Moles!

Whatcha talkn’ about a mole?First…a dozen.

12 cookies = 12 bagels =12 cans =12 footballs =12 dance moves =

As scientist we need to determine something to quantify atoms, molecules, particles, or something very, very small

1 dozen1 dozen

1 dozen

1 dozen1 dozen

As scientist we don't work with dozen....we work with something called a mole!

Back to original question: "What the heck is a mole?!!"

Same concept as a dozen but larger....much larger.

If I want to have a mole of cookies….

6.022 x 1023 times!!

I mole of cookies = 6.022 x 1023 cookies

What about a mole of dance moves?

6.022 x 1023 times!!

I mole of dance moves= 6.022 x 1023 dance moves

Mole in Chemistry?Obviously in chemistry you don’t

have cookies and dance moves….we have

AtomsMoleculesParticlesFormula units (ionic compounds)Electrons

Abbreviation for mole is mol

Moles!1 mole of atoms =

1 mole of molecules =

1 mole of particles =

1 mole of formula units =

1 mole of electrons =

6.022 x 1023 atoms

6.022 x 1023 molecules

6.022 x 1023 particles

6.022 x 1023 formula units

6.022 x 1023 electrons

6.022 x 1023 is called Avogadro number (named after Amadeo Avogadro)

So what does this mean?Converting….DA!!! Yeah!

1. How many moles are in 4.56 x 1024 atoms of hydrogen?

4.56 x 1024 atoms 6.022 x 1023 atoms

1 mole= 7.57 moles

2. How many formula units are in 8.92 moles of NaCl?8.92 moles6.022 x 1023 formula units

1 mole = 5.37 x 1024 formula units

More moley practice…1. How many atoms of oxygen are

in 5.00 moles of carbon dioxide (CO2)?5.00 mol CO2

1 mole of CO2

6.022 x 1023 molecules CO2

1 molecule CO2

2 atoms O

= 6.02 x 1024 atoms of O

Why moles, not atoms?1 atom of Al = 4.48 x 10-23 gramsUsing moles to work with easier

numbers

1 mole of Al = 26.98 gramsWhich is called molar mass

◦Mass (in grams) of a moleGuess what….another conversionYeah!! Boom! No doubt!

Molar Mass1. Determine the molar mass of Sucrose

(C12H22O11)C – 12(12.01)H – 22(1.008)O – 11(15.999)

2. Determine the molar mass of Ca(NO3)2

Ca – 1(40.078)N – 2(14.0067)O – 6(15.999)

= 342.29 g/mol

= 164.09 g/mol

Converting Moles to grams or grams to moles

All conversions come from molar mass…aka periodic table

1. How many grams of aluminum are in 4.89 moles of Al?

2. How many moles of CaCl2 are in 13.5 g of CaCl2?

4.89 mol of Al mol of Al grams of Al26.98

1= 132 g of Al

13.5 g of CaCl2 g of CaCl2

mol of CaCl2110.984

1 = 0.122 mol of CaCl2

Molar mass of Al

Molar mass of CaCl2

Practice1. How many moles are in 25.67

grams of nickel?0.4374 mol of Ni

2. How many grams are in 0.234 moles of lead(II) iodide (PbI2)?

108 grams of PbI2

More Practice1. How many atoms are in 14.0

grams of manganese?1.53 x 1023 atoms of Mn

2. How many molecules are in 46.7 grams of nitrogen gas?

1.00 x 1024 molecules of N2

3. How many oxygen atoms are in 3.45 grams of CO2?

9.44 x 1022 atoms of O

MolarityWhat are these bottles we keep

working with numbers and a big M?

Well, Molarity is how scientist describe concentration of a solution

Molarity = how many moles per 1 liter of solution

MolarityWhat does this mean?

For example: 2.00 M HCl

2.00 moles of HCl per 1 liter OR2.00 moles of HCl = 1 L

That’s right another conversion factor!! Yeah!

