationship as a law of nature.

V ∝ 1P→PV=k

Note: It is an approximation and most accurate in low pressure and high temperature. Ideal Gas: gas which yield with constant P and V as product no matter what the pressure is. It behaves perfectly.

P1V1 = P2V2

Applications: Breathing Mechanisms

Example 13. A sample of gas occupies a volume of 250 mL at a pressure of 760 torr. What will the gas occupy if its pressure is reduced to 700 torr.

P1V1 = P2V2

(760 torr) (250 mL) = (700 torr) (V2)

V 2=(760 torr )(250mL)

700torr=271.46mL

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II. CHARLES’ LAW: by Jacques Charles, first to fill a balloon with Hydrogen and solo balloon fly in ParisAt Constant Pressure and no. of moles, Volume TemperatureAs Volume Increases, Temp, inc. (T KE)

Absolute Zero: lowest possible temp.; represents the zero point in OKKelvin Scale: Absolute Temperature Scale.

00C = 273 K K = 0C + 273

V1 =

V2

T1

T2

Standard Temperature and Pressure (STP): 00C or 273 K ; 1 atm or 760 torr

SOLVE:1. A 10.0 mL sample at 250C is cooled at constant pressure at 00C. What is the new

volume?2. At what pressure must a 300 mL sample of oxygen at 200C be cooled at constant

pressure so that the volume will be 200 mL.

CHEMICAL REACTIONS BETWEEN GASES2 V H2 + 1 V O2 = 2 V H2O1 V N2 + 3 V H2 = 2 V NH3

1 V H2 + 1 V Cl2 = 1 V HCl

Volume is not always conserved. Reactant volume ≠ Volume of the ProductVolume combined are in small whole numbers.

III. GAY LUSSAC’S LAW OF COMBINING VOLUMESThe volumes of gaseous substances that are conserved and produced in a chemical reaction are in ratios of small whole numbers, provided the volumes are measured under, the same conditions of temperature and pressure

Example 14. What volume of Oxygen will measure 250C and 760 torr is required to rest 2.0 L of Methane measured under the same condition of T and P.

CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)

Let x = equal to the volume of Oxygen

x2.0Lof Methane

= 2 LOxy gen1.0 Lof Methane

x=(2.0 LMethane )(2Loxygen)

(1.0LMethane)=4.0LOxygen

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IV. AVOGADRO’S PRINCIPLEUnder conditions of constant T and P, equal volume of gases contain equal number of molecules.

Since equal number of moles = equal number of moles VV n, at constant T and PV = kn

V1 =

V2

N1

N2

Constant V at STP: 22.4 L, Molar Volume

Example 15. What volume will 14.0 g of Nitrogen gas occupy at STP?

n=14.0g N2×1mole N2

28.0 g/mol=0.500molesof N2

Let y = volume of Nitrogen gas

y0.500moles

= 22.4 L1mole

y=(22.4 L )(0.500moles)

(1mole)=11.2moles

V. IDEAL GAS EQUATION: P, V, T, nBoyle’s Law: V 1/P at constant n and TCharles’ Law: V T at constant n and PAvogadro’s Law: V n at constant T and P

Combining:

V ∝n 1pT∨PV

T=nR

Ideal Gas EquationPV = nRT, R as a universal gas constant

R=0.082 atm∙ Lmole ∙ K

at STP

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Using the Ideal Gas Equation, derive R, the universal gas constant.

SOLVE:1. What volume will 0.89 moles of Nitrogen gas occupy at 300C and a pressure of

0.72 atm?2. Calculate the no. of moles of gas in 50.0 L at STP.

Combined Gas Law:P

1V1 =P

2V2

T1 T2

SOLVE:A sample of He is at 760 torr, 200 mL at 350C. What must be the temperature if

pressure is 740 torr, and volume is 300 mL?

¿ the IdealGas Equation :PV=nRT

Since:

moles=mass∈gramsMolar Mass

PV=mass∈gramsMolar Mass

RT

Molar Mass=Mass RTPV

SOLVE:A 2.00 L flask contains 3.11 grams of gas at 684 torr and 230C.

Find the Molar Mass.

Molar Mass=Mass RTPV

Molar Mass=MassV

RTV

Molar Mass=d RTV

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d=Molar MassPRT

SOLVE:Find the density of a gas is 1.34 g/L at 250C and 760 torr.

VI. GRAHAM’S LAW OF EFFUSIONDiffusion: Physical Properties of Gases (Ex.: Fragrance of a perfume)Effusion: Gas pressure escape from high to low pressure and escape through a small opening.

