How much Iron (II) in 100 grams of Spinach Oleracea?

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How much Iron (II) in 100 grams of Spinach Oleracea?

Spinach Oleracea

Name : Jade Taylor

Candidate Number : 3689

Centre Number : 58203

Year of Entry : 2004

 

Contents

Plan

Aim

To found out how much Iron (II) is present in 100 grams of Spinach Oleracea.  

The factors that I am going to investigated in this experiment include finding the best method to determine the concentration of an Iron (II) Ammonium Sulphate (aq) by trying colorimetry, an electrochemical cells experiment and a redox titration with Potassium Manganate (aq).  After this I will extract Iron (II) from Spinach Oleracea using various methods (i.e. boiling the Spinach Oleracea for a range of times in different solutions) and use this spinach extract solution to determine the volume of iron extracted.  I will take into account the presences of Oxalate ions and change my experiment accordingly (i.e. heating the spinach extract solution before titration’s).

Introduction

It is important to know the Iron (II) content in 100 grams of Spinach Oleracea, as this allows people to calculate how much Spinach Oleracea needs to be eaten in order to obtain the Recommended Daily Allowance (RDA) of Iron (II).  As shown in the table below the Recommended Daily Allowance for Iron (II) varies with age and sex (2)  

The Recommended Daily Allowance of Iron (II) (2)

 

As the table shows, the volume of Iron (II) needed by the human body increases with age; woman generally require more Iron (II) than men as this is needed to replenish the Iron (II) that is lost during menstruation.  As different age groups and sexes require different volumes of Iron (II) they will need to consume different volumes Spinach Oleracea (2).

It is important to obtain the correct amount of Iron (II) as it is a necessary component in the body.  It is present in Haemoglobin, which is the most efficient respiratory pigment.  Haemoglobin, which is present in most animals and even a few plants, is made up of a “haem” group, which contains an iron porphyrin compound, and a “globin” group, which consists of a protein.  The iron atom that is present is capable of carrying a single oxygen molecule; human haemoglobin contains four iron atoms and therefore it is capable of carrying four molecules of oxygen. (1)


A molecule of haemoglobin made up of four sub units (1)

Haemoglobin is component of Red Blood Cells. Red Blood Cells have a biconcave-disc shape, which produces a compromise between a large surface area and a large volume; this allows the cell to contain a large quantity of haemoglobin while still allowing efficient diffusion through the cell.  The cells are only about 7.5 micrometers in diameter; this small size ensures that all the haemoglobin that the cell contains is close to the surface of the cell and thus allows for rapid diffusion of gases in and out of the cell.  Another advantage of Red Blood Cells is that they do not contain a nucleus or any other organelles, which allows for more Haemoglobin to be compacted into it. (8)

The shape of a Red Blood Cell (1)

Haemoglobins role is to pick up oxygen breathed into the lungs and transport it through the blood stream to every cell in the body.  This oxygen is then used to release the maximum energy content from food; this energy is used to make the heartbeat and the body grow and stay warm (2).

Without iron the body cannot produce normal amounts of Haemoglobin and as a result the blood will transport less oxygen.  With less oxygen reaching the cells and muscles cells the body becomes weak and tired (2).

In this experiment I will be required to make up solutions from hydrated solids and trying to work out the concentrations of solutions and thus it will be important to know the molar mass (Mr) of the substances that I will be using.  The molar mass of a substance is the mass of that substance, which contains 1 mol of that substance.  It is worked out by adding up the Atomic Number (Ar) of all the elements present in the substance.

The molar mass of Potassium Manganate (VII)(s) 

Potassium Manganate (VII) = 158

The molar mass of Iron (II) Ammonium Sulphate(s)

Iron (II) Ammonium Sulphate = 392.14

The molar mass of Oxalic Acid(aq)

Oxalic Acid = 126.07

With the Molar mass of the substances I will be using now known I am able to work out how much solid needs to be weighed out and diluted into certain amounts of acid or distilled water in order to obtain the concentrations of the solutions required for the experiments.  The equation for this exercise is

Mass = Mr x Moles

After working out the moles or mass, I can further the equation to work out concentration or volume needed by using this equation:

Concentration = Mass x Volume


Experiment One - How accurate is a Potassium Manganate titration in determining the concentration of an Iron (II) Ammonium Sulphate (aq)

Aim

To find out how accurate a Redox Titration between Potassium Manganate (VII) (aq) and Iron (II) Ammonium Sulphate (aq) is in determining the concentration of Iron (II) Ammonium Sulphate (aq).

