Chapter 2: Introduction to Analytical Chemistry
In the 1830s an English chemist named James Marsh was asked to help settle a suspected poisoning case. Arsenic was suspected, so Marsh carried out the standard chemical test of his day on the samples he was given. He did obtain a precipitate — but by the time the case came before the jury the sample had deteriorated, the evidence no longer looked convincing, and the accused man was acquitted.
Marsh was not satisfied. A test that gave the right answer in the laboratory but could not be trusted in front of others was, to him, not good enough. So he designed a better one. The suspect material was treated with zinc and acid, which converted any arsenic present into a gas; the gas was then heated, and it deposited a bright metallic film on a cold porcelain surface. The film could be seen, kept, and shown to anyone. Even very small amounts of arsenic gave a clear result. Within a few years the Marsh test was being used in criminal trials across Europe.
Notice what Marsh actually improved. He did not discover a new element. He improved the reliability of a measurement — how surely a substance could be detected, how small an amount could be found, and how honestly the result could be reported. That is the whole business of analytical chemistry, and it is what this chapter is about.
Diagram
Fig. 2.0 – A 19th-century laboratory bench: flask containing the sample with zinc and acid, delivery tube leading to a flame, and a cold porcelain dish showing a dark arsenic mirror; beside it a modern burette and a weighing balance, labelled “detection = qualitative” and “how much = quantitative”.
Keep these questions in mind as you read. The chapter gives you everything you need to reason your way to each answer.
2.1 Introduction
Every substance a chemist handles carries a question with it: what is inside this material, and in what proportion? Analytical chemistry is the branch of chemistry that exists to answer that question. It makes the investigation of the chemical composition of substances possible, using carefully chosen instruments and methods to separate, identify and quantify the matter under study.
The word analysis describes this entire exercise. The result of an analysis is chemical or physical information about the material examined — which substances it contains, how much of each is present, what its melting point or boiling point is, how it behaves towards a particular reagent. Without such information, most of the rest of chemistry would have nothing solid to stand on.
Two Kinds of Questions
An analysis may be qualitative or quantitative, and the difference between the two is worth fixing firmly in mind, because almost everything else in this chapter sits on one side of that line or the other.
Qualitative analysis is concerned with the detection of the presence or absence of elements in compounds and in mixtures of compounds. It answers the question “what is present?” Its result is a name, not a number — chloride is present, lead is absent, this powder contains sulphur.
Quantitative analysis deals with the determination of the relative proportions of elements in compounds and mixtures of compounds. It answers the question “how much is present?” Its result always carries a number, and with that number a unit.
In practice the qualitative question is settled first. You cannot sensibly measure how much of a substance a sample contains until you know that the substance is there at all, and know what it is you are measuring.
Why Analytical Chemistry Matters
A course in analytical chemistry extends the knowledge a student has already gained in general, inorganic and organic chemistry. Those subjects tell you how substances are built and how they react; analysis tells you what you are actually holding in your hand.
Chemical analysis is one of the most important methods of monitoring the composition of raw materials, intermediates and finished products. It is also used to monitor the composition of air in streets and in the premises of industrial plants, where a slow rise in the level of a pollutant would otherwise go unnoticed.
In agriculture, chemical analysis is used to determine the composition of soils and fertilizers, so that what a field lacks can be supplied and what it does not need is not wasted. In medicine, it is used to determine the composition of medicinal preparations, where the difference between a correct dose and a harmful one is a matter of measurement.
Analytical chemistry also has applications in forensic science, engineering and industry. An industrial process as a whole, and the production of new kinds of materials, are closely associated with analytical chemistry — a process cannot be controlled unless the composition of what goes in and what comes out is known at every stage.
Classical and Instrumental Methods
Analytical chemistry consists of classical, wet chemical methods and modern instrumental methods. The classical, wet methods depend on chemical reactions carried out in solution — a precipitate that forms, a colour that appears, a volume of reagent that must be added — and need little more than glassware and reagents.
Modern instrumental methods instead measure a physical property of the sample with an instrument, and read the composition from that measurement. They are generally faster, and can work with far smaller amounts of material, which matters when the whole sample available is a single crystal or a drop of blood.
The two families are not rivals. A wet chemical test may identify what is present, and an instrument may then measure how much. The scheme of the subject is set out in FC 2.1.
Flowchart
FC 2.1 – How analytical chemistry divides up: Analytical chemistry → branch by question asked: (i) qualitative analysis → detection of presence/absence of elements and ions; (ii) quantitative analysis → determination of relative proportions. Second branch by method used: (i) classical wet chemical methods → precipitation, extraction, distillation, volumetric, gravimetric; (ii) modern instrumental methods → measurement of a physical property by an instrument. Note: qualitative step normally precedes quantitative step.
2.2 Analysis
Analysis is carried out on a small sample of the material to be tested, and not on the entire bulk. A chemist testing a wagon-load of ore, a tank of acid or a field of soil does not put the whole of it through the laboratory. A small portion is taken, and the answer obtained from that portion is taken to hold for the whole consignment.
When the amount of a solid or liquid sample used is a few grams, the analysis is called semi-microanalysis. The name simply records the scale of working: not the large quantities of older methods, and not the very small quantities of true micro-scale work, but something in between.
Analysis is of two types: qualitative and quantitative. Both are applied to the same small sample, and the qualitative work is normally done first.
Classical Methods of Analysis
Classical qualitative analysis methods include separations such as precipitation, extraction and distillation. Each of these first isolates the substance of interest from the rest of the mixture, because a component is far easier to identify once it has been separated from everything that was masking it.
Once separated, identification may be based on differences in colour, odour, melting point, boiling point, and reactivity. These are properties that belong to a substance and stay with it, so a substance can be recognised by the set of values it shows.
Classical quantitative methods consist of volumetric analysis, gravimetric analysis, etc. In volumetric analysis the measured quantity is a volume of solution; in gravimetric analysis it is a mass. Both are described later in this chapter in the form of the calculations they lead to.
2.2.1 Chemical Methods of Qualitative Analysis
Chemical analysis of a sample is carried out mainly in two stages: by the dry method, in which the sample under test is not dissolved, and by the wet method, in which the sample under test is first dissolved and then analyzed to determine its composition.
The dry method is usually used as preliminary tests in the qualitative analysis. Preliminary here means exactly what it says — these tests narrow the field. Heating a substance, or noting its colour and its behaviour in a flame, quickly rules out many possibilities before the sample is dissolved, so that the wet tests that follow can be chosen sensibly instead of tried at random. The two stages are set out in order in FC 2.2.
Flowchart
FC 2.2 – Stages of chemical qualitative analysis: sample taken → DRY METHOD (sample not dissolved) → preliminary tests → possibilities narrowed → WET METHOD (sample dissolved in a suitable solvent) → systematic tests → detection and confirmation of cations (basic radicals) and anions (acidic radicals) → composition reported.
