chemical quantities: the mole chem i / ih : chapter 10

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CHEMICAL QUANTITIES: THE MOLE chem i / Ih : Chapter 10. MEASURING MASS. A mole is a quantity of things, just as… 1 dozen= 12 things 1 gross = 144 things 1 mole= 6.02 x 10 23 things “Things” usually measured in moles are atoms, molecules, ions, and formula units. - PowerPoint PPT Presentation

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MEASURING MASS

A mole is a quantity of things, just as…

1 dozen = 12 things1 gross = 144 things1 mole = 6.02 x 1023

things “Things” usually measured in moles

are atoms, molecules, ions, and formula units

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You can measure mass, or volume, or you can count pieces

We measure mass in grams

We measure volume in liters

We count pieces in MOLES

3

A MOLE… is an amount, defined as the

number of carbon atoms in exactly 12 grams of carbon-12

1 mole = 6.02 x 1023 of the representative particles

Treat it like a very large dozen 6.02 x 1023 is called:

Avogadro’s number

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Similar Words for an amount: Pair: 1 pair of shoelaces = 2 shoelaces

Dozen: 1 dozen oranges = 12 oranges

Gross: 1 gross of pencils= 144 pencils

Ream: 1 ream of paper= 500 sheets of paper

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What are Representative Particles (“RP”)?

The smallest pieces of a substance:1. For a molecular compound: it is

the molecule.2. For an ionic compound: it is the

formula unit (made of ions)3. For an element: it is the atom

Remember the 7 diatomic elements? (made of molecules)

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Practice Counting Particles How many oxygen atoms in the following?1. CaCO3 3 atoms of oxygen2. Al2(SO4)3 12 (4 x 3) atoms of oxygen

How many ions in the following?1. CaCl2

3 total ions (1 Ca2+ ion and 2 Cl1- ions)2. NaOH

2 total ions (1 Na1+ ion and 1 OH1- ion)3. Al2(SO4)3

5 total ions (2 Al3+ + 3 SO4 ions)

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CONVERSION FACTOR

RPs x ____1 mole___ = MOLES

6.02 x 1023 RPs

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EXAMPLES: ATOMS MOLES

How many moles of B are in 3.15 x 1023 atoms of B? Conversion: 1 mole B = 6.02

x 1023 atoms B(b/c the atom is the RP of

boron)

1 mole B

3.15 x 1023 atoms of B

6.02 x 1023 atoms B

= 0.523 mole

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EXAMPLES: MOLES ATOMS

How many atoms of Al are in 1.5 mol of Al? Conversion: 1 mole = 6.02 x 1023

atoms

1.5 mol of Al6.02 x 1023 atoms Al

1 mole Al

= 9.03 x 1023 atoms of Al

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CAUTION: Identify RPs Carefully!

See next slide!

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EXAMPLES: MOLECULES MOLESHow many atoms of H are there in 3 moles of H2O? (HINT: Are atoms the RP for water?)Conversions:1 mole = 6.02 x 1023 molecules

(b/c molecules are the RP for H2O)

3 moles of H2O 6.02 x 1023 molec H2O

1 mole H2O

2 atoms H

1 H2O molecule

= 3.612 x 1024 atoms H

H2O molecule = 2 atoms of Hydrogen

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MOLAR MASSDef: The mass of a mole of representative particles of a substance.

Each element & compound has a molar mass.

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MOLAR MASS OF AN ELEMENTDetermined simply by looking at the periodic table

Molar mass (g) = Atomic Mass (amu)

Ca20

40.08

* Thus, 1 mol Ca = 40 g

1 atom of Ca weighs 40.08 amu1 mole of Ca atoms weighs 40.08 grams

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MOLAR MASS FOR COMPOUNDS

To calculate the molar mass of a compound, find the number of grams of each element in one mole of the compound

Then add the masses within the compound

Example: H2O

H= 1.01 2 (1.01) + 1 (15.999)= 18.02 g/mol

O= 15.99915

SOME PRACTICE PROBLEMS How many atoms of O are in 3.7 mol of

O? 2.2 X 1024 atoms of oxygen

How many atoms of P are in 2.3 mol of P? 1.4 x 1024 atoms of phosphorus

How many atoms of Ca are there in 2.5 moles of CaCl2? 1.5 x 1024 atoms Ca

How many atoms of O are there in 1.7 moles of SO4? 4.1 x 1024 atoms of oxygen

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Remember!!!! The molar mass of any substance (in

grams) equals 1 mole This applies to ALL substance: elements,

molecular compounds, ionic compounds Use molar mass to convert between mass

and moles Ex: Mass, in grams, of 6 mol of MgCl2 ?

mass of MgCl2 = 6 mol MgCl2 92.21 g MgCl2 1 mol

MgCl2

= 571.26 g MgCl2

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VOLUME AND THE MOLE

Volume varies with changes in temperature & pressure

Gases are predictable, under the same physical conditions

Avogadro’s hypothesis helps explain:equal volume of gases, at the same temp and pressure contains equal number of particles

Ex: helium balloon

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MOLAR VOLUME Gases vary at different

temperatures, makes it hard to measure

To make it easier to measure gases, we use something called “STP” “Standard Temperature and Pressure” Temperature = 0° C Pressure = 1 atm (atmosphere)

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Molar Volume

At STP:1 mole, 6.02 x 1023 atoms, of any gas has a volume of 22.4 L

1 mole gas = 22.4 L gas

Use it to convert between # of moles and vol of a gas @ STP Ex: what is the volume of 1.25 mol of sulfur

gas?

