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1.1 Exercises

1. What is the difference between the words qualitative and quantitative in the context of chemistry?

Qualitative refers to the identity of a compound or element whereas quantitative refers to the amount of a known

compound or element present in terms of mass, concentration or number of moles etc.

2. Define the following terms:

a) element

A substance that cannot be divided into simpler, smaller substances. In an element, all the atoms have the

same number of protons or electrons, but the number of neutrons may vary (more about this Topic 2)

b) atom

The smallest part of an element that can exist. An atom consists of an extremely tiny central nucleus,

surrounded by electrons. For more information see Topic 2, Atomic Theory.

c) molecule

A combination of two or more atoms chemically combined together. For example elemental oxygen, O2 and

water, H2O are both molecules.

d) compound

A substance that is made up of more than one element chemically bonded together. For example, water H2O, or

common table salt, NaCl.

e) cation

A species that carries a positive electric charge, for example Na+ and Mg

2+, which have fewer electrons than

their neutral parent atoms. Polyatomic cations also exist, such as the ammonium ion, NH4+.

f) anion

A species that carries a negative electric charge, for example Cl- or O

2- which have more electrons than their

neutral parent atoms. Polyatomic anions also exist, such as sulfate, SO2-2

.

g) formula unit

The group of ions that is the smallest unit of an ionic compound, for example common table salt has the formula

unit of NaCl, which tells us that in every crystal of salt for every Na+ ion there is one Cl

- ion present.

h) molecular formula

A combination of chemical symbols showing the actual numbers of atoms of each element present in a

molecule of a compound. For example, the molecular formulae for water and acetic acid are H2O and

CH3COOH.

i) mole

The mole is the SI unit for amount of substance, has the symbol mol. In one mole of a substance there are 6.02

x 1023

particles, where 6.02 x 1023

is the Avogadro number or constant.

j) Avogadro’s constant

Or Avogadro’s number, has symbol L, is the number of species per mole of substance, the value is 6.02 x 1023

.

The species may be particles such as ions, electrons, atoms, or molecules.

3. How many molecules are there in 2 mol of H2O?

There is Avogadro’s number in one mole, so in two mol:

two moles x 6.02 x 1023

= 1.20 x 1024

molecules

1

4. How many oxygen atoms are there in:

a) one molecule of glucose, C6H12O6?

According to the molecular formula, C6H12O6, there are six oxygen atoms in every molecule of glucose.

b) 1.00 mol of glucose?

There is Avogadro’s number of glucose molecules in one mol of glucose so if there are 6 oxygen atoms for

every molecule present then:

number of oxygen atoms = 6 x 6.02 x 1023

= 3.61 x 1024

oxygen atoms

5. How many ions are there in 0.200 mol of NaCl?

In 0.200 mol of Na+ and Cl

-, there are 0.400 mol of ions, 0.200 mol of Na

+ and 0.200 mol of Cl

-. In one mole,

there are an Avogadro’s number, so:

0.400 mol x 6.02 x 1023

= 2.40 x 1023

ions

6. How many moles of Mg2+

cations and Cl -

anions are there in 1.00 mol of magnesium chloride,

MgCl2?

From the formula unit, there is one mol of magnesium cations Mg2+

, for every mol of MgCl2.

Therefore there are 6.02 x 1023

ions of Mg2+

.

From the formula unit, there are two mol of chloride anions Cl -, for every one mol of MgCl2.

Therefore there are 2 mol x 6.02 x 1023

= 1.20 x 1024

anions of Cl -.

7. How many mol of Na+ cations and CO3

2- anions are there in 1.00 mol of sodium carbonate, Na2CO3?

From the formula unit, there is one mol of carbonate anions CO32 -

, for every mol of Na2CO3.

Therefore there are 6.02 x 1023

anions of CO32 -

.

From the formula unit, there are two mol of sodium cations Na+ for every one mol of Na2CO3.

Therefore there are 2 mol x 6.02 x 1023

= 1.20 x 1024

cations of Na+.

8. How many hydrogen atoms are there in:

a) one molecule of acetic acid, CH3COOH?

From the molecular formula, CH3COOH there are four atoms of hydrogen atoms present for each mol of acetic

acid.

b) 1.00 mol of acetic acid?

There is Avogadro’s number of acetic acid molecules in one mol of acetic acid so if there are 4 hydrogen atoms

for every molecule present then:

number of hydrogen atoms = 4 x 6.02 x 1023

= 2.41 x 1024

hydrogen atoms

9. One mol of hydrogen gas at room temperature contains:

A atoms

B 1 molecule

C 6 x 1023

atoms

D 6 x 1023

molecules

Answer: D (by definition)

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10. How many mol of hydrogen atoms are there in one mol of methane, CH4?

From the molecular formula, there are four hydrogen atoms per mol of CH4.

11. Pure silicon and germanium are two semi-conductors used in computer circuitry. In a 12.0 g sample

of germanium it was found that 1 atom in 109 was not geranium. How many non-germanium atoms

would be present in the sample?

A 1x1014

B 1x10-14

C 1x10-9

D 109

Answer: A

How many mol present? For Ge 12/73 = 0.16 mol (2 sig figs). Therefore number of atoms

= 0.16 x 6 x 1023

. But 1 in 109 not Ge. Therefore not Ge = 0.96 x 10

23/1 x 10

9 = 1 x 10

14

12. Why is the statement ‘one mol of oxygen’ imprecise?

It has not been specified whether it is a mole of O atoms or a mole of O2 molecules.

1.2 Exercises

1. Define the following terms giving the symbol, units and an example where appropriate:

a) relative atomic mass

The relative atomic mass, symbol Ar, of an element is the ratio of the average of its naturally occurring isotopic

masses of its atoms relative to 1/12 of the mass of a carbon-12 atom. There are no units as Ar values are

relative. For example, the Ar of oxygen is 16.00. The relative atomic masses of all elements known can be found

on most periodic tables and in your data booklet.

b) relative molecular mass

The relative molecular mass, symbol Mr otherwise known as the molecular weight, is the sum of the relative

atomic masses of all the atoms that comprise the molecule. As for relative atomic mass there are no units. For

example, the Mr of water H2O is (H: 2 x 1.01) + (O: 16.00) = 18.02.

c) molar mass

The collective name for properties that are expressed with units of mass per mole is known as molar mass.

Symbol M or Mm. It is the mass per mole of a substance. The molar mass has the same numerical value as the

atomic mass or molecular mass, but has the units g mol -1

. For example water has a Mr of (H: 2 x 1.01) + (O:

16.00) = 18.02 g, so its molar mass, M(H2O) is 18.02 g mol -1

.

d) relative formula mass

This is more precise term for relative molecular mass as it applies to ionic compounds as well as molecules. It is

much less commonly used though, and relative molecular mass, although technically incorrect, is commonly

used for ionic compounds. For example NaCl has a relative formula mass of (Na: 22.99) + (Cl: 35.45) = 58.44 g.

e) empirical formula

The empirical formula of a compound gives the lowest ratio of elements present, uses the smallest whole

numbers of atoms. For example the empirical formula for acetylene, C2H2 is CH.

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f) molecular formula

A combination of chemical symbols showing the actual numbers of atoms of each element present in a

molecule. For example, the molecular formulae for water and acetic acid are H2O and CH3COOH.

2. Explain why the relative atomic masses of the elements not exactly whole numbers? (You can find

the values in the IB Chemistry Data Booklet)

The relative atomic masses are average masses involving all the naturally occurring isotopes of that element.

The isotopes all have slightly different masses, so the average is not always a whole number.

3. Write the formula for the relationship between number of moles, mass and molar mass. Label the

equation with symbols for the quantity and symbols for the units of each quantity.

n =M

m

mass of substanceunits: grams symbol for units: g

molar mass of substance

units: grams per mole

symbol for units: g mol-1

amount of substanceunits: molesymbol for units: mol

NB: The symbols used for the quantities are as written in the equation.

4. Which of the following are correct? The molecular formula of hydrogen sulphide, H2S along with the

periodic table, can be used to determine:

A the number of atoms in the molecule,

B the molar mass of H2S,

C the number of moles in a given mass of H2S,

D the percentage of sulfur present in this compound.

E all of the above

Answer: E (as all of the above information can be obtained using the data available)

5. Explain why the statement: ‘The molar mass of sulfuric acid is 98.08 g’, is incorrect.

The units for molar mass are g mol-1

. This statement excludes the “mol” term from the unit. This 98.08g is simply

a unit of mass.

