Chapter 3:Mass Relationships in Chemical Reactions

Example no:3.1,Page no:81

In [33]:
#Variable declaration
Cu63=69.09 #percent of Cu 62.93 amu
Cu65=30.91 #percent of Cu 64.9278 amu

#Calculation
AverageAMU=62.93*Cu63/100+64.9278*Cu65/100 # average amu

#Result
print"The average atomic mass of Copper is :",round(AverageAMU,2),"amu\n"

The average atomic mass of Copper is : 63.55 amu

Example no:3.2,Page no:84

In [28]:
#Variable declaration
mass=6.46 #mass of He, g

#Calculation
moles=mass/4.003 #no. of moles of He, as mol. mass of He is 4.003 amu

#Result
print"The no. of moles of He is :",round(moles,2),"mol He\n"

The no. of moles of He is : 1.61 mol He

Example no:3.3,Page no:84

In [30]:
#Variable declaration
moles=0.356 #moles of Zn

#Calculation
mass=moles*65.39 #mass of Zn, g, 1 mole=65.39 g

#Result
print"The mass of Zn is :",round(mass,1),"g Zn\n"

The mass of Zn is : 23.3 g Zn

Example no:3.4,Page no:84

In [32]:
#Variable declaration
Na=6.022*10**23 # Avogadro number, atoms/mol
mass=16.3 #mass of sulfur, g


#Calculation
moles=mass/32.07 #moles of S
atoms=moles*Na #number of atoms of S

#Result
print"The no. of atoms of S is :%.2e"%atoms,"S atoms"

The no. of atoms of S is :3.06e+23 S atoms

Example no:3.5,Page no:86

In [34]:
#Variable declaration
MassO=16.0 #mass of O, amu
MassS=32.07 #mass of S, amu
MassN=14.01 #mass of N, amu
MassH=1.008 #mass of H, amu
MassC=12.01 #mass of C, amu

#Calculation
#(a)
MassSO2=MassS+MassO*2 #mass of SO2, amu
#(b)
MassC8H10N4O2=8*MassC+10*MassH+4*MassN+2*MassO 

#Result
print"(a).The molecular mass of SO2 is :",MassSO2,"amu\n"
print"(b).The molecular mass of C8H10N4O2 is :",MassC8H10N4O2,"amu\n"

(a).The molecular mass of SO2 is : 64.07 amu

(b).The molecular mass of C8H10N4O2 is : 194.2 amu

Example no:3.6,Page no:86

In [35]:
#Variable declaration
Mass=6.07 #mass of CH4, g
MassC=12.01 #mol. mass of C, amu
MassH=1.008 #mol. mass of H, amu

#Calculation
MassCH4=MassC+4*MassH #mol. mass of CH4, amu
Moles=Mass/MassCH4 #no. of moles of CH4

#Result
print"The no. of moles of CH4 is :",round(Moles,3),"mol CH4\n"

The no. of moles of CH4 is : 0.378 mol CH4

Example no:3.7,Page no:87

In [36]:
#Variable declaration
Na=6.07*10**23 # Avogadro number, atoms/mol
Mass=25.6 #mass of Urea, g
MolMass=60.06 #mol. mass of Urea, g

#Calculation
moles=Mass/MolMass #moles of Urea, mol
Atoms=moles*Na*4 #No. of atoms of Hydrogen

#Result
print"The no. of atoms of hydrogen are :%.2e"%Atoms,"H atoms\n"

The no. of atoms of hydrogen are :1.03e+24 H atoms

Example no:3.8,Page no:89

In [37]:
#Variable declaration
H=1.008 #molar mass of H, g
P=30.97 #molar mass of P, g
O=16 #molar mass of O, g
MolMass=97.99 #mol. mass of H3PO4, g

#Calculation
percentH=3*H/MolMass*100 #percent of H
percentP=P/MolMass*100 #percent of P
percentO=4*O/MolMass*100 #percent of O

#Result
print"The percent by mass of Hydrogen is :",round(percentH,3),"%\n"
print"The percent by mass of Phosphorus is :",round(percentP,2),"%\n"
print"The percent by mass of Oxygen is :",round(percentO,2),"%\n"

The percent by mass of Hydrogen is : 3.086 %

The percent by mass of Phosphorus is : 31.61 %

The percent by mass of Oxygen is : 65.31 %

Example no:3.9,Page no:90

In [6]:
#Variable declaration
H=1.008 #molar mass of H, g
C=12.01 #molar mass of C, g
O=16.0 #molar mass of O, g
percentC=40.92 #percent of C

#Calculation
nC=percentC/C 
percentH=4.58 #percent of H
nH=percentH/H 
percentO=54.5 #percent of O
nO=percentO/O 
if(nC>nH):# determining the smallest subscript
    small=nH
else:
    small=nC 
    if(small>nO):
        small=nO 

nC=nC/small #dividing by the smallest subscript
nH=nH/small 
nO=nO/small 
#the approximate values of these variables are to be multiplied by appropriate number to make it an integer by trial and error method
#in this case we need to multiply with 3 to get integer values
nC=nC*3 
nH=nH*3 
nO=nO*3 

#Result
print"The empirical formula of ascorbic acid is : C%.f"%nC,"H%.f"%nH,"O%.f"%nO

