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Q uantitative chemistry
Quantitative chemistry
some fundamental concepts
C
hemistry is a science that deals with the composition, structure and reactions of matter. It is involved with looking at the properties of materials and interpreting these in terms of models on a sub-microscopic scale. Investigations form an important part of any study of chemistry. This involves making observations, and using these in the solution of problems. A typical investigation requires choosing a problem, working out a way of attempting to solve it, and then describing both the method, the results and the manner in which these are interpreted. Namely, “a scientist chooses, imagines, does and describes”. Along with many other syllabuses, practical investigations are a requirement of IB Chemistry.
Matter occupies space and has mass. It can be subdivided into mixtures and pure substances. Mixtures consist of a number of different substances, not chemically combined together. Thus the ratio of these components is not constant from one sample of mixture to another. The different components of a mixture often have different physical properties (such as melting point and density) and chemical properties (such as flammability and acidity). The properties of the mixture are similar to those of the components (e.g. a match burns in both air and pure oxygen), though they will vary with its exact composition. The fact that the different components of the mixture have different physical properties means that the mixture can be separated by physical means, for example by dissolving one component whilst the other remains as a solid. A pure substance cannot be separated in this way because its physical properties are constant throughout all samples of that substance. Similarly all samples of a pure substance have identical chemical properties, for example pure water from any source freezes at 0
o C.
Pure substances may be further subdivided into elements and compounds. The difference between these is that an element cannot be split up into simpler substances by chemical means, whilst a compound can be changed into these more basic components.
The interpretation on a sub-microscopic scale is that all substances are made up of very tiny particles called atoms. Atoms are the smallest particles present in an element which can take part in a chemical change and they cannot be split by ordinary chemical means.
An element is a substance that only contains one type of atom, so it cannot be converted into anything simpler by chemical means. (note; ‘type’ does not imply that all atoms of an element are identical. Some elements are composed of a mixture of closely related atoms called isotopes (refer to Section 2.1). All elements have distinct names and symbols. Atoms can join together by chemical bonds to form compounds. Compounds are therefore made up of particles (of the same type), but these particles are made up of different types of atoms chemically bonded together. This means that in a compound , the constituent elements will be present in fixed proportions such as H 2 O (water), H 2 SO 4 (sulfuric acid), CO 2 carbon dioxide) and NH 3 (ammonia). The only way to separate a compound into its component elements is by a chemical change that breaks some bonds and forms new ones, resulting in new substances. The physical and chemical properties of a compound are usually totally unrelated to those of its component elements. For example a match will not burn in water even though it is a compound of oxygen.
. The mole concept and Avogadro’s constant
.2 Formulas
.3 Chemical equations
.4 Mass and gaseous volume relationships in chemical reactions
.5 Solutions
Chapter 1
Substance Proportions Properties separation
Element
Copper - a pure element
Contains only one type of atom.
These will depend on the forces between the atoms of the element.
Cannot be converted to a simpler substance by chemical means.
Compound
Water - a compound of oxygen and hydrogen
Always contains two hydrogen atoms for every oxygen atom.
Totally different from its elements, e.g. water is a liquid, but hydrogen and oxygen are gases.
Requires a chemical change, e.g. reacting with sodium will produce hydrogen gas.
Mixture
Air - a mixture of nitrogen, oxygen, argon, carbon dioxide etc.
The proportions of the gases in air, especially carbon dioxide and water vapour, can vary.
Similar to its constituents, e.g. supports combustion like oxygen.
Can be carried out by physical means, e.g. by the fractional distillation of liquid air.
Figure 102 The properties of a typical element, compound and mixture
Figure 101 The particles in an element, a compound and a mixture
Element Compound Mixture
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the types of atoms There are 92 kinds of atoms, and hence 92 chemical elements, that occur naturally and about another seventeen that have been produced artificially. Only about thirty of these elements are usually encountered in school chemistry and most of this would deal with about half of these, shown in bold type in Figure 103. Each element is given a symbol that is used to write the formulas of the compounds that it forms. The significance of the atomic number and relative atomic mass of the elements will be explained in Sections 2.1 and 1.3).
Figure 103 shows the common elements and some of their characteristics.
Parts of the names where there are common spelling difficulties have been underlined. You should know the symbols for the elements, especially those in bold type. Most of them are closely related to the name of the element (e.g. chlorine is Cl). Elements that were known in early times have symbols that relate to their Latin names (e.g. Ag, silver, comes from Argentium). Note that the first letter is always an upper case letter and the second one a lower case, so that, for example Co (cobalt) and CO (carbon monoxide) refer to very different substances.
If a substance contains different types of particles, then it is a mixture. These concepts in terms of particles are illustrated in Figure 101. Copper, water and air provide good examples of an element, a compound and a mixture respectively.
The term molecule refers to a small group of atoms joined together by covalent bonds (refer to Section 4.2). If the atoms are of the same kind, then it is a molecule of an element, if different it is a molecule of a compound. Most elements that are gases are diatomic (composed of molecules containing two atoms). Examples are hydrogen gas (H 2 ), nitrogen gas (N 2 ) and oxygen gas (O 2 ). The halogens (F 2 , Cl 2 , Br 2 and I 2 ) are also diatomic in all physical states. The noble gases (He, Ne, Ar, Kr, Xe and Rn) however are monatomic (i.e. exist as single atoms).
