Oxidation Number

Oxidation Number

In chemistry, the oxidation number is used to keep track of the positive or negative character of atoms or ions. When electrons are removed or shifted away from an atom during a chemical reaction, the atom is given a more positive oxidation number.

When electrons are gained or shifted toward an atom during a reaction, the atom is given a more negative oxidation number. The actual numerical value of the oxidation number depends on the number of electrons shifted or transferred.

For simple ions the oxidation number is equal to the charge on the ion. In NaF, the sodium ion has an oxidation number of +1 because it lost one electron. The fluoride ion has an oxidation number of -1 because it gained one electron. The elements of Group I of the periodic table have oxidation numbers of +1 since they have one valence electron which they can lose. In compounds involving only two elements (binary compounds) the elements of Group VII of the periodic table usually have oxidation number of -1 since they have seven valence electrons and can gain one more electron to complete a stable octet.

In compounds the Group II elements have oxidation numbers of +2 and the elements of Group III often have oxidation numbers of +3. In binary compounds the elements of Group VI usually have oxidation number of -2 and the elements of Group V can have oxidation number of -3

In covalent substances, the more electronegative element is assigned a negative oxidation number, and the less electronegative element is assigned a positive oxidation number. In HCl, the hydrogen atom has an oxidation number of +1 due to a shift of the valence electron of the hydrogen atom towards the more electronegative chlorine atom. The chlorine atom has an oxidation number of -1.

In H2O, the oxygen atom is more electronegative than the hydrogen atoms. Each hydrogen atom has an oxidation number of +1 and the oxygen is -2. In NH3 the oxidation number of each hydrogen is +1 and the oxidation number is -3.

Many elements have more than one oxidation number depending on their various compounds. Iron has an oxidation number of +2 in FeCl2 and +3 in FeCl3.

Tin has an oxidation number of +2 and +4 in SnCl2 and SnCl4, respectively.

 

USUAL OXIDATION NUMBERS OF SOME COMMON ELEMENTS

ELEMENTS PERIODIC TABLE PLACEMENTS OXIDATION NUMBER
H   +1 (except -1 with metals)
Li Group I +1
Na Group I +1
K Group I +1
Ag Transition Metal +1
Mg Group II +2
Ba Group II +2
Ca Group II +2
Ni Transition Metal +2
Zn Transition Metal +2
Pb Group IV +2
Al Group III +2
Cl Group VII -1 (with metals and many nonmetals)
Br Group VII -1(with metals and many nonmetals)
I Group VII -1 (with metals and many nonmetals)
F Group VII -1
O Group VI -2 (except -1 in peroxides)
S Group VI -2 (also +4 and +6)
Hg Transition Metal +1 and +2
Cu Transition Metal +1 and +2
Fe Transition Metal +2 and +3
Co Transition Metal +2 and +3
Sn Group VI +2 and +4

 

Types of Solutions

Solutions

When a lump of sugar is placed in a cup of tea, the sugar disappears but it remains unchanged. We know that sugar is unchanged when it mixes with the water of the tea because we can still taste it. This is an example of a solution. The sugar has dissolved in the hot water of the tea.

A solution is a homogenous (same through-out) mixture of two or more substances. The composition of a solution can vary within certain limits. Two words that we should consider are solvent and solute. Generally we refer to the substance that is present in the larger quantity as the solvent. The solute is present in smaller amounts. In our example the sugar is the solute and the water is the solvent. The sugar dissolves in the water. In solutions where one substance is a liquid the other substance is a gas or solid, the liquid is usually the solvent.

Types of Solutions

Liquid solutions are made by dissolving solids, liquids or gases in liquids. Sugar in water is an example of a solution of a solid dissolved in a liquid. Salt water is another example of a common type of solution.

Gaseous solutions are made by dissolving a gas in another gas. All gases mix in all proportions to produce a homogenous solution. Air is an example of a gaseous solution, which is a solution of oxygen and many other gases dissolved in nitrogen.

A solid solution is formed when one solid substance is mixed with another solid substance to produce a homogenous mixture. Many alloys are solid solutions. For example brass is a solid solution in which zinc atoms have been mixed into solid crystal of copper atoms.

The enthalpy change for a reaction between a strong acid and a strong alkali

Aim / Objective: 

To determine the enthalpy change for a reaction between a strong acid and a strong alkali.

