Section 6.5: Mass and Number of Entities
Tutorial 1 Practice, page 280
1. Given: mPb = 8.16 kg
Required: number of atoms of lead, N Pb
Solution:
Step 1. Look up the molar mass of lead.
M Pb = 207.20 g/mol
Step 2. Calculate the amount by multiplying the mass of lead by a conversion factor derived
from the molar mass.
mPb = 8.16 kg = 8.16 ! 103 g
" 1 mol %
nPb = (8.16 ! 103 g) $
'
# 207.2 g &
nPb = 39.382 mol (2 extra digits carried)
Step 3. Calculate the number of atoms using an appropriate conversion factor.
" 6.02 ! 1023 atoms %
N Pb = (39.382 mol ) $
'
#
&
1 mol
N Pb = 2.37 ! 1025 atoms
Statement: A typical car battery contains 2.37 × 1025 atoms of lead.
2. Given: mC = 2.00 ! 10"1 g
Required: number of atoms of carbon, N C
Solution:
Step 1. Look up the molar mass of carbon.
M C = 12.01 g /mol
Step 2. Calculate the amount by multiplying the mass of carbon by a conversion factor derived
from the molar mass.
# 1 mol &
nC = (2.00 ! 10"1 g) %
(
$ 12.01 g '
nC = 0.016 653 mol (2 extra digits carried)
Step 3. Calculate the number of atoms using an appropriate conversion factor.
" 6.02 ! 1023 atoms %
N C = (0.016 653 mol ) $
'
#
&
1 mol
N C = 1.00 ! 1022 atoms
Statement: A one-carat diamond contains 1.00 × 1022 atoms of carbon.
Copyright © 2011 Nelson Education Ltd.
Chapter 6: Quantities in Chemical Formulas
6.5-1
3. Given: mC H = 7.3 g
2
4
Required: number of molecules of ethylene, N C H
2
4
Solution:
Step 1. Calculate the molar mass of ethylene.
!
!
g $
g $
M C H = 2 # 12.01
+ 4 # 1.01
&
2 4
mol %
mol &%
"
"
g
2 4
mol
Step 2. Calculate the amount by multiplying the mass of ethylene by a conversion factor derived
from the molar mass.
! 1 mol $
nC H = (7.3 g) #
&
2 4
" 28.06 g %
M C H = 28.06
nC H = 0.2602 mol [2 extra digits carried]
2
4
Step 3. Calculate the number of molecules using an appropriate conversion factor.
" 6.02 ! 1023 molecules %
N C H = (0.2602 mol ) $
'
2 4
#
&
1 mol
N C H = 1.6 ! 1023 molecules
2
4
Statement: A 7.3 g sample of ethylene contains 1.6 × 1023 molecules.
4. Given: mCaCl = 1.5 kg
2
Required: number of formula units of calcium chloride, N CaCl
2
Solution:
Step 1. Calculate the molar mass of calcium chloride.
!
!
g $
g $
M CaCl = # 40.08
+ 2 # 35.45
&
2
mol %
mol &%
"
"
g
2
mol
Step 2. Calculate the amount by multiplying the mass of calcium chloride by a conversion factor
derived from the molar mass.
mCaCl = 1.5 kg = 1.5 ! 103 g
M CaCl = 110.98
2
" 1 mol %
nCaCl = (1.5 ! 103 g) $
'
2
# 110.98 g &
nCaCl = 13.52 mol [2 extra digits carried]
2
Copyright © 2011 Nelson Education Ltd.
Chapter 6: Quantities in Chemical Formulas
6.5-2
Step 3. Calculate the number of formula units using an appropriate conversion factor.
" 6.02 ! 1023 formula units %
N CaCl = (13.52 mol ) $
'
2
#
&
1 mol
N CaCl = 8.1 ! 1024 formula units
2
Statement: A 1.5 kg pail of calcium chloride contains 8.1 × 1024 formula units of calcium
chloride.
Tutorial 2 Practice, page 282
1. Given: mH O = 15.0 g
2
2
Required: number of atoms of hydrogen, N H ; number of atoms of oxygen, N O
Solution:
Step 1. Calculate the molar mass of hydrogen peroxide.
!
!
g $
g $
M H O = 2 # 1.01
+
2
16.00
#"
2 2
mol &%
mol &%
"
g
mol
Step 2. Calculate the amount by multiplying the mass of hydrogen peroxide by a conversion
factor derived from the molar mass.