6. Calculate molar mass of:

a) sucrose, C12H22O11

M(C12H22O11) = (C: 12 x 12.01) + (H: 12 x 1.01) + (O: 11 x 16.00) = 332.24 g mol-1

b) ammonium chloride, NH4Cl

M(NH4Cl) = (N: 1 x 14.00) + (H: 4 x 1.01) + (Cl: 1 x 35.45) = 53.49 g mol-1

c) silver sulfate, Ag2SO4

M(Ag2SO4) = (Ag: 2 x 107.87) + (S: 1 x 32.06) + (O: 4 x 16.00) = 311.8 g mol-1

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7. Calculate the number moles that make up:

a) 74.6 g of potassium chloride, KCl

M(KCl) = (K: 39.10) + (Cl: 35.45) = 74.55 g mol-1

n = m/M = 74.6 g / 74.55 = 1.00 mol of molecules (3 s.f.)

b) 49.1 g of pure sulfuric acid H2SO4

M(H2SO4) = (H: 2 x 1.01) + (S: 1 x 32.06) + (O: 4 x 16.00) = 98.08 g mol-1

n = m/M = 49.1 g / 98.08 g mol-1

= 0.500 mol (3.s.f)

c) 72.1 g of water

M(H2O) = (H: 2 x 1.01) + (O: 16.00) = 18.02 g mol-1

n = m/M = 72.1 g / 18 g mol-1

= 4.00 moles of molecules (3 s.f.)

d) 5.61 g of calcium oxide, CaO

M(CaO) = (Ca: 40.08) + (O: 16.00) = 56.08 g mol-1

n = m/M = 5.61 g / 56.08 g mol-1

= 0.100 mol of molecules (3 s.f.)

8. Calculate the mass of each of the following:

a) 1.50 mol of calcium carbonate, CaCO3

M(CaCO3) = (Ca: 40.08) + (C: 12.01) + (O: 3 x 16.00) = 100.09 g mol-1

m = n x M = 1.50 mol x 100.09 g mol-1

= 150.135 g = 150 g (3 s.f.)

b) 0.200 mol of potassium sulfate, K2SO4

M(K2SO4) = (K: 2 x 39.10) + (S: 32.06) + (O: 4 x 16.00) = 174.26 g mol-1

m = n x M = 0.200 mol x 174.26 g mol-1

= 34.852 = 35.9 g (3 s.f.)

c) 1.25 mol of copper(II) sulfate pentahydrate, CuSO4.5H2O

M(CuSO4.5H2O) = (Cu: 63.54) +(S: 32.06) + (O: 4 x 16.00) + (H2O: 5 x 18.02)

= 249.7 g mol-1

m = n x M = 1.25 mol x 249.6 g mol-1

= 312.125 g = 312 g (3 s.f.)

9. Determine the identity of the following gaseous elements:

a) 0.200 mole of which has a mass 7.6 g

M = m/n = 7.6 g / 0.200 mol = 38 g mol-1

.

This is the molar mass of the gas, where each molecule contains TWO atoms of the element. So, do determine

the element involved, we must divide the answer by two!

So the element has the molar mass of: M = 38 g.mol-1

/2 = 19 g.mol-1

The element with the closest molar mass

is Fluorine.

b) 6.02 x 1022

molecules of which has a mass 140 g

In 1 mole there are 6.02 x 1023

molecules, so in 6.02 x 1022

molecules there are

6.02 x 1022

/6.02 x 1023

mol = 0.1 mol

M = m/n = 0.1 mol x 140 g = 14 g mol-1

The element is nitrogen.

5

The following questions are designed to test your ability to round off numbers to the correct number of

significant figures, and quote answers in scientific notation where appropriate. Refer to the introduction

section if you need practice with these skills before continuing with this section.

10. Calculate the number of moles from the following masses of substances:

a) 73 mg NaOH

M(NaOH) = (Na: 22.99) + (O: 16.00) + (H: 1.01) = 40 g mol-1

73 mg /1000 = 0.073 g

n = m/M = 0.073 g / 40 g mol-1

= 0.001825 = 1.8 x 10-3

mol of molecules (2 s.f.)

b) 50 mg H2O

M(H2O) = (H: 2 x 1.01) + (O: 16.00) = 18.02 g mol-1

500 mg /1000 = 0.050 g

n = m/M = 0.050 g / 18.02 g mol-1

= 0.0027746 = 2.8 x 10-3


c) 1.00 g Na2S2O3.5H2O

M(Na2S2O3.5H2O) = (Na: 2 x 22.99) + (S: 2 x 32.06) + (O: 3 x 16.00) + (H2O: 5 x 18.02)

= 248.2 g mol-1

n = m/M = 1 g / 248.2 g mol-1

= 0.004029 = 4.03 x 10-3


11. What is the mass of 2.5 x 10-3

mol of water?

M(H2O) = (H: 2 x 1.01) + (O: 16.00) = 18.02 g mol-1

m = n x M = 2.5 x 10-3

mol x 18.02 g mol-1

= 0.04505 = 4.5 x 10-2

g or 45 mg (2 s.f.)

12. What is the amount of substance present in each of the following:

a) 0.250 g of calcium carbonate

M(CaCO3) = (Ca: 40.08) + (C: 12.01) + (O: 3 x 16.00) = 100.09 g mol-1

n = m/M = 0.250 g /100.09 g mol-1

= 0.002497 = 2.50 x 10-3


b) 5.30 g of anhydrous sodium bicarbonate

M(NaHCO3) = (Na: 22.99) + (H: 1.01) + (C: 12.01) + (O: 3 x 16.00) = 84.01 g mol-1

n = m/M = 5.30 g / 84.01 g mol-1

= 0.0630877 = 6.31 x 10-2


c) 5.72 g of sodium carbonate decahydrate (Na2CO3.10H2O)

M(Na2CO3.10H2O) = (Na: 2 x 22.99) + (C: 12.01) + (O: 3 x 16.00) + (H2O: 10 x 18.02)

= 286.19 g mol-1

n = m/M = 5.72 g / 286.19 g mol-1

= 0.01998 = 2.00 x 10-2


13. Natural gas an alternative to petrol as a source of fuel. The main constituent of natural gas is

methane, CH4. For one mole of methane:

a) What is the mass?

M(CH4) = (C: 12.01) + (H: 4 x 1.01) = 16.05 g mol-1

m = n x M = 1 mol x 16.05 g mol-1

= 16.05 g

b) What is the mass of the carbon component?

m(C) = 12.01 g

c) How many mol of hydrogen atoms are there?

n(H) = 4

6

d) How many atoms of hydrogen are there?

n(H) = 4 therefore number of atoms in 4 mol = 4 x 6.02 x 1023

= 2.408 x 1024

atoms

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

%mass carbon = mass of carbon/total mass methane = 12.01 g /16.05 g (x 100) = 75%

14. Chloroform, an organic compound once used as an anaesthetic, has the composition of 10.04% C,

0.837% H and 89.12% Cl. The molar mass of chloroform is 119.37 g mol-1

.

a) Calculate the empirical formula of chloroform

100 g of chloroform contains 10.4 g of C, 0.837 g of H and 89.12 g of Cl.

n = m/M

n(C) = 10.04 g / 12.01 g mol-1

= 0.836 mol

n(H) = 0.837 g / 1.01 g mol-1

= 0.829 mol

n(Cl) = 89.12 g / 35.45 g mol-1

= 2.51 mol

Composition C:H:Cl =0.836:0.829:2.51

Simplest Ratio:

C: 0.836 / 0.829 = 1 mol

H: 0.829 / 0.829 = 1 mol

Cl: 2.51 / 0.829 = 3 mol

Ratio = 1:1:3 therefore empirical formula = CHCl3

b) What is the molecular formula of chloroform?

Molar mass of formula unit CHCl3:

M(CHCl3) = (C: 12.01) + (H: 1.01) + (Cl: 3 x 35.45) = 119.37 g mol-1

The molar mass of chloroform is 119.37 g mol-1

, therefore there is one formula unit per molecule, hence the

molecular formula is the same as the empirical formula, CHCl3.

15. An organic compound was found to have the composition 40.0% C, 6.70% H

and 53.3% O.

a) Calculate the empirical formula of the substance.

