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Mole Calculations
West Midlands Chemistry Teachers Centre
Tuesday 23rd November 2010
Presenter: Dr Janice Perkins
Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)
Reacting ratio in equations
1 ‘formula’ 3 molecules 2 atoms 3 molecules
10 ‘formulae’ 30 molecules 20 atoms 30 molecules
1106 ‘formulae’
3106 molecules
2106 atoms
3106 molecules
1dozen ‘formulae’
3 dozen molecules
2 dozen atoms
3 dozen molecules
Funny numbers
Dozen = 12 Gross = 12 12 = 144
6.023 1023
602300000000000000000000
Mole = Score = 20
That’s
602300000000000000000000
The Avogadro Constant (L)
6.02 1023
Or
It is just a number – no more special than a ton, a score or a dozen – its
just a bit bigger!
Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)
Reacting Ratio
1 formula 3 molecules 2 atoms 3 molecules
10 ‘formulae’ 30 molecules 20 atoms 30 molecules
1106 ‘formulae’
3106 molecules
2106 atoms
3106 molecules
1 mole 3 moles 2 moles 3 moles
6.021023 ‘formulae’
18.06 1023
molecules12.04 1023
atoms18.06 1023 molecules
3:2
Fe2O3 : Fe = 1:2
Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)
1:2
Mole Ratio (Reacting Ratio)
CO : Fe = 3:2
CO : CO2 = 1:1
3:3
Fe2O3 : CO = 1:3
1:3
Sufficient data to calculate moles?
Calculation based on
Data needed
Mass
Solution
Gases
Mass Mass Mass Mass and Mr Mass Mass and Mr or Formula Mass Mass and Mr or Formula or Name
Solution Volume Solution Volume and Concentration
Gases PGases P and T Gases P and T and VGases P and T and V and R
Mole ratio
from equation
Moles of known
substance
Moles of unknown substance
Mass
Volume of solution
Volume of gas
Mass
Mr
Mole Calculations
Moles = MassMr
n = mMr
n = v c
n = pVRT
Moles = volume concMoles = P(in Pa) V(in m3)
R T(in Kelvin)Ideal Gas Equation pV = nRT
Mass = Moles Mr
Mass
Mr = MassMoles
Rearranging the formula
Moles = MassMr
Moles Mr
Mass
Moles Mr
Mass
Moles Mr
Mole ratio
from equation
Moles of known
substance
Moles of unknown substance
Mass
Volume of solution
Volume of gas
Mass
Mr
Vol
Conc
Mole Calculationsn = m
Mr
n = v c
n = pVRT
m = n Mr
Mr = m n
Vol = MolesConc
Moles
Rearranging the formula
Moles = Volume Concentration
Vol Conc
Conc = Moles Vol
Moles
Vol Conc
Moles
Vol Conc
Mole ratio
from equation
Moles of known
substance
Moles of unknown substance
Mass
Volume of solution
Volume of gas
Mass
Mr
Vol
Conc
V
P
T
Mole Calculationsn = m
Mr
n = v c
n = pVRT
m = n Mr
Mr = m n
c = n
v
v = n c
Volume = nRTp
Units are vital:
‘V’ always in m3
‘P’ always in Pa
‘T’ always in Kelvin
Rearranging the formula
Moles = pVRT
Pressure = nRTV
Temperature = pVnR
Mole ratio
from equation
Moles of known
substance
Moles of unknown substance
Mass
Volume of solution
Volume of gas
Mass
Mr
Vol
Conc
V
P
T
Mole Calculationsn = m
Mr
n = v c
n = pVRT
m = n Mr
Mr = m n
c = n
v
v = n c
V = nRTp
p = nRTV
T = pVnR
Example 1: Calculating moles from masses
Use the equation Mass = Mr x moles
Mr = 100
Rearrange equation
Moles = mass/Mr
= 300 100 = 3 moles
Calculate the number of moles in 300g of CaCO3
First we need to calculate the number of moles of HCl.
Moles = mass/Mr
Moles HCl = 19.6
Conc of HCl(aq) = moles =volume (in dm3)
=
Calculate the concentration, in mol dm-3, of the solution formed when 19.6 g of hydrogen chloride, HCl, are dissolved in water and the volume made up to 250 cm3.
Conc of HCl(aq) = moles = 0.537volume (in dm3)
=
Example 2: Calculating concentration of solution
Calculate the concentration, in mol dm-3, of the solution formed when 19.6 g of hydrogen chloride, HCl, are dissolved in water and the volume made up to 250 cm3.