Apparatus Used
The semi-micro qualitative analysis is carried out using apparatus such as: test tubes, beakers, evaporating dish, crucible, spot plate, watch glass, wire gauze, water bath, burner, blow pipe, pair of tongs, centrifuge, etc. The common pieces are shown together in Fig. 2.3.
Diagram
Fig. 2.3 – Apparatus of semi-micro qualitative analysis, drawn and labelled: test tube, beaker, evaporating dish, crucible with lid, spot plate, watch glass, wire gauze, water bath, burner, blow pipe, pair of tongs, centrifuge.
Each item earns its place. A spot plate lets a test be done in a single drop; a centrifuge throws a precipitate down in seconds where settling would take much longer; a water bath heats gently and evenly to a temperature that cannot exceed the boiling point of water; tongs let a hot crucible be moved without being touched.
Tests for Organic and Inorganic Compounds
The qualitative analysis of organic and inorganic compounds involves different types of tests, because the two classes of compound raise different questions.
The majority of organic compounds are composed of a relatively small number of elements. The most important ones are: carbon, hydrogen, oxygen, nitrogen, sulphur, halogen, phosphorus. Elementary qualitative analysis is concerned with the detection of the presence of these elements. Since so few elements are involved, simply listing them does not distinguish one organic compound from another.
For that reason the identification of an organic compound involves tests such as detection of functional group, determination of melting/boiling point, etc. The functional group — the reactive group of atoms that gives a compound its characteristic chemical behaviour — tells the chemist what family the compound belongs to, and a sharp melting or boiling point then points to the individual member of that family.
The qualitative analysis of simple inorganic compounds involves detection and confirmation of cationic and anionic species, that is, the basic and acidic radicals in them. A cation is a positively charged ion and an anion is a negatively charged ion; in the older language of salt analysis, the cation is called the basic radical and the anion the acidic radical. Identify both, and the salt itself is identified.
2.2.2 Chemical Methods of Quantitative Analysis
Once identity is settled, the question becomes one of amount, and the methods change accordingly.
Quantitative analysis of organic compounds involves methods such as (i) determination of percentage constituent element, and (ii) determination of the concentration of a known compound in the given sample, etc. The first tells how much carbon, hydrogen or nitrogen the compound itself contains; the second tells how much of that compound is present in the mixture handed to the laboratory.
Quantitative analysis of simple inorganic compounds involves methods based on (i) decomposition reaction, which is gravimetric analysis, and (ii) the progress of reaction between two solutions till its completion, which is titrimetric or volumetric analysis, etc.
In the gravimetric route the substance is converted into a product that can be isolated and weighed, and the mass of that product is worked back to the quantity originally present. In the titrimetric route a solution of known concentration is added to the sample solution until the reaction between them is just complete, and the volume needed carries the answer.
The quantitative analytical methods therefore involve measurement of quantities such as mass and volume, by means of some equipment or apparatus such as a weighing machine and a burette. This is worth noticing carefully: every quantitative result in this chapter finally rests on a mass read from a balance or a volume read from graduated glassware. If those readings are poor, no calculation performed afterwards can repair them — which is exactly why the next section turns to the accuracy of measurement.
2.3 Mathematical Operation and Error Analysis
The accuracy of measurement is of great concern in analytical chemistry. A result is only as trustworthy as the readings it was built from, so a chemist has to be as careful about the numbers as about the reactions.
There can also be intrinsic errors in the analytical measurement — errors that belong to the method or the apparatus itself and cannot be removed simply by being careful. An instrument can only read to a certain fineness, and no amount of patience will make it read finer.
The numerical data obtained experimentally are treated mathematically to reach some quantitative conclusion. Raw readings are rarely the answer; they are multiplied, divided, averaged and compared before the answer appears.
Therefore, an analytical chemist has to know how to report the quantitative analytical data, indicating the extent of the accuracy of measurement, how to perform the mathematical operation, and how to properly express the quantitative error in the result. In the following subsections we will consider these aspects related to measurements and calculation.
2.3.1 Scientific Notation (Exponential Notation)
A chemist has to deal with numbers as large as 602,200,000,000,000,000,000,000 for the molecules of 2 g of hydrogen gas, or as small as 0.00000000000000000000000166 g, that is, the mass of a hydrogen atom.
Writing such numbers out in full is not merely tedious; it invites mistakes, because a single dropped zero changes the answer by a factor of ten. To avoid the writing of so many zeros in mathematical operations, scientific notations, that is, exponential notations, are used.
Here, any number can be represented in the form \(N \times 10^n\), where \(n\) is an exponent having positive or negative values and \(N\) can vary as \(1 \leq N < 10\). The condition on \(N\) is what makes the notation standard: it forces exactly one non-zero digit to stand before the decimal point, so that every number has one and only one correct form.
Thus, we can write the above values as \(6.022 \times 10^{23}\) and \(1.66 \times 10^{-24}\ \text{g}\).
Shifting the Decimal Point
The number 123.546 becomes \(1.23546 \times 10^{2}\) in scientific notation. Note that while writing it, we have moved the decimal to the left by two places, and the same is the exponent (2) of 10 in the scientific notation.
Similarly, 0.00015 can be written as \(1.5 \times 10^{-4}\). Here the decimal has to be moved four places to the right, and \((-4)\) is the exponent in the scientific notation.
The pattern behind both cases is worth stating once. Moving the decimal point to the left makes the coefficient smaller, so the exponent must be made positive to compensate; moving it to the right makes the coefficient larger, so the exponent must be made negative. The value of the number never changes — only the way it is written.
Now let us perform mathematical operations on numbers expressed in scientific notation.
Addition and Subtraction
To perform addition, first the numbers are written in such a way that they have the same exponent. The coefficients are then added. The reason is simple: \(10^4\) and \(10^3\) are different-sized units, and quantities can only be added when they are expressed in the same unit. Subtraction follows exactly the same preparation.
Multiplication
The rule for the multiplication of exponential numbers can be well explained from the solved problems 2.5 and 2.6. In multiplication the coefficients are multiplied together and the exponents are added, because multiplying powers of the same base means adding their indices.
2.3.2 Precision and Accuracy of Measurement
The aim of any measurement is to get the actual value, called the true value or accepted value, of a quantity. Nearness of the measured value to the true value is called the accuracy of measurement. The larger the accuracy, the smaller the error.
Accuracy depends upon the sensitivity or least count — the smallest quantity that can be measured — of the measuring equipment. An instrument cannot report what it cannot resolve.
Consider, for example, a burette reading of 10.2 mL. For all the three situations shown in Fig. 2.1 the reading would be noted as 10.2 mL.
Diagram
Fig. 2.1 – Three possibilities of a burette reading 10.2 mL: three burette sections graduated from 9.0 to 11.0 with the liquid level at three slightly different positions between 10.1 and 10.3, each recorded as 10.2 mL; mark the least count as 0.1 mL.