100 g of the organic compound contains 40.0 g of C, 6.70 g of H and 53.3 g of O.

n = m/M

n(C) = 40.0 g / 12.01 g mol-1

= 3.33 mol

n(H) = 6.70 g / 1.01 g mol-1

= 6.63 mol

n(Cl) = 53.3 g / 16.00 g mol-1

= 3.33 mol


Simplest Ratio:

C: 3.33 / 3.33 = 1 mol

H: 6.63 / 3.33 = 2 mol

O: 3.33 / 3.33 = 1 mol

Ratio = 1:2:1 therefore the empirical formula = CH2O

b) If the relative molecular mass is 60.06, what is its molecular formula?

Molar mass of formula unit CH2O:

M(CHO2) = (C: 12.01) + (H: 2 x 1.01) + (O: 16.00) = 30.03 g mol-1

7

60.06/30.03 = 2.

Therefore there are two formula units per molecule, so molecular formula = C2H4O2

c) How many mole of the organic compound is contained in a mass of 12 g?

n = m/M = 12.0 g / 60.06 g mol-1

= 0.1998 = 0.20 mol of molecules (2 s.f.)

16. The compound methyl butanoate smells like apples. An experiment determined the percentage

composition as C, 58.8%; H, 9.8% and O, 31.4%, and its molar mass 102.15 g mol-1

. Calculate the

molecular formula for methyl butanoate.

100 g of methyl butanoate contains 58.8 g of C, 9.8 g of H and 31.4 g of O.

n = m/M

n(C) = 58.8 g / 12.01 g mol-1

= 4.9 mol

n(H) = 9.8 g / 1.01 g mol-1

= 9.7 mol

n(O) = 31.4 g / 16.00 g mol-1

= 1.96 mol


Simplest Ratio: C: 4.9 / 1.96 = 2.5 mol

H: 9.7 / 1.96 = 5

O: 1.96 / 1.96= 1 mol

Simplest whole number ratio = 5:10:2 therefore the empirical formula = C5H10O2

Molar mass of formula unit C5H10O2:

M(C5H10O2) = (C:5 x 12.01) + (H: 10 x 1.01) + (O: 2 x 16.00) = 102.15 g mol-1

Therefore there is one formula unit per molecule, so molecular formula = C5H10O2 also.

17. An organic compound was analysed and it was found to contain 54.53% of carbon, 9.15% of

hydrogen and 36.32% of oxygen. How many hydrogen atoms will be in the empirical formula for this

compound?

A 2

B 4

C 6

D 9

Answer: B

100 g of the organic compound contains 54.53 g of C, 9.15 g of H and 36.32 g of O.

n = m/M

n(C) = 54.53 g / 12.01 g mol-1

= 4.54 mol

n(H) = 9.15 g / 1.01 g mol-1

= 9.06 mol

n(O) = 36.32 g / 16.00 g mol-1

= 2.27 mol


Simplest Ratio:

C: 4.54 / 2.27 = 2 mol

H: 9.06 / 2.27 = 4 mol

O: 2.27 / 2.27= 1 mol

Simplest whole number ratio = 2:4:1 therefore the empirical formula = C2H4O

Therefore there are 4 hydrogen atoms in the empirical formula.

8

18. An experiment similar to one that John Dalton may have used to establish the existence of atoms

involves determining the formula for magnesium oxide by using a crucible with lid to burn a known

mass of magnesium in air. The resulting mass of magnesium oxide produced can then be

measured. Here are the results from such an experiment.

mass of crucible with lid 24.12 g

mass of crucible, lid and magnesium 24.70 g

mass of crucible, lid and magnesium oxide 25.50 g

a) Calculate the empirical formula for magnesium oxide.

m(Mg) = 24.70 g – 24.12 g = 0.58 g

m(Magnesium Oxide) = 25.50 g – 24.12 g = 1.38 g

Therefore, the % mass of magnesium in magnesium oxide is:

Mass of magnesium/mass of magnesium oxide (x100) = 0.58 g / 1.38 g (x100) = 42%.

Therefore the % mass of oxygen = 100% - 42% = 58%

100 g of Magnesium oxide contains 42 g of magnesium and 58 g of oxygen.

n = m/M

n(Mg) = 42 g / 24.31 g mol-1

= 1.72 mol

n(O) = 58 g / 16.00 g mol-1

= 3.625 mol

Composition = Mg:O = 1.72:3.625

Simplest ratio

Mg: 1.72/1.72 = 1 mol

O: 3.625/1.72 = 2 mol

Ratio = 1:2, therefore the empirical formula = MgO2

b) The results may not match with the actual empirical formula and even if they are correct, we

make assumptions about the experiment in calculating the results. List four assumptions made.

The sample of magnesium is pure.

All of the magnesium is being completely oxidised upon burning.

There was no heat loss from the system.

Only MgO was formed, no by-products (oxygen is not the only gas present in air, some Mg3N2 forms)

1.3 Exercises

1. What is a chemical reaction?

A chemical change in which one or more elements or compounds (the reactants) form new compound/s (the

product/s).

2. Define chemical equation. Include in your definition an explanation of the use of coefficients and

subscripts.

A way of representing a chemical reaction using symbols for the participating particles (atoms, molecules, ions,

electrons etc) where an arrow points from the reactants to the products.

For example: xA + yB � zC + wD. The numbers x, y, z, and w are known as the stoichiometric coefficients and

they show the relative number of molecules (or mole ratios) of the reactants and products. Any subscripts used

are part of the formula for that species and should never be altered when balancing equations.

9

3. What is the meaning of the symbols (s), (l), (g), (aq) when used in chemical equations?

When reactions involve different phases it may be indicated in brackets after the symbol using the following

symbols: (s) = solid, (l) = liquid, (g) = gas, (aq) = aqueous (meaning that the substance is dissolved in a solution

of water)

4. Why are chemical equations balanced?

Chemical equations are balanced as a requirement of the law of conservation of matter. If they are not

balanced, the equation may indicate the creation or destruction of matter; not possible according to the law!

Balancing ensures the mass of the compounds and the elements are equal on either side of the arrow.

5. Explain in words what the following chemical equation tells us:

O2(g) + 2H2(g) �� 2H2O(l)

1 mole of dioxygen gas reacts with 2 moles of dihydrogen gas to give 2 moles of liquid water.

6. Balance each of the following skeletal chemical equations:

a) CO(g) + O2(g) �� CO2(g)

2CO(g) + O2(g) � 2CO2(g)

b) Cd(NO3)2(aq) + Na2S(aq) �� CdS(s) + NaNO3(aq)

Cd(NO3)2(aq) + Na2S(aq) � CdS(s) + 2NaNO3(aq)

c) C2H2 + O2 �� CO2 + H2O

2C2H2 + 5O2 � 4CO2 + 2H2O

d) P4O10(s) + H2O(l) �� H3PO4(l)

P4O10(s) + 6H2O(l) � 4H3PO4(l)

7. In the following unbalanced equation, what are the values of x and y?

MnO2 + xHCl �� MnCl2 + Cl2 + yH2O

A 1,2

B 4,1

C 2,4

D 4,2

Answer: D (equations must balance, 4H LHS, 2H2 RHS)

8. Balance each of the following skeletal chemical equations:

a) Br - + Cl2 �� Cl

- + Br2

2Br - + Cl2 � 2Cl

- + Br2

b) AgNO3 + BaCl2 �� AgCl + Ba(NO3)2

2AgNO3 + BaCl2 � 2AgCl + Ba(NO3)2

c) KClO3 �� KCl + O2

2KClO3 � 2KCl + 3O2

d) H2SO4 + HI �� H2S + H2O + I2

H2SO4 + 8HI � H2S + 4H2O + 4I2

e) SO2 + H2S �� H2O + S (the reaction thought to occur when volcanoes erupt, as deposits of sulfur

are found near volcanoes)

SO2 + 2H2S � 2H2O + 3S

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9. Which of the following can be determined from the reaction equation:

AgNO3(aq) + HCl(aq) �� AgCl(s) + HNO3(aq)

A the reaction is neutralization.

B the molar proportions of reactants to products.

C the amount of matter destroyed during the reaction.

D silver chloride is insoluble.

Answer: B and D (the reaction is not neutralisation, nor is matter destroyed in a chemical reaction)

10. Write an balanced equations including state symbols for the following reactions:

a) Sodium metal reacting with water to form sodium hydroxide and hydrogen gas.

2Na(s) + 2H2O(l) � 2NaOH(aq) + H2(g)

b) Hot Lithium metal reacts in an atmosphere of nitrogen to from lithium nitride, Li3N.

6Li(s) + N2(g) � 2Li3N(s)

c) Calcium reacts with water leads to the formation of calcium hydroxide Ca(OH)2 and the evolution

of hydrogen gas.