Introduction:  Enthalpy is defined as the total energy in a system.  The change in energy ∆H can be positive in heat absorbing (endothermic reactions) or negative in heat releasing (exothermic reactions). This experiment focuses on one form of enthalpy change which is enthalpy of neutralization (∆Hn).  Enthalpy of Neutralization is the enthalpy change observed when one mole of water is formed when a base reacts with an acid in a thermodynamic system.

The literature standard enthalpy for a strong acid-base reaction is -57.1kJ/mol. For weak acids and bases the heat of neutralization is different as they are not fully dissociated and hence some heat will be absorbed.

Materials/ Apparatus:

1.0M HCl solution, 1.0M NaOH solution, 2 measuring cylinders, 2 Styrofoam cups, 2 beakers, 1 thermometer, 1 glass stirring rod.

Method / Procedure:

  1. Use a measuring cylinder to measure 50cm3 of sodium hydroxide (NaOH) and pour into a Styrofoam cup.
  2. Use a thermometer to measure the temperature of the NaOH. Record the reading.
  3. Use a measuring cylinder to measure 50cm3 of hydrochloric acid (HCl) and pour into a Styrofoam cup.
  4. Use a thermometer to measure the temperature of the HCl. Record the reading.
  5. Mix the contents of both cups in a beaker and stir the contents using the glass stirring rod.
  6. Take the temperature after stirring the mixture then record the reading.
  7. Repeat steps 1 to 6 TWO times and then tabulate the results.

Suggested Results:

Experiment number Initial temperature / °C Final temperature / °C
NaOH HCl NaOH+ HCl
1 32 33.5 46
2 30 30.5 46
3 30 30.5 46

Discussion / Calculations:

Heat released = mcΔT

  1. Calculate mass

1000g = 1kg

1000g ÷1000

x = 1kg

50cm3 of acid and 50cm3 alkali = 100cm3 = 1kg

  1. Average temperature = (32+30+30+33.5+30.5+30.5) / 6

= 186.5/6

= 31.08°C

  1. Change in temperature (ΔT) = final temperature – initial temperature

= (46 – 31.08) °C

= 14.92°C

Change °C to K = 14.92 + 273 =287.92

  1. Specific heat capacity = 4.187JK-1kg-1
  1. Heat energy released = mcΔT

= 1kg x 4.187JK-1kg-1 x 287.92K

= 1205.52104J

  1. Enthalpy change = mcΔT / number of moles

50cm3 of acid & base contains (2/1000) x 50

=0.1 moles

0.1 moles H20 = 727.92J

1.0 moles H2O = 727.92J / 0.1 moles

= 7279.2J

Change from J to kJ

7279.2J / 1000 = 7.279kJ

Write the molecular and ionic equation for the reaction

NaOH(aq)   +  HCL(aq)                NaCl(aq) +  H2O(l)

H+(aq)   +    OH(aq)                 H2O(l)

Explain whether the reaction was exothermic or endothermic?

Heat was lost from the mixture to the environment can be said to be exothermic.

What was the use of the Styrofoam cups?

They were used to minimize heat loss of the mixture to the environment.

Redox Reactions

Metals in Contact With Aqueous Solutions

When a metal rod is placed in a beaker containing a solution of its own metal ions, two solutions can arise.

1. Metal atoms leave the rod and become metal ions in solution

M = M + e

2. Metal ions leave the solution and become metal atoms on the surface of the rod.

M + e = M

In the first case the electrons are released when the metal ions are formed stay on the rod, so the rod carries a negative charge. We say it has a negative potential. Zinc and Magnesium are examples of metals that behave in this way.
Screen shot 2014-08-01 at 8.17.37 PM
In the second instance, electrons are attracted out of the rod into the solution so the rod carries a positive charge. We say it has a positive potential. Copper and Silver are examples of metals that behave in this way.

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Half Cells
The examples shown above each exhibits a half cell. This is because any two of them can be joined to form a whole cell.
This figure shows an attempt to link tow half cells to make a whole cell.
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– The electrons repel each other from election rich zinc to electron deficient copper
– Ions (electrons) transfer negative charge through the electrolyte in the salt bridge without the two solutions mixing together.
– The salt bridge completes the circuit. This is called a Daniell Cell. The salt bridge is a tube containing (or a strip of filter paper soaked in) an aqueous solution of a good electrolyte example. Potassium Nitrate KNO3 or Potassium Chloride KCl. The salt bridge is usually made up with Potassium Nitrate solution because the salt does not react with other ions commonly used in electro chemical cells.