Moles HCl = 19.6 36.5
Moles HCl = 19.6 = 0.537 mol36.5
Conc of HCl(aq) = moles = 0.537volume (in dm3) 250 10-3
=
Conc of HCl(aq) = moles = 0.537volume (in dm3) 250 10-3
= 2.15 mol dm-3
A metal carbonate MCO3 reacts with HCl as in the following equation.
MCO3 + 2HCl MCl2 + H2O + CO2
A 0.548 g sample of MCO3 reacted completely with 30.7 cm3 of 0.424 mol dm-3 HCl.
(a) Calculate the amount, in moles, of HCl which reacted with the 0.548 g of MCO3
Moles = vol x conc (in dm3)
= 30.7/1000 x 0.424
Example 3: Calculating moles of solution and solid, then Mr and Ar (Jan 09 chem1)
= 0.0130 mol
(b) Calculate the amount, in moles, of MCO3 in 0.548 g
Look at equation again
MCO3 + 2HCl MCl2 + H2O + CO2
There is a 2:1 ratio of reactants.
We have calculated that there are 0.0130 mol of HCl
so there must be half that amount of MCO3
= 0.0130/2
= 0.0065 mol = 6.50 x 10-3 mol
(d) Use your answer to deduce the Ar of M
We have just calculated the Mr of MCO3 to be 84.3
Since there is 1xC and 3xO this makes 12 + 48 = 60.
So this means M must have an Ar of 84.3 – 60 = 24.3
This is Magnesium.
(c) Calculate the Mr of MCO3
A mass of 0.548g of MCO3 contains 0.0065 mol.
Use the equation Mr = mass/moles
Mr = 0.548/0.0065 = 84.3
Example 4: Identity unknown – less structured calculationThe carbonate of metal M has the formula M2CO3. The
equation for the reaction of this carbonate with hydrochloric acid is given below.
M2CO3 + 2HCl 2MCl + CO2 + H2O
A sample of M2CO3, of mass 0.394 g, required the addition
of 21.7 cm3 of 0.263 mol dm-3 solution of hydrochloric acid for complete reaction.
Calculate the Ar of metal M and deduce its identity.(a) Find moles of known substance - in this case HCl
Moles HCl(aq) = volume concentration==
The carbonate of metal M has the formula M2CO3. The
equation for the reaction of this carbonate with hydrochloric acid is given below.
M2CO3 + 2HCl 2MCl + CO2 + H2O
A sample of M2CO3, of mass 0.394 g, required the addition
of 21.7 cm3 of 0.263 mol dm-3 solution of hydrochloric acid for complete reaction.
Calculate the Ar of metal M and deduce its identity.
Moles HCl(aq) = volume concentration= 21.7=
The carbonate of metal M has the formula M2CO3. The
equation for the reaction of this carbonate with hydrochloric acid is given below.
M2CO3 + 2HCl 2MCl + CO2 + H2O
A sample of M2CO3, of mass 0.394 g, required the addition
of 21.7 cm3 of 0.263 mol dm-3 solution of hydrochloric acid for complete reaction.
Calculate the Ar of metal M and deduce its identity.