Other examples1. 0.500 M of Pb(NO3)2

0.500 moles of Pb(NO3)2 = 1 L

2. 1.56 M of NaCl1.56 moles of NaCl = 1 L

Using molarity…All about the DA1. A 250. mL container has a solution of NaCl

with a molarity of 1.54 M. How many moles of NaCl is present?

2. How many grams of NaCl would that contain?

250. mL 1000

mL

1 LLmol1.54

1 0.385 mol of NaCl=

molg58.44

31 = 22.5 g of NaCl

Molar mass of NaCl

Why finding mass?You need the mass…

- how much you need to measure in the lab (because you can’t measure moles on a scale)

- make accurate concentrations of your solutions

Practice1. How many moles are present in

500.0 mL of 0.800 M solution of HCl?

2. How many grams are present in 250. mL of 0.500 M solution of Pb(NO3)2?

500. 0 mL 1000

mL

1 L 0.800 mol1 L = 0.400 mol of

HCl

250. 0 mL 1000

mL

1 L 0.500 mol1 L

331.21 g Pb(NO3)21 mol

= 41.4 g of Pb(NO3)2

Molar mass of Pb(NO3)2

More Practice…1. How many grams are needed to

make 4.00 L of 0.250 M solution of NaCl?

2. How many grams are needed to make 2.50 L of 0.500 M solution of Na2CO3?

= 58.4 g of NaCl

= 132 g of Na2CO3

More Practice…Complete the table:

Formula Grams Dissolved

Moles Dissolved

Volume of Solution

Molarity

NaCl 4.00 grams

? 1.50 L ?

KOH ? 2.56 x 10-3

1.80 L ?Moles NaCl = 0.0684 molMolarity of NaCl = 0.0456 M

Grams KOH = 0.144 gMolarity of KOH= 0.00142 M

Making Solutions1. Obtain correct volumetric flask

2. Determine the correct number of grams needed to measure

3. Add the massed out compound to flask4. Rinse weigh boat with DI into flask5. Add DI to etched line on flask

Example1. How many grams are needed to

make 250 mL of 0.250 M solution of CuSO4?250. 0

mL 1000 mL

1 L 0.250 mol1 L

159.608 g CuSO41 mol

= 9.98 g of CuSO4

Molar mass of CuSO4

-Mass out 9.98 grams of copper(II) sulfate-Hot dog weigh boat and use DI water to add CuSO4 to 250 mL volumetric flask-Add DI water to 250 mL mark-Add stir bar and stir till all dissolves-Add solution to appropriate containers

Percent Mass and Determining Compound FormulasFind percent of each element in

the compound

Define Molecular and Empirical Formulas

Mass PercentWhat percent of each element in a

compound?Example:

Find the percent of each element in ethanol (C2H5OH).

C – (2)(12.01) = 24.02H – (6)(1.008) = 6.048O – (1)(15.999) = 15.99946.07

g/mol% C = 24.02 g/mol

46.07 g/mol

x 100 =52.14% of C

Mass PercentWhat percent of each element in a

compound?Example:

Find the percent of each element in ethanol (C2H5OH).

C – (2)(12.01) = 24.02H – (6)(1.008) = 6.048O – (1)(15.999) = 15.99946.07

g/mol% H = 6.048 g/mol

46.07 g/mol

x 100 =13.13% of H

Mass PercentWhat percent of each element in a

compound?Example:

Find the percent of each element in ethanol (C2H5OH).

C – (2)(12.01) = 24.02H – (6)(1.008) = 6.048O – (1)(15.999) = 15.99946.07

g/mol% O = 15.999 g/mol

46.07 g/mol

x 100 =37.73% of O

Mass Percent Practice1. Find the percent of each element in CuSO4.

2. Find the percent of each element in C2H2.

Cu = (1)63.546 = 63.546S = (1)32.065 = 32.065O = (4)15.999 = 63.996

159.607 g/mol

%Cu = 39.81%%S = 20.09%%O = 40.10%

H = (2)1.008 = 2.016C = (2)12.01 = 24.02

26.04 g/mol

%H = 7.743%%C = 92.26%

Empirical and Molecular FormulasLook at C2H2 (Acetylene gas)

Acetylene + O2 + pumpkin = awesome!

