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VCE Chemistry -Unit 3 -2013 Area of Study 1- Chemical Analysis Quantitative analysis = Analysis to find out: _________________________________________ ______________________________________________________________ _____________. Qualitative analysis = Analysis to find out: _________________________________________ ______________________________________________________________ _____________. Measuring Water Content Water is a component of most consumer products whether it is added or has occurred naturally. Measuring the water content in a product can be achieved by heating a sample of the product to evaporate the water and then comparing the change in mass. Steps for measuring water content: 1 | Page Assessment = an extended investigation on the analysis of a commercial lawn fertilizer. 3 experimental activities that focus on analysing and determining the amounts of sulphates, nitrogen, iron and phosphates in fertilizer. A summary and conclusion of the findings of the investigation as a whole.

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Page 1: VCE Chemistry Unit 3 - fitzyatnorwood - home · Web viewVCE Chemistry -Unit 3 -2013 17 | Page Area of Study 1- Chemical Analysis Assessment = an extended investigation on the analysis

VCE Chemistry -Unit 3 -2013

Area of Study 1- Chemical Analysis

Quantitative analysis= Analysis to find out: _________________________________________

___________________________________________________________________________.

Qualitative analysis = Analysis to find out: _________________________________________

___________________________________________________________________________.

Measuring Water Content

Water is a component of most consumer products whether it is added or has occurred naturally.

Measuring the water content in a product can be achieved by heating a sample of the product to evaporate the water and then comparing the change in mass.

Steps for measuring water content:

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Assessment = an extended investigation on the analysis of a commercial lawn fertilizer.

3 experimental activities that focus on analysing and determining the amounts of sulphates, nitrogen, iron and phosphates in fertilizer.

A summary and conclusion of the findings of the investigation as a whole.

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VCE Chemistry -Unit 3 -2013

Calculations by mass

When given the mass of a particular substance, the amount (in mol) can be calculated by:

Where n=

m=M=

Molar volume of gas

The volume of one mole of gas will depend on the temperature and pressure.

At Standard Temperature and Pressure (STP, 0°C and 1atm) the molar volume, Vm, is 22.4.

At Standard Laboratory Conditions (SLC, 25°C and 1atm) the molar volume, Vm, is 24.5.

To calculate the amount of a gas at these conditions we use the formula:

Where n =

V=

Vm=

Example: Calculate the amount of sulfur dioxide in 10.0L of the gas measured at STP.

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VCE Chemistry -Unit 3 -2013

Amounts of gases

When the substance is in a gaseous state we use the General Gas Equation to calculate the amount of gas present.

Where P= V=

n= R=

T=

Example: A steel cylinder with a volume of 30.0L is filled with nitrogen gas to a pressure of 2atm at 25°C. What mass of nitrogen does the cylinder contain?

Finding the Composition of a Compound

Using chemical analysis we can find the simplest whole number ratio of atoms in a compound.

This is called the empirical formula.

Empirical formulas are determined experimentally, usually by finding the mass of each element in a given compound.

Example: Chemical analysis of an oxide of sulfur present in the gaseous emissions from a factory shows that it contains 40% sulfur. Find its empirical formula.

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VCE Chemistry -Unit 3 -2013

Molecular formula

The molecular formula of a compound gives the actual number of atoms present in a molecule of a compound.

For example, the empirical formula of glucose is CH2O. Its molecular formula is 6 times larger – C6H12O6.

A molecular formula can be deduced from the empirical formula of a compound if the molar mass is known.

Calculating masses of reactants and products

Using a balanced equation we can use ratios to determine the mass of product produced or the mass of reactants consumed.

The following steps outline this process:

Write balanced equation. Calculate the number of mol from the known quantities Find the ratio of amounts. Calculate n of unknown. Calculate the unknown mass.

Example: Calculate the mass of lead iodide that can be made from a reaction using 30.0g of potassium iodide and lead nitrate.

Calculations involving excess reactants

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VCE Chemistry -Unit 3 -2013

In most reactions, one of the reactants is usually in excess.

When doing calculations on these reactions, the calculations must be based around the reactant that is totally consumed.

This reactant is called the limiting reactant.

Example: Calculate the mass of silver bromide that can be formed if a solution containing 15.0g of silver nitrate is allowed to react with a solution containing 10.0g of calcium bromide.