This enables charge to be transferred without the solution zinc sulphate and copper sulphate being mixed.
NB. The electrons pass through the wire while the ions move along the salt bridge.
Electrolytes are solutions, which decompose at the electrode when an electric current passes through. However, there is no decomposition in the salt bridge.

Functions Of The Salt Bridge
– One obvious function of the salt bridge is to complete the circuit without allowing the two solutions to mix.
– It has another important function. Without it the zinc half-cell would slowly become positively charged as electrons leave it, and the copper half cell would become negatively charged. With the salt bridge in place ions from the salt bridge are able to move in and out of the solutions to neutralize any build up of charge.

Non-Metals in Contact With Aqueous Solutions

It is convenient to make half cells of non-metals such as chlorine and hydrogen. Hydrogen half cell would consist of a layer of hydrogen atoms in contact with hydrogen ions in solution just as a copper electrode consists of copper atoms with in contact with copper ions in solutions.
A layer of immobile hydrogen atoms is created by solidifying it at negative 260 degrees celcius which will be in contact with hydrogen ions in solution.
Eg. HCL, H+, Cl –
Platinum is used to hold the layer of hydrogen atoms. Platinum absorbed molecules of gases onto its surface that is, it holds them in place when they come in contact with the metal.
In a commercial hydrogen electrode the platinum is not shinny but black. The surface of the electrode is porous and pitted. This creates a very large surface area for the hydrogen to be absorbed into.

The Standard Hydrogen Electrode (S.H.E)

The electromotive force (E.M.F) of the daniell cell is the maximum potential difference between the potential of the zinc half-cell and the potential of the copper half -cell. The question is how do we find the potential of each half-cell individually?
To do this we use the standard hydrogen electrode, which is given a potential of zero volts. When it is connected to another half cell the E.M.F between the S.H.E and the second half-cell is equal to the potential of the second half-cell.
The S.H.E is actually the standard hydrogen half-cell.
Screen shot 2014-08-01 at 8.56.39 PM

It consist of hydrogen gas bubbling over a platinum electrode immersed in a solution of hydrochloric acid supplying hydrogen ions.
The Reaction that takes place is    2 H+ + 2 e- = H2
The platinum electrode simply produces an inert metal connection between hydrogen gas and hydrogen ions in solution.

When the S.H.E is being used to establish the potential of a Redox system, standard conditions must apply. These are:
– The hydrogen must be at a pressure of 1ATM
– The condition of the hydrogen ions must be 1mol dm3
– The temperature must be 25 degrees Celsius (298k)

Under these conditions the potential is defined as exactly zero volts.

Rate Of Reaction

The rate of reaction or sometimes called the speed of reaction refers to how slow or fast a reaction takes place. The following concepts are associated with reaction rate and that you will see often.

Reaction Rate

Reaction rate is defined as the change in concentration of a substance divided by the time taken for that change to take place it is measured in mol dm-3 s-1 .

Rate constant

Proportionality constant in the relationship between reaction rate and reactant concentration

Rate = k [A] [B]

k = rate constant

Order of Reaction

Reactions have an order with respect to each reactant. E.g.The rate equation has been determined experimentally.

Rate = k [NO] O3

The reaction is first order with respect to NO, which means that the rate depends on the concentration of NO raised to the first power [NO]1.

It is also first order with respect to O3and that is [O3]1 . The sum of the individual orders gives the overall reaction order.

Rate – Determining Step

This is the slowest step in a reaction mechanism and is the one that determines the overall rate.

Half Life

Half life of a reaction is the time required for a reactant to reach half of its original concentration.

Activation Energy

Activation Energy Ea, is the minimum collision energy required to activate the molecules into a state from which reactant bonds can change into product bonds.
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Collision Theory

For a chemical reaction to occur the reacting molecules must collide with each other.
This forms the basis of the collision theory of chemical kinetics.
Basically, this theory states that the rate of a reaction is proportional to the number of collisions occurring each second between reacting molecules.
Rate is prpportional to Number Of Collisions divided by Seconds

It is important that you make note of the following:
– Molecules will only react if they collide with each other.
– Reactions will occur if there is enough energy in the collision
– Increased concentration increases the likely hood of collision, which increases reaction rate.
– Increased temperature increases the average energy of collisions, which increases the reaction rate.
– NB. The molecules must collide in the correct orientation to produce a chemical change.
Screen shot 2014-08-01 at 7.16.30 PM This reaction cannot produce any net chemical change
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This collision that can lead to a net reaction is known as an effective collision.

