Section 6.4: Molar Mass
Tutorial 1 Practice, page 275
1. (a) Given: substance with formula S8
Required: molar mass of S8, nNaCl
2
Solution:
Step 1. Look up the molar mass of the element.
g
M S = 32.07
mol
Step 2. Multiply the molar mass of the element by the number of atoms of that element in the
substance.
M S = 8M S
8
!
g $
= 8 # 32.07
mol &%
"
g
mol
Statement: The molar mass of S8 is 256.56 g/mol.
(b) Given: compound with formula H2S
Required: molar mass of H2S, M H S
M S = 256.56
8
2
Solution:
Step 1. Look up the molar masses of the elements.
g
g
M H = 1.01
; M S = 32.07
mol
mol
Step 2. Add the molar masses of the elements, multiplying the molar mass of each element by
the number of atoms of that element in the compound.
M H S = 2 M H + MS
2
!
g $ !
g $
= 2 # 1.01
+ # 32.07
&
mol % "
mol &%
"
g
2
mol
Statement: The molar mass of H2S is 34.09 g/mol.
(c) Given: compound with formula NaOH
Required: molar mass of NaOH, M NaOH
Solution:
Step 1. Look up the molar masses of the elements.
g
g
g
M Na = 22.99
; M O = 16.00
; M H = 1.01
mol
mol
mol
M H S = 34.09
Copyright © 2011 Nelson Education Ltd.
Chapter 6: Quantities in Chemical Formulas
6.4-1
Step 2. Add the molar masses of the elements, multiplying the molar mass of each element by
the number of atoms of that element in the compound.
M NaOH = M Na + M O + M H
!
g $ !
g $ !
g $
= # 22.99
+ # 16.00
+ # 1.01
&
&
mol % "
mol % "
mol &%
"
g
mol
Statement: The molar mass of NaOH is 40.00 g/mol.
(d) Given: compound with formula Fe(OH)3
Required: molar mass of Fe(OH)3, M Fe(OH)
M NaOH = 40.00
3
Solution:
Step 1. Look up the molar masses of the elements.
g
g
g
M Fe = 55.85
; M O = 16.00
; M H = 1.01
mol
mol
mol
Step 2. Add the molar masses of the elements, multiplying the molar mass of each element by
the number of atoms of that element in the compound.
M Fe(OH) = M Fe + 3( M O + M H )
3
!
!
g $
g
g $
= # 55.85
+
3
16.00
+
1.01
#"
mol &%
mol
mol &%
"
g
mol
Statement: The molar mass of Fe(OH)3 is 106.88 g/mol.
(e) Given: compound with formula (NH4)2S
Required: molar mass of (NH4)2S, M (NH ) S
M Fe(OH) = 106.88
3
4 2
Solution:
Step 1. Look up the molar masses of the elements.
g
g
g
M N = 14.01
; M H = 1.01
; M S = 32.07
mol
mol
mol
Step 2. Add the molar masses of the elements, multiplying the molar mass of each element by
the number of atoms of that element in the compound.
M (NH ) S = 2( M N + 4 M H ) + M S
4 2
"
g
g % "
g %
= 2 $ 14.01
+ 4 ! 1.01
+ $ 32.07
'
mol
mol & #
mol '&
#
g
mol
Statement: The molar mass of (NH4)2S is 68.17 g/mol.
M (NH
4 )2 S
= 68.17
Copyright © 2011 Nelson Education Ltd.
Chapter 6: Quantities in Chemical Formulas
6.4-2
(f) Given: compound with formula Ca3(PO4)2
Required: molar mass of Ca3(PO4)2, M Ca (PO )
3
4 2
Solution:
Step 1. Look up the molar masses of the elements.
g
g
g
M Ca = 40.08
; M P = 30.97
; M O = 16.00
mol
mol
mol
Step 2. Add the molar masses of the elements, multiplying the molar mass of each element by
the number of atoms of that element in the compound.
M Ca (PO ) = 3M Ca + 2( M P + 4 M O )
3
4 2
!
!
g $
g
g $
= 3# 40.08
+
2
30.97
+
4
'
16.00
#"
mol &%
mol
mol &%
"
g
mol
Statement: The molar mass of Ca3(PO4)2 is 310.18 g/mol.
