ch4

Worked solutions to textbook questions 1

Heinemann Chemistry 1 (4th edition)

Reed International Books Australia Pty Ltd

Chapter 4 Relative atomic mass and the mole

E1.

A sample of argon (Ar) known to contain isotopes of mass numbers 36, 39 and 40 is

introduced in a mass spectrometer. The sample is bombarded with electrons to form

positively charged ions.

a Which ion, Ar36

18+ or Ar40

18+, is likely to be deflected most in the magnetic field?

Give a reason for your answer.

b Some atoms lose two electrons in the ionisation chamber. Which ion, Ar36

18+ or

Ar36

182+, is likely to be deflected most in the magnetic field? Give a reason for

your answer.

AE1.

a Ar36

18+ has the lighter mass, so will be deflected the most.

b Ar36

182+ has the higher positive charge and so will be deflected the most.

Q1.m

Use the data in Table 4.2 on page 55 to calculate the relative atomic mass of:

a oxygen

b silver

c hydrogen

A1.

a Ar(O) = 100

)20.0999.17()04.0999.16()76.99995.15(

= 15.999

b Ar(Ag) = 100

)2.489.108()8.519.106(

= 108

c Ar(H) = 100

)0001.0016.3()014.0014.2()986.99008.1(

= 1.008




Q2.

The element lithium has two isotopes:

• 6Li has a relative isotopic mass of 6.02.

• 7Li has a relative isotopic mass of 7.02.

The relative atomic mass of lithium is 6.94. Calculate the percentage abundance of the

lighter isotope.

A2.

Let the percentage abundance of the lighter isotope be x%.

percentage abundance of the heavier isotope will be (100 – x)%.

6.94 = 100

)02.7)100(()02.6( xx

x = 8.0

proportion of the lighter isotope = 8.0%

Q3.

Calculate the relative molecular masses of:

a sulfuric acid (H2SO4)

b ammonia (NH3)

c ethane (C2H6)

A3.

a Mr(H2SO4) = 2 × Ar(H) + Ar(S) + 4 × Ar(O)

= 2 × 1.008 + 32.06 + 4 × 16.00

= 98.1

b Mr(NH3) = Ar(N) + 3 × Ar(H)

= 14.007 + 3 × 1.008

= 17.0

c Mr(C2H6) = 2 × Ar(C) + 6 × Ar(H)

= 2 × 12.01 + 6 × 1.008

= 30.1

Q4.

Calculate the relative formula mass of:

a potassium chloride (KCl)

b sodium carbonate (Na2CO3)

c aluminium sulfate (Al2(SO4)3)

A4.

a RFM(KCl) = Ar(K) + Ar(Cl)

= 39.01 + 35.45

= 74.5

b RFM(Na2CO3) = 2 × Ar(Na) + Ar(C) + 3 × Ar(O)

= 2 × 22.99 + 12.01 + 3 × 16.00

= 106

c RFM(Al2(SO4)3) = 2 × Ar(Al) + 3 × Ar(S) + 12 × Ar(O)

= 2 × 26.98 + 3 × 32.06 + 12 × 16.00

= 342




Q5.

Calculate the number of:

a atoms in 2.0 mol of sodium atoms (Na)

b molecules in 0.10 mol of nitrogen molecules (N2)

c atoms in 20.0 mol of carbon atoms (C)

d molecules in 4.2 mol of water molecules (H2O)

e atoms in 1.0 × 10–2 mol of iron atoms (Fe)

f molecules in 4.62 × 10–5 mol of CO2 molecules

A5.

Number of particles = amount (mol) NA

a Number of sodium atoms (Na) = 2.0 6.02 1023

= 1.2 1024

b Number of nitrogen molecules (N2) = 0.10 6.02 1023

= 6.02 1022

c Number of carbon atoms (C) = 20.0 6.02 1023

= 1.20 1025

d Number of water molecules (H2O) = 4.2 6.02 1023

= 2.5 1024

e Number of iron atoms (Fe) = 1.0 10–2 6.02 1023

= 6.0 1021

f Number of CO2 molecules = 4.62 10–5 6.02 1023

= 2.78 1019




Q6.

Calculate the amount of substance (in mol) represented by:

a 3.0 × 1023 molecules of water (H2O)

b 1.5 × 1023 atoms of neon (Ne)

c 4.2 × 1025 atoms of iron (Fe)

d 4.2 × 1025 molecules of ethanol (C2H5OH)

A6.

Remember: Avogadro’s number, NA, is 6.02 1023.

Amount (mol) =

A

particles ofnumber

N

a Amount (H2O molecules) =

A

23100.3

N

= 0.5 mol

b Amount (Ne atoms) =

A

23105.1

N

= 0.25 mol

c Amount (Fe atoms) =

A

25102.4

N

= 70 mol

d Amount (C2H5OH molecules) =

A

25102.4

N

= 70 mol




Q7.

Calculate the amount (in mol) of:

a sodium atoms represented by 1.0 × 1020 sodium atoms

b aluminium represented by 1.0 × 1020 aluminium atoms

c chlorine molecules represented by 1.0 × 1020 chlorine molecules

A7.

Remember: Avogadro’s number, NA, is 6.02 1023.

Amount (mol) =

A

particles ofnumber

N

a Amount (Na atoms) =

A

20100.1

N

= 1.7 10–4 mol

b Amount (Al atoms) =

A

20100.1

N

= 1.7 10–4 mol

c Amount (Cl2 molecules) =

A

20100.1

N

= 1.7 10–4 mol

Q8.

Calculate the amount (in mol) of:

a chlorine atoms in 0.4 mol of chlorine (Cl2)

b hydrogen atoms in 1.2 mol of methane (CH4)

c hydrogen atoms in 0.12 mol of ethane (C2H6)

d oxygen atoms in 1.5 mol of sodium sulfate (Na2SO4)

A8.

a Each Cl2 molecule has 2 Cl atoms.

amount (Cl atoms) = 0.4 2

= 0.8 mol

b Each methane molecule has 4 H atoms.

amount (H atoms) = 4 1.2

= 4.8 mol

c Each ethane molecule has 6 H atoms.

amount (H atoms) = 6 0.12

= 0.72 mol

d Each sulfate ion has 4 O atoms.

amount (O atoms) = 4 1.5

= 6.0 mol




Q9.

Calculate the molar mass of:

a nitrogen (N2)

b ammonia (NH3)

c sulfuric acid (H2SO4)

d iron(III) nitrate (Fe(NO3)3)

e acetic acid (CH3COOH)

f sulfur atoms (S)

g vitamin C (ascorbic acid C6H8O6)

h hydrated copper(II) sulfate (CuSO4.5H2O)

A9.

a M(N2) = 2 14.007

= 28.0 g mol–1

b M(NH3) = (1 × 14.007) + (3 × 1.008)

= 17.0 g mol–1

c M(H2SO4) = (2 1.008) + (1 × 32.064) + (4 15.994)

= 98.1 g mol–1

d M(Fe(NO3)3) = (1 55.847) + (3 × 14.007) + (9 15.994)

= 242 g mol–1

e M(CH3COOH) = (4 1.008) + (2 × 12.012) + (2 15.994)

= 60.0 g mol–1

f M(S) = (1 32.1)

= 32.1 g mol–1

g M(C6H8O6) = (6 12.012) + (8 1.008) + (6 15.994)

= 176.1 g mol–1

h M(CuSO4.5H2O) = (1 63.54) + (1 × 32.064) + (4 15.994) + (5 × 18.01)

= 250 g mol–1

Q10.

Calculate the mass of:

a 1.0 mol of sodium atoms (Na)

b 2.0 mol of oxygen molecules (O2)

c 0.10 mol of methane molecules (CH4)

d 0.25 mol of aluminium oxide (Al2O3)

A10.

It is useful to remember the formula m = nM, where m is the mass in grams, n the

amount of substance in mol, and M the molar mass. Use a periodic table to work out

the molar masses.

a m(Na atoms) = 1.0 mol × 23 g mol–1

= 23.0 g

b m(O2) = 2.0 mol × 32.0 g mol–1

= 64.0 g

c m(CH4) = 0.10 mol × 16.0 g mol–1

= 1.60 g

d m(Al2O3) = 0.25 mol × 102 g mol–1

= 25.5 g




Q11.

Calculate the amount, in mol, of:

a H atoms in 5 g of hydrogen

b H2 molecules in 5 g of hydrogen

c Al atoms in 2.7 g of aluminium

d CH4 molecules in 0.4 g of methane

e O atoms in 0.10 g of oxygen

f O2 molecules in 0.10 g of oxygen

g P atoms in 1.2 × 10–3 g of phosphorus

h P4 molecules in 1.2 × 10–3 g of phosphorus

A11.