 

NB. Molecules must possess enough energy when they collide to produce a chemical change.
When two slow moving molecules collide, their electron clouds cannot inter-penetrate much and they just bounce on each other, chemically unchanged.
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When fast moving molecules collide atoms approach each other much more closely and their electron clouds inter-penetrate. This can lead to bond breaking and bond making. The net change here is  A B + C = A + B C

Catalysis
A catalyst is something added to a reaction that increases its rate, but does not change in concentration. This means the same amount of catalyst remains after the reaction as before.
Catalyst increases reaction rate by lowering the activation barrier for the reaction without altering the products of the reaction.
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Enzymes

A vast variety of significant catalysts are provided by enzymes. They occur in living systems and allow reactions to occur at relatively low temperatures like 37 degrees Celsius) our body temperature) and at relatively low pressure.
They are organic compounds that catalyze the reactions involved in the vital processes within animals and plants.
Enzymes are proteins, long chains of amino acids linked together by peptide bonds. Enzymes are involved in almost every reaction in your body. For example they help you to digest food and protect you from dangerous waste products that form in your body.
Enzymes are often used in industrial processes to catalyze biochemical reactions.

 

 

 

 

Gravimetric Methods of Analysis

Gravimetric Methods of Analysis

Gravimetric analysis is based on the measurement of the weight of the analyte or of a compound of known composition that contains the analyte. There are two general types of gravimetric methods: precipitation and volatilization.

 

Precipitation

Gravimetric analysis of a substance by precipitation involves isolation of an ion in a solution as a sparingly soluble precipitate that either has a known composition or can be converted to a product of known composition.

The precipitate is filtered and washed free of contaminants, dried or ignited and weighed. The amount of the original ion can be determined from the mass and known composition of the precipitate.

 

For a successful determinations, the precipitate must

– have a sufficiently low solubility so that losses from dissolving are negligible

-be of known composition after drying and igniting

– be easily filtered

– be readily washed free of impurities

– not react in the atmosphere

– have a high purity. It difficult to obtain a product which is pure , but careful precipitation and sufficient washing helps reduce the level of impurity.

 

Unwanted precipitation may occur with the desired product. This is called co-precipitation. It cannot be avoided but can be minimized by slow precipitation and thorough washing.

The precipitate should consist of crystals large enough to be easily washed and filtered. Large crystals also have smaller surface areas for surface adsorption of foreign materials.

 

Vacuum Filtration

Vacuum filtration is used to increase the filtration rate. A Buchner or suction funnel is fitted to a suction flask with an adaptor. A sheet of filter paper of a suitable size just to cover all the holes in the funnel is placed in the funnel and moistened with a little of the solvent to be used in the filtration.

The filter flask is then connected to a vacuum source. The mixture to be filter is poured onto the filter paper and the vacuum rapidly pulls the liquid through the funnel. The filtrate is collected in the suction flask. The suction flask is made of thick glass to prevent breakage when a vacuum is applied.

 

funnel

 

Drying to a constant mass

The filter paper with the product is transferred to a weighed crucible and the filter paper is ignited.  The sample is dried to a constant, reproducible mass. This is especially important in gravimetric analysis because solids may absorb water from the air and increase in weight. The sample is usually dried in a low temperature oven. The oven is electrically heated and can maintain a constant temperature. The sample is allowed to cool then weighed. It is heated again cooled and weighed a second time. The process is repeated until a constant mass is found. It is important to allow the sample to return to room temperature before weighing to prevent convection currents around the balance which can lead to a steady increase in mass while the sample is on the balance. This can severely disrupt the accuracy of the method.  A calculation should be able to help you see how this works.

Calculation

The treatment of a 0.800g sample of impure potassium chloride with excess aqueous silver nitrate resulted in precipitation of 1.460g of silver chloride. Calculate the percentage of potassium chloride in the sample.

Number of Moles in 1.46g of AgCl = 1.460/ 143.5

= 0.01 mol

Since 1 mole of AgCl is formed from 1 mole of Cl- ions

Ag+ (aq) + Cl- = AgCl

And 1 mole of Cl- ions is formed from mole of KCl

KCl = K+ + Cl-

The number of moles of KCl reacted = 0.01moles

The mass of  0.01 moles of KCl = 0.01 x 74.5 = 0.745g

Thus the % KCl in the sample = 0.745/0.800 = 93.13%

 

Volatilization

Gravimetric analysis of a substance by volatilization involves heating an analyte or its decomposition products into the gaseous form. The gaseous products can be collected and weighed or the mass of the products can be determined indirectly from the loss in mass of the sample.

