NWTC General Chemistry Ch 07

Chapter 7
Quantitative Composition of
These black
Compounds
pearls are made
of layers of
calcium
carbonate.
They can be
measured by
counting or
weighing. Introduction to General, Organic, and Biochemistry 10e
John Wiley & Sons, Inc
Morris Hein, Scott Pattison, and Susan Arena

Chapter Outline

7.1 The Mole 7.5 Calculating Empirical
7.2 Molar Mass of Compounds Formulas
7.3 Percent Composition of 7.6 Calculating the Molecular
Compounds Formula from the Empirical
Formula
7.4 Empirical Formula versus
Molecular Formula

Copyright 2012 John Wiley & Sons, Inc

Convenient Ways of Counting Items

• 1 dozen eggs = 12 eggs
• 1 ream paper = 500 sheets
• 1 gross pencils = 144 pencils
1×10 -10 m
How do chemists count really
small things, like atoms?
1 mole atoms =
Review Question 1
6.022×1023 atoms
This number is known as
Avogadro’s number. 2012 John Wiley & Sons, Inc
Copyright

The Mole
Review Question 7 & 8
1 mole of anything contains Avogadro’s number
(6.022×1023 ) of particles.
• 1 mol atoms = 6.022×1023 atoms
• 1 mol molecules = 6.022×1023 molecules
• 1 mol ions = 6.022×1023 ions
Avogadro’s number can be used as a conversion factor:

6.022 1023 particles 1 mol
1 mol 6.022 1023 particles


The Mole

Calculate the number of atoms in 2.4 mol Na.

Plan 2.5 mol Na  Na atoms
6.022 x 1023 Na atoms 1 mol Na
1 mol Na 6.022 x 1023 Na atoms
Calculate

6.022 x 1023 Na atoms
2.4 mol Na = 1.4 1024 Na atoms
1 mol Na


Your Turn!

How many moles of HCl are present in 4.3×1023
molecules?
a. 2.6×1047 mol
b. 0.71 mol
c. 2.6 mol
d. 7.1×1045 mol

4.3 x10 23 molecules 1 mol = 0.71 mol
6.022 x10 23 molecules


Convenient Ways of Counting

It is often convenient to count using mass.
For example, a canning recipe calls for 150 apples to be
peeled and cored.
The average mass of an individual apple is 235 g. How
many kg are needed to complete this recipe?

235 g 1 kg
150 apples × = 35 kg
1 apple 1000 g

The canner should buy 35 kg of apple.

Counting Atoms with Mass

• Chemists count atoms by using mass since individual
atoms are too small to count.
• Average mass of an atom = atomic mass in amu
• Average mass of 1 mole of atoms = atomic mass
expressed in grams (molar mass).
Review Question 5
1 mol atoms = atomic mass in grams
1 mol atoms = 6.022×1023 atoms
atomic mass in grams = 6.022×1023 atoms


The Mole


Your Turn!

Which of these is not correct? 6
a. The mass of 1 atom of C is 12.01 amu. C
b. The mass of 1 mole of C atoms is 12.01 g. 12.01
c. Avogadro’s number of atoms has a mass of 12.01
amu.
d. Avogadro’s number of atoms has a mass of 12.01 g.


The Mole

Calculate the mass of 2.4 mol C.
atomic mass
Plan 2.4 mol C  g C 12.01

Molar mass is a conversion factor:
12.01g C 1 mol C
1 mol C 12.01g C

Calculate
12.01g C
2.4 mol C = 29 g C
1 mol C


Your Turn!

What is the correct set up to calculate number of moles
of atoms contained in 3.52 g Al? atomic mass
a. 1 mol Al Al 26.98
3.52g Al
26.98 g Al
b. 26.98g Al
3.52g Al
1 mol Al
c. 6.022 x 10 23 Al atoms
3.52g Al
1 g Al


The Mole

Calculate the number of atoms in 36.0 g C.
atomic mass
Plan 36.0 g C  moles C  atoms C C 12.01

1 mol C 6.022 1023 atoms C
12.01 g C 1 mol C
1 mol C 6.022 1023 atoms C
Calculate 36.0 g C
12.01 g C 1 mol C

= 1.8 1024 atoms C


Your Turn!

What is the mass of 3.01 ×1023 atoms of lead?
a. 104 g atomic mass

b. 414 g Pb 207.2

c. 0.500 g
d. 1.04×1048 g

3.01 x10 23 atoms 1 mol 207.2 gram = 104 gram
6.022 x10 23 atoms 1 mol


Your Turn!