The properties of a typical element, compound and mixture are shown in Figure 102.
Chapter 1
Exercise
Figure 105 Avogadro’s number
6.02 (^10) 12g ×^23 (^12) C atoms
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.. Apply the mole concept to substances.
..2 Determine the number of particles and the amount of substance (in moles).
© IBO 2007
1.1 the mole
concePt and
avogadro’s
constant
. A grey solid when heated vapourised to form pure white crystals on the cooler parts of the test tube leaving a black solid as the residue. It is likely that the original solid was
A an element. B a metal. C a pure compound. D a mixture.
- Which one of the following is a chemical property rather than a physical property?
A Boiling point B Density C Flammability D Hardness
- Which one of the following is a physical change rather than a chemical change?
A Combustion B Distillation C Decomposition D Neutralisation
- State whether the sketches below represent elements, compounds or mixtures.
A B C D E
- State whether the following refer to an element, a compound or a mixture:
(a) Easily separated into two substances by distillation. (b) Its components are always present in the same proportions. (c) Its properties are similar to those of its components. (d) Cannot be broken down by chemical means. (e) Very different properties to its components.
Atoms and molecules are inconceivably minute, with equally small masses. The masses of all atoms are not however the same and it is often convenient to be able to weigh out amounts of substances that contain the same number of atoms or molecules. The same amount of any substance will therefore contain the same number of particles and we measure the amount of substance in moles. It is for this reason that the mole concept is important.
Amount of substance , n (the number of moles), is a quantity that is proportional to the number of particles in a sample of substance: its units are moles (mol). A mole of a substance contains 6.02 × 10 23 particles of the substance (this very large number is expressed in scientific notation; for further explanation of scientific notation refer to Appendix 1A). This is the same number of particles as there are atoms in exactly 12 g of the C-12 ( 126 C) isotope.
Since a mole of carbon atoms weighs 12 g, an atom of carbon (C) weighs only: 1.995×10−23^ g atom−1.
The value 6.02 × 10 23 mol –1^ is called the Avogadro’s Constant (L or NA). The particles may be atoms, molecules, ions, formula units, etc., but should be specified. For example, 1 mol of carbon contains 6.02 × 10 23 carbon (C) atoms (and weighs 12 g), where as 1 mol of water, H 2 O, contains 6.02 × 10 23 H 2 O molecules or 3 × 6.02 × 10 23 atoms since each water molecule contains a total of 3 atoms.
Number of moles = _________________Number of particles 6.02 × 1023
Q uantitative chemistry
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TOK The scale of chemistry
‘Chemistry deals with enormous differences in scale. The magnitude of Avogadro’s constant is beyond the scale of our everyday experience’. © IBO 2007 Dealing with very large and very small scales is almost impossible for the human brain - maybe it would disconcert our egos to be continually aware that we are infinitesimal dots living on an infinitesimal dot! So how can we try to come to terms with the enormous and minute numbers we meet in science? One approach is to try to apply the scale to things we are familiar with. For example if sand were made up of cubic grains / 0 th^ mm on each side, then what length of a beach 60 m deep and km wide would mole (i.e. 6 × 023 ) of these form? The answer is about 0,000 km of it!!! Coming from the other end, suppose the population of the whole world (let’s say 4000 million people) were turned into atoms of gold (quite big, heavy atoms) our combined mass would be about one-millionth of a microgram, too small for any balance to detect (our grains of sand above would weigh about 2 μg) and we would form a cube with a side of about / 50 μm, a factor of ten below the theoretical range of the best optical microscope. Hence we would not be able to detect ourselves! Fortunately we have mathematics to rely on, so we do not have to depend on being able to imagine these scales!
This can be written as n = ________ 6.02 ×N 1023 or N = n × 6.02 × 1023.
A sample of water that contains 3.01 × 1025 water
molecules therefore contains 3.01 ×^10
__________^25
6.02 × 1023
= 50 moles of
water molecules.
This is very similar to saying that 36 oranges is equivalent
to 3 dozen oranges (___^3612 = 3).
The formula may be rearranged to calculate the number of particles. For example 0.020 moles of carbon dioxide will contain 0.020 × 6.02 × 1023 = 1.2 × 1022 molecules of CO 2.
This number of molecules of carbon dioxide will contain 1.2 × 1022 atoms of carbon but 2.4 × 1022 (i.e. 2 × 1.2 × 1022 ) atoms of oxygen, because each molecule contains one atom of carbon, but two atoms of oxygen and the total number of atoms is 3.6 × 1022 (i.e. 3 × 1.2 × 1022 ). Note, therefore, how important it is to state what particles are being referred to.
Exercise .
(Take the value of Avogadro’s constant as 6.02 × 1023 mol–1.)
. Calculate how many atoms are there in 5 moles of sulfur atoms.
A 1.20 × 10^23 B 6.02 × 10^23 C 6.02 × 10^115 D 3.01 × 10^24
- Which one of the following is not the same number as the rest?
A The number of molecules in 4 moles of CO 2. B The number of hydrogen atoms in 2 moles of H 2 O. C The number of chloride ions in 4 moles of CaCl 2. D The number of hydrogen atoms in 1 mole of C 3 H 8.
