Introduction to Quantitative Chemistry

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1. Chemical Reactions and Stoichiometry

● conduct practical investigations to observe and measure the quantitative relationships of chemical reactions, including but not limited to:

- masses of solids and/or liquids in chemical reactions- volumes of gases in chemical reactions (ACSCH046) - relate stoichiometry to the law of conservation of mass in chemical reactions by

investigating:- balancing chemical equations (ACSCH039) - solving problems regarding mass changes in chemical reactions

(ACSCH046

Chemical Reactions and Stoichiometry

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Chemical reaction, a process in which one or more substances, the reactants, are converted to one or more different substances, the products. Substances are either chemical elements or compounds. A chemical reaction rearranges the constituent atoms of the reactants to create different substances as products.

Investigation Mass Changes in Chemical Reactions 1:

Law Of Conversation of Mass

The Law of Conservation of Mass states that matter cannot be created or destroyed in a chemical reaction, but merely changed from one form to another.

Experiment: Conservation of Mass p142 Chemistry in Focus. (ACSCH046)

Investigation 7.1

For this investigation use the working scientifically scaffold:

● Problem – does mass change during a chemical reaction?

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● Aim – to observe mass before and after a reaction.

● Hypothesis – the mass will not change because atoms are just rearranged in a chemical reaction.

● Equipment – beakers, silver nitrate and sodium chloride, electronic balance

● Risk assessment – skin and eye irritants; appropriate disposal of waste CH11/12-2a, 2b

● Procedure – 10mL each solution, weigh mass before and after, compare masses. CH11/12-3a, 3b

● Results – record masses, photos before and after

● Discussion

● Conclusion – mass before and after were the same, law of conservation of mass.. Validity? Reliability? CH11/12-4a, 4b

Investigation Mass Changes in Chemical Reactions 2:

In small groups, use their previous scientific evidence and critical thinking to design an investigation related to conservation of mass. Using vinegar, baking soda and a balloon. CH11/12-2 CH11/12-3 CH11/12-4 CH11/12-6b

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Chemical Equations

Write a balanced chemical equation and net ionic equation for the reaction between sodium chloride and silver nitrate. (ACSCH039)

1. The Atom revisited.

An atom is made up of electrons, protons and neutrons.

Protons and neutrons are in the nucleus of an atom.

Protons are positively charged, neutrons have no charge.

Electrons have a negative charge.

The Atomic number, Z, of an atom is the number of protons in the nucleus of that

element.

For example, Calcium has 20 protons in its nucleus and has the atomic number 20.

The atomic number also equals the number of electrons in a neutral atom.

The Mass Number, A, of an element is the number of protons plus neutrons of an

atom of that element.

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For example:

Calcium has 20 protons and 20 neutrons, so the mass number of calcium is 40.

The mass number and atomic number are represented as follows:

A Mass Number

X Element

Z Atomic Number

Therefore, Calcium would be represented as:

40

Ca

20

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Writing Chemical Equations

There are essentially three steps to the process:

1. Write the unbalanced equation.

Chemical formulas of reactants are listed on the left hand side of the equation.

Products are listed on the right hand side of the equation.

Reactants and products are separated by putting an arrow between them to show the direction of the reaction.

Write correct chemical formulas for the reactants and products.

2. Balance the equation.

Apply the Law of Conservation of Mass to get the same number of atoms of every element on each side of the equation. Tip: Start by balancing an element that appears in only one reactant and product.

Once one element is balanced, proceed to balance another, and another, until all elements are balanced.

Balance chemical formulas by placing coefficients in front of them. Do not add subscripts, because this will change the formulas.

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3. Indicate the states of matter of the reactants and products.

Use (g) for gaseous substances.

Use (s) for solids.

Use (l) for liquids.

Use (aq) for species in solution in water.

Write the state of matter immediately following the formula of the substance it describes.

● conduct a practical investigation to demonstrate and calculate the molar mass (mass of

one mole) of:- an element- a compound (ACSCH046)

Define Stoichiometry.

The study of the quantitative aspects of formulae and equations is called stoichiometry.

Complete a worksheet of problems regarding mass changes in chemical reactions and write balanced chemical equations for each problem. (ACSCH046) (ACSCH039) CH11/12-6a, 6b

Avogadro’s NumberAvogadro’s number, NA, is the number of atoms in exactly 12 grams of carbon.

