Qualitative & Quantitative Analysis (Chemistry): Types & More - Embibe
• Written By Sahana Soma Kodarkar
• Written By Sahana Soma Kodarkar

# Qualitative and Quantitative Analysis: Detection of Elements

We know that organic compounds are either hydrocarbons or their derivatives. Almost every organic substance contains carbon and hydrogen. Organic molecules may also contain nitrogen, sulphur, halogens, phosphorus, and oxygen, among other elements. Qualitative and Quantitative Analysis are two methods for detecting these elements.

In brief, qualitative analysis implies evaluating anything based on its quality rather than its quantity. Quantitative analysis is the polar opposite of qualitative analysis; it measures quantity rather than quality. We look at facts, measures, statistics, and percentages when we undertake quantitative analysis.

The main distinction between qualitative and quantitative chemistry is that qualitative analysis determines whether or not different chemical components are present in a sample, whereas quantitative analysis determines the amount of different chemical components present in a sample. To read more about qualitative and quantitative analysis, read the below article.

## Define Qualitative and Quantitative Analysis

The qualitative and quantitative analysis definition has been explained below:

#### Qualitative Analysis

The presence or absence of a compound is determined by a qualitative analysis, but not the mass or concentration. Qualitative analysis, by definition, do not measure quantity. Carbon, hydrogen, oxygen, nitrogen, phosphorus, sulphur, and halogens are the elements that make up organic molecules.

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#### Quantitative Analysis

The measuring of the quantities of certain chemical constituents contained in a sample is known as quantitative analysis. Gravimetric analysis is the quantitative determination of a substance by precipitation, followed by isolation and weighing of the precipitate. The percentage of a certain element or ion in a sample is determined via quantitative analysis.

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### Difference between Qualitative and Quantitative Analysis

The differences between Qualitative and Quantitative Analysis are tabulated below:

### Quantitative and Qualitative Analysis in Organic Chemistry

The quantitative and qualitative analysis in organic chemistry are explained below:

#### Qualitative Analysis

Carbon and hydrogen are the elements that make up organic molecules. They may also contain oxygen, nitrogen, sulphur, halogens, and phosphorus in addition to these elements.

#### 1. Detection of Carbon and Hydrogen

Carbon and hydrogen are detected by heating the compound with copper (ll) oxide. The carbon in the compound is oxidised to carbon dioxide. The presence of carbon dioxide can be detected by passing the gas through lime water which becomes milky and hydrogen is oxidised to water (as evidenced by the blue colour of anhydrous copper sulphate).

$${\text{C}} + 2{\text{CuO}}\xrightarrow{\Delta }2{\text{Cu}} + {\text{C}}{{\text{O}}_2}$$

$${\text{2H}} + {\text{CuO}}\xrightarrow{\Delta }{\text{Cu}} + {{\text{H}}_{\text{2}}}{\text{O}}$$

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##### Detection of carbon dioxide and water

$${\text{C}}{{\text{O}}_2} + {\text{Ca}}{({\text{OH}})_2} \to {\text{CaC}}{{\text{O}}_3} \downarrow + {{\text{H}}_2}{\text{O}}$$

$$5{{\text{H}}_2}{\text{O}} + \underset{{{\text{White}}}}{\mathop {{\text{CuS}}{{\text{O}}_{\text{4}}}}} \to \underset{{{\text{Blue}}}}{\mathop {{\text{CuS}}{{\text{O}}_{\text{4}}}}} .5{{\text{H}}_2}{\text{O}}$$

#### 2. Detection of other elements

Nitrogen, sulphur, halogens, and phosphorus present in an organic compound are detected by “Lassaigne’s test”.
Preparation of Lassaigne’s extract: A little piece of dry sodium metal is gradually heated in a fusion tube until it melts into a shining globule. A little amount of organic substance is added at this point, and the tube is heated vigorously for two to three minutes. In a china dish, the red-hot tube is submerged in distilled water. After boiling for a few minutes, the contents of the dish are cooled and filtered. The filtrate is known as Lassaigne’s extract or sodium extract.

