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Lab Title:

Behavior of Gases: Molar Mass

of a VaporPurpose:

To determine the molar mass of a gas

from a knowledge of its mass,

temperature, pressure, and volume.

Pre-Lab Notes

• Scientists represent atoms by using

different colored circles, called a model.

• Each element is unique. Elements have their own physical and chemical characteristics.

• Below is an example of each element in “model format.” An element in model form is a single, colored circle.

• Elements are organized in a table called the periodic table of the elements.

• Different element have a different number of protonsand thus different physical & chemical characteristics.

• The atomic number indicates the number of protons.

• Most substances are chemical combinations of

elements (atoms) , called compounds.

• Below are some models of different compounds.

• Compounds are more than one colored circle

connected by a “stick” known as a bond.

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Diatomic Elements• Some elements are

naturally compounds!

• Poly (many)

• atomic (atoms)

• Polyatomic elements consist of several “like” atoms bonded together

• Memorize these:

– 7th Heaven Rule

– 7 Diatomic elements:H2, N2 , O2, F2, Cl2Br2, I2

ReviewElement or Compound?

Polyatomic Sulfur & Sulfur Polymer A)Element B)Compound C)Mixture

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A)Element B)Compound C)Mixture A)Element B)Compound C)Mixture

A)Element B)Compound C)MixtureCalculating AMU

• As you learned last year, there are different

types of atoms known as elements.

• Each element has a different mass and it is

called the amu = atomic mass units

• Another name for amu is molar mass

• Molar mass is obtained by summing the masses of the component atoms.

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Remember

7th Heaven Rule

On Your Periodic Table

Color Code Those 7 Elements

For example: NH3 has the following amu:

N = 1 atom x 14.01 amu = 14.01 amu

H = 3 atoms x 1.01 amu = 3.03 amu

total amu = 17.04 amu

or 17.04 molar mass

Ch 3.3-3.7

(Brown & LeMay)

• To find the percent of “anything”.

• First find the total “part” wanted.

• Next count the total number, the “whole”.

• Then divide the “part” over the “whole” and multiply by 100 %

Percent composition = Part x 100%

whole

This can be used for calculating different types

of percentage compositions.

What is the % N to % H of

Windex, which is ammonia, (NH3)?

Notice no numbers are given,

therefore use the compound’s amu.

part x 100%

whole

N = 1 atom x 14.01 amu = 14.01 x 100%

17.03

= 82.21 % N

H = 3 atoms x 1.01 amu = 3.03 x 100%

17.03

= 17.79 % H

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• The percentages may not always total to 100% due to rounding,

for example if you go to “1 decimal spot”

82.2% N

+ 17.7% H

99.9 % total

• But if you go to “2 decimal spots” you get closer to 100%

82.21% N

+ 17.79% H

100.00 % total

• Overall your numbers should

add up close to 100%

Calculating % Composition By

Given Masses• According to the Law of Constant

Composition, any sample of a pure

compound always consists of the same

elements combined in the same proportion

by mass.

Percent composition can be determined of each

element in a compound by its mass:

% Composition = part mass x 100%

whole mass

This formula can be applied by using the

formula of the compound or by experimental

mass analysis of the compound

A sample of butane (C4H10)--lighter fluid--

contains 288 g carbon and 60 g hydrogen.

Find %C and %H in butane

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Percent composition = Part x 100%

whole

288 g C + 60 g O = 348 g

Part Part whole

288 g C (C part) x 100 % = 82.8 % C

348 g (whole)

60 g O (O part) x 100% = 17.2 % O

348 g (whole) + ---------------

100.0 %

The Mole

Molar Mass

• The mass of one mole of atoms of any element is the molar mass which is numerically equal to the atomic mass unit (amu), but in grams…

• Molar mass = __g = 1 amu

1 mole

• Therefore CO2 has an amu = 44.01

or 44.01 g

1 mol

• Your and your lab partner will be assigned

either an element or a compound.