- The number of atoms present in 0.10 mol of ethene (C 2 H 4 ) is:
A 3.61 × 10^22 B 6.02 × 10^22 C 3.61 × 10^23 D 6.02 × 10^23
- One mole of water contains
A 6.02 × 10^23 atoms of hydrogen. B 2.01 × 10^23 atoms of oxygen. C 6.02 × 10^23 atoms in total. D 6.02 × 10^23 molecules of water.
- The number of atoms present in 36 molecules of glucose (C 6 H 12 O 6 ) is
A 24 B 36 C 24 × 36 D 24 × 36 × 6.02 × 10^23
- The mass of one atom of carbon -12 is:
A 1 g. B 12 g. C 12 × 6.02 × 10^23 g. D 12 / 6.02 × 10^23 g.
Q uantitative chemistry
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Exercise .
. The molar mass of iron(III) sulfate Fe 2 (SO 4 ) 3 will be
A 191.76 g mol– B 207.76 g mol– C 344.03 g mol– D 399.88 g mol–
- A certain substance has a molar mass (to 2 significant figures) of 28 g mol –1. Which of the following is not a possible formula?
A CH 2 O B Si C C 2 H 4 D CO
- Calculate the molar mass of the following substances (correct to 1 decimal place).
(a) HI (b) NaClO 3 (c) (NH 4 ) 2 HPO 4 (d) (CO 2 H) 2 .2H 2 O (e) Chromium(III) oxide (f) Iodine trichloride
moles and mass
.2.3 Solve problems involving the relationship between the amount of substance in moles, mass and molar mass.
© IBO 2007
It follows from the above that the mass of one mole of any substance will be equal to its molar mass in grams. If one mole of fluorine atoms has a mass of 19 g, then 76 g of fluorine atoms will contain four moles of fluorine atoms, hence the amount of substance may be calculated from its molar mass using the formula
Amount of substance,
n = __________________Mass (g) Molar Mass (g mol–1)
n = __m M
4.904 g of sulfuric acid will therefore contain:
__^ m M =^
___________4.904 g 98.08 g mol– = 0.05000 moles of sulfuric acid
Exercise .2.
[Note - It is important to pay attention to significant figures in calculations. A short description of the scientific notation and significant figures is given later in the chapter.]
The equation may be rearranged to calculate the molar mass from the mass and the amount of substance, or to find the mass from the amount and the molar mass. For example the mass of 3.00 moles of carbon dioxide is
n × M = 3.00 mol × 44.01 g mol- = 132 g of carbon dioxide
Similarly if 0.200 moles of a substance has a mass of 27. g, then its molar mass (M) will be
M = __m n = _________27.8 g 0.200 mol
= 139 g mol–
Knowing the mass of a given number of atoms or molecules means that the mass of one atom or molecule can be calculated. For example as the molar mass of the hydrogen atom is 1.01 g mol–1, 6.02 × 10^23 atoms of hydrogen have a mass of 1.01 g, hence the mass of a single atom is:
___________________^ 1.01 g^ mol– 6.02 × 1023 atoms mol– = 1.68 × 10 –24^ g atom–
Similarly the mass of one molecule of glucose (C 6 H 12 O 6 , M = 180.18 g mol–1) is:
______________________^ 180.18 g^ mol– 6.02 × 1023 molecules mol– = 2.99 × 10 –22^ g molecule–
. Determine the mass of 0.700 moles of Li 2 SO 4 taking its molar mass as exactly 110 g mol -1.
A 15.4 g B 77 g C 110 g D 157 g
- 0.200 moles of a substance has a mass of 27.0 g. Determine the molar mass of the substance.
A 13.5 g mol- B 27 g mol- C 54 g mol- D 135 g mol-
Chapter 1
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Knowing the formula of a substance and the molar masses of the elements, then the percentage composition may be found by calculating the proportion by mass of each element and converting it to a percentage. For example the molar mass of carbon dioxide is
12.01 + (2 × 16.00) = 44.01 g mol-
Oxygen constitutes 32.00 g mol-1^ of this (2 × 16.00), so that the percentage of oxygen by mass in carbon dioxide is:
32.00_____ 44.01 × 100 = 72.71% oxygen by mass.
The empirical formula, sometimes called the simplest formula, of a compound indicates:
the elements present in the compound the simplest whole number ratio of these elements
.2.4 Distinguish between the terms empirical formula and molecular formula.
.2.5 Determine the empirical formula from the percentage composition or from other experimental data.
.2.6 Determine the molecular formula when given the empirical formula and experimental data.
© IBO 2007
percentage composition and
empirical formula
- One drop of water weighs 0.040 g. Calculate how many molecules are there in one drop, taking the molar mass of water as exactly 18 g þmol-1.
A 1.3 × 10^21 B 2.4 × 10^22 C 3.3 × 10^22 D 3.9 × 10^22
- Determine the mass (in g) of one molecule of sulfuric acid (H 2 SO 4 ).
A 98. B 98.08 ÷ (6.02 × 10^23 ) C 98.08 ÷ 7 D 98.08 ÷ (7 × 6.02 × 10^23 )
- A polymer molecule has a mass of 2.5 × 10-20^ g. Determine the molar mass of the polymer.