Background Theory

The reaction that will occur in this experiment is a redox reaction, which involves electron transfer.  It can be split into two half reactions, one producing electrons and the other accepting them (4). In most redox reactions an indicator needs to be added to the experiment to show the end point of the reaction but this is not necessary as the reaction that occurs produces a very obvious colour change (7).

The Fe2+ ion present in the Iron (II) Ammonium Sulphate (aq) reacts with the MnO4- ions in the Potassium Manganate (VII) (aq) (7).

5Fe2+ (aq) +  MnO4 (aq) + 8H+ (aq)                         5Fe3+ (aq) + Mn2+ (aq) + 4H2O (l)

The colour of the Potassium Manganate (VII) disappears as it reacts with the Fe2+ (aq) ion.  This provides a way of determining when the titration is complete because when all the Fe2+ (aq) has been used up, just one more drop of Potassium Manganate (VII) (aq) will turn the solution a pale purple (5).  This end point is reached when Iron (II) Ammonium Sulphate (aq) is completely oxidised by the Potassium Manganate (VII) (aq)(7). 

Oxidation numbers provide an accurate way of following shifts in electron density during a reaction.  Each atom in a molecule or ion is assigned an oxidation state at the beginning of the reaction and at the end of the reaction, and by comparing these it is possible to work out how much it has been oxidised or reduced. There are a few simple rules that need to be followed in order to assign oxidation states to elements and the ones that are necessary for this reaction include:

  • The oxidation numbers of any combined element is zero.
  • The oxidation number of an ion of an element is the same as its charge.
  • The sum of all oxidation numbers in a molecule (or completed ion) is equal to the charge on the particle.
  • Combined Oxygen always has an oxidation number of –2.
  • Combined Hydrogen has an oxidation number of +1 (9).

5Fe2+ (aq) + MnO4 (aq) + 8H+ (aq)                         5Fe3+ (aq) + Mn2+ (aq) + 4H2O (l)

Fe2+                   Fe3+                                        MnO4                    Mn2+

                

+2                   +3                                        +8                     +2                                                                        

Fe2+ + e-           Fe3+                                        MnO4                    Mn2+ + 6e-

The half reaction involving a loss of electrons is known as the oxidation reaction and the reaction that involves a gain in electrons is known as a reduction reaction.  

In order to produce a titration involving Iron (II) Ammonium Sulphate (aq) it will be necessary to dissolve hydrated Iron (II) Ammonium Sulphate (s).  It will be important to know the exact concentration of the solution that I make up so that I can compare this to the results of the titration.  As I will be making the solutions up myself I will need to know how much hydrated Iron (II) Ammonium Sulphate (s) to weigh out and dissolve and for this I need to know the molecular mass of the substance.

Iron (II) Ammonium Sulphate

Mass to be weighed out  = Molecular Mass x Moles wanted

Molecular Mass = 392.14

Mass to be weighed out = 392.14 x 0.1

Mass = 39.214 g  (per 1000 dm-3)

39.214 g

      10

= 3.9214 g (per 100 dm-3)

The Potassium Manganate (VII) (s) is also supplied as a solid and will need to be dissolved.  

Potassium Manganate (VII)

Mass to be weighed out = Molecular Mass x Moles wanted

Molecular Mass = 158.03

Mass = 158.03 x 0.1

Mass = 15.803 g (per 1000 dm-3)

15.803 g

      4  

= 3.95 g (per 250 dm-3) (4)

Method (adapted from EL 2.1- How much iron is in a sample of an iron compound?)

                                                                                                                                                                                                                                                                                                                                                             

  1. Make up a solution of 0.1 mol dm-3 Iron (II) Ammonium Sulphate by completing the following steps:

 

  1. Accurately weigh out 3.92 grams of hydrated Iron (II) Ammonium Sulphate (s) using a balance and transfer it into a 100cm3 beaker.
  2. Using a 50 cm3 measuring cylinder measure out approximately 25cm3 of Sulphuric Acid (aq) and pour it into the beaker, using a glass rod stir the acid and solid together until the entire solid has dissolved.
  3. Transfer the contents of the beaker through a small funnel into a 100cm3 volumetric flask.
  4. Rinse the beaker and glass rod twice using small quantities of Sulphuric Acid (aq) and transfer all the washings through the funnel to the volumetric flask.
  5. Rinse the funnel with a small volume of Sulphuric Acid (aq) and transfer it into the volumetric flask.
  6. Add Sulphuric Acid (aq) to the volumetric flask until it is about 1 cm below the graduation mark.
  7. Slowly add the acid using a clean dropping pipette until the bottom of the meniscus is touching the graduation mark.
  8. Stopper the flask and invert it several times.

