IB Chemistry/Stoichiometry

Topic 1 - Stoichiometry

1.1 Mole concept & Avogadro's constant

1.1.1: Apply the mole concept to substances.

A mole is equivalent to 6.022 x 10²³ (Avogadro's constant) units. Chemists refer to a mole of something much as we refer to a dozen eggs; it is a convenient unit for counting. The periodic table provides molar masses, i.e. the number of grams of an element equivalent to one mole of atoms of that specific element. This can be extrapolated to molecules of known molecular formula.

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

Number of moles = mass / molar mass (Usually found on periodic table). The coefficients in chemical equations give the molar ratios of reactants and products i.e. 2A + 3B → C. There is 2/3 as much A as B, and 3 times more B than C involved in the reaction. Assuming the reaction goes to completion, there must be 3/2 times as much B as A for neither to remain. If this ratio is not followed, one will be a limiting reactant, and so the reaction will have some of the other reactant left over when it completes.

1.2 Formula

1.2.1: Define the term molar mass (M) and calculate the mass of one mole of a species.

The molar mass (M) is the mass of one mole's worth of a substance. To find the atomic/molecular mass, multiply the amount of moles by the molar mass.

1.2.2: Atomic mass, Molecular Mass, Formula Mass

The molar mass can be found in the periodic table, and will give the mass for 1 mol of the species (or rather the average accounting for different isotopes and their relative abundance).

1.2.3: Define the terms relative molecular mass (Mr) and relative atomic mass (Ar)

M_r is the ratio between the molar masses of two species. A_r is the ratio of the number of atoms between two species.

1.2.4: The difference between moles and mass

The number of moles refers to the amount of the substance, every mol being 6.02 x 10²³ individual elements. Mass is the property which results in 'weight' in the presence of gravity. Given a molar mass, M a mass m and a number of mols n then n = m / M.

1.2.5: The difference between the molecular formula and the empirical formula

An 'empirical formula' is the formula describing the different atoms present in a molecules, and their ratios, but not the actual number present. For example, A_xB_yC_z could be an empirical formula if x, y, and z are in lowest common terms. The molar mass can then be used to calculate the actual numbers of each atom present per molecule. The empirical formula can be determined by percentage composition, or anything else which gives the ratios of atoms present. A 'molecular formula', on the other hand, has the actual number of atoms present in each molecule. It will be an integer multiple of the empirical formula. For example, A_2xB_2yC_2z.

Example:

If you are given the empirical formula with an equation CH₃, and a mass of 30.08 g mol^-1. What is the molecular formula of this compound?

We will compare the molecular mass of the molecular formula, 30.08 g mol^-1, to that of the empirical formula, found to be 15.04 g mol^-1. By dividing, we find the ratio to be 2, meaning that the molecular formula must be 2 times as large as the empirical formula. The molecular formula is therefore C₂H₆.

1.3 Chemical Equations

Chemical equations are a convenient, standardised system for describing chemical reactions. They contain the following information.

• The type of reactants consumed and products formed
• The relative amounts of reactants and products
• The electrical charges on ions
• The physical state of each species (e.g. solid, liquid, gas)
• The reaction conditions (e.g. temperature, catalysts)

The final two points are optional and sometimes omitted.

Anatomy of an Equation

${\hbox{H}}_{2(g)}+{\hbox{Cl}}_{2(g)}\to 2{\hbox{HCl}}_{(g)}$

Hydrogen gas and chlorine gas will react vigorously to produce hydrogen chloride gas. The equation above illustrates this reaction. The reactants, hydrogen and chlorine, are written on the left and the products (hydrogen chloride) on the right. The large number 2 in front of HCl indicates that two molecules of HCl are produced for each 1 molecule of hydrogen and chlorine gas consumed. The 2 in subscript below H indicates that there are two hydrogen atoms in each molecule of hydrogen gas. Finally, the (g) symbols subscript to each species indicates that they are gases.

Reacting Species

Species in a chemical reaction is a general term used to mean atoms, molecules or ions. A species can contain more than one chemical element (HCl, for example, contains hydrogen and chlorine). Each species in a chemical equation is written:

${\hbox{E}}_{x}^{y}$

E is the chemical symbol for the element, x is the number of atoms of that element in the species and y is the charge (if it is an ion).

For example, ethanol would be written ${\hbox{C}}_{2}{\hbox{H}}_{6}{\hbox{O}}_{}$ because each molecule contains 2 carbon, 6 hydrogen and 1 oxygen atom. A magnesium ion would be written ${\hbox{Mg}}^{2+}$ because it has a double positive charge. Finally, an ammonium ion would be written $[{\hbox{NH}}_{4}]^{+}$ because each molecule contains 1 nitrogen and 4 hydrogen atoms and has a charge of 1+.

Coefficients

The numbers in front of each species have a very important meaning - they indicate the relative amounts of the atoms that react. The number infront of each species is called a coefficient. In the above equation, for example, one H₂ molecule reacts with one Cl₂ molecule to produce two molecules of HCl. This can also be interpreted as moles (i.e. 1 mol H₂ and 1 mol Cl₂ produces 2 mol HCl).

Other Information

Occasionally, other information about a chemical reaction will be supplied in an equation (such as temperature or other reaction conditions). This information is often written above the reaction arrow. We will ignore this for now, as it only complicates matters (and it's hard to draw in TeX).

1.3.1 : The mole ratio of two species in a chemical equation is the ratio of their coefficients...ie aX + bY → cZ : The ratio of X/Y is a/b, Y/Z = b/c etc.

Chemical equations are useful because they give the relative amounts of the substances that react in a chemical equation. For example, from the chemical equation for the formation of ammonia, we can see that one mole of nitrogen gas will combine with three moles of hydrogen gas to form two moles of ammonia gas.