Finding the composition of a mixture :-GRAVIMETRIC ANALYSIS

When trying to determine the amount of a particular ingredient in a substance we use a process called Gravimetric Analysis.

The substance in question can be isolated by dissolving it water and filtering away all excess solids.

The substance can then be isolated by reacting it with a chemical that will form a precipitate.

The precipitate can then be filtered and washed thoroughly to remove any excess ions

The precipitate can then be dried and weighed to determine the mass of the product formed from the reaction.

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VCE Chemistry -Unit 3 -2013

Draw a flowchart showing the steps of Gravimetric Analysis

When doing a Gravimetric Analysis it is important that you make the correct choices when attempting to form a precipitate.

The precipitate should:

Have a known formula Have low solubility Be stable when heated Not form other precipitates with other ions that may be present

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VCE Chemistry -Unit 3 -2013

Example: A 7.802g sample of baby cereal was blended with water and filtered. The solution obtained was mixed with excess silver nitrate solution, causing silver chloride solution to precipitate. The precipitate was collected by filtration, dried and weighed. A mass of 0.112g was obtained. What is the percentage of salt in the baby food?

Full equation for the precipitation reaction is:

Concentration

Concentration is how much solute dissolved in a solvent.

Molar concentration or molarity is how many mole of solute in one L of solvent given in the units mol L-1(moles per litre) or M.

When given the concentration of a particular substance, the amount can be calculated by the formula:

Where n= c= V=

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VCE Chemistry -Unit 3 -2013

Example: Calculate the molarity (concentration) of a 250mL solution containing 0.5mol of HCl.

When converting molarity to gL-1 you simply multiply by the molar mass

Eg. A 2.0M solution of HCl (molar mass of HCl is 1+ 35.5 = 36.5)

2.0 x 36.5 = 73gL-1

When converting gL-1 to molarity you simply divide by the molar mass

Eg. A solution containing 18.25gL-1 of HCl

18.25/36.5 = 0.5M

Standard Solutions

Solutions called Primary Standards can be to test the contents of unknown substances.

For Primary standards to be useful they should:

Have a known formula Be readily obtainable in pure form Be easy to store

Draw a diagram showing the steps to prepare a Primary Standard

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VCE Chemistry -Unit 3 -2013

Volumetric Analysis

Volumetric analysis is used to find the concentration of an unknown solution by reacting a measured volume of a standard solution with a measured volume of the unknown solution until it reaches equivalence point (just reacted completely or neutralised).

An acid-base indicator is used to show when the equivalence point is reached.

Draw a diagram showing the steps to a Volumetric Analysis

Figure 3.6 A pipette must always be filled using a safety pipette filler. When the pipette is filled correctly, the meniscus sits on the graduation line.

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VCE Chemistry -Unit 3 -2013

Analysing Acids and Bases

Acids and bases can be defined in the following ways:

Acids are proton donors Bases are proton acceptors Acid-Base reactions involve the transfer of a proton from an acid to a base.

Consider the following Acid-Base reaction:

HCl is a strong acid as it ionises completely in water to for H3O+ and Cl-.

During the reaction there is a proton (H+) transfer between the HCl and the H2O.

HCl acts as the acid and donates the proton to form Cl-.

H2O acts as a base and accepts the proton to form H3O+.

Choosing Indicators

The choice of an appropriate acid-base indicator is very important when conducting a volumetric analysis.

An indicator will change colour at a particular pH. This is called the end point.

The end point of the indicator must closely match the equivalence point of the acid-base reaction.

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Figure 3.7 The volume of solution in a burette is read going down from the top of the burette.

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VCE Chemistry -Unit 3 -2013

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Figure 4.4 pH curves showing change of pH during a titration of a a strong base with a strong acid, and b a weak base with a strong acid. Phenolphthalein, which changes colour in the pH range 8.2–10, givesa sharp end point in a but a broad end point in b. Methyl orange, which changes colour between pH 3.1 and 4.5, would be a more suitable indicator for the second titration.

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VCE Chemistry -Unit 3 -2013

Example: A commercial concrete cleaner contains concentrated hydrochloric acid. A 25.0mL volume of cleaner was diluted to 250.0mL in a volumetric flask.