(g) Given: compound with formula MgSO4•7H2O
Required: molar mass of MgSO4•7H2O, M MgSO •7H O
M Ca
3 (PO 4 )2
= 310.18
4
2
Solution:
Step 1. Look up the molar masses of the elements.
g
g
g
g
M Mg = 24.31
; M S = 32.07
; M O = 16.00
; M H = 1.01
mol
mol
mol
mol
Step 2. Add the molar masses of the elements, multiplying the molar mass of each element by
the number of atoms of that element in the compound.
M MgSO •7H O = M Mg + M S + 4 M O + 7(2 M H + M O )
4
2
!
!
!
g $ !
g $
g $
g
g $
= # 24.31
+ # 32.07
+ 4 # 16.00
+ 7 # 2 ' 1.01
+ 16.00
&
&
&
mol % "
mol %
mol %
mol
mol &%
"
"
"
g
mol
Statement: The molar mass of MgSO4•7H2O is 246.52 g/mol.
M MgSO
4 •7H 2 O
= 246.52
Mini Investigation: The Mole Exhibit, page 275
A. Answers may vary. Answers are determined from experimental data. Molar mass of a
compound is calculated by adding the molar masses of the elements, multiplying the molar mass
of each element by the number of atoms of that element in the compound.
Sample answer:
Given: amount of calcium chloride, nCaCl = 1.00 mol
2
Required: molar mass of calcium chloride,
Solution:
Step 1. Look up the molar masses of the elements that are in the substance.
M Ca = 40.08 g/mol; M Cl = 35.45 g/mol
Copyright © 2011 Nelson Education Ltd.
Chapter 6: Quantities in Chemical Formulas
6.4-3
Step 2. Determine the sum of the molar masses of the elements in the substance.
M CaCl = M Ca + 2 M Cl
2
= 40.08
!
g
g $
+ 2 # 35.45
mol
mol &%
"
g
2
mol
Statement: The molar mass of 1.00 mol of calcium chloride is 111 g/mol.
I will have to measure approximately 111 g of calcium chloride for 1.00 mol of the substance.
B. Each molar quantity contains 6.02 × 1023 entities. Therefore there 6.02 × 1023 entities of
calcium chloride in the beaker.
M CaCl = 110.98
Tutorial 2: Practice, page 277
1. (a) Given: mNaCl = 500 g
Required: amount of table salt, nNaCl
Solution:
Step 1. Calculate the molar mass of table salt, M NaCl .
!
g $ !
g $
M NaCl = # 22.99
+
35.45
mol &% #"
mol &%
"
g
mol
Step 2. Use the mass of table salt and its molar mass to calculate the amount of table salt.
! 1 mol $
nNaCl = (500 g) #
&
" 58.44 g %
M NaCl = 58.44
nNaCl = 8.56 mol
Statement: The amount of sodium chloride in a 500 g of table salt is 8.56 mol.
(b) Given: mAl = 14.2 g
Required: amount of aluminum, nAl
Solution:
Use the molar mass of aluminum to calculate the amount of aluminum.
! 1 mol $
nAl = (14.2 g) #
&
" 26.98 g %
nAl = 0.526 mol
Statement: The amount of aluminum in a 14.2 g sample is 0.526 mol.
Copyright © 2011 Nelson Education Ltd.
Chapter 6: Quantities in Chemical Formulas
6.4-4
(c) Given: mCaO = 1.00 kg
Required: amount of calcium oxide, nCaO
Solution:
Step 1. Calculate the molar mass of calcium oxide, M CaO .
!
g $ !
g $
M CaO = # 40.08
+
16.00
mol &% #"
mol &%
"
g
mol
Step 2. Use the mass of calcium oxide and its molar mass to calculate the amount of calcium
oxide.
mCaO = 1.00 kg = 1.00 ! 103 g
M CaO = 56.08
" 1 mol %
nCaO = (1.00 ! 103 g) $
'
# 56.08 g &
nCaO = 17.8 mol
Statement: The amount of calcium oxide in a 1.00 kg sample is 17.8 mol.
(d) Given: mHCl = 1.75 kg
Required: amount of hydrogen chloride, nHCl
Solution:
Step 1. Calculate the molar mass of hydrogen chloride, M HCl .