Moles HCl(aq) = volume concentration= 21.7 10-3
=
Moles HCl(aq) = volume concentration= 21.7 10-3 0.263=
Moles HCl(aq) = volume concentration= 21.7 10-3 0.263= 5.71 10-3 mol
(b) Calculate the moles of the other substance- in this case M2CO3
The mole ratio is 2:1
We have calculated the moles HCl = 5.71 10-3 mol
So the moles M2CO3 = 5.71 10-3 /2 = 2.85 x 10-3 mol
M2CO3 + 2HCl 2MCl + CO2 + H2O
(c) Now find Mr of M2CO3
Mr of M2CO3 = mass = moles
=
Mr of M2CO3 = mass = 0.394 moles
=
Mr of M2CO3 = mass = 0.394 moles 2.85 10-3
=
Mr of M2CO3 = mass = 0.394 moles 2.85 10-3
= 138(d) find Ar of metal M and hence deduce its identity
2 Ar(M) = Mr(M2CO3) - Mr(‘CO3’)
= 138 – 60 = 78
Ar(M) = 78/2 = 39
M = Potassium
Use pV = nRTUse pV = nRT
V = nRT = p
=
=
Use pV = nRT
V = nRT = 0.0296 8.31 366 p 100000
=
=
Use pV = nRT
V = nRT = 0.0296 8.31 366 p 100000
= 8.992 10-4
=
Example 5: Gases – calculating the volume
A sample of ethanol vapour, C2H5OH (Mr = 46), was
maintained at a pressure of 100 kPa and at a temperature of 366 K. Use the ideal gas equation to calculate the volume, in cm3, that 1.36 g of ethanol vapour would occupy under these conditions. (The gas constant R = 8.31 J K-1 mol-1)
Moles ethanol = 1.36
A sample of ethanol vapour, C2H5OH (Mr = 46), was
maintained at a pressure of 100 kPa and at a temperature of 366 K. Use the ideal gas equation to calculate the volume, in cm3, that 1.36 g of ethanol vapour would occupy under these conditions. (The gas constant R = 8.31 J K-1 mol-1)
A sample of ethanol vapour, C2H5OH (Mr = 46), was
maintained at a pressure of 100 kPa and at a temperature of 366 K. Use the ideal gas equation to calculate the volume, in cm3, that 1.36 g of ethanol vapour would occupy under these conditions. (The gas constant R = 8.31 J K-1 mol-1)
Moles ethanol = 1.3646
Moles ethanol = 1.36 = n = 0.0296 mol46
Use pV = nRT
V = nRT = 0.0296 8.31 366 p 100000
= 8.992 10-4 m3
=
Use pV = nRT
V = nRT = 0.0296 8.31 366 p 100000
= 8.992 10-4 m3
= 8.992 10-4 106
Use pV = nRT
V = nRT = 0.0296 8.31 366 p 100000
= 8.992 10-4 m3
= 8.992 10-4 106 = 899 cm3
An oxide of nitrogen contains 30.4% by mass of nitrogen, Calculate the empirical formula.First calculate the % of O 100 – 30.4 = 69.6%Now put in columns N Omass 30.4 69.6 Ar 14 16= moles 2.17 4.35Ratio 2.17/2.17 = 1 4.35/2.17= 2
Empirical formula = NO2
Example 6: Empirical formula Jun 09
Mr(Na2CO3 • xH2O) = 250
Example 7: Water of crystallisation
The Mr of a hydrated sodium carbonate was found to be 250. Use this value to calculate the number of molecules of water of crystallisation, x, in this hydrated sodium carbonate, Na2CO3
• xH2O.
The Mr of a hydrated sodium carbonate was found to be 250. Use this value to calculate the number of molecules of water of crystallisation, x, in this hydrated sodium carbonate, Na2CO3
• xH2O.
Mr(Na2CO3 • xH2O) = 250 Mr(Na2CO3) = 106
Mr(Na2CO3 • xH2O) = 250 Mr(Na2CO3) = 106
x Mr(H2O) = Mr(Na2CO3 • xH2O)
Mr(Na2CO3 • xH2O) = 250 Mr(Na2CO3) = 106
x Mr(H2O) = Mr(Na2CO3 • xH2O) - Mr(Na2CO3)
Mr(Na2CO3 • xH2O) = 250 Mr(Na2CO3) = 106
x Mr(H2O) = Mr(Na2CO3 • xH2O) - Mr(Na2CO3)
= 250 – 106
Mr(Na2CO3 • xH2O) = 250 Mr(Na2CO3) = 106
x Mr(H2O) = Mr(Na2CO3 • xH2O) - Mr(Na2CO3)
= 250 – 106 = 144
Mr(Na2CO3 • xH2O) = 250 Mr(Na2CO3) = 106
x Mr(H2O) = Mr(Na2CO3 • xH2O) - Mr(Na2CO3)
= 250 – 106 = 144
x 18 = 144
Mr(Na2CO3 • xH2O) = 250 Mr(Na2CO3) = 106
x Mr(H2O) = Mr(Na2CO3 • xH2O) - Mr(Na2CO3)
= 250 – 106 = 144
x 18 = 144
x = 144/18
Mr(Na2CO3 • xH2O) = 250 Mr(Na2CO3) = 106
x Mr(H2O) = Mr(Na2CO3 • xH2O) - Mr(Na2CO3)
= 250 – 106 = 144
x 18 = 144
x = 144/18 = 8
Mr(Na2CO3 • xH2O) = 250 Mr(Na2CO3) = 106
x Mr(H2O) = Mr(Na2CO3 • xH2O) - Mr(Na2CO3)
= 250 – 106 = 144
x 18 = 144
x = 144/18 = 8
Example 8: Water of crystallisation – less structured
Sodium carbonate forms a number of hydrates of general formula Na2CO3
• xH2O. A 3.01 g sample of one of these hydrates was dissolved in water and the solution made up to 250 cm3.In titration, a 25 cm3 portion of this solution required 24.3 cm3 of 0.200 mol dm-3 hydrochloric acid for complete reaction
Na2CO3 + 2HCl 2NaCl + H2O + CO2
Calculate the Mr of Na2CO3 • xH2O
Sodium carbonate forms a number of hydrates of general formula Na2CO3
• xH2O. A 3.01 g sample of one of these hydrates was dissolved in water and the solution made up to 250 cm3.In titration, a 25 cm3 portion of this solution required 24.3 cm3 of 0.200 mol dm-3 hydrochloric acid for complete reaction
Na2CO3 + 2HCl 2NaCl + H2O + CO2
Calculate the Mr of Na2CO3 • xH2O(a) Calculate moles HCl since you can do this
Moles HCl =volume concentration = 24.3 10-3 0.200 = 4.86 10-3 mol
(b) Now calculate the moles of Na2CO3 in the 25 cm3 sample of solution.