• You will need to place 1 mole of the

element or compound into a container.

• Make sure you do the following:

– label the bottle with the correct formula

– Label the bottle 1 mole

– You have a permanent marker in your lab

drawer.

Now you & your lab partner

draw “molecular level” pictures

of the mole containers.

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Using The Mole Map Converting

from grams to moles

EXAMPLE: A student weighs out 88 grams

of solid CO2 (dry ice), how many moles

does the student have?

First, find the amu of CO2 :

C = 1 atom x 12.01 amu = 12.01 amu

O = 2 atoms x 16.00 amu = 32.00 amu

CO2 total amu = 44.01 amu

CO2 = 44g

1 mol

• Then convert from grams to moles

• Use your mole map…

Using The Mole Map

88 grams CO2

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(What’s Given) x 1 mole

(molar mass)

88 g CO2 x 1 mole = 2 moles CO2

44 g

Converting To Particles

• A particle can

be defined as:

– an atom,

– a molecule,

– or formula unit

(ionic

compounds)

• Next, convert from mol to molecules:

• Use your mole map…

2 mole CO2

(moles calculated) x (Avogadro’s #)

1 mole

2 mole CO2 x 6.02 x 1023 molecules

1 mole

= 1.204 x 1024 molecules CO2

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• Chalk is calcium carbonate (CaCO3).

• I want you to calculate the number of moles and

the number of particles of CaCO3 it takes to

write your FULL name (First, Middle, & Last)

on a piece of paper.

Avogadro’s Law

Ch 10.3

Avogadro’s

Law

• In 1811 Italian chemist

Avogadro postulated equal

volumes of gases at the

same temperature and

pressure contains the same

number of “particles”

V = k n

n = moles,

k = proportionality constant

• At constant temp and press,

volume is proportional to

moles

• Again: PV = nRT

Avogadro’s Law

• One mole (6.02x1023

particles) of ANY ideal

gas at standard

temperature and

pressure, (STP = 0o C,

1 atm) occupies a

volume of 22.4 L

• 22.4 L is about the

volume of a basketball.

• You and your partner calculate the volume (L) at

STP (0 degrees C and 1 atm pressure) for 1

mole.

PV = nRT Or V = nRT

P

• Note: You have use the units Kelvin

• Note: R constants on your AP “Cheat Sheet”

• Which one to use?

• Hint: look at units

V = nRT =

P

(1.000mol)(0.08206 L•atm/K•mol)(273.2 K)

1.000 atm

V = 22.42 L

Does this look familiar???

• This volume is the molar volume of an ideal

gas (0°C and 1 atm)

• The conditions 0°C and 1 atm, are called

standard temperature and pressure

(abbreviated STP)

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It Does NOT Matter The Type of Gas

(at STP)• AT STP:

• To go from

moles � L or

L � moles we will

be using the “top

section” of the

mole map.

• A reaction produces 893

Liters of air, how many

moles at STP?

• At STP how many Liters

in 3.45 moles of O2?

The Ideal Gas Equation

10.4

The Ideal Gas Law

• The ideal gas law: PV = nRT

• Where R is the combined proportionality constant is called the Universal Gas Constant:

• R = 0.08206 L • atm

K • mol

• A gas that obeys this equation is said to behave ideally.

• The ideal gas equation is best regarded as a limiting law—it expresses behavior that real gases approach at low pressures and high temperatures (an ideal gas is a hypothetical substance).

• Most gases obey the ideal gas equation closely enough at pressures below 1 atm.

• The Universal Gas Constant can be calculated by

using various units, therefore it is important that

you note the units in the question being asked

and therefore the correct gas constant.

• R = 0.0821 L • atm

K • mol

• R = 62,400 mL • mm Hg

K • mol

• Some conversions to consider:

• L � mL, atm �mmHg, degrees C �K

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• Conversions: 1000 mL = 1 L, 1 atm = 760 mmHg,

you either add 273 or subtract 273 for

Celsius/Kelvin.