#### A. Test for Nitrogen

The sodium fusion extract is boiled with iron (II) sulphate and then acidified with concentrated sulphuric acid. The formation of Prussian blue confirms the presence of nitrogen. When sodium cyanide interacts with iron(ll) sulphate, sodium hexacyanidoferrate is formed (ll). Some iron(ll) ions are oxidised to iron (lll) ions when heated with concentrated sulphuric acid, which combines with sodium hexacyanidoferrate (ll) to produce iron (lll) hexacyanidoferrate (ll), which is Prussian blue.

$$6{\text{C}}{{\text{N}}^ – } + {\text{F}}{{\text{e}}^{2 + }} \to {\left[{{\text{Fe}}{{({\text{CN}})}_6}} \right]^{4 – }}$$

$$3{\left[{{\text{Fe}}{{({\text{CN}})}_6}} \right]^{4 – }} + 4{\text{F}}{{\text{e}}^{3 + }}\xrightarrow{{{\text{x}}{{\text{H}}_2}{\text{O}}}}\underset{{{\text{Prussian}}\,{\text{blue}}}}{\mathop {{\text{F}}{{\text{e}}_{\text{4}}}{{\left[{{\text{Fe(CN}}{{\text{)}}_{\text{6}}}} \right]}_{\text{3}}}{\text{.x}}{{\text{H}}_{\text{2}}}{\text{O}}}}$$

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#### B. Test for Sulphur

The following test can detect the presence of sulphur:

a. Lead acetate test: When the sodium fusion extract is acidified with acetic acid before being mixed with lead acetate, a dark precipitate of lead sulphide is formed that confirms the presence of sulphur.

$${\text{N}}{{\text{a}}_2}{\text{S}} + \underset{{{\text{Lead}}\,{\text{acetate}}}}{\mathop {{{\left({{\text{C}}{{\text{H}}_{\text{3}}}{\text{COO}}} \right)}_{\text{2}}}}} {\text{Pb}} \to \underset {{{(\text{Black}}\,{\text{ppt.})}}}{\mathop{\text{PbS}}} + \underset{{{\text{Sod}}{\text{.}}\,{\text{acetate}}}}{\mathop {{\text{2C}}{{\text{H}}_{\text{3}}}{\text{COONa}}}}$$

b. Sodium nitroprusside test: When a few drops of sodium nitroprusside solution are added to another part of Lassaigne’s extract, the solution turns violet. The appearance of violet colour indicates the presence of sulphur in the compound.

$${\text{N}}{{\text{a}}_2}{\text{S}} + \underset{{{\text{Sod}}{\text{.}}\,{\text{nitroprusside}}}}{\mathop {{\text{N}}{{\text{a}}_{\text{2}}}\left[{{\text{Fe(CN}}{{\text{)}}_{\text{5}}}{\text{NO}}} \right]}} \to \underset{{\left({{\text{Violet}}\,{\text{colour}}} \right)}}{\mathop {{\text{N}}{{\text{a}}_{\text{4}}}\left[{{\text{Fe(CN}}{{\text{)}}_{\text{5}}}{\text{NOS}}} \right]}}$$

#### C. Test For Halogens

Nitric acid is used to acidify the sodium fusion extract, which is then treated with silver nitrate. A white precipitate that is soluble in ammonium hydroxide indicates the presence of chlorine. In contrast, a yellowish precipitate that is sparingly soluble in ammonium hydroxide indicates the presence of bromine. A yellow precipitate that is insoluble in ammonium hydroxide indicates the presence of iodine.

$${{\text{X}}^ – } + {\text{A}}{{\text{g}}^ + } \to {\text{AgX}}$$
$$\rm{X}$$ represents a halogen $$-\rm{Cl}$$, $$\rm{Br}$$ or $$\rm{I}$$.

#### D. Test for Phosphorus

When a phosphorus-containing compound is heated with an oxidising agent (sodium peroxide), the phosphorus in the compound is oxidised to form phosphate. The solution is boiled with nitric acid and then treated with ammonium molybdate. The formation of a yellow precipitate indicates the presence of phosphorus.