Na2CO3 + 2HCl 2NaCl + H2O + CO2
Moles Na2CO3 =Mole Ratio Moles HCl
(c) deduce the moles of Na2CO3 in 250 cm3 of solution
Moles Na2CO3 (250) = Moles Na2CO3 (25)
Moles Na2CO3 (250) = Moles Na2CO3 (25) 250
25
Moles Na2CO3 (250) = Moles Na2CO3 (25) 250
25= 2.43 10-2
Moles Na2CO3 =Mole Ratio Moles HCl
= 1/2 4.86 10-3
Moles Na2CO3 =Mole Ratio Moles HCl
= ½ 4.86 10-3 = 2.43 10-3
(d) Calculate the Mr of hydrated Na2CO3 [Mass = 3.01 g]
Mr= Mass / MolesMr = 3.01/2.43 10-2 = 124
(a) Calculate the amount, in moles, of TiCl4 in 165 g
Moles = mass = 165
Mr 189.9 = 0.869
Example 9: Percentage yield Jan 09
(b) Calculate the maximum amount, in moles, of TiO2 which can be formed in this experiment.
Look at the equation.
1 mole of TiCl4 gives 1 mole of TiO2
So 0.869 moles of TiCl4 gives 0.869 moles of TiO2
In an experiment 165 g TiCl4 were added to an excess of water.
The equation for the reaction is as follows
TiCl4 + 2H2O TiO2 + 4HCl
(c) Calculate the maximum mass of TiO2 formed in this experiment.
mass = Mr x moles
= 79.9 x 0.869
= 69.4 g
(d) In this experiment only 63.0 g of TiO2 were produced. Calculate the percentage yield.
Percentage yield = actual (experimental) mass of product theoretical ( calculated) mass of product = 63.0 69.4 = 90.8%
% atom economy = mass of desired product x 100 total mass of reactants
Example 10 : % atom economy
You should learn this formula and be able to use it correctly.
Calculate the % atom economy for the formation of CH2Cl2 in this reaction.
CH4 + 2Cl2 CH2Cl2 +2HCl
There are no numbers given
You only need the Mr values
Desired product = CH2Cl2 Mr =(12+ 2 +71) = 85
Reactants =CH4+2Cl2 Mr = (12+4)+(2 x71)= 158
%atom economy = 85 x 100 = 53.8% 158
The chloride of an element Z reacts with water according to the following equation.
ZCl4(l) + 2H2O(l) ZO2(s) + 4HCl(aq)
A 1.304 g sample of ZCl4 was added to water. The solid ZO2 was removed by filtration and the resulting solution was made up to 250 cm3 in a volumetric flask.
25.0 cm3 of this solution was titrated against 0.112 mol dm-3 NaOH(aq), and 21.7 cm3 were needed to reach the end point.
Use this information to calculate the number of moles of HCl produced and hence the number of moles of ZCl4 present in the sample.
Calculate the Mr of ZCl4. From your answer deduce the Ar of element Z and hence its identity.
Example 11: Moles HCl, moles & identity ‘Z’The chloride of an element Z reacts with water according to the following equation.
ZCl4(l) + 2H2O(l) ZO2(s) + 4HCl(aq)
A 1.304 g sample of ZCl4 was added to water. The solid ZO2 was removed by filtration and the resulting solution was made up to 250 cm3 in a volumetric flask.