It is useful to remember the formula n = M

m, where m is the mass in grams, n the


the molar masses.

a n(H atoms) = 1

5

= 5 mol

b n(H2) = 2

5

= 2.5 mol

c n(Al atoms) = 27

7.2

= 0.10 mol

d n(CH4) = 16

4.0

= 0.025 mol

e n(O atoms) = 16

10.0

= 0.0063 mol

f n(O2) = 32

10.0

= 0.0031 mol

g n(P atoms) = 31

102.1 3

= 3.9 10–5 mol

h n(P4) = 124

102.1 3

= 9.7 10–6 mol




Q12.

Calculate the number of atoms in:

a 23 g of sodium (Na)

b 4.0 g of argon (Ar)

c 0.243 g of magnesium (Mg)

d 10.0 g of gold (Au)

A12.

Use the formulas: Number of particles = n × NA, where NA = 6.02 1023 and n = M

m

,

where m is the mass in grams, n the amount of substance in mol, and M the molar

mass. Use a periodic table to work out the molar masses.

a n(Na) = 23

23 = 1.0 mol

Number of Na atoms = 1.0 × 6.0 1023 = 6.0 1023 atoms

b n(Ar) = 95.39

0.4 = 0.10 mol

Number of Ar atoms = 0.10 × 6.0 1023 = 6.0 1022 atoms

c n(Mg) = 3.24

243.0 = 0.01 mol

Number of Mg atoms = 0.01 × 6.0 1023 = 6.0 1021 atoms

d n(Au) = 97.196

0.10 = 0.051 mol

Number of Mg atoms = 0.051 × 6.0 1023 = 3.0 1022 atoms

Q13.

Calculate:

a the number of molecules in:

i 16 g of oxygen (O2)

ii 2.8 g of nitrogen (N2)

b the number of oxygen atoms in 3.2 g of sulfur dioxide (SO2)

c the total number of atoms in 288 g of ammonia (NH3)




A13.

Use the formulas: Number of particles = n × NA, where NA = 6.02 1023, and n = M

m,

where m is the mass in grams, n the amount of substance in mol, and M the molar

mass. Use a periodic table to work out the molar masses.

a i n(O2) = 32

16 = 0.5 mol

Number of O2 molecules = 0.5 × 6.0 1023 = 3.0 1023 molecules

ii n(N2) = 28

8.2 = 0.1 mol

Number of N2 molecules = 0.1 × 6.0 1023 = 6.0 1022 molecules

b n(SO2) = 64

2.3 = 0.05 mol

Number of SO2 molecules = 0.05 × 6.0 1023 = 3.0 1022 molecules

Each molecule contains 2 oxygen atoms.

So, number of oxygen atoms = 6.0 1022 atoms.

c n(NH3) = 17

288 = 16.9 mol

Number of NH3 molecules = 16.9 × 6.0 1023 = 1.0 1025 molecules

Each molecule contains 4 atoms (1 of N and 3 of H).

So, total number of atoms = 4.0 1025 atoms.

Q14.

Calculate the percentage by mass of:

a iron in iron(III) oxide (Fe2O3)

b uranium in uranium oxide (U3O8)

c nitrogen in ammonium chloride (NH4Cl)

d oxygen in copper nitrate (Cu(NO3)2)




A14.

a Percentage by mass of an element

= compound theof mol 1 of mass

compound of mol 1in element of mass × 100%

Use a periodic table to work out the molar masses.

For example, M(Fe2O3) = 159.6 g mol–1.

%(Fe) = 6.159

1008.552

= 69.9%

b %(U) = 09.842

10003.2383

= 84.8%

c %(N) = 5.53

10014

= 26.2%

d %(O) = 5.187

100166

= 51.2%

Q15.

Determine the empirical formula of the compounds with the following compositions:

a 2.74% hydrogen, 97.26% chlorine

b 42.9% carbon, 57.1% oxygen

c 10.0 g of a compound of magnesium and oxygen that contains 6.03 g of

magnesium

d 3.2 g of a hydrocarbon that contains 2.4 g of carbon

A15.

a The empirical formula provides the simplest whole-number ratio of atoms in a

compound. The number of moles of each atom is found by using n = M

m, where

m is the mass in grams, and M is the molar mass in g mol–1.

H Cl

Mass 2.74 g 97.26 g

Molar mass 1 g mol–1 35.5 g mol–1

Amount using n = M

m n =

1

74.2 = 2.74 mol n =

5.35

26.97 = 2.74 mol

Divide all by the smallest

amount 74.2

74.2 = 1

74.2

74.2 = 1

Round off to whole numbers 1 1

empirical formula is HCl




b

C O

Mass 42.9 g 57.1 g

Molar mass 12 g mol–1 16 g mol–1

Amount, using n = M

m n =

12

9.42 = 3.575 mol n =

16

1.57 = 3.57 mol


amount 57.3

575.3 = 1

57.3

57.3 = 1


empirical formula is CO

c When 6.03 g is the mass of magnesium in 10.0 g of the compound, the mass of

oxygen is (10.0 – 6.03) = 3.97 g.

Mg O

Mass 6.03 g 3.97 g

Molar mass 24.3 g mol–1 16 g mol–1

Amount, using n = M

m n =

3.24

03.6 = 0.248 mol n =

16

97.3 = 0.248 mol


amount 248.0

248.0 = 1

248.0

248.0 = 1


empirical formula is MgO

d When 2.4 g is the mass of carbon in 3.2 g of the hydrocarbon, the mass of

hydrogen is (3.2 – 2.4) = 0.8 g.

C H

Mass 2.4 g 0.8 g


Amount, using n = M

m n =

12

4.2 = 0.2 mol n =

1

8.0 = 0.8 mol


amount 2.0

2.0 = 1

2.0

8.0 = 4


empirical formula is CH4




Q16.

Determine the molecular formula of each compound in the table below.

Table 4.8

Empirical formula Relative molecular mass

a CH 78

b HO 34

c CH2O 90

d NO2 46

e CH2 154

A16.

a The molecular formula is always a whole-number multiple of the empirical

formula. The empirical formula provides the simplest whole-number ratio of

atoms in a compound. The number of moles of each atom is found by using

n = M

m, where m is the mass in grams and M is the molar mass.

Molar mass of a CH unit (empirical formula) = 12 + 1 = 13 g mol–1

Molar mass of the compound (molecular formula) = 78 g mol–1

number of CH units in one molecule = 13

78 = 6

molecular formula is C6H6 (which is benzene)

b Molar mass of an HO unit (empirical formula) = 1 + 16 = 17 g mol–1


number of OH units in one molecule = 17

34 = 2

molecular formula is H2O2

c Molar mass of a CH2O unit (empirical formula) = 12 + (2 × 1) + 16 = 30 g mol–1


number of CH2O units in one molecule = 30

90 = 3

molecular formula is C3H6O3

d Molar mass of an NO2 unit (empirical formula) = 14 + (16 × 2) = 46 g mol–1


number of NO2 units in one molecule = 46

46 = 1

molecular formula is NO2

e Molar mass of a CH2 unit (empirical formula) = 12 + (1 × 2) = 14 g mol–1


number of CH2 units in one molecule = 14

154 = 11

molecular formula C11H22




Q17.

A hydrocarbon contains 85.7% carbon. Its relative molecular mass is 70.

a Determine the empirical formula of the hydrocarbon.

b Determine the molecular formula of the hydrocarbon.

A17.



atoms in a compound. The number of moles of each atom is found by using

n = M

m, where m is the mass in grams and M is the molar mass in g mol–1.

A hydrocarbon contains only carbon and hydrogen.

Hence, %H = 100 – 85.7 = 14.3%

H C

Mass 14.3 g 85.7 g


Amount, using n = M

m n =

1

3.14 = 14.3 mol n =

12

7.85 = 7.14 mol


amount 14.7

3.14 = 2

14.7

14.7 = 1


empirical formula is CH2

b Molar mass of a CH2 unit (empirical formula) = 12 + 2 = 14 g mol–1


number of CH2 units in one molecule = 14

70 = 5

molecular formula is C5H10

Q18.

A sample of the carbohydrate glucose contains 1.8 g carbon, 0.3 g hydrogen and 2.4 g

oxygen. Calculate the empirical formula of the compound. Deduce its molecular

formula, given that its relative molecular mass is 180.




A18.

The molecular formula is always a whole-number multiple of the empirical formula.

The empirical formula provides the simplest whole-number ratio of atoms in a


m, where m is

the mass in grams and M is the molar mass in g mol–1.

C H O

Mass 1.8 g 0.3 g 2.4 g

Molar mass 12 g mol–1 1 g mol–1 16 g mol–1

Amount, using n = M

m n =

12

8.1 = 0.15 mol n =

1

3.0 = 0.30 mol n =

16

4.2 = 0.15 mol

Divide all by the

smallest amount 15.0

15.0 = 1

15.0

30.0 = 2

15.0

15.0 = 1

Round off to whole

numbers 1 2 1

empirical formula is CH2O

Molar mass of a CH2O unit (empirical formula) = 12 + (2 × 1) + 16 = 30 g mol–1



180 = 6


Chapter review

Q19.