It means that there is an uncertainty about the digit appearing after the decimal point in the reading 10.2 mL. This is because the least count of the burette is 0.1 mL.
The meaning of the reading 10.2 mL is that the true value of the reading lies between 10.1 mL and 10.3 mL. This is indicated by writing \(10.2 \pm 0.1\) mL. Here, the burette reading has an error of \(\pm 0.1\) mL (Fig. 2.1).
Errors: Absolute and Relative
Errors may be expressed as absolute or relative error.
\[\text{Absolute error} = \text{Observed value} - \text{True value}\]
Relative error is generally a more useful quantity than absolute error. Relative error is the ratio of an absolute error to the true value. It is expressed as a percentage.
\[\text{Relative error} = \frac{\text{Absolute error}}{\text{True value}} \times 100\ \%\]
Why is the relative form more useful? Because an absolute error of 0.1 g means very different things in different experiments. On a 0.5 g sample it is a serious error; on a 500 g sample it is negligible. Dividing by the true value makes the error speak in proportion to what was being measured, so that errors from different experiments can be compared.
There can be error in a measurement due to a number of reasons, including inefficiency of the person doing the measurement. A parallax error while reading a burette, a balance not levelled, a wet crucible — all of these push the reading away from the true value.
Precision and Deviation
Multiple readings of the same quantity are noted to minimize the error. If the readings match closely, they are said to have high precision. High precision implies reproducibility of the readings.
That last point deserves care, because it is the single most commonly tested idea in this section. Precision describes how the readings sit with respect to each other; accuracy describes how they sit with respect to the true value. A balance that reads 0.5 g too high every time will give beautifully reproducible — and consistently wrong — results.
Diagram
Fig. 2.4 – Four target boards, each with three shots marked, illustrating: (i) shots close together and on the bull’s-eye, (ii) shots close together but well off the bull’s-eye, (iii) shots scattered but grouped around the bull’s-eye, (iv) shots scattered and off the bull’s-eye; leave the four boards unlabelled and number them (i)–(iv).
Precision is expressed in terms of deviation. An absolute deviation is the modulus of the difference between an observed value and the arithmetic mean for the set of several measurements made in the same way. It is a measure of absolute error in the repeated observation.
\[\text{Absolute deviation} = \left|\,\text{Observed value} - \text{Mean}\,\right|\]
The modulus signs matter. Deviations are taken as positive quantities, because otherwise the positive and negative departures from the mean would cancel and every set of readings would appear perfect.
The arithmetic mean of all the absolute deviations is called the mean absolute deviation in the measurements. The ratio of mean absolute deviation to its arithmetic mean is called relative deviation.
\[\text{Relative deviation} = \frac{\text{Mean absolute deviation}}{\text{Mean}} \times 100\ \%\]
Notice how closely this mirrors the pair of error formulas above. Error is measured from the true value; deviation is measured from the mean. That is the only difference, and it is exactly the difference between accuracy and precision.
2.3.3 Significant Figures
Uncertainty in a measured value leads to uncertainty in the calculated result. Uncertainty in a value is indicated by mentioning the number of significant figures in that value.
Consider the column reading \(10.2 \pm 0.1\) mL recorded on a burette having the least count of 0.1 mL. Here it is said that the last digit ‘2’ in the reading is uncertain, its uncertainty being \(\pm 0.1\) mL. On the other hand, the figure ‘10’ is certain.
The significant figures in a measurement or result are the number of digits known with certainty plus one uncertain digit.
This is why the number of significant figures is not a matter of taste. It is a statement about the instrument that produced the reading. Writing 10.2 mL announces that the measurement came from an instrument reading to a tenth of a millilitre; writing 10.20 mL would announce something the burette never said.
In a scientific experiment a result is obtained by doing a calculation in which values of a number of quantities, measured with equipment of different least counts, are used. Following rules are to be followed during such calculation.
2.3.4 Rules for Deciding Significant Figures
All non-zero digits are significant; e.g. 127.34 g contains five significant figures, which are 1, 2, 7, 3 and 4.
All zeros between two non-zero digits are significant; e.g. 120.007 m contains six significant figures.
Zeros on the left of the first non-zero digit are not significant. Such a zero indicates the position of the decimal point. For example, 0.025 has two significant figures, and 0.005 has one significant figure.
Zeros at the end of a number are significant if they are on the right side of the decimal point. Terminal zeros are not significant if there is no decimal point. (This is because the least count of an instrument contains a decimal point.) For example, 0.400 g has three significant figures. The measurement here indicates that it is made on a weighing machine having a least count of 0.001 g.
Significant figures are also indicated in scientific notation by means of the decimal point. For example, the measurement 400 g has one significant figure. The measurement \(4.0 \times 10^2\) g has two significant figures, whereas the measurement \(4.00 \times 10^2\) g has three significant figures. The zeros after the decimal points in these cases indicate that the least counts of the weighing machines are 1 g, 0.1 g and 0.01 g, respectively.
In numbers written in scientific notation, all digits are significant. For example, \(2.035 \times 10^2\) has four significant figures, and \(3.25 \times 10^{-5}\) has three significant figures.
The scheme for applying these rules to any given number is set out in FC 2.3.
Flowchart
FC 2.3 – Counting significant figures: take the number → is the digit non-zero? → yes: significant. Is it a zero? → branch: zero lying between two non-zero digits → significant; zero to the left of the first non-zero digit → not significant (only fixes the decimal point); zero at the end WITH a decimal point present → significant; zero at the end WITHOUT a decimal point → not significant. Separate branch: number already in scientific notation N × 10ⁿ → all digits of N are significant.
2.3.5 Calculations with Significant Figures
When performing calculations with measured quantities, the rule is that the accuracy of the final result is limited to the accuracy of the least accurate measurement. In other words, the final result cannot be more accurate than the least accurate number involved in the calculation.
The logic is that of a chain. A calculated answer inherits the weaknesses of every value that went into it, so the weakest measurement sets the standard for the result. A calculator will happily print eight digits; most of them are not measurements at all, and reporting them would claim a precision the experiment never had.
Rounding Off
The final result of a calculation often contains figures that are not significant. When this occurs, the final result is rounded off. The following rules are used to round off a number to the required number of significant figures.
If the digit following the last digit to be kept is less than five, the last digit is left unchanged; e.g. 46.32 rounded off to two significant figures is 46.
If the digit following the last digit to be kept is five or more, the last digit to be kept is increased by one; e.g. 52.87 rounded to three significant figures is 52.9.
Flowchart
FC 2.4 – Rounding off to a required number of significant figures: decide how many significant figures are to be kept → look at the first digit to be dropped → branch: digit is less than 5 → leave the last kept digit unchanged and drop everything to its right; digit is 5 or more → increase the last kept digit by 1 and drop everything to its right → write the rounded value.