A 1.4 × 10^4 g mol- B 2.4 × 10^43 g mol- C 6.7 × 10-5^ g mol- D 4.2 × 10-44^ g mol-
- Calculate (correct to 3 significant figures) the mass of
(a) 3.00 moles of ammonia. (b) ¼ mole of Li 2 O. (c) 0.0500 moles of aluminium nitrate. (d) 3.01 × 10^23 molecules of PCl 3. (e) 2.60 × 10^22 molecules of dinitrogen monoxide.
- Calculate how many moles are there in
(a) 28.1 g of silicon. (b) 303 g of KNO 3. (c) 4000 g of nickel sulfate. (d) 87.3 g of methane.
- a) 0.30 moles of a substance has a mass of 45 g. Determine its molar mass.
b) 3.01 × 10^25 molecules of a gas has a mass of 6.40 kg. Determine its molar mass.
- Some types of freon are used as the propellant in spray cans of paint, hair spray, and other consumer products. However, the use of freons is being curtailed, because there is some suspicion that they may cause environmental damage. If there are 25.00 g of the freon CCl 2 F 2 in a spray can, calculate
how many molecules are you releasing to the air when you empty the can.
0. Vitamin C, ascorbic acid, has the formula C 6 H 8 O 6.
a) The recommended daily dose of vitamin C is 60.0 milligrams. Calculate how many moles are you consuming if you ingest 60 milligrams of the vitamin. b) A typical tablet contains 1.00 g of vitamin C. Calculate how many moles of vitamin C does this represents. c) When you consume 1.00 g of vitamin C, calculate how many oxygen atoms are you eating.
Chapter 1
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2.80 g of an organic compound, containing only carbon and hydrogen forms 8.80 g of carbon dioxide and 3.60 g of water when it undergoes complete combustion. Determine its empirical formula.
Amount of CO 2 = amount of C
Amount of C = __ Mm
= __________8.80 g 44.0 g mol– = 0.200 mol
Amount of H 2 O = ½ amount of H
Amount of H 2 O = m__ M
= __________3.60 g 18.0 g mol– = 0.200 mol
Therefore amount of H = 2 × 0. = 0.400 mol
Ratio C : H = 0.200 : 0.400 = 1 : 2; Therefore the empirical formula is CH 2.
Example
Consider vitamin C, a compound that contains carbon, hydrogen and oxygen only. On combustion of 1.00 g vitamin C, 1.50 g CO 2 and 0.408 g H 2 O are produced. Determine the empirical formula of vitamin C.
All the carbon in the CO 2 came from the vitamin C. The mass of carbon in 1.50 g CO 2 can be easily calculated.
Amount of CO 2 = __ Mm
= __________1.50 g 44.0 g mol– = 0.03408 mol
Since each CO 2 contains 1 C
mC = nC × A(C) = 0.0340(8) mol × 12.01 g mol– = 0.409 g C
1.50 g CO 2 contains 0.409 g of C all of which came from vitamin C.
Similarly, all the H in the H 2 O is from the vitamin C:
18.02 g H 2 O contains 2.02 g H ∴ 0.408 g H 2 O contains 0.408 g × _____2.
= 0.0457 g of hydrogen
The rest must therefore be oxygen: ∴ Mass of oxygen = (1.00 – 0.409 – 0.0457) g = 0.54(5) g
Once the proportion by mass of the elements is known, the mole ratios can be calculated: C H O
m (g): 0.409 0.0457 0.54(5) Ar: 12.01 1.01 16. n (mol): 0.0341 0.0452 0.
divide by smallest number:
1.00 1.33 1. 1 : 1⅓ : 1 3 : 4 : 3 Therefore the empirical formula of vitamin C is C 3 H 4 O 3
eXPerimental methods Empirical formulas can often be found by direct determination, for example converting a weighed sample of one element to the compound and then weighing the compound to find the mass of the second element that combined with the first (see exercise 1.5, Q 13). Another method is to decompose a weighed sample of a compound containing only two elements, so that only one element remains and then finding the mass of that element. This second method is similar to the one that is usually used to determine the formula of a hydrated salt (see exercise 1.5, Q 14). There are also many other methods for determining percentage composition data, too numerous to mention.
The percentage composition of organic compounds is usually found by burning a known mass of the compound in excess oxygen, then finding the masses of both carbon dioxide (all the carbon is converted to carbon dioxide) and water (all the hydrogen is converted to water) formed. The mass of oxygen can be found by subtracting the mass of these two elements from the initial mass, assuming that this is the only other element present.
Solution
Example 2
Solution
Q uantitative chemistry
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Example
A hydrocarbon contains 92.24% by mass of carbon and its Mr = 78.1. Determine its molecular formula.
molecular formula
The molecular formula of a compound indicates:
the elements present in the compound the actual number of atoms of these elements in one molecule
Hence the molar mass can be calculated from the molecular formula, as can the percentage composition by mass and the empirical formula. Note that the molecular formula of a compound is a multiple of its empirical formula, e.g., for butane, its molecular formula is C 4 H 10 and its empirical formula is C 2 H 5. Other examples are shown in Figure 106. It therefore follows the relative molecular divided by the relative empirical mass is a whole number.
n.b.The relative empirical mass is the sum of the relative atomic masses of the atoms in the empirical (simplest) formula.