The standard on which all relative masses are based is the 12C isotope, which is given

a mass of 12 exactly. Explain why then, in the table of relative atomic masses in

Appendix 4 on page 489, the relative atomic mass of carbon is listed as 12.011.

A19.

The relative atomic mass of carbon is the weighted average of the isotopic masses of

all carbon isotopes (i.e. 12C, 13C and 14C). Small amounts of 13C and 14C make this

average slightly greater than 12, the relative isotopic mass of the 12C isotope.




Q20.

When a sample of palladium is placed in a mass spectrometer, the following peaks are

recorded at the relative atomic masses and corresponding percentage abundances

given in the table.

Relative isotopic mass Percentage abundance

101.9049 0.9600%

103.9036 10.97%

104.9046 22.23%

105.9032 27.33%

107.9039 26.71%

109.9044 11.80%

Calculate the relative atomic mass of palladium.

A20.

Ar(Pd) =

100

11.80109.9044 26.71107.9039 27.33105.9032 22.23104.9046 10.97103.9036 0.96101.9049

= 106.4

Q21.

The following table gives isotopic composition data for argon and potassium.

Element

Atomic

number Relative isotopic mass Relative abundance (%)

Argon 18 35.978 0.307

37.974 0.060

39.974 99.633

Potassium 19 38.975 93.3

39.976 0.011

40.974 6.69

a Determine the relative atomic masses of argon and potassium.

b Explain why the relative atomic mass of argon is greater than that of potassium,

although potassium has a larger atomic number.

A21.

a Ar(Ar) = 100

)633.99974.39()060.0974.37()307.0978.35( = 39.96

Ar(K) = 100

)69.6974.40()011.0976.39()3.93975.38( = 39.11

b Although potassium atoms have one more proton than argon atoms, the most

abundant isotope of argon has 22 neutrons, giving it a relative atomic mass close

to 40. The most abundant isotope of potassium has only 20 neutrons, giving it a

relative atomic mass close to 39.




Q22.

The relative atomic mass of rubidium is 85.47. The relative isotopic masses of its two

isotopes are 84.94 and 86.94. Calculate the relative abundances of the isotopes in

naturally occurring rubidium.

A22.

Let the percentage abundance of the lighter isotope be x%.


85.47 = 100

)94.86)100(()94.84( xx

x = 26.5

proportion of the isotopes are 26.5% and 73.5%.

Q23.

Determine the percentage abundance of the lighter isotope of each of the following

elements.

a Gallium: relative isotopic masses 68.95 and 70.95, respectively; Ar = 69.72

b Boron: relative isotopic masses 10.02 and 11.01, respectively; Ar = 10.81

A23.

a Let the percentage abundance of the lighter isotope be x%.


69.72 = 100

)95.70)100(()95.68( xx

x = 61.5

proportion of the lighter isotope = 61.5%

b 20.2% (calculated by using the same process as for part a)

Q24.

What is the relative molecular mass (Mr) of the following?

a water (H2O)

b white phosphorus (P4)

c carbon monoxide (CO)

A24.

a The relative molecular mass, Mr, is the sum of the relative atomic masses, Ar, of

the elements in the compound.

Mr(H2O) = 2 + 16

= 18

b Mr(P4) = 4 + 31

= 124

c Mr(CO) = 12 + 16

= 28




Q25.

How would the molar mass (M) of a compound differ from its relative molecular mass

(Mr)?

A25.

The molar mass, M, has the same numerical value as the relative molecular mass, Mr,

which is the sum of the relative atomic masses, Ar, of the elements in the compound.

The molar mass, M, is the actual mass of one mole and so has the unit g mol–1.

Q26.

What is the molar mass (M) of each of the following?

a iron (Fe)

b sulfuric acid (H2SO4)

c sodium oxide (Na2O)

d zinc nitrate (Zn(NO3)2)

e glycine (H2NCH2COOH)

f aluminium sulfate (Al2(SO4)3)

g hydrated iron(III) chloride (FeCl3.6H2O)

A26.

a 55.8 g mol–1

b Mr(H2SO4) = 2 + 32 + 64 = 98

M(H2SO4) = 98 g mol–1

c Mr(Na2O) = (2 × 23) + 16 = 62

M(Na2O) = 62 g mol–1

d Mr(Zn(NO3)2) = 65.4 + (2 × 14) + (6 × 16) = 189.4

M(Zn(NO3)2) = 189.4 g mol–1

e Mr(H2NCH2COOH) = (5 × 1) + 14 + (2 × 12) + (2 × 16) = 75

M(H2NCH2COOH) = 75.0 g mol–1

f Mr(Al2(SO4)3) = (2 × 27) + (3 × 32) + (12 × 16) = 342

M(Al2(SO4)3) = 342 g mol–1

g Mr(FeCl3.6H2O) = 55.9 + (3 × 35.5) + (12 × 1.01) + (6 × 16) = 271

M(FeCl3.6H2O) = 271 g mol–1

Q27.

What is the mass of:

a 0.060 mol of ethane (C2H6)?

b 0.32 mol of glucose (C6H12O6)?

c 6.8 × 10–3 mol of urea ((NH2)2CO)?

d 6.12 mol of copper atoms (Cu)?

A27.


amount of substance in mol, and M the molar mass.

a m(C2H6) = 0.060 (24 + 6) = 1.8 g

b m(C6H12O6) = 0.32 ((6 × 12) + (12 × 1) + (6 × 16)) = 58 g

c m((NH2)2CO) = 6.8 × 10–3 ((2 × 14) + (4 × 1) + 12 + 16)) = 0.41 g

d m(Cu) = 0.6.12 63.5 = 389 g




Q28.

What is the amount (in mol) of each of the following?

a carbon atoms in 1.201 g carbon

b sulfur molecules (S8) in 10.0 g sulfur

c methane molecules (CH4) in 20.0 g methane

d aspirin molecules (C6H4(OCOCH3)COOH) in 300 mg aspirin

e aluminium oxide (Al2O3) in 3.5 tonnes of aluminium oxide (1 tonne = 1000 kg)

A28.

a It is useful to remember the formula n = M


amount of substance in mol, and M the molar mass in g mol–1. Use a periodic

table to work out the molar masses.

n(C) = 12

201.1 = 0.10 mol

b n(S8) = 256

0.10 = 0.0391 mol

c n(CH4) = 16

0.20 = 1.25 mol

d n(C6H4(OCOCH3)COOH) = 180

300.0 = 0.001 67 mol

e n(Al2O3) = 102

0005003 = 3.4 104 mol

Q29.

a If 6.0 10 23 atoms of calcium have a mass of 40.1 g, what is the mass of one

calcium atom?

b If 1 mol of water molecules has a mass of 18 g, what is the mass of one water

molecule?

c What is the mass of one molecule of carbon dioxide?

A29.

a Mass of one atom = mole ain part icles ofnumber

mole 1 of mass =

AN

massmolar

Mass of one calcium atom = 23100.6

1.40

= 6.67 10–23 g

b Mass of one water molecule = 23100.6

18

= 3.0 10–23 g

c Mass of one CO2 molecule = 23100.6

44

= 7.3 10–23 g




Q30.

For each of the following molecular substances, calculate:

a the amount of substance, in moles

b the number of molecules

c the total number of atoms

i 4.2 g of phosphorus (P4)

ii 75.0 g of sulfur (S8)

iii 0.32 g of hydrogen chloride (HCl)

iv 2.2 × 10–2 g of glucose (C6H12O6)

A30.

i a n(P4) = m/M = 4.2/(4 × 31.0) = 0.034 mol

b N(P4) = n × 6.02 × 1023 = 2.04 × 1022 molecules

c Total number of atoms = 4 × 2.04 × 1022 = 8.2 × 1022 atoms

ii a n(S8) = m/M = 75.0/(8 × 32.1) = 0.292 mol

b N(S8) = n × 6.02 × 1023 = 1.75 × 1023 molecules

c Total number of atoms = 0.292 × 8 × 6.02 × 1023 = 1.41 × 1024 atoms

iii a n(HCl) = m/M = 0.32/(1.01 + 35.5) = 0.0088 mol

b N(HCl) = n × 6.02 × 1023 = 5.3 × 1021 molecules

c Total number of atoms = 0.0088 × 2 × 6.02 × 1023 = 1.1 × 1022 atoms

iv a n(C6H12O6) = m/M = 2.2 × 10–2/(6 × 12.01) + (12 × 1.01) + (6 × 16.0)

= 1.22 × 10–4 mol

b N(C6H12O6) = n × 6.02 × 1023 = 7.3 × 1019 molecules

c Total number of atoms = 1.22 × 10–4 × 24 × 6.02 × 1023 = 1.8 × 1021 atoms

Q31.

What mass of iron (Fe) would contain as many iron atoms as there are molecules in

20.0 g of water (H2O)?

