THE MOLEThe Chemist’s Package

HOW DO WE GROUP THINGS?

Pairs, Dozen, ???

Chemists group chemicals into moles.

A mole of any chemical is the same amount. Just like a dozen is always 12…

FROM ONE TO ANOTHER…MOLES TO MOLES

We can change from one substance to another using a ratio. Usually a part to a whole

This ratio is a mole ratio. It converts from moles of one thing to moles of

another.

____moles X = ____moles Y

Ex. H2O : Part to Whole 2 moles H = 1 mole H2O 1 mole O = 1 mole H2O

HOW MANY ARE THERE?MOLES TO PARTICLES

Moles can also be converted to tell us how many particles there are. Particles can be atoms (atom), ions (ion),

molecules(molec), and formula units (fmu). There are always 6.022 x 1023 particles in a

mole. This number is known as Avagadro’s Number.

Ex. H2O: 1 mole H2O = 6.022 x 1023 molec H2O

Ex. Na: 1 mole Na = 6.022 x 1023 atoms Na

HOW MUCH IS IT?MOLES TO GRAMS

The mass of one mole is a chemical substance’s molar mass.

Each substance has its own molar mass that can be determined using the periodic table.

The molar mass of an element is equal to its atomic mass. Ex. Na: 1 mol Na = 22.99 g Na

The molar mass of a compound is equal to the sum of the atomic masses of the elements that are in the compound. Ex. H2O: 1 mole H2O = 18.02 g H2O (2H +

1O)

HOW BIG IS IT?MOLES TO VOLUME

The volume of a mole of gas under standard conditions is always the same.

This is not true under nonstandard conditions or for liquids and solids.

Standard Conditions of Temperature and Pressure (STP) are: 273 K and 101.3 kPa

At STP, 1 mole of gas has a volume of 22.4 L. This is called molar volume.

Ex. O2 at STP: 1 mole O2 = 22.4 L O2

Ex. CO2 at STP: 1 mole CO2 = 22.4 L CO2

CONVERSION FACTORS

These are all conversion factors. Mole to Mole – Mole Ratio Mole to Particle – Avagadro’s Number Mole to Mass – Molar Mass Mole to Volume – Molar Volume

MOLE (mol)

VOLUME (L)

MOLES OF ANOTHER

MASS (g)

PARTICLES(atoms, ions, molec, fmu)

Mole Ratio

Molar Volume

Avagadro’s Number

Molar Mass

MOLE MAP

PRACTICE PROBLEMSMOLE TO MOLE

How many moles of carbon are in 2.0 moles of sucrose, C12H22O11? Known: 2.0 moles sucrose Unknown: ? moles carbon Conversion Factor: Mole Ratio 1 mol C12H22O11 = 12 mol

C

 2.0 mol C12H22O11 12 mol C = 24 mol C

1 mol C12H22O11

How many moles of sucrose will contain 12 moles of hydrogen? K: 12 mol H UK: ? Mol C12H22O11

CF: Mole Ratio - 22 mol H = 1 mol C12H22O11

12 mol H 1 mol C12H22O11 = 0.55 mol C12H22O11

22 mol H

PRACTICE PROBLEMSMOLE TO PARTICLE

How many moles are equal to 2.41 x 1024 formula units of sodium chloride? K: 2.41 x 1024 fmu NaCl UK: ? mol NaCl CF: Avagadro’s Number - 1 mol NaCl = 6.022 x 1023 fmu NaCl

How many moles are equal to 9.03 x 1024 atoms of mercury? K: 9.03 x 1024 atoms Hg UK: ? mol Hg CF: 1 mol Hg = 6.022 x 1023 atoms Hg

How many atoms are equal to 4.50 moles of copper? K: 4.50 mol Cu UK: ? atoms Cu  CF: 1 mol Cu = 6.022 x 1023 atoms Cu

How many molecules are equal to 100.0 moles of carbon dioxide? K: 100.0 moles CO2 UK: ? molec CO2

CF: 1 mol CO2 = 6.022 x 1023 molec CO2

PRACTICE PROBLEMSMOLES TO PARTICLES (2 STEPS)

How many atoms of oxygen are in 3.65 moles of sucrose? K: 3.65 mol C12H22O11 UK: ? atoms O CF: Mole Ratio 1 mol C12H22O11 = 11 mol O CF: Avagadro’s Number 1 mol O= 6.022 x 1023 atoms O

 3.65 mol C12H22O11 11 mol O 6.022 x 1023 atoms O =

1 mol C12H22O11 1 mol O

= 2.42 x 1025 atoms O

How many atoms are in 1.00 mole of sucrose?

PRACTICE PROBLEMSMOLAR MASS

Determine the molar mass of the following compounds. Carbon dioxide

CO2 : 1 Carbon + 2 Oxygen = 12.01g + 2(16.00 g) = 44.01 g CO2

Sulfur trioxideSO3 : 1 Sulfur + 3 Oxygen =

BromineBr2 : 2 Bromine =

Sodium hydroxideNaOH : 1 Sodium + 1 Oxygen + 1 Hydrogen =

Barium nitrateBa(NO3)2 : 1 Barium + 2 Nitrogen + 6 Oxygen =

PRACTICE PROBLEMSMOLES TO MASS

Calculate the mass in grams of 0.250 moles of the following compounds.

K: 0.250 moles UK: ? grams Sucrose

  Sodium chloride

  

Potassium permanganate  

Calcium sulfide  

Lithium chlorate

PRACTICE PROBLEMSMOLES TO MASS

Calculate the number of moles in 100.0 grams of the following compounds.