2.4 Determination of Molecular Formula
The molecular formula of a compound is the formula which indicates the actual number of atoms of the constituent elements in a molecule. When you write \(\mathrm{H_2SO_4}\), you are stating that one molecule truly contains two hydrogen atoms, one sulphur atom and four oxygen atoms — not a ratio, but a count.
The molecular formula can be obtained from the experimentally determined values of the percent elemental composition and the molar mass of that compound. Two pieces of information are therefore needed, and neither will do alone: the composition tells you the proportion in which the atoms are combined, and the molar mass tells you how many of them a single molecule actually holds.
2.4.1 Percent Composition and Empirical Formula
Compounds are formed by chemical combination of different elements. Quantitative determination of the constituent element by suitable methods provides the percent elemental composition of a compound — that is, the mass of each element present in 100 parts by mass of the compound.
If the percent total is not 100, the difference is considered as percent oxygen. This is a practical convention rather than a guess: oxygen is difficult to determine directly in the usual analytical procedures, so it is taken to be whatever mass is left unaccounted for once every other element has been measured.
From the percent composition, the ratio of the atoms of the constituent elements in the molecule is calculated. The simplest ratio of atoms of the constituent elements in a molecule is called the empirical formula of that compound.
The molecular formula can be obtained from the empirical formula if the molar mass is known. The molar mass of the substance under examination is determined by some convenient method.
Why the Ratio Comes from Moles, Not Masses
Before working through the steps, one point must be clear. The percent composition is a ratio of masses, but a formula is a ratio of numbers of atoms, and the two are not the same thing because atoms of different elements have different masses.
The bridge between them is the mole. Dividing the mass of an element by its atomic mass converts grams into moles, and moles are directly proportional to numbers of atoms. Only after that conversion does the ratio mean anything about a formula.
The Sequence of Steps
The complete route from laboratory data to molecular formula is set out in FC 2.5, and the following example illustrates this sequence.
Flowchart
FC 2.5 – From percent composition to molecular formula: percent composition of each element → check whether the percentages add up to 100 → if the total falls short, treat the difference as percent oxygen → take 100 g of the compound, so each percent becomes a mass in grams → divide each mass by the atomic mass of that element to get moles → divide all mole values by the smallest of them → if the ratio is whole numbers, write the empirical formula; if not, multiply all values by a suitable coefficient to make them whole numbers → add up atomic masses to get the empirical formula mass → r = molar mass ÷ empirical formula mass → molecular formula = r × empirical formula.
When the Ratio Is Not a Whole Number
The step that most often causes trouble is Step IV, because division by the smallest value does not always land on whole numbers. When it does not, the whole set is multiplied by the smallest coefficient that clears the fractions — and the values must be multiplied together, never rounded off individually, since a ratio of 1 : 1.33 is not 1 : 1.
Questions of this type are a standing feature of board and entrance papers, usually in one of three forms: percent composition to empirical formula, empirical formula plus molar mass to molecular formula, or the reverse question — given a formula, calculate the percentage of one element in it. All three are the same six steps read in different directions.
2.5 Chemical Reactions and Stoichiometric Calculations
Calculations based on a balanced chemical equation are known as stoichiometric calculations. Everything in this section rests on one idea: a balanced equation is not merely a record of what turns into what, it is a set of fixed numerical relationships that can be read and used.
A balanced chemical equation is a symbolic representation of a chemical reaction. It supplies the following information, which is useful in solving problems based on chemical equations:
It indicates the number of moles of the reactants involved in a chemical reaction and the number of moles of the products formed.
It indicates the relative masses of the reactants and products linked with a chemical change, and
It indicates the relationship between the volume or volumes of the gaseous reactants and products, at STP.
Reading the Three Kinds of Information
These three readings of the same equation are worth separating clearly, because a problem will ask for one of them and give you another.
The coefficients in a balanced equation are, first and foremost, mole ratios. When the equation for the combustion of methane carries a coefficient 2 before oxygen, it says that two moles of oxygen are consumed for every one mole of methane.
Multiply each mole figure by the corresponding molar mass and the same equation now reads as a mass relationship — so many grams of this reacting with so many grams of that. Mass is conserved through a chemical change, so the total mass on the left always equals the total on the right.
For gases there is a third reading. One mole of any ideal gas occupies 22.4 L at STP, so for gaseous species the mole ratio is also a volume ratio, provided the volumes are compared at the same temperature and pressure.
2.5.1 Stoichiometric Problems
Generally, problems based on stoichiometry are of the following types:
Problems based on mass-mass relationship;
Problems based on mass-volume relationship, and
Problems based on volume-volume relationship.
The three types differ only in what is given and what is asked; the equation and the reasoning behind it stay the same.
Steps in a Stoichiometric Calculation
Steps involved in problems based on stoichiometric calculations:
Write down the balanced chemical equation representing the chemical reaction.
Write the number of moles and the relative masses or volumes of the reactants and products below the respective formulae.
Relative masses or volumes should be calculated from the respective formula mass, referring to the condition of STP.
Apply the unitary method to calculate the unknown factor or factors as required by the problem.
The unitary method in Step 4 is the ordinary rule of proportion you have used since middle school: find what one unit corresponds to, then multiply up to the quantity asked for. The whole sequence is summarised in FC 2.6.
Flowchart
FC 2.6 – Solving a stoichiometric problem: read the problem and identify the substance given and the substance asked for → write the balanced chemical equation → write moles and relative masses (or volumes at STP) under each formula → branch: both given and asked are masses → mass–mass working; one is a mass and the other a gas volume → use 22.4 L per mole for the gas; both are gas volumes → use the coefficients directly as a volume ratio → apply the unitary method → write the answer with its unit.
In board and entrance papers these calculations appear constantly, most often as “find the mass of the product”, “find the volume of gas at STP” or “find the percentage purity of the sample”. The single most common cause of a lost mark is not the arithmetic but the first step — writing an equation that is not balanced, after which every proportion built on it is wrong.
2.6 Limiting Reagent
When a chemist carries out a reaction, the reactants are not usually present in exact stoichiometric amounts, that is, in the proportions indicated by the balanced equation. The equation says what the reaction needs; the flask contains whatever was weighed out and poured in.
Because the goal of a reaction is to produce the maximum quantity of a useful compound from the starting materials, frequently a large excess of one reactant is supplied. This is done to ensure that the more expensive reactant is completely converted into the desired product. It is a deliberate choice: the cheaper substance is wasted so that none of the costly one is left behind unreacted.
The reactant which is present in lesser amount gets consumed after some time and subsequently, no further reaction takes place, whatever be the amount left of the other reactant present. Once one partner has run out, the reaction simply stops, no matter how much of the other is still sitting in the flask.
Hence, the reactant which gets consumed limits the amount of product formed and is therefore called the limiting reagent.
Which Reactant Runs Out First?
The point that has to be grasped here is that “present in lesser amount” is not decided by mass, and not decided by the number of moles taken either. It is decided by comparing what each reactant can produce.