A31.



amount of substance in mol, and M the molar mass in g mol–1. Use a periodic table to

work out the molar masses of iron and water.

M(Fe) = 55.8 g mol–1 and 18.0 g mol–1

n(H2O) = 18

20 = 1.11 mol

n(Fe) needed = 1.11 mol

m(Fe) needed = 1.11 55.8 = 62.0 g




Q32.

For each of the following ionic substances, calculate:


b the amount of each ion, in moles

i 5.85 g of NaCl

ii 45.0 g of CaCl2

iii 1.68 g of Fe2(SO4)3

A32.




the molar masses.

i a n(NaCl) = 8.58

85.5 = 0.100 mol

b n(Na+) = n(NaCl) = 0.100 mol Na+

n(Cl–1) = n(NaCl) = 0.100 mol Cl–

ii a n(CaCl2) = 111

0.45 = 0.405 mol

b n(Ca2+) = n(CaCl2) = 0.405 mol

n(Cl–) = 2 n(CaCl2) = 2 × 0.405 mol = 0.81 mol

iii a n(Fe2(SO4)3) = 6.399

68.1 = 0.004 20 mol

b n(Fe3+) = 2 × n(Fe2(SO4)3) = 0.008 40 mol Fe3+

n(SO42–) = 3 × n(Fe2(SO4)3) = 0.0126 mol SO4

2–

Q33.

For each of the following numbers of molecules, calculate:


b the mass, in grams, of substance

i 4.50 1023 molecules of water (H2O)

ii 9.00 1024 molecules of methane (CH4)

iii 2.3 1028 molecules of chlorine (Cl2)

iv 1 molecule of sucrose (C12H22O11)




A33.

i It is useful to remember the formula n = M


amount of substance in mol, and M the molar mass. Remember also that the

number of particles in 1 mol, Avogadro’s number, NA = 6.02 1023.

Use the formula: n = A

particles ofnumber

N. Use a periodic table to work out the

molar masses.

a n(H2O) = 23

23

1002.6

1050.4

= 0.75 mol

b m(H2O) = 0.75 18 = 13.5 g

ii a 15.0 mol

b 240 g

iii a 3.8 104 mol

b 2.7 106 g

iv a n(C12H22O11) = 231002.6

1

= 1.7 10–24 mol

b m(C12H22O11) = 1.7 10–24 342 = 5.7 10–22 g

Q34.

a If 0.50 mol of a substance has a mass of 72 g, what is the mass of 1.0 mol of the

substance?

b If 6.0 1022 molecules of a substance have a mass of 10 g, what is the molar

mass of the substance?

A34.

It is useful to remember the formula M = n


amount of substance in mol, and M the molar mass in g mol–1. Remember also that the

number of particles in 1 mol is Avogadro’s number, NA = 6.02 1023.

Use the formula: n =.A

particles ofnumber

N

a M(substance) = 5.0

72 = 144 g mol–1

b n(substance) = AN

22100.6 = 0.1 mol

M(substance) = 1.0

10 = 100 g mol–1




Q35.

Calculate the molar mass of a substance if:

a 2 mol of the substance has a mass of 80 g

b 0.1 mol of the substance has a mass of 9.8 g

c 1.7 mol of the substance has a mass of 74.8 g

d 3.5 mol of the substance has a mass of 371 g

A35.

It is useful to remember the formula M = n


amount of substance in mol, and M the molar mass in g mol–1.

a M(substance) = 2

80 = 40 g mol–1

All other parts follow the same process.

b 98 g mol–1

c 44 g mol–1

d 106 g mol–1

Q36.

Which of the following metal samples has the greatest mass?

a 100 g copper

b 4.0 mol of iron atoms

c 1.2 1024 atoms of silver

A36.


amount of substance in mol, and M the molar mass. Use a periodic table to find the

molar masses of iron and silver. M = 55.8 g mol–1 and 108 g mol–1, respectively.

Remember also that the number of particles in 1 mol is NA.

m(Fe) = 4.0 55.8 = 223 g

n(Ag) = A

24102.1

N

= 2.0 mol

m(Ag) = 2.0 108 = 216 g

the mass of the iron is the greatest




Q37.

A new antibiotic has been isolated and only 2.0 mg is available. The molar mass is

found to be 12.5 kg mol–1.

a Express the molar mass in g mol–1.

b Calculate the amount of antibiotic (in mol).

c How many molecules of antibiotic have been isolated?

A37.


amount of substance in mol, and M the molar mass in g mol–1. Remember also that the

number of particles in 1 mol is, Avogadro’s number, NA = 6.02 1023.

a M(antibiotic) = 12 500 g mol–1 = 1.25 104 g mol–1

b n(antibiotic) = 4

3

1025.1

100.2

= 1.6 10–7 mol

c number of molecules = nNA

= 1.6 10–7 6.02 1023

= 9.6 1016 molecules

Q38.

Calculate the percentage by mass of each element in the following compounds:

a Al2O3

b Cu(OH)2

c MgCl2.6H2O

d Fe2(SO4)3

e perchloric acid (HClO4)

A38.

Percentage by mass of an element = compound of mol 1 of mass

compound of mol 1in element of mass × 100.

Use a periodic table to work out the molar masses. A useful check of these answers is

provided by seeing that they add up to 100%, or somewhere close to that value.

a M(Al) = 27 g mol–1, M(O) = 16 g mol–1, M(Al2O3) = 102 g mol–1.

%(Al) = 102

272 × 100 = 52.9%

%(O) = 102

163 × 100 = 47.1%

b Cu 65.1%; O 32.8%; H 2.1%




c M(Mg) = 24.3 g mol–1, M(Cl) = 35.5 g mol–1, M(MgCl2.6H2O) = 203.3 g mol–1

%(Mg) = 3.203

3.24 × 100 = 12.0%

%(Cl) = 3.203

5.352 × 100 = 34.9%

%(H) = 3.203

112 × 100 = 5.9%

%(O) = 3.203

166 × 100 = 47.2%

d Fe 27.9%; S 24.1%; O 48.0%

e H 1.0%; Cl 35.3%; O 63.7%

Q39.

Determine the percentage of carbon in the following compounds:

a naphthalene (C10H8)

b acetic acid (CH3COOH)

c urea (NH2CONH2)

d aspirin (C6H4(OCOCH3)COOH)

A39.

Percentage by mass of an element = compound of mol 1 of mass

compound of mol 1in element of mass × 100.

Use a periodic table to work out the molar masses. A useful check of these answers is

provided by seeing that they add up to 100%, or somewhere close to that value.

a M(C10H8) = 128 g mol–1, M(H) = 1 g mol–1, M(C) = 12 g mol–1

%(C) = 128

1210 × 100 = 93.8%

b %C = 40%

c %C = 19.9%

d %C = 60.0%




Q40.

Determine the empirical formulas of the compounds with the following compositions:

a 42.9% carbon, 57.1% oxygen

b 27.2% carbon, 72.8% oxygen

c 54.5% carbon, 9.1% hydrogen, 36.4% oxygen

d 1.72 g iron, 1.48 g sulfur, 3.02 g oxygen

e 9.6 g carbon, 0.67 g hydrogen, 4.7 g chlorine

f 4.42 g carbon, 0.842 g hydrogen

A40.



m, where m is

the mass in grams and M is the molar mass.

a

C O

Mass 42.9 g 57.1 g


Amount, using n = M

m n =

12

9.42 = 3.575 mol n =

16

1.57 = 3.569 mol

Divide all by smallest

amount 3.569

3.575 = 1

3.569

3.569 = 1

Round off to whole

numbers 1 1

empirical formula is CO

b

C O

Mass 27.2 g 72.8 g


Amount, using n = M

m n =

12

2.27 = 2.267 mol n =

16

8.72 = 4.55 mol


amount 2.267

2.267 = 1

2.267

4.55 = 2

Round off to whole

numbers 1 2

empirical formula is CO2




c

C O H

Mass 54.5 g 36.4 g 9.1 g


Amount, using n = M

m

n = 12

5.54

= 4.54 mol

n = 16

4.36

= 2.275 mol

n = 1

1.9

= 9.1 mol


amount 2.275

4.54 = 1.99

2.275

2.275 = 1

2.275

9.1 = 4

Round off to whole

numbers 2 1 4

empirical formula is C2H4O

d

Fe S O

Mass 1.72 g 1.48 g 3.02 g

Molar mass 55.8 g mol–1 32 g mol–1 16 g mol–1

Amount, using n = M

m

n = 8.55

72.1

= 0.031 mol

n = 32

48.1

= 0.046 mol

n = 16

02.3

= 0.189 mol


amount 0.031

0.031 = 1

0.031

0.046 = 1.5

0.031

0.189 = 6

Round off to whole

numbers 2 3 12

empirical formula is Fe2S3O12, which is Fe2(SO4)3

e

C Cl H

Mass 9.6 g 4.7 g 0.67 g

Molar mass 12 g mol–1 35.5 g mol–1 1 g mol–1

Amount, using n = M

m

n = 12

6.9

= 0.800 mol

n = 5.35

7.4

= 0.132 mol

n = 1

67.0

= 0.67 mol


amount 0.132

0.800 = 6

0.132

0.132 = 1

0.132

0.67 = 5

Round off to whole

numbers 6 1 5

empirical formula is C6H5Cl




f

C H

Mass 4.42 g 0.842 g


Amount, using n = M

m n =

12

42.4 = 0.368 mol n =

1

842.0 = 0.842 mol


amount 0.368

0.368 = 1

0.368

0.842 = 2.3

Round off to whole

numbers 1 × 7 = 7 2.3 × 7 = 16

empirical formula is C7H16

Q41.

