Chapter 01 - Quantitative Chemistry.pdf 16473r

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Chapter 01 – Quantitative Chemistry

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Chapter 01 – Quantitative Chemistry Topic 01 from the IB HL Chemistry Curriculum

1.1

The Mole and Avogadro’s constant (2 Hours)

1.1.1

Assessment Statement Apply the mole concept to substances

1.1.2

Determine the number of particles and the amount of substance (in moles).

Obj 2

3

Teacher’s Notes The mole concept applies to all kinds of particles: atoms, molecules, ions, electrons, formula units, and so on. The amount of substance is measured in moles (mol). The approximate value of Avogadro’s 23 −1 constant (L), 6.02 × 10 mol , should be known. Convert between the amount of substance (in moles) and the number of atoms, molecules, ions, electrons and formula units.

Measurement and units Scientists search for order in their observations of the world. Measurement is a vital tool in this search. It makes our observations more objective and helps us find relationships between different properties. The standardization of measurement of mass and length began thousands of years ago when kings and emperors used units of length based on the length of their arms or feet. Because modern science is an international endeavor, a more reliable system of standards, the Systeme International is needed. A lot of experimental chemistry relies on the accurate measurement and recording of the physical quantities of mass, time, temperature, volume and pressure. The SI units for these are given below. Property

Unit

Symbol for Unit

mass time temperature volume pressure

kilogram second kelvin cubic meter pascal

kg s K m3 Pa or N∙m−2

These units are, however, not always convenient for the quantities typically used in the laboratory. Volumes of liquids and gases, for example, are measured in cubic centimeters.

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Other units used in chemistry are shown in the table below. Property

Unit

Symbol for Unit

mass time temperature volume pressure

gram minute degree celsius cubic centimeter atmosphere

g min °C cm3 atm

Amounts of substance Chemists need to measure quantities of substances for many purposes. Pharmaceutical companies need to check that a tablet contains the correct amount of the drug. Food manufacturers check levels of purity. In the laboratory, reactants need to be mixed in the correct ratios to prepare the desired product. We measure mass and volume routinely in the laboratory but they are not direct measures of amount. Equal quantities of apples and oranges do not have equal masses or equal volumes, but equal numbers. The chemist adopts the same approach. As all matter is made up from small particles, we measure amount by counting particles. If the substance is an element we usually count atoms, if it is a compound we count molecules or ions. A standard unit of amount can be defined in of a sample amount of any substance. Shoes and socks are counted in pairs, eggs in dozens and atoms in moles. A mole is the amount of a substance which contains the same number of chemical species as there are atoms in exactly 12 grams of the isotope carbon-12. The mole is a SI unit with the symbol mol. The word derives from the Latin for heap or pile. Naturally occurring carbon is, however, made up from a mixture of different types (called isotopes) of carbon atom. The presence of these isotopes increases the mass of one mole from 12.00 g (for a pure sample of carbon-12) to 12.01 g. More generally, the mass of one mole of atoms of an element is simply the relative atomic mass (Ar) expressed in grams. One mole of hydrogen atoms has a mass of 1.01 g, one mole of helium 4.00 g and so on. Some elements exist as molecules, and not as individual atoms. The composition of a molecule is given by its molecular formula. Hydrogen gas, for example, is made from diatomic molecules and so has the molecular formula H2. Water molecules are made from two hydrogen atoms and one oxygen atom and have the molecular formula H2O. The relative molecular mass (Mr) is calculated by adding the relative atomic masses of the atoms making up the molecule. Worked example Calculate the relative molecular mass of ethanol, C2H5OH.

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Solution: The compound is made from three elements, carbon, hydrogen and oxygen. Find the number of atoms of each element and their relative atomic masses from the Periodic Table, as shown in the following table.

Relative atomic mass

C

H

O

12.01

1.01

16.00

2

5+1=6

1

Number of atoms in one molecule of the compound Calculate the relative molecular mass of the molecule.

Relative molecular mass = (2 × 12.01) + (6 × 1.01) + 16.00 = 46.08 The molar mass of ethanol is 46.08 g mol−1. The molar mass of a compound is calculated in the same way as that of the elements. It is the relative molecular mass in grams. It is incorrect to use the term relative molecular mass for ionic compounds such as sodium chloride, as they are made from ions (e.g. Na1+ and Cl1−) not molecules. The term relative formula mass is used. It is calculated in the same way. Once the molar mass is calculated and the mass is measured, the number of moles can be determined. Worked example Calculate the number of moles in 4.00 g of sodium hydroxide, NaOH. Solution: The relative atomic masses are Na: 22.99, O: 16.00 and H: 1.01. The relative formula mass = 22.99 + 16.00 + 1.01 = 40.00 M = 40.00 g mol−1 n = m ÷ M = 4.00 ÷ 40.0 = 0.100 mol It is important to be precise when calculating amounts. One mole of hydrogen atoms has a molar mass of 1.01 g mol−1 but one mole of hydrogen molecules, H2 has a molar mass of 2 × 1.01 = 2.02gmol−1.

Counting particles The number of atoms, ions or molecules in a sample can be determined from mass measurements.

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As the mass of an individual atom can be measured using a mass spectrometer, the mole is a counting unit. Whereas socks are counted in pairs, sheets of paper in reams (1 ream = 500 sheets), atoms are counted in moles. From mass spectrometer measurements: mass of 1 atom of 12C = 1.99265 × 10−23 g Mass of 1 mole of 12C = 12 g Number of atoms in one mole = 12 ÷ 1.99265 × 10−23 = 602214000000000000000000. This is a big number. It is called Avogadro's number and is more compactly written in scientific notation as 6.02 × 1023. It is the number of atoms in one mole of an element and the number of molecules in one mole of a covalent compound. Although we could never count to Avogadro’s Number, even with the most powerful computer, we can prepare samples with this number of atoms. The atoms are counted in the same way as coins are counted in a bank; we use a balance: 3.01 × 1023 atoms of C = ½ mol = 0.5 × 12.01 g = 6.005 g. 3.01 X 1023 carbon atoms are 'counted out' when we prepare a sample of 6.005 g. Worked example Calculate the amount of water, H2O, that contains 1.80 × 1024 molecules. Solution: n = number of particles ÷ 6.02 × 1023 n = (1.80 × 1024) ÷ (6.02 × 1023) = 2.99 mol Note the answer should be given to 3 significant figures - the same precision as the data given in the question. If the amount given was 1.8 × 1024, the correct answer would be 3.0. Worked example Calculate how many hydrogen atoms are present in 3.0 moles of ethanol, C2H5OH. Solution: In 1 molecule of ethanol there are 6 H atoms. In 1 mole of ethanol molecules there are 6 moles of H atoms. In 3 moles of ethanol there are 3 × 6 = 18 moles of H atoms. Number of H atoms = 18 × 6.02 × 1023 = 1.08 × 1025

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Zumdahl, 5th Edition contains information relevant to this topic in Chapter 01. You should take some time to review the approach outlined in that textbook. IB Examiner Hints:    

Pay attention to decimal places. When adding or subtracting, the number of decimal figures in the result should be the same as the least precise value given in the data. Pay attention to significant figures. When multiplying or dividing, the number of significant figures in the result should be the same as the least precise value in the data. Use the accepted shorthand to solve problems more quickly in exams, for example n for moles, m for mass and M for molar mass. In some resources, Avogadro’s constant is shortened to L. You might find this helpful. Although Avogadro's constant is given in the IB Data booklet, you need to memorize its value for Paper 1.

Exercises: Show all work in the space provided. For multiple choice answers, briefly explain your reasoning. Calculate how many hydrogen atoms are present in 0.040 moles of C2H6.

Calculate the molar mass of magnesium nitrate, Mg(NO3)2.

Calculate how many hydrogen atoms are contained in 2.3 g of C2H5OH (Mr = 46).

The relative molecular mass of a compound is 98.0. Calculate the number of molecules in a 4.90 g sample.

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97N109 Which one of the following ionic compounds contains the greatest number of ions per mole of compound? A. Al2(SO4)3 B. Ca(HCO3)2 C. (NH4)2Cr2O7 D. Li3PO4

98M101 How many atoms are present in 0.10 mole of propyne, C3H4? A. 4.2 × 1022 B. 6.0 × 1022 C. 4.2 × 1023 D. 6.0 × 1023

02M102 Which sample contains the smallest amount of oxygen? A. 0.3 mol H2SO4 B. 0.6 mol O3 C. 0.7 mol HCOOH D. 0.8 mol H2O

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1.2

Formulas (3 Hours)

1.2.1

Assessment Statement Define the relative atomic mass (Ar) and relative molecular mass (Mr).

Obj 1

1.2.2

Calculate the mass of one mole of a species from its formula.

2

1.2.3

Solve problems involving the relationship between the amount of substance in moles, mass, and molar mass.

3

1.2.4

Distinguish between the empirical formula and molecular formula.

2

1.2.5

Determine the empirical formula from the other percentage composition or from other experimental data.

3

1.2.6

Determine the molecular formula when given both the empirical formula and experimental data.

3

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Teacher’s Notes

-1

The term molar mass (in g mol ) will be used.

Aim 7: Virtual experiments can be used to demonstrate this.