25.0 cm3 of this solution was titrated against 0.112 mol dm-3 NaOH(aq), and 21.7 cm3 were needed to reach the end point.
Use this information to calculate the number of moles of HCl produced and hence the number of moles of ZCl4 present in the sample.
Calculate the Mr of ZCl4. From your answer deduce the Ar of element Z and hence its identity.
The chloride of an element Z reacts with water according to the following equation.
ZCl4(l) + 2H2O(l) ZO2(s) + 4HCl(aq)
A 1.304 g sample of ZCl4 was added to water. The solid ZO2 was removed by filtration and the resulting solution was made up to 250 cm3 in a volumetric flask.
25.0 cm3 of this solution was titrated against 0.112 mol dm-3 NaOH(aq), and 21.7 cm3 were needed to reach the end point.
Use this information to calculate the number of moles of HCl produced and hence the number of moles of ZCl4 present in the sample.
Calculate the Mr of ZCl4. From your answer deduce the Ar of element Z and hence its identity.
The chloride of an element Z reacts with water according to the following equation.
ZCl4(l) + 2H2O(l) ZO2(s) + 4HCl(aq)
A 1.304 g sample of ZCl4 was added to water. The solid ZO2 was removed by filtration and the resulting solution was made up to 250 cm3 in a volumetric flask.
25.0 cm3 of this solution was titrated against 0.112 mol dm-3 NaOH(aq), and 21.7 cm3 were needed to reach the end point.
Use this information to calculate the number of moles of HCl produced and hence the number of moles of ZCl4 present in the sample.
Calculate the Mr of ZCl4. From your answer deduce the Ar of element Z and hence its identity.
The chloride of an element Z reacts with water according to the following equation.
ZCl4(l) + 2H2O(l) ZO2(s) + 4HCl(aq)
A 1.304 g sample of ZCl4 was added to water. The solid ZO2 was removed by filtration and the resulting solution was made up to 250 cm3 in a volumetric flask.
25.0 cm3 of this solution was titrated against 0.112 mol dm-3 NaOH(aq), and 21.7 cm3 were needed to reach the end point.
Use this information to calculate the number of moles of HCl produced and hence the number of moles of ZCl4 present in the sample.
Calculate the Mr of ZCl4. From your answer deduce the Ar of element Z and hence its identity.
The chloride of an element Z reacts with water according to the following equation.
ZCl4(l) + 2H2O(l) ZO2(s) + 4HCl(aq)
A 1.304 g sample of ZCl4 was added to water. The solid ZO2 was removed by filtration and the resulting solution was made up to 250 cm3 in a volumetric flask.
25.0 cm3 of this solution was titrated against 0.112 mol dm-3 NaOH(aq), and 21.7 cm3 were needed to reach the end point.
Use this information to calculate the number of moles of HCl produced and hence the number of moles of ZCl4 present in the sample.
Calculate the Mr of ZCl4. From your answer deduce the Ar of element Z and hence its identity.
The chloride of an element Z reacts with water according to the following equation.
ZCl4(l) + 2H2O(l) ZO2(s) + 4HCl(aq)
A 1.304 g sample of ZCl4 was added to water. The solid ZO2 was removed by filtration and the resulting solution was made up to 250 cm3 in a volumetric flask.
25.0 cm3 of this solution was titrated against 0.112 mol dm-3 NaOH(aq), and 21.7 cm3 were needed to reach the end point.
Use this information to calculate the number of moles of HCl produced and hence the number of moles of ZCl4 present in the sample.
Calculate the Mr of ZCl4. From your answer deduce the Ar of element Z and hence its identity.
The chloride of an element Z reacts with water according to the following equation.
ZCl4(l) + 2H2O(l) ZO2(s) + 4HCl(aq)
A 1.304 g sample of ZCl4 was added to water. The solid ZO2 was removed by filtration and the resulting solution was made up to 250 cm3 in a volumetric flask.
25.0 cm3 of this solution was titrated against 0.112 mol dm-3 NaOH(aq), and 21.7 cm3 were needed to reach the end point.
Use this information to calculate the number of moles of HCl produced and hence the number of moles of ZCl4 present in the sample.
Calculate the Mr of ZCl4. From your answer deduce the Ar of element Z and hence its identity.
The chloride of an element Z reacts with water according to the following equation.
ZCl4(l) + 2H2O(l) ZO2(s) + 4HCl(aq)
A 1.304 g sample of ZCl4 was added to water. The solid ZO2 was removed by filtration and the resulting solution was made up to 250 cm3 in a volumetric flask.