A compound used as a solvent for nitrocellulose, resins and dyes has the following

composition by mass: 32% carbon, 6.7% hydrogen, 18.7% nitrogen and 42.6%

oxygen. Find the empirical formula of the compound.

A41.


compound. The amount of each atom is found by using n = M

m, where m is the mass

in grams and M is the molar mass in g mol–1. Use a periodic table to find the molar

masses of C, H, N and O.

C H N O

Mass 32 g 6.7 g 18.7 g 42.6 g

Molar mass 12 g mol–1 1 g mol–1 14 g mol–1 16 g mol–1

Amount, using

n = M

m

n = 12

32

= 2.67 mol

n = 1

7.6

= 6.7 mol

n = 14

7.18

= 1.34 mol

n = 16

6.42

= 2.66 mol

Divide all by


2.67 = 1.99

1.34

6.7 = 5

1.34

1.34 = 1

1.34

2.66 =1.99

Round off to

whole numbers 2 5 1 2

empirical formula is C2H5NO2




Q42.

A compound of tungsten and sulfur is a useful solid lubricant. Deduce the empirical

formula of this compound if a particular sample is formed when 1.84 g of tungsten

reacts exactly with 0.64 g of sulfur.

A42.





masses of W and S.

W S

Mass 1.84 g 0.64 g


Amount, using n = M

m n =

184

84.1 = 0.01 mol n =

32

64.0= 0.020 mol


amount 0.01

0.01 = 1

0.10

0.20 = 2

Round off to whole

numbers

1 2

empirical formula is WS2




Q43.

A clear liquid extracted from fermented lemons was found to consist of carbon,

hydrogen and oxygen. Analysis showed it to be 52.2% carbon and 34.8% oxygen.

a Find the empirical formula of the substance.

b If 2.17 mol of the compound has a mass of 100 g, find the molecular formula of

the compound.

A43.

a Step 1: Calculate mass of each element present in 100 g.

m(C) = 52.2 g

m(O) = 34.8 g

m(H) = 100 – 52.2 – 34.8

= 13 g

Step 2: Calculate amount, in mol, of each element present.

n(C) = M

m

= 12

52.2

= 4.35 mol

n(O) = 16

34.8

= 2.175 mol

n(H) = 1

13

= 13 mol

Step 3: Convert to whole-number ratios.

n(C) = 2.175

4.35

= 2

n(O) = 2.175

2.2175

= 1

n(H) = 2.175

13

= 6

Step 4: Write as empirical formula.

C2H6O




b Step 1: Calculate molar mass of one empirical formula unit.

M = 2 12 + 6 1 + 16

= 46 g mol–1

Step 2: Calculate molar mass of compound.

M = n

m

= 2.17

100

= 46 g mol–1

Step 3: Calculate number of empirical formula units in compound.

no. of units = 46

46

= 1

Step 4: Write molecular formula.

C2H6O

Q44.

When 0.200 g of white phosphorus is burnt in oxygen, 0.456 g of an oxide of

phosphorus is formed. Deduce the empirical formula of this oxide.

A44.





masses of P and O.

O P

Mass 0.456 – 0.200 = 0.256 g 0.200 g


Amount, using n = M

m n =

16

256.0= 0.016 mol n =

31

200.0 = 0.006 45 mol


amount 45006.0

016.0= 2.5

45006.0

43006.0= 1

Round off to whole

numbers 5 2

empirical formula is P2O5




Q45.

A hydrocarbon is a compound that contains carbon and hydrogen only. Determine the

empirical formula of a hydrocarbon that is used as a specialty fuel and contains 90.0%

carbon.

A45.





masses of C and H.

C H

Mass 90.0 g 100 – 90.0 = 10.0 g


Amount, using n = M

m n =

12

90 = 7.5 mol n =

1

0.10 = 10 mol


amount 7.5

7.5 = 1

7.5

10 = 1.3

Round off to whole

numbers 1 × 3 = 3 1.3 × 3 = 4


Q46.

Find the relative atomic mass of nickel if 3.370 g nickel was obtained by reduction of

4.286 g of the oxide (NiO).

A46.





mass of O.

Ni O

Mass 3.370 g 4.286 – 3.370 = 0.916 g

Molar mass ? g mol–1 16 g mol–1

Amount, using n = M

m n =

(Ni)

370.3

M n =

16

916.0= 0.0573 mol

(O)

)Ni(

n

n =

(Ni)

370.3

M

0.0573

1

As the empirical formula is NiO, (Ni)

370.3

M

0.0573

1 =

1

1

M(Ni) = 0573.0

370.3 = 58.9 g mol–1




Q47.

4.150 g tungsten was burned in chlorine and 8.950 g tungsten chloride (WCl6) was

formed. Find the relative atomic mass of tungsten.

A47.





mass of Cl.

W Cl

Mass 4.150 g 8.950 – 4.150 = 4.800 g

Molar mass ? g mol–1 35.5 g mol–1

Amount, using n = M

m n =

(W)

150.4

M n =

5.35

800.4= 0.135 mol

(Cl)

)W(

n

n =

135.0 (W)

150.4

M

As the empirical formula is WCl6, 135.0 (W)

150.4

M =

6

1

M(W) = 4.150 135.0

6 = 184.2 g mol–1

Q48.

If 3.72 g of element X reacts with exactly 4.80 g of oxygen to form a compound

whose molecular formula is shown, from other experiments, to be X4O10, what is the

relative atomic mass of X?

A48.





mass of O.

X O

Mass 3.72 g 4.80 g


Amount, using n = M

m n =

(X)

72.3

M n =

16

80.4= 0.3 mol

(O)

)X(

n

n =

3.0 (X)

72.3

M

As the empirical formula is X4O10, 3.0 (X)

72.3

M =

10

4

M(X) = 43.0

1072.3

= 31.0 g mol–1




Q49.

Determine the molecular formulas of compounds with the following compositions and

relative molecular masses:

a 82.75% carbon, 17.25% hydrogen; Mr = 58

b 43.66% phosphorus, 56.34% oxygen; Mr = 284

c 40.0% carbon, 6.7% hydrogen, 53.3% oxygen; Mr = 180

d 0.164 g hydrogen, 5.25 g sulfur, 9.18 g oxygen; Mr = 178

A49.



atoms in a compound. The amount of each atom is found by using n = M

m,

where m is the mass in grams and M is the molar mass in g mol–1. Use a periodic

table to find the molar masses of C and H.

C H

Mass 82.75 g 17.25 g


Amount, using n = M

m n =

12

75.82 = 6.89 mol n =

1

25.17 = 17.25 mol


amount 6.89

6.89 = 1

6.89

17.25 = 2.5

Round off to whole

numbers 1 × 2 = 2 2.5 × 2 = 5


Molar mass of a C2H5 unit (empirical formula) = 24 + 5 = 29 g mol–1


number of C2H5 units in one molecule = 29

58 = 2


b P4O10

c C6H12O6

d The molecular formula is always a whole-number multiple of the empirical



m,


table to find the molar masses of O, S and H.




H S O

Mass 0.164 g 5.25 g 9.18 g


Amount, using

n = M

m

n = 1

164.0

= 0.164 mol

n = 32

25.5

= 0.164 mol

n = 16

18.9

= 0.574 mol

Divide all by


0.164 = 1

0.164

0.164 = 1

164.0

574.0 = 3.5

Round off to

whole numbers 1 × 2 = 2 1 × 2 = 2 3.5 × 2 = 7

empirical formula is H2S2O7

Molar mass of a H3S3O7 unit (empirical formula) = (2 × 1) + (2 × 32) + (7 × 16)

= 178 g mol–1


number of C2H5 units in one molecule = 178

178 = 1

molecular formula is H2S2O7

Q50.

Using suitable examples, clearly distinguish between the following terms:

a relative isotopic mass

b relative atomic mass

c relative molecular mass

d relative formula mass

e molar mass

A50.

The relative isotopic mass (Ir) of an isotope is the mass of an atom of that isotope relative

to the mass of an atom of 12C, taken as 12 units exactly. For example, the relative isotopic

mass of the lighter of the two chlorine isotopes (see Table 4.3, page 56) is 34.969.