1 gram of which of the following elements would
contain the largest number of atoms? atomic mass
a. nitrogen
Review Question 3 N 14.01
H 1.01
b. hydrogen
P 30.97
c. phosphorus O 16.00
d. oxygen
1 gram H 1 mol 6.022 x10 23 atom = 5.96 x10 23 atoms
1.01 g 1 mol

1 gram P 1 mol 6.022 x10 23 atom = 1.94 x10 22 atoms
30.97 g 1 mol

Your Turn!

Which has a higher mass, mol of K or mol of Au?
a. potassium
Review Question 2 atomic mass

b. gold K 39.0983
Au 196.96654

1 Mol of K 39.0983 gram = 39.0983 gram
1 mol

1 Mol of Au 196.9665 gram = 196.9665 gram
1 mol


Your Turn!

Which has more electrons, mol of K or mol of Au?
a. potassium
Review Question 4 atomic number
b. gold K 19
Au 79

1 Mol K 6.022 x10 23 atom 19 e- = 1.144 x10 25 E-
1 mol 1 atom

1 Mole Au 6.022 x10 23 atom 79 e- = 4.757 x10 25 E-
1 mol atom

Review Question 9

a. Mole of O atoms = ? Atoms 6.022 x10 23
b. Mole of O2 molecules = ? Molecules 6.022 x10 23
c. Mole of O2 molecules = ? atoms 1.204 x10 24
d. Mole of O atoms= ? grams 15.99 g
e. Mole of O2 molecules = ? grams 31.98 g
1 Mole of O 15.99 gram = 15.99 gram
1 mole

1 Mole O 2 2 Mole of O 15.99 gram = 31.98 gram
1 Mole O 2 1 mole

Review Question 9 Copyright 2012 John Wiley & Sons, Inc

Molar Mass of Compounds

1 mol compound = 6.022×1023 formula units compound
Molar mass of compound = mass of 1 mol compound
The molar mass of a compound is the sum of the atomic
masses of each atom in the compound.

What is the molar mass of CO2?
atomic mass
1C 1(12.01g) C 12.01
2O 2(16.00g) O 16.00
CO2 44.01g/mol


Molar Mass of Compounds

What is the molar mass of Al2(CO3)3?
2Al 2(26.98g) atomic mass

3C 3(12.01g) Al 26.98
C 12.01
9O 9(16.00g) O 16.00
Al2(CO3)3 233.99g/mol


Your Turn!

What is the molar mass of (NH4)3PO4?
a. 141.04g/mol atomic mass

b. 144.07g/mol N 14.01
H 1.01
c. 146.09g/mol P 30.97
d. 149.12g/mol O 16.00

3*1 N 3(14.01g)
3*4 H 12(1.01g)
1P 1(30.97g)
4O 4(16.00g)

Your Turn!

What is the molar mass of Mg(ClO4)2?
a. 301.01g/mol atomic mass

b. 191.21g/mol Mg 24.31
Cl 35.45
c. 223.21g/mol O 16.00
d. 123.76g/mol

__ Mg __(____g)
__ Cl __(____g)
__ O __(____g)

Using Molar Masses of Compounds

Molar mass = 1 mol = 6.022×1023 formula units
Calculate the mass of 0.150 mol Mg(ClO4)2
Plan 0.150 mol Mg(ClO4)2  g Mg(ClO4)2
1 mol Mg(ClO 4 ) 2 223.21 g Mg(ClO 4 ) 2
223.21 g Mg(ClO 4 ) 2 1 mol Mg(ClO 4 ) 2
Calculate
223.21 g Mg(ClO 4 ) 2
0.150 mol Mg(ClO4 ) 2 = 33.5 g Mg(ClO 4 ) 2
1 mol Mg(ClO 4 ) 2



Calculate the number of moles in 35 g H2O.
Plan 35 g H2O  moles H2O atomic mass

1 mol H 2 O H 1.01
18.02 g H 2 O
O 16.00
18.02 g H 2 O 1 mol H 2 O

1 mol H 2 O
Calculate 35 g H 2 O = 1.9 g H 2 O
18.02 g H 2O



Calculate the number of molecules in 35 g H2O.
Plan 35 g H2O  mol H2O molecules H2O
18.02 g H 2 O 6.022 1023 molecules H2O
1 mol H 2 O 1 mol H2O
Calculate
1 mol H 2O 6.022 1023 molecules H 2O
35.0 g H 2O
18.02 g H 2O 1 mol H 2O

= 1.2 10 24 molecules H 2 O


Preview Question 10
How many molecules are present in 1 molar mass of
sulfuric acid (H2SO4)?
atomic mass
1. What is the molar mass of H2SO4?
S 32.06
2 H 2(1.01g) H 1.01
1 S 1(32.06g) O 16.00

4 O 4(16.00g) = 98.08g/mol
2. How many atoms in one molecule? 2+1+4=7
How many atoms in 1 molar mass? ___
98.08 g 1 mol 6.022 x10 23 molecules = 6.022 x10 23 molecules
98.08 g 1 mol

Your Turn!