${\hbox{N}}_{2(g)}+3{\hbox{H}}_{2(g)}\to 2{\hbox{NH}}_{3(g)}$

1.3.2 : Balancing equations...change only the coefficients, not the subscripts to make sure all atoms, and charge is conserved (half equations can be balanced by addition of electrons to either side...2 half equations can be added by making the number of electrons equal in each, then vertically adding.)

In some cases, however, we may not know the relative amounts of each substance that reacts. Fortunately, we can always find the correct coefficients of an equation (the relative amounts of each reactant and product) by applying the law of conservation of matter. Because matter can neither be created nor destroyed, the total number of each atom on one side of the equation must be the same as the total on the other. This process of finding the coefficients is known as balancing the equation.

For example, assume in the above equation that we do not know how many moles of ammonia gas will be produced:

${\hbox{N}}_{2(g)}+3{\hbox{H}}_{2(g)}\to ?{\hbox{NH}}_{3(g)}$

From the left side of this equation, we see that there are 2 atoms of nitrogen gas in the molecule N₂ (2 atoms per molecule x 1 molecule), and 6 atoms of hydrogen gas in the 3 H₂ molecules (2 atoms per molecule x 3 molecules). Because of the law of conservation of matter, there must also be 2 atoms nitrogen gas and 6 atoms of hydrogen gas on the right side. Since each molecule of the resultant ammonia gas (NH₃) contains 1 atom of nitrogen and 3 atoms of hydrogen, 2 molecules are needed to obtain 2 atoms of nitrogen and 6 atoms of hydrogen.

In a similar manner, you can use the law of conservation of matter to solve equations containing a greater number of unknown coefficients (the relative amounts of each reactant and product), or even subscripts (the number of each element in a molecule) on either side of the equation:

${\hbox{N}}_{2(g)}+3{\hbox{H}}_{2(g)}\to ?{\hbox{NH}}_{3(g)}$

1.3.3 : State symbols -- (s)-Solid , (l)-liquid, (g)-gas, (aq)-aqueous solution...i.e. something dissolved in water. Should be included in all chemical reactions (but won't be penalized).

${\hbox{E}}_{(state)}\,$

E is the chemical symbol for the element and (state) is the physical state.

The symbols in parentheses (in subscript below each species) indicate the physical state of each reactant or product. (s) means solid (l) means liquid (g) means gas (aq) means aqueous solution (i.e. dissolved in water) (CH3CH2OH) means dissolved in ethanol - any solvent can be noted this way.

Examples

${\hbox{CH}}_{4}+2{\hbox{O}}_{2}\to {\hbox{CO}}_{2}+2{\hbox{H}}_{2}{\hbox{O}}$

Above is the equation for burning methane gas (CH₄) in the presence of oxygen (O₂) to form carbon dioxide and water: CO₂ and H₂O respectively.

${\hbox{Pb}}_{(aq)}^{+2}+2{\hbox{I}}_{(aq)}^{-}\to {\hbox{PbI}}_{2(s)}$

This is a precipitation reaction in which dissolved lead cations and iodide anions combine to form a solid yellow precipitate of lead iodide (an ionic solid).

Simple Formulas

Moles to Particles: number of moles x (6.02 x 10²³/1 mole)= number of Particles

ex: 2.50 mol Zn x (6.02*10²³ atoms of Zn/1 mol Zn)= 1.51 x 10²⁴ atoms of Zn

Particles to Moles: number of Particles * (1 mole/6.02 x 10²³)= number of moles

ex: 1.204 x 10²⁴ atoms Ag * (1 mol/6.02 x 10²³ atoms Ag)= 2 moles of Ag

Moles to Mass: number of moles x (molar mass/ 1 mole)= number of grams (or other unit of measurement)

ex: 3.57 mol Al x (26.982 g Al/ 1 mol Al)= 96.3 g

Mass to Moles: number of grams * (1 mole/ molar mass)= number of moles

ex: 25.5 g Ag * (1 mol Ag/ 107.868 g Ag)= 0.236 mol

• • • to find molar mass

ex: Na₂CO₃ = 2(mass of Na)+ mass of C + 3(mass of O)

=2(22.990)+12.011+3(15.999)

=45.98+12.001+47.997

=105.99 g Na₂CO₃

1.4 Mass relationships in chemical reactions

1.4.1 : Mass is conserved throughout reactions. This fact allows masses to be calculated based on other masses in the reaction eg burning Mg in air to produce MgO and so to find the mass or Mg present in the original sample (ie purity)...can be extended to concentrations...ie titration.

1.4.2 : When a reaction contains several reactants, some may be in excess...is more is present that can be used in the reaction. The first reactant to run out is the limiting reagent (or reactant). Knowing the number of mols of the limiting reagent allows all other species to be calculated, and so the yield, and remaining quantities of other reactants.

1.5 Solutions

1.5.1

• Solvent - the fluid you're dissolving in e.g. water.
• Solute - something which has been dissolved to form a solution e.g. an ionic compound such as sodium hydroxide.
• Solution - a mixture of solvent and solute where the solute has dissolved into the solvent to form an indistinguishable liquid.
• Concentration - the amount of solute per volume of solution. Measured in moles per dm³ (i.e. moles per liter) or grams per liter.

1.5.2 : Apply the equation concentration = moles/volume...rather obvious from the units of concentration, but remember to convert everything into the same units.

1.5.3 : Use chemical equations to relate the amount of one species to the amounts of others.