!
g $ !
g $
M HCl = # 1.01
+
35.45
mol &% #"
mol &%
"
g
mol
Step 2. Use the mass of hydrogen chloride and its molar mass to calculate the amount of
hydrogen chloride.
mHCl = 1.75 kg = 1.75 ! 103 g
M HCl = 36.46
" 1 mol %
nHCl = (1.75 ! 103 g) $
'
# 36.46 g &
nHCl = 48.0 mol
Statement: The amount of hydrogen chloride in a 1.75 kg sample is 48.0 mol.
Copyright © 2011 Nelson Education Ltd.
Chapter 6: Quantities in Chemical Formulas
6.4-5
(e) Given: mC
13 H18 O 2
= 200.0 mg
Required: amount of ibuprofen, nC
13 H18 O 2
Solution:
Step 1. Calculate the molar mass of ibuprofen, M C
13 H18 O 2
MC
13 H18 O 2
.
!
!
!
g $
g $
g $
= 13# 12.01
+ 18 # 1.01
+ 2 # 16.00
&
&
mol %
mol %
mol &%
"
"
"
g
mol
Step 2. Use the mass of ibuprofen and its molar mass to calculate the amount of ibuprofen.
mC H O = 200.0 mg
MC
13 H18 O 2
13
18
= 206.31
2
= 200.0 ! 10"3 g
mC
= 2.000 ! 10"1 g
nC
# 1 mol &
= (2.000 ! 10"1 g) %
(
$ 206.31 g '
nC
= 9.694 ! 10"4 mol
13 H18 O 2
13 H18 O 2
13 H18 O 2
Statement: The amount of ibuprofen in a 200.0 mg tablet is 9.694 × 10−4 mol.
2. (a) Given: nH O = 0.80 mol
2
2
Required: mass of hydrogen peroxide, mH O
2
2
Solution:
Step 1. Calculate the molar mass of hydrogen peroxide, M H O .
2
2
!
!
g $
g $
M H O = 2 # 1.01
+ 2 # 16.00
&
2 2
mol %
mol &%
"
"
g
2 2
mol
Step 2. Use the amount of hydrogen peroxide and its molar mass to calculate the mass of
hydrogen peroxide.
! 34.02 g $
mH O = (0.80 mol ) #
&
2 2
" 1 mol %
M H O = 34.02
mH O = 27 g
2
2
Statement: The mass of 0.80 mol of hydrogen peroxide is 27 g.
Copyright © 2011 Nelson Education Ltd.
Chapter 6: Quantities in Chemical Formulas
6.4-6
(b) Given: nNaHSO = 3.25 mol
4
Required: mass of sodium hydrogen sulfate, mNaHSO
4
Solution:
Step 1. Calculate the molar mass of sodium hydrogen sulfate, M NaHSO .
4
!
!
g $ !
g $ !
g $
g $
M NaHSO = # 22.99
+ # 1.01
+ # 32.07
+ 4 # 16.00
&
&
&
4
mol % "
mol % "
mol %
mol &%
"
"
g
4
mol
Step 2. Use the amount of sodium hydrogen sulfate and its molar mass to calculate the mass of
sodium hydrogen sulfate.
! 120.07 g $
mNaHSO = (3.25 mol ) #
&
4
" 1 mol %
M NaHSO = 120.07
mNaHSO = 3.90 ' 102 g
4
Statement: The mass of 3.25 mol of sodium hydrogen sulfate is 3.90 × 102 g.
(c) Given: nCaCO = 5.0 mmol
3
Required: mass of calcium carbonate, mCaCO
3
Solution:
Step 1. Calculate the molar mass of calcium carbonate, M CaCO .
3
!
!
g $ !
g $
g $
M CaCO = # 40.08
+
12.01
+
3
16.00
#"
3
mol &% #"
mol &%
mol &%
"
g
mol
Step 2. Use the amount of calcium carbonate and its molar mass to calculate the mass of calcium
carbonate.
nCaCO = 5.0 mmol = 5.0 ! 10"3 mol
M CaCO = 100.09
3
3
# 100.09 g &
mCaCO = (5.0 ! 10"3 mol ) %
(
3
$ 1 mol '
mCaCO = 0.50 g
3
Statement: The mass of 5.0 mmol of calcium carbonate is 0.50 g.