How many moles of molecules are present in 146 g of
glucose (C6H12O6)? atomic mass
a. 180. mol C 12.01
H 1.01
b. 0.810 mol
__ C __(____g) O 16.00
c. 26300 mol __ H __(____g)
d. 4.88×1023 mol __ O __(____g) = _____ g/mol

g = molecules


Your Turn!

What is the mass of 1.20 ×1023 molecules of CH3OH?
a. 161g atomic mass

b. 38.5g C 12.01
H 1.01
c. 32.1g __ C __(____g) O 16.00
d. 6.39g __ H __(____g)
__ O __(____g) = _____ g/mol

molecules = g


Your Turn!

How many molecules are present in 4.21 moles of HBr?
a. 2.53×1023 molecules atomic mass

b. 2.53×1024 molecules H 1.01

-24 molecules
Br 79.90
c. 6.99×10
d. 3.97×102 molecules
e. 6.99×1024 molecules


Percent Composition of Compounds

Percent composition is a list of the mass percent of each
element in a compound.
Na2CO3 is
 43.38% Na
 11.33% C
 45.29% O
How do we calculate the mass percent of Na2CO3?


Calculating Percent Composition

First determine the molar mass of Na2CO3
2(22.99g Na) + 1(12.01g C) + 3(16.00 g O) = 105.99g/mol Na2CO3

Then find ratio of the mass of each element to the mass of
the compound. 2 22.99 g Na
x100 43.38% Na
105.99 g Na2CO3

1 12.01 g C
x100 11.33% C
105.99 g Na2 CO3

3 16.00 g O
x100 45.29% O
105.99 g N a2 CO3

Calculating Percent Composition

A compound is found to consist of 2.74g of iron and
5.24g of chlorine. What is the percent composition of
the compound?
1. Calculate the mass of the product formed:
2.74g Fe + 5.24g Cl = 7.98g product
2. Calculate the percent for each element.

2.74 g Fe 5.24 g Cl
x100 = 34.3% Fe x100 = 65.7% Cl
7.98 g 7.98 g


First determine the molar mass of Na2CO3

Your Turn!

What is the percent carbon in acetic acid, HC2H3O2?
a. 41.01% C __ C __(____g) atomic mass

b. 20.00% C __ H __(____g) C 12.01
__ O __(____g) H 1.01
c. 6.73% C = _____ g/mol O 16.00
d. 39.99% C


Your Turn!

A 6.00g sample of calcium sulfide is found to contain
3.33g of calcium. What is the percent by mass of
sulfur in the compound?
a. 80.2% S
b. 55.5% S
c. 44.5% S
d. 28.6% S


Empirical Formula versus
Molecular Formula
Review Question 11
The molecular formula for a substance is the
C6H12O6
actual number of atoms of each element.
The empirical formula is the lowest whole
CH2O
number ratio of atoms in a compound.
Note that the molecular formula is a whole (CH2O)6
number multiple of the empirical formula.


Empirical Formula versus
Molecular Formula
It is possible for several different molecules to have the
same empirical formula.


Your Turn!

What is the empirical formula for the compound P4O10?
a. P4O10
b. P2O5
c. PO2.5
d. PO3


Calculating Empirical Formulas

Steps for calculating an empirical formula:
Review Question 12 100g of compound and express the mass of
1. Assume
each element in grams.
2. Convert the grams of each element to moles.
3. Find the mole ratio of each element. Round to
nearest whole number if it is close to the whole
number.
4. If necessary, multiply the ratios by the smallest
whole number that will convert them to a whole
number.