  1. Make up a solution of 0.1 mol dm–3 Potassium Manganate (VII) by following the following steps:

  1. Accurately weigh out 3.95 grams of Potassium Manganate (VII) (aq) using a balance and transfer it to a 100cm3 beaker.
  2. Using a 50 cm3 measuring cylinder measure out approximately 50 cm3 of distilled water (l) and pour it into the beaker, using a glass rod stir the water and solid together until the entire solid has dissolved.
  3. Transfer the contents of the beaker through a small funnel into a 250cm3 volumetric flask.
  4. Rinse the beaker and glass rod twice using small quantities of distilled water (l) and transfer the washings to the volumetric flask.
  5. Rinse the funnel with a small volume of distilled water (l) and transfer it to the volumetric flask.
  6. Add distilled water (l) to the volumetric flask until it is about 1cm below the graduation mark.
  7. Slowly add the distilled water (l) using a clean dropping pipette until the bottom of the meniscus is touching the graduation mark.
  8. Stopper the flask and invert it several times.

  1. Fill a 50 cm3 burette with the Potassium Manganate (VII) (aq).

  1. Using a 10cm3 pipette transfer 10 cm3 of the Iron (II) Ammonium Sulphate (aq) into a conical flask.

  1. Add 10cm3 of Sulphuric Acid (aq) to the conical flask using a 10 cm3 measuring cylinder.

  1.  Place the conical flask on a white tile.

  1. While swirling the conical flask, titrate the Potassium Manganate (VII) (aq) into the conical flask containing Iron (II) Ammonium Sulphate (aq) and Sulphuric Acid (aq) until the first trace of a pink colour appears.

  1. Continue to add the Potassium Manganate (VII) (aq) to the conical flask drop by drop, swirling between each drop, until the solution remains a pink colour for 30 seconds.

  1. Repeat this experiment until 3 values within 0.1 of each other have been obtained.

The set-up of the apparatus for this experiment

Modifications

An improvement that I feel is necessary to improve the accuracy of this experiment is changing the concentration of the Potassium Manganate (VII)(aq) from 0.1 mol dm-3 to 0.01 mol dm-3.  This is because the 3.95 grams of solute needed to produce the 0.1 mol dm-3 was very difficult to dissolve fully in the beaker before being washed into the volumetric flask.  This meant that some of the solute was left behind in the beaker when the volumetric flask was full up.  

I tried using hot water from a kettle to dissolve the solute in and this worked well, enabling me to get the concentration of solution that I wanted.  But when I put this concentration of Potassium Manganate (aq) into the burette I found that it covered the walls in a dark purple colour, which made it difficult to tell the exact level of the Potassium Manganate (aq).  

Another reason why I feel the concentration of the Potassium Manganate (VII) (aq) needs to be changed is because when I was adding it to the Iron (II) Ammonium Sulphate (aq) the end point of the reaction was reached with only a very small quantity of Potassium Manganate (VII) (aq) being added, which could cause the burette reading to be inaccurate.

Realising that this would dramatically affect my results I decided to decrease the concentration of the solution.

Mass to be weighed out = Molecular Mass x Moles wanted

Molecular Mass = 158.03

Mass = 158.03 x 0.01 mol dm-3

Mass = 1.5803 g (per 1000 dm-3)

1.5803 g

      4  

= 0.395 g (per 250 dm-3)

I now need to weigh out 0.395 grams to obtain a Potassium Manganate (VII) (aq) of 0.01 mol dm-3.

Justifications

This solution of Iron (II) Ammonium Sulphate was made up using Sulphuric Acid (aq) as adding Distilled water (l) would cause the Fe2+ ions to be oxidised to Fe3+ which is a light brown colour and is not what is required for this experiment.

As the volume of Sulphuric Acid (aq) added to the Iron (II) Ammonium Sulphate (aq) had to be in excess to supply H+ ions I felt it would not be necessary to have an exact volume so I used a measuring cylinder to measure the quantity required instead of the more accurate equipment like a pipette which I used to measure out the volumes of the other solutions.