Volatization is commonly used to determine the amount of water and carbon dioxide in a sample. Heating the sample or collecting the water vapour in a solid desiccant can determine water. The amount of water can be determined from the increase in the mass of the desiccant. Alternatively, the amount of water can also be determined indirectly from the mass loss from the sample as a result of heating.

This method assumes that water is the only component volatilized.

Carbon dioxide can be determined from decomposition of carbonates. The mass of carbon dioxide is determined from the increase in weight of a solid absorbent.

 

Example:

A 2.5g sample of impure calcium carbonate was decomposed with excess hydrochloric acid. The liberated carbon dioxide collected in an absorbent was found to weigh 0.88g. Calculate the percentage of calcium carbonate in the sample.

 

Number of moles in 0.88g of CO2 = 0.88/44

= 0.02mole

CaCO3  +   2HCl   =    CaCl2  +   CO2    +   H2O

 

1 mole of CO2 is formed from 1 mole of CaCO3, therefore 0.02moles of CaCO3 was used in the reaction

Mass of CaCO3 used = 0.02 x 100

= 2g

 

% CaCO3 in sample = 2×100/ 2.5

= 80%

Uncertainty in measurement – Accuracy|Precision|Mean|Standard Deviation|Errors|

Any measurement, no matter how accurate or precise, has some amount of error. The true value of a quantity cannot be measured with infinite precision as there are always variations in measurements that come about from errors.  These errors are small and uncontrollable but do cause variations in measurements. The difference between a measured quantity and what is considered to be the true value is known as uncertainty in the measurement. Two concepts that deal with measurements are accuracy and precision.

 

Accuracy and Precision

The accuracy of the measurement refers to how close the measured value is to the true or accepted value. Precision refers to the agreement between two or more measurements that have been carried out exactly the same way.

It must be noted that precision has nothing to do with the true or accepted value of a measurement. Precision is determined with replicate measurements. Replicate measurements are obtained when a number of sample are analyzed in exactly the same way. Therefore, it is quite possible to be very precise and totally inaccurate.

 

accuracy1

accuracy2

accuracy3

Accuracy can be expressed in terms of either absolute or relative error. The absolute error (E) is found by subtracting the true or accepted value (Xt) from the measured value (Xm).

E = Xt – Xm

The value of the absolute error may be positive or negative. The relative error (Er) is a measurement of the absolute error relative to the rue or accepted value.

Er = (Xt – Xm) / Xt

Random or Indeterminate error & Systemic or determinate error

Random errors are the existing fluctuations of any measuring apparatus usually resulting from the experimenter’s inability to take the same measurement in exactly the same way to get the exact value. Even the process itself may introduce variables that may cause measurements to fluctuate.

 

There are many sources of random errors associated in the calibration. These are small and uncontrollable variables such as

–          visual judgment with respect to reading the marking on the glassware and the thermometer

–          temperature fluctuations which affects the volume of the glassware, the viscoscity of the liquid and the performance of the balance

–          wind that cause variations in the balance readings

Random errors affect the precision of a measurement. Precision os usually measured in terms of the deviation of a set of results from the mean value of the set. This is measured using the standard deviation (s)

 

Average or mean value

The mean is calculated by dividing the sum of the replicate measurements by the number of measurements in the set.

Example 1

table2

Calculate the mean value from the data in table 1

Mean = 22.10 + 23.09 + 20.01 + 24.00 + 21.11 +  20.20 / 6

= 21.75 cm3

Standard Deviation

The standard deviation (s) is a measure of the variation of a set of measurement about its mean value. It is typically called the uncertainty in a measurement. It tells how values bunch together from the mean set of data.

Calculate the standard deviation for the data in table 2

table 1

Systematic Errors

A systematic error is a consistent difference between a measurement and its true value that is not due to random chance. It affects all the data set in the same way each time a measurement is made. There are three sources of systematic errors. These are:

  1. Instrument errors – these are caused by errors such as faulty calibrations, instruments being used under different conditions from which they were calibrated or unstable power supply. These errors can be eliminated by calibration or checking the instrument against a standard.
  1. Method errors – these arise from behaviours of reagents and reaction such as incompleteness of reaction or occurrence of side reactions which interefers with the measuring process. These errors are difficult to detect and correct.
  1. Personal error – these result from personal judgement such as the end point in a titration or estimating measurements between two scale markings. These errors can be minimized by care and self-discipline.