1.25 mol S 22.4 L S = 28.0 L

1 mol S 20

MOLAR MASS FROM DENSITY Different gases have different densities Density of a gas measured in g/L @ a specific

temperature Can use the following formula to solve :

grams = grams X 22.4 Lmole L 1 mole Ex: Density of gaseous compound containing

oxygen and carbon is 1.964 g/ L, what is the molar mass?

grams = 1.964 g X 22.4 L then you solve

mole 1 L 1 mole = 44.o g/mol

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Atoms, molecules, etc. 22

Molarity

Def: the concentration of a solution. How many moles/liter

Can be used to calculate # of moles of a solute

Ex: Household laundry bleach is a dilute aqueous solution of sodium hypochlorite (NaClO). How many moles of solute are present in 1.5 L of 0.70 M NaClO?

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Calculating Percent Composition of a Compound Like all percent problems: a part ÷ the

whole1. Find the mass of each of the components

(the elements)2. Next, divide by the total mass of the

compound3. Then X 100 % = percent

Formula:% Composition = Mass of element X 100%

Mass of compound

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Method #1: % Comp When Actual Masses are Given

A compound is formed when 9.03 g of Mg combines completely with 3.48 g of N.

What is the percent composition of the compound?1. First add the 2 mass of the 2 compounds to

reach the total mass 9.03 g Mg + 3.48 g N = 12.51 g Mg3N2

2. Find the % of each compound

% Mg= 9.03 g Mg X 100 = 72.2 % 12.51 g Mg3N2

% N= 3.48 g N X 100 = 27.8 % 12.51 g Mg3N2

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Method 2: % Comp When Only The Formula is Known Can find the percent composition of a

compound using just the molar mass of the compound and the element

% mass=mass of the element 1 mol cmpd X100%

molar mass of the compound Example:

Find the percent of C in CO2

12.01 g C X 100 = 27.30% C

44.01 g CO2 26

Using % Composition Can use % composition as a conversion factor

just like the mole After finding the % comp. of each element in

a cmpd. can assume the total compound = 100g

Example: C= 27.3% 27.3 g C O= 72.7 % 72.7 g O

In 100 g sample of compound there is 27.3 g of C & 72.7 g of O

How much C would be contained in 73 g of CO2?

73 g CO2 27.3 g C = 19.93 g C

100 g CO2

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EMPIRICAL FORMULAS Empirical formulas are the

lowest WHOLE number ratios of elements contained in a compound

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REMEMBER… Molecular formulas tells the actual

number of of each kind of atom present in a molecule of the compound

Ex:H2O2 HOMolecular Empirical Formula Formula

CO2 CO2

Molecular Empirical For CO2 they are the same

Formula Formula

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Formulas for ionic compounds are ALWAYS empirical (the lowest whole number ratio = can not be reduced)

Examples: NaCl MgCl2 Al2(SO4)3 K2CO3

Simplest whole number ratio for NaCl

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A formula is not just the ratio of atoms, it is also the ratio of moles

In 1 mole of CO2 there is 1 mole of carbon and 2 moles of oxygen

In one molecule of CO2 there is 1 atom of C and 2 atoms of O

Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio)

Molecular: H2O C6H12O6 C12H22O11

(Correct formula)

Empirical:

(Lowest whole H2O CH2O C12H22O11

number ratio)

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STOP HERE!!!

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CALCULATING EMPIRICAL We can get a ratio from the percent

composition 1.Assume you have a 100 g sample the

percentage become grams (75.1% = 75.1 grams)

2.Convert grams to moles3.Find lowest whole number ratio by

dividing each number of moles by the smallest value

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Example calculations Calculate the empirical formula of a

compound composed of 38.67 % C, 16.22 % H, and 45.11 %N

Assume 100 g sample, so 38.67 g C x 1 mol C =

12.0 g C 16.22 g H x 1 mol H =

1.0 g H 45.11 g N x 1 mol N =

14.0 g N *Now divide each value by the smallest value

3.22 mole C

3.22 mole N

16.22 mole H

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…Example 1

The ratio is 3.22 mol C = 1 mol C 3.22 mol N 1

mol N The ratio is 16.22 mol H = 5 mol H

3.22 mol N 1 mol N

C1H5N1 which is = CH5N

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MORE PRACTICE

A compound is 43.64 % P and 56.36 % O

What is the empirical formula? PO3

Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O

What is its empirical formula?

C4H5N2O

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EMPIRICAL TO MOLECULAR

Since the empirical formula is the lowest ratio, the actual molecule would weigh more

Divide the actual molar mass by the empirical formula mass – you get a whole number to increase each coefficient in the empirical formula

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EXAMPLE Caffeine has a molar mass of 194 g, what

is its molecular formula?1.Find the mass of the empirical formula,

C4H5N2O

2.Divide the molar mass by the empirical mass:194.0 g/mol =97.1 g/mol

3.Now multiply the entire empirical formula by 2

2(C4H5N2O) =

final molecular formula

2

C8H10N4O2

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