## What is Qualitative Analysis?

Quantitative analysis is an analysis method used to determine the number of elements or molecules produced during a chemical reaction. Organic compounds are comprised of carbon, hydrogen, oxygen, nitrogen, phosphorus, sulphur and halogens. The various methods used for the measurement of percentage composition of elements in an organic compound are explained here.

### Table of Content

- Detection of C and H
- Test for Phosphorous
- Liebig’s Combustion Method
- Carius Method
- Estimation of Sulphur
- Estimation of Phosphorus
- Estimation of Nitrogen by Dumas Method
- Estimation of Nitrogen by Kjeldahl Method
- Estimation of Oxygen by Aluise’s method

Qualitative analysis is the analysis of the species present in a given compound. For example, if a compound is taken, the qualitative analysis would be more focused on finding the elements and the ions present in the compound rather than study as to how much they are present.

## Detection of C and H

C and H are detected by heating the compound with CuO in a dry test tube. They are oxidised to CO_{2} and H_{2}O respectively. If the CO_{2} turns lime water milky, and H_{2}O turns anhydrous CuSO_{4} blue, then the presence of C and H is confirmed.

## Test for Phosphorous

The organic compound is heated with an oxidising agent to oxidise phosphorous to phosphate. The solution is then boiled with concentrated HNO_{3} and treated with ammonium molybdate. Yellow precipitate confirms the presence of phosphorous.

The reaction is given below,

Na_{3}PO_{4} + 3HNO_{3} → H_{3}PO_{4} + 3NaNO_{3}

H_{3}PO_{4} + 12(NH_{4})_{2}MoO_{4} + 21HNO_{3} → (NH_{4})_{3}PO_{4}.12MoO_{3} + 21NH_{4}NO_{3} + 12H_{2}O

Quantitative analysis is more towards finding out how much of the elements are present, that is, their amounts.

#### Also Read

- Salt Analysis
- Organic Chemistry
- Inductive Effect
- Purification of Organic Compounds
- Lassaigne’s test
- Determine Molecular Masses

## Estimation of C and H

### Liebig’s Combustion Method

A known mass of the compound is heated with CuO. The carbon present is oxidised to CO_{2} and hydrogen to H_{2}O. The CO_{2} is absorbed in KOH solution, while H_{2}O is absorbed by anhydrous CaCl_{2} and they are weighed.

Percentage of C = 12/44 x ((Mass of CO_{2})/(Mass of compound)) x 100

Percentage of H = 2/18 x ((Mass of H_{2}O)/(Mass of compound)) x 100

## Estimation of Halogens

### Carius Method

A known mass of the compound is heated with Conc. HNO3 in the presence of AgNO3 in a hard glass tube called Carius tube. C and H are oxidised to CO2 and H2O. The halogen forms the corresponding AgX. It is filtered, dried and weighed.

Percentage of X =

((Atomic mass of X)/(Molecular mass of AgX))x((Mass of AgX)/(Mass of the compound))x100

**Calculations:**

Let the mass of the given organic compound be m g.

Suppose the mass of AgX formed = m_{1} g.

We know that 1 mol of AgX consists of 1 mol of X.

So, in m_{1} g of AgX , mass of halogen =

Percentage of halogen =

## Estimation of Sulphur

A known mass of the compound is heated with conc. HNO_{3} in the presence of BaCl_{2} solution in Carius tube. Sulphur is oxidised to H_{2}SO_{4} and precipitated as BaSO_{4}. It is then dried and weighed.

Percentage of S= ((Atomic mass of S)/(Molecular mass of BaSO_{4})) x (( Mass of BaSO_{4})/(Mass of the compound)) x 100

**Calculations:**

Suppose the mass of organic compound = mg

Let the mass of barium sulphate formed = m_{1 }g

We know that 32 g sulphur is present in 1 mol of BaSO_{4}

Therefore, 233 g BaSO_{4} contains 32 g sulphur:

⇒ M_{1} g of BaSO_{4 }contains

Percentage of sulphur =

## Estimation of Phosphorus

A known mass of the compound is heated with HNO_{3} in a Carius tube, which oxidises phosphorous to phosphoric acid. It is then precipitated as ammonium phosphomolybdate ((NH_{4})_{3}PO_{4}.12MoO_{3}) by adding NH_{3} and ammonium molybdate ((NH_{4})_{2}MoO_{4}). It is filtered, dried and weighed.