Finding chemical formulas in the laboratory When magnesium is burned in air, its mass increases as it is combining with oxygen. The mass change can be investigated experimentally. Item

Mass / g (±0.001)

empty crucible

25.000

crucible with magnesium before heating crucible with solid after heating

25.050 25.084

The masses of the magnesium and oxygen are then calculated as shown below. The precision of the calculated value is limited by the precision of the mass measurements to 2 significant figures. Element magnesium oxygen

Mass / g (±0.002) 25.050 – 25.000 = 0.050 25.084 – 25.050 = 0.034

The ratio of magnesium: oxygen atoms = 0.0021 : 0.0021 = 1 : 1. This is the simplest ratio known and is called as the empirical formula.

Moles 0.050 ÷ 24.31 = 0.0021 0.034 ÷ 16.00 = 0.0021

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Worked example A 2.765 g sample of a lead oxide was heated in a stream of hydrogen gas and completely converted to elemental lead with a mass of 2.401 g. What is the empirical formula of the oxide?

Solution: The mass loss is caused by a loss of oxygen. Set out the calculation in a table. Pb

O

Mass / g

2.401

2.765 − 2.401 = 0.364

Moles

= 2.401 ÷ 207.19 = 0.01159

= 0.364 ÷ 16.00 = 0.0228

Simplest ratio

= 0.01159 ÷ 0.01159 = 1

= 0.0228 ÷ 0.01159 = 2

Empirical formula = PbO2

Worked example A hydrocarbon contains 85.7% by mass of carbon. Deduce the empirical formula.

Solution: As a hydrocarbon contains only carbon and hydrogen, the compound contains 14.3% (100 − 85.7) by mass of hydrogen, as shown below. C

H

Mass / g

85.7

100 − 85.7 = 14.3

Moles

= 85.7 ÷ 12.01 = 7.14

= 14.3 ÷ 1.01 = 14.16

Simplest ratio

= 7.14 ÷ 7.14 = 1

= 14.16 ÷ 7.14 = 2

Empirical formula = CH2

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Exercises: An oxide of sulfur contains 60% by mass of oxygen. Deduce the empirical formula.

Pure nickel was discovered in 1751. It was named from the German word “kupfernlckel” meaning “devil's copper”. A compound of nickel was analyzed and shown to have the following composition by mass: Ni: 37.9%, S: 20.7%, O: 41.4%. Deduce the empirical formula.

Molecular formula The empirical formula does not give the actual number of atoms in the molecule. The hydrocarbon in the previous worked example had an empirical formula of CH2 but no stable molecule with this formula exists. The molecular formula, which is a multiple of the empirical formula, can only be determined once the relative molecular mass is known. This can either be measured by a mass spectrometer or calculated from the ideal gas equation. Some molecular formulas are shown below. Substance

Formula

Substance

Formula

Hydrogen Oxygen Nitrogen water

H2(g) O2(g) N2(g) H2O(l)

Carbon Dioxide Ammonia Methane Glucose

CO2(g) NH3(g) CH4(g) C6H12O6(g)

Worked example What is the empirical formula of glucose?

Solution: From the table above, the molecular formula = C6H12O6 Express this as the simplest ratio: CH2O.

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Worked example The compound with the empirical formula of CH2 is analyzed by a mass spectrometer and its relative molecular mass found to be 42.09. Deduce its molecular formula. Solution: Empirical formula = CH2 Molecular formula = CnH2n (where n is an integer) Mr = 42.09 = (12.01n) + (2n × 1.01) = 14.03n n = 42.09 ÷ 14.03 = 3 Molecular formula: C3H6

Sometimes chemists need to know the percentage of an element in a compound. They can calculate this from the molecular formula. Worked example Nitrogen is an important constituent in fertilizers. Calculate the percentage by mass of the element in ammonium sulfate, (NH4)2SO4. Solution: Molar mass = (2 × 14.01) + (8 × 1.01) + 32.06 + (4 × 16.00) g mol−1 Mass of N in one mole = 2 × 14.01 % of N (in one mole) = (2 × 14.01) ÷ ((2 × 14.01) + (8 × 1.01) + 32.06 + (4 × 16.00)) = 21.20%

Zumdahl, 5th Edition contains information relevant to this topic in Chapter 03. You should take some time to review the approach outlined in that textbook. IB Examiner Hints:  

The uncertainties in all the measurements should be included in all data tables. This is discussed in Chapter 02. Practice empirical formula calculations. All steps in the calculation must be shown. “Keep going” as errors are carried forward so that a correct method in a later part of the question is rewarded even if you have made earlier mistakes. One common problem is the use of too few significant figures in intermediate answers.

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Exercises: Which formula can be determined by only using the percent mass composition data of an unknown compound? I. Molecular formula II. Empirical (simplest) formula A. B. C. D.

I only II only Both I and II Neither I nor II

CFCs are compounds of carbon, hydrogen, chlorine and fluorine which catalyze the depletion of the ozone layer. The composition of one CFC is shown below. Carbon

Hydrogen

Chlorine

Fluorine

17.8 %

1.5 %

52.6 %

28.1 %

The value of its Mr is 135. Determine the molecular formula of the CFC.

Calculate the percentage by mass of nitrogen in ammonium nitrate.

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97M106 1.0 g samples of each of the following compounds were dehydrated. Which sample would lose the greatest mass percentage due to the removal of water? A. B. C. D.

MgCl2∙6H2O Mg(NO3)2∙6H2O NiSO4∙6H2O Fe(NO3)3∙6H2O

Molar mass = 203 g/mol Molar mass = 256 g/mol Molar mass = 263 g/mol Molar mass = 305 g/mol

05N102 The relative molecular mass (Mr) of a compound is 60. Which formulas are possible for this compound? I. CH3 CH2CH2NH2 II. CH3 CH2CH2OH III. CH3 CH(OH)CH3 A. B. C. D.

I and II only I and III only II and III only I, II, and III

02N102 Formation of polyethene from calcium carbide, CaC2, can take place as follows: CaC2 + 2 H2O → Ca(OH)2 + C2H2 C2H2 + H2 → C2H4 nC2H4 → −(−CH2−CH2−)n− What mass of polyethene is obtained from 64 kg of CaC2? A. B. C. D.

7 kg 14 kg 21 kg 28 kg

00N102 A certain compound has a relative molar mass of 88. A possible empirical formula for this compound is A. B. C. D.

CH2 CH2O CH3O C2H4O

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97N123 When 10 cm3 of a gaseous hydrocarbon is reacted with excess oxygen, the gaseous products consist of 40 cm3 of CO2 and 40 cm3 of H2O measured under the same conditions of pressure and temperature. What is the molecular formula of the hydrocarbon? A. B. C. D.

CH4 C2H4 C4H4 C4H8

97N106 1.64 grams of a compound are produced when 1.08 g of A reacts completely with excess B. What is the simplest formula of the compound? (Ar of A = 27, B = 14). A. B. C. D.

AB A2B AB2 A3B2

97M107 Complete combustion of a sample of a hydrocarbon produces 0.66 g of CO2 and 0.36 g of H2O. What is the empirical formula of this hydrocarbon? A. B. C. D.

CH2 CH4 C3H4 C3H8

02M101 A compound that contains only carbon, hydrogen, and oxygen has the following percentage by mass: carbon 60%, hydrogen 8%, oxygen 32% What is a possible molecular formula? A. B. C. D.

C5H8O2 C5H4O C6HO3 C7HO4

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98M102 A 26.0 g sample of an unknown compound containing only carbon and hydrogen was burned in excess oxygen and 88.0 g of CO2 were produced. What is a possible molecular formula of this compound? A. B. C. D.

CH C2H2 C3H6 C6H6

1.3a Chemical equations (1 Hour) Assessment Statement Deduce chemical equations when all reactants and products are given.

Obj 3

1.3.2

Identify the mole ratio of any two species in a chemical equation.

2

1.3.3

Apply the state symbols (s), (l), (g) and (aq).

2

1.3.1

Teacher’s Notes Students should be aware of the difference between coefficients and subscripts.

Chemical equations: the language of chemistry Atoms cannot be created or destroyed during a chemical reaction; they are simply rearranged. A chemical equation provides a balance sheet which allows us to monitor these changes as reactants are transformed into products. The number of atoms of each element must be the same on both sides of the equation. For example, the formation of liquid water involves two molecules of hydrogen gas combining with one molecule of oxygen gas to produce two molecules of liquid water. This information can be expressed in a more concise form: 2 H2(g) + O2(g) → 2 H2O(l) The reactants H2 and O2 are on the left-hand side and the product H2O is on the right-hand side. There are four atoms of H and two atoms of O on both sides. The only change is in how these atoms are bonded to each other. Since the mole is a counting unit, equations can also be interpreted in of moles: 2 H2(g) + O2(g) → 2 H2O(l) 2 mol 1 mol 2 mol Two moles of water can be formed from one mole of oxygen and two moles of hydrogen. One mole of water is formed from half a mole of oxygen and one mole of hydrogen. The coefficients in front of each of the molecules give the molar ratios of the reactants and products. As the physical states of the reactants and products can affect the energy change and rate of reaction, it is good practice to include state symbols in chemical equations as shown above.