The relative molecular mass, Mr, for a compound can be determined experimentally using mass spectrometry
(refer to Section 2.2), or from physical properties such as the density of the substance in the gas phase (refer to Section 1.4) making it possible to determine the molecular formula of the compound provided the empirical formula is known.
It can be seen from Figure 106 that ethyne and benzene both have the same empirical formula (CH), but the relative molar mass of ethyne is ≈ 26, whereas that of benzene is ≈ 78, so the molecular formula of ethyne must be C 2 H 2 (2 × CH) whereas that of benzene is C 6 H 6.
In the case of the compound with the empirical formula P 2 O 5 , the molecular formula could actually be P 2 O 5 or it could be P 4 O 10 , P 6 O 15 , P 8 O 20 etc. The relative empirical mass of P 2 O 5 is 141..
In order to know which molecular formula is correct, it is necessary to have information about the approximate molar mass of the substance. The molar mass of this oxide of phosphorus is found to be ≈ 280 g mol -1. It must therefore contain two P 2 O 5 units, hence the molecular formula is P 4 O 10.
C H
m (g): 92.24 (100-92.24) = 7.
Mr: 12.01 1.
n (mol): 1 : 1
Simplest formula is CH, and its relative empirical mass = 12.01 + 1.01 = 13.
Since Mr = 78.1; molecular formula is C 6 H 6.
Solution
Compound Percentage composition
Empirical Formula Mr
Molecular Formula
Ethane %H = 6 ×^ 1.01⁄30.08 = 20.1%;^ %C = 2 ×^ 12.01⁄30.08 = 79.9%^ CH 3 30.08^ C 2 H 6
Hexene %H = 12 ×^ 1.01⁄84.18 = 14.4%;^ %C = 6 ×^ 12.01⁄84.18 = 85.6%^ CH 2 84.18^ C 6 H 12
Benzene %H = 6 × 1.01⁄78.12 = 7.8%; %C = 6 × 12.01⁄78.12 = 92.2% CH 78.12 C 6 H 6
Ethyne %H = 2 ×^ 1.01⁄26.04 = 7.8%;^ %C = 2 ×^ 12.01⁄26.04 = 92.2%^ CH^ 26.04^ C^2 H 2
Figure 106 The percentage composition and empirical formulas of some hydrocarbons
Q uantitative chemistry
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- The empirical formula of a compound with the molecular formula C 6 H 12 O 3 is
A C 6 H 12 O 3 B C 3 H 6 O 2 C C 2 H 3 O D C 2 H 4 O
0. What percentage of ‘chrome alum’ [KCr(SO 4 )2•12H 2 O; M = 499.4 g mol-1] is water.
A 18.01_____ 499.4 × 100
B 12 × 18.01_________ 499.4 × 100
C ___________ 499.4 – 18.0118.01 × 100
D _________________12 × 18. 499.4 – (12 × 18.01)
× 100
. What percentage by mass of sodium thiosulfate pentahydrate (Na 2 S 2 O 3 •5H 2 O) is water.
2. a) 2.0 g of an oxide of iron contains approximately 0.60 g oxygen and 1.4 g iron. Determine the empirical formula of the oxide.
b) A compound of silicon and fluorine contains about 73% fluorine by mass. Determine its empirical formula. c) A compound of carbon, hydrogen and oxygen only, with a molar mass of ≈90 g mol- contains 26.6% carbon and 2.2% hydrogen by mass. Determine its molecular formula.
3. 1.000 g of tin metal burns in air to give 1.270 g of tin oxide. Determine the empirical formula of the oxide.
4. A 1.39 g sample of hydrated copper(II) sulfate (CuSO 4 •xH 2 O) is heated until all the water of hydration is driven off. The anhydrous salt has a mass of 0.89 g. Determine the formula of the hydrate.
5. The red colour of blood is due to haemoglobin. It contains 0.335% by mass of iron. Four atoms of iron are present in each molecule of haemoglobin. If the molar mass of iron is 55.84 g mol–1, estimate the molar mass of haemoglobin.
6. A 200.0 mg sample of a compound containing potassium, chromium, and oxygen was analyzed and found to contain 70.8 mg chromium and 53.2 mg potassium. Calculate the empirical formula of the sample.
7. The molecular formula of the insecticide DDT is C 14 H 9 Cl 5. Calculate the molar mass of the compound and the percent by mass of each element.
8. The percentages of carbon, hydrogen, and oxygen in vitamin C are determined by burning a sample of vitamin C weighing 1.000 g. The masses of CO 2 and H 2 O formed are 1.500 g and 0.408 g, respectively.
a) Calculate the masses and amounts of carbon and hydrogen in the sample. b) Determine the amount of oxygen in the sample. c) From the above data, determine the empirical formula of vitamin C.
9. The percentages by mass of carbon, hydrogen and nitrogen in an unknown compound are found to be 23.30%, 4.85%, and 40.78%, respectively. (Why do. these not add up to 100%?). Determine the empirical formula of the compound. If the molar mass of the compound is 206 g mol -1, determine its molecular formula.
- Efflorescence is the process by which some hydrated salts lose water of crystallisation when exposed to the air. ‘Washing soda’ (Na 2 CO 3 •10H 2 O) is converted to the monohydrate (Na 2 CO 3 • H 2 O) when exposed to the air. Determine the percentage loss in mass of the crystals.
