Moles and Formula Mass in Chemistry

 
Moles and Formula Mass
Moles and Formula Mass
The Mole
The Mole
1 dozen =
1 gross = 
1 ream =
 
1 mole =
 
12
144
500
6.022 x 10
23
 
There are 
exactly
 12 grams of
carbon-12 in one mole of carbon-12.
 
Avogadro’s Number
Avogadro’s Number
 
6.022 x 10
23
 is called “Avogadro’s Number” in
honor of the Italian chemist Amadeo Avogadro
(1776-1855).
 
Amadeo Avogadro
I didn’t discover it. Its
just named after me!
Calculations with Moles:
Calculations with Moles:
Converting moles to grams
Converting moles to grams
How many grams of lithium are in 3.50 moles
of lithium?
 
3.50 mol Li
 
=          g Li
 
1 mol Li
 
6.94 g Li
 
45.1
45.1
Calculations with Moles:
Calculations with Moles:
Converting grams to moles
Converting grams to moles
How many moles of lithium are in 18.2 grams
of lithium?
 
18.2 g Li
 
=        mol Li
 
6.94 g Li
 
1 mol Li
 
2.62
2.62
Calculations with Moles:
Calculations with Moles:
Using Avogadro’s Number
Using Avogadro’s Number
How many 
atoms
 of lithium are in 3.50
moles of lithium?
 
3.50 mol
 
  
=              atoms
 
1 mol
 
6.02 x 10
23
 atoms
 
2.07 x 10
2.07 x 10
24
24
Calculations with Moles:
Calculations with Moles:
Using Avogadro’s Number
Using Avogadro’s Number
How many 
atoms
 of lithium are in 18.2 g of
lithium?
 
18.2 g Li
 
  
=              atoms Li
 
1 mol Li
 
6.022 x 10
23
 atoms Li
 
1.58 x 10
24
 
6.94 g Li
 
1 mol Li
 
(18.2)(6.022 x 10
23
)/6.94
Calculating Formula Mass
Calculating Formula Mass
Calculate the formula mass of magnesium carbonate,
Calculate the formula mass of magnesium carbonate,
MgCO
MgCO
3
3
.
.
 
24.31
24.31
 
 
g  +  12.01 g  +  3(16.00 g) =
g  +  12.01 g  +  3(16.00 g) =
 
84.32 g
84.32 g
Calculating Percentage Composition
Calculating Percentage Composition
Calculate the percentage composition of magnesium
Calculate the percentage composition of magnesium
carbonate, MgCO
carbonate, MgCO
3
3
.
.
 
From previous slide:
From previous slide:
24.31 g  +  12.01 g  +  3(16.00 g) = 
24.31 g  +  12.01 g  +  3(16.00 g) = 
84.32 g
84.32 g
 
100.00
Formulas
Formulas
 
 molecular formula  =  (empirical
 molecular formula  =  (empirical
formula)
formula)
n
n
 [
 [
n
n
 = integer]
 = integer]
 molecular formula  =  C
 molecular formula  =  C
6
6
H
H
6
6
  =  (CH)
  =  (CH)
6
6
 
 
empirical formula  =  CH
empirical formula  =  CH
 
Empirical formula
: the lowest whole number
ratio of atoms in a compound.
 
Molecular formula
: the true number of
atoms of each element in the formula of a
compound.
Formulas
Formulas
 
 
(continued)
(continued)
Formulas for 
Formulas for 
ionic compounds 
ionic compounds 
are 
are 
ALWAYS
ALWAYS
empirical (lowest whole number ratio).
empirical (lowest whole number ratio).
 
Examples:
Examples:
 
NaCl
 
MgCl
2
 
Al
2
(SO
4
)
3
 
K
2
CO
3
Formulas
Formulas
 
 
(continued)
(continued)
Formulas for 
Formulas for 
molecular compounds
molecular compounds
 
 
MIGHT
MIGHT
be empirical (lowest whole number ratio).
be empirical (lowest whole number ratio).
 
Molecular:
Molecular:
 
H
2
O
 
C
6
H
12
O
6
 
C
12
H
22
O
11
 
Empirical:
Empirical:
 
H
2
O
 
CH
2
O
 
C
12
H
22
O
11
Empirical Formula Determination
Empirical Formula Determination
 
1.
Base calculation on 100 grams of
Base calculation on 100 grams of
compound.
compound.
2.
Determine moles of each element in
Determine moles of each element in
100 grams of compound.
100 grams of compound.
3.
Divide each value of moles by the
Divide each value of moles by the
smallest of the values.
smallest of the values.
4.
Multiply each number by an integer
Multiply each number by an integer
to obtain all whole numbers.
to obtain all whole numbers.
Empirical Formula Determination
Empirical Formula Determination
Adipic acid contains 49.32% C, 43.84% O, and
6.85% H by mass. What is the empirical formula
of adipic acid?
Empirical Formula Determination
Empirical Formula Determination
(part 2)
(part 2)
Divide each value of moles by the smallest
Divide each value of moles by the smallest
of the values.
of the values.
 
Carbon:
Carbon:
 
Hydrogen:
Hydrogen:
 
Oxygen:
Oxygen:
Empirical Formula Determination
Empirical Formula Determination
(part 3)
(part 3)
Multiply each number by an integer to
Multiply each number by an integer to
obtain all whole numbers.
obtain all whole numbers.
 
Carbon: 1.50
Carbon: 1.50
 
Hydrogen: 2.50
Hydrogen: 2.50
 
Oxygen: 1.00
Oxygen: 1.00
 
x  2
 
x  2
 
x  2
 
3
3
 
5
5
 
2
2
 
Empirical formula:
 
C
3
H
5
O
2
Finding the Molecular Formula
Finding the Molecular Formula
The empirical formula for adipic acid is
The empirical formula for adipic acid is
C
C
3
3
H
H
5
5
O
O
2
2
. The molecular mass of adipic acid
. The molecular mass of adipic acid
is 146 g/mol. What is the molecular
is 146 g/mol. What is the molecular
formula of adipic acid?
formula of adipic acid?
 