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Balancing equations Trying to balance chemical equations can be very frustrating so it is important to follow a systematic method. Consider the unbalanced equation for the reaction between methane and oxygen to form carbon dioxide and water: ___CH4(g) + ___O2(g) → ___CO2(g) + ___H2O(l) It is a good idea to start with the elements that are present in the least number of substances: in this case C and H. Balance the C: _1_CH4(g) + ___O2(g) → _1_CO2(g) + ___H2O(l)

1 mole of C atoms on both sides

Balance the H: _1_CH4(g) + ___O2(g) → _1_CO2(g) + _2_H2O(l)

4 moles of H atoms on both sides

Balance the element which occurs in the most substances last: in this case, O. Changing the product side would change the C or H which are already balanced, so we change the reactant side. _1_CH4(g) + _2_O2(g) → _1_CO2(g) + _2_H2O(l)

4 moles of O atoms on both sides

Worked example Balance the equation for the combustion of ethane shown below: ___C2H6(g) + ___O2(g) → ___CO2(g) + ___H2O(l) Solution: Balance the C: 2 mol of atoms are needed on the product side. _1_C2H6(g) + ___O2(g) → _2_CO2(g) + ___H2O(l) Balance the H: 6 mol of atoms needed on the product side which gives 3 H2O molecules. _1_C2H6(g) + ___O2(g) → _2_CO2(g) + _3_H2O(l) Balance the O: (4 + 3) mol on the product side; 7 mol of O atoms needed on the reactant side, which gives 3½ mol of O2 molecules. _1_C2H6(g) + _3½_O2(g) → _2_CO2(g) + _3_H2O(l) Sometimes it is more convenient to deal with whole numbers so we multiply the equation by 2. _2_C2H6(g) + _7_O2(g) → _4_CO2(g) + _6_H2O(l)

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Zumdahl, 5th Edition contains information relevant to this topic in Chapter 03. You should take some time to review the approach outlined in that textbook. IB Examiner Hints:   

Practice writing and balancing a wide range of equations. Make sure that you do not change any formulas. Check that everything balances as it is very easy to make careless mistakes. Balance each element in turn, but if an element appears more than once on each side of the equation, leave it until last. Treat ions such as NO31− and SO42− as complete units when the nitrogen only appears as a nitrate, and the S as a sulfate.

Exercises: Nitrogen and oxygen react in the cylinders of car engines to form nitrogen monoxide (NO). Give a balanced equation for this reaction.

Nitrogen monoxide is a primary pollutant. After it escapes into the atmosphere, it reacts with oxygen to produce nitrogen dioxide. Give a balanced equation for this reaction.

Nitrogen dioxide is a secondary pollutant which can react further with oxygen and water in the atmosphere to produce nitric acid, HNO3(aq), one of the ingredients of acid rain. Give a balanced equation for the formation of nitric acid from nitrogen dioxide.

Balance the following equation: ___KClO(s) → ___KCl(s) + ___KClO3(s) Balance the following equation: ___Fe2O3(s) + ___H2SO4(aq) → ___Fe2(SO4)3(aq) + ___H2O(l) 99N102 Arsenic, As4, reacts with oxygen to produce the oxide As4O10. What is the sum of the coefficients for the reactants in the balanced equation? ___As4 + ___O2 → ___As4O10 A. 4 B. 5 C. 6 D. 7 04M103

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What is the coefficient for O2(g) when the equation below is balanced? ___C3H8(g) + ___O2(g) → ___CO2(g) + ___H2O(g) A. B. C. D.

2 3 5 7

03N103 Lithium hydroxide reacts with carbon dioxide as follows. 2 LiOH + CO2 → Li2CO3 + H2O What mass (in grams) of lithium hydroxide is needed to react with 11 g of carbon dioxide? A. B. C. D.

6 12 24 48

97N108 Ethyne, C2H2, can be reacted with oxygen to produce very high temperatures for cutting metal. The unbalanced equation for this reaction is given below. ___C2H2 + ___O2 → ___CO2 + ___H2O What should the C2H2 : O2 ratio be in order to have the proper stoichiometry (and the hottest flame)? A. B. C. D.

1:3 2:5 1:2 2:3

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1.3b Patterns in Chemical Equations (2 Hours) Classifying Reactions As you learned during your sophomore chemistry course, there are 5 basic types of chemical reactions.  Synthesis or composition o A + B  AB o Two or more reactants are combined to form one compound  Decomposition o AB  A + B o One reactions is broken into two or more products o Decomposition reactions are endothermic – the reaction absorbs heat.  Single Replacement or Single Displacement o Element + Compound  Element + Compound o A + BC  AC + B o Single replacement reactions are reactions that involve an element replacing one part of a compound. The products include the displaced element and a new compound. An element can only replace another element that is less active than itself.  Double Replacement, Double Displacement, or Metathesis o AB + CD  AD + BC o In many reactions between two compounds in aqueous solutions, the cations and anions appear to switch partners. The two compounds form two new compounds. No changes in oxidation numbers. o All double replacement rations must have a “driving force” or a reason why the reaction occurs. The driving force in a double replacement reaction is the removal of at least one pair of ions from solution. o Removal of ions can occur in one of three ways:  Formation of a precipitate – a solid has formed that is insoluble in water.  Formation of a gas – gases may form directly from the reaction or a side decomposition reaction occurs causing the gas to form. Reactions that produce three of the gases (CO2, SO3, NH3) involve the initial formation of a substance that breaks down to give the gas and water.  H2SO3    water + SO2  H2CO3    water + CO2  NH4OH    water + NH3  Formation of water – water is formed from an acid/base reaction.  Combustion reactions o Chemical + O2  chemical oxides o CxHy + O2  CO2 + H2O o S8 + O2  SO3 o Combustion reactions are exothermic reactions that release heat.

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The Law of Conservation of Mass states that in a chemical reaction there is no loss of mass. Each type of element will have the same mass before the reaction and after the reaction, or as reactant and product. But you can’t change the materials that participate in the reaction, so you must write an integer coefficient in front of (to the left of) each material in the reaction to make sure every type of atom has the same number on each side of the reaction. Tips for Balancing Equations     

Be sure each molecular formula is written correctly and each compound is neutral (meaning no charge). Mentally count or tally how many of each type of atom is present on each side of the equation. Begin balancing elements that are only found in one substance on each side. Balance oxygen and hydrogen last. If there is an odd number of an element on one side and an even number on the other, the odd will need to be evened out—so use a coefficient of 2 for that substance.  When tallying, be sure to adjust the count for each and every element that an added coefficient affects.  Combustion reactions that don’t seem to balance will often turn out better if a coefficient of 2 is used for the hydrocarbon.  Keep polyatomic ions grouped together in double replacement reactions.

When prediction reactions you will need to follow steps in order to get the correct final balanced reaction. 1. Write out the reactants. ionic compounds need to have their charges balanced. your diatomic elements (H2, O2, N2, Cl2, Br2, I2, F2) 2. Classify the reaction according to the outline on the following three pages. 3. Use the prompts to figure out the products. ionic compounds need to have their charges balanced. 4. Balance the reaction.

 Synthesis Reactions o Element + Element  Zn + Cl2  ZnCl2 o Compound + Compound  Formation of an Acid  Gas + Water  Acid  CO2 + H2O  H2CO3  Formation of a Base  Metal Oxide + Water  Metal Hydroxide  CaO + H2O  Ca(OH)2

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 Decomposition Reactions o Binary Compound—a compound with two elements  Compound  Element + Element  2AlCl3  2Al + 3Cl2 o Metal Carbonate  Metal Carbonate  Metal Oxide + Carbon Dioxide  Na2CO3  Na2O + CO2 o Metal Oxide  Metal Oxide  Metal + Oxygen Gas  2FeO  2Fe + O2 o Metal Peroxide  Metal Peroxide  Metal Oxide + Oxygen Gas  2PbO2  2PbO + O2 o Metal Hydroxide  Metal Hydroxide  Metal Oxide + Water  2LiOH  Li2O + H2O o Acid  Acid  Water + Gas Left Over  H2SO4  H2O + SO3 o Metal Chlorate  Metal Chlorate  Metal Chloride + Oxygen Gas  2KClO3  2KCl + 3O2

 Single Replacement Reactions o Active metals replace less active metals from their compounds in aqueous solutions.  Magnesium turnings are added to a solution of iron (III) chloride.  Reactants: Mg + FeCl3  Mg is higher on the activity Series than Fe so it replaces the iron  Equation: 3Mg + 2FeCl3  2Fe + 3MgCl2 o Active metals replace hydrogen in water.  Sodium is added to water.  Reactants: Na + H2O  Equations: 2 Na + 2 H2O  H2 + NaOH o Active metals replace hydrogen in acids  Lithium is added to hydrochloric acid  Reactants: Li + HCl  Equation: Li + HCl  H2 + LiCl o Active nonmetals replace less active nonmetals from their compounds in aqueous solutions.  Chlorine gas is bubbled into a solution of potassium iodide.  Cl2 + KI  KCl + I2

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 Double Replacement Reactions o AB + CD  AD + BC o Metals will interchange and new products will be formed. Use solubility rules to find if a precipitate will form. o AgNO3 + NaCl  NaNO3 + AgCl  All chlorides are soluble except silver chloride. AgCl is the solid.