Consider the formation of nitrogen dioxide (\(\mathrm{NO_2}\)) from nitric oxide (NO) and oxygen.
\[2\,\mathrm{NO}\text{(g)} + \mathrm{O_2}\text{(g)} \longrightarrow 2\,\mathrm{NO_2}\text{(g)}\]
Suppose initially we have 8 moles of NO and 7 moles of \(\mathrm{O_2}\). One way to determine which of the two reactants is the limiting reagent is to calculate the number of moles of \(\mathrm{NO_2}\) obtainable from the given initial quantities of NO and \(\mathrm{O_2}\).
From the preceding definition, we see that the limiting reagent will yield the smaller amount of the product. Starting with 8 moles of NO, we find the number of moles of \(\mathrm{NO_2}\) produced is
\[8\ \text{mol NO} \times \frac{2\ \text{mol } \mathrm{NO_2}}{2\ \text{mol NO}} = 8\ \text{mol } \mathrm{NO_2}\]
and starting with 7 moles of \(\mathrm{O_2}\), the number of moles of \(\mathrm{NO_2}\) formed is
\[7\ \text{mol } \mathrm{O_2} \times \frac{2\ \text{mol } \mathrm{NO_2}}{1\ \text{mol } \mathrm{O_2}} = 14\ \text{mol } \mathrm{NO_2}\]
Because 8 moles of NO result in a smaller amount of \(\mathrm{NO_2}\), NO must be the limiting reagent, and \(\mathrm{O_2}\) is the excess reagent, before the reaction has started.
Look carefully at what has just been done. Each calculation asks a “what if” question — what if this reactant were the one to be used up completely? Only one of those two answers can actually happen, and it must be the smaller one, because the reaction cannot go on producing \(\mathrm{NO_2}\) after one of its ingredients has vanished. The larger figure of 14 mol is not a possible yield at all; it is what the oxygen could have given if enough NO had been supplied.
Note also that fewer moles of oxygen were taken than of nitric oxide, and yet the oxygen is the one in excess. The mole ratio in the equation is 2 : 1, so a mole of oxygen goes twice as far as a mole of NO. This is exactly why a limiting reagent cannot be identified by inspection.
Flowchart
FC 2.7 – Identifying the limiting reagent: write the balanced equation → convert the given quantity of EACH reactant into moles → for reactant A, calculate moles of product using the mole ratio from the equation → for reactant B, calculate moles of product the same way → compare the two answers → the reactant giving the SMALLER amount of product is the limiting reagent; the other is the excess reagent → use the smaller value to calculate the mass of product → to find the excess left: moles of excess reagent actually consumed → mass consumed → subtract from the mass originally taken.
Diagram
Fig. 2.5 – Graph of moles of NO₂ formed (y-axis, 0 to 10) against moles of O₂ added (x-axis, 0 to 8), for a fixed 8 mol of NO: a straight line rising from the origin up to the point (4, 8), then a horizontal plateau at 8 mol NO₂ for all larger amounts of O₂; mark the point (4, 8) and leave the two regions unlabelled.
In examinations, limiting-reagent questions are recognised by a single feature: quantities are given for two reactants instead of one. Whenever that happens, the first move is not to start calculating the product, but to test both reactants and find out which one the reaction will run out of.
2.7 Concentration of Solution
A majority of reactions in the laboratory are carried out in solutions. Solids react slowly and unevenly when simply mixed as powders, but once dissolved their particles move freely and meet one another readily, so reactions in solution are faster and far easier to control.
Therefore, it is important to understand how the amount of substance is expressed when it is present in the form of a solution. Two words are needed throughout this section: the solute is the substance that dissolves, and the solvent is the liquid it dissolves in. Together they make the solution, so the mass of the solution is the mass of the solute plus the mass of the solvent.
The concentration of a solution, or the amount of substance present in a given volume of a solution, can be expressed in any of the following ways:
- Mass percent or weight percent (w/w %)
- Mole fraction
- Molarity (M)
- Molality (m)
All four answer the same question — how much solute is present in how much of the mixture — and differ only in the units chosen for “how much” in each half of that sentence.
2.7.1 Mass Percent
Mass percent is obtained by using the following relation:
\[\text{Mass percent} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100\ \%\]
Notice that the denominator is the mass of the solution, not of the solvent. A common slip is to divide by the solvent alone, which gives a different and larger number.
2.7.2 Mole Fraction
Mole fraction is the ratio of the number of moles of a particular component of a solution to the total number of moles of the solution.
If a substance ‘A’ dissolves in substance ‘B’ and their numbers of moles are \(n_A\) and \(n_B\) respectively, then the mole fractions of A and B are given as:
\[\text{Mole fraction of A} = \frac{\text{No. of moles of A}}{\text{No. of moles of solution}}\]
\[\therefore\ \text{Mole fraction of A} = \frac{n_A}{n_A + n_B}\]
\[\text{Mole fraction of B} = \frac{\text{No. of moles of B}}{\text{No. of moles of solution}}\]
\[= \frac{n_B}{n_A + n_B}\]
Mole fraction is the only one of the four measures that is a pure number: moles are divided by moles, so the units cancel and no unit is written after the answer.
2.7.3 Molarity
Molarity is the most widely used unit and is denoted by M. It is defined as the number of moles of the solute present in 1 litre of the solution. Thus,
\[\text{Molarity (M)} = \frac{\text{No. of moles of solute}}{\text{Volume of solution in litres}}\]
Molarity is popular for a practical reason. Solutions are handled by volume in the laboratory — measured out with a pipette or run in from a burette — so a concentration expressed per litre of solution converts a measured volume into moles at once.
Diagram
Fig. 2.6 – Preparing a solution of known molarity: (i) the weighed solute on a watch glass being transferred to a volumetric flask through a funnel, (ii) the flask being swirled with a little water to dissolve the solute, (iii) water being added from a wash bottle up to the calibration mark on the neck; label the calibration mark and the stated capacity (250 mL) on the flask.
Note carefully, as shown in Fig. 2.6, that the solute is made up to the mark with solvent rather than being added to a litre of solvent. The volume in the definition is the volume of the final solution, not of the water used.
2.7.4 Molality
Molality is defined as the number of moles of solute present in 1 kg of solvent. It is denoted by m.
\[\text{Molality (m)} = \frac{\text{No. of moles of solute}}{\text{Mass of solvent in kg}}\]
Note that molality of a solution does not change with temperature, since mass remains unaffected with temperature.
Two differences from molarity are packed into that short definition, and both matter. The denominator is a mass, not a volume; and it is the mass of the solvent alone, not of the whole solution.
Stock Solutions and Dilution
Often in the chemistry laboratory, a solution of desired concentration is prepared by diluting a solution of known higher concentration. The solution of higher concentration is also known as stock solution.
Working this way saves both time and accuracy. A concentrated solution can be prepared and standardised once, and many weaker solutions drawn from it afterwards, without weighing out a fresh small quantity of solute — which is exactly the kind of small weighing that carries the largest relative error.