TOPIC 1: QUANTITATIVE CHEMISTRY (12.5 HOURS)

Notes below are based on the specification for exams in 2009

Formula and Valency of Ions

SOME COMMON CATIONS

Ammonium	NH4+	Barium	Ba+2
Hydrogen	H+	Calcium	Ca+2
Lithium	Li+	Copper	Cu+2
Potassium	K+	Lead	Pb+2
Sodium	Na+	Magnesium	Mg+2
Silver	Ag+	Mercury	Hg+2
		Nickel	Ni+2
Aluminium	Al+3	Zinc	Zn+2

N.B. Many metals, including some of those above, show more than one valency. The valency is then always given with the name of the compound. It is written in Roman Numerals after the name of the metal.

e.g. Iron may be Fe⁺² or Fe⁺³

Iron II Chloride is Fe⁺²(Cl^-)₂

Iron III Chloride is Fe⁺³(Cl^-)₃

Other examples which you may meet are

Chromium II or Chromium III Cr⁺² or Cr⁺³

Manganese II or Manganese IV Mn⁺² or Mn⁺⁴

Tin II or Tin IV Sn⁺² or Sn⁺⁴

Lead II or Lead IV Pb⁺² or Pb⁺⁴

Some Common Anions

Aluminate	Al(OH)4- (aqueous) or AlO2- (crystalline)	TetraHydroxoZincate	Zn(OH)4-2 (aqueous) or ZnO2-2 (crystalline)
Bromide	Br-	Chromate (VI)	CrO4-2
Chlorate (V)	ClO3-	DiChromate (VI)	Cr2O7-2
Chloride	Cl-	Ethanedioate (oxalate)	C2O4-2
Cyanide	CN-	Oxide	O-2
Ethanoate (acetate)	CH3CO2-	Peroxide	O2-2
Ethoxide	C2H5O-	Sulphate	SO4-2
Fluoride	F-	Sulphide	S-2
Hydride	H-	Sulphite	SO3-2
HydrogenCarbonate	HCO3-	ThioSulphate	S2O3-2
HydrogenSulphate	HSO4-	HexaCyanoFerrate III	[Fe(CN)6]-3
Hydroxide	OH-	Carbonate	CO3-2
HypoChlorite	ClO-
Iodide	I-	Nitride	N-3
Methanoate	HCO2-	Phosphate	PO4-3
Nitrate	NO3-	HexaCyanoFerrate II	[Fe(CN)6]-4
Nitrite	NO2-
Manganate (VII) (PerManganate)	MnO4-
ThioCyanate	SCN-

Rules of thumb: Anions consisting of just one element have names ending -ide.

• Anions consisting of an element plus oxygen have names ending in -ate.

• Notable exception: OH^- is hydroxide, not 'hydrate'.

• If an element has two anions whose names should end in -ate, the one with fewer oxygens ends in -ite.

EXERCISE 1

Work out the formula for the following ionic chemicals

1. Zinc Chloride ZnCl₂
2. Sodium Oxide Na₂O
3. Aluminium Fluoride AlF₃
4. Iron III Chloride FeCl₃
5. Aluminium Oxide Al₂O₃
6. Zinc Hydroxide Zn(OH)₂
7. Copper (II) Nitrate Cu(NO₃)₂
8. Barium Sulphate BaSO₄
9. Sodium Carbonate Na₂SO₄
10. Calcium Hydrogencarbonate Ca(HCO₃)₂
11. Lead Iodide PbI₂
12. Nickel (II) Sulphate NiSO₄
13. Ammonium Carbonate (NH₄)₂CO₃
14. Manganese IV Oxide MnO₂
15. Sodium Bromide NaBr
16. Magnesium Ethanoate Mg(CH₃COO)₂
17. Silver Nitrate AgNO₃
18. Mercury II Chloride HgCl₂
19. Magnesium Nitride Mg₃N₂
20. Potassium Phosphate K₃PO₄

What are the names of the following chemicals?

1. CuBr₂ Copper II Bromide
2. K₂O Potassium Oxide
3. K₂Cr₂O₇ Potassium DiChromate (VI)
4. KNO₃ Potasium Nitrate
5. AlCl₃ Aluminium Chloride
6. FeCl₂ Iron II Chloride
7. FeS Iron II Sulphide
8. Na₂S₂O₃Sodium ThioSulphate
9. Al₂(SO₄)₃ Aluminium Sulphate
10. Mg(OH)₂ Magnesium Hydroxide
11. Pb(CH₃CO₂)₂ Lead (II) Ethanoate
12. Na₂Zn(OH)₄ Sodium Zincate
13. Pb(NO₃)₂ Lead (II) Nitrate
14. KMnO₄ Potassium Manganate (VII)
15. PbO₂ Lead (IV) Oxide
16. Ni(NO₃)₂ Nickel Nitrate
17. Na₂SO₃ Sodium Sulphite/Sodium Sulphate (IV)
18. Ca(HCO₃)₂ Calcium HydrogenCarbonate
19. CuO Copper II Oxide
20. MgCO₃ Magnesium Carbonate

1.1 MOLE CONCEPT AND AVOGADRO’S CONSTANT (2H) 1.2 FORMULAS (3H)

Relative molecular mass (Mr) and relative atomic mass (Ar).

Definition: The relative isotopic mass of ¹²C = 12

All other relative isotopic masses are measured relative to this standard and are not whole numbers e.g.

¹³C 13.003 ¹⁹F 18.998 ³⁵Cl 34.97 ³⁷Cl 36.97

The IB requires you to use whole numbers as an approximation to the relative isotopic mass of every isotope.

The relative atomic mass (A_r) is the average value of an element’s relative isotopic masses e.g. A_r (C) = 12 * 0.9889 + 13.0033 * 0.011 = 12.0098

What is A_r (Cl) if 75.77 % is ³⁵Cl and the rest is ³⁷Cl?

Relative atomic masses are in your data booklet, written beneath each element’s symbol in Table 5. e.g.

The relative molecular mass (M_r) is the sum of a compound’s relative atomic masses e.g.

M_r (CO) = 12.01 + 16.00

What is M_r (CO₂)?

What is M_r (Al₂(CO₃)₃)?

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

Another way of saying the same thing is to use the term ‘Molar mass’

The molar mass (M) of carbon-12 atoms is 12 g mol^-1

NB Molar mass has units (it is not ‘relative’)

You must specify if it refers to atoms, isotopes, molecules, etc.