 Combustion Reactions o Chemical + O2  chemical oxides o C3H8 + 5O2  3CO2 + 4H2O o S8 + O2  SO3 o 2C8H18 + 25O2  16CO2 + 18H2O

 Net Ionic Equations In overall ionic equations, formulas of the reactants and products are written to show the predominant form of each substance as it exists in aqueous solution. Soluble salts, strong acids, and strong bases are written as separate ions. Insoluble salts, suspensions, solids, weak acids, weak bases, gases, water and organic compounds are always written as compounds. Strong Acids—HCl, HI, HBr, HNO3, H2SO4. HClO3 Writing Net ionic Reactions 1. 2. 3. 4. 5. 6.

Write the equation and put the phases (s, l, g, aq) of each chemical. Balance the equation Rewrite the equations splitting the (aq) substances—Overall ionic equation. Cancel any ions that are found on both sides of the equation—spectator ions. Rewrite what is left in the reaction—net ionic equation. Reduce if needed

Example: Aqueous solutions of sulfuric acid and excess sodium hydroxide are combined. 1. 2. 3. 4. 5. 6.

H2SO4 (aq) + NaOH (aq)  H2O (l) + Na2SO4 (aq) H2SO4 (aq) + 2NaOH (aq)  2H2O (l) + Na2SO4 (aq) 2H+1(aq) + SO4-2 (aq) + 2Na+1(aq) + 2OH-1(aq)  2H2O (l) + 2Na+1(aq) + SO4-2 (aq) Na+1 and SO4-2 are spectator ions and are canceled 2H+1(aq) + 2OH-1(aq)  2H2O (l) Reduce: H+1(aq) + OH-1(aq)  H2O (l)

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Balance and classify the following reactions using the information on the preceding three pages. If the products are not identified, use context, the activity series, and the solubility rules to predict them. If possible, include state symbols. Write a net ionic equation for any double replacement reactions. The first two problems are done for you.

_1_Zn(s) + _2_HCl(aq)  _1_ZnCl2(aq) + _1_H2(g) _2_Fe(OH)3 + _3_H2SO4  _1_Fe2(SO4)3 + _6_H2O ___Na + ___O2  ___Na2O ___C9H20 + ___O2  ___CO2 + ___H2O ___NH3  ___N2 + ___H2 ___S8 + ___O2  ___SO3 ___AgNO3 + ___H2S  ___Ag2S + ___HNO3 ___H2O + ___Na2O  ___NaOH ___PbCO3  ___CO2 + ___PbO ___Mg + ___N2  ___Mg3N2 ___KClO3  ___KCl + ___O2 ___Na + ___H2O  ___NaOH + ___H2 ___MgCl2 + ___NaOH  ___Mg(OH)2 + ___NaCl ___HNO3  ___H2O + ___N2O5 ___H2O + ___Cl2O7  ___HClO4

Single Replacement: Active metals replace hydrogen in acids Double Replacement: H+(aq) + OH−(aq) → H2O(l)

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___KOH  ___Zn + ___CuCl2  ___Cu + ___ZnCl2  ___NaNO3 + ___K2SO3  ___NaClO3  ___C6H14 + ___O2  ___PbCO3  ___Cl2 + ___KI  ___I2 + ___KCl  ___K2SO4 + ___BaCl2  ___C6H6 + ___O2  ___AgMnO4 + ___KCl  ___Al(OH)3 + ___HC2H3O2 

For each of the following problems, write and balance a chemical equation. Include state symbols, and wherever appropriate, a net ionic equation. Solid beryllium is left in a container with liquid bromine.

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Solid bismuth is burned in air.

Solid gold is added to fluorine gas.

Solid calcium oxide is added to water.

Dinitrogen trioxide gas is bubbled through water.

Hydrogen and oxygen gases are mixed.

Potassium hydroxide is heated.

Sulfur trioxide is bubbled through water.

Solid magnesium oxide is placed in water.

Phosphorus trichloride is placed in a chlorine rich atmosphere.

Cesium is put in a container with chlorine gas.

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Sodium carbonate is heated.

Sulfurous acid is boiled.

Molten aluminum chloride is electrolyzed.

A sample of magnesium carbonate is heated.

A sample of ammonium carbonate is heated.

Carbonic acid is boiled.

Hydrogen peroxide decomposes.

Ammonium hydroxide is heated.

Liquid bromine is added to a container of sodium iodide crystals.

Lithium chlorate decomposes by heating.

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1.4

Mass and gaseous volume relationships in chemical reactions (4.5 Hours)

1.4.1

Assessment Statement Calculate theoretical yields from chemical equations.

Obj 2

Teacher’s Notes Given a chemical equation and the mass or (in moles) of one species, calculate the mass or amount of another species.

1.4.2

Determine the limiting reactant and the reactant in excess when quantities or reacting substances are given.

3

Aim 7: Virtual experiments can be used here.

1.4.3

Solve problems involving theoretical, experimental and percentage yield.

3

1.4.4

Apply Avogadro’s law to calculate reacting volume of gases.

2

1.4.5

Apply the concept of molar volume at standard temperature and pressure in calculations.

2

The molar volume of an ideal gas under standard conditions is −2 3 −1 3 −1 2.24 x 10 m mol (22.4 dm mol ).

1.4.6

Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas.

3

Aim 7: Simulations can be used to demonstrate this.

1.4.7

Solve problems using the ideal gas equation, PV = nRT.

3

TOK: The distinction between the Celsius and Kelvin scales as an example of an artificial and natural scale could be discussed.

1.4.8

Analyze graphs relating to the ideal gas equation.

3

General strategies A balanced chemical reaction is a quantitative description of a chemical reaction and can be used to make numerical predictions. Consider the thermal decomposition of limestone (CaCO3) to make lime (CaO): CaCO3(s) Initial amounts

1 mol

Initial masses

100.09 g

Final amounts

0 mol

Final masses

→

CaO(s)

+

CO2(g)

0 mol

0 mol

1 mol

1 mol

56.08 g

44.01 g

This equation shows that one mole of calcium carbonate will produce one mole of calcium oxide, or expressing the relationship in of mass, 100 g of calcium carbonate will produce 56 g of calcium oxide. The interpretation of the coefficient of a balanced equation as the number of moles opens the door to a wide range of calculations discussed in this section. The general strategy for the solution of these problems is outlined on the next page.

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1. Write the equation for the reaction. 2. Write the amounts in moles of the relevant reactants and products of interest from the equation and show the relationship between them. 3. Convert the known data given into moles to find moles of the substance required. If the amounts of all reactants are given, work out which reactant is in excess and which is the limiting reagent. 4. Convert the number of moles to the required quantities (mass, volumes, etc). Express the answer to the correct number of significant figures and include units. You will need to carry out conversions between moles and masses and volumes.

Worked example: Ethyne is used in welding as its combustion gives a lot of heat. The reaction can be described by the equation: 2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(l) Calculate the mass of CO2 produced from the complete combustion of 1.00 g of C2H2.

Solution 1.00 g C2H2

×

1 mol C2H2 26.04 g C2H2

×

4 mol CO2 2 mol C2H2

×

44.01 g CO2 1 mol CO2

= 3.38 g CO2

Exercise: The combustion of hydrocarbon fuels is an environmental concern as it adds to the carbon dioxide levels in the atmosphere. Calculate the mass of CO2 produced when 100 g of propane is burned according to the equation: C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)

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Using chemical equations: the theoretical yield When you plan a meal you need to check that you have the correct amount of ingredients to prepare the food in the required amounts. The chemist faces the same problem when planning the synthesis of a new compound. A balanced chemical reaction provides the recipe. You can use the strategy in the previous worked example to predict how much product will be produced from given masses of starting materials. Exercise: Iron is produced in the blast furnace by reduction of iron(III)oxide: Fe2O3(s) + 3 CO(g) → 2 Fe(l) + 3 CO2(g) Calculate the minimum mass of iron(lll) oxide needed to produce 800 g of iron.

Limiting reactants The equation for the reduction of iron (III) oxide in the previous exercise shows that one mole of iron (III) oxide reacts with three moles of carbon monoxide to produce two moles of iron. The same amount of iron would be produced if we had doubled the amount of iron oxide and kept the amount of carbon monoxide the same because there is insufficient carbon monoxide to reduce the additional iron (III) oxide. The iron oxide is said to be in excess. The amount of iron produced is limited by the amount of carbon monoxide. The carbon monoxide is the limiting reagent.

Worked example: A reaction vessel is filled with 4.04 g of hydrogen gas and 16.00 g of oxygen gas and the mixture is exploded. Identify the limiting reagent and deduce the mass of water produced.

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Solution Step 1 2 H2(g) + 1 O2(g) → 2 H2O(l) Step 2 2 moles of H2 react with 1 mole of O2 to produce 2 moles of H2O. Step 3 4.04 g H2

×

1 mol H2 2.02 g H2

×

2 mol H2O 2 mol H2

×

18.02 g H2O 1 mol H2O

= 36.04 g H2O

16.00 g O2

×

1 mol O2 32.00 g O2

×

2 mol H2O 1 mol O2

×

18.02 g H2O 1 mol H2O

= 18.02 g H2O

If you solve the problem from the perspectives of each reactant, you will get two different answers. The smaller of the two is called the theoretical yield. The theoretical yield is the maximum quantity of product that can be obtained, according to the balanced equation, from given quantities of reactants. The other answer is what should be made by the reagent in excess, and is useless because there is not enough of the limiting reagent to create that much product.