The relative atomic mass of an element is the weighted average of the relative masses

of the isotopes of the element on the 12C scale. For example, the relative atomic mass

of boron (see Table 4.4, on page 57) is 10.81.

The relative molecular mass (Mr) of a compound is the mass of one molecule of that

substance relative to the mass of a 12C atom, which is 12 exactly. For example, the

relative molecular mass of carbon dioxide is 44.0.

Relative formula mass is calculated by taking the sum of the relative atomic masses of

the elements in the formula. Relative formula mass (rather than relative molecular

mass) is the appropriate term to use for ionic compounds as these do not contain

molecules. For example, the relative formula mass of sodium chloride is 58.5.

The molar mass of an element is the mass of one mole of the element. It is equal to

the relative atomic mass of the element expressed in grams. For example, the molar

mass of magnesium (see Table 4.5, on page 62) is 24.3 g mol–1. Note that relative

atomic mass and molar mass of an element are numerically equal. However, relative

atomic mass has no units because it is the mass of one atom of the element compared

with the mass of one atom of the carbon-12 isotope.

The molar mass of a compound is the mass of one mole of the compound. It is equal to

the relative molecular or relative formula mass of the compound expressed in grams.

For example, the molar mass of sodium chloride (Table 4.5, page 62) is 58.5 g mol–1.




Q51.

Caffeine, which is a stimulant found in coffee, tea and cola drinks, contains 49.48%

carbon, 5.15% hydrogen, 28.87% nitrogen and the rest oxygen.

a Determine the empirical formula of caffeine.

b If 0.20 mol of caffeine has a mass of 38.8 g, what is the molar mass of a caffeine

molecule?

c Determine the molecular formula of caffeine.

d How many moles of caffeine molecules are in 1.00 g caffeine?

e How many molecules of caffeine are in 1.00 g caffeine?

f How many atoms altogether are in 1.00 g caffeine?

A51.




m,


table to find the molar masses of C, H, N and O.

Take one mole of caffeine.

Mass of oxygen can be found by subtraction = 100 – (49.48 + 5.15 + 28.87)

= 16.5 g

C H N O

Mass (g) 49.48 5.15 28.87 16.5

Molar mass 12.0 g mol–1 1.01 g mol–1 14.0 g mol–1 16.0

Amount,

using n = M

m

n = 0.12

98.49

= 4.17 mol

n = 01.1

15.5

= 5.10 mol

n = 0.14

87.28

= 2.06 mol

n = 0.16

5.16

= 1.02 mol

Divide all by

smallest

amount 1.02

4.17 = 4.08

1.02

5.10 = 5.0

02.1

06.2 = 2.02

02.1

02.1 = 1.0

Round off to

whole

numbers

4 5 2 1

empirical formula is C4H5N2O

b Molar mass of caffeine = 38.8 × 1/0.2 = 194 g mol–1

c The molecule must contain a whole number of (C4H5N2O) units.

Molar mass of a C4H5N2O unit is (4 × 12.0) + ( 3 × 1.01) + (2 × 14.0) + (16.0)

= 97 g mol–1.

If the compound has a molar mass of 194 g mol–1, then the number of (C4H5N2O)

units in a molecule = molar mass of compound/molar mass of one unit

= 194 g mol–1/97 g mol–1

= 2

The molecular formula of caffeine is therefore 2 × (C4H5N2O), that is,

C8H10N4O2.




d n(caffeine) = m/Mr

= 1.00/194

= 5.15 × 10–3 mol

e N(caffeine) molecules = n(caffeine) × NA

= 5.15 × 10–3 × 6.02 × 1023

= 3.10 × 1020 molecules of caffeine

f N(caffeine) atoms = n(caffeine) × 24 × 6.02 × 1023

= 7.44 × 1025 atoms altogether

Q52.

The empirical formula of a metal oxide can be found by experimentation (see figure

below). The mass of the metal and the mass of the oxygen that reacts with it must be

determined. The six boxes below each contain one step in the experimental method.

A Ignite a burner and heat the metal. B Allow the crucible to cool, then

weigh it.

C Continue the reaction until no

further change occurs.

D Clean a piece of metal with the

emery paper to remove any oxide

layer.

E Place the metal in a clean,

weighed crucible and cover with

a lid.

F Weigh the metal and record its

mass.

a Place the steps in the correct order by letter.

b Wan and Eric collected the following data:

Mass of the metal = 0.542 g

Mass of the empty crucible = 20.310 g

Mass of the crucible and metal oxide = 21.068 g

They found from the data that the metal oxide had a 1 : 1 formula (i.e. MO, where

M = metal). Complete the table, using the data given.

Metal Oxygen

Mass (g)

Relative atomic mass 16.0

Moles

Ratio

c What metal was used in the experiment?

Equipment that can be used to find the empirical formula of a metal oxide.




A52.

a D, F, E, A, C, B

b Step 1: Calculate mass of compound after reaction.

m = 21.068 – 20.310

= 0.758 g

Step 2: Calculate mass of oxygen.

m = 0.758 – 0.542

= 0.216 g

Step 3: Calculate amount, in mol, of oxygen.

n = M

m

= 16

0.126

= 0.0135 mol

Step 4: Use mole ratios to determine amount of metal.

Ratio is 1 : 1, so n = 0.0135 mol

Step 5: Calculate molar mass of metal.

M = n

m

= 0.0135

0.542

= 40.1

Step 6: Fill in table.

Metal Oxygen

Mass (g) 0.542 0.216

Relative atomic mass 40.1 16.0

Moles 0.0135 mol 0.0135 mol

Ratio 1 1

c Refer to periodic table to identify metal as calcium.




Unit 1 Area of Study 1 Review

Multiple-choice questions

Q1.

Which of these proposals corresponds to the ideas put forward by John Dalton in

1804?

I Matter consists of indivisible particles.

II Atoms of a particular element can vary in their mass.

III The proportion and type of atom is always the same in a particular compound.

A I only

B I and II

C I and III

D I, II and III

A1.

C. Dalton did not say that atoms of a particular element could vary in their mass. He

had no concept of the existence of isotopes.

Q2.

When he first constructed his periodic table, Mendeleev arranged the known

elements:

A in order of their atomic number

B according to their electronic configuration

C into vertical groups according to their mass number

D into horizontal periods according to their atomic mass

A2.

D. Atomic number; electronic configuration and mass number were not known at that

time.

Q3.

Ernest Rutherford contributed to knowledge about the structure of the atom by:

A discovering the composition of alpha particles

B discovering that protons are found in the nucleus of an atom

C proving the existence of neutrons in the nucleus of an atom

D proposing that electrons move in circular orbits around the nucleus

A3.

B. This was part of the research that Rutherford and his team of co-workers did that

showed that most of the mass, and the positive charge, of an atom is concentrated in

the nucleus.




Q4.

Which of the following elements would have atoms with the largest atomic radius?

A Li

B Na

C F

D Cl

A4.

B. Sodium is on the left-hand side of period 3; hence, its outer-shell electron is subject

to the lowest core charge for that period.

Q5.

Which of the following elements would be expected to show greatest similarity in

chemical properties to the element that has 14 electrons in its neutral atoms?

A Al

B P

C Ga

D Ge

A5.

D. Ge is in the same group of the periodic table, with four outer-shell electrons.

Q6.

The Pauli exclusion principle states that:

A all atomic orbitals must hold two electrons

B an atomic orbital must hold a minimum of two electrons

C a new subshell is started whenever an atomic orbital holds two electrons

D an atomic orbital may hold a maximum of two electrons

A6.

D

Q7.

A double negatively charged ion has eight protons. The number of electrons in the ion

is:

A 10

B 8

C 6

D 2

A7.

A. The double negative charge means there are two more electrons than protons,

giving ten electrons in the ion.




Q8.

A sample of chlorine was thought to consist of the isotopes 3517 Cl and 37

17 Cl. If the

relative atomic mass of this sample of chlorine was found to be 35.5, it can be said

that:

A there are equal amounts of 3517 Cl and 37

17 Cl

B there is a greater abundance of 3517 Cl than 37

17 Cl

C there is a greater abundance of 3717 Cl than 35

17 Cl

D the sample consists of a different isotope with a relative isotopic mass of 35.5

A8.

B. There is a greater abundance of Cl3517

than Cl3717 , as the relative atomic mass is

closer to 35 than to 37.

Q9.

The number of neutrons in 3919 K+ is:

A 18

B 19

C 20

D 39

A9.

C. The number of neutrons is the mass number (39) minus the atomic number (19).

Q10.

Which one of the following has a different electronic configuration from the others?

A Na+

B K+

C Ne

D F–

A10.

B. All have ten electrons except K+, which has eighteen electrons.

Q11.

In which groups of the periodic table would you not expect to find a metal?

A groups 14 and 15

B groups 14 and 18

C groups 13 and 17

D groups 17 and 18

A11.

D. These groups have a large number of electrons in the outer shell and cannot lose

electrons easily.




Q12.