Avogadro’s number:

NA = 6.022 x 1023

This means that 12g of carbon has 6.022 x 1023 atoms.

Homework:

Answer the following questions:

1. How many atoms are there in 24.312g of magnesium?

2. How many atoms are there in 20.04g of calcium?

3. How many atoms are there in 127.08g of copper?

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The MOLEA mole of an element is the mass which in grams is numerically equal to the relative

atomic mass.

That is:

12g of carbon = 1 mole

40.08g of calcium = 1 mole

16g oxygen = 1 mole

Therefore:

1 mole of any element = 6.022 x 1023 atoms.

Number of moles = number of particles

Number of particles in 1 mole

n = __N_____NA

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solve problems and analyse information from secondary sources to perform calculations involving Avogadro’s number and the equation for calculating the number of moles of a substance:

Number of moles = ______mass_______Mass of 1 mole

n = _m_ M

Therefore:

20.04g of calcium = ½ mole or 3.011 x 1023 atoms

Homework:

Answer the following questions:

1. How many:

a. Atoms

b. Moles

In

- 19.55g of potassium

- 36.048g of Beryllium

- 8.016g of sulfur

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n mM

Relative Atomic MassThe mass that appears on a periodic table is the Relative Atomic Mass (Ar). It is the

mass of that element relative to the Carbon 12 atom.

For example:

The relative atomic mass of Calcium is 40.08. This means that the Calcium atom is

3.34 (40.08 12) times heavier than Carbon.

Relative Molecular MassThe molecular mass is the mass of a molecule of the compound relative to the mass of

a carbon 12 atom.

The relative molecular mass of a compound is the sum of the relative atomic masses

of the atoms as given in the molecular formula.

To calculate the molecular mass of a substance, add up the mass of all the elements in

the compound.

For example: Methane has the chemical formula CH4 .

This means that there is 1 carbon atom and 4 hydrogen atoms in each molecule.

Therefore, methane has a molecular mass of:

C = 12.01g

H = 1.008g

Molecular mass = (12.01 x 1) + (1.008 x 4) = 16.072g

Therefore:

16.072g methane = 1 mole methane = 6.022 x 1023 molecules

Homework:

Calculate the molecular mass of the following compounds:

1. Carbon dioxide CO2

2. Ethanol C2H5OH

3. Glucose C6H12O6

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Relative Formula MassRemember: the formula for an ionic compound, is expressed as the ratio of elements

that form the ionic lattice.

The formula mass for common table salt, sodium chloride, NaCl is

Atomic mass Na = 22.99

Cl = 35.45

Therefore the formula mass for NaCl = 22.99 + 35.45 = 58.44

Hence, there is 58.44g in 1 mole of NaCl.

Homework:

Calculate the formula mass for the following ionic compounds:

1. magnesium oxide

2. potassium fluoride

3. calcium carbonate

4. sodium sulfate

5. magnesium nitrate

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recount Avogadro’s law and describe its importance in developing the mole concept

Avogadro’s LawAvogadro’s Law states that under the same conditions of temperature and pressure,

the same volume of gases contain equal numbers of molecules. (Or the same number

of moles).

distinguish between empirical formulae and molecular formulae

complete text exercises on empirical and molecular formula. CH11/12-6 deduce why the formulae of all metal-containing compounds are considered empirical.

Empirical FormulaThe empirical formula of a compound is the formula which expresses the atoms in the

compound in the simplest whole number ration. For example:

The empirical formula for magnesium oxide is MgO.

Glucose has the empirical formula CH2O

Molecular FormulaThe molecular formula is the actual number ratio of the atoms in a compound.

The molecular formula is always a multiple of the empirical formula.

Glucose, has the molecular formula - C6H12O6

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describe the contribution of Gay-Lussac to the understanding of gaseous reactions

and apply this to an understanding of the mole concept

Gay-Lussac’s Law of Combining Gas Volumes.

The ratios of the volumes of gases involved in a reaction, if measured at the same

temperature and pressure, are expressed by simple, whole number ratios.

For example:

1. If 100mL of hydrogen reacts with 100mL of chlorine to form 200mL of

hydrogen chloride then we say that: one volume reacts with one volume to

form 2 volumes.

Diagrams p194 text

2. If 100mL of hydrogen reacts with 50mL of oxygen gas to form 100mL steam

(gaseous water) then we say that: two volumes react with one volume to form 2

volumes.