Ca(s) + 2H2O(l) � Ca(OH)2 + H2(g)

11. Aluminium saucepans do not react with atmospheric oxygen, or with the food being cooked in them

because of the very hard surface coating of aluminium oxide that forms as soon as aluminium is

exposed to the oxygen in air. The balanced chemical equation for the reaction between aluminium

and oxygen is:

A 2Al(s) + O2(g) �� 2AlO(s)

B Al(s) + O2(g) �� 2AlO2(s)

C 4Al(s) + 3O2(g) �� 2Al2O3(s)

D 3Al(s) + 2O2(g) �� Al3O2(s)

Answer: C (as the correct formula for aluminium oxide is Al2O3)

12. The equation for the reaction in a blast furnace for obtaining iron from its ore, haematite, is shown:

Fe2O3 + 3CO �� 2Fe + 3CO2

The reaction takes place at temperatures high enough (about 1000 °°°°C) to produce molten iron which

is collected.

a) What information can be obtained from the equation in terms of the reacting species?

For every 1 mole of haematite reacted with 3 moles of carbon monoxide, 2 moles of iron are obtained along with

the production of 3 moles of carbon dioxide.

b) What mass of haematite would need to be mined to produce 1000 kg of iron?

n(Fe) = m/M = 1 x 106 g / 55.85 g mol

-1 = 17 905 mol

Mole ratio of haematite to iron = 1:2

n(Fe2O3) = 17 905 mol / 2 = 8952 mol

M(Fe2O3) = (Fe: 2 x 55.85) + (O: 3 x 16.00) = 159.7 g mol -1

m = n x M = 8952 mol x 159.7 g mol -1

= 1.4 x 106 g or 1430 kg haematite required

11

13. Write balanced equations for the following combustion reactions:

a) ethanol, C2H5OH(l), burning in oxygen to form carbon dioxide and water.

C2H5OH(l) + O2(g) � 2CO2(g) + 3H2O(l)

b) acetylene, C2H2(g), burning in oxygen to form carbon dioxide and water.

2C2H2(g) + 5O2(g) � 4CO2(g) + 2H2O(l)

14. Octane, C8H18(l), is a constituent of petrol. When it burns in air carbon dioxide and water vapour

form. It releases a considerable amount of energy when burnt.

a) Write a balanced equation for the complete combustion of octane.

2C8H18(g) + 25O2(g) � 16CO2(g) + 18H2O(l)

b) What mass of water is produced when 1.0 kg of octane is completely burnt?

M(C8H18) = (C: 8 x 12.01) + (H: 18 x 1.01) = 114.26 g mol-1

n(C8H18) = m/M = 1000 g / 114.26 g mol-1

= 8.75 mol

n(C8H18) : n(H2O) = 2:18 or 1:9

Therefore, n(H2O) = n(C8H18) x 9 = 8.75 mol x 9 = 78.75 mol

M(H2O) = 18.02 g mol-1

m(H2O) = n x M = 78.77 mol x 18.02 g mol-1

= 1419 g = 1.4 kg (2 s.f.)

c) What mass of carbon dioxide is produced when 1.0 kg of octane is completely burnt?

M(C8H18) = (C: 8 x 12.01) + (H: 18 x 1.01) = 114.26 g mol-1

n(C8H18) = m/M = 1000 g / 114.26 g mol-1

= 8.75 mol

n(C8H18) : n(CO2) = 2:16 or 1:8

Therefore, n(CO2) = n(C8H18) x 8 = 8.75 mol x 8= 70 mol

M(CO2) = 44.01 g mol-1

m(CO2) = n x M = 70 mol x 44.01 g mol-1

= 3080.7 g = 3.1 kg (2 s.f.)

d) What mass of oxygen is required for the complete combustion of 10 g of octane, C8H18?

M(C8H18) = (C: 8 x 12.01) + (H: 18 x 1.01) = 114.26 g mol-1

n(C8H18) = m/M = 10 g / 114.26 g mol-1

= 8.75 x 10-2

mol

n(C8H18) : n(O2) = 2:25 or 1:12.5

Therefore, n(O2) = n(C8H18) x 12.5 = 8.75 x 10-2

mol x 12.5 = 1.09 mol

M(O2) = 32 g mol-1

m(O2) = n x M = 1.09 mol x 32 g mol-1

= 35 g (2 s.f.)

15. How many moles of dioxygen are produced when 58.4 g of water is electrolysed and decomposes

according to the equation: 2H2O(l) �� 2H2(g) + O2(g)

From the equation, mole ratio of H2O to O2 is 2:1

n(H2O) = m/M = 58.4 g / 18.02 g mol-1

= 3.24 mol molecules H2O

therefore n(O2) = 3.24 mol / 2 = 1.62 mol dioxygen molecules.

12

1.4 Exercises

1. Define the following:

a) theoretical yield

The maximum mass of a product that can be obtained from a given amount of reactant/s. It is based on the

stoichiometry of a chemical equation.

b) limiting reactant

The reactant that is present in an amount of moles smaller than that required by the stoichiometry of the

equation is called the limiting reactant, it dictates the maximum yield (theoretical yield) of a product in a reaction.

c) reactant in excess

The reactant that is present in an amount of moles larger than that required by the stoichiometry of the equation

is called the reactant in excess, there will be leftover of that reactant (or excess) at the end of the reaction.

d) experimental yield

The experimental yield is the actual mass of product obtained from a chemical reaction in practice.

e) percentage yield

The percentage yield is the percentage of the theoretical yield actually achieved. It is calculated by dividing the

actual yield obtained (experimental yield) by the calculated yield based on the limiting reactant (theoretical yield)

x 100%

f) Avogadro’s law

The volume of a sample of gas at constant temperature and pressure is proportional to the amount of gas

molecules in the sample, V ∝ n.

g) Boyle’s law

The volume of a fixed amount of gas inversely proportional to its pressure at a constant termperature.

h) Charles’ law

The volume of a fixed amount of gas at a constant pressure is proportional to the absolute temperature.

i) An ideal gas

A hypothetical gas that consists of identical particles of zero volume with no intermolecular forces and the

constituent atoms or molecules of the gas undergo perfect elastic collisions (collisions with no loss of energy).

j) An ideal gas equation

The combination of Charles’ law, Avogadro’s law and Boyle’s law. These laws combined, give the following

relationship:

With the inclusion of the Ideal gas constant to this expression, the Ideal gas equation is given by

PV= nRT

P

TnV

×∝

From Avogadro’s

Law

From

Charles’ Law

From Boyle’s Law

13

k) molar volume, Vm

The volume of a sample divided by the number of moles it contains. The molar volume of an ideal gas at

standard temperature and pressure is 22.4 dm3 mol

-1 (2.24 x 10

-2 m

3 mol

-1). That is, one mole of a gas at STP

occupies 22.4 dm3 (22.4 L).

l) STP

Refers to conditions of standard temperature and pressure. IUPAC defines these values as 1.01 x 105 Pa (or

760 mmHg, atmospheric pressure) and 273 K (or 0°C). You may find slightly a slightly different set of values for

STP quoted by other organizations and textbooks, the most common deviation is the use of a temperature of

25°C instead of 0 °C. For this reason, the actual temperature should be quoted.

2. A sample of pure iron, mass 1.4 g, was placed in excess dilute sulfuric acid. There was a steady

effervescence of gas, the solution became warm and turned pale green.

a) Write a balanced equation (with state symbols) for the above reaction.

Fe(s) + H2SO4(aq) � FeSO4(aq) + H2(g)

b) Calculate the number of moles of iron reacting.

n = m/M = 1.4 g / 55.85 g mol-1

= 0.025067 = 2.5 x 10-2

mol

c) What is the volume of gas produced, assuming it is an ideal gas at STP?

Since n(Fe) : n(H2) = 1:1, n(H2) = 2.5 x 10-2

mol

According to Avogadro’s law, the volume of the gas is proportional to the number of moles in the sample, and

the molar volume of a sample at STP is 22.4 dm3 mol

-1.

Therefore 22.4 dm3 mol

-1 x 2.5 x 10

-2 = 0.56 dm

3.

d) What information do you need to know to ensure that the sulfuric acid is in excess?

The number of moles of sulfuric acid is required, so for this to be calculated we must know the concentration of

the acid.

3. 100 g of sodium carbonate was mixed with 2 mol of hydrogen chloride in solution.

a) Write the equation for the reaction of sodium carbonate and hydrogen chloride.