Copyright © 2011 Nelson Education Ltd.
Chapter 6: Quantities in Chemical Formulas
6.4-7
(d) Given: nNaOCl = 1.2 ! 103 mol
Required: mass of sodium hypochlorite, mNaOCl
Solution:
Step 1. Calculate the molar mass of sodium hypochlorite, M NaOCl .
!
g $ !
g $ !
g $
M NaOCl = # 22.99
+ # 16.00
+ # 35.45
&
&
mol % "
mol % "
mol &%
"
g
NaOCl
mol
Step 2. Use the amount of sodium hypochlorite and its molar mass to calculate the mass of
sodium hypochlorite.
" 74.44 g %
mNaOCl = (1.2 ! 103 mol ) $
'
# 1 mol &
M
= 74.44
mNaOCl = 8.9 ! 104 g
Statement: The mass of 1.2 × 103 mol of sodium hypochlorite is 8.9 × 104 g.
(e) Given: nHe = 45 mmol
Required: mass of helium, mHe
Solution:
Use the molar mass of helium to calculate the mass of helium.
nHe = 45 mmol = 4.5 ! 10"2 mol
# 4.00 g &
mHe = (4.5 ! 10"2 mol ) %
(
$ 1 mol '
mHe = 1.8 ! 10"1 g
Statement: The mass of 45 mmol of helium is 1.8 × 10−1 g.
Section 6.4 Questions, page 277
1. One mole samples of different compounds have different masses because the molar mass of
the compounds are different.
2. A precise estimate of the number of entities in a sample is determined by dividing the mass of
the sample by the molar mass of the substance in the sample. In this way, the mass of the sample
is an indirect way to count the number of entities present.
3. The amount obtained using this equation is always an estimate because mass is a measured
quantity. There is some degree of uncertainty in all measured quantities.
Copyright © 2011 Nelson Education Ltd.
Chapter 6: Quantities in Chemical Formulas
6.4-8
4. (a) Given: iron(III) oxide, Fe2O3
Required: molar mass of Fe2O3, M Fe O
2
3
Solution:
Step 1. Look up the molar masses of the elements.
g
g
M Fe = 55.85
; M O = 16.00
mol
mol
Step 2. Add the molar masses of the elements, multiplying the molar mass of each element by
the number of atoms of that element in the compound.
M Fe O = 2 M Fe + 3M O
2
3
!
!
g $
g $
= 2 # 55.85
+
3
16.00
#"
mol &%
mol &%
"
g
mol
Statement: The molar mass of iron(III) oxide is 159.70 g/mol.
(b) Given: calcium carbonate, CaCO3
Required: molar mass of CaCO3, M CaCO
M Fe O = 159.70
2
3
3
Solution:
Step 1. Look up the molar masses of the elements.
g
g
g
M Ca = 40.08
; M C = 12.01
; M O = 16.00
mol
mol
mol
Step 2. Add the molar masses of the elements, multiplying the molar mass of each element by
the number of atoms of that element in the compound.
M CaCO = M Ca + M C + 3M O
3
!
!
g $ !
g $
g $
= # 40.08
+ # 12.01
+ 3# 16.00
&
&
mol % "
mol %
mol &%
"
"
g
3
mol
Statement: The molar mass of calcium carbonate is 100.09 g/mol.
(c) Given: octane, C8H18
Required: molar mass of C8H18, M C H
M CaCO = 100.09
8
18
Solution:
Step 1. Look up the molar masses of the elements.
g
g
M C = 12.01
; M H = 1.01
mol
mol
Copyright © 2011 Nelson Education Ltd.
Chapter 6: Quantities in Chemical Formulas
6.4-9
Step 2. Add the molar masses of the elements, multiplying the molar mass of each element by
the number of atoms of that element in the compound.
M C H = 8M C + 18M H
8
18
!
!
g $
g $
= 8 # 12.01
+
18
1.01
#"
mol &%
mol &%
"
g
mol
Statement: The molar mass of octane is 114.26 g/mol.