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process information from secondary sources to interpret balanced chemical equations in terms of mole ratios

● explore the concept of the mole and relate this to Avogadro’s constant to describe,

calculate and manipulate masses, chemical amounts and numbers of particles in: (ACSCH007, ACSCH039)

- moles of elements and compounds (n = chemical amount in moles, m = mass in grams, MM = molar mass in gmol-1)

- percentage composition calculations and empirical formulae- limiting reagent reactions

Stoichiometric CalculationsA chemical equation tells us the ratios by weight in which substances react or are

formed in a reaction.

Chemical equations indicate the number of moles if reactants and products in a

chemical reaction.

Stoichiometric calculations involve the calculation of the masses and volumes of

gases of products formed in chemical reactions.

Mass – Mass calculations

In carrying out stoichiometric calculations it is useful to remember the following:

n = m from n = mM M

equation

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Mass

of

Known

Moles

of

Known

Moles

of

Unknown

Mass

of Unknown

For example:

If 26.13g of barium nitrate is reacted with excess potassium sulfate, what is the mass of

barium sulfate formed?

Step 1:

Write a balanced chemical equation:

Ba(NO3)2 + K2SO4 BaSO4 + 2KNO3

1 1 1 2

This shows that 1 mole of barium nitrate reacts with 1 mole of potassium sulfate to produce 1

mole of barium sulfate and 2 moles of potassium nitrate.

Calculations:

1. Mass of known = 26.13g

2. Moles of known = n = mM

M = formula (molecular mass)

m = mass of reactant

Formula mass = sum of atomic masses of the elements in the compound

Barium = 137.3

Nitrogen = 14

Oxygen = 16

Therefore for the compound Ba(NO3)2

the formula mass = 137.3 + (14 x 2) + (16 x 3 x 2) = 261.3g

n = m = 26.13 = 0.1 mole

M 261.3

Moles of unknown from stoichiometric ratio = 0.1 mole

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M (formula mass of barium sulfate) BaSO4 =

Barium = 137.3g

Sulfur = 32.1g

O = 16g

Therefore for the compound BaSO4

the formula mass = 137.3 + 32.1 + (16 x 4) = 233.4g

Mass of unknown

n = m M

m = n x M

= 0.1 x 233.4

= 23.34g

Example 2:

What mass of sodium carbonate is needed to react with 14.1g of nitric acid, and how much

sodium nitrate is formed?

Follow the processes as in example 1.

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process information from secondary sources to investigate the relationship between the volumes of gases involved in reactions involving a metal and relate this to an understanding of the mole

Mass - Volume CalculationsThe calculations for mass – volume calculations follows the same processes as in

mass calculations.

n = m from n = mM M

equation

Restating Avogadro’s Law in terms of volumes of gases:

Equal numbers of molecules of different gases occupy the same volume, at the same

temperature and pressure.

Thus one mole of any gas at the same temperature and pressure occupies the same

volume.

When calculating volumes of gases standard conditions are used. These are:

One mole of any gas occupies:

1. At Standard Temperature and Pressure, (STP), 0C and 100 kPa

1 mole of any gas occupies = 22.71 Lmol-1

2. At Standard Laboratory Conditions, (SLC), 25C and 100 kPa

1 mole of any gas occupies = 24.79 Lmol-1

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Mass

of

Known

Moles

of

Known

Moles

of

Unknown

Volume

of

Unknown

For example:

1. What is the volume of hydrogen gas produced, at STP, when 4.86g of magnesium is

reacted with excess Hydrochloric acid.

Step 1:

Write a balanced chemical equation:

Mg + 2HCl MgCl2 + H2

1 2 1 1

This shows that 1 mole of magnesium reacts with 2 moles of hydrochloric acid to produce 1

mole of magnesium chloride and 1 mole of hydrogen.

Calculations:

2. Mass of known = 4.86 g

Moles of known = n = mM

M = formula (molecular mass)

m = mass of reactant

Atomic mass magnesium =

Magnesium = 24.3g

Number of moles of magnesium=

n = m = 4.86 = 0.2 moleM 24.3

Moles of hydrogen produced from stoichiometric ratio = 0.2 mole

When calculating volume of gas produced the following equation is used:

Volume of gas produced = number of moles x molar volume (MV)

V = n x MV (In this example molar volume is at STP = 22.4 Lmol-1)

V(hydrogen) = 0.2 x 22.4 (mol x Lmol-1)

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= 4.48 L

Limiting Reagent Reactions

When given values for the reactants, either as a mass or a volume and concentration of solution, then the reactant that limits how far the reaction can proceed must be identified. This is called the Limiting Reagent. The other reagent (reactant) is in excess.