Alternatively, you might perform the initial gram-to-mole coversions, and then use a “start-change-finish” box to solve the problem. Consider the following solution. 2 H2(g)

+

1 O2(g)

→

2 H2O(l)

Start

2.0 mol

0.5 mol

0.0 mol

Change

−1.0 mol

−0.5 mol

+1.0 mol

Finish

1.0 mol

0.0 mol

1.0 mol

Amount of product produced = 1.0 mol H2O × 18.02 g H2O/mol = 18.02 g H2O The “start-change-finish” box illustrates how the quantities of the reactants and products change over time. The rate of change has to mirror the mole ratio from the balanced equation. The reaction proceeds until one of the reactants is completely used up – that reactant is the theoretical yield. The “start-change-finish” box only works with moles. If you are using grams, you will have to do additional work. Numerous methods can be employed to solve stoichiometry problems. Whichever method you use, that you must always show your work.

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Percentage yield Few chemical reactions are completely efficient. The experimental yield (i.e. the amount actually produced) is generally less than the theoretical yield predicted from the equation. There are several reasons for this: • • • •

The reaction is incomplete. There are side reactions in which unwanted substances are produced. Complete separation of the product from reaction mixture is impossible. Product is lost during transfers of chemicals during the preparation.

The efficiency of the procedure can be quantified by the percentage yield. Percentage yield =

experimental yield theoretical yield

× 100%

Worked example: Aspirin, C9H8O4, is made by reacting ethanoic anhydride, C4H6O3, with 2-hydroxybenzoic acid, C7H6O3, according to the equation: 2 C7H6O3 + C4H6O3 → 2 C9H8O4 + H2O 13.80 g of 2-hydroxybenzoic acid is reacted with 10.26 g of ethanoic anhydride. (a) Determine the limiting reagent in this reaction. (b) The mass obtained in this experiment was 10.90 g. Calculate the percentage yield of aspirin.

Solution 13.80 g C7H6O3

10.26 g C4H6O3

×

×

1 mol C7H6O3 138.13 g C7H6O3

×

1 mol C4H6O3 102.64 g C4H6O3

×

2 mol C9H8O4 2 mol C7H6O3

180.17 g C9H8O4 1 mol C9H8O4

×

2 mol C9H8O4 1 mol C4H6O3

×

180.17 g C9H8O4 1 mol C9H8O4

= 18.00 g C9H8O4

= 36.02 g C9H8O4

The limiting reagent is the one that produces the smaller value for theoretical yield. Therefore, C7H6O3 is limiting. Using the above equation, we can calculate percentage yield. 10.90 g C9H8O4 18.00 g C9H8O4

× 100%

=

60.56%

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Exercises: The Haber process, which provides ammonia needed in the manufacture of fertilizers, has enabled us to increase food production to cater for the world's growing population. Ammonia is produced by the synthesis of nitrogen and hydrogen: N2(g) + 3 H2(g) → 2 NH3(g) 400 kg of N2 is mixed with 200 kg of H2 to produce 220 kg of NH3. Calculate the percentage yield of ammonia.

Write and balance an equation for the formation of aluminum iodide from aluminum and iodine.

5.00 g of aluminium is mixed with 30.00 g of iodine. Calculate the theoretical yield of aluminum iodide from this reaction mixture, identify the limiting reagent.

20.00 g of aluminum iodide was produced. Calculate the experimental yield.

Suggest why production of purple vapor during the reaction leads to low experimental yield.

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States of matter If you were hit with 180 g of solid water (ice) you could be seriously injured, but you would be only annoyed if it was 180 g of liquid water. 180 g of gaseous water (steam) could also be harmful. These three samples are all made from the same particles – 10 moles of water molecules. The difference in physical properties is explained by kinetic theory. The basic ideas are: • all matter consists of particles (atoms or molecules) in motion • as the temperature increases, the movement of the particles increases. The three states can be characterized in of the arrangement and movement of the particles and the forces between them.

Most substances can exist in all three states. The state at a given temperature and pressure is determined by the strength of the interparticle forces.

Absolute zero and the kelvin scale The movement or kinetic energy of the particles of a substance depends on the temperature. If the temperature of a substance is decreased, the average kinetic energy of the particles also decreases. Absolute zero (−273°C) is the lowest possible temperature attainable as this is the temperature at which all movement has stopped. The kelvin scale emphasizes this relationship between average kinetic energy and temperature, as the absolute temperature, measured in kelvin, is directly proportional to the average kinetic energy of its particles. Temperature can be converted from Celsius to the kelvin scale by the relation: T (K) = T (°C) + 273 The kelvin is the SI unit of temperature.

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Changes of state The movement or kinetic energy of the particles depends on the temperature. When the temperature increases enough for the particles to have sufficient energy to overcome the interparticle forces, a change of state occurs. The following heating curve shows how the temperature changes as ice is heated from −40°C to steam at 140°C.

Consider a sample of ice at −40°C = 233 (−40 + 273) K. The water molecules vibrate at this temperature about their fixed positions.  

   

As the ice is heated, the vibrational energy of its particles increases and so the temperature increases. At the melting point of 0°C (273 K), the vibrations are sufficiently energetic for the molecules to move away from their fixed positions and liquid water starts to form. The added energy is needed to break the bonds between the molecules – the intermolecular bonds. There is no increase in kinetic energy so there is no increase in temperature. As the water is heated, the particles move faster and so the temperature increases. Some molecules will have sufficient energy to break away from the surface of the liquid so some water evaporates. At the boiling point of water, there is sufficient energy to break all the intermolecular bonds. The added energy is used for this process, not to increase the kinetic energy, and so the temperature remains constant. As steam is heated, the average kinetic energy of the molecules increases and so the temperature increases.

Worked example: In which sample do molecules have the greatest average kinetic energy? A. He at 100 K B. H2 at 200 K C. O2 at 300 K D. H2O at 400 K

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Solution Answer = D. The sample at the highest temperature has the greatest kinetic energy. Some substances change directly from a solid to gas at atmospheric pressure. This change is called sublimation. Exercise: When a small quantity of perfume is released into the air, it can be detected several meters away in a short time. Use the kinetic theory to explain why this happens.

Worked example: A flask contains water and steam at boiling point. Distinguish between the two states on a molecular level by referring to the average speed of the molecules and the relative intermolecular distances. Solution As the two states are at the same temperature, they have the same average kinetic energy and are moving at the same speed. The separation between the particles in a gas is significantly larger than that in a liquid. Exercise: Which of the following occur when a solid sublimes? I. The molecules increase in size. II. The distances between the molecules increase. A. B. C. D.

I only II only Both I and II Neither I nor II

Reacting gases The analytical balance used to measure mass is not always the most convenient instrument to measure quantity. Volume is often used for liquids and gases. Investigations into the relationship between the volumes of reacting gases were carried out by the French chemist Joseph Gay-Lussac (1778-1850) at the beginning of the 19th century. He observed that when gases react, their volumes and that of any products formed were in simple whole number ratios .

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A modern version of one of Gay-Lussac's experiments is described in the worked example. Worked example: Nitrogen monoxide, NO(g), reacts with oxygen, O2(g), to form one product. This is a brown gas, nitrogen dioxide, NO2(g). Consider the following apparatus.

Syringe A contains 50 cm3 of nitrogen monoxide. Syringe B contains 50 cm3 of oxygen gas. In the experiment, 5.0 cm3 portions of oxygen were pushed from syringe B into A. After each addition the tap was closed. After the gases had returned to their original temperature, the total volume of gases remaining was measured. The results are shown graphically in the following figure.

(a) Deduce a balanced equation for the reaction. (b) State the total volume of gases when the reaction is complete. (c) Deduce the volume of oxygen that reacts with 50 cm3 of nitrogen monoxide. (d) Identify the limiting reagent in the reaction. (e) Deduce the volume of nitrogen dioxide formed. (f) Compare the volume ratios of the three gases involved in the reaction with their molar ratios. Suggest a reason for any relationships you find.

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Solution (a) The unbalanced equation: ___NO(g) + ___O2(g) → ___NO2(g) The N atoms are balanced. The only way to balance the O atoms without changing the N atoms is to change the coefficient of O2. Eventually, the equation will balance as: _2_NO(g) + _1_O2(g) → _2_NO2(g) (b) The reaction is complete when the volume stops decreasing. Reading from the graph: total volume = 75 cm3. (c) The reaction stops after 25 cm3 of O2 is added. (d) The limiting reactant is nitrogen monoxide because the oxygen is left in excess. (e) Volume of nitrogen dioxide = total volume − volume of oxygen = 75 − 25 cm3 = 50 cm3. (f) The ratio of the volumes of the gases NO : O2 : NO2 is 50 : 25 : 50 = 2 : 1 : 2. This is the same as the molar ratios expressed in the balanced equation. This implies that equal volumes correspond to equal amounts (NO and NO2 in this example). The volume of NO is double the volume of oxygen as there is double the amount of NO compared to O2. One mole of each of the gases has the same volume . This explanation of Gay-Lussac's results was first proposed by the Italian scientist Amedeo Avogadro. Avogadro's hypothesis states that equal volumes of different gases contain equal numbers of particles at the same temperature and pressure. Worked example: 40 cm3 of carbon monoxide is reacted with 40 cm3 of oxygen. _2_CO(g) + _1_O2(g) → _2_CO2(g) What volume of carbon dioxide is produced? Assume all volumes are measured at the same temperature and pressure. Solution 2 CO(g) 3

+

1 O2(g) 40 cm

3

→

2 CO2(g) 0 cm3

Start

40 cm

Change

−40 cm3

−20 cm3

+40 cm3

Finish

0 cm3

20 cm3

40 cm3

Avogadro’s hypothesis states that volume is proportional to moles. The oxygen is in excess. 40 cm3 CO2 is made.