Chapter 1
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Exercise .
. The formula of the cadmium ion is Cd 2+^ and that of the benzoate ion is C 6 H 5 COO-. Determine the formula of cadmium benzoate.
A Cd(C 6 H 5 COO) 2 B CdC 6 H 5 COO C Cd 2 (C 6 H 5 COO) 2 D Cd 2 C 6 H 5 COO
- An ore contains calcium hydroxide, Ca(OH) 2 in association with calcium phosphate, Ca 3 (PO 4 ) 2. Analysis shows that calcium and phosphorus are present in a mole ratio of 5:3. Which of the following best represents the composition of the ore?
A Ca(OH 2 ) · Ca 3 (PO 4 ) 2 B Ca(OH) 2 · 2 Ca 3 (PO 4 ) 2 C Ca(OH) 2 · 3 Ca 3 (PO 4 ) 2 D Ca(OH) 2 · 4 Ca 3 (PO 4 ) 2
- If the formula of praseodymium oxide is PrO 2 , Determine the formula of praseodymium sulfate.
A Pr 2 SO 4 B PrSO 4 C Pr 2 (SO 4 ) 3 D Pr(SO 4 ) 2
chemical formulas
Chemical formulas are a shorthand notation for elements, ions and compounds. They show the ratio of the number of atoms of each element present and, in the case of molecules or ions held together by covalent bonds, it gives the actual number of atoms of each element present in the molecule or ion. For example, the formula for magnesium chloride, which is ionically bonded, is MgCl 2. This tells us that in magnesium chloride there are twice as many chloride ions as there are magnesium ions. The formulas of ionic compounds can be deduced from the electrical charges of the ions involved (refer to Section 4.1).The formula for glucose, which is a molecular covalent compound, is C 6 H 12 O 6. This tells us that a molecule of glucose contains six carbon atoms, twelve hydrogen atoms and six oxygen atoms. The formulas of covalent compounds have to be memorised or deduced from their names.
The carbonate ion, which is a covalently bonded ion, has the formula CO 3 2-. This tells us that the carbonate ion consists of a carbon atom bonded to three oxygen atoms, that has also gained two electrons (hence the charge). Brackets are used to show that the subscript affects a group of atoms, for example the formula of magnesium nitrate is Mg(NO 3 ) 2 , showing that there are two nitrate ions (NO 3 - ) for every magnesium ion (Mg2+). Sometimes brackets are also used to indicate the structure of the compound, for example urea is usually written as (NH 2 ) 2 CO rather than CN 2 H 4 O, to show that it consists of a carbon joined to two -NH (^2) groups and an oxygen. The ending of the names of ions often indicate their composition. For example the ending –ide usually indicates just the element with an appropriate negative charge (e.g. sulfide is S 2-). The ending ‑ate usually indicates the the ion contains the element and oxygen atoms (e.g. sulfate is SO 4 2-). The ending –ite, also indicates an ion containing oxygen, but less oxygen than the –ate (e.g. sulfite is SO 3 2-).
Sometimes compounds are hydrated, that is they contain water molecules chemically bonded into the structure of the crystals. This is known as water of crystallisation or hydration and it is indicated by the formula for water following the formula of the substance and separated from it by a dot. For example in hydrated sodium sulfate crystals seven molecules of water of crystallisation are present for every sulfate ion and every two sodium ions, so its formula is written as Na 2 SO 4 · 7H 2 O. When the crystals are heated this water is frequently given off to leave the anhydrous salt (Na 2 SO 4 ). Similarly blue hydrated copper(II) sulfate crystals (CuSO 4 ·5H 2 O) forms white anhydrous copper(II) sulfate (CuSO 4 ) when strongly heated.
.3. Deduce chemical equations when all reactants and products are given.
.3.2 Identify the mole ratio of any two species in a chemical equation.
.3.3 Apply the state symbols (s), (l), (g) and (aq).
© IBO 2007
1.3 chemical eQuations
Chapter 1
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produces calcium chloride, carbon dioxide and water can be represented as
Calcium carbonate + Hydrochloric acid Calcium chloride + Carbon dioxide + Water
The next stage is to replace the names of the compounds with their formulas, so that this equation becomes:
CaCO 3 + HCl CaCl 2 + CO 2 + H 2 O
Finally, because matter cannot be created or destroyed (at least in chemical reactions) and the charge of the products must be equal to that of the reactants, the equation must be balanced with respect to both number of atoms of each element and the charge by placing coefficients, also called stoichiometric coefficients, in front of some of the formulas. These multiply the number of atoms of the elements in the formula by that factor and represent the number of moles of the species required. In the example above, there are two chlorines on the right hand side, but only one on the left hand side. Similarly the hydrogen atoms do not balance. This can be corrected by putting a ‘2’ in front of the hydrochloric acid, so the final balanced equation is
CaCO 3 + 2 HCl CaCl 2 + CO 2 + H 2 O
This means that one formula unit of calcium carbonate will just react completely with two formula units of hydrochloric acid to produce one formula unit of calcium chloride, one molecule of carbon dioxide and one molecule of water. Scaling this up means that one mole of calcium carbonate reacts with two moles of hydrochloric acid to produce one mole of calcium chloride, one mole of carbon dioxide and one mole of water. The amounts of substances in a balanced equation are known as the stoichiometry of the reaction, hence these equations are sometimes referred to as stoichiometric equations. One corollary of balancing chemical equations is that as a result mass is conserved.