1. Find the formula mass of C
1. Find the formula mass of C
3
3
H
H
5
5
O
O
2
2
 
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
Finding the Molecular Formula
Finding the Molecular Formula
The empirical formula for adipic acid is
The empirical formula for adipic acid is
C
C
3
3
H
H
5
5
O
O
2
2
. The molecular mass of adipic acid
. The molecular mass of adipic acid
is 146 g/mol. What is the molecular
is 146 g/mol. What is the molecular
formula of adipic acid?
formula of adipic acid?
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
 
2. Divide the molecular mass by the
2. Divide the molecular mass by the
mass given by the emipirical formula.
mass given by the emipirical formula.
Finding the Molecular Formula
Finding the Molecular Formula
The empirical formula for adipic acid is
The empirical formula for adipic acid is
C
C
3
3
H
H
5
5
O
O
2
2
. The molecular mass of adipic acid
. The molecular mass of adipic acid
is 146 g/mol. What is the molecular
is 146 g/mol. What is the molecular
formula of adipic acid?
formula of adipic acid?
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
 
3. Multiply the empirical formula by this
3. Multiply the empirical formula by this
number to get the molecular formula.
number to get the molecular formula.
 
(C
(C
3
3
H
H
5
5
O
O
2
2
) x 2 =
) x 2 =
 
C
C
6
6
H
H
10
10
O
O
4
4
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Dive into the world of moles, Avogadro's number, and formula mass calculations in chemistry. Learn how to convert moles to grams, grams to moles, and determine the number of atoms using Avogadro's number. Explore the concepts of formula mass and percentage composition to understand the composition of compounds like magnesium carbonate. Unravel the difference between empirical and molecular formulas, essential in understanding the atomic composition of compounds.

  • Chemistry
  • Moles
  • Avogadro
  • Formula Mass
  • Empirical Formula

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  1. Moles and Formula Mass

  2. The Mole 1 dozen = 1 gross = 1 ream = 1 mole = 12 144 500 6.022 x 1023 There are exactly 12 grams of carbon-12 in one mole of carbon-12.

  3. Avogadros Number 6.022 x 1023is called Avogadro s Number in honor of the Italian chemist Amadeo Avogadro (1776-1855). I didn t discover it. Its just named after me! Amadeo Avogadro

  4. Calculations with Moles: Converting moles to grams How many grams of lithium are in 3.50 moles of lithium? 3.50 mol Li 6.94 g Li = g Li 45.1 1 mol Li

  5. Calculations with Moles: Converting grams to moles How many moles of lithium are in 18.2 grams of lithium? 18.2 g Li 1 mol Li 2.62 = mol Li 6.94 g Li

  6. Calculations with Moles: Using Avogadro s Number How many atoms of lithium are in 3.50 moles of lithium? 6.02 x 1023 atoms 3.50 mol = atoms 2.07 x 1024 1 mol

  7. Calculations with Moles: Using Avogadro s Number How many atoms of lithium are in 18.2 g of lithium? 18.2 g Li 6.022 x 1023 atoms Li 1 mol Li 1 mol Li 6.94 g Li (18.2)(6.022 x 1023)/6.94 = atoms Li 1.58 x 1024

  8. Calculating Formula Mass Calculate the formula mass of magnesium carbonate, MgCO3. 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g

  9. Calculating Percentage Composition Calculate the percentage composition of magnesium carbonate, MgCO3. From previous slide: 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g 24.31 84.32 84.32 84.32 = 12.01 48.00 = 100 100 14.24% = 100 56.93% = 28.83% Mg C O 100.00 = =

  10. Formulas Empirical formula: the lowest whole number ratio of atoms in a compound. Molecular formula: the true number of atoms of each element in the formula of a compound. molecular formula = (empirical formula)n [n = integer] molecular formula = C6H6 = (CH)6 empirical formula = CH

  11. Formulas (continued) Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio). Examples: NaCl MgCl2 Al2(SO4)3 K2CO3

  12. Formulas (continued) Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio). Molecular: C6H12O6 H2O C12H22O11 Empirical: H2O CH2O C12H22O11

  13. Empirical Formula Determination 1. Base calculation on 100 grams of compound. 2. Determine moles of each element in 100 grams of compound. 3. Divide each value of moles by the smallest of the values. 4. Multiply each number by an integer to obtain all whole numbers.

  14. Empirical Formula Determination ( ( ( Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid? )( ( ( ( ) ) 49.32g C 1 mol C )( 6.85 g H 43.84 g O )( ) 1 mol H 1 mol O =4.107 mol C 6.78 = ) ) ) mol H mol O = 2.74 12.01 g C 1.01 16.00 g H g O

  15. Empirical Formula Determination (part 2) Divide each value of moles by the smallest of the values. Carbon: Hydrogen: Oxygen:

  16. Empirical Formula Determination (part 3) Multiply each number by an integer to obtain all whole numbers. Carbon: 1.50 Hydrogen: 2.50 Oxygen: 1.00 x 2 x 2 x 2 3 5 2 Empirical formula: C3H5O2

  17. Finding the Molecular Formula The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 1. Find the formula mass of C3H5O2 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

  18. Finding the Molecular Formula The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 2. Divide the molecular mass by the mass given by the emipirical formula. 146 73 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g = 2

  19. Finding the Molecular Formula The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 3. Multiply the empirical formula by this number to get the molecular formula. 146 73 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g (C3H5O2) x 2 =C6H10O4 = 2

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