What is the ratio between hydrogen and carbon atoms?

What about the mole ratio between hydrogen and carbon?

2 atoms of H2 atoms of C

= 1 H : 1 C

Same: 1 H : 1 C

Why?

2 atoms H6.022 x 1023 atoms

1 mol

3.32 x 10-24 moles H

2 atoms O6.022 x 1023 atoms

1 mol

3.32 x 10-24 moles O

What does this tell us?A chemical compound’s formula can be

determined by finding the moles of each element.

Yeah!! Moles!I Moles

This formula is called Empirical (strikes back) formula

The empirical formula only tells you the ratio between elements…

For example:C2H2 and C6H6 both have the same empirical formula… CH

To find the correct formula (molecular formula)…you need the molar mass of the compound to be given.

Example:Given molar mass of 78.1 g/mol and empirical formula of CH…Molar mass of CH = 13.02 g/mol (Periodic table)It takes 6 times of 13.02 to get 78.1… therefore the empirical formula (CH) gets multiplied by 6 and thus C6H6

Molecular Formula Empirical Formula

The exact formula for the compound.

Ex. C4H10 or P4O8

The reduced formula for the compound.

Ex. C2H5 or PO2

The empirical can be the molecular just as MgCl2 or Fe2O3

Determine Empirical Formula1. Must find moles of each element (molar

mass)2. Find ratios between each element (large

over smallest)Problem: A compound is made up of sulfur

and oxygen. 25.0 grams each of sulfur and oxygen are present. Find the empirical formula.25.0 g S 1 mol of S

32.06 g S= 0.780 mol of S

25.0 g O 1 mol of O15.999 g O

1.56 mol of O

= 1.56 mol of O

0.780 mol of S

= 2 O : 1 S

Formula: SO2

Determine Molecular Formula1. Must find molar mass of empirical formula (periodic table)

2. Find ratio between given molecular molar mass and empirical molar mass (large over small)

Problem: A compound is made up of sulfur and oxygen. 25.0 grams each of sulfur and oxygen are present. The molar mass of the compound is 128.12 g/mol. Determine the molecular formula.

Empirical Formula: SO2

Molar Mass of SO2 = 64.06 g/mol

128.12 g/mol64.06 g/mol

= 2 : 1

Molecular Formula: S2O4

Determine Empirical/Molecular Formula1. Must find moles of each element (molar

mass)2. Find ratios between each element (large

over smallest) to get empirical formula3. Find molar mass of the empirical

formula4. Find the ratio between molar mass5. Multiply the ratio to each element in the

empirical formula

Example 1An unknown compound contains both hydrogen

and carbon has a molecular molar mass of 26.04 g/mol. The compound contains 1.85 grams of hydrogen and 22.15 grams of carbon. Determine the empirical and molecular formulas.1.85 g H 1 mol of H1.008 g H

= 1.84 mol of H

22.15 g C 1 mol of C12.01 g C

1.84 mol of H

= 1.84 mol of C

1.84 mol of C= 1 H : 1 C

Emp Formula: CH Molar mass of CH = 13.02 g/mol

26.04 g/mol13.02 g/mol

= 2:1 Mol. Formula: C2H2

Example 2An unknown compound contains both sulfur and

oxygen has a molecular molar mass of 160.1 g/mol. The compound contains 40.0% sulfur and 60.0% oxygen Determine the empirical and molecular formulas.40.0 g S 1 mol of S32.06 g S

= 1.25 mol of S

60.0 g O 1 mol of O15.999 g O

3.75 mol of O

= 3.75 mol of O

1.25 mol of S= 3 O : 1 S

Emp. Formula: SO3 Molar mass of SO3 = 80.06 g/mol

*****Math Alert!!*****

Attention: This is a math ALERT!!!

Treat percentage just like grams….it doesn’t matter how much sample you

have!!

Therefore if you have %...make it grams!160.1 g/mol = 2:1 Mol. Formula:

S2O680.06 g/mol