EXERCISE 2:

Calculate the Relative Molecular Mass, Mr, of the following

1. Sodium Chloride

2. Copper Sulphate

3. Lead Nitrate

4. Magnesium Oxide

5. Potassium Hydroxide

6. Sulphuric Acid

7. Carbon Dioxide

8. Calcium Carbonate

9. Aluminium Sulphate

10. Sulphur Trioxide

11. Ammonium Nitrate

12. Zinc Chloride

13. Iron II Bromide

14. Iron III Iodide

15. Oxygen

16. Aluminium Zincate

17. Nitric Acid

18. Dilead II lead IV Oxide Pb3O4.

19. Magnesium Sulphate - 7 – water MgSO4.7H2O

20. Ammonium Iron II Sulphate - 6 – water (NH4)2SO4.FeSO4.6H2O

Calculate M for :-

1. Copper Carbonate

2. Potassium Dichromate

3. Magnesium Hydroxide

4. Sodium Phosphate

5. Calcium Hydrogencarbonate

6. Potassium Oxalate

7. Sodium Cyanide

8. Nitrogen Dioxide

9. Ethanol

10. Lead Chromate

11. Potassium Permanganate

12. Tin IV Chloride

13. Ethane

14. Lithium Carbonate

15. Ammonia

16. Butane

17. Sodium Chromate

18. Hydrochloric Acid

Applying the mole concept to substances.

Definition: A mole of particles contains the same number of particles as there are atoms in 12 g of carbon-12.

It turns out that there are 6.02 x 10²³ particles per mole.

6.02 x 10²³ mol^-1 is Avogadro’s constant

Solving problems involving amount, mass and molar mass.

Maths (learn this thoroughly):

Mass = Amount x Molar mass

or

m = nM

or

Grams = Moles x Grams per mole

EXERCISE 3

1. What is the mass of 0.5 moles of calcium carbonate (CaCO3)?

2. What is the amount of 27 g of water?

3. What is the relative molecular mass of fullerene if 2/3 mole has a mass of 480 g?

4. What mass of magnesium contains the same number of atoms as there are in 4 grams of calcium?

5. Calculate the mass of iron which contains the same number of atoms as there are molecules in 16 grams of oxygen.

6. What is the mass of copper sulphate (CuSO4 ) which contains the same number of copper atoms as there are atoms in 6.5 grams of zinc?

7. How much nitrogen contains the same number of atoms as there are sulphur atoms in 3.96 grams of ammonium sulphate ([NH4]2SO4)?

8. What mass of manganese contains twice the number of atoms that there in 0.6 grams of carbon?

Empirical formula and molecular formula.

Empirical formula: the simplest expression of the proportions of the elements in a compound. The empirical formula can be obtained from the proportions of masses of each element.

Molecular formula: the numbers of atoms of each element in a molecule of a compound.

Sometimes the two formulae are the same: H₂O, for example, is both the molecular and empirical formula.

Hydrogen peroxide has an empirical formula of HO but its molar mass is 34 g mol^-1. Its molecular formula is H₂O₂

EXERCISE 4 1. Analysis of two compounds showed that they contained the following proportions of elements: A: Carbon 26.1% Oxygen 69.6% Hydrogen 4.3% B: Carbon 52.2% Oxygen 34.8% Hydrogen 13.0% Calculate the empirical formula of each compound

2. If the empirical formula of methane is CH₄, and its relative molecular mass is 16, what is the molecular formula?

3. If the empirical formula of ethane is CH₃, and its relative molecular mass is 30, what is the molecular formula?

4. What is the mass of carbon dioxide that is produced when 50 g of calcium carbonate decomposes?

CaCO₃ → CaO + CO₂

5. If 12 g of magnesium react with oxygen to produce 20 g of magnesium oxide: a. What is the formula of magnesium oxide? b. What is the equation for the reaction?

6. Determine the empirical formula and the molecular formula of the following compounds from the given information. a. 0.36 mol S are combined with 0.72 mol oxygen atoms. M_r of this compound is measured to be 64. b. 0.193 mol Fe combined with 0.2895 mol O and its molar mass is 160 g mol^-1 c. 0.3285 mol Pb combined with 0.438 mol O. M_r = 685 d. 0.309 mol carbon combined with 0.618 mol H. Molar mass is 42 g mol^-1 e. 0.379 mol H combined with 0.379 mol O. M_r = 34

7. A compound was shown by qualitative analysis to contain nitrogen and hydrogen as the only elements. It was found by quantitative analysis that 0.02 mol of nitrogen are combined with 0.04 mol of hydrogen. Its relative molecular mass is approximately 30. Determine its empirical formula and its molecular formula.

8. The relative molecular mass of a compound of boron and hydrogen was measured as just under 28. Every 0.25 mol of boron are combined with 0.75 mol hydrogen atoms. Find its empirical and its molecular formula.

9. Calculate the empirical formulae of the following substances whose compositions are given below: a. 5g of an oxide of magnesium which contains 3g magnesium b. 7g nitrogen combined with 16g oxygen c. 0.6g magnesium combined with 1.2g oxygen and 0.3g of carbon d. 3.5g nitrogen combined with 2g oxygen e. 4.8 g oxygen combined with 3.2 g copper and 1.4 g nitrogen f. 28.2 g of an oxide of potassium which contains 23.4 g potassium g. 0.08 mol uranium combined with 3.413 g oxygen h. 1 mol zinc combined with 32 g oxygen and 2 g hydrogen i. A chloride of lead in which 1.302 g of the compound contains 0.530 g chlorine j. Aluminium bromide which contains 2.4 g bromine in every 2.67 g of bromide

10. Determine the molecular formula of the following from the given information: a. 8 g carbon are combined with 2 g hydrogen and the molar mass is 30 g mol-1. b. 1.56 g of a hydrocarbon contains 1.44 g carbon and its relative molecular mass is 78 c. 0.2 mol nitrogen atoms are combined with every 6.4 g oxygen and the molar mass of this compound is 92 g mol-1. e. 3.55 g of an oxide of phosphorus contain 1.55 g phosphorus. Its relative molecular mass is estimated to be a little below 300.