K: 100.0 grams UK: ? moles Sucrose CF:

Sodium chloride CF:

   Potassium permanganate CF:

  

Calcium sulfide CF:  

Lithium chlorate CF:

PRACTICE PROBLEMSMOLES TO VOLUME

What is the volume in liters of 8.35 moles of sulfur trioxide gas at STP?

  K: 8.35 mol SO3 UK: ? L SO3

CF: 1 mol SO3 = 22.4 L SO3

What is the volume in liters of 0.750 moles of carbon dioxide gas at STP?

   K: 0.750 mol CO2 UK: ? L CO2

CF: 1 mol CO2 = 22.4 L CO2

   How many moles are there in 52.5 liters of oxygen gas at STP?  K: 52.5 L O2 UK: ? mol O2

CF: 1 mol O2 = 22.4 L O2

  How many moles are there in 15.0 liters of nitrogen gas at STP?

 K: 15.0 L N2 UK: ? mol N2

CF: 1 mol N2 = 22.4 L N2

MULTI-STEP PROBLEMS

Same conversions…just more than one at a time…

Don’t solve them any differently… Identify your known and unknown. Identify the conversion factors you need.

Let the Mole Map do the work for you… Solve.

MULTI-STEP PRACTICE How many grams of carbon dioxide are there in 4.20 L

of carbon dioxide at STP?  

What is the volume of 245 grams of water vapor at STP?   

How many formula units are there in 63.4 grams of barium hydroxide?

 

MULTI-STEP PRACTICE What is the volume of 6.21 x 1031 molecules of sulfur

trioxide gas at STP?

   What is the mass of 5.23 x 1018 atoms of gold?  

  How many molecules of silicon dioxide gas does it take

to fill a 6.0 L container at STP? 

MULTI-STEP PRACTICE How many atoms of carbon are there in 4.24 grams of carbon

tetrabromide?

   How many liters of carbon dioxide gas are there if you have

4.7 x 1020 atoms of oxygen?    If you have 3.62 x 1024 atoms of nitrogen, then what is the

maximum number of grams of dinitrogen pentoxide that can be formed?

EMPIRICAL FORMULAS

Empirical Formulas are the LOWEST whole number ratio of elements in a compound. Sometimes this is the same as the chemical

formula Examples:

C2H6 → CH3 C6H12O6 → CH2O

Empirical Formulas can be determined experimentally from the percent composition of a compound.

EMPIRICAL FORMULA DETERMINATION

STEPS

Assume you have 100 g of the substance (makes the math easier because everything is a straight percent). Consider the amounts you are given as being in

units of grams.

Convert the grams to moles for each element.

Find the smallest whole number ratio of moles for each element by dividing all values by the smallest value.

EXAMPLE: FIND THE EMPIRICAL FORMULA FOR A COMPOUND CONSISTING OF 63% MN AND 37% O

Solution for Finding the Empirical Formula Assuming 100 g of the compound, there would be 63 g Mn

and 37 g O Convert grams to moles. (Use molar mass from the periodic

table.) 63 g Mn × (1 mol Mn)/(54.94 g Mn) = 1.1 mol Mn 37 g O × (1 mol O)/(16.00 g O) = 2.3 mol O

Find the smallest whole number ratio by dividing the number of moles of each element by the number of moles for the element present in the smallest molar amount. (In this case there is less Mn than O, so divide by the number of moles of Mn) 1.1 mol Mn/1.1 = 1 mol Mn 2.3 mol O/1.1 = 2.1 mol O The best ratio is Mn:O of 1:2 and the formula is MnO2

The empirical formula is MnO2

EMPIRICAL FORMULA CALCULATIONWORKSHEET

1. 88.8% Cu and 11.2% O Assume you have 100 grams.

88.8 g Cu and 11.2 g O Convert grams to moles

88.8g Cu 1 mol Cu = 1.40 mol Cu63.55g Cu

11.2g O 1 mol O = 0.700 mol O16.00g O

Find the smallest whole number ratio. 1.40 mol Cu / 0.700 = 2 0.70 mol O /0.700 = 1

So the empirical formula is Cu2O.

MOLECULAR FORMULA FROM EMPIRICAL FORMULA The empirical formula is the lowest ratio of elements

and not always the actual ratio of elements in a compound.

The actual ratio (molecular formula) can be determined using the molar mass of the actual compound and the molar mass of the empirical formula. The molar mass of the molecular formula is

experimentally determined (Given). The molar mass of the empirical formula can be

determined from the periodic table. The molecular mass is divided by the empirical mass

and equals a whole number. The whole number is multiplied by the subscripts in

the molecular formula to give the empirical formula.

EXAMPLE: FIND THE MOLECULAR FORMULA FOR A COMPOUND THAT HAS AN EMPIRICAL FORMULA OF CH2 AND A MOLECULAR MASS OF 41.5G.

The molecular mass is 41.5 grams. Given

The empirical mass is 14.03 grams. C + 2H = 12.01 + 2(1.008) = (14.026)14.03

41.5 / 14.03 = (2.96) 3

3 (CH2) = C3H6

The molecular formula is C3H6 .

MOLECULAR FORMULA PRACTICE

Use Empirical Formulas from Worksheet with the following Molecular Masses.

1. 290 g/mol2. 178 g/mol3. 26 g/mol4. 160 g/mol5. 34 g/mol6. 160 g/mol7. 138 g/mol8. 240 g/mol9. 245 g/mol10. 230 g/mol