● conduct an investigation to determine that chemicals react in simple whole number ratios by moles

Investigation: The Molar Mass of an Element – magnesium and oxygen reaction

2. Concentration and Molarity

● conduct practical investigations to determine the concentrations of solutions and

investigate the different ways in which concentrations are measured (ACSCH046, ACSCH063)

● manipulate variables and solve problems to calculate concentration, mass or volume

using:- c=n/v (molarity formula) (ACSCH063) - dilutions (number of moles before dilution = number of moles of sample after

dilution)

Quantitative methods of expressing solution concentrations are needed so that the

precise quantity of solute in a given volume of solutions can be determined.

The use of solution concentrations is important in chemical analysis, chemical

manufacturing, the prescription of drugs and medicines , and other situations

requiring quantitative work.

Four methods of expressing solution concentrations are:

1. grams per litre

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2. percentage by composition

3. parts per million

4. moles per litre

1. Percentage by mass:

Percentage by mass is used extensively in industry

Percentage by mass = mass (solute) x 100 %mass (solution)

For example:

What would the percentage by mass be by mixing 10g of NaCl in 90g of water give.

(Express answer as %m/m)

2. Percentage by Volume

Percentage by volume is used for mixtures of liquids, e.g. alcoholic drinks.

Percentage by volume = volume (solute) x 100 %volume(solution)

For example:

What would the percentage by volume be by mixing 5 mL of ethanol in 95 mL of water

give. (Express answer as %v/v)

3. Mass per Volume

In pharmacy and medicine mass per volume is normally used.

Mass per volume of solution (m/v) is used to measure the blood alcohol level of drivers. A

blood level of 0.02 refers to 0.02 g/100 mL of blood.

For example:

An 18 year old driver has 5 L of blood in her body. What would be the blood alcohol

level if they consumed 1.5g of alcohol. (Express answer as %m/v)

Percentage by mass/volume = mass (solute) x 100 %volume(solution)

4. Parts per Million

Parts per million (ppm) are often used to measure environmental pollutants.

The ppm can be m/m or v/v (usually used for gases).

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Water pollution measurements are often reported as milligrams per litre, (mg/L)

which is usually considered equivalent to ppm; 1L of water has a mass close to 1 kg,

thus:

1mg/L = 1mg/kg = 10-3g/103g = 1/106 = 1ppm

Investigation – concentrations of solutions 1:

In small groups, follow a procedure to use gravimetric analysis to determine the concentration of potassium iodide in a solution. Evaluate and manage risks, ensure and evaluate accuracy, apply quantitative processes and evaluate the quality of the data collected. CH11/12-3a,3 b, CH11/12-4

Calculate the concentration in grams per litre and per cent composition.

Investigation – concentrations of solutions 2:

In small groups, follow a procedure to dilute food colouring in water from 100% composition to 0.001% composition. CH11/12-3a,3 b

Convert the percentage composition ratios to parts per million (ppm). CH11/12-4b

● conduct an investigation to make a standard solution and perform a dilution

Manipulate variables and solve problems to calculate concentration, mass or volume using:

– c=n/v (molarity formula) (ACSCH063)

– dilutions (number of moles before dilution = number of moles of sample after dilution) CH11/12-4b, CH11/12-6

Standard Solutions

A solution of accurately known concentration is called a standard solution.

A volumetric flask is used to make up the standard solution.

For a chemical to be suitable to prepare as a standard solution, it must:

be a water soluble solid

 have high purity - usually Analytical Reagent (A.R.) grade

 have an accurately known formula

 be stable in air, i.e. it does not lose or gain water or react with oxygen or

carbon dioxide in air.

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The solution is prepared by:

accurately weighing a calculated amount of solid dissolving it in de-ionised water

transferring all of the dissolved solid to a volumetric flask

adding water to the flask to prepare a fixed volume of solution.

E.g. deliquescent

The concentration is calculated in mol L-1

Dilutions

The equation used for dilution calculations is:

C1V1 = C2V2

Where

C1 = Initial concentration

V1 = Initial volume

C2 = Final concentration

V2 = Final Volume

In groups, conduct an investigation to make a standard solution and perform a dilution. CH11/12-2, CH11/12-3, CH11/12-4, CH11/12-6

Prepare a 250mL standard solution of 0.010mol/L potassium dichromate. Devise and perform an investigation to accurately dilute the standard solution to produce 20mL of 0.5g/L solution.