Rate of Effusion∝√ 1dRate of Effusionof Gas ARate of Effusionof GasB

=√ Density of Gas BDensity of Gas A

Rate of Effusionof Gas ARate of Effusionof GasB

=√ Molar Mass of Gas BMolar Mass of Gas A

Example 16. Gas A, Carbon Dioxide has a molar mass of 44 g/mol while Gas B, NH3 has 17 g/mol.

√ 4417=1.6 ,NH 3 effuse faster by1.6×thanCO2

QUIZ: What volume will 1.02 g of ammonia (NH3) occupy at 30 0C if its pressure is 720 torr.

VII. DALTON’S LAW OF PARTIAL PRESSUREDeveloped by John Dalton. In a mixture of gas, each gas in the mixture will contribute to the pressure. It exerts a partial pressure independent of the other pressure of the different gases.

Pt=Pa+Pb+Pc…+Pn

Where: Pt is the total partial pressure

Pn is the partial pressure of the component gases.

Example 17. A misture of N2, O2 and CO2 are confined in a container. What is the total pressure given the partial pressure of N2, 200 torr; O2, 250 torr; CO2, 300 torr?

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Solution: (200 + 250 + 300) torr = 750 torr

P1=n1RT

V, P2=

n2RT

V, P3=

n3 RT

V; Pressure depends uponthe number of moles present

Example 18. A mixture of gases in a 5.0 L vessel at 30 0C consisted of a 0.025 moles N2, 0.030 moles H2 and 0.050 moles O2. Calculate the total and partial pressures.

Given: V = 5.0 LT = (30 + 273) K = 303 Kn1 of N2 = 0.025 molesn2 of H2 = 0.030 molesn3 of O2 = 0.053 moles

Pt=(0.025+0.030+0.053 )moles (0.082 L∗atm

mole∗K )(303K )

5.0 L=0.54 atm

PN2=

(0.025moles )(0.082 L∗atmmole∗K )(303K )

5.0 L=0.124atm

PH 2=

(0.030moles)(0.082 L∗atmmole∗K )(303K )

5.0 L=0.149atm

PO2=

(0.053moles )(0.082 L∗atmmole∗K )(303K )

5.0L=0.263atm

VIII. MOLE FRACTION (x)Mole fraction is the fraction of the total number of moles of a gas

X a=Na

N total

Where: X ais themole fraction N total isthe total numberof moles of a gas

Na is theno .of moles of A

X N2= 0.025moles

(0.025+0.030+0.053 )moles=0.0250.108

=0.23

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X H 2=0.0300.108

=0.28

XO 2=0.0530.108

=0.49

TotalMole Fraction s hould be≅ 1.

Pa=Xa Pt

PN2=0.23 (0.54 atm )=0.124atm

PH 2=0.28 (0.54atm )=0.151atm

PO2=0.49 (0.54atm )=0.264atm

Pt=0.539∨0.54atm

IX. VAPOR PRESSUREExample 19. A student generates oxygen gas in the lab and collected it by water displacement method. She collects the gas at 25 0C until the level of water inside and outside the flask are equal. If the volume of the flask is 245 mL and atmospheric pressure is 758 torr. What is the partial pressure of oxygen in the “wet” gas mixture at 25 0C.

Patm=Poxygen+Pwater

Poxygen=Patm−Pwater

Given:P atm = 758 torrP water at 25 0C = 23.8 torr

Poxygen=758−23.8 torr=734 torr

Calculate the no. of moles and mass given the data above.

X. TEMPERATURE AND PRESSURE RELATIONSHIP: GAY – LUSSAC’S LAW

Pressure∝Kelvin temperatureP=ktPT

=k , constant

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P1 =

P2

T1

T2

SOLVE:The pressure in a hair spray can is 4.0 atm at 27 0C. If the can is thrown into fire,

what is the gas pressure when the can temperature increases to 402 0C?

XI. KINETIC MOLECULAR THEORY OF GASESBased on the ideal gas law, developed by Daniel Bernoulli in 1938.

Postulates:1. Gases are composed of particles such as atom or molecules.2. The particles are every small.3. The particles move very rapidly4. The forces of attraction between particles are extremely weak (moving

independently)5. Collisions are perfectly elastic. Kinetic Energy before and after collision is

constant at a given temperature. 6. As temp. of a gas increases, particles move very rapidly (T ∝KE)

PropertySTATES

Solid Liquid Gas1. Relative volumes of a

sample at different states

Large volume(Low density)

Small Volume(High density)

Small Volume(High density)

2. Dependence of a sample in the volume of a container

Volume changes to container

Fixed volume Fixed volume

3. Dependence of a shape of container

Shape changes to shape of container

Shape changes to filled part of container.

Fixed Shape

4. Ease of Easily compressed Nearly Nearly

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compression incompressible incompressible

5. Ease of moving Rapidly mixedMix more slowly

than gasesNegligible mixing

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chem 16 (unit 2 lecture)

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