Flowchart
FC 2.8 – Converting molarity to molality: take exactly 1 L of the solution → moles of solute = molarity × 1 L → mass of solute = moles × molar mass → mass of 1 L of solution = 1000 mL × density → mass of solvent = mass of solution − mass of solute → convert mass of solvent from g to kg → molality = moles of solute ÷ mass of solvent in kg.
Concentration questions appear in board and entrance papers both as straight one-step calculations and as conversions between two of these units — most often molarity to molality using density, as in Solved Example 2.25. The conversions all turn on keeping strict track of one thing: whether the quantity in the denominator refers to the solution or to the solvent.
2.8 Use of Graph in Analysis
Analytical chemistry also involves deducing some relation, if any, between two or more properties of matter under study. Measuring a quantity is only the first task; the second is to find out whether the numbers obtained hang together in a pattern, and a graph is the instrument for that.
One of the classic examples is the relation between the temperature and the volume of a given amount of gas. A set of experimentally measured values of volume and temperature of a definite mass of a gas, upon plotting on a graph paper, appeared as in the figure (Fig. 2.2 (a)).
Diagram
Fig. 2.2 – Drawing an average curve through the points on a graph, in four panels sharing axes V (vertical) and T (horizontal): (a) the plotted experimental points alone, scattered about a rising line; (b) the same points joined directly to one another, giving a zig-zag; (c) a single smooth straight line drawn through the scatter as an average curve; (d) the same straight line with a short perpendicular dropped from each point to the line, perpendiculars on one side drawn in red and those on the other side in blue.
Why the Points Scatter
When the points are directly connected, a zig zag pattern results (Fig. 2.2 (b)). From this pattern no meaningful result can be deduced.
A zig zag pattern results due to the many types of errors that occur in the many measurements involved in an experiment. Every point on the graph is the outcome of two separate readings — one of volume, one of temperature — and each of those readings carries the uncertainty of the instrument that produced it, as discussed in Section 2.3.
Joining the points therefore draws a picture of the errors rather than of the relationship. The line rises and falls in a way that has no physical meaning at all, and any slope read from a segment of it would be an accident of where a particular reading happened to fall.
Drawing the Average Curve
Figure 2.2 (c) shows a smooth curve which may be called an average curve passing through these points. In the above example it happens to be a straight line, and the inference is that
\[V \propto T\]
Read that inference carefully. It is drawn not from any single point but from the shape of the whole set, which is exactly why the average curve is worth the trouble. Individual points may sit above or below it; the trend they share is what the line records. This particular proportionality holds for a fixed mass of gas at constant pressure, with the temperature taken on the kelvin scale, and you will meet it again in the study of the gaseous state.
Testing the Curve: Even Distribution
While fitting it to a smooth curve, care is taken that the plotted points are evenly distributed about it. A line that leaves most of the points on one side of it has not represented them fairly, however smooth it may look.
Mathematically, ‘even distribution’ is understood as follows. From each point, draw a perpendicular to the curve. The perpendicular represents the deviation of each point from the curve (Fig. 2.2 (d)).
The positive deviations are shown in red and the negative deviations are shown in blue. Take the sum of all the red perpendiculars and all the blue perpendiculars separately. If the two sums are equal (or nearly equal), the curve drawn shows the experimental points in the best possible representation.
This test is the graphical twin of an idea already met in Section 2.3. There, deviations of repeated readings were measured from the arithmetic mean, and they cancelled out around it. Here, deviations of plotted points are measured from the curve, and the well-drawn curve is the one they cancel around. In both cases the aim is the same: to let the errors balance instead of letting one of them decide the answer.
Flowchart
FC 2.9 – Using a graph in analysis: measure pairs of values of the two properties → plot the points on graph paper with suitable scales → do NOT join the points to one another → draw a single smooth curve or straight line passing among the points → drop a perpendicular from each point to the curve → sum the positive deviations and the negative deviations separately → branch: the two sums are equal or nearly equal → accept the curve as the best representation; the sums differ appreciably → redraw the curve and test again → read the relation between the two properties from the shape of the accepted curve.
In examinations, graph-based questions rarely ask you to draw anything. They show a plot and ask what it means: which curve represents the data best, what relation between the two quantities the shape implies, what the slope or intercept stands for, and what a point lying well off the line indicates. All of these are answered from the reasoning in this section — the shape of the whole set carries the relation, and the scatter about it carries the error.
2.9 Summary
Analytical chemistry is the branch of chemistry dealing with the separation, identification, and qualitative and quantitative determination of the compositions of different substances. It uses instruments and methods to separate, identify and quantify the matter under study.
Analysis gives chemical or physical information about a sample. Qualitative analysis detects the presence or absence of elements in compounds and mixtures of compounds; quantitative analysis determines the relative proportions of elements in them. The qualitative question is normally settled first.
Chemical analysis is used to monitor raw materials, intermediates and finished products, and the air in streets and industrial premises. It is applied in agriculture (soils and fertilizers), medicine (medicinal preparations), forensic science, engineering and industry. It consists of classical, wet chemical methods and modern instrumental methods.
Analysis is carried out on a small sample and not on the entire bulk. When the sample of a solid or liquid is a few grams, the analysis is called semi-microanalysis.
Classical qualitative methods include separations such as precipitation, extraction and distillation; identification may be based on differences in colour, odour, melting point, boiling point and reactivity. Classical quantitative methods consist of volumetric and gravimetric analysis.
Chemical qualitative analysis is carried out in two stages: the dry method, in which the sample is not dissolved and which serves as the preliminary test, and the wet method, in which the sample is first dissolved and then analyzed. Apparatus used includes test tubes, beakers, evaporating dish, crucible, spot plate, watch glass, wire gauze, water bath, burner, blow pipe, pair of tongs and centrifuge.
Organic compounds are made up of a small number of elements — carbon, hydrogen, oxygen, nitrogen, sulphur, halogen, phosphorus — so their identification also needs detection of the functional group and determination of melting or boiling point. Qualitative analysis of simple inorganic compounds involves detection and confirmation of the cationic and anionic species, that is, the basic and acidic radicals.
Quantitative analysis of organic compounds includes determination of the percentage of a constituent element and of the concentration of a known compound in a sample. For simple inorganic compounds it is based on decomposition reactions (gravimetric analysis) and on the progress of a reaction between two solutions till completion (titrimetric or volumetric analysis). Such methods measure mass and volume, using apparatus such as a weighing machine and a burette.
Very large and very small numbers are written in scientific (exponential) notation as \(N \times 10^n\), where \(n\) may be positive or negative and \(1 \leq N < 10\). For addition and subtraction the numbers are first written with the same exponent and the coefficients are then added or subtracted. For multiplication the coefficients are multiplied and the exponents are added.
