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