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3 Gas Laws

● conduct investigations and solve problems to determine the relationship between the Ideal Gas Law and:

- Gay-Lussac’s Law (temperature)- Boyle’s Law- Charles’s Law- Avogadro’s Law

When gases are produced in chemical reactions, it is often more important to know

the volume of gas that is produced during a reaction than its mass.

At STP, 1 mole of any gas will occupy a volume of 22.41 L.

At non-STP conditions, the ideal gas law must be used.

STP is defined as a temperature of 273.15 K (0 °C) and an absolute pressure of

exactly 105 Pa (100 kPa).

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Gay-Lussac’s Law of Combining Volumes

The ratios of the volumes of gases involved in a reaction, if measured at the same

temperature and pressure, are expressed by simple, whole number ratios.

For example:

1. If 100mL of hydrogen reacts with 100mL of chlorine to form 200mL of hydrogen

chloride then we say that: one volume reacts with one volume to form 2 volumes.

Investigation – Gas Laws 1:

Use a stoppered syringe to investigate how pressure relates to volume (Boyle’s Law) and temperature relates to volume (Charles’s Law).

Control variables, collect data and graph results. CH11/12-4a, CH11/12-6a

Make conclusions about the relationship between temperature and volume and between volume and pressure in gases. CH11/12-6b

Check the validity of the conclusions with Stoichiometry and Gases – it’s what you don’t see that counts!

When gases are produced in chemical reactions, it is often more important to know the volume of gas that is produced during a reaction than its mass.

At STP, 1 mole of any gas will occupy a volume of 22.414 L.

At non-STP conditions, the ideal gas law must be used.

secondary sources. CH11/12-3c

Use the following simulations to further understand Boyle’s and Charles’s Laws:

http://www.grc.nasa.gov/WWW/K-12/airplane/boyle.html

http://www.grc.nasa.gov/WWW/K-12/airplane/glussac.html

Describe the contribution of Gay-Lussac to the understanding of gaseous reactions and apply this to an understanding of the mole concept.

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Boyles Law.For a given quantity of gas at a constant temperature, the product of its volume, V, and its

pressure, P, is constant.

PV = k

Volume is inversely proportional to pressure. If pressure is doubled the volume is halved. If

the pressure decreases to one fifth of its initial value, the volume increase to 5 times its initial

value.

An alternative form of Boyles Law is:

P1V1 = P2V2

Worked example 9.3 and Try it yourself p204

Investigation 9.3

Charles Law

Charles Law states that for a fixed quantity of gas at constant pressure, volume increases

linearly with temperature.

Introduce Kelvin temperature scale

When data as show ae plotted against the Kelvin scale, all lines pass through the origin.

Fig 9.9

Then V = kT

The most common form of this equation is:

V1 = V2

T1 T2

Combining Boyles and Charles Gas Laws

Boyles Law and Charles law can be combined into the following equation. Pressure and

volume can be in any units by temperature must be in Kelvin

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P1V1 = P2V2

T1 T2

Pressure and Temperature Relationship

If the volume of a sample of gas is held constant then the above equation becomes;

P1 = P2

T1 T2

Recount Avogadro’s law and describe its importance in developing the mole concept.

Avogadro's law states that, "equal volumes of all gases, at the same temperature and

pressure, have the same number of molecules".

Alternatively:

The volume of a gas is proportional is proportional to the number of moles present constant

temperature and pressure and is independent of the nature of the gas.

This can be written as:

V = k n at constant T and P

The Ideal Gas Laws (Combining the three gas laws)

This is useful when dealing with real life situations, e.g. scuba diving, where all of the conditions of temperature, pressure, quantity and volume of gas can be varied.

This can be written as:

PV = nRT

Where R = the universal gas constant and is the same for all gases.

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Investigation:

Conduct the crushing can demonstration to show the relationship between pressure and temperature in gases. CH11/12-2a, CH11/12-2c, CH11/12-3a,

Research the Gas Laws explain the movement of the gas molecules in the crushing can demonstration.

Research to show that Charles’s Law, Boyle’s Law, Gay-Lussac’s Law and Avogadro’s Law are summed up in CH11/12-6the Ideal Gas Law: PV = nRT. CH11/12-3c, CH11/12-6

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