Which one of the following pairs of elements is most likely to combine to form a

compound with properties similar to those of sodium chloride?

A calcium and bromine

B carbon and oxygen

C copper and nickel

D phosphorus and chlorine

A12.

A. Calcium is a metal with a small number of electrons in the outer shell. Bromine is a

non-metal with a large number of electrons in the outer shell. Sodium chloride is also

a compound of a metal and a non-metal, so it will have similar properties to a

compound of calcium and bromine.

Q13.

Isotopes of an element contain:

A the same atomic number and the same mass number

B the same atomic number and a different mass number

C a different atomic number and the same mass number

D a different atomic number and a different mass number

A13.

B. Isotopes have the same number of protons (atomic number) and a different number

of neutrons (mass number is protons plus neutrons).

Q14.

Going down group 7 of the periodic table, the electronegativity:

A decreases because the atomic radius increases

B increases because the atomic number increases

C decreases because the atomic number increases

D increases because the number of subshells increases

A14.

A. The atomic radius increases; hence, the attraction for an additional electron

decreases.

Q15.

The relative atomic mass of magnesium, Ar(Mg), is 24.31. The most important reason

why it is not a whole number is that:

A magnesium atoms lose electrons when they react

B the relative atomic mass given is only an approximation

C not all atoms of magnesium have the same number of neutrons

D the mass of the magnesium atom is compared to the mass of the 12C isotope

A15.

C. Relative atomic mass is an average of the relative isotopic masses for an element.

Isotopes are atoms of magnesium with different numbers of neutrons.




Q16.

Which of the following lists contain empirical formulas only?

A C2H2, CBr4, Ca(OH)2, KMnO4

B NH3, N2H4, C6H5CH3, HCOOH

C H2SO4, Al2(SO4)3, C6H5CH3, CH3Cl

D HCOOH, C2H6O, Fe2O3.xH2O, C6H12O6

A16.

C. The following are not empirical formulas: H2SO4, Al2(SO4)3, C6H5CH3, CH3Cl.

Q17.

A compound has an empirical formula of CH. A 0.25 mol sample of the compound

weighs 13 g. The molecular formula is:

A C8H8

B C6H6

C C4H4

D C2H2

A17.

C.

Molar mass of hydrocarbon is mol 25.0

g 13 = 52 g mol–1

Molar mass of CH unit is 13 g mol–1. Number of units = 13

52 = 4


Short-answer questions

Q18.

a In what part of the periodic table are the metals found? Why are they found there?

b The helium atom contains two electrons in the outer shell. Why is helium not

placed in group 2 with the other elements also containing two electrons in the

outer shell?

c The heavier elements, atomic numbers 87–112, are all metals. Explain this fact.

d Some metallic elements are very reactive. Which group of the periodic table

contains the most reactive metals?

A18.

a Metals are found at the left and bottom of the periodic table. These elements have

a small number of electrons in their outer shell.

b Helium has a full outer shell and cannot easily donate these electrons as metals

do.

c Elements 87–112 all have just one or two outer-shell electrons, which can be

easily lost.

d group 1




Q19.

a Explain, using suitable examples, the relationship between the electronic

configuration of an element and its position in the periodic table.

b Consider the following: Na, Cl, Mn, Ca2+, O2–, Al3+. For each:

i give its electronic configuration

ii indicate its position in the periodic table (e.g. group 17, period 2, or transition

series, period 4)

A19.

a For any element, the shell of the highest order of an element containing electrons

determines the period of an element. For example, in calcium the fourth shell

contains electrons and, being the highest order in which electrons are found,

makes calcium a period 4 element.

The number of electrons in the outermost shell of an element determines the

group number of the element. For example, calcium has two electrons in its

outermost shell and so belongs to group 2.

Transition metals are those that have a ‘d’ subshell being filled.

Lanthanides and actinides have ‘f’ subshells being filled.

b Na 1s22s22p63s1 group 1 period 3

Cl 1s22s22p63s23p5 group 17 period 3

Mn 1s22s22p63s23p63d54s2 transition series period 4

Ca2+ 1s22s22p63s23p6 group 2 period 4

O2– 1s22s22p6 group 16 period 2

Al3+ 1s22s22p6 group 13 period 3

Q20.

Select your answers to the questions below from the following list of elements:

Cl, C, Na, Mg, K, O, F, Al, N, Ca.

Which elements:

a are in period 2 of the periodic table?

b are in period 3 of the periodic table?

c are in group 1 of the periodic table?

d are in group 2 of the periodic table?

e are in group 13 of the periodic table?

f are in group 16 of the periodic table?

g are classified as metals?

h have one valence electron?

i have three valence electrons?

j has the highest molar mass?

A20.

a C, N, O and F

b Na, Mg, Al and Cl

c Na and K

d Mg and Ca

e Al

f O

g Na, Mg, Al, K and Ca

h Na and K

i Al

j Ca




Q21.

a Supply the missing information in the table.

Atomic

number

Neutron

number

Mass

number Symbol

4 (i) 9 9

4 Be

8 (ii) (iii) 17

8 O

(iv) (v) (vi) 31

15 P

34 (vii) 78 (viii)

b The neutron was not discovered until more than 30 years after the discovery of

the proton and the electron. Why was the neutron more difficult to detect?

A21.

a i 5

ii 9

iii 17

iv 15

v 16

vi 31

vii 44

viii Se78

34

b Most of the instruments used for investigating the structure of the atom are based

on the use or measurement of electric charge. As the neutron is an uncharged

particle, it was not detected by these instruments.

Q22.

a Explain the meanings of the following terms:

i relative atomic mass

ii relative molecular mass

iii mole

iv Avogadro’s number

v molar mass

b Write the electronic configuration for the element phosphorus.

c When 0.100 g of white phosphorus is burned in oxygen, 0.228 g of an oxide of

phosphorus is produced. The molar mass of the oxide is 284 g mol–1.

i Determine the empirical formula of the phosphorus oxide.

ii Determine the molecular formula of the phosphorus oxide.

d Would you expect the properties of the oxide of phosphorus to be more similar to

those of sodium chloride or those of water? Explain your answer.




A22.

a i Relative atomic mass—the weighted mean of the relative masses of the

isotopes of an element on the 12C scale.

ii Relative molecular mass—the relative mass of a molecule on the 12C scale.

iii Mole—the amount of substance that contains the same number of specified

particles as there are atoms in exactly 12 g of 12C.

iv Avogadro’s number—the number of carbon atoms in exactly 12 g of 12C

(approximately 6.02 1023).

v Molar mass—mass in grams of a mole of a substance.

b 1s22s22p63s23p3

c i Step 1: Write the ratio by mass.

P : O

0.100 g : (0.228 – 0.100) g

0.100 g : 0.128 g

Step 2: Calculate the ratio by amount (in moles).

1molg974.30

g100.0

: 1molg0.16

g128.0

0.003 229 mol : 0.008 mol

Step 3: Divide by the smaller amount.

008.0

229 003.0 :

008.0

008.0

0.4036 : 1

Step 4: Express as integers by multiplying by 5.

2 : 5

empirical formula of the compound is P2O5

ii As the empirical formula is P2O5, the molecule must contain a whole number

of P2O5 units.

The molar mass of one of these units is

((2 30.974) + (5 16)) = 141.948 g mol–1.

The number of units in a molecule = molar mass of the compound/molar

mass of one unit

= 1

1

molg948.141

mol g284

= 2

molecular formula of the compound is P4O10

d Properties are more similar to those of water. Water is a compound of two non-

metals (hydrogen and oxygen), as is the oxide of phosphorus, since both

phosphorus and oxygen are non-metals.




Q23.

The original version of the periodic table was devised by the Russian chemist Dmitri

Mendeleev in 1869.

a What were the two pieces of information that he used to devise the table?

In the modern form of the periodic table, the elements are arranged in order of atomic

number.

b What information about the structure of the atom is given by its atomic number?

c What is the link between the atomic number of an element and the block in the

periodic table in which it would be found?

d Consider the element with atomic number 14.

i Write the full electronic configuration.

ii In which group, period and block of the periodic table would it be found?

A23.

a He arranged the known elements:

in order of increasing atomic mass;

with elements having similar chemical properties in vertical groups.

b the number of protons

c For a neutral atom, the number of protons equals the number of electrons, and the

electrons are arranged into shells and subshells of increasing energy. The block of

the periodic table corresponds to the highest energy subshell of electrons in the

particular atom.

d i 1s22s22p63s23p2

ii group 4, period 3, p-block

Q24.

The electronic configurations of seven elements (A–G) are given below.