Find the formula for each of the following substances whose percentage composition is given

11. 40% calcium, 12% carbon & 48% oxygen

12. 40% copper, 40% oxygen & sulphur

13. the oxide of copper which contains 88.9% Cu by mass

14. the hydrocarbon which contains 85.7% carbon and has a relative molecular mass of 96

15. the carbohydrate which has a molar mass of 180 g mol-1 and contains 53.33% oxygen and 6.67% hydrogen.

1.3 CHEMICAL EQUATIONS (1H)

Balancing chemical equations

Once you know the formulae of the chemicals in a reaction, you need to make a balanced chemical equation.

You cannot change the formulae of the chemicals.

You can only add coefficients to indicate how many molecules of each chemical react.

Example 1

C₁₀H₁₆ + Cl₂ → C + HCl

This is unbalanced. Let’s balance the carbons first:

1 C₁₀H₁₆ + Cl₂ → 10 C + HCl

Now let’s try the hydrogens:

1 C₁₀H₁₆ + Cl₂ → 10 C + 16 HCl

Finally, time for the chlorines:

C₁₀H₁₆ + 8 Cl₂ → 10 C + 16 HCl

Example 2

Si₂H₃ + O₂ → SiO₂ + H₂O

This is unbalanced. Let’s balance the silicon first:

1 Si₂H₃ + O₂ → 2 SiO₂ + H₂O

Now let’s try the hydrogens:

2 Si₂H₃ + O₂ → 4 SiO₂ + 3 H₂O

Finally, time for the oxygens:

4 Si₂H₃ + 11 O₂ → 8 SiO₂ + 6 H₂O

Here is a trick you can use to guide you in steps 2 and 3:

1 Si₂H₃ + O₂ → 2 SiO₂ + ³/₂ H₂O

Example 3

Ca₁₀F₂(PO₄)₆ + H₂SO₄ → Ca(H₂PO₄)₂ + CaSO₄ + HF

I don’t like the calcium. It appears in more than two compounds here. Next in line is fluorine:

1 Ca₁₀F₂(PO₄)₆ + H₂SO₄ → Ca(H₂PO₄)₂ + CaSO₄ + 2 HF

The phosphate and sulphate can be treated as units:

1 Ca₁₀F₂(PO₄)₆ + H₂SO₄ → 3 Ca(H₂PO₄)₂ + CaSO₄ + 2 HF

1 Ca₁₀F₂(PO₄)₆ + 1 H₂SO₄ → 3 Ca(H₂PO₄)₂ + 1 CaSO₄ + 2 HF

Note that although we have coefficients for every substance, the coefficients 1, 3 and 2 are not balanced with the coefficients 1 and 1 – so I write them in two different colours.

The hydrogens look tricky too, but it’s either them or the calcium now:

1 Ca₁₀F₂(PO₄)₆ + 7 H₂SO₄ → 3 Ca(H₂PO₄)₂ + 7 CaSO₄ + 2 HF

Better do the calcium now…

Ca₁₀F₂(PO₄)₆ + 7 H₂SO₄ → 3 Ca(H₂PO₄)₂ + 7 CaSO₄ + 2 HF

EXERCISE 5 Balance these equations: 1. Cu + O2 → CuO 2. Fe + O2 → Fe3O4 3. Mg + HCl → MgCl2 + H2 4. Al + HCl → AlCl3 + H2 5. P + O2 → P2O5 6. NaHCO3 → Na2CO3 + CO2 + H2O 7. KClO3 → KCl + O2 8. NaNO3 → NaNO2 + O2 9. Cu(NO3)2 → CuO + NO2 + O2 10. PbO + C → Pb + CO2 11. C4H8 + O2 → CO2 + H2O 12. C4H10 + O2 → CO2 + H2O 13. Na2CO3 + HCl → NaCl + H2O + CO2 14. KOH + H2SO4 → K2SO 4 + H2O 15. (NH4)2SO4 + NaOH → NH3 + Na2SO4 + H2O 16. Mg + H2O → MgO + H2 17. Na + H2O → NaOH + H2 18. Pb3O 4 + H2 → Pb + H2O 19. Fe2O3 + CO → Fe + CO2 20. Fe(OH)3 → Fe2O 3 + H2O 21. CuO + H2 → Cu + H2O 22. Zn + HCl → ZnCl2 + H2 23. KClO3 → KCl + O2 24. S8 + F2 → SF6 25. Fe + O2 → Fe2O3 26. C2H6 + O2 → CO2 + H2O

The mole ratios of any species in a chemical equation.

EXERCISE 6

1. What is the amount (unit : mole) of hydrochloric acid needed to react with 1 mol Mg and how much magnesium chloride is formed?

2. What is the amount of chlorine which will react with 4 mol Fe? How much iron III chloride is formed?

3. How much carbon dioxide can be produced by reacting 0.1 mol CuCO3 with hydrochloric acid? What is the minimum amount of acid required?

4. What is the amount of sodium hydrogencarbonate which should be heated to produce 0.2 mol Na2CO3?

5. What amount of barium sulphate can be precipitated from a solution containing 0.01 mol barium chloride? What amount of potassium sulphate will be required to react with this barium chloride? What would be the effect of adding twice the amount of potassium sulphate?

The state symbols (s), (l), (g) and (aq).

EXERCISE 7 Can you add state symbols to the equations above? Aqueous solutions and Ionic equations Many substances dissolve in water to form aqueous solutions. Ionic substances are usually soluble, separating into their component ions: Al2(SO4)3 (s)  2 Al+3(aq) + 3 SO4-2(aq) The mobility of the ions in solution means that these solutions will conduct electricity (though the solid substance will not). Covalent substances do not usually dissolve in water. If they do dissolve, the molecules remain intact: C6H12O6(s)  C6H12O6(aq) When no more solute can be dissolved in a solvent, the solution is said to be saturated.