(d) Given: calcium chlorate, Ca(ClO3)2
Required: molar mass of Ca(ClO3)2, M Ca(ClO )
M C H = 114.26
8
18
3 2
Solution:
Step 1. Look up the molar masses of the elements.
g
g
g
M Ca = 40.08
; M Cl = 35.45
; M O = 16.00
mol
mol
mol
Step 2. Add the molar masses of the elements, multiplying the molar mass of each element by
the number of atoms of that element in the compound.
M Ca(ClO ) = M Ca + 2( M Cl + 3M O )
3 2
!
!
g $
g
g $
= # 40.08
+ 2 # 35.45
+ 3 ' 16.00
&
mol %
mol
mol &%
"
"
g
mol
Statement: The molar mass of calcium chlorate is 206.98 g/mol.
(e) Given: ammonium carbonate, (NH4)2CO3
Required: molar mass of (NH4)2CO3, M (NH ) CO
M Ca(ClO
= 206.98
3 )2
4 2
3
Solution:
Step 1. Look up the molar masses of the elements.
g
g
g
g
M N = 14.01
; M H = 1.01
; M C = 12.01
; M O = 16.00
mol
mol
mol
mol
Step 2. Add the molar masses of the elements, multiplying the molar mass of each element by
the number of atoms of that element in the compound.
M (NH ) CO = 2( M N + 4 M H ) + M C + 3M O
4 2
3
"
"
g
g % "
g %
g %
= 2 $ 14.01
+ 4 ! 1.01
+ $ 12.01
+ 3$ 16.00
'
'
mol
mol & #
mol &
mol '&
#
#
g
mol
Statement: The molar mass of ammonium carbonate is 96.11 g/mol.
M (NH
4 )2 CO3
= 96.11
Copyright © 2011 Nelson Education Ltd.
Chapter 6: Quantities in Chemical Formulas
6.4-10
5. Given: mCaCl = 10.0 g
2
Required: amount of calcium chloride, nCaCl
2
Solution:
Step 1. Calculate the molar mass of calcium chloride, M CaCl .
2
!
!
g $
g $
M CaCl = # 40.08
+ 2 # 35.45
&
2
mol %
mol &%
"
"
g
2
mol
Step 2. Use the mass of calcium chloride and its molar mass to calculate the amount of calcium
chloride.
! 1 mol $
nCaCl = (10.0 g) #
&
2
" 110.98 g %
M CaCl = 110.98
nCaCl = 0.0901 mol
2
Statement: The amount of calcium chloride in a 10.0 sample is 0.0901 mol.
6. Given: mC H N O = 80.0 mg
8
10
4
2
Required: amount of caffeine, nC H
8
10 N 4 O 2
Solution:
Step 1. Calculate the molar mass of caffeine, M C H
8
MC H
8
10 N 4 O 2
10 N 4 O 2
.
!
!
!
!
g $
g $
g $
g $
= 8 # 12.01
+
10
1.01
+
4
14.01
+
2
16.00
#"
#"
#"
mol &%
mol &%
mol &%
mol &%
"
g
mol
Step 2. Use the mass of caffeine and its molar mass to calculate the amount of caffeine.
mC H N O = 80.0 mg = 8.00 ! 10"2 g
MC H
8
8
10 N 4 O 2
10
4
2
nC H
10 N 4 O 2
nC H
10 N 4 O 2
8
8
= 194.22
# 1 mol &
= (8.00 ! 10"2 g) %
(
$ 194.22 g '
= 4.12 ! 10"4 mol
Statement: The amount of caffeine in an 80.0 mg sample is 4.12 × 10−4 mol.
Copyright © 2011 Nelson Education Ltd.
Chapter 6: Quantities in Chemical Formulas
6.4-11
7. Given: nH O = 1.22 ! 10"3 mol
2
Required: mass of water, mH O
2
Solution:
Step 1. Calculate the molar mass of water, M H O .
2
!
g $ !
g $
M H O = 2 # 1.01
+ # 16.00
&
2
mol % "
mol &%
"
g
2
mol
Step 2. Use the amount of water and its molar mass to calculate the mass of water.