• A gas sample of 0.896g occupies a volume of 524

mL at 730 mm Hg and 28 degrees Celsius. What

is the number of moles moles? What is the Molar

Mass (MM)?

NOTE: Which constant to use?

• R = 0.0821 L • atm

K • mol

• R = 62,400 mL • mm Hg

K • mol

• You can use either one, but the second one

involves less conversions…(note mol = n)

• R = 62,400 mL • mm Hg

K • mol

V = 524 mL, P =730. mm Hg,T =28 degrees Celsius

PV = nRT, thus PV =n AND 28 + 273 = 301 K

RT

(730. mm Hg)(524 mL) = n

(62,400 mL • mm Hg/ K • mol)(301 K)

0.0204 mol = n

• MM = g/mol

• One can identify an unknown volatile gas by

using the Universal Gas Law and the mass of the

unknown.

• A gas sample of 0.896g occupies a volume of

524 mL at 730. mm Hg and 28 degrees Celsius.

What is the number of moles moles? What is the

Molar Mass (MM)?

• 0.0204 mol = n and 0.896 g of the sample:

• 0.896 g /0.0204 mol = 43.9 g/mol

• PV = nRT and Density = mass/volume

• A short cut, “Moles put dirt over their pee”

• MM = DRT = mass RT

P V P

Ch 3.6

Mole-Mole Relationships Stoichiometry is stoicheion meaning

elements and metron meaning to

measure...

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Mole-Mole Relationships

• In order to calculate an amount of reactant or

product at the end of a reaction you will need to

do the following:

1. Balance the

reaction.

2. Use the mole ratio

from the balanced

equation to

compare the moles

“known” to the

moles “unknown.”

3. Multiple/divide.

Using Mole RatiosPhosphorus is placed in a flask of chlorine gas, heat

and light is given off forming phosphorus trichloride:

If 0.06 moles of Cl2 gas how many moles of PCl3

produced?

P4 + 6 Cl2 --> 4 PCl3

Use Your Stoichiometry Map

If 0.06 mol of Cl2 gas how many mol of PCl3 produced?

P4 + 6 Cl2 --> 4 PCl3

0.06 mols ? mols ?

If 0.06 mol of Cl2 gas how many mol of PCl3 produced?

P4 + 6 Cl2 --> 4 PCl3

0.06 mols ? mols ?

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Now You Try

Ammonia (NH3) is used in huge quantities as a fertilizer. It is manufactured by combining nitrogen and hydrogen according to the following equation:

N2 + 3 H2 � 2 NH3

How many moles of NH3 can be made from 1.30 mol H2?

N2 + 3 H2 � 2 NH3

1.30mol ? mol

1.30 mol H2 x 2 mol NH3 = 0.867 mol NH3

3 mol H2

Mass-Mass Calculations

Steps for Calculating the Masses of Reactants & Products in Chemical Reactions

STEP 1 Balance the equation for the reaction.

STEP 2 Convert the masses of reactants or products to

moles

STEP 3 Use the balanced equation to set up the appropriate

mole ratio(s).

STEP 4 Use the mole ratio(s) to calculate the number

of

moles of the desired reactant or product.

STEP 5 Convert from moles back to mass.

How many grams of Cl2 will react with 1.24 g of

P4?

P4 + 6 Cl2 � 4 PCl3

1.24 g ? g

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P4 + 6 Cl2 � 4 PCl3

1.24 g ? g

g P4 � mol P4 � mol: mol ratio � g Cl2

1.24 g P4 x 1 mol P4 = 0.0100 mol P4

123.88 g

0.0100 mol P4 x 6 mol Cl2 = 0.0601 mol Cl2

1 mol P4

0.0601 mol Cl2 x 70.90 g Cl2 = 4.26 g Cl2

1 mol Cl2