Na2CO3 + 2HCl � 2NaCl + H2O + CO2

b) Determine which reactant is in excess and which is the limiting reagent and hence calculate the

theoretical yield of sodium chloride.

n(HCl) = 2 mol

n(Na2CO3) = m/M = 100 g / 105.99 g mol -1

= 0.943 mol = 9.43 x 10-1

mol

n(NaCl) from HCl:

mole ratio = n(HCl) : n(NaCl) = 1:1, therefore n(NaCl) = 2 mol

n(NaCl) from Na2CO3:

mole ratio = n(Na2CO3) : n(NaCl) = 1:2, therefore n(NaCl) = 2 x 9.43 x 10-1

mol = 1.886 mol

Therefore Na2CO3 is the limiting reactant and HCl is in excess.

m(NaCl) = n x M = 1.886 mol x 58.44 g mol-1

= 110.21 g = 1.10 x 103 g (3 s.f.)

14

4. Carborundum or silicon carbide SiC, is used as an abrasive due to its hardness. It is made from

sand and coke, according to the following equation: SiO2 + C �� SiC + CO2.

A problem the industrial chemist frequently encounters is that the reactants to be used are not

always pure. To overcome this, amounts in excess of the mole ratio given by the equation for the

reaction may be used.

a) Calculate the number of kg of coke, 90% pure, required to yield 15.0 kg of silicon carbide.

First, calculate the number of moles of silicon carbide in 15 kg.

M(SiC) = (Si: 28.09) + (C: 12.01) = 40.1 g mol-1

n(SiC) = m/M = 15 000 g / 40.1 g mol -1

= 374.06 = 374 mol (3 s.f.)

From the equation, the mole ratio of C to SiC is 1:1 therefore n(SiC) = n(C).

So, 374 mol of coke are required in the reaction.

The molar mass of coke is 12.01 g mol-1

.

So, if the coke was 100% pure, the mass of coke required would be:

m = n x M = 374 mol x 12.01 g. mol

-1 = 4491.74 g

So, the mass of carbon required taking into account it is only 90% pure would be:

(mass of 90% pure coke) x 0.9 = 4491. 74 g

(mass of 90% pure coke) = 4990.8 g

This is equivalent to 4.99 kg of coke.

b) How many kg of pure sand, SiO2, would be required for this?

From equation: n(SiC) = n(SiO2)

M(SiO2) = (Si: 28.09) + (O: 2 x 16.00) = 60.09 g mol-1

m(SiO2) = n x M = 374 mol x 60.09 g mol -1

= 22.5 kg (3 s.f.) of pure SiO2 required.

5. Ammonia, NH3, is used to make fertilizers and explosives. It is made from gaseous nitrogen and

hydrogen in the industrial reaction called the Haber process. Fritz Haber was the German chemist who

first developed the process. The equation for the reaction is:

N2(g) + 3H2(g) �� 2NH3(g).

a) If at a given temperature and pressure, 10 dm3 of N2 is reacted with 30 dm

3 of H2 how many dm

3

of ammonia are produced?

Using Avogadro’s law and the equation given:

n(NH3) = 2 x n(N2) or 2/3 x n(H2) and since volumes are proportional to the number of moles, therefore n(NH3) =

20 dm3

b) In actual fact, a yield of about 55% is obtained. How much N2 and H2, in dm3, must be initially

present in order to produce 100 dm3 of NH3.

Use Avogadro’s law. From the equation the volume relationship for 100% yield = 100 dm3 is 50 dm

3 N2 + 150

dm3 H2 � 100 dm

3 NH3. With a 55% yield 50 dm

3 N2 � 55 dm

3 NH3. Therefore for 100 dm

3 NH3 need 50 x 100/

55 = 91 dm3 N2 and 3 x 182 dm

3 H2 = 273 dm

3.

c) Calculate the number of moles of reactants required to make 7.24 mol of ammonia based on a

yield of 55%?

The equation gives mole ratio N2(g) + 3H2(g) � 2NH3(g). Therefore for 100% yield for 7.24 mol need 3.62 mol

N2 and 10.86 mol H2. 3.62 mol N2 only � 55% NH3. ? mol N2 � 100% NH3.

? = 100 x 3.62/55 = 6.58 mol N2. Therefore required mol H2 = 3 x 6.82 = 19.7 mol (3 sig fig.)

15

6. The combustion of ethane is shown by the equations:

complete combustion: C2H4(g) + 3O2(g) �� 2CO2(g) + 2H2O(g)

incomplete combustion :C2H4(g) + 2O2(g) �� 2CO(g) + 2H2O(g)

Each reaction begins with 2.70 mol of C2H4 and 6.30 mol of O2. For both reactions:

a) identify the limiting reagent

For complete combustion:

Number of moles CO2 from C2H4:

From equation: n(C2H4) : n(CO2) = 1:2

Therefore n(CO2) = n(C2H4) x 2 = 2.70 mol x 2 = 5.40 mol

Number of moles of CO2 from O2:

From equation: n(O2) : n(CO2) = 3:2

Therefore n(CO2)= n(O2) x 2/3 = 6.30 mol x 2/3 = 4.20 mol

Therefore O2 is the limiting reactant (the least amount of either product is produced from O2).

For incomplete combustion:

Number of moles CO from C2H4:

From equation: n(C2H4) : n(CO) = 1:2

Therefore n(CO) = n(C2H4) x 2 = 2.70 mol x 2 = 5.4 mol

Number of moles of CO from O2:

From equation: n(O2) : n(CO) = 1:1

Therefore n(CO2)= 6.30 mol x 1 = 6.30 mol

Therefore C2H4 is the limiting reactant (the least amount of product is produced from C2H4).

b) calculate the moles of water produced (theoretical yield)


From equation= n(LR) : n(H2O) = 3:2 (LR = limiting reagent)

Therefore n(H2O) = n(LR) x 2/3 = 6.30 mol x 2/3 = 4.20 mol


From equation= n(LR) : n(H2O) = 1:2

Therefore n(H2O) = n(LR) x 2 = 2.70 mol x 2 = 5.40 mol

c) calculate the moles of excess reactant remaining


Excess reactant is C2H4

Excess C2H4 = C2H4 given - C2H4 used.

From equation n(C2H4) : n(H2O) = 1:2

Therefore n(C2H4) used = n(H2O)/2 = 4.20 mol/2 = 2.10

n(C2H4) given = 2.70 mol

Excess = 2.70 mol -2.10 mol = 0.60 mol C2H4 remaining


Excess reactant is O2

Excess O2 = O2 given – O2 used.

16

From equation n(O2) : n(H2O) = 1:1

Therefore n(O2) used = n(H2O) x 1 = 5.40 mol

n(O2) given = 6.30 mol

Excess = 6.30 mol – 5.40 mol = 0.9 mol O2 remaining

7. Calcium compounds are often used as structural materials. Mortar consists of about one part lime

and three parts sand (silica, SiO2). Lime is obtained from naturally occurring limestone containing

calcium carbonate, CaCO3, which decomposes upon heating to produce lime CaO, and CO2.

a) Write the balanced equation for this reaction.

CaCO3(s) � CaO(s) + CO2(s)

b) What is the theoretical yield of lime if 49.6 g of CaCO3 is decomposed?

From equation, since mole ratio of n(CaCO3):n(CaO) is 1:1 therefore n(CaO) = n(CaCO3)

n(CaCO3) = m/M = 49.9 g / 100.09 g mol-1

= 0.4985 mol

n(CaO) = 0.500 mol

m(CaO) = n x M = 0.4985 x 56.08 g mol-1

= 27.7343 = 27.7 g (3 s.f.)

c) What is the percentage yield if only 23.7 g of lime is obtained?

Percentage yield = experimental yield/ theoretical yield x 100 = 23.7 g/27.7 g x 100 = 86%

d) Suggest a reason for the difference between the theoretical yield and the actual yield.

Crude limestone contains varying amounts of impurities such as clay, silica (sand), iron oxide, organic materials

etc.

8. In the production of pure nickel, the gas carbon monoxide, CO(g), is passed over the impure product

at 60 oC and volatile liquid nickel carbonyl, Ni(CO)4, forms. This is then heated to 180

oC and pure nickel

is deposited. The CO is recycled. If 5.9 g of Ni are heated with 14 g of CO how much Ni(CO)4 is made?