! 1 mol $
nH O = (15.0 g) #
&
2 2
" 34.02 g %
M H O = 34.02
2
2
nH O = 0.440 92 mol [2 extra digits carried]
2
2
Step 3. Calculate the number of molecules of hydrogen peroxide using an appropriate conversion
factor.
"
molecules %
N H O = 0.440 92 mol $ 6.02 ! 1023
'
2 2
#
1 mol &
(
)
N H O = 2.6543 ! 1023 molecules [2 extra digits carried]
2
2
Step 4. Determine the number of each type of atom present.
" 2 atoms %
N H = (2.6543 ! 1023 molecules ) $
'
# 1 molecule &
N H = 5.31 ! 1023 atoms
" 2 atoms %
N O = (2.6543 ! 1023 molecule s) $
'
# 1 molecule &
N O = 5.31 ! 1023 atoms
Statement: A 15.0 g sample of hydrogen peroxide contains 5.31 × 1023 atoms of hydrogen and
5.31 × 1023 atoms of oxygen.
Copyright © 2011 Nelson Education Ltd.
Chapter 6: Quantities in Chemical Formulas
6.5-3
2. Given: mCa
3 (PO 4 )2
= 2.5 g
Required: number of ions calcium, N Ca 2+
Solution:
Step 1. Calculate the molar mass of calcium phosphate.
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
Step 2. Calculate the amount by multiplying the mass of calcium phosphate by a conversion
factor derived from the molar mass.
! 1 mol $
nCa (PO ) = (2.5 g) #
&
3
4 2
" 310.18 g %
M Ca
nCa
3 (PO 4 )2
3 (PO 4 )2
= 310.18
= 0.008 060 mol [2 extra digits carried]
Step 3. Calculate the number of formula units of calcium phosphate using an appropriate
conversion factor.
"
formula units %
N Ca (PO ) = (0.008 060 mol ) $ 6.02 ! 1023
'
3
4 2
#
&
1 mol
N Ca
3 (PO 4 )2
= 4.8521 ! 1021 formula units [2 extra digits carried]
Step 4. Determine the number of calcium ions present.
"
%
3 ions
N Ca 2+ = (4.8521 ! 1021 formula units ) $
'
# 1 formula unit &
N Ca 2+ = 1.5 ! 1022 ions
Statement: A 2.5 g sample of calcium phosphate contains 1.5 × 1022 calcium ions.
3. Given: N CO = 8.4 ! 1024 molecules
2
Required: mass of carbon dioxide, mCO
2
Solution:
Step 1. Calculate the amount of carbon dioxide using an appropriate conversion factor.
"
%
1 mol
nCO = (8.4 ! 1024 molecules ) $
'
2
# 6.02 ! 1023 molecules &
nCO = 13.95 mol [2 extra digits carried]
2
Copyright © 2011 Nelson Education Ltd.
Chapter 6: Quantities in Chemical Formulas
6.5-4
Step 2. Calculate the molar mass of carbon dioxide.
M CO = M C + 2 M O
2
!
!
g $
g $
= # 12.01
+ 2 # 16.00
&
mol %
mol &%
"
"
g
2
mol
Step 3. Use the amount of carbon dioxide and its molar mass to calculate the mass of carbon
dioxide.
! 44.01 g $
mCO = (13.95 mol ) #
&
2
" 1 mol %
M CO = 44.01
mCO = 6.1 ' 102 g
2
Statement: The mass of 8.4 × 1024 molecules of carbon dioxide is 6.1 × 102 g.
Mini Investigation: Determining Numbers in Everyday Entities, page 282
Answers are determined from experimental data. Sample answer:
Mass of chalk required to write your name: 0.08 g
The mass of water that you can hold in your hand: 125.45 g
The mass of water drops that will fit on a penny without overflowing: 2.50 g
A. (i) Given: mCaCO = 0.08 g
3
Required: amount of calcium carbonate, nCaCO
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 mass of calcium carbonate and its molar mass to calculate the amount of calcium
carbonate.
! 1 mol $
nCaCO = (0.08 g) #
&
3
" 100.09 g %
M CaCO = 100.09
3
nCaCO = 7.99 ' 10(4 mol [2 extra digits carried]
3
Statement: The amount of calcium carbonate in a 0.08 g piece of chalk is 8 × 10−4 mol.
Copyright © 2011 Nelson Education Ltd.