A 20.0mL aliquot of 0.4480M sodium carbonate solution was placed in a conical flask. Methyl Orange indicator was added and the solution was titrated with the diluted cleaner.

The indicator changed permanently from yellow to pink when 19.84mL of the cleaner was added. Calculate the concentration of hydrochloric acid in the concrete cleaner

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Figure 4.7 pH Change in pH during a titration of:

a a strong acid with a strong base; b a strong acid with a weak basec weak acid with a strong base; d weak acid with a weak base.

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VCE Chemistry -Unit 3 -2013

Back Titrations

Back titrations are used to find the end point of very weak acids and bases that do not give definite colour changes.

The weak acid is first reacted with excess base

A titration can then be carried out to determine the amount of base that didn’t react.

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VCE Chemistry -Unit 3 -2013

The difference between the amount that reacted and the original amount of excess base added will be the amount that reacted with the weak acid.

The amount of the weak acid can then be determined.

Complete a flowchart containing diagrams showing a back titration used to find the concentration of ammonium in fertilizer

Example: A 1.50g sample of fertiliser was boiled with 25.00mL of 0.9987M sodium hydroxide solution. Once no further ammonia gas was evolved from the mixture, it was cooled and titrated with 0.2132M hydrochloric acid, using phenolphthalein as an indicator. A titre of 19.78mL was required.

Calculate the percentage of ammonium ions in the fertiliser.

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VCE Chemistry -Unit 3 -2013

Analysing Oxidants and Reactants

Oxidation/Reduction reactions (REDOX) are two reactions that occur simultaneously.

The oxidation part of a redox reaction is where a substance loses electrons (OIL – Oxidation Is Loss)

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VCE Chemistry -Unit 3 -2013

The reduction part of a redox reaction is where a substance gains electrons (RIG – Reduction Is Gain)

Example: Consider the Redox reaction:

The magnesium will undergo oxidation by losing electrons to form magnesium ions:

The oxygen gas will undergo reduction by gaining electrons to form oxide ions:

Classifying Redox Reactions

Redox reactions can be identified by observing a change in Oxidation Numbers of the substances involved.

Oxidation Number Rules

Oxidation numbers are determined using the following rules:

1. Free elements have an oxidation number equal to 0. Eg Na(s), C(s), Cl2(g).2. In ionic compounds the oxidation number s equal to the charge on the ion. Eg

CaCl2: Ca2+=+2 and Cl- = -1.3. Oxygen usually has an oxidation number of -2 and hydrogen has +1 but there are

a few exceptions.4. The sum of oxidation numbers in a neutral compound is 0 and in a polyatomic

ion is equal to the charge of the ion.

Using Oxidation Numbers to Identify Redox Reactions

By defining oxidation numbers for the atoms involved in a reaction we can look for increases and decreases in oxidation numbers.

An increase in oxidation number means the element has undergone oxidation.

A decrease in oxidation number means the element has undergone reduction.

Example:

For the reaction:

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VCE Chemistry -Unit 3 -2013

Carbon in the carbon monoxide has gone from +2 to +4 which means that oxidation has occurred.

Oxygen in the oxygen gas has gone from 0 to -2 which means that reduction has occurred.

As both oxidation and reduction have occurred the reaction is a redox reaction.

Writing Half Equations

Although most half equations are quite easy to write, some involving polyatomic ions can be more difficult.

The following steps will make balancing these half equations easier:

1. Balance all elements except O and H in the half equation.2. Balance the O atoms by adding water. 3. Balance the H atoms by adding H+ ions. 4. Balance the charge by adding electrons (e) and then add states.

Example: A green solution containing Fe2+ ions is mixed with a purple solution containing MnO4- ions. Fe3+ and Mn2+ ions are formed.

Write a balanced equation for this reaction.

Volumetric Analysis of Redox Reactions

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VCE Chemistry -Unit 3 -2013

Volumetric analysis to determine unknown concentration or mass can be performed in the same way that it is with acid-base reactions.

Equivalence point of most redox reactions will be indicated by a colour change in the reacting solutions.

Volumetric analysis involving redox reactions are used to determine the composition of substances such as Fruit Juice, Wine, Bleach and Hair Bleach.

Draw a flow chart of a redox titration

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