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Exercises: Assume all volumes are measured at the same temperature and pressure. (a) What volume of nitrogen forms when 100 cm3 of ammonia, NH3, decomposes completely into its elements. _2_NH3(g) → _1_N2(g) + _3_H2(g)

(b) What volume of oxygen is needed to react with 40 cm3 of butane, C4H10, and what volume of carbon dioxide is produced? _2_C4H10(g) + _13_O2(g) → _8_CO2(g) + _10_H2O(l)

The molar volume of a gas All gases have the same molar volume at the same temperature and pressure. The standard conditions of temperature and pressure (STP) are 273 K (0°C) and 100 kPa pressure. This pressure replaces the previous standard of 1 atm which is 101.3 kPa. One mole of gas occupies 22400 cm3 under these conditions. As this is such a large number it is often quoted in dm3 (1 dm = 10 cm, so 1 dm3 = 103 cm3 = 1000 cm3). At the higher temperature of 298 K (room temperature) the molar volume is 24dm3 (298 K and 100 kPa is called RTP). The molar volume can be used to calculate the amount of gases in the same way as molar mass. Calculations are simpler as all gases have the same molar volume. Worked example: Calculate the amount of chlorine in 44.8 cm3 of the gas at STP. Solution n = V/Vmol = 44.8 ÷ 22400 = 0.00200 mole Exercise: Calculate the volume occupied by 4.40 g of carbon dioxide at standard temperature and pressure.

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The volume of gaseous reactants and products in chemical reactions can be calculated using a similar strategy to that outlined earlier to calculate masses. Worked example: What volume of hydrogen (H2) is produced when 0.056 g of lithium (Li) reacts completely with water (H2O): _2_Li(s) + _2_H2O(l) → _2_LiOH(aq) + _1_H2(g) Assume the volume is measured at STP. Solution 0.056 g Li

×

1 mol Li 6.94 g Li

×

1 mol H2 2 mol Li

×

22400 cm3 H2 1 mol H2

= 90 cm3 H2

Exercises: Calcium reacts with water to produce hydrogen: Ca(s) + 2 H2O(l) → Ca(OH)2(aq) + H2(g) Calculate the volume of gas, measured at STP, produced when 0.200 g of calcium reacts completely with water.

Dinitrogen oxide, N2O, is a greenhouse gas produced from the decomposition of artificial nitrate fertilizers. Calculate the volume (at STP)of N2O produced from 1.0 g of ammonium nitrate, when it reacts according to the equation: NH4NO3(s) → N2O(g) + 2 H2O(l)

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The gas laws The gaseous state is the simplest state as all gases have the same molar volume and respond in similar ways to changes in temperature, pressure and volume. The gas laws describe this behavior. Pressure If you have ever pumped a bicycle tire or squeezed an inflated balloon you have experienced the pressure of a gas. A gas produces a pressure when its particles collide with the walls of its container. An increase in the frequency or energy of these collisions will increase the pressure. Relationship between volume and pressure for a gas An increase in volume reduces the frequency of the collisions with the walls of the container and so the pressure decreases. The relationship was studied experimentally by Robert Boyle in the 17th century. He found that if the temperature and amount of gas is kept constant, the pressure halves if the volume is doubled. The graphs in the following figure show that the pressure of a gas is inversely proportional to the volume.

Relationship between temperature and pressure for a gas You may have noticed that balloons have an increased tendency to 'pop' on hot summer days. An increase in temperature increases the average kinetic energy of the particles. The particles move faster and collide with the walls of the balloon with more energy and more frequency. Both factors lead to an increase in pressure. When the relationship is studied experimentally at constant volume, the graphs in the following figure are produced.

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Effect of temperature on the gas volume Combining the two previous relationships we can predict how the volume changes with absolute temperature. Consider the following sequence. 1. The temperature is doubled at fixed volume. 2. The volume is doubled at fixed temperature. The changes are summarized below. Feature

Step 1

Step 2

Overall change

Temperature

Doubled

Fixed

Doubled

Volume

Constant

Doubled

Doubled

Pressure

Doubled due to increase in temperature

Halved due to increase in volume

No change

The volume and the temperature of the gas have both doubled at fixed volume. This relationship is sometimes called Charles' law.

The volume of a gas is proportional to the absolute temperature . V = kT, where k is a constant. The combined gas law We can combine the three gas laws for a fixed mass of gas into one expression: VαT PαT P α 1/V

→

PV α T PV/T = constant

The response of a gas to a change in conditions can be predicted by a more convenient form of the expression: P1V1 T1

=

P2V2 T2

where 1 refers to the initial conditions and 2 the final conditions.

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Worked example: What happens to the volume of a fixed mass of gas when its pressure and its temperature (in kelvin) are both doubled? Solution The pressure and temperature are both doubled: P2 = 2P1, T2 = 2T1: P1V1 T1

=

P2V2 T2

P1V1 T1

=

2P1V2 2T1

V1

=

2V2 2

Substitute for P2 and T2:

P1 and T1 cancel from both sides: =

V2

The volume does not change. Exercises: The temperature in kelvin of 4.0 dm3 of hydrogen gas is increased by a factor of three and the pressure is increased by a factor of four. Deduce the final volume of the gas.

The molar volume of a gas at STPis 22.4 dm3. Use the combined gas equation to show that the molar volume of gas is 24 dm3 at RTP.

The ideal gas equation The combined gas equation refers to a fixed mass of gas. When you blow into a balloon you increase the number of particles and this increases the volume. When you pump up a bicycle tire the added gases cause the pressure to increase. The number of moles can be included in the combined gas equation to give the ideal gas equation. PV = nRT, where R is the gas constant. When SI units are used R has the value 8.31 J K−1mol−1.

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Gases which follow this equation exactly are called ideal gases. Real gases deviate from the equation at high pressure and low temperature owing to the effects of inter-particle forces. Worked example: A helium party balloon has a volume of 18.0 dm3. At room temperature (25°C) the internal pressure is 1.05 atm. Calculate the number of moles of helium (Ar = 4) in the balloon and the mass needed to inflate it. Solution PV = nRT n

=

PV RT

Convert data into SI units: P = 1.05 atm = 1.05 × 1.01 × 105 Pa V = 18.0 dm3 = 18.0 × 10−3 m3 T = 25°C = (25 + 273) K = 298 K Plug the numbers into the equation: n

=

(1.05 × 1.01 × 105)( 18.0 × 10−3) (8.31)(298)

= 0.771 mole

Mass of helium in the balloon = 0.771 mole × 4.00 = 3.08 g

Measuring the molar mass The ideal gas equation can be used to find the molar mass of gases or volatile liquids. Density can be used with this equation, as well. Useful permutations of these equations are as follows.

Ideal gas law: PV = nRT Molar mass: Mr =

Density: ρ =

mass n

mass V

Useful formula modifications: Mr =

mass∙R∙T P∙V

Mr =

ρ∙R∙T , when density is in g m−3 P

Worked example: A sample of gas has a volume of 432 cm3 and a mass of 1.500 g at a pressure of 0.974 atm and a temperature of 28°C. Calculate the molar mass of the gas.

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Solution Mr =

mass∙R∙T P∙V

Convert into SI units (the mass should be kept in g). T = 273 + 28 K, P = 0.974 × 1.01 × 105 Pa, V = 432 × 10−6 m3 Mr =

1.500 ∙ 8.31 ∙ (273 + 28) 0.974 × 1.01 × 105 ∙ 432 × 10−6

= 88.3 g mol−1

Zumdahl, 5th Edition contains information relevant to this topic in Chapter 05. You should take some time to review the approach outlined in that textbook. IB Examiner Hints:     

Whenever you multiply or divide data, quote the answer to the same number of significant figures as the least precise data. Practice setting out calculations in a logical way, including a few words to indicate what process is being used. Practice data response questions which involve the manipulation and interpretation of unfamiliar data - particularly if presented in graphical form. Problems involving volumes of gases can often be solved directly. There is no need for intermediate steps which calculate the number of moles. Make sure that you use the correct units when using the ideal gas equation. SI units should be used when R = 8.31 J K−1 mol−1. P should be in units of N m−2 (Pa), V in units of m3 and T in units of K.

Exercises: The density of a gaseous hydrocarbon with the empirical formula C3H7 is found to be 2.81 g dm−3 at 100°C and 1.00 atm. Calculate the molar mass of the hydrocarbon and find its molecular formula.