Chapter 6: Quantities in Chemical Formulas
6.5-5
(ii) Given: mH O = 125.45 g
2
Required: amount of water, nH 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 mass of water and its molar mass to calculate the amount of water.
! 1 mol $
nH O = (125.45 g) #
&
2
" 18.02 g %
M H O = 18.02
nH O = 6.96171 mol [2 extra digits carried]
2
Statement: The amount of water in 125.45 g of water is 6.962 mol.
(iii) Given: mH O = 2.50 g
2
Required: amount of water, nH O
2
Solution:
Use the mass of water and its molar mass to calculate the amount of water.
! 1 mol $
nH O = (2.50 g) #
&
2
" 18.02 g %
nH O = 0.13873 mol [2 extra digits carried]
2
Statement: The amount of water in 2.50 g of water is 0.139 mol.
B. (i) Given: nCaCO = 7.99 ! 10"4 mol
3
Required: number of formula units of calcium carbonate, N CaCO
3
Solution:
Calculate the number of formula units of calcium carbonate using an appropriate conversion
factor.
"
formula units %
N CaCO = (7.99 ! 10-4 mol ) $ 6.02 ! 1023
'
3
#
&
1 mol
N CaCO = 5 ! 1020 formula units
3
Statement: A 0.08 g piece of chalk contains 5 × 1020 formula units of calcium carbonate.
Copyright © 2011 Nelson Education Ltd.
Chapter 6: Quantities in Chemical Formulas
6.5-6
(ii) Given: nH O = 6.96171 mol
2
Required: number of molecules of water, N H O
2
Solution:
Calculate the number of molecules of water using an appropriate conversion factor.
"
molecules %
N H O = (6.96171 mol ) $ 6.02 ! 1023
'
2
#
1 mol &
N H O = 4.191 ! 1024 molecules
2
Statement: 125.45 g of water contains 4.191 × 1024 molecules of water.
(iii) Given: nH O = 0.138 73 mol
2
Required: number of molecules of water, N H O
2
Solution:
Calculate the number of molecules of water using an appropriate conversion factor.
"
molecules %
N H O = (0.138 73 mol ) $ 6.02 ! 1023
'
2
#
1 mol &
N H O = 8.35 ! 1022 molecules
2
Statement: 2.50 g of water contains 8.35 × 1022 molecules of water.
Section 6.5 Questions, page 283
1. (a) Given: nNH = 2.0 mol
3
Required: number of molecules of ammonia, N NH
3
Solution:
Calculate the number of molecules of ammonia using an appropriate conversion factor.
"
molecules %
N NH = (2.0 mol ) $ 6.02 ! 1023
'
3
#
1 mol &
N NH = 1.2 ! 1024 molecules
3
Statement: A 2.0 mol sample of ammonia contains 1.2 × 1024 molecules.
(b) Given: mH O = 1.6 mg
2
Required: number of molecules of water, N H 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 &%
"
M H O = 18.02
2
g
mol
Copyright © 2011 Nelson Education Ltd.
Chapter 6: Quantities in Chemical Formulas
6.5-7
Step 2. Use the mass of water and its molar mass to calculate the amount of water.
mH O = 1.6 mg = 1.6 ! 10-3 g
2
# 1 mol &
nH O = (1.6 ! 10"3 g) %
(
2
$ 18.02 g '
nH O = 8.879 ! 10"5 mol [2 extra digits carried]
2
Step 3. Calculate the number of molecules of water using an appropriate conversion factor.
"
molecules %
N H O = (8.879 ! 10-5 mol ) $ 6.02 ! 1023
'
2
#
1 mol &
N H O = 5.3 ! 1019 molecules
2
Statement: A 1.6 mg sample of water contains 5.3 × 1019 molecules.
2. (a) Given: N Cu = 3.01 ! 1023 atoms
Required: amount of copper, nCu
Solution:
Calculate the amount of copper using an appropriate conversion factor.
"
%
1 mol
nCu = (3.01 ! 1023 atoms ) $
'
# 6.02 ! 1023 atoms &
nCu = 0.500 mol
Statement: The amount of copper present is 0.500 mol.
(b) Given: N HCl = 6.02 ! 1025 molecules
Required: amount of hydrochloric acid, nHCl
Solution:
Calculate the amount of hydrochloric acid using an appropriate conversion factor.