A 1s22s22p5

B 1s22s22p63s1

C 1s22s22p63s13p5

D 1s22s22p63s23p64s2

E 1s22s22p63s23p63d64s1

F 1s22s22p63s23p63d84s2

G 1s22s22p63s23p63d104s24p5

Indicate which one or more are likely to be:

a metals

b d-block elements

c group 17 elements

d period 3 elements

e elements not in the ground state

A24.

a B, D, E, F

b E, F

c A, G

d B, C

e C, E




Q25.

a Give the ground-state electronic configuration of calcium (Ca).

b Explain, giving an example of the new electronic configuration, what happens to

the energy levels of the electrons when the atom in the ground state is provided

with sufficient energy to become:

i an excited atom

ii a charged ion

c The stable ion of calcium is Ca2+. What would be the difference in atomic radius

of a Ca atom and a Ca2+ ion?

d Why would a Ca atom and a Ca2+ ion have almost the same mass?

e Give the symbol of two elements that would have chemical properties similar to

those of calcium.

A25.

a 1s22s22p63s23p64s2

b i Electrons are promoted to a higher energy level. An example of an electronic

configuration of an excited calcium atom could be (several answers

possible): 1s22s22p63s23p64p2.

ii With sufficient energy, a calcium atom can lose its valence electrons to form

a cation e.g. Ca2+ with electronic configuration 1s22s22p63s23p6.

c The ion is much smaller as electrons occupy most of the atom’s volume and the

ion has one less electron shell than the atom.

d The mass of electrons is negligible in comparison to that of the protons and

neutrons in an atom. The number of protons and neutrons is the same in the atom

and ion.

e e.g. Mg, Sr

Q26.

Give explanations for the following:

a A sodium atom has a larger atomic radius than a chlorine atom, even though both

belong to period 3.

b A sodium ion (Na+) is much smaller than the sodium atom.

c Fluorine is more electronegative than iodine.

A26.

a Members of period 3 all have their outer-shell electrons in the third shell but the

core charge increases across the period from +1 for sodium to +7 for chlorine.

Hence, the outer-shell electrons in the chlorine are attracted more strongly to the

nucleus, reducing the radius of the species.

b From the electronic configuration for sodium, 1s22s22p63s1, it can be seen that the

Na atom has three shells of electrons but the Na+ ion has lost the single outer-

shell electron and the remaining eight electrons of the second shell are then

attracted to a greater core charge.

c The fluorine atom is smaller than that of iodine. Electrons are attracted to fluorine

atoms more strongly than to those of iodine because they are closer to the

positively charged nucleus in fluorine. So, fluorine is more electronegative.




Q27.

The following is a list of atoms: 2713 A,

4020 B,

3216 C, 14

7 D, 83 E, 15

7 F, 199 G,

209 H, 4

2I.

a Which pairs of atoms are isotopes?

b Which atoms have equal numbers of protons and neutrons in the nucleus?

c Which is an isotope of sulfur?

d Which has one more electron than a magnesium atom?

e Which is a group 2 element?

f How many different elements are shown?

A27.

a D and F, G and H

b B, C, D and I

c C

d A

e B

f 7

Q28.

A sample of magnesium carbonate weighs 21.8 g.

a Calculate the amount (mol) of magnesium carbonate present.

b Calculate the amount (mol) of oxygen atoms present.

c Calculate the number of carbon atoms present.

d Calculate the total number of atoms present.

e Calculate the percentage, by mass, of magnesium in magnesium carbonate.

A28.

a Step 1: Calculate the molar mass of MgCO3.

M = 24.3 + 12 + 3 16

= 84.3 g mol–1

Step 2: Calculate the amount (mol) of MgCO3.

n = M

m

= 84.3

21.8

= 0.2586 mol

Step 3: Give answers with correct number of significant figures.

n = 0.259 mol

b Step 1: Calculate number of oxygen atoms per molecule.

No. atoms per molecule = 3

Step 2: Calculate amount (mol) of oxygen atoms.

n = 3 0.259

= 0.776 mol




c Step 1: Calculate number of carbon atoms per molecule.


Step 2: Calculate amount (mol) of carbon atoms.

n = 1 0.259

= 0.259 mol

Step 3: Calculate number of carbon atoms.

No. atoms = 0.259 6.02 1023

= 1.56 1023 atoms

d Step 1: Calculate number of atoms per molecule.


Step 2: Calculate amount (mol) of atoms.

n = 5 0.259

= 1.30 mol

Step 3: Calculate total number of atoms.

No. atoms = 1.30 6.02 1023

= 7.80 1023 atoms

e % of Mg in MgCO3 %100)(MgCO

(Mg)

3

M

M

= %1003.84

3.24

= 28.8%

Q29.

Ethylene glycol is a compound often used as an antifreeze in cars in cold weather. Its

molar mass is 62 g mol–1. It has a percentage composition of 38.7% carbon, 9.7%

hydrogen and the rest oxygen. Determine:

a the empirical formula of ethylene glycol

b the molecular formula of ethylene glycol




A29.

The molecular formula is always a whole-number multiple of the empirical formula.





masses of C, H and O.

a

C O H

Mass 38.7 g 100 – (38.7 + 9.7) = 51.6 g 9.7 g


Amount, using

n = M

m

n = 12

7.38

= 3.225 mol

n = 16

6.51

= 3.225 mol

n = 1

7.9

= 9.7 mol

Divide all by


3.225 = 1

3.225

3.225 = 1

3.225

9.7 = 3

Round off to

whole numbers

1 1 3

empirical formula is CH3O

b Molar mass of a CH3O unit (empirical formula) = 12 + 3 + 16 = 31 g mol–1



62 = 2


Q30.

Methane (CH4) is the major component of natural gas.

a What is the mass of 0.50 mol methane?

b How many molecules are there in 0.100 g methane?

c How many atoms are there altogether in 0.10 g methane?

d How many protons are there in 0.10 g methane?

e What is the percentage by mass of carbon in methane?

f What mass of carbon would be present in 34 g methane?

A30.


amount of substance in mol, and M the molar mass in g mol–1. Use a periodic table to

find the molar masses. M(C) = 12 g mol–1 and M(CH4) = 16 g mol–1. Remember also

that the number of particles in 1 mol is Avogadro’s number, NA = 6.02 1023.

a m(CH4) = 0.50 16 = 8.0 g

b n(CH4) = 16

100.0

= 0.006 25 mol

number of molecules = nNA

= 0.006 25 6.02 1023

= 3.76 1021 molecules




c There are 5 atoms in each molecule of methane (1 of C and 4 of O).

So, total number of atoms in 3.76 1021 molecules = 5 3.76 1021 atoms

= 1.88 1022 atoms

d There are 10 protons in each molecule of methane (6 in each C and 1 in each of

the 4 H atoms).

So, total number of protons in 3.76 1021 molecules = 10 3.76 1021 protons

= 3.76 1022 protons

e %(C) = 16

12 × 100

= 75%

f m(C in 34 g) = 75% of 34 = 34 100

75

= 26 g

Q31.

Cobalt (Co) has an atomic number of 27.

a Give the ground state electronic configuration of cobalt.

b To what section of the periodic table does cobalt belong?

Cobalt reacts with oxygen to form a compound, cobalt oxide, formula CoO. Cobalt

oxide contains 78.6% by mass of cobalt.

c Use the above information to calculate the relative atomic mass of cobalt.

A31.

a 1s22s22p63s23p63d74s2

b first transition metal series or d-block

c The empirical formula provides the simplest whole-number ratio of atoms in a


m, where m is the

mass in grams and M is the molar mass in g mol–1. Use a periodic table to find the

molar mass of O.

Co O

Mass 78.6 g 21.4 g


Amount, using n = M

m n =

(Co)

6.78

M n =

16

4.21= 1.34 mol

As the empirical formula is CoO, the ratio is 1 : 1 = (Co)

6.78

M : 1.34

M(Co) = 34.1

6.78

= 58.7 g mol–1




Q32.

a Calculate the relative atomic mass of silicon from the data given below.

Relative isotopic mass Relative abundance (%)

27.98 92.2

28.98 4.7

29.98 3.1

b To what:

i group

ii period

of the periodic table does silicon belong?

A32.

a Substitute into the formula for relative atomic mass.

Ar(Si) = abundance relative total

abundance) relativemass (isotopic

= 100

)1.397.29()7.498.28()2.92 (27.98

= 100

92.907 136.206 2579.756

= 100

2808.869

= 28.1

b i 14

ii 3




Q33.

The successive ionisation energies for an element have been plotted in the diagram.

a How many shells of electrons exist for this atom?

b Use this information to write the electronic configuration for this element in its

ground state.

c What is the likely charge on the stable cation of the element?

A33.

a 3

b 1s22s22p63s1

c 1+

Q34.

The graphs show trends in some of the properties of elements as you go from top to

bottom down group 17 of the periodic table.

Which graph best represents the trend for:

a electronegativities?

b first ionisation energies?

c number of outer-shell electrons?

d atomic radii?

A B C

Figure 3.13

A34.

a C. Electronegativity decreases as atomic radius increases down a group.

b C. First ionisation energy decreases as atomic radius increases down a group.

c A. Number of outer-shell electrons is constant within a group.

d B: Atomic radius increases down a group as nth number of electron shells

increases.

ch4

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