25.0 cm3 of this solution was titrated against 0.112 mol dm-3 NaOH(aq), and 21.7 cm3 were needed to reach the end point.
Use this information to calculate the number of moles of HCl produced and hence the number of moles of ZCl4 present in the sample.
Calculate the Mr of ZCl4. From your answer deduce the Ar of element Z and hence its identity.
The chloride of an element Z reacts with water according to the following equation.
ZCl4(l) + 2H2O(l) ZO2(s) + 4HCl(aq)
A 1.304 g sample of ZCl4 was added to water. The solid ZO2 was removed by filtration and the resulting solution was made up to 250 cm3 in a volumetric flask.
25.0 cm3 of this solution was titrated against 0.112 mol dm-3 NaOH(aq), and 21.7 cm3 were needed to reach the end point.
Use this information to calculate the number of moles of HCl produced and hence the number of moles of ZCl4 present in the sample.
Calculate the Mr of ZCl4. From your answer deduce the Ar of element Z and hence its identity.
• The only complete data we have is for NaOH(aq) • From this we can calculate the moles of HCl in 25.0 cm3 • We scale this to give the number of moles of HCl, in
250 cm3, formed in the reaction
Moles HCl(25) = mole ratio moles of NaOH
Moles NaOH(aq) = volume concentrationMoles NaOH(aq) = volume concentration= 21.7
Moles NaOH(aq) = volume concentration= 21.7 10-3
Moles NaOH(aq) = volume concentration= 21.7 10-3 0.112
Moles NaOH(aq) = volume concentration= 21.7 10-3 0.112= 2.43 10-3 mol
Moles HCl(25) = mole ratio moles of NaOH= 2.43 10-3
Moles HCl(25) = mole ratio moles of NaOH= 1 2.43 10-3
Moles HCl(25) = mole ratio moles of NaOH= 1 2.43 10-3 = 2.43 10-3
Moles HCl(25) = mole ratio moles of NaOH= 1 2.43 10-3 = 2.43 10-3
Moles HCl(250) = 2.43 10-3 250/25
Moles HCl(25) = mole ratio moles of NaOH= 1 2.43 10-3 = 2.43 10-3
Moles HCl(250) = 2.43 10-3 250/25 = 0.0243 mol
Find HCl : ZCl4 mole ratio and hence find moles ZCl4
ZCl4(l) + 2H2O(l) ZO2(s) + 4HCl(aq)
Find Mr of ZCl4 – then Ar of Z and hence identify Z
Moles ZCl4 = Mole Ratio Moles HCl
Mr of ZCl4 = mass = moles
ZCl4(l) + 2H2O(l) ZO2(s) + 4HCl(aq)
Moles ZCl4 = Mole Ratio Moles HCl
= 0.0243
Moles ZCl4 = Mole Ratio Moles HCl
= 1/4 or 4/1 0.0243
Moles ZCl4 = Mole Ratio Moles HCl
= 1/4 0.0243
Moles ZCl4 = Mole Ratio Moles HCl
= 1/4 0.0243 = 6.076 10-3
Mr of ZCl4 = mass = 1.304 moles
Mr of ZCl4 = mass = 1.304 moles 6.076 10-3
Mr of ZCl4 = mass = 1.304 = 214.6 moles 6.076 10-3
Mr of ZCl4 = mass = 1.304 = 214.6 moles 6.076 10-3
Ar of Z = Mr (ZCl4)
Mr of ZCl4 = mass = 1.304 = 214.6 moles 6.076 10-3
Ar of Z = Mr (ZCl4) – 4 Ar (Cl)
Mr of ZCl4 = mass = 1.304 = 214.6 moles 6.076 10-3
Ar of Z = Mr (ZCl4) - 4 Ar (Cl) = 214.6 - 142
Mr of ZCl4 = mass = 1.304 = 214.6 moles 6.076 10-3
Ar of Z = Mr (ZCl4) - 4 Ar (Cl) = 214.6 - 142 = 72.6
Mr of ZCl4 = mass = 1.304 = 214.6 moles 6.076 10-3
Ar of Z = Mr (ZCl4) - 4 Ar (Cl) = 214.6 - 142 = 72.6
Element Z = Ge
Mr of ZCl4 = mass = 1.304 (given in q) = 214.6 moles 6.076 10-3
Ar of Z = Mr (ZCl4) - 4 Ar (Cl) = 214.6 - 142 = 72.6
Element Z = Ge So, ZCl4 = GeCl4