Ionic equations describe only the ions that react, omitting other (‘spectator’) ions. AgNO3 (aq) + NaCl(aq) ® AgCl(s) + NaNO3 (aq) Each aqueous compound can be treated as a pair of ions: Ag+(aq) + NO-3 (aq) + Na+(aq) + Cl-(aq) ® AgCl(s) + Na+(aq) + NO-3 (aq) The sodium and nitrate ions do not take part in the reaction Ag+(aq) + Cl-(aq) ® AgCl(s) This treatment simplifies the equation and makes it more generally applicable. E.g. What if silver fluoride and potassium chloride solutions were mixed? When an acid and an alkali react, ionic equations can be very useful: Sodium hydroxide (NaOH) and hydrochloric acid (HCl) react to form sodium chloride and water Magnesium hydroxide (Mg(OH)2) and sulphuric acid (H2SO4) react to form magnesium sulphate (MgSO4) and water Treating Na+, Mg2+, Cl- and SO42- as spectator ions, what is the common ionic equation when a hydroxide reacts with an acid? EXERCISE 8 Write balanced equations for the following reactions. Put in the physical states in the appropriate manner and write the ionic equation as well where this is appropriate (numbers 1, 2, 5, 11, & 15). Occasionally, oxygen appears as reactant but is not in any of the named products. In these cases, assume it is converted into water. 1. Iron filings reacting with dilute hydrochloric acid to give hydrogen plus iron II chloride solution. 2. Zinc oxide powder dissolves in dilute hydrochloric acid to give zinc chloride and water. 3. In the Contact process, sulphur dioxide and oxygen combine to form sulphur trioxide in the presence of vanadium V oxide catalyst and at a temperature of 450°C. 4. When crystalline potassium nitrate [nitrate (V)] is heated, oxygen is given off and a residue of potassium nitrite [nitrate (III)] remains. 5. When chlorine is bubbled into aqueous potassium iodide, a ppt. of iodine is formed along with a solution of potassium chloride. 6. The effect of heat on dilead II lead IV oxide (Pb3O4) is to produce oxygen, leaving lead II oxide behind. 7. Baking soda, sodium hydrogen carbonate, is used in cooking and when heated it decomposes to form sodium carbonate, carbon dioxide and steam. 8. When carbon dioxide is bubbled through lime water (calcium hydroxide solution), milkiness is observed due to the formation of a ppt of calcium carbonate. 9. If carbon dioxide is allowed to continue to bubble through the solution from (8) for a long time, the ppt redissolves, forming calcium hydrogencarbonate. 10. In the blast furnace, carbon monoxide reduces iron III oxide to molten iron while being oxidised itself into carbon dioxide. 11. The chemical test for a sulphate is to add barium chloride solution to a solution of the test sample. A white ppt. of barium sulphate indicates a positive result. 12. Magnesium burns in an extremely exothermic reaction in which magnesium oxide is the product. 13. When conc. sulphuric acid (a liquid) is poured onto common salt, fumes of hydrogen chloride are observed, while sodium hydrogensulphate remains in the test tube. 14. A solution of hydrogen peroxide will decompose spontaneously, in the presence of manganese IV oxide catalyst, to liberate oxygen. 15. When solutions of ammonium chloride and sodium hydroxide were heated together, the smell of ammonia was observed. Sodium chloride was one of the other products.

1.4 MASS AND GASEOUS VOLUME RELATIONSHIPS IN CHEMICAL REACTIONS (3H)

Calculating theoretical yields from chemical equations.

EXERCISE 9

1 What is the amount of sodium nitrate (nitrate V) (NaNO3) in 17.0g ? What mass of sodium nitrite (nitrate III) will be obtained by heating this if it decomposes according to the equation :- 2 NaNO3  2 NaNO2 + O2

2. What mass of zinc oxide remains after heating 9.45g zinc nitrate? 2 Zn(NO3)2  2 ZnO + 4 NO2 + O2

3. What mass of sodium carbonate remains after heating 4.2g sodium hydrogencarbonate?

4. What mass of powdered zinc would react with 0.96 g of sulphur? What is the expected yield of zinc sulphide?

5. What will be the loss in mass on heating 6.2g copper carbonate until no further reaction takes place? CuCO3 (s)  CuO (s) + CO2 (g)

6. A solution contained 6.62g of lead nitrate. A solution of sodium chloride was added in order to precipitate lead chloride. What is the expected yield of lead chloride?

7. What is the amount of Mg contained in 4.8g magnesium? If this reacts with hydrochloric acid, what is the amount of hydrogen gas which will be produced?

8. An aqueous solution contains 12.6 g HNO3. What amount of HNO3 is this? Find the amount and the mass of potassium hydroxide which will just neutralise this. What mass of potassium nitrate could you expect as product?

9. 5.05 g of copper nitrate decomposes according to the following equation when it is heated 2 Cu(NO3)2  2 CuO + 4 NO2 + O2 (a) What mass of copper oxide is produced? (b) What amount of nitrogen dioxide is formed?

The limiting reactant and the reactant in excess: Theoretical, experimental and percentage yield.

EXERCISE 10

1. What mass of zinc will be required to react completely with 5g of copper sulphate and what mass of copper will be formed?

2. How much iron II sulphate will be required to react with 4g sodium hydroxide and what mass of iron II hydroxide will be formed?

3. 8g of calcium are dissolved in excess hydrochloric acid. How many molar masses of acid will actually react? What amount of calcium chloride can you expect? If this crystallises out of solution as CaCl2.6H2O, what is the maximum mass of crystals that can be produced?