An unknown noble gas has a density of 5.84 g dm-3 at STP. Calculate its molar mass, and hence identify the gas.

An oxide of sulfur has a density of 3.60g dm-3 at STP. Calculate its molar mass, and hence identify the gas.

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99N103 What is the minimum number of grams of O2 (Mr = 32) required to burn 1.6 grams of CH4 (Mr = 16) according to the equation below? CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) A. 1.6

B. 3.2

C. 6.4

D. 32

98N103 Aluminum reacts with hydrochloric acid to produce hydrogen gas according to the equation below: 2 Al(s) + 6 HCl(aq) → 3 H2(g) + 2 AlCl3(aq) Which expression gives the number of moles of hydrogen that can be produced from 0.24 moles of Al and excess hydrochloric acid? A. B. C. D.

0.24 × 3/2 0.24 × 2/3 0.24 × 3/6 0.24 × 6/2

02N103 Ammonia is manufactured by the synthesis of nitrogen and hydrogen as follows: N2(g) + 3 H2(g) → 2 NH3(g) 56.0 g of N2 produces 34.0 g of NH3. What is the percentage yield of ammonia? A. 50

B. 68

C. 74

D. 100

97M125 A certain gas has a density of 2.35 g dm−3 at 30°C and 96 kPa (0.95 atm). The molar mass of the gas will be closest to which of the following? A. 50

B. 60

C. 70

D. 80

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1.5 1.5.1

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Solutions (2 Hours) Assessment Statement Distinguish between the solute, solvent, solution and concentration −3 −3 (g dm and mol dm ). Solve problems involving concentration, amount of solute and volume of solution.

Obj 2

Teacher’s Notes −3 Concentration in mol dm is often represented by square brackets around the substance under consideration, for example, [HCl].

3

Liquids Liquids, like gases, can also be conveniently quantified by measuring their volume rather than their mass. Unlike gases, however, there is no direct relationship between the volume of a liquid and its amount. The mass can be calculated from the volume if the density is known. mass Density: ρ = V In the laboratory, the volume of a liquid can be measured using different pieces of apparatus depending on the precision required. When the volume is known, volumetric flasks or pipettes are used. A 25 cm3 pipette has a typical uncertainty of ±0.06 cm3 and a 250 cm3 volumetric flask has an uncertainty of ±0.30 cm3. A buret is used when the volume is unknown. A 50 cm3 buret has a typical uncertainty of ±0.1 cm3. Solutions The discussion so far has focused on pure substances but chemists often carry out reactions in solution. Solutions are mixtures of two components. The less abundant component is the solute and the more abundant the solvent. The solute can be solid, liquid, or gas but the solvent is generally a liquid. Salt water is a solution with salt as the solute and water the solvent. Solutions in water are particularly important. These are called aqueous solutions and are given the state symbol (aq). Concentration The composition of a solution is generally expressed in of its concentration. As more and more solute dissolves in the solvent, the solution becomes more and more concentrated. When the solvent cannot dissolve any more solute, it is saturated.

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The concentration is generally expressed in of the mass or amount of solute dissolved in 1dm3 of solution. The units are either g dm−3 or mol dm−3. For example, one mole of sodium chloride has a mass of 22.99 + 35.45 g = 58.44 g. When this amount of solute is added to water to make a 1 dm3 solution, the concentration can either be expressed as 58.44 g dm−3 or 1.00 mol dm−3. Square brackets are used to represent concentrations, so this can be written down as [NaCI] = 1.00 mol dm−3. Concentrations in mol dm−3 are generally used in solving problems involving chemical equations. Worked example: A solution of sodium hydroxide has a concentration of 8.00 g dm−3. What is its concentration in mol dm−3? Solution Find the number of moles. A concentration of 8.00 g dm−3 means that there are 8.00 g of solute dissolved in 1 dm3 of solution. Sodium hydroxide has a molar mass of 40.00 g. 8.00 g is 1/5 of 40.00. Therefore, the concentration is 0.200 mol dm−3.

Standard solutions A solution of known concentration is called a standard solution. The amount of solute needed can be calculated from the concentration and the volume required: n concentration = (dm3) V Worked example: Calculate the mass of copper (II) sulfate pentahydrate, CuSO4∙5H2O, required to prepare 500 cm3 of a 0.400 mol dm−3 solution. Solution Concentration = 0.400 mol dm−3 = n mol ÷ 0.500 dm−3 n = 0.200 mol mass = 0.200 mol × (63.55 + 32.06 + 4(16.00) + 5(18.02)) = 49.9 g CuSO4∙5H2O Exercises: Calculate the mass of potassium hydroxide needed to prepare 250 cm3 of a 0.200 mol dm−3 solution.

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Calculate the mass of magnesium sulfate heptahydrate, MgSO4·7H2O, required to prepare 0.100 dm−3 of a 0.20 mol dm−3 solution.

Calculate the number of moles of chloride ions in 0.250 dm−3 of 0.020 mol dm−3 of zinc chloride (ZnCI2) solution.

Titrations Standard solutions are used to find the concentration of other solutions. The analysis of composition by measuring the volume of one solution needed to react with a given volume of another solution is called volumetric analysis. One of the most important techniques is titration. Typically, a known volume of one solution is measured into a conical flask using a pipette. The other solution is then added from a burette to find the equivalence point – the volume when the reaction is just complete. In acid-base reactions the equivalence point can be detected by the color change of an indicator. Worked example: What volume of hydrochloric acid with a concentration of 2.00 mol dm−3 would have to be added to 25.0 cm−3 of a 0.500 mol dm−3 sodium carbonate solution to produce a neutral solution of sodium chloride? Solution Na2CO3(aq) + 2 HCl(aq) → 2 NaCl(aq) + H2O(l) + CO2(g) 1 mole 2 moles Molesacid = Molesbase Concentrationacid × Volumeacid × Protonsacid = Concentrationbase × Volumebase × Protonsbase * (2.00 mol dm−3) × (Vacid) × (1) = (0.500 mol dm−3) × (0.025 dm3) × (2) Vacid = 0.0125 dm3 *If this equation does not look familiar, ask your teacher for a recap. You have seen something like it before.

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Zumdahl, 5th Edition contains information relevant to this topic in Chapters 04 and 11. You should take some time to review the approach outlined in that textbook. IB Examiner Hint: 

In Paper 1 you can find the best solution with less precise relative atomic mass values. This makes the calculation easier and saves time. For papers 2 and 3 estimate the answer before you use a calculator. This will help you spot careless mistakes.

Exercises: 25.00 cm3 of 0.100 mol dm−3 sodium hydrogen carbonate solution were titrated with dilute sulfuric acid: 2 NaHCO3(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l) + 2 CO2(g) 15.2 cm3 of the acid were needed to neutralize the solution. (a) Calculate the concentration of the sulfuric acid.

(b) Calculate the volume of carbon dioxide, measured at STP, produced during the titration.

When 0.025 g of an unknown Group 1 metal were added to excess water, 40 cm3 of hydrogen gas were collected at STP. Calculate the molar mass of the metal.

Nitroglycerine C3H5(NO3)3 is a liquid which explodes to produce a mixture of gases and water. Balance the equation for the detonation of the explosive below: ___C3H5(NO3)3(l) → ___CO2(g) + ___H2O(I) + ___N2(g) + ___O2(g) Deduce how many moles of gas are produced when one mole of nitroglycerine is detonated.

Calculate the total volume of gas produced at STP when 1.00 g of the nitroglycerine explodes.

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97M131 When equal volumes of 1 × 10−2 M Ba(NO3)2 and 4 × 10−3 M Na2SO4 are mixed, a precipitate of BaSO4 is formed. What is the Ba2+ ion concentration (in mol dm−3) in the resulting solution? A. 0.008

B. 0.006

C. 0.004

D. 0.003

02M104 2.02 g of KNO3 (Mr = 101) is dissolved in sufficient water to prepare 0.500 dm3 of solution. What is the concentration of this solution in mol dm−3? A. 0.02

B. 0.04

C. 0.10

D. 0.20

01N103 Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) Powdered zinc reacts with Cu2+ ions according to the equation above. What will be the result of adding 3.25 g of Zn to 100 cm3 of 0.25 mol dm−3 CuSO4 solution? A. B. C. D.

All the Cu2+ ions react and some solid zinc remains. All the Cu2+ ions react and no solid zinc remains. All the solid zinc reacts and Cu2+ ions remain. Neither solid zinc nor Cu2+ ions remain.

01M101 10.0 cm3 of 0.200 mol dm−3 H3PO4(aq) is converted into Na2HPO4(aq). What volume (in cm3) of 0.200 mol dm−3 NaOH(aq) is required? A. 10.0

B. 13.3

C. 20.0

D. 30.0

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97M108 The reaction of magnesium metal with aqueous silver nitrate, represented by the equation below, is exothermic. Mg(s) + 2 AgNO3(aq) → 2 Ag(s) + Mg(NO3)2(aq) Which of the following experiments would exhibit the greatest temperature increase? A. B. C. D.

