MOLES

Mole: A collection of objects

A collection of Avogadro’s number of objects

Avogadro’s number 6.022 * 1023 AtomsParticlesIonsCationsAnionsMoleculesFormula Units

Mole Ratio

The coefficient in front of each component in a balanced equation

2 Al (s) + 6 HCl (aq) ----- 2 AlCl 3 (aq) + 3 H2 (g)

This states: 2 moles of Al reacts w/ 6 moles HCl to produce2 moles AlCl3 and 3 moles H2 gas

MOLE CALCULATIONS

One mole of helium atoms …. ….. is a number

6.022 * 1023 Helium atoms

One mole of argon atoms …. ….. is a number

6.022 * 1023 Argon atoms

….. has a mass ….. has a mass

4.00 grams 39.9 grams

RECAP: 1 mol of “anything” contains 6.022*1023 “parts” Elements on p.table = I mol

Molar mass (mass/1 mol)1. Elements - decimal # on p.table Ti 47.87 amu or g2. Molecules/Cmpd - sum of mass of all elements

MOLES

MASS

Mul

tiply

by D

ivid

e

by

mass = 1 mole

Given mass, find moles

How many moles are in 345.6 g NaNO3 ?

Step 1: find formula wt. of NaNO3

1 Na = 23.0 g1 N = 14.0 g3 O = 3 * 16.0 = 48.0 g

1 mole of NaNO3 = 85.0 g

Step 2 : Find # of moles Use factor-label method

Conversion * 1

Given g 85.0

mole 1 * 1

g 345.64.07 moles NaNO3

Given moles, find mass

How many grams are in 0.6 moles N2O ?

Step 1: find molecular wt. of N2O

2 N = 2 * 14.0 = 28.0 g1 O = 16.0 g

1 mole of N2O = 44.0 g

Step 2 : Find mass Use factor-label method

mol 1

g 44.0 * 1moles 0.6 26.4 g N2O

MOLES

NUMBER of PARTICLES

6.02*1023

Avogadro’s N

umber

Given moles, find # molecules

How many molecules are in 1.6 moles oxygen?

Step 1: Recognize that mass does not apply here. But Avogadro’s # is used

Step 2 : Find # molecules Use factor-label method

mol 1

10 * 6.02 * 1

moles 1.6 molecules239.63 * 1023 molecules O2

Given # atoms, find moles

How many moles are in 3.01 *1012 atoms of Chromium?

Step 1: Recognize that mass does not apply here. But Avogadro’s # is used

Step 2 : Find # moles Use factor-label method

10 * 6.02

mole 1 * 1

atoms 10 * 3.01atoms23

125.00 * 10-12 mols Cr

MOLES

MASSNUMBER of PARTICLES

Mul

tiply

by D

ivid

e

by

mass = 1 mole

6.02*1023

There is not a direct relationshipbetween MASS and PARTICLES

Avogadro’s N

umber

Given # formula units, find grams

2 Step Conversion ProblemWill need to use MASS at some point in the problem

How many grams are in 1.208 * 1024 formula units AgCl ?

Step 1: find formula wt. of AgCl

1 Ag = 107.9 g1 Cl = 35.5 g

1 mole of AgCl = 143.4 g

Step 2 : Find moles Use factor-label method

mol 1

g 143.4 * units 10 * 6.02

mol 1 * 1

units 10*1.20823

24288 g AgCl

Converts formula units to MOLES THEN

Converts moles to MASS

Step 3 : Find mass

Given grams, find # atomsWill need to use MASS at some point in the problem

2 Step Conversion Problem

How many atoms are in 128.0 grams of Mercury ?

Step 1: find molar mass of Hg

1 mole of Hg = 200.6 g

Step 2 : Find moles Use factor-label method

mol 1

atoms 10 * 6.02 *

g 200.6mol 1

* 1

g 128.0 23

3.84*1023 atoms Hg

Converts mass to MOLES THEN

Converts moles to ATOMS

Step 3 : Find atoms

PROBLEM What is the formula weight of sodium carbonate, Na2CO3. This is an industrial chemical used inmaking glass.

SOLUTION

2 Na

1 C

3 O

2 * 23.0 = 46.0

1 * 12.0 = 12.0

3 * 16.0 = 48.0

106.0 g

Also, equivalent to 1mole of a substance

Sodium Phosphate (aq) + Barium Nitrate (aq) ------ Barium Phosphate (s) + Sodium Nitrate (aq)

Na3PO4 (aq) + Ba(NO3)2 (aq) ----- Ba3(PO4)2 (s) + NaNO3 (aq) 3 632

BALANCE EQUATIONS “STOICHIOMETRY”

Al + HCl ---- AlCl3 + H23 36 22

Aluminum (s) + Hydrochloric Acid (aq) ----------> Aluminum Chloride (aq) + Hydrogen (g)

MASS - MASS CALCULATIONS & LIMITING REAGENT

Solid calcium metal burns in air (oxygen) to form a calcium oxide cmpd. Using 4.20 g Ca & 2.80 g O2 how much pdt is made? Find 1) grams of pdt made from g of Ca 2) grams of pdt made from g of O2

3) limiting reagent 4) how much pdt can be made 5) % yield if 3.85 g produced

PLAN: Need ------> balanced chem. rxn

g Ca -----> mols Ca -----> mols pdt ------> g pdtM g Ca coeff X pdt/Y react M g pdt

STEPS:

Ca(s) + O2(g) -------> CaO(s)2 2

M g Ca M g O2 coeff X pdt/Y react M g pdt40.1 g Ca:CaO 2:2 56.1 g 32.0 g O2:CaO 1:2

(same steps converting oxygen)

(amts of 2 reactants given: limiting reactants)

2 Ca(s) + O2(g) -------> 2 CaO(s)

4.20 g 2.80 g X g

M g : 40.1g 56.1 gCoeff : 2 1 2mols :

CaO g 5.89 mol 1

g 56.1mol) 105.0(CaO mol 105.0

Ca mol 2

CaO mol 2

g 40.1

mol 1Ca g 20.4 )1

0.105 0.105

CaO g 82.9mol 1

g 1.56mol 175.0CaO mol 175.0

O mol 1

CaO mol 2

g 32.0

mol 1O g 80.2 )2

22

0.0875 0.175

5.89 g 9.82 g

3) limiting reagent: Which produced the least amt-- Ca or O2?Ca ---> CaO = 5.89 g O2 ---> CaO = 9.82 gCa, limiting reagent

4) how much pdt can be made: 5.89 g CaO

5) % yield= 100

amountltheoretica

amountactual % 65.4 100g 89.5

g 85.3

RECAP: -- Balanced chem. equation -- then, GIVEN USE FIND grams Molar Mass mols mols coeff mols mols Molar Mass gramslimiting reagent: the reactant that produces the least amt of moles or mass of a specific pdt