Calculate the empirical formula of a compound that is
63.19% Mn and 36.81% O. atomic mass
1. Assume 100 g of material. Mn 54.94
0 16.00
63.19 g Mn
36.81 g O
2. Convert grams of each element to moles:
1 mol Mn
63.19 g Mn× =1.1502 mol Mn
54.94 g Mn
1 mol O
36.81 g O× = 2.3006 mol O
16.00 g O


3. Change the numbers of atoms to whole numbers by
dividing by the smallest number.
1.1502 mol Mn 2.3006 mol O
Mn = = 1.000 O= = 2.000
1.1502 mol Mn 1.1502 mol Mn

The simplest ratio of Mn:O is 1:2.

Empirical formula = MnO2



Calculate the empirical formula of a compound that is
72.2% Mg and 27.8% N. atomic mass
1. Assume 100 g of material. Mg 24.31
N 14.01
72.2 g Mg
27.8 g O
2. Convert grams of each element to moles:
1mol Mg
72.2g Mg× =2.970 mol Mg
24.31g Mg
1mol N
27.8g N× = 1.984 mol N
14.01g N


3. Change the numbers of atoms to whole numbers by
dividing by the smallest number.
2.970 mol Mg 1.984 mol N
Mg = = 1.500 N= = 1.000
1.984 mol N 1.984 mol N
4. Multiply by a number that will give whole numbers.

Mg: (1.500)2 = 3.00 N: (1.000)2 = 2.00
Empirical formula = Mg3N2


Your Turn!
What is the empirical formula of an alcohol that is
52.13% C, 13.15% H and 34.72% O.
atomic mass
a. CH2O 1. Assume 100 g so
52.13g C, 13.15g H and 34.72g O C 12.01
b. C4HO3 2. Moles H 1.01
52.13g C/12.01= 4.34 mol C
c. C2H6O 13.15g H/1.01= 13.02 mol H
O 16.00

d. C2H3O2 34.72g O/16.00= 2.17mol O
3. Fractional part
4.34 mol C /2.17= 2
13.02 mol H /2.17= 6
2.17mol O /2.17= 1
4. Whole numubers
Already done
5. Empirical formula: 2 C & 6 H & 1 O, so …
Give it a try

Calculating the Molecular Formula
from the Empirical Formula
The molecular formula will be either equal to the empirical
formula or some integer multiple of it.

The ratio of the molecular mass to the mass predicted by
the empirical formula tells us how many times larger the
molecular formula is.
molecular formula mass
n= = number of empirical formula units
empirical formula mass


Determine the molecular formula for glyceraldehyde
which has a molar mass of 90.08 g/mol and an empirical
formula of CH2O.

n=

90.09 g
n= = 3 (CH2O)3 = C3H6O3
30.03 g CH 2O


Determine the molecular formula of a nitrogen oxide
compound (NxOy) with a molar mass of 92.011 g/mol
and a empirical formula of NO2.

n=

92.011 g N x O y (NO2)2 = N2O4
n 2
46.0055 g NO2


Your Turn!
What is the molecular formula of a compound with the
empirical formula CH2Cl and molar mass of 197.92 g/mol?
a. CH2Cl PLAN THE STEPS
b. C2H4Cl2 atomic mass
A. Empirical formula mass
C 12.01
c. C3H6Cl3
B. Divide molar by empirical H 1.01
d. C4H8Cl4 Cl 35.45
C. Multiple result by formula

A. 1(12.01g C) + 2(1.01g H) + 1(35.45 g Cl) = 49.48g/mol CH2Cl
B. 197.92 g/mol / 49.48 g/mol = 4

Plan – Set up - Calcualte
C. (CH Cl) *4=
2


A disinfectant is known to be 76.57% C , 6.43% H, and
17.00% O. It has a molar mass of 188.24 g/mol
Determine its molecular formula.
Determine the mass and moles of C, H and O.
C: 76.57%(188.24g) = 144.1 g C/(12.01g/mol) = 12
H: 6.43% (188.24g) = 12.1 g H/(1.01g/mol) = 12
O: 17.00%(188.24g) = 32.00 g O/(16.00 g/mol) = 2
Molecular formula: C12H12O2


Your Turn!

What is the molecular formula of a substance that
consists of 85.60% C and 14.40% H and has a molar
mass of 28.08 g/mol?
atomic mass
a. CH2
C 12.01
b. C2H2 H 1.01
c. CH3
d. C2H4
e. C2H6


Questions

Review Questions
– Did in class
Paired Questions (pg 139)
– Do 1, 3, 7, 9, 11, 15, 21, 27, 29, 31, 35 , 39, 43
– Practice later 2, 6, 12, 16, 20, 24, 28, 32, 36, 40, 44

Copyright 2012 John Wiley & Sons, Inc 1-50

NWTC General Chemistry Ch 07

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