Equation for the reaction:

Ni(s) + 4CO(g) � Ni(CO)4

Limiting reactant:

Moles of Ni(CO)4 from Ni

mole ratio Ni:Ni(CO)4 = 1:1

n(Ni) = 5.9 g/58.71 g mol -1

= 0.10 mol

n(Ni(CO)4) = 0.10 mol x 1 = 0.10 mol

Moles of Ni(CO)4 from CO

Mole ratio CO:Ni(CO)4 = 4:1

n(CO)= 14 g/28.01 g mol -1

= 0.4999 mol

n(Ni(CO)4) = = 0.4999 mol /4 = 0.1249 = 0.12 mol (2 s.f.)

Therefore, Ni is limiting reactant.

Theoretical yield = n(limiting reactant) x M(product) = 0.10 mol x 170.75 g mol -1

= 17.075 = 17 g Ni(CO)4 (2 s.f.)

17

9. How many mol of water will be produced when 2 mol of hydrogen, H2, are reacted with 2 mol of

oxygen, O2?

Equation for the reaction:

2H2(g) + O2(g) � 2H2O(l)

Since 2 moles of H2 are required to produce 2 mole H2O but only 1 mole of O2 produces the same amount, H2 is

the limiting reactant and only 2 mole of H2O can possibly be produced according to the mole ratio.

Extended answer:

Limiting reactant:

Moles of H2O from H2:

Mole ratio H2O:H2 = 1:1

n(H2) = 2 , therefore n(H2O) produced = 2.

Moles of H2O from O2:

Mole ratio O2: H2O = 1:2

n(O2) = 2 , therefore n(H2O) produced = 4.

therefore H2 is limiting reactant, so based on the number of moles of H2 present, 2 moles of water will be

produced.

10. What is the volume of oxygen required to react with 100 cm3 of hydrogen to produce water at

constant temperature and pressure?

Equation for the reaction: see Q9

Using Avogadro’s law and the equation:

n(O2) = 1/2 x n(H2) and since volumes are proportional to the number of moles

therefore n(O2) = 50 cm3

11. What is the volume in dm3, of 1.20 mol of SO2(g) at STP?

Vm = 22.4 dm3 mol

-1 therefore, 1.2 mol x 22.4 dm

3 mol

-1 = 26.88 dm

3 of SO2

12. Calculate the number of mol in 64 dm3 of oxygen gas at STP.

Vm = 22.4 dm3 mol

-1 therefore, 64 dm

3/22.4 dm

3 mol

-1 = 2.857 = 2.9 mol (3 s.f.)

13. Pentane, C5H12, is a constituent of petrol. In a well tuned car engine the complete combustion of

pentane produces CO2(g) and H2O(l).

a) Write the equation for this reaction in the engine.

C5H12(l) + 8O2(g) � 5CO2(g) + 6H2O(l)

In travelling 1 km a car uses 100 g of pentane. Calculate:

b) the moles of pentane used.

M(C5H12) = 72.17 g mol -1

n = m/M = 100 g / 72.17 g mol -1

= 1.38561 = 1.39 mol (3 s.f.)

c) the volume of air required at STP, assuming air is 20% oxygen.

From equation, n(C5H12) : n(O2) = 1:8

Therefore n(O2) required = 8 x n(C5H12) = 1.39 mol x 8 = 11.12 mol

18

However given that air is only 20% oxygen, (need 5 times as much) 11.12 mol x 100/20 = 55.6 mol of “air”

Volume = 22.4 dm3 mol

-1 x 55.6 mol = 1245.44 = 1250 dm

3 air required (3 s.f)

14. A weather balloon contains 12.0 dm3 of hydrogen gas at STP.

a) How many molecules of hydrogen are present?

Number of moles = 12.0 dm3 / 22.4 dm

3 mol

-1 = 0.5357 = 0.536 mol (3 s.f.)

Number of molecules = 0.536 mol x 6.02 x 1023

= 3.23 x 1023

molecules H2.

b) Would your answer change if helium gas were used instead of hydrogen? Give a reason for your

answer.

The answer would not change as Avogadro’s law states that the molar volume, Vm = 22.4 dm3 mol

-1, at STP,

regardless of what substance is present.

15. This question is about the ideal gas equation.

a) State the ideal gas equation.

PV = nRT

b) What is R?

c) What are the SI units for R?

R is the gas constant = 8.31 J K-1

mol-1

. You will not need to remember this value. It is given in the IB Chemistry

data booklet.

The value and units of the gas constant, R

The gas constant (otherwise known as the universal or ideal gas constant) is denoted by the symbol R and expressed in units of energy per Kelvin per mole it has the value: R = 8.31 J K

-1 mol

-1

You will use this value and units for the gas constant, R for the IB course (given in the IB Data Booklet), however there are many other units that can be used to express the value of R. The gas constant, R is a physical constant which forms a part of many fundamental equations in physical science. One important equation that you will use is the ideal gas equation: PV = nRT So where does this value and units of R = 8.31 J K

-1 mol

-1 actually come from?

To find the value: substitute values for one mole of an ideal gas at atmospheric pressure and 273K into the ideal gas equation: P = 1 atm = 1.01 x 10

5 Pa

V = 2.24 m3 (molar volume of an ideal gas at 273 K and 1.01 × 10

5 Pa = 2.24 m

3 mol

–1)

n = 1 mol T = 273 K R = PV/nT R = 1.01 x 10

5 Pa x 2.24 m

3 / 1 x 273

R = 8.31 Pa m3 K

-1 mol

-1

To find the units: take a look at the units used in the ideal gas equation. SI units are used, therefore volume is in m

3 (1 litre = 1 dm

3 = 1 x 10

-3 m

3), pressure in Pascals, and temperature is

in K. The unit analysis for R is shown: R units = Pa x m

3 / mol x K

This gives the units for R as shown above as: Pa m3 K

-1 mol

-1

The units for pressure can be simplified to include units for energy, joules, J. Here is how. Pressure is simply force (newtons, N) per square metre (m

-2) which = N m

-2. One N m

-2 is one

Pascal of pressure. Therefore the P x V term = N m-2

x m3 which simplifies to N m. Moving a

force of one N over 1 metre requires energy of one joule (J), therefore the units N m can be replaced with J to give the units of R as J K

-1 mol

-1

19

16. Explosives get their damaging effect because the heat generated causes a rapid increase in the

volume of gas produced. During an explosion 200 cm3 of a gas were heated from 20

oC to 1000

oC.

Calculate the new volume, in litres at the highest temperature.

This is an application of Charles’ Law: volume proportional to absolute temperature,

V = constant x T

T1 = 293 K

T2 = 1273 K

V1 = 0.2 dm3

V2 = ?

T1/V1 = T2/V2 (multiply both sides by 1/T2 to isolate V2)

Therefore, V2 = T1/V1 x (1/T2) = 293 K/ 0.2 dm3 x (1/1273 K) = 1.15 dm

3.

17. In the Ferrari car engine the maximum cylinder volume is about 500 cm3. If air enters the cylinder at

50°C and 1 atm pressure, how many grams of octane, C8H18, should the fuel injection system send into

the cylinder, if it is to burn completely in that air when the spark plug fires? Assume air is 21% oxygen

by volume.

Equation for reaction:

C8H18(l) + 12.5O2(g) � 8CO2(g) + 9H2O(l)

or

2C8H18(l) + 25O2(g) � 16CO2(g) + 18H2O(l)

Use molar volume 1 mol = 22.4 dm3 at STP

500 cm3 air = 0.005 dm

3 = 21% oxygen = 21 x 0.500/100 dm

3 at 50

OC (323 K)

therefore volume at 273 K (standard T) = 0.105 x 273/323 dm3 = 0.0887 dm

3

therefore mol oxygen = 0.089/22.4 = 0.00396 mol

mol ratio octane to oxygen = 1 : 12.5

therefore mol octane required = 0.00396/12.5

molar mass octane = (8 x 12 for C = 96 + 18 x 1 for H = 18) = 114 g, therefore mass octane injected = 114 x

0.00396/12.5 g

= 0.036 g

Using the ideal gas equation PV = nRT to calculate n:

Vol of O2 in 500 cm3 air = 500 x 21/100 = 105 cm

3 = 0.105 dm

3 = 0.105/1000 m

3

n = PV/RT = 101000 x 0.000105/ 8.3 x 323 = 0.00396 mol O2

mass = molar mass x mol = 114 x 0.00396 = 0.036 g

18. A tyre has a volume of 2.6 dm3, and the recommended pressure is two atmospheres at 20°C. What is

the tyre pressure on a hot day when the road temperature and the tyre temperature reach 50°C?