0.24 g of Mg are added to 100 cm3 of 0.10 M AgNO3. 0.48 g of Mg are added to 100 cm3 of 0.20 M AgNO3. 0.24 g of Mg are added to 100 cm3 of 0.40 M AgNO3. 0.96 g of Mg are added to 100 cm3 of 0.20 M AgNO3.

05N103 Which aqueous solution contains the most hydrogen ions? A. 20 cm3 of 2 mol dm−3 sulfuric acid B. 10 cm3 of 4 mol dm−3 nitric acid C. 80 cm3 of 0.5 mol dm−3 hydrochloric acid D. 40 cm3 of 0.5 mol dm−3 sulfuric acid

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Chapter 01 Review Questions: What amount of oxygen, O2, (in moles) contains 1.8 × 1022 molecules? A. 0.0030 B. 0.030 C. 0.30 D. 3.0

___C2H2(g) + ___O2(g) → ___CO2(g) + ___H2O(g) When the above equation is balanced, what is the coefficient for oxygen? A. 2 B. 3 C. 4 D. 5

3.0 dm3 of sulfur dioxide is reacted with 2.0 dm3 of oxygen according to the equation below. 2 SO2(g) + O2(g) → 2 SO3(g) What volume of sulfur trioxide (in dm3) is formed? (Assume the reaction goes to completion and all gases are measured at the same temperature and pressure.) A. 5.0

B. 4.0

C. 3.0

D. 2.0

What volume of 0.500 mol dm−3 HCI(aq) is required to react completely with 10.0 g of calcium carbonate according to the equation below? CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g) A. 100 cm3

B. 200 cm3

C. 300 cm3

D. 400 cm3

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00N116 125 cm3 of an unknown gas has a mass of 0.725 g at 25°C and 0.97 atmospheres. Which expression will give the relative molar mass of the gas? (R = 82.05 cm3 atm K−1 mol−1) A.

0.725 × 82.05 × 25 0.97 × 125

B.

125 × 0.97 0.725 × 82.05 × 298

C.

0.725 × 82.05 × 298 0.97 × 0.125

D.

0.725 × 82.05 × 298 0.97 × 125

The relative molecular mass of aluminium chloride is 267 and its composition by mass is 20.3% AI and 79.7% chlorine. Determine the empirical and molecular formulas of aluminium chloride.

27.82 g of hydrated sodium carbonate crystals, Na2CO3·xH2O, were dissolved in water and made up to 1.000 dm3. 25.00 cm3 of this solution were neutralized by 48.80 cm3 of hydrochloric acid of concentration 0.1000 mol dm−3. (a) Write an equation for the reaction between sodium carbonate and hydrochloric acid. (2)

(b) Calculate the molar concentration of the sodium carbonate solution neutralized by the hydrochloric acid. (3)

(c) Determine the mass of sodium carbonate neutralized by the hydrochloric acid and hence the mass of sodium carbonate present in 1.000 dm3 of solution. (3)

(d) Calculate the mass of water in the hydrated crystals and hence find the value of x. (4)

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Describe in molecular the processes that occur when: (a) a mixture of ice and water is maintained at the melting point. (2)

(b) a sample of a very volatile liquid (such as ethoxyethane) is placed on a person's skin. (2)

The percentage composition by mass of a hydrocarbon is C = 85.6% and H = 14.4%. (a) Calculate the empirical formula of the hydrocarbon. (2)

(b) A 1 g sample of the hydrocarbon at a temperature of 273 K and a pressure of 1.01 × 105 Pa (1.00 atm) has a volume of 0.399 dm3. (i) Calculate the molar mass of the hydrocarbon. (2)

(ii) Deduce the molecular formula of the hydrocarbon. (1)

When a small quantity of strongly smelling gas such as ammonia is released into the air, it can be detected several meters away in a short time. (a) Use the kinetic molecular theory to explain why this happens. (2)

(b) State and explain how the time taken to detect the gas changes when the temperature is increased. (2)

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Sodium reacts with water as follows. 2 Na(s) + 2 H2O(l) → 2 NaOH(aq) + H2(g) 1.15 g of sodium is allowed to react completely with water. The resulting solution is diluted to 250 cm3. Calculate the concentration, in mol dm−3, of the resulting sodium hydroxide solution. (3)

(a) Write an equation for the reaction between hydrochloric acid and calcium carbonate. (2)

(b) Determine the volume of 1.50 mol dm−3 of hydrochloric acid that would react with exactly 1.25 g of calcium carbonate. (3)

(c) Calculate the volume of carbon dioxide, measured at 273 K and 1.01 × 105 Pa, which would be produced when 1.25 g of calcium carbonate reacts completely with the hydrochloric acid. (2)

An oxide of copper was placed in an unreactive porcelain dish and reduced in a stream of hydrogen. After heating, the stream of hydrogen gas was maintained until the apparatus had cooled. The following results were obtained. Mass of empty dish = 13.80 g Mass of dish and contents before heating = 21.75 g Mass of dish and contents after heating and leaving to cool = 20.15 g (a) Explain why the stream of hydrogen gas was maintained until the apparatus cooled. (1)

(b) Calculate the empirical formula of the oxide of copper using the data above, assuming complete reduction of the oxide. (3)

(c) Write an equation for the reaction that occurred. (1)

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An organic compound, A, containing only the elements carbon, hydrogen and oxygen was analyzed. (a) A was found to contain 54.5% C and 9.1% H by mass, the remainder being oxygen. Determine the empirical formula of the compound. (3)

(b) A 0.230 g sample of A, when vaporized, had a volume of 0.0785 dm3 at 95°C and 102 kPa. Determine the relative molecular mass of A. (3)

(c) Determine the molecular formula of A using your answers from parts (a) and (b). (1)

An element X reacts with oxygen to form the oxide X2O3. (a) Write a balanced equation for the reaction. (1)

(b) If 2.199 g of the oxide was obtained from 1.239 g of X, calculate the relative atomic mass of X and identify the element. (5)

(c) Nitrogen also forms an oxide on reaction with oxygen. This oxide contains 25.9% of nitrogen and 74.1% of oxygen by mass. Calculate the empirical formula of this second oxide. (3)

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Actual IB Questions with Complete Grading Markschemes: The following questions are followed by the actual rubrics used to grade them. Answer each question to the best of your ability, and then look at the rubric, to see where you lost points. Question A: A student is asked to prepare some copper (II) nitrate by reacting nitric acid with copper (II) oxide. (a) Write a balanced equation for this reaction. (1)

(b) The student carries out this reaction by adding 0.0345 mol of copper (II) oxide to 36.0 cm 3 of 1.15 mol dm−3 nitric acid solution. Calculate the amount (in mol) of nitric acid. (1)

(c) Use the information in (a) and (b) to identify the limiting reagent and determine the amount (in mol) of copper (II) nitrate formed. (2)

(d) The product of this reaction is isolated as copper (II) nitrate trihydrate. Calculate the molar mass of copper (II) nitrate trihydrate and the mass of product obtained. (2)

Question B: (a) Aqueous XO43− ions form a precipitate with aqueous silver ions, Ag1+. Write a balanced equation for the reaction, including state symbols. (2) (b) When 41.18 cm3 of a solution of aqueous silver ions with a concentration of 0.2040 mol dm−3 is added to a solution of XO43− ions, 1.172 g of the precipitate is formed. (i) Calculate the amount (in moles) of Ag1+ ions used in the reaction. (1)

(ii) Calculate the amount (in moles) of the precipitate formed. (1)

(iii) Calculate the molar mass of the precipitate. (2)

(iv) Determine the relative atomic mass of X and identify the element. (2)

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Rubric A: (a) CuO + 2 HNO3

→ Cu(NO3)2 + H2O

(b) 0.036 dm3 × 1.15 mol dm−3 = 0.0414 mol

(1) (1)

(c) 0.0414 mol HNO3 × 1 mol Cu(NO3)2/2 mol HNO3 = 0.0207 mol Cu(NO3)2 0.0345 mol CuO × 1 mol Cu(NO3)2/1 mol CuO = 0.0345 mol Cu(NO3)2 HNO3 is limiting reagent (Must be justified, not guessed. Allow ECF.)

(1)

Therefore 0.0207 mol Cu(NO3)2 formed. (Allow ECF)

(1)

(d) 63.55 + 124.02 + 54.06 = 241.63 g mol−1 241.63 g mol−1 × 0.0207 mol = 5.00 g (allow ECF from (c) and from molar mass)

(1) (1)

Rubric B: (a) 3 Ag1+(aq) + XO43−(aq) → Ag3XO4(s) (1) For balanced equation and (1) for states. (b) (i)

nAg+ = cV = 0.2040 mol dm−3 × 0.04118 dm3 = 0.008401 mol (−1 s.f.)

(ii) nAg3XO4 = 1/3 nAg+ = 1/3 × 0.008401 = 0.002800 mol

(2)

(1) (1)

(iii) 0.002800 mol weighs 1.172 g 1 mol weighs 1.172 g / 0.002800 mol = 418.6 g mol−1 Accept answer in range 418 to 419. No penalty for too many sig figs. ECF from (b)(ii)( g mol−1); Do not accept g. (iv) (3 × 107.87) + x + 4(16.0) = 418.6 Therefore, x = 30.99 (accept 31 or 31.0); P / phosphorus

(2)

(2)

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