Percentage of P=

((Atomic mass of P)/(Molecular mass of (NH_{4})_{3}PO_{4}.12MoO_{3})) x ((Mass of (NH_{4})_{3}PO_{4}.12MoO_{3})/(Mass of compound)) x 100

## Estimation of Nitrogen

### Estimation of Nitrogen by Dumas Method

A known mass of the compound is heated with CuO in an atmosphere of CO_{2}, which yields free nitrogen along with CO_{2} and H_{2}O.

C_{x}H_{y}N_{z} + (2x+ 0.5y) CuO → xCO_{2} + 0.5y H_{2}O + 0.5z (N_{2}) + (2x+ 0.5y)Cu

The gases are passed over a hot copper gauze to convert trace amounts of nitrogen oxides to N_{2}. The gaseous mixture is collected over a solution of KOH which absorbs CO_{2}, and nitrogen is collected in the upper part of the graduated tube.

Let the volume of N_{2} collected be V_{1} mL.

Then, volume of N_{2} at STP = (P_{1}V_{1} x 273)/ (760 x T_{1}) = V mL

Where P_{1} and V_{1} are the pressure and volume of N_{2}.

P_{1}= Atmospheric pressure – aqueous tension

22.4 L of N_{2} weighs 28 g,

Therefore, V ml of N_{2} weighs

=(28 x V)/22400 grams

Percentage of N would be,

= (28/22400) x (V/ Mass of compound) x 100

### Estimation of Nitrogen by Kjeldahl Method

A known mass of an organic compound is (0.5 g) is mixed with K_{2}SO_{4} (10 g) and CuSO_{4} (1.0 g) and conc.H_{2}SO_{4} (25 mL), and heated in a Kjeldahl’s flask.

CuSO_{4} acts as a catalyst, while K_{2}SO_{4} raises the boiling point of sulphuric acid. The nitrogen in the compound is quantitatively converted to (NH_{4})_{2}SO_{4}. The resulting mixture is reacted with an excess of NaOH solution and the NH_{3} so evolved, is passed into a known but the excess volume of standard acid.

The acid left unreacted is estimated by titration with some standard alkali. Thus the percentage of nitrogen can be calculated.

**The reactions are given below:**

- C + H + S → CO
_{2}+ H_{2}O + SO_{2} - N → (NH
_{4})_{2}SO_{4} - (NH
_{4})_{2}SO_{4}+ 2NaOH → Na_{2}SO_{4}+ 2NH_{3}+ 2H_{2}O - 2NH
_{3}+ H_{2}SO_{4}→ (NH_{4})_{2}SO_{4}

**Calculation of the percentage of N**

Let the mass of the organic compound be mg.

Volume of H_{2}SO_{4} (Molarity M) = V mL

The volume of NaOH of molarity M used for titration excess of H_{2}SO_{4}= V_{1} mL

mEq of excess H_{2}SO_{4} = mEq of NaOH

= MV_{1} mEq

Total mEq of H_{2}SO_{4} taken = 2MV

mEq of H_{2}SO_{4} used for neutralisation of NH_{3} = (2MV – MV_{1})

Therefore,

mEq of NH_{3} = (2MV-MV_{1})

1000 mEq or 1000 mL of NH_{3} solution contains = 17 g of NH_{3} (or) 14 g of N

Therefore,

(2MV-MV_{1}) mEq of NH_{3} contains

=(14 x (2MV-MV_{1})) / 1000 g of N

Percentage of N= ((14x(2MV-MV_{1})) x (100/(1000xm)))

## Estimation of Oxygen – Aluise’s method

A known mass of the compound is decomposed by heating it in the presence of N_{2} gas. The mixture of gases so produced is passed over red hot coke. This is done so that all the O_{2} is converted to CO. This mixture is heated with I_{2}O_{5} in which CO is oxidised to CO_{2} liberating I_{2}.

The reactions are given below,

Organic compound → Other gaseous products + O_{2}

2C + O_{2} → 2CO

I_{2}O_{5} + 5CO → 5CO_{2} + I_{2}

Percentage of O = ((Molecular mass of O_{2}/Molecular mass of CO_{2}) x (Mass of CO_{2}/Mass of the compound) x 100