PV=nRT

V = 2.6 dm3 = constant, n = constant R = constant

P = nRT/V

P/T = nR/V

Therefore, P1/T1 = P2/T2

20

P2 = P1/T1 x T2

T1 = 293 K

P1 = 2 atm

T2 = 323 K

P2 = 2/ 293 K x 323 K = 2.2 atm

19. Draw simple graphs to show the following:

a) Volume in m3 against pressure in Pa, T constant.

Vo

lum

e (

m3)

Pressure (Pa)

b) Volume against temperature in K, pressure constant.

Temperature (K)

Vo

lum

e (

m3)

c) Pressure against temperature in K, volume constant.

Temperature (K)

Pre

ssu

re (

Pa)

The graphs show the direct or inverse proportionality relationships between the variables.

21

20. A clay compound with the formula X.1/2H2O when mixed with water, forms the hard dihydrate

X.2H2O, when dried. It is used to make ceramics.

If 2.90 g of X.1/2H2O yields 3.44 g of X.2H2O, the calculated molar mass of X.2H2O is:

A 860 g mol-1

B 688 g mol-1

C 344 g mol-1

D 172 g mol--1

Answer: D

Mass change due to the addition of 1 ½ H2O = 3.44 – 2.90 g = 0.54 g

therefore total mass H2O (2H2O) present = 0.54 + 0.18 g (1 ½ + ½) = 0.72 g

therefore mass of X in the sample = 2.72 g

mol ratio of X : H2O in dihydrate = 1 : 2

mol present in 0.72 g of H2O = 0.72/18 = 0.040 mol

therefore mol X present = 0.020

2.72 g � 0.020 mol, therefore ? g (the molar mass) � 1.00 mol

therefore ? = 2.72/0.020 = 136 g

add 2 x 18 for 2H2O gives 36 + 136 = 172 g mol-1

1.5 Exercises

1. Define the terms:

a) homogeneous

A material that has uniform properties and composition.

b) solution

A homogeneous mixture. It is a sample of matter that comprises more than one pure substance. The properties

of the mixture does not vary within the sample.

c) solute

A substance being dissolved in a liquid to form a solution.

d) solvent

The most abundant component of a solution. A liquid used to dissolve a solute in.

e) decimetre

One tenth of a metre. A measure of length.

f) g dm – 3

Grams per decimetre. A way of quoting the amount of grams of a solute per dm3 of solvent. Equivalent to grams

per litre. Can be rewritten as g/dm 3. One dm

3 is equivalent in volume to one litre.

g) mol dm – 3

Moles per decimetre cubed. A way of quoting the amount of moles of a solute per dm3 of solvent. Equivalent to

moles per litre.

2. What is the meaning of [NH4+(aq)]?

NH4+ is the formula for the ammonium ion. The (aq) following the formula indicates that the NH4

+ ions are in an

aqueous solution (water). The square brackets indicate “moles per dm3”. So a [NH4

+(aq)] = 0.1 means: the

concentration of ammonium ions in an aqueous solution is 0.1 mol dm-3

.

22

3. A student prepared a solution by dissolving 1.55 g of NaNO3 in water to prepare a 25.00 cm

3

solution. What is the molarity of the solution?

First calculate the number of moles of NaNO3 added. Then use this value to determine the concentration of the

solution in mol dm-3

.

3

3729.0

02500.0

0182.0

0182.016301.1499.22

55.1

−===

=×++

==

dmmoldm

mol

V

n

molg

M

mn

4. Silver nitrate, AgNO3 is used in photography in the development process. What mass of silver

nitrate must be placed in a 25.00 cm3 volumetric flask and diluted to the mark in order to prepare a 0.25

mol dm – 3

solution of AgNO3 (aq)?

First, use the formula n = cV to determine the number of moles of AgNO3 required.

molLdmmolcVn33 1025.602500.025.0 −−

×=×==

Then, use the formula, m = nM to determine the mass that this many moles corresponds with.

figures)t significan (21.1062.1)16301.1487.107(1025.6 13 ggmolgmolnMm ==×++××==−−

5. A mass and concentration question.

a) What mass of sodium hydroxide pellets is contained in a 1.0 dm3

stock solution with a

concentration of 5.0 mol dm – 3

?

N = CV = 5.0 mol dm-3

x 1.0 dm3 = 5.0 mol

M = nM = 5.0 mol x (22.99+16.00+1.01) g mol-1

= 200 g

b) What is the concentration of the sodium hydroxide in g dm – 3

?

200 g dm-3

6. What is the volume of a stock solution of 2 mol dm – 3

potassium manganate (VII) (also known as

potassium permanganate) required to make up 500 mL of a 0.015 mol dm – 3

solution?

Use: C1V1 = C2V2

2 mol dm – 3

x V1 = 0.015 mol dm – 3

x 0.500 dm3

V1 = 3.75 x 10-3

dm3

of the stock solution is required

7. 100 cm3 of a stock solution of concentrated HCl (aq) (12.0 mol dm

– 3) was diluted with 275 cm

3 of

water. What is the resultant concentration?

Use: C1V1 = C2V2

12.0 mol dm-3

x 0.100 dm3 = C2 x 0.375 dm

3

therefore C2 = 3.2 mol dm-3

8. What would be the concentration of the solution formed by mixing 50 cm3 of 0.15 mol dm

– 3

CH3COOH (aq) with 125 cm3 of 2.0 mol dm

– 3 CH3COOH (aq)? (Hint: you will need to find out the total

number of moles as well as the total volume)

Number of moles of CH3COOH added from the 0.15 mol dm – 3

solution:

n = CV = 0.15 mol x 0.050 dm3 = 7.5 x 10

-3 mol

23

Number of moles of CH3COOH added from the 2.0 mol dm

– 3 solution:

n = CV = 2 mol x 0.125 dm3 = 0.25 mol

Total volume of new solution= 0.05 dm3 + 0.125 dm

3 = 0.175 dm

3

So, new concentration

3

3

3

47.1175.0

25.0105.7

V

−

−

=+×

== dmmoldm

molmolnC

9. Wine has about a 12% alcohol volume/total volume, while beer and premix spirits range from 1% to

8% v/v. If a 375 mL can of beer has label that reads ‘5.5% alcohol v/v’ what is the molarity of alcohol in

the can? (The alcohol is ethanol CH3CH2OH, density 0.785 g cm-3

).

v/v means volume/volume ratio

therefore 5.5 cm3 of alcohol in 100 cm

3 of the beer

therefore ? cm3 of alcohol in 373 cm

3 of the beer

? = 5.5 x 375 /100 = 20.6 cm3

mass of alcohol = 20.6 x 0.785 g = 16.2 g

molar mass alcohol = 24 (2C) + 6 (6H) + O (16) = 46 g

therefore mol alcohol = 16.2/46 = 0.352 mol

molarity = n/V = 0.352/0.375 = 0.939 = 0.94 mol dm-3

(2 sig. fig.)

10. Our blood contains sodium ions. It is important for our health that the concentration of sodium ions

in our blood remains within an acceptable range. Calculate the molarity of sodium ions in the blood

given that each dm3 of human blood contains about 3.4 g of Na

+ ions.

m = 3.4 g in one dm3

In one litre, how many moles?

molgmol

g

M

mn 15.0

99.22

4.31

===−

So, the concentration is 0.15 mol dm-3

11. Seawater also contains many solutes. The three most abundant are chloride Cl-, sodium Na

+ and

magnesium Mg2+

. If 106

dm3

of seawater is found to contain 16600 kg of Cl-, 9200 kg of Na

+ and 1180 kg

of Mg2+

calculate the molarity of each.

V =106 dm

3

Cl –

m = 16600000g

How many moles?

molmolg

g

M

mn 1622.4682625

45.35

166000001

===−

So, concentration can now be calculated.

3-

36dm mol468.0

101

1622.4682625

V===

dmx

molnC

Na+

24

m = 9200000 g

How many moles?

molgmol

g

M

mn 9887.400173

99.22

92000001

===−


3-

36400.0

101

9887.400173

Vdmmol

dmx

molnC ===

Mg2+

m = 1180000 g

How many moles?

molmolg

g

M

mn 7.48539

31.24

11800001

===−


3

360485.0

101

7.48539

V

−=== dmmol

dmx

molnC

12. The amount of ascorbic acid (vitamin C, Mr = 180.18) in a fruit juice is 5.0 x 10 – 4

mol.

a) if the pulp of one kiwifruit was used to make the juice, what was the mass of vitamin C (in mg) in

the kiwi fruit?

m = nM

m = 5.0 x 10-4

mol x 180.18 g mol-1

m = 0.090 g

b) what is the concentration of vitamin C in g dm – 3

if the fruit juice has a volume of 300 mL?