The solutions that I was required to make myself were made up into volumetric flasks as these could provide a more accurate measurement for the volume of sulphuric acid and distilled water that needed to be added.

Three values within 0.1 cm3 of each other need to recorded in this experiment, as an average titre needs to be obtained.  If points within 0.1 cm3 of each other are used the average results will be more accurate.  This will also ensure that any random results are noted and not included in the equation that I will use to work out the average.  


Experiment Two - How accurate is a Calibration Graph in determining the concentration of an Iron (II) Ammonium Sulphate (aq)

Aim

To find out how accurate a Calibration Graph is in determining the concentration of Iron (II) Ammonium Sulphate (aq).

Background Theory

We see object because it reflects light, if all the light is reflected it appears white, however if wavelengths of light corresponding to particular colours are absorbed the object will appear coloured in the complementary colour (4)

A colour wheel showing complementary colours opposite one another

In order for a substance such as Iron (II) Ammonium Sulphate (aq) to absorb particular wavelengths, it must have a pair of electrons whose energy levels are separated by an energy gap that corresponds to a wavelength of visible light.  When this occurs an electron is promoted from an orbital of lower to one of higher energy so that the atom absorbing the radiation changes from its ground state to its excited state (4).

Iron (II) Ammonium Sulphate appears coloured because it absorbs light at more than one wavelength in the visible part of the electromagnetic spectrum and reflects and transmits the others.  Each wavelength of light in this region appears as a different colour.  From looking in textbooks it is possible to predict that the Iron (II) Ammonium Sulphate will absorb wavelengths of about 600 to 650 nm (7).

As Iron is a d-block transition metal, the colour that it appears depends on the number of d-electrons present in the transition metal Iron; and the arrangement of ligands around the ion, as they effect the splitting of the d-sub shell; and the nature of the ligands, as different ligands have a different effect on the relative energies of the d-orbital of a particular ion (4).

The presence of ligands, with their lone pair of electrons, affects the electrons in the d-orbital of the central ion as orbitals close to the ligands are pushed to a higher energy level than those further away.  This results in the d-orbital being split into 2 groups at different energy levels (4).

As Iron (II) Ammonium Sulphate contains ligands it is known as a complex.  A complex is a central ion or molecule surrounded by a number of negatively charged ions or molecules containing a lone pair of electrons.  Bonding in complexes is complicated, involving the electron pair form the ligand being shared with the central ion, thus ligands are known as electron donors (4)

A colorimeter can be used in this experiment as Iron (II) Ammonium Sulphate (aq) appears coloured.

How a colorimeter is used to detect concentration (6)

A colorimeter consists of a light source with filters to select a suitable colour (wavelengths) of light, which is absorbed by the Iron (II) Ammonium Sulphate (aq).  The light passes through the solution onto a detector whose output goes to a meter or a recording device. (6)

Iron (II) Ammonium Sulphate

Mass to be weighed out  = Molecular Mass x Moles wanted

Molecular Mass = 392.14

Mass to be weighed out = 392.14 x 0.1

Mass = 39.214 g  (per 1000 dm-3)

39.214 g

      10

= 3.9214 g (per 100 dm-3)


Method (Adapted from SS 1.1- How much manganese in a paper clip?)

  1. Make up a set of various concentrations of Iron (II) Ammonium Sulphate by completing the following steps:

 

  1. Firstly make up a Iron (II) Ammonium Sulphate (aq) at a concentration of 0.1 mol dm-3.  
  2.  Accurately weigh out 3.92 grams of hydrated Iron (II) Ammonium Sulphate using a balance and place it in a 100cm3 beaker.
  3. Using a 50 cm3 measuring cylinder measure out 25cm3 of Sulphuric Acid (aq) and transfer it to the beaker and using a glass rod stir the acid and solid together until the entire solid has dissolved.
  4. Transfer the contents of the beaker through a small funnel into a 100cm3 volumetric flask.
  5. Rinse the beaker and glass rod twice using small quantities of Sulphuric Acid (aq) and transfer all the washings to the volumetric flask.
  6. Rinse the funnel with a small amount of Sulphuric Acid (aq) and transfer that into the volumetric flask.
  7. Add Sulphuric Acid (aq) to the volumetric flask until it is about 1 cm below the graduation mark.
  8. Slowly add the acid using a clean dropping pipette until the bottom of the meniscus is just touching the graduation mark.
  9. Stopper the flask and invert it several times.
  10. Pipette 9 cm3 of the already made up Iron (II) Ammonium Sulphate (aq) into 10 cm3 volumetric flask.
  11. Using a dropping pipette add Sulphuric Acid (aq) to the volumetric flask until the bottom of the meniscus is touching the graduation mark.
  12. Stopper the flask and invert it several times.  This will make up a solution with the concentration of 0.09 mol dm-3.
  13. Repeat points j through to l varying the volume of 0.1 mol dm-3 Iron Ammonium Sulphate (aq) added to the volumetric flask to make up solutions at the concentrations of 0.08 mol dm-3,  0.07 mol dm-3, 0.06 mol dm-3, 0.05 mol dm-3, 0.04 mol dm-3 , 0.03 mol dm-3, 0.02 mol dm-2 and 0.01 mol dm-3.