$${\text{N}}{{\text{a}}_3}{\text{P}}{{\text{O}}_4} + 3{\text{HN}}{{\text{O}}_3} \to {{\text{H}}_3}{\text{P}}{{\text{O}}_4} + 3{\text{NaN}}{{\text{O}}_3}$$

$${{\text{H}}_3}{\text{P}}{{\text{O}}_4} + \underset{{{\text{Ammonium}}\,{\text{molybdate}}}}{\mathop {{\text{12}}{{\left({{\text{N}}{{\text{H}}_{\text{4}}}} \right)}_{\text{2}}}{\text{Mo}}{{\text{O}}_{\text{4}}}}} + 21{\text{HN}}{{\text{O}}_3} \to \underset{{{\text{Ammonium}}\,{\text{phosphomolybdate}}}}{\mathop {{{\left({{\text{N}}{{\text{H}}_{\text{4}}}} \right)}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}{\text{. 12Mo}}{{\text{O}}_{\text{3}}}}} + 21{\text{N}}{{\text{H}}_4}{\text{N}}{{\text{O}}_3} + 12{{\text{H}}_2}{\text{O}}$$

#### Quantitative Analysis

In organic chemistry, quantitative analysis of molecules is important. It helps scientists in calculating the mass per cent of elements in a compound. The following procedures are used to determine the percentage composition of elements contained in an organic compound:

#### 1. Estimation of Carbon and Hydrogen

Carbon and hydrogen are estimated in Liebeg’s method. In this method, A known mass of an organic compound is burnt in excess of oxygen and copper (II) oxide. Then the carbon and hydrogen present in the organic compound oxidise to carbon dioxide and water, respectively. The percentage of carbon and hydrogen in the compound is calculated using the below formula.

$${\rm{Percentage}}\,{\rm{of}}\,{\rm{C}} = \frac{{12}}{{44}} \times \frac{{{\rm{Mass}}\,{\rm{of}}\,{\rm{C}}{{\rm{O}}_2}{\mkern 1mu} {\rm{formed}}}}{{{\rm{Mass}}\,{\rm{of}}\,{\rm{the}}\,{\rm{compound}}}} \times 100$$

$${\text{Percentage}}\,{\text{of}}\,{\text{H}} = \frac{{2}}{{18}} \times \frac{{{\text{Mass}}\,{\text{of}}\,{\text{H}_2}{{\text{O}}}\,{\text{formed}}}}{{{\text{Mass}}\,{\text{of}}\,{\text{the}}\,{\text{compound}}}} \times 100$$

#### 2. Estimation of Nitrogen

i. Dumas Method: In this method, the nitrogen-containing organic compound heated with copper oxide in an atmosphere of carbon dioxide to yield free nitrogen in addition to carbon dioxide and water. The small amount of nitrogen oxides formed is converted into nitrogen gas by passing over a heated copper wire gauze. The percentage of nitrogen formed is calculated using the below formula.

$${\text{Percentage}}\,{\text{of}}\,{\mathbf{N}} = \frac{{28}}{{22400}} \times \frac{{{\text{Volume}}\,{\text{of}}\,{{\text{N}}_2}\,{\text{at}}\,{\text{S}}{\text{.T}}{\text{.P}}}}{{{\text{Mass}}\,{\text{of}}\,{\text{the}}\,{\text{compound}}}} \times 100$$

ii. Kjeldahl’s method: The compound containing nitrogen is heated with concentrated sulphuric acid. Ammonium sulphate is formed when nitrogen in the compound is converted. After that, an excess of sodium hydroxide is added to the acid mixture. In excess of the standard sulphuric acid solution, the released ammonia gas is absorbed. The amount of ammonia produced can be calculated by measuring the amount of sulphuric acid consumed in the reaction.

$${\text{Percentage}}\,{\text{of}}\,{\mathbf{N}} = \frac{{1.4 \times {\text{Normality}}\,{\text{of}}\,{\text{acid}}\,{\text{used}} \times {\text{Volume}}\,{\text{of}}\,{\text{acid}}\,{\text{used}}}}{{{\text{Mass}}\,{\text{of}}\,{\text{the}}\,{\text{compound}}}}$$

#### 3. Estimation of Halogens

Carius Method: In the presence of silver nitrate, a known mass of an organic compound is heated with fuming nitric acid. The resulting silver halide is filtered, washed, and dried. The percentage of halogen is calculated using the below formula.

$${\text{Percentage}}\,{\text{of}}\,{\text{halogen}}({\mathbf{X}}) = \frac{{{\text{Atomic}}\,{\text{mass}}\,{\text{of}}\,{\text{X}}}}{{(108 + {\text{Atomic}}\,{\text{mass}}\,{\text{of}}\,{\text{X}})}} \times \frac{{{\text{Mass}}\,{\text{of}}\,{\text{AgX}}\,{\text{formed}}}}{{{\text{Mass}}\,{\text{of}}\,{\text{the}}\,{\text{compound}}}} \times 100$$