3

33.0

300.0

090.0

V

−=== dmg

dm

gmC

c) how much of the fruit juice must consumed to take in 45 mg of vitamin C?

There are 0.3 g in one dm3 (Litre) of the fruit juice so:

3

315.0

3.0

045.0dm

dmg

g==

−

13. Caffeine (Mr = 194.18) has a concentration of around 1.3 x 10–3

mol dm– 3

in most common energy

drinks.

a) What is the volume of an energy drink given that there is about 80 mg of caffeine per can?

To convert units of concentration of mol dm– 3

to grams dm– 3

, multiply by the molar mass.

1.3 x 10-3

mol dm – 3

x 194.18 g mol-1

= 0.25 g dm– 3

In each drink there is 0.08 g, so the volume of the drink must be: 3

3-32.0

25.0

08.0dm

dmg

g==

b) what is the concentration of caffeine in the energy drink in g dm– 3

?

25

3-

325.0

32.0

08.0

Vdmg

dm

gmC ===

14. 2.0 dm3 of concentrated hydrochloric acid, HCl (12.0 mol dm

- 3) was spilled in a laboratory accident.

Solid sodium bicarbonate (NaHCO3) was used to neutralise the spilled acid.

a) Write a balanced equation for the neutralisation reaction.

HCl + NaHCO3 � NaCl + CO2 + H2O

b) Calculate the minimum mass of sodium bicarbonate needed to completely neutralise the spilled

acid.

The ratio of HCl to NaHCO3 is 1:1, so the same number of moles of NaHCO3 will be needed as that of HCl.

How many moles of HCl were spilled?

n = ?

V = 2 dm - 3

C = 12.0 mol dm- 3

n = CV

n = 2 dm – 3

x 12.0 mol dm– 3

= 24 mol

24 mol of HCl were spilled, so 24 mol of NaHCO3 will be needed to completely neutralise the spilled acid. The

mass of 24 mol of NaHCO3 is:

M = 22.99 + 1.01 + 12.01 + (3x 16) = 84.01 g mol-1

m = nM

m = 24 mol x 84.01 g mol-1

m = 2016.24 g

15. Aspirin, a common analgesic (Mr = 180.17) may be taken for treatment of headaches, fever or

arthritis. Given a typical dose of Aspirin is 5.55 mmol:

a) how many 500 mg tablets constitute a single dose?

The aspirin dose is 5.55 x 10-3

mol. How many grams is this?

m = nM

m = (5.55 x 10-3

mol) x 180.17 g mol-1

m = 1.0 g = 1000 mg

So, the dose is 2 tablets

b) if dissolvable tablets were used, what would be the molarity of a 250 mL glass of water

containing a single dose of aspirin?

3-

3

3

0222.025.0

1055.5

Vdmmol

dm

molnC =

×==

−

16. Oleic acid (Mr = 282.52) is a monounsaturated fatty acid found in olive oil. If 100 cm3 of a typical olive

oil is found to contain 0.25 mol of oleic acid, calculate:

a) the amount of oleic acid in the olive oil in g dm – 3

Concentration in mol/ dm3 :

26

3-

3025.0

100.0

25.0

Vdmmol

dm

molnC ===

C = n/V = 0.25mol/0.100 dm3 = 2.50 mol dm

-3

To get this in g dm-3

, you need to multiply by the molar mass.

= 2.50 mol dm-3

x 282.52 g mol-1

= 706.3 g dm-3

b) the amount in grams of oleic acid in a 20 mL serve of olive oil

n = CV = 706.3 g dm-3

x 0.020 dm3 = 14.13 grams

c) the percentage weight per volume of oleic acid in the olive oil

Using results in b) 14.13/ 20 x 100 = 70.6%

17. Hydrogen peroxide H2O2, is a commonly used bleach. In washing powders the compound sodium

perborate NaBO3.4H2O, produces hydrogen peroxide in solution. Hairdressers use hydrogen peroxide to

change dark hair into blonde. Concentrated hydrogen peroxide is diluted before use. If 10 g of a 50%

w/v solution of hydrogen peroxide is made up to 100 cm3 of solution calculate the molarity of this

solution (% w/v means the percentage weight of solute per total volume of solution).

The original solution is 50% w/v of hydrogen peroxide, so on dilution of a factor of 10 (assuming 10 g

approximates to 10 cm3) it becomes 5% w/v, i.e. 5 g in 100 cm

3.

molar mass H2O2 = 2H + 2O = 2 + 32 = 34 g

therefore mol H2O2 in 100 cm3 = 5/34 mol

therefore mol H2O2 in 1000 cm3 = 5/34 x 10 = 1.47 = 1.5 mol dm

3 (2 sig.fig.)

18. An analysis of a sample of sea water gave the concentration of the bromide ion, Br -, as being 8.44 x

10-4

mol dm-3

. Which of the following calculations would give the theoretical yield of bromine, Br2 that

could be obtained from 1 m3 of sea water?

A 8.44 x 10- 4

x 80 g

B 8.44 x 10- 4

x 80 x 103 g

C 8.44 x 10- 4

x 160 x 103 g

D ½ x 8.44 x 10- 4

x 80 x 103 g

Answer: C (molar mass Br2 = 2 x 80 = 160 g, therefore does C fit? Mass = n x C = 8.44 x 10- 4

x 160 g per dm3.

There are 1000 (103) dm

3 in one m

3, therefore C is the answer.

19. Cane sugar solutions are widely used in soft drinks, sweets, cordial and jam. What is the resulting

molarity of the mixture made from 50 cm3

of 0.12 mol dm -3

cane sugar and 20 cm3 of 0.25 mol dm

-3 cane

sugar?

Number of moles of cane sugar added from the 0.12 mol dm – 3

solution:

n = CV = 0.12 mol x 0.050 dm3 = 6 x 10

-3 mol

Number of moles of cane sugar added from the 0.25 mol dm – 3

solution:

n = CV = 0.25 mol dm-3

x 0.020 dm3 = 5 x 10

-3 mol

Total volume of new solution= 0.05 dm3 + 0.02 dm

3 = 0.07 dm

3

So, new concentration

27

3-

3

33

157.007.0

105106

Vdmmol

dm

molmolnC =

×+×==

−−

20. The concentration of sulphur dioxide, SO2(aq), present in wine can be determined by titrating with a

standard iodine solution. The reaction is:

SO2(aq) + I2(aq) + 2H2O �� SO4(aq) + 2I -(aq) + 4H

+(aq)

One experiment using 100.0 cm3 samples of wine reacted with the 0.010 mol dm

-3 I2 (aq) as follows:

titre 1 = 33.6 cm3

titre 2 = 32.8 cm3

titre 3 = 32.9 cm3

titre 4 = 32.7 cm3

a) what is the average titre?

Average = (32.8 + 32.9 + 32.7)/3 = 32.8 cm3 Titre 1 is not used, the 3 consistent titres are used.

b) what is the mole ratio of SO2 and I2?

1:1

c) how many moles of SO2 are present in the sample of wine?

The number of moles of SO2 in the wine is the same as the number of moles of I2 added.

The number of moles of I2 added was:

n = CV = 0.010 mol dm-3

x 0.0328 dm-3

= 3.3 x 10-4

mol (2 sig fig.)

d) what is the concentration in mol dm-3

of the SO2 present?

3-3

3

4

103.3100.0

103.3

Vdmmol

dm

molnC

−

−

×=×

==

e) what is the concentration in g dm -3

of the SO2 present?

Convert concentration in mol dm-3

to g dm-3

by multiplying by the molar mass of SO2.

M(SO2) = S + 2O = 32.06 + (2x16) = 64.06 g mol-1

So, the concentration is:

3.3 x 10-3

mol dm-3

x 64.06 g mol-1

= 0.211 g dm-3

f) the acceptable levels of SO2 present are a maximum of 0.35 g dm-3

and a minimum of 0.10 g dm

-3. As the wine chemist what would you report to the wine maker about the SO2 concentration

calculated in (e)?

The concentration in the wine is within the acceptable range.

Why chemistry?

Major issues today, such as flu epidemics, nanotechnology, safe and secure water supplies, use of nuclear power, developing renewable energy sources, climate change, health, disease, diet and exercise require an understanding of chemistry. The major challenge for today's scientists is finding a way to communicate chemistry knowledge to our decision-makers so that effective policy to address these issues can be developed.

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