These errors affect the accuracy of a measurement. When the magnitude of the error is independent of the size of the sample being measured, the error is referred to as a constant error. This means that whether a small or large sample is used for analysis, the magnitude of the error is the same. Constant errors are minimized by using a large as possible sample. When errors vary with the size of the sample, they are referred to as proportional errors. This means that the magnitude of the error increases or decreases as the size of the same increases or decreases.

Empirical, Molecular and Structural Formulae

Empirical Formula

The empirical formula shows the simplest whole number ratio of the atoms in a molecule. For example: the empirical formula of hydrogen peroxide is HO.

The determination of the empirical formula of an organic compound involves combustion. A known mass of the compound is burned completely in excess oxygen.

There are several crucial steps in determining the empirical formula of a compound.

Steps for determining an empirical formula

Step 1 – Start with the number of grams of each element, given in the problem. If percentages are given, assume that the total mass is 100g therefore the percent given is equal to the mass of each element present.

Step 2 – Convert the mass of each element to moles using the molar mass in the periodic table

Step 3 – Divide each mole value by the smallest mole that was calculated

Step 4 – Round off to the nearest whole number

Now lets use an example to help us understand exactly how to do this.

Question Compound X contains 45.9% carbon, 32% hydrogen 13.5% nitrogen and 8.6% oxygen. Calculate the empirical formula of compound X.

Step 1 – The mass of each element based on the question

C – 45.9g

H – 32g

N – 13.5g

O – 8.6g

Step 2 – Convert the mass to moles using the molar masses in the periodic table

C          45.9 / 12 = 3.825

H         32 / 1 = 32

N         13.5 / 14 = 0.9643

O         8.6 / 16 = 0.5375

Step 3 – Divide each mole by the smallest number of moles calculated. Remember to round off to the nearest whole number.

C          3.825/0.5375 = 7

H         32 / 0.5375 = 2

N         0.9643 / 0.5375 = 2

O         0.5375 / 0.5375 = 1

 

Empirical Formula of Compound X – C7H2N2O

 

Molecular Formula

The molecular formula shows the actual number of atoms of each element in a molecule of the compound. E.g. the molecular formula of the compound hydrogen peroxide is H2O2

Molecular formula can be determined if the molar mass of the compound is known. To find this, calculate the mass of the empirical formula and divide the molar mass of the compound by the empirical formula. Use the number calculated and multiply all the atoms by this ratio to find the molecular formula.

Using an example should make this much easier to understand.

Using the same question above:

Compound X contains 45.9% carbon, 32% hydrogen 13.5% nitrogen and 8.6% oxygen. Calculate the empirical formula of compound X. The molar mass of compound X is 482g/mol

With the empirical formula calculated to be C7H2N2O. We can now go ahead and calculate the molar mass.

C7H2N2O = (7x12g) + (2×1) + (14×2) + (16×1)

84 + 2+ 98 + 16 = 200g/mol

 

Molar Mass / empirical formula =

482g/mol / 200 g/mol = 2

 

Therefore multiply the atoms in C7H2N2O by 2

The molecular formula of C7H2N2O is C14H4N4O2

 

Structural Formula

The structural formula shows the actual number of atoms and the bonds between them; that is the arrangement of atoms in the molecule. The structural formula of hydrogen peroxide is H-O-O-H. The structural formula shows how the various atoms are bonded.

structural formula

Periodic Table Trends

Periodic Table Trends

Atomic Radius

The atomic radius of an element is half of the distance between the centers of two atoms of that element that are touching each other. Generally, the atomic radius decreases across a period from left to right and increases down a given group. The atoms with the largest atomic radii are located in group 1 and at the bottom of groups.

Moving from left to right across a period, electrons are added one at a time to the outer energy shell. Electrons within a shell cannot shield each other from the attraction of protons. Since the number of protons is also increasing, the effective nuclear charge increases across the period. This causes the atomic radius to decrease. Moving down a group in the periodic table, the number of electrons and filed electron shells increases, but the number of valence electrons remains the same. The outermost electrons in a group are exposed to the same effective nuclear charge, but electrons are found farther from the nucleus as the number of filled energy shells increases. Therefore, the atomic radius increases.