Accuracy is the nearness of the measured value to the true or accepted value; greater accuracy means smaller error. It depends on the least count, the smallest quantity an instrument can measure. A burette of least count 0.1 mL gives a reading such as \(10.2 \pm 0.1\) mL, meaning the true value lies between 10.1 mL and 10.3 mL.
Errors are expressed as absolute or relative:
\[\text{Absolute error} = \text{Observed value} - \text{True value}\]
\[\text{Relative error} = \frac{\text{Absolute error}}{\text{True value}} \times 100\ \%\]
- Repeated readings that match closely are said to have high precision, which means reproducibility. High precision is a prerequisite for high accuracy, but does not by itself guarantee it. Precision is expressed through deviations:
\[\text{Absolute deviation} = \left|\,\text{Observed value} - \text{Mean}\,\right|\]
\[\text{Relative deviation} = \frac{\text{Mean absolute deviation}}{\text{Mean}} \times 100\ \%\]
Significant figures in a measurement or result are the digits known with certainty plus one uncertain digit. In deciding them: all non-zero digits are significant; zeros between two non-zero digits are significant; zeros to the left of the first non-zero digit are not; terminal zeros are significant only if a decimal point is present; and in scientific notation all digits are significant.
In calculations, the accuracy of the final result is limited to the accuracy of the least accurate measurement. Results are rounded off: if the digit following the last digit to be kept is less than five, the last digit is left unchanged; if it is five or more, the last digit kept is increased by one. A smaller least count gives more significant figures and greater accuracy.
The molecular formula gives the actual number of atoms of each element in a molecule; the empirical formula gives the simplest ratio of those atoms. Percent composition gives the empirical formula; if the percentages fall short of 100, the difference is taken as percent oxygen. The molecular formula follows once the molar mass is known:
\[r = \frac{\text{Molar mass}}{\text{Empirical formula mass}}, \qquad \text{Molecular formula} = r \times \text{Empirical formula}\]
Stoichiometric calculations are calculations based on a balanced chemical equation. A balanced equation gives the number of moles of reactants and products, their relative masses, and the relationship between the volumes of gaseous reactants and products at STP, where one mole of an ideal gas occupies 22.4 L. Problems are of the mass–mass, mass–volume and volume–volume types, and are solved by writing the balanced equation, entering the moles and relative masses or volumes below the formulae, and applying the unitary method.
The reactant that is completely consumed first is the limiting reagent; it limits the amount of product formed, while the other reactant is the excess reagent. It is identified by calculating the amount of product obtainable from each reactant separately and taking the smaller value.
The concentration of a solution may be expressed as mass percent, mole fraction, molarity or molality:
\[\text{Mass percent} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100\ \%\]
\[\text{Mole fraction of A} = \frac{n_A}{n_A + n_B}, \qquad \text{Mole fraction of B} = \frac{n_B}{n_A + n_B}\]
\[\text{Molarity (M)} = \frac{\text{No. of moles of solute}}{\text{Volume of solution in litres}}\]
\[\text{Molality (m)} = \frac{\text{No. of moles of solute}}{\text{Mass of solvent in kg}}\]
Molarity depends on temperature, because the volume of a solution is temperature dependent; molality does not, because mass is unaffected by temperature. A solution of desired concentration is often prepared by diluting a solution of higher concentration, called the stock solution.
Analysis also deduces relations between two or more properties by plotting a graph. Joining the plotted points directly gives a zig-zag from which nothing can be deduced, because it traces the experimental errors. A smooth average curve is drawn instead; for the volume and temperature of a definite mass of gas it is a straight line, giving \(V \propto T\). The curve is best drawn when the points are evenly distributed about it — that is, when the sum of the positive perpendicular deviations equals, or nearly equals, the sum of the negative ones.
Flowchart
FC 2.10 – The chapter as one sequence: sample drawn from the bulk (semi-micro, a few grams) → QUALITATIVE ANALYSIS: dry method (preliminary tests) → wet method → identity established (elements and functional group for organic; cationic and anionic radicals for inorganic) → QUANTITATIVE ANALYSIS: measure mass and volume (gravimetric, volumetric) → treat the readings: scientific notation → error and deviation → significant figures and rounding off → CALCULATE: percent composition → empirical formula → molecular formula (using molar mass); balanced equation → stoichiometric calculation (mass–mass, mass–volume, volume–volume) → limiting reagent where two reactants are given; express solution strength as mass percent, mole fraction, molarity or molality → PRESENT: plot the results and draw the average curve → report the relation between the properties.
2.10 Check Your Concepts
Analysis is deliberately carried out on a small sample and not on the entire bulk of a material. Explain why this is justified, and describe one situation in which it would give a misleading result.
Why is the dry method carried out before the wet method in qualitative analysis, when it is the wet method that gives the confirmatory result?
The qualitative analysis of an organic compound does not stop at detecting carbon, hydrogen, oxygen and nitrogen, whereas the analysis of a simple inorganic salt is essentially complete once the cation and the anion are identified. Explain the reason for this difference.
Assertion: A set of measurements can be highly precise and yet highly inaccurate. Reason: Precision is judged from the agreement of repeated readings with one another, while accuracy is judged from their nearness to the true value. State whether the assertion and the reason are true, and whether the reason correctly explains the assertion.
“High precision is a prerequisite for high accuracy.” Explain what this statement means, and explain why the converse — that high precision guarantees high accuracy — is not true.
A student reports a burette reading as 15.24 mL using a burette of least count 0.1 mL. Identify the error in the way the result has been reported, and state what the student should have written instead.
The measurements 400 g, \(4.0 \times 10^2\) g and \(4.00 \times 10^2\) g are numerically equal. Explain what each one tells you about the balance on which it was measured.
Why is relative error generally more useful than absolute error when comparing the results of two different experiments? Illustrate your answer with an example of your own.
While calculating the mean absolute deviation of a set of readings, the deviations are taken as positive quantities. Explain what would happen to the calculated deviation if the signs were retained, and why this would be misleading.
In determining an empirical formula from percent composition, why must the mass percentages be converted into moles before the ratio is taken? What error would result from taking the ratio of the masses directly?
While working out an empirical formula, a student obtains the mole ratio 1 : 1.5 and rounds it to 1 : 2. Explain why this is incorrect and state the correct procedure.
Two compounds have exactly the same percentage composition but different molar masses. State what this tells you about their empirical and molecular formulas, and explain how the two compounds could be distinguished analytically.
In an analysis, the percentages of all the elements determined add up to 87.5 %. State what is assumed about the remaining 12.5 %, and explain why this assumption is made rather than treating the shortfall as experimental error.
Match the following:
| Column A | Column B |
|---|---|
| a. Gravimetric analysis | i. Number of moles of solute per kilogram of solvent |
| b. Volumetric analysis | ii. Simplest whole-number ratio of atoms |
| c. Empirical formula | iii. Based on measurement of mass |
| d. Molality | iv. Reactant completely consumed first |
| e. Limiting reagent | v. Based on the volume of a solution required for complete reaction |
In an industrial reaction, one reactant is deliberately taken in large excess. State which reactant is normally chosen for this and why, and describe one disadvantage of using a large excess of a costly reactant.