4. A technician used 2.4g of magnesium ribbon to produce crystals of magnesium sulphate - 7 - water by reaction with dil. sulphuric acid. What should his yield (in grams) have been? If he managed to make only 16.4 g, calculate his percentage yield.

5. A portable gas stove burns butane at the rate of 116 g per hour. What mass of water will be produced after one hour? In what physical state will this water first appear? What will eventually happen to it? If the appliance is used in a confined space, state what problems or dangers might result.

6. 8.0 g of copper oxide were dissolved in dilute sulphuric acid and after suitable treatment 20 g copper II sulphate - 5 - water were obtained. What is the percentage yield?

7. Ethanol (20 cm3) was heated with an equal volume of ethanoic acid in the presence of concentrated sulphuric acid catalyst. The yield of ethyl ethanoate was 8 cm3. (a) Calculate the mass and the amount of each reagent. (b) Which of the two reagents, ethanol or ethanoic acid is the limiting reagent i.e. which is not in excess? (c) Calculate the percentage yield of ester. [ densities in g cm-3 :- ethanol 0.79; ethanoic acid 1.05; ethyl ethanoate 0.9]

8. 8.4g of sodium hydrogen carbonate are mixed with 15g of hydrochloric acid. Explain which reagent is in excess. Find the amount of carbon dioxide produced in the reaction. Would more or less carbon dioxide have been produced by heating the hydrogencarbonate?

9. On heating 4.2g of a metal carbonate, XCO 3, 0.05 mol carbon dioxide was produced. Calculate the relative atomic mass, A r, of the metal X. XCO 3  XO + CO 3

10. Bath salts (sodium carbonate crystals) (9.53g) were gently heated in a crucible in order to drive off all their water of crystallisation. 3.53g of anhydrous salt remained. Calculate the number of molecules of water of crystallisation in the sodium carbonate and hence its formula. However when 12 g of the crystals were left in a hot room, the loss in mass was only 6.80g. What can you deduce about the formula of the sodium carbonate after it had undergone efflorescense in these conditions?

Avogadro’s law of volumes of gases.

Avogadro’s law: Equal volumes of gases, at the same pressure and temperature, contain equal amounts of molecules. The law can be used to find the volumes of gases in a reaction: e.g. 1 dm3 of hydrogen will react with 1 dm3 of chlorine to give 2 dm3 of hydrogen chloride. EXERCISE 11 What volume of CO2 will be produced when 1 dm3 of methane (CH4) combines with 3 dm3 of oxygen? What volume of CO2 will be produced when 1 dm3 of ethene (C2H4) combines with 3 dm3 of oxygen?

Graphs relating to the ideal gas equation.

This section meets many criteria for Topic 11.3 Graphical techniques

EXERCISE 12

1. A sample of gas was heated while the pressure was kept constant. The changes in volume were recorded. Graph the following data:

Temperature (°C)	Volume (dm3)
0	1.37
100	1.84
200	2.33
300	2.79
400	3.34
500	3.81
600	4.28

Use a scale which starts at -300 °C and 0 dm³.

Estimate the temperature required for the gas to have a zero volume.

2. Another gas sample was pressurized and the effect on its volume was recorded. The temperature was kept constant. Graph the following data:

Pressure (bar)	Volume (dm3)
1	4.64
2	2.40
3	1.60
4	1.19
5	0.94
6	0.82
7	0.67

What is the relationship between pressure and volume?

3. A gas was gradually introduced to a rigid, constant temperature, vacuum-filled container, and the pressure recorded as more gas was added. Graph the following data:

Amount (mol)	Pressure (bar)
0	0
1	0.500
2	0.998
3	1.510
4	1.988
5	2.517
6	3.118
7	3.441

What is the relationship between pressure and amount?

4. One mole of a gas was kept in a 1 m³ (1000 dm³) container as the temperature was varied. The effect on the pressure was recorded:

Temperature (K)	Pressure (Pa)
253	2037
273	2280
293	2501
313	2615
333	2753
353	2918
373	2925

What is the relationship between pressure and temperature?

What is the gradient of the graph?

Use the gradient to estimate the temperature required for the pressure to be 101325 Pa (1 atm).

Molar volume of a gas

A mole of any gas at 298 K and 1 atmosphere pressure occupies 24 dm3. At 273 K the volume is 22.4 dm3. E.g. 2 moles of argon atoms occupy 48 dm3 0.5 moles of oxygen molecules occupy 12 dm3 What is the volume of carbon dioxide that is produced when 50 g of calcium carbonate decomposes? CaCO3 ® CaO + CO2

The relationship between temperature, pressure and volume for a fixed mass of an ideal gas.

We can summarize the behavior of an ideal gas with this equation:

PV = nRT

P is the pressure (in Pa or N m-2)

V is the volume (m3)

n is the number of moles of the gas

R is the ideal gas constant (=8.31 J K-1 mol-1)

T is the absolute temperature (K)

You can see from this that if pressure increase, either the volume must decrease or the temperature increases. Similarly, increasing temperature causes increased volume and/or pressure.

EXERCISE 12

Use PV = nRT to calculate: a) The volume occupied by 1 mol of a gas at 25 oC (298 K) and 100000 Pa (1 bar) pressure.

b) The pressure required to make 5 mol of gas occupy 50 dm3 (50 x 10-3 m3) at 35 oC.

c) The number of moles of gas which occupy 25 dm3 at 85 oC and 1.05 bar.

d) The temperature of a 3 mol gas sample which has a volume of 18 dm3 at 1 atm (1.013 bar) pressure.

If we change the conditions for a given sample of gas, we can simplify the equation considerably. Call the initial pressure, temperature and volume P1, T1 and V1, and the values after the conditions have changed are called P2, T2 and V2. The amount of gas, n, and the constant, R, are the same in both cases. This gives us the formula:

 =  P1*V1 / T1 = P2*V2 / T2

The ‘1’ and ‘2’ refer to the volume (V) of the gas under different conditions of Pressure (P) and Absolute Temperature (T - measured in kelvins) Helpfully, the pressure and volume do not now have to measured in Pa and m3. Any units are suitable, so long as the same units are used for P1and P2, and for T1 and T2.