#### 4. Estimation of Sulphur

In a Carius tube, a known mass of an organic compound is heated with sodium peroxide or fuming nitric acid. Sulphur in the compound is oxidised to sulphuric acid. When an excess of barium chloride solution is added to water, it precipitates as barium sulphate. Filter, wash, dry, and weigh the precipitate. The mass of barium sulphate can be used to calculate the sulphur percentage.

$${\text{Percentage}}\,{\text{of}}\,{\text{S}} = \frac{{32}}{{233}} \times \frac{{{\text{Mass}}\,{\text{of}}\,{\text{BaS}}{{\text{O}}_4}\,{\text{formed}}}}{{{\text{Mass}}\,{\text{of}}\,{\text{the}}\,{\text{compound}}}} \times 100$$

#### 5. Estimation of Phosphorus

When an organic compound with a known mass is heated with fuming nitric acid, the phosphorus in the compound is oxidised to phosphoric acid. By combining ammonia with ammonium molybdate, it forms ammonium phosphomolybdate. Alternatively, phosphoric acid can be converted to $$\rm{MgNH}_4 \rm{PO}_4$$ by adding a magnesia mixture, which produces $$\rm{Mg}_2 \rm{P}_2 \rm{O}_7$$ when ignited. The percentage of phosphorus is calculated using the below formula.

$${\text{Percentage}}\,{\text{of}}\,{\text{P}} = \frac{{62}}{{222}} \times \frac{{{\text{Mass}}\,{\text{of}}\,{\text{M}}{{\text{g}}_2}{{\text{P}}_2}{{\text{O}}_7}\,{\text{formed}}}}{{{\text{Mass}}\,{\text{of}}\,{\text{the}}\,{\text{compound}}}} \times 100$$

### Qualitative and Quantitative Analysis in Biochemistry

A qualitative study allows us to make inferences about the biochemical pathway, such as the impact of specific processes, metabolites, or pathway segments on the overall system.

A quantitative study allows us to understand how much compound is present in the biomolecules. People with diabetes can use quantitative whole blood glucose measurements to track their blood glucose levels. The quantitative method is useful in the analysis of urea, protein, carbohydrate, etc.

### Summary

In chemistry, quantitative analysis determines the presence or absence of chemical components in a sample. The qualitative analysis does not measure quantity. The qualitative analysis includes distillation, extraction, colour change, chromatography, and so on. The percentage of a certain element or ion is determined via quantitative analysis.

### FAQ’s

We have provided some frequently asked questions about Qualitative and Quantitative Analysis here:

Q.1. What is an example of quantitative analysis?
Ans: The analysis of the species contained in a compound is known as qualitative analysis. Suppose a compound is taken, for example. In that case, the qualitative analysis will be more focused on identifying the elements and ions present in the compound rather than determining how much of each is there. That is elemental analysis, identification of sample structure to be analysed, etc.

Q.2. What are the types of qualitative analysis?
Ans: Types of qualitative analysis are as follows
a) Classical Wet Analysis
b) Instrumental Analysis

Q.3. What is qualitative and quantitative analysis in gas chromatography?
Ans: Gas chromatography can be used for both qualitative and quantitative analysis.
Qualitative: By comparing the retention times of peaks in a sample to the retention times of standards, it is possible to identify elements in the sample.
Quantitative: It is based on the fact that the area under a single component peak is proportional to the quantity of the detected component.

Q.4. What is qualitative and quantitative data analysis?
Ans:
Quantitative data analysis is based on the classification of data based on computable values, whereas qualitative data analysis is focused on the classification of objects (participants) based on properties and attributes.

Q.5. What is qualitative analysis?
Ans: Qualitative chemical analysis is a branch of chemistry concerned with identifying and grouping components present in the sample.

Q.6. What is an example of qualitative analysis?
Ans: The presence or absence of a compound is determined by a qualitative analysis, but not the mass or concentration. Qualitative analysis, by definition, do not measure quantity.  For example, Observing that a reaction produces gas that bubbles out of solution or that a reaction results in a colour change, flame test.

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