 

Ionization energy

The ionization energy or ionization potential is the energy required to completely remove an electron from a gaseous atom or ion. The closer and more tightly bound an electron is to the nucleus, the more difficult it will be to remove, and thus the higher its ionization energy will be. The first ionization energy is the energy required to remove one electron from its parent atom. The second ionization energy is the energy required to remove a second valence electron. The second ionization energy is always greater than the first ionization energy.

Ionization energies increase moving from left to right across a period (deceasing atomic radius). Ionization energy decreases moving down a group (increasing atomic radius). Group 1 elements have low ionization energies because the loss of an electron forms a stable octet.

 

Electron Affinity

Electron affinity reflects the ability of an atom to accept an electron. It is the energy change that occurs when an electrons is added to a gaseous atom. Atoms with stronger effective nuclear charge have greater electron affinity. Group IIA elements and alkaline earth metals have low electron affinity values. These elements are relatively stable because they have filled “s” subshells. Group VIIA elements, the halogens have high electron affinities because the addition of an electron to an atom results in a completely filled shell. Group VIII elements, noble gases, have electron affinities near zero, since each atom possesses a stable octet and will not accept an electron readily. Elements of other groups have low electron affinities.

In a period, the halogens will have the highest electron affinity; while the noble gas will have the lowest electron affinity. Electron affinity decreases moving down a group because a new electro would be further from the nucleus of a large atom.

 

Electro-negativity

Electro-negativity is a measure of the attraction of an atom for the electrons in a chemical bond. The higher the electro-negativity of an atom, the greater its attraction will be for bonding electrons. Electro-negativity is related to ionization energy. Electrons with low ionization energies have low electro-negativities because their nuclei exert a strong attractive force on electrons. Elements with high ionization energies have high electro-negatives due to the strong pull exerted on electrons by the nucleus. in a group, the electro-negativity decreases as atomic number increases, as a result of increased distance between the valence electron and nucleus (greater atomic radius).

Summary of trends

 

Moving left to right

  1. Atomic radius decreases
  2. Ionization energy increases
  3. Electron affinity generally increases
  4. Electro-negativity increases

Moving top to bottom

  1. atomic radius increases
  2. ionization energy decreases
  3. electron affinity generally decreases
  4. electro-negativity decreases

chem

 

Anions and Cations – Common ions and their charges, test to identify cations and anions and observations of test carried out on positive and negative ions

Anions and Cations – Common ions and their charges, test to identify cations and anions and observations of test carried out on positive and negative ions

Anions

Anions are atoms that have gained electrons and since they now have more electrons than protons, anions have a negative charge.

Common Anions and their Charges

Mono-atomic Anions

Names

F

Fluoride

Cl

Chloride

Br

Bromide

I

Iodide

O2

Oxide

S2

Sulphide

Se2-

Selenide

Te2-

Telluride

N3-

Nitride

P3-

Phosphide

A3-

Arsenide

Polyatomic Anions

Name

NH4

Ammonium

NO2

Nitrite

N03

Nitrate

S032-

Sulphite

SO42-

Sulphate

OH

Hydroxide

PO43-

Phosphate

CO32-

Carbonate

ClO3

Chlorate

C2H3O2

Acetate

CH3COO

Ethanoate

MnO4

Manganate (VII)

Cr2O72-

Dichromate (VI)

PO43-

Phosphate

S2O32-

Thiosulphate

Test for the identification of Anions

Anion

Symbol

Test

Results

Bromide

Br

Add silver nitrate to a  solution of  the substance in dilute nitric acid

Cream precipitate, dissolves slightly in ammonia solution

Carbonate

CO32-

  • Add dilute hydrochloric acid to the substance

  • Add a drop of phenolphthalein to a solution of the substance
  • Carbon dioxide gas is given off

  • Turns bright pink

Chloride

Cl

Add silver nitrate to a solution of substance in dilute nitric acid

Thick white precipitate dissolves in ammonia solution

Hydrogen-Carbonate

HCO32-

  • Add dilute hydrochloric acid to the substance

  • Add a drop of phenolphthalein to a solution of the substance.
  • Carbon dioxide gas is given off.

  • Turns light pink

Iodide

I

Add silver nitrate to a solution of the substance in nitric acid

Pale yellow precipitate, does not dissolve in ammonia solution

Nitrate

NO3

Add iron (II) sulphate solution followed by concentrated sulphuric acid to the solution.

Brown rings forms at the junction of the two liquids

Sulphate

SO42-

Add solution of barium chloride to the solution

White precipitate does not dissolve in dilute hydrochloric acid

Sulphite

SO32-

Add solution of barium chloride to the solution

White precipitate dissolves in dilute hydrochloric acid

Sulphide

S2

Add lead (II) ethanoate solution to the solution

Black precipitate.