"
%
1 mol
nHCl = (6.02 ! 1025 molecules ) $
'
23
# 6.02 ! 10 molecules &
nHCl = 1.00 ! 102 mol
Statement: The amount of hydrochloric acid present is 1.00 × 102 mol.
(c) Given: mCaCO = 42.0 g
3
Required: amount of calcium carbonate, nCaCO
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 &%
"
"
M CaCO = 100.09
3
g
mol
Copyright © 2011 Nelson Education Ltd.
Chapter 6: Quantities in Chemical Formulas
6.5-8
Step 2. Use the mass of calcium carbonate and its molar mass to calculate the amount of calcium
carbonate.
! 1 mol $
nCaCO = (42.0 g) #
&
3
" 100.09 g %
nCaCO = 0.420 mol
3
Statement: The amount of calcium carbonate in a 42.0 g sample is 0.420 mol.
3. (a) Given: N CO = 4.2 ! 1024 molecules
2
Required: number of atoms of carbon, N C ; number of atoms of oxygen, N O
Solution:
Multiply the number of molecules by the number of each type of atom per molecule.
" 1 atom %
N C = (4.2 ! 1024 molecules ) $
'
# 1 molecule &
N C = 4.2 ! 1024 atoms
" 2 atoms %
N O = (4.2 ! 1024 molecules ) $
'
# 1 molecule &
N O = 8.4 ! 1024 atoms
Statement: A sample of 4.2 × 1024 carbon dioxide molecules contains 4.2 × 1024 atoms of
carbon and 8.4 × 1024 atoms of oxygen.
(b) Given: N CO = 4.2 ! 1024 molecules
2
Required: mass of carbon dioxide, mCO
2
Solution:
Step 1. Calculate the amount of carbon dioxide using an appropriate conversion factor.
"
%
1 mol
nCO = (4.2 ! 1024 molecules ) $
'
2
# 6.02 ! 1023 molecules &
nCO = 6.977 mol [2 extra digits carried]
2
Step 2. Calculate the molar mass of carbon dioxide.
!
!
g $
g $
M CO = # 12.01
+
2
16.00
#"
2
mol &%
mol &%
"
M CO = 44.01
2
g
mol
Copyright © 2011 Nelson Education Ltd.
Chapter 6: Quantities in Chemical Formulas
6.5-9
Step 3. Use the amount of carbon dioxide and its molar mass to calculate the mass of carbon
dioxide.
! 44.01 g $
mCO = (6.977 mol ) #
&
2
" 1 mol %
mCO = 310 g
2
Statement: The mass of 4.2 × 1024 molecules of carbon dioxide is 310 g.
4. Given: mNaOCl = 120.0 g
Required: number of formula units of sodium hypochlorite, N NaOCl
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
mol
Step 2. Use the mass of sodium hypochlorite and its molar mass to calculate the amount of
sodium hypochlorite.
! 1 mol $
nNaClO = (120.0 g) #
&
" 74.44 g %
M NaOCl = 74.44
nNaClO = 1.612 04 mol [2 extra digits carried]
Step 3. Calculate the number of formula units of sodium hypochlorite using an appropriate
conversion factor.
"
formula units %
N NaOCl = (1.612 04 mol ) $ 6.02 ! 1023
'
#
&
1 mol
N NaOCl = 9.70 ! 1023 formula units
Statement: A 120.0 g sample of sodium hypochlorite contains 9.70 × 1023 formula units.
5. (a) Given: mC H O = 250.0 mg
9
8
4
Required: amount of ASA in one tablet, nC H O
9
8
4
Solution:
Step 1. Calculate the molar mass of ASA, M C H O
9
8
4
!
!
!
g $
g $
g $
M C H O = 9 # 12.01
+ 8 # 1.01
+ 4 # 16.00
&
&
9 8 4
mol %
mol %
mol &%
"
"
"
M C H O = 180.17
9
8
4
g
mol
Copyright © 2011 Nelson Education Ltd.
Chapter 6: Quantities in Chemical Formulas
6.5-10
Step 2. Use the mass of ASA and its molar mass to calculate the amount of ASA.
mC H O = 250.0 mg = 2.500 ! 10"1 g
9
8
4
# 1 mol &
nC H O = (2.500 ! 10"1 g) %
(
9 8 4
$ 180.17 g '
nC H O = 1.387 58 ! 10"3 mol [2 extra digits carried]
9
8
4
Statement: A 250.0 mg sample of ASA contains 1.388 × 10−3 mol.