Note that the formulas of compounds can never be changed, so balancing the equation by altering the subscripts, for example changing calcium chloride to CaCl 3 or water to H 3 O, is incorrect.
It is sometimes helpful to show the physical state of the substances involved and this can be done by a suffix, known as a state symbol placed after the formula. The state symbols used are; (s) - solid, (l) - liquid, (g) - gas and (aq) - aqueous solution. State symbols should be used as a matter of course as it gives more information about a reaction and in some cases, such as when studying thermochemistry, their use is vital. Adding these, the
equation for the reaction between calcium carbonate and hydrochloric acid becomes
CaCO 3 (s) + 2 HCl (aq) CaCl 2 (aq) + CO 2 (g) + H 2 O (l)
It is often better to write the equation for a reaction occurring in aqueous solution as an ionic equation. This is particularly true for precipitation reactions, acid-base reactions and redox reactions. An example would be the reaction between aqueous lead nitrate and aqueous sodium chloride to precipitate lead chloride and leave a solution of sodium nitrate.
Pb(NO 3 ) 2 (aq) + 2 NaCl(aq) PbCl 2 (s) + 2 NaNO 3 (aq)
When soluble ionic compounds, as well as strong acids and bases, dissolve in water they totally dissociate into their component ions and so the equation above would be more correctly written as:
Pb2+^ (aq) + 2 NO 3 −^ (aq) + 2 Na+^ (aq) + 2 Cl−^ (aq) PbCl 2 (s) + 2 Na+^ (aq) + 2 NO 3 −^ (aq)
This shows that the reaction actually involves just the lead ions and chloride ions. The hydrated nitrate ions and sodium ions are present in both the reactants and products and so do not take part in the reaction. They are known as spectator ions. The spectator ions can therefore be cancelled from both sides so that the net ionic equation becomes:
Pb2+(aq) + 2 Cl−(aq) PbCl 2 (s)
Ionic equations are far more general than normal equations. This ionic equation, for example, states that any soluble lead compound will react with any soluble chloride to form a precipitate of lead chloride. For example the reaction
Pb(CH 3 COO) 2 (aq) + MgCl 2 (aq) PbCl 2 (s) + Mg(CH 3 COO) 2 (aq)
would have exactly the same ionic equation:
Pb2+^ (aq) + 2 Cl−^ (aq) PbCl 2 (s)
Q uantitative chemistry
core
Exercise .3.
. Which one of the following equations best represents the reaction between iron and hydrochloric acid?
A Fe + HCl FeCl + H B Fe + HCl FeCl + H 2 C Fe + 2 HCl FeCl 2 + H 2 D Fe + 2 HCl FeCl 2 + 2 H
- What numerical value of Q is required to balance the equation below?
2 H 2 S + Q O 2 2 SO 2 + 2 H 2 O
A 2 B 3 C 4 D 6
- The equation for the reaction of sodium sulfate with barium nitrate to form a precipitate of barium sulfate is
Ba(NO 3 ) 2 + Na 2 SO 4 BaSO 4 + 2 NaNO 3
Which one of the following is the correct ionic equation for this reaction?
A Ba2+^ + SO 4 2-^ BaSO 4 B Na+^ + NO 3 -^ NaNO 3 C Ba2+^ + Na 2 SO 4 BaSO 4 + 2 Na+ D Ba(NO 3 ) 2 + SO 4 2-^ BaSO 4 + 2 NO 3 -
In order to know which salts are soluble there are certain simple rules that it is useful to remember:
Always soluble – salts of Na+, K+, NH 4 +^ and NO 3 -
Usually soluble - salts of Cl -^ and SO 4 2-, but AgCl, PbCl 2 , PbSO 4 and BaSO 4 are insoluble
Usually insoluble - salts of OH - , O 2-, CO 3 2-^ and PO 4 3-, but Na+, K+, NH 4 +^ salts soluble
Common slightly soluble substances – Ca(OH) 2 and Ca SO 4
- Insert coefficients, to balance the following equations.
(a) CaO + HNO 3 Ca(NO 3 ) 2 + H 2 O (b) NH 3 + H 2 SO 4 (NH 4 ) 2 SO 4 (c) HCl + ZnCO 3 ZnCl 2 + H 2 O + CO 2 (d) SO 2 + Mg S + MgO (e) Fe 3 O 4 + H 2 Fe + H 2 O (f) K + C 2 H 5 OH KC 2 H 5 O + H 2 (g) Fe(OH) 3 Fe 2 O 3 + H 2 O (h) CH 3 CO 2 H + O 2 CO 2 + H 2 O (i) Pb(NO 3 ) 2 PbO + NO 2 + O 2 (j) NaMnO 4 + HCl NaCl + MnCl 2 + Cl 2 + H 2 O
- Write balanced equations for the following reactions.