Cations

Cations are atoms that have lost electrons. Since they now have more protons than electrons, cations have a positive charge.

Common Cations and their Charges

Mono-atomic Cations

Name

H+

Hydrogen

Li+

Lithium

Na+

Sodium

K+

Potassium

Rb+

Rubidium

Cs+

Cesium

Be2+

Beryllium

Mg2+

Magnesium

Sr2+

Calcium

Ba2+

Barium

Al3+

Aluminum

Ga3+

Gallium

Ag+

Silver

Zn2+

Zinc

NH4+

Ammonium

Fe3+

Iron(III)

Fe2+

Iron(II)

Cu2+

Copper(II)

Cu+

Copper(I)

Cr3+

Chromium(III)

Ni2+

Nickel (II)

Pb4+

Lead(IV)

Pb2+

Lead(II)

Hg2+

Mercury (II)

Sn2+

Tin

(Hint: Roman numerals give the positive charge)

Test for the identification of Cations

Cation

Symbol

Test

Results

Aluminum

Al3+

  • Add dilute sodium hydroxide solution to a solution of the substance

  • Add dilute ammonia solution to a solution of the substance
  • White precipitate that dissolves as more sodium hydroxide is added

  • White precipitate that remains insoluble as more ammonia is added

Ammonium

NH4+

Add sodium hydroxide solution to a solution of the substance and gently heat

Ammonia gas is given off

Calcium

Ca2+

Add dilute sulphuric acid to a solution of the substance

White precipitate is formed.

Copper (II)

Cu2+

  • Add dilute sodium hydroxide solution to a solution of the substance.

  • Add dilute ammonia solution to a solution of the substance

  • Pale blue precipitate that does not dissolve as more sodium hydroxide is added

  • Pale blue precipitate, changing to deep blue as more ammonia is added

Iron (II)

Fe2+

  • Add dilute sodium hydroxide solution to a solution of the substance.

  • Add dilute ammonia solution to a solution of the substance

  • Pale green precipitate is formed

  • Pale green precipitate is formed

Iron (III)

Fe3+

  • Add dilute sodium hydroxide solution to a solution of the substance.

  • Add dilute ammonia solution to a solution of the substance

  • Red -Brown precipitate is formed

  • Red-Brown precipitate is formed

Lead (III)

Pb2+

  • Add dilute sodium hydroxide solution to a solution of the substance.

  • Add dilute ammonia solution to a solution of the substance

  • White precipitate that dissolves as more sodium is added

  • White precipitate that does not dissolve as more ammonia is added

Magnesium

Mg2+

  • Add dilute sodium hydroxide solution to a solution of the substance.

  • Add dilute ammonia solution to a solution of the substance

  • White precipitate that does not dissolve as more sodium hydroxide is added.

  • White precipitate that does not dissolve as more ammonia is added.

Zinc

Zn2+

  • Add dilute sodium hydroxide solution to a solution of the substance.

  • Add dilute ammonia solution to a solution of the substance

  • White precipitate that dissolves as more sodium hydroxide is added

  • White precipitate that dissolves as more ammonia is added.

Observations of test carried out on Positive and Negative Ions

Positive Ions

Discharge of Positive Ion

Observation

Na+ + e = Na

Grey deposit is formed

Al3+ + 3e = Al

Grey deposit is formed

Pb2+ + 2e = Pb

Grey deposit is formed

Cu2+ + 2e = Cu

Brown deposit is formed

Ag+ + e = Ag

Silver deposit is formed

2H+ + 2e = H2

Gas bubble is formed. a popping sound is produced when a lighted splinter is placed near the mouth of the test-tube

Negative Ion

Discharge of Negative Ion

Observation

2Cl  =  Cl2 + 2e

Bubbles of pungent yellow-green gas are produced. The gas turns moist litmus paper to red and then bleaches it.

2Br =  Br2 + 2e

  • Molten Electrolyte: brown color gas is produced.
  • Aqueous Solution: light brown solution is formed

2I  =  I2 + 2e

  • Molten Electrolyte: brown color gas is produced.
  • Aqueous Solution: light brown solution is formed. the solution turns blue when a few drops of starch solution is added in.

4OH  =  O2 + 2H2O + 4e

Gas bubble is formed. Gas produces light up wooden splinter.