(b) Given: mC H O = 250.0 mg ; nC H O = 1.387 58 mol
9
8
4
9
8
4
Required: number of hydrogen atoms, N H
Solution:
Step 1. Calculate the number of molecules of ASA using an appropriate conversion factor.
"
molecules %
N C H O = (1.387 58 ! 10-3 mol ) $ 6.02 ! 1023
'
9 8 4
#
1 mol &
N C H O = 8.3532 ! 1020 molecules [2 extra digits carried]
9
8
4
Step 2. Determine the number of hydrogen atoms present.
" 8 atoms %
N H = (8.3532 ! 1020 molecules ) $
'
# 1 molecule &
N H = 6.68 ! 1021 atoms
Statement: There are 6.68 × 1021 hydrogen atoms in 250.0 mg of ASA.
6. (a) Given: mHe = 0.327 g ; VHe = 2.00 L
Required: number of atoms of helium, N He
Solution:
Step 1. Use the mass of helium and its molar mass to calculate the amount of helium.
g
M He = 4.00
mol
! 1 mol $
nHe = (0.327 g) #
&
" 4.00 g %
nHe = 0.081 750 mol [2 extra digits carried]
Step 2. Calculate the number of helium atoms present.
"
atoms %
N He = (0.081 750 mol ) $ 6.02 ! 1023
'
#
1 mol &
N He = 4.9214 ! 1022 atoms [2 extra digits carried]
Statement: A 0.327 g sample of helium contains 4.92 × 1022 atoms.
Copyright © 2011 Nelson Education Ltd.
Chapter 6: Quantities in Chemical Formulas
6.5-11
(b) Given: VHe = 2.00 L ; Vblimp = 5.74 ! 106 L ; N He = 4.9214 ! 1022 atoms
Required: number of helium atoms in blimp, N HeBlimp
Solution:
Use the volume of the blimp and the number of helium atoms per 2.00 L to calculate the number
of helium atoms in the blimp.
" 4.9214 ! 1022 atoms %
N HeBlimp = (5.74 ! 106 L) $
'
2.00 L
#
&
N HeBlimp = 1.41 ! 1029 atoms [2 extra digits carried]
Statement: There are 1.41 × 1029 atoms of helium in the blimp.
7. (a) Given: mAs O = 14 mg
2
3
Required: number of formula units of arsenic(III) oxide, N As O
2
Solution:
Step 1. Calculate the molar mass of As2O3, M As O
2
3
3
!
!
g $
g $
M As O = 2 # 74.92
+
3
16.00
#"
2 3
mol &%
mol &%
"
g
mol
Step 2. Use the mass of As2O3 and its molar mass to calculate the amount of As2O3.
mAs O = 14 mg = 1.4 ! 10"2 g
M As O = 197.84
2
2
3
3
# 1 mol &
nAs O = (1.4 ! 10"2 g) %
(
2 3
$ 197.84 g '
nAs O = 7.076 ! 10"5 mol (2 extra digits carried)
2
3
Step 3. Calculate the number of formula units of As2O3 using an appropriate conversion factor.
"
formula units %
N As O = (7.076 ! 10-5 mol ) $ 6.02 ! 1023
'
2 3
#
&
1 mol
N As O = 4.260 ! 1019 formula units [2 extra digits carried]
2
3
Statement: There are 4.3 × 1019 formula units of As2O3 in a 14 mg sample of the compound.
Copyright © 2011 Nelson Education Ltd.
Chapter 6: Quantities in Chemical Formulas
6.5-12
(b) Given: N As O = 4.260 ! 1019 formula units
2
3
Required: number of oxide ions, N O2Solution:
Calculate the number of oxide ions present using an appropriate conversion factor.
"
%
3 ions
N O2- = (4.260 ! 1019 formula units ) $
# 1 formula units '&
N O2- = 1.3 ! 1020 ions
Statement: There are 1.3 × 1020 oxide ions present in a 14 mg sample of As2O3.
8. (a) Given: mass of toothpaste containing Na2FPO3 = 120.0 g;
percentage of Na2FPO3 in toothpaste = 0.76 %
Required: amount of sodium monofluorophosphate, nNa FPO
2
3
Solution:
Step 1. Calculate the mass of Na2FPO3 in the toothpaste sample.