# 18.02 g &
mH O = (1.22 ! 10"3 mol ) %
(
2
$ 1 mol '
M H O = 18.02
mH O = 2.20 ! 10"2 g
2
Statement: The mass of 1.22 × 10−3 mol of water is 2.20 × 10−2 g.
8. Given: mSi = 5.8 mg
Required: amount of silicon, nSi
Solution:
Use the molar mass of silicon to calculate the amount of silicon.
# 1 mol &
nSi = (5.8 ! 10"3 g) %
(
$ 28.09 g '
nSi = 2.1 ! 10"4 mol
Statement: The amount of silicon in a 5.8 mg sample is 2.1 × 10−4 mol.
9. (a) Given: VH SO = 34 000 L ; d H SO = 1.84 kg/L
2
4
2
4
Required: mass of sulfuric acid, mH SO
2
Solution: Use the formula d =
m
V
m = Vd
mH SO = VH SO d H SO
4
m
to calculate the mass of sulfuric acid.
V
d=
2
4
2
4
2
4
! 1.84 kg $
= (34 000 L) #
" L &%
mH SO = 62 560 kg [2 extra digits carried]
2
4
Statement: The mass of sulfuric acid in the truck is 63 000 kg.
Copyright © 2011 Nelson Education Ltd.
Chapter 6: Quantities in Chemical Formulas
6.4-12
(b) Given: mH SO = 62 560 kg
2
4
Required: amount of sulfuric acid, nH SO
2
4
Solution:
Step 1. Calculate the molar mass of sulfuric acid, M H SO .
2
4
!
!
g $ !
g $
g $
M H SO = 2 # 1.01
+ # 32.07
+ 4 # 16.00
&
&
2
4
mol % "
mol %
mol &%
"
"
g
2
4
mol
Step 2. Use the mass of sulfuric acid and its molar mass to calculate the amount of sulfuric acid.
mH SO = 62 560 kg
M H SO = 98.09
2
4
= 62 560 ! 103 g
mH SO = 6.256 ! 107 g
2
4
" 1 mol %
nH SO = (6.256 ! 107 g) $
'
2
4
# 98.08 g &
nH SO = 6.4 ! 105 mol
2
4
Statement: The amount of sulfuric acid in the 34 000 L sample is 6.4 × 105 mol.
10. (a) Given: nNaCl = 0.154 mol
Required: mass of sodium chloride, mNaCl
Solution:
Step 1. Calculate the molar mass of sodium chloride, M NaCl .
!
g $ !
g $
M NaCl = # 22.99
+ # 35.45
&
mol % "
mol &%
"
g
mol
Step 2. Use the amount of sodium chloride and its molar mass to calculate the mass of sodium
chloride.
! 58.44 g $
mNaCl = (0.154 mol ) #
&
" 1 mol %
M NaCl = 58.44
mNaCl = 9.00 g
Statement: The mass amount of 0.154 mol of sodium chloride is 9.00 g.
Copyright © 2011 Nelson Education Ltd.
Chapter 6: Quantities in Chemical Formulas
6.4-13
11. Given: mcompound = 2.9 g ; ncompound = 0.050 mol
Required: molar mass of compound, M compound
Solution:
Use the formula n =
m
to calculate the molar mass of the compound.
M
m
M
m
M=
n
n=
M compound =
mcompound
ncompound
2.9 g
0.050 mol
g
M compound = 58
mol
Statement: The molar mass of the compound is 58 g/mol.
=
12. Given: nNa = 1.00 mol ; matom = 22.99 u ; conversion factor = 1.1661 ! 10"24
g
;
u
atoms
mol
Required: mass of 1.00 mol sodium atom, mNa
Solution: Calculate the number of atoms in 1.00 mol, then the mass of this number of atoms in
atomic mass units, u, then convert the mass of this number of atoms from atomic mass units, u,
to grams, g.
"
atoms % "
u %"
(24 g %
mNa = (1.00 mol ) $ 6.02 ! 1023
' $ 22.99
' $# 1.1661 ! 10 u '&
mol & #
atom &
#
N A = 6.02 ! 1023
mNa = 23.0 g
Statement: The mass of 1.00 mol sodium atoms is 23.0 g.
Copyright © 2011 Nelson Education Ltd.
Chapter 6: Quantities in Chemical Formulas
6.4-14