The empirical formula of ascorbic acid is : C3 H4 O3

Example no:3.10,Page no:91

In [1]:
#Variable declaration
massCuFeS2=3.71*10**3 #given mass of CuFeS2, kg
CuFeS2=183.5 #mol. mass of CuFeS2, g
Cu=63.55 #mol. mass of Cu, g

#Calculation
percentCu=Cu/CuFeS2*100 #percent Cu in CuFeS2
massCu=percentCu*massCuFeS2/100#mass of Cu in given CuFeS2, kg

#Result
print"The mass of Cu in CuFeS2 is :%.2e"%massCu,"kg\n"

The mass of Cu in CuFeS2 is :1.28e+03 kg

Example no:3.11,Page no:93

In [9]:
#Variable declaration
M_N=1.52      #Mass of nitrogen in g
M_O=3.47      #Mass of oxygen in g
N=14.01       #Atomic mass of N
Molar1=95     #Molar mass in g 
O=16          #Atomic mass of O

#Calculation
n_N=M_N/N       #No of moles of N
n_O=M_O/O       #No of moles of O
emp=N+2*(O)     #Empirical molar massof NO2
ratio=Molar1/emp
ratio=round(ratio)
actual=ratio*emp

#Result
print"Molecular formula is N%.3g"%n_N,"O%.3g"%n_O
print "Actual molar mass is",actual,"g,which is between 90 g and 95 g"

Molecular formula is N0.108 O0.217
Actual molar mass is 92.02 g,which is between 90 g and 95 g

Example no:3.13,Page no:101

In [3]:
#Variable declaration
CO2=44.01 #mol. mass of CO2, g
Glucose=180.2 #mol. mass of Glucose, g
massGlucose=856 #given mass of Glucose, g

#Calculation
moleGlucose=massGlucose/Glucose # moles of glucose
moleCO2=moleGlucose*6 #1 mole glucose gives 6 moles of CO2
massCO2=moleCO2*CO2 # mass of CO2, g

#Result
print"The mass of CO2 is :%.2e"%massCO2,"g CO2\n"

The mass of CO2 is :1.25e+03 g CO2

Example no:3.14,Page no:102

In [4]:
#Variable declaration
H2=2.016 #mol. mass of H2, g
Li=6.941 #mol. mass of Li, g
mH2=9.89 #mass of H2, g

#Calculation
nH2=mH2/H2 #moles of H2
nLi=2*nH2 #moles of Li, 1mol H2 given by 2mol Li
mLi=Li*nLi ##mass of Li, g

#Result
print"The mass of Li is :",round(mLi,1),"g Li\n"

The mass of Li is : 68.1 g Li

Example no:3.15,Page no:104

In [5]:
#Variable declaration
Urea=60.06 #mol. mass of Urea, g
NH3=17.03 #mol. mass of NH3, g
CO2=44.01 #mol. mass of CO2, g

#(a)
#for NH3
mNH3=637.2 #mass of NH3, g

#Calculation
nNH3=mNH3/NH3 #moles of NH3
nUrea1=nNH3/2#moles of Urea
#for CO2
mCO2=1142#mol. mass of CO2, g
nCO2=mCO2/CO2 #moles of CO2
nUrea2=nCO2 #moles of Urea
if(nUrea1>nUrea2): #finding limiting reagent
    nUrea=nUrea2 
    limiting="CO2" 
else:
    limiting="NH3" 
    nUrea=nUrea1 
    
#Result
print"(a).The limiting reagent is :",limiting
#(b)
mUrea=nUrea*Urea #mass of urea produced
print"(b).The mass of the Urea produced is :",round(mUrea),"g (NH2)2CO"

#(c)
if(limiting=="NH3") :#finding excess reagent
    nCO2excess=nCO2-nNH3/2 
    mCO2excess=nCO2excess*CO2 
    print"(c).The mass of excess CO2 is :",round(mCO2excess),"g \n"
else: 
    nNH3excess=nNH3-2*nCO2 
    mNH3excess=nNH3excess*NH3 
    print"The mass of excess NH3 is :",mNH3excess,"g\n"

(a).The limiting reagent is : NH3
(b).The mass of the Urea produced is : 1124.0 g (NH2)2CO
(c).The mass of excess CO2 is : 319.0 g

Example no:3.16,Page no:106

In [27]:
#(a)
#for TiCl4

#Variable declaration
mTiCl4=3.54*10**7 #mass of TiCl4, g
nTiCl4=mTiCl4/189.7 #moles of TiCl4
nTi1=nTiCl4*1.0 #moles of Ti

#for Mg
mMg=1.13*10**7 #mass of Mg, g
nMg=mMg/24.31 #moles of Mg
nTi2=nMg/2.0 #moles of Ti


#Calculation
if(nTi1>nTi2):   #finding imiting reagent
     nTi=nTi2 
else:
    nTi=nTi1 
    mTi=nTi*47.88 
    print"(a).The theoretical yield is :%.2e"%mTi,"g\n"
    print"\nNOTE:Variation from answer due to approx value of mTi taken in book\n"

#(b)

mTiactual=7.91*10**6 #given, actual Ti produced
p_yield=mTiactual/mTi*100.0 
print"(b).The percent yield is :",round(p_yield,1),"%\n"

(a).The theoretical yield is :8.93e+06 g


NOTE:Variation from answer due to approx value of mTi taken in book

(b).The percent yield is : 88.5 %