0.76
mNa FPO =
! 120.0 g
2
3
100
= 0.9120 g [2 extra digits carried]
Step 2. Calculate the molar mass of Na2FPO3, M Na FPO
2
M Na
2 FPO3
3
!
!
g $ !
g $ !
g $
g $
= 2 # 22.99
+ # 19.00
+ # 30.97
+ 3# 16.00
&
&
&
mol % "
mol % "
mol %
mol &%
"
"
g
mol
Step 3. Use the mass of Na2FPO3 and its molar mass to calculate the amount of Na2FPO3.
! 1 mol $
nNa FPO = (0.9120 g) #
&
2
3
" 143.95 g %
M Na
nNa
2 FPO3
2 FPO3
= 143.95
= 6.336 ' 10(3 mol [2 extra digits carried]
Statement: A 120.0 g tube of toothpaste contains 6.3 × 10−3 mol of sodium
monofluorophosphate.
(b) Given: nNa FPO = 6.336 ! 10"3 mol
2
3
Required: number of sodium ions, N Na + , number of monofluorophosphate ions, N FPO 2!
3
Solution:
Step 1. Calculate the number of formula units of Na2FPO3 using an appropriate conversion
factor.
"
formula units %
N Na FPO = (6.336 ! 10 –3 mol ) $ 6.02 ! 1023
'
2
3
#
&
1 mol
N Na
2 FPO3
= 3.814 ! 1021 formula units [2 extra digits carried]
Copyright © 2011 Nelson Education Ltd.
Chapter 6: Quantities in Chemical Formulas
6.5-13
Step 2. Determine the number of sodium ions and monofluorophosphate ions present.
"
%
2 ions
N Na + = (3.814 ! 1021 formula units ) $
'
# 1 formula units &
N Na + = 7.6 ! 1021 ions
#
&
1 ion
N FPO 2! = (3.814 " 1021 formula units ) %
(
3
$ 1 formula units '
N FPO 2! = 3.8 " 1021 ions
3
Statement: A 120.0 g tube of toothpaste contains 7.6 × 1021 sodium ions and
3.8 × 1021 monoflurophosphate ions.
9. (a) Given: mAg = 509.0 g ; mAu = 6.0 g
Required: number of silver atoms, N Ag ; number of gold atoms, N Au
Solution:
Step 1. Look up the molar mass of silver, M Ag , and the molar mass of gold, M Au .
g
mol
g
M Au = 196.97
mol
Step 2. Use the mass and the molar mass to calculate the amount.
! 1 mol $
nAg = (509.0 g) #
&
" 107.87 g %
M Ag = 107.87
nAg = 4.718 64 mol [2 extra digits carried]
! 1 mol $
nAu = (6.0 g) #
&
" 196.97 g %
nAu = 3.046 ' 10(2 mol [2 extra digits carried]
Step 3. Calculate the number of each type of atoms present.
"
atoms %
N Ag = (4.718 64 mol ) $ 6.02 ! 1023
'
1 mol &
#
N Ag = 2.84 ! 1024 atoms
"
atoms %
N Au = (3.046 ! 10 –2 mol ) $ 6.02 ! 1023
'
#
1 mol &
N Au = 1.8 ! 1022 atoms
Statement: The gold medal contains 2.84 × 1024 atoms of silver and 1.8 × 1022 atoms of gold.
Copyright © 2011 Nelson Education Ltd.
Chapter 6: Quantities in Chemical Formulas
6.5-14
(b) Answers may vary. Answers are based on $44.53 per gram of gold and $0.943 per gram of
silver.
$44.53
$ 0.943
Given: cost of Au =
; cost of Ag =
; mmedal = 515.0 g ; mAg = 509.0 g ; mAu = 6.0 g
1g
1g
Required: cost of pure gold medal, CAuMedal ; cost of actual medal, Cmedal
Solution:
Cost of gold medal:
! $44.53 $
CAuMedal = (515 g) #
&
" 1g %
CAuMedal = $22 900
Cost of actual medal:
! $44.53 $
! $0.943 $
Cmedal = (6.0 g) #
+ (509.0 6.0 g) #
&
&
" 1g %
" 1 6.0 g %
Cmedal = $750
Statement: A medal made of pure gold would cost $22 930 while the actual medal would cost
$750, based on $44.53 per gram of gold and $0.943 per gram of silver.
Copyright © 2011 Nelson Education Ltd.
Chapter 6: Quantities in Chemical Formulas
6.5-15