(a) Copper(II) carbonate forming copper (II) oxide and carbon dioxide. (b) Nickel oxide reacting with sulfuric acid to form nickel sulfate and water. (c) Iron and bromine reacting to give iron(III) bromide. (d) Lead(IV) oxide and carbon monoxide forming lead metal and carbon dioxide. (e) Iron(II) chloride reacting with chlorine to form iron(III) chloride. (f) Ethanol burning in air to form carbon dioxide and water. (g) Silver reacting with nitric acid to form silver nitrate, nitrogen dioxide and water. (h) Manganese(IV) oxide reacting with hydrochloric acid to form manganese(II) chloride, chlorine and water. (i) Sulfur dioxide reacting with hydrogen sulfide to form sulfur and water. (j) Ammonia reacting with oxygen to form nitrogen monoxide and water.
Q uantitative chemistry
core
What mass of sodium hydrogencarbonate must be heated to give 8.80 g of carbon dioxide?
Stage One
Amount of CO 2 = ___________8.80 g 44.01 g mol– = 0.200 mol
Stage Two
2 NaHCO 3 Na 2 CO 3 + CO 2 + H 2 O
2 mol 1 mol
2 × 0.200 = 0.400 mol 0.200 mol
Stage Three
Mass of NaHCO 3 = 0.400 mol × 84.01 g mol-
= 33.6 g
Calculate the mass of O 2 required for the combustion of 0.250 mol propane gas, C 3 H8 (g).
Stage One
Aleady completed; we are given the moles of propane (0.250)
Stage Two
C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (l) 1 mol C 3 H 8 5 mol O 2
0.250 mol C 3 H 8 0.25 × 5 = 1.25 mol O 2
Stage Three
Mass of O 2 required = 1.25 mol × 32.0 g mol−^1 = 40.0 g
. When butane is burnt in excess air, the following reaction takes place
2 C 4 H10 (g) + 13 O2 (g) 8 CO2 (g) + 10 H 2 O(l)
Calculate how many moles of oxygen are required to react with five moles of butane.
A 6. B 13 C 32. D 65
- When magnesium is added to aqueous silver nitrate, the following reaction takes place
Mg(s) + 2 AgNO3 (aq) 2 Ag(s) + Mg(NO 3 )2 (aq)
What mass of silver is formed when 2.43 g of magnesium is added to an excess of aqueous silver nitrate.
A 107.9 g B 21.6 g C 10.8 g D 5.4 g
- When heated potassium chlorate(V) decomposes to form potassium chloride and oxygen. The unbalanced equation is:
KClO3 (s) KCl (^) (s) + O2 (g)
(a) Balance the equation. (b) Calculate how many moles of KClO 3 are needed to produce 0.60 moles of oxygen. (c) What mass of KClO 3 is needed to produce 0.0200 moles of KCl?
Exercise .
in summary Calculate the amount of the substance whose mass is given. Use the balanced equation to calculate the amount of the required substance. Calculate the mass of the required substance from the amount of it.
Example 2
Solution
Example 3
Solution
Chapter 1
core
Consider the reaction:
H 2 O2 (aq) + 2KI(aq) + H 2 SO4 (aq) I2 (s) + K 2 SO4 (aq) + 2H 2 O(l)
What mass of iodine is produced when 100.00 g of KI is added to a solution containing 12.00 g of H 2 O 2 and 50.00 g H 2 SO 4?
The mole ratio from the equation is
H 2 O 2 : KI : H 2 SO 4 1 : 2 : 1
The actual mole ratio of reagents present is
12.00_____
______100.
50.00_____
Ratio divided by coefficient = 1 : 0.854 : 1.
It can be seen that even though there is the greatest mass of potassium iodide, it is still the limiting reagent, owing to its large molar mass and the 2:1 mole ratio. The maximum yield of iodine will therefore be ½ × 0.6024 = 0.3012 moles. This is the theoretical yield, which can, if required, be converted into a mass:
Theoretical yield = n × M = 0.3012 mol × 253.8 g mol- = 76.44 g
There is an excess of both hydrogen peroxide and sulfuric acid. The amounts in excess can be calculated:
0.6024 moles of potassium iodide will react with:
½ × 0.6024 = 0.3012 moles of both H 2 O 2 and H 2 SO 4 (both a 2:1 mole ratio).
Mass of H 2 O 2 reacting = 0.3012 mol × 34.02 g mol - = 10.24 g,
Mass of H 2 SO 4 reacting = 0.3012 × 98. = 29.54 g,
Example
limiting reagents The quantities of reactants related in the section above are the precise amounts, or stoichiometric amounts, required to just react with each other. More commonly there will be an excess of all of the reagents except one, so that all of this last reagent will be consumed. This reagent is known as the limiting reagent because it is the amount of this that limits the quantity of product formed. It may be identified by calculating the amount of each reagent present and then dividing by the relevant coefficient from the equation. The reagent corresponding to the smallest number is the limiting reagent.
- What mass of copper(II) sulfate pentahydrate (CuSO 4 · 5H 2 O) can be produced by reacting 12.00 g of copper(II) oxide with an excess of sulfuric acid?
- Pure compound A contains 63.3% manganese and 36.7% oxygen by mass. Upon heating compound A, oxygen is evolved and pure compound B is formed which contains 72.0% manganese and 28.0% oxygen by mass.
(a) Determine the empirical formula for compounds A and B. (b) Write a balanced equation which represents the reaction that took place. (c) Calculate how many grams of oxygen would be evolved when 2.876 g of A is heated to form pure B.
Solution