  1. Calibrate the colorimeter by setting the 0% absorbance by using a curvet filled with distilled water.

  1. Set the 100% absorbance by using a curvet filled with black solid (eg wax).

  1. Choose a range of filters and place one in the colorimeter.

  1. Place each solution in turn in the colorimeter and note the colorimeter readings.

  1. Place each filter in the colorimeter separately and each solution in there in turn and read of the results.

  1. Find out which filter absorbs the most light and then use this filter to gain results to plot a graph of absorption against concentration.  This is known as the calibration graph.

Modifications

I found that the concentrations of these solutions were very low causing them to appear colourless, thus the colorimeter was unable to detect an absorption of light in the solutions. I have decided to increase the concentrations of the solutions to 1.0 mol dm-3, 0.8 mol dm-3, 0.6 mol dm-3, 0.4 mol dm-3 and 0.2 mol dm-3.

As the solutions are going to require more solid to be dissolved in the same volume of sulphuric acid it will be necessary to heat the solutions over a Bunsen burner, which will help the solid dissolve.

Justifications

In order to dilute down the Iron (II) Ammonium Sulphate (aq) into various solutions of different concentrations a certain amount had to be removed from the original solution and placed in a volumetric flask, which was filled up to the graduation mark with Sulphuric Acid (aq).  If volumes of Sulphuric Acid (aq) and Iron (II) Ammonium Sulphate (aq) were added together they might not have produced 10cm3 of solution as the solutions contained different densities thus the required concentration of solution would not have been produced.  For this reason the solutions were made up into the graduation mark in 10cm3 volumetric flasks.

Only a small volume of solution is needed to use in the colorimeter and thus I am only making up 10 cm3 of each concentrated solution.

The colorimeter was calibrated using a black curvet and a curvet containing distilled water (aq) to establish a relationship between its reading and the concentration of the Iron (II) Ammonium Sulphate (aq) being observed.

The solutions that I was required to make myself were made up into volumetric flasks as these could provide a more accurate measurement for the volume of sulphuric acid that needed to be added to make the solutions of varying concentrations.


Experiment Three -How accurate is an Electrochemical Cell in determining the concentration of an Iron(II) Ammonium Sulphate (aq)

Aim

To find out how accurate an Electrochemical Cell is in determining the concentration of an Iron (II) Ammonium Sulphate (aq)

Background Theory

Electrochemical cells are the basis of electrical batteries (6).  In one half of the cell an oxidation reaction occurs and electrons are produced and transferred through an external circuit to the other part of the cell where a reduction reaction takes place, accepting the electrons (4).

Half-cells help determine the potential measure of how readily electrons are released by the Iron (s).   This would then tell us how good a reducing agent the metal is, because reducing agents release electrons.  It is impossible to measure the electrical potential only the potential difference or voltage, by connecting together 2 different half-cells and measuring the difference as a voltage (6).

The standard electrode potential or redox potential can be arranged in an electrochemical series with the most negative values at the top of table as these half-cells donate electrons to the other half-cells below them (6).

Join now!

Electrode potentials vary with temperature and concentrations of solutions therefore this experiment must be carried out under standard conditions (i.e. at a temperature of 298K and using a 1.0 mol dm –3 Copper Sulphate (aq)) (4).  The Iron (II) Ammonium Sulphate (aq) concentration will be varied, as I want to see the effect of varying the concentration of solutions.

In this experiment Iron will be acting as the negative terminal as it has a more negative electrode potential.  Thus an oxidation reaction will be taking place and the electrons produced will be supplied to the reduction reaction which is occurring ...

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