EXERCISE 13

1. Use the simplified equation (and answers to the previous exercise) to calculate:

a) The volume occupied by 1 mol of a gas at 0 oC and 100000 Pa (1 bar) pressure.

b) The pressure required to make 5 mol of gas occupy 40 dm3 (50 x 10-3 m3) at 35 oC.

c) The temperature of a 3 mol gas sample which has a volume of 25 dm3 at 1 atm (1.013 bar) pressure.

d) If a sample of gas which occupies 26.25 dm3 at 120.8 oC and 1.10 bar could be the same as a sample which occupies 25 dm3 at 85 oC and 1.05 bar.

2. If 1 dm3 of gas is warmed from 298 to 745 K, while the pressure is kept constant at 1 atm, what is its new volume?

3. What pressure is now required to pressurize the hot gas until it reaches its original volume?

1.5 SOLUTIONS (2H)

Solute, solvent, solution and concentration

Concentration (c) is measured as the quantity of solute per unit volume (usually per dm3 i.e. dm-3) of solution (not solvent). Remember that 1 dm3 = 1000 cm3 The solute quantity can be measured as mass, so the units are g dm-3. cm = m/v mass concentration = mass / volume The solute quantity can be measured as amount, so the units are mol dm-3. cn = n/v molar concentration = amount / volume The molar concentration of a substance is often written in the shorthand form [substance]. E.g. Instead of writing ‘The molar concentration of NaCl’ we can write [NaCl].

EXERCISE 14

1) Write down all the possible mathematical expressions which might be needed when doing calculations based on volumetric analysis (titrations).

2) What is the mass concentration if 5 g of NaCl are dissolved in 500 cm³ of solution?

3) What is the amount of 5 g of NaCl?

4) What is the molar concentration if 5 g of NaCl are dissolved in 500 cm³ of solution?

5) 25 cm³ of hydrochloric acid (0.12 mol dm^-3) were just neutralized by 28.4 cm³ of sodium hydroxide solution. Calculate the concentration of the base.

6) 23.2 cm³ of hydrochloric acid were required to cause the formation of the very pale grey colour - almost colourless - of screened methyl orange indicator in 25.0 cm³ of potassium hydroxide solution. Determine the concentration of the acid given that the potassium hydroxide was made by dissolving 1.372 g in 250.0 cm³ of solution.

7) A sample of lithium hydroxide was weighed as 0.62 g. It was dissolved in water & made up to 250.0 cm³. 25.00 cm³ aliquots were titrated and 21.25 cm³ of standard hydrochloric acid (0.100 mol dm^-3) were required. The lithium hydroxide is thought to be contaminated with lithium chloride. Calculate the mass of lithium hydroxide in the sample and hence its percentage purity.

8) Butanedioic acid is a dibasic acid (diprotic acid) of molecular formula C4H6O4. 1.0073 g were weighed, dissolved in water and the solution made up to 250 cm³. 25 cm³ portions required 23.95 cm³ of potassium hydroxide solution to cause the colour change of a suitable indicator. Calculate the concentration of the alkali.

a) What amount of butanedioic acid was used (how many moles)?

b) What amount was used in each titration?

c) What amount (mol) of potassium hydroxide was present in each titration?

d) What is the concentration of potassium hydroxide?

9) A technician was asked to analyse a solution of commercial caustic soda for its sodium hydroxide & sodium carbonate content. She did this by titrating a 25.00 cm³ portion with hydrochloric acid solution (0.100 mol dm^-3.) using firstly phenolphthalein and then, as part of the same titration, by adding methyl orange.

Results obtained were 16.90 cm³ of acid to bring about the colour change of the phenolphthalein and a total end point of 24.55 cm³.

At the first colour change (phenolphthalein goes pink to colourless), the reactions taking place are:

OH^- + H⁺  H2O & also CO3^-2 + H⁺  HCO3^-2.

At the second colour change (methyl orange turns green to red), the reaction taking place is

HCO3^-2 + H⁺  CO2 + H2O

Thus the first part of the titration relates to the concentration of hydroxide plus carbonate. The second part refers to the concentration of the carbonate (in its hydrogencarbonate form) only.

a) Calculate the number of moles of acid used in the first part and thus the total number of moles of hydroxide plus carbonate in the 25.00 cm³ portion.

b) Calculate the volume and hence the number of moles of acid used in the second part of the titration and thus the number of moles of hydrogencarbonate (and thus carbonate).

c) From a) & b) calculate the number of moles of hydroxide and of carbonate in the 25.00 cm³ portion. From these results calculate the concentrations of the sodium hydroxide & sodium carbonate in the solution. Give your answer in each case in mol dm^-3 & also in g dm^-3.

10) 1.483 g of the mineral Dolomite were added to excess hydrochloric acid (50 cm³, 1.0 mol dm^-3). After reaction was complete, the solution was filtered and made up to 250 cm³. 25 cm³ portions of this solution containing unreacted acid were titrated against 0.1 mol dm^-3 KOH when a mean titre of 26.3 cm³ was obtained. Assuming that Dolomite is essentially magnesium carbonate plus insoluble silica, calculate its percentage composition.

a) What amount of potassium hydroxide was in the 26.3 cm³?

b) What amount of hydrochloric acid would have reacted with this?

c) What amount of hydrocloric acid was in the 250 cm³ solution?

d) What amount of acid was taken at the start?

e) What amount of acid reacted with the Dolomite?

f) Calculate the amount and hence the mass of magnesium carbonate which would have reacted with the hydrochloric acid.

g) Calculate the percentage by mass of magnesium carbonate in Dolomite.