Equal numbers of moles of two reactants are taken, and yet one of them turns out to be the limiting reagent. Explain how this is possible, and state what information is needed before the limiting reagent can be identified.
A vessel contains 8 mol of NO and 7 mol of \(\mathrm{O_2}\), which react according to \(2\,\mathrm{NO}\text{(g)} + \mathrm{O_2}\text{(g)} \rightarrow 2\,\mathrm{NO_2}\text{(g)}\). Without any calculation of masses, explain how you would decide which gas limits the amount of \(\mathrm{NO_2}\) formed.
A solution is warmed from 298 K to 318 K in a closed flask. State, with reasons, which of its mass percent, mole fraction, molarity and molality would change.
Why is molarity the concentration unit most often used in laboratory work, even though it varies with temperature?
A dilute solution of an acid is to be prepared from a stock solution of the concentrated acid. State the order in which the acid and the water must be mixed, and explain the reason behind the rule.
When a set of experimental points is plotted, joining the points one to the next gives a zig-zag from which no conclusion can be drawn. Explain what the zig-zag actually represents, and what is done instead.
State the test used to decide whether an average curve drawn through a set of plotted points is the best representation of those points, and explain why a curve that leaves most of the points on one side of it is unsatisfactory.
While reading a value from a graph, why is it better to take the value from the average curve than from a nearby plotted point?
Define: Least count.
What do you mean by significant figures? State the rules for deciding significant figures.
Distinguish between accuracy and precision.
Explain the terms percentage composition, empirical formula and molecular formula.
What is a limiting reagent? Explain.
What do you mean by SI units? What is the SI unit of mass?
Explain the following terms: (a) Mole fraction (b) Molarity (c) Molality
Define: Stoichiometry.
Why is there a need of rounding off figures during calculation?
Why does molarity of a solution depend upon temperature?
Define analytical chemistry. Why is accurate measurement crucial in science?
2.11 Practice With Numericals
In an experiment, the mass of oxygen obtained from a sample is 2.45 g, while the accepted value is 2.50 g. Calculate the absolute error and the relative error, and state what the sign of your answer indicates.
Four determinations of the same quantity give 5.12 g, 5.16 g, 5.14 g and 5.18 g. Calculate the mean, the absolute deviation of each reading, the mean absolute deviation and the relative deviation.
State the number of significant figures in each of the following: (a) 0.00620 kg, (b) 2500 mL, (c) \(2.500 \times 10^3\) mL, (d) 90.007 g, (e) 0.0105 L, (f) 6.0200 cm.
Round off each of the following as indicated: (a) 8.746 to three significant figures, (b) 0.03451 to two significant figures, (c) 145.5 to three significant figures, (d) \(7.2649 \times 10^{-3}\) to three significant figures.
Carry out the following operations and express each answer in correct scientific notation:
- \(2.4 \times 10^5 + 3.6 \times 10^4\)
- \(8.5 \times 10^{-3} - 6.2 \times 10^{-4}\)
- \((6.4 \times 10^{-3}) \times (2.5 \times 10^{2})\)
- \(\dfrac{8.4 \times 10^6}{2.1 \times 10^2}\)
A metal block has a mass of 12.48 g measured on a balance of least count 0.01 g, and a volume of 4.5 mL measured in a cylinder of least count 0.1 mL. Calculate its density and report the answer to the correct number of significant figures, stating which measurement limits the result.
A compound contains 52.2 % carbon, 13.0 % hydrogen and 34.8 % oxygen by mass, and its molar mass is 46 g mol⁻¹. Determine its empirical formula and its molecular formula. (At. mass: C = 12, H = 1, O = 16)
Calculate the percentage composition of each element in calcium hydroxide, \(\mathrm{Ca(OH)_2}\). (At. mass: Ca = 40, O = 16, H = 1)
An oxide of nitrogen contains 25.9 % nitrogen by mass and has a molar mass of 108 g mol⁻¹. Find its empirical formula, the value of \(r\), and its molecular formula. (At. mass: N = 14, O = 16)
Magnesium burns in oxygen according to the equation
\[2\,\mathrm{Mg}\text{(s)} + \mathrm{O_2}\text{(g)} \rightarrow 2\,\mathrm{MgO}\text{(s)}\]
Calculate the mass of magnesium oxide formed when 4.8 g of magnesium is burnt completely. (At. mass: Mg = 24, O = 16)
What volume of carbon dioxide, measured at STP, is obtained by the complete decomposition of 10 g of calcium carbonate? (At. mass: Ca = 40, C = 12, O = 16; molar volume at STP = 22.4 L)
10 L of methane is burnt completely in oxygen, all volumes being measured at the same temperature and pressure:
\[\mathrm{CH_4}\text{(g)} + 2\,\mathrm{O_2}\text{(g)} \rightarrow \mathrm{CO_2}\text{(g)} + 2\,\mathrm{H_2O}\text{(g)}\]
Calculate the volume of oxygen required and the volume of carbon dioxide formed.
2.5 g of impure limestone, on complete decomposition, gives 0.448 L of carbon dioxide measured at STP. Calculate the percentage purity of the limestone sample. (At. mass: Ca = 40, C = 12, O = 16)
14 g of nitrogen is allowed to react with 6 g of hydrogen according to
\[\mathrm{N_2}\text{(g)} + 3\,\mathrm{H_2}\text{(g)} \rightarrow 2\,\mathrm{NH_3}\text{(g)}\]
- Identify the limiting reagent. (b) Calculate the mass of ammonia formed. (c) Calculate the mass of the excess reagent left unreacted. (At. mass: N = 14, H = 1)
- 5.6 g of calcium oxide is treated with 8.8 g of carbon dioxide:
\[\mathrm{CaO}\text{(s)} + \mathrm{CO_2}\text{(g)} \rightarrow \mathrm{CaCO_3}\text{(s)}\]
Determine the limiting reagent, the mass of calcium carbonate formed, and the mass of the excess reagent remaining. (At. mass: Ca = 40, C = 12, O = 16)
Calculate the molarity of a solution prepared by dissolving 4.9 g of \(\mathrm{H_2SO_4}\) in enough water to make 500 mL of solution. (Molar mass of \(\mathrm{H_2SO_4}\) = 98 g mol⁻¹)
12 g of a solute of molar mass 60 g mol⁻¹ is dissolved in 250 g of water. Calculate the molality of the solution.
A solution is prepared by dissolving 40 g of sodium hydroxide in 90 g of water. Calculate (a) the mass percent of the solute and (b) the mole fraction of each component. (Molar masses: NaOH = 40 g mol⁻¹, \(\mathrm{H_2O}\) = 18 g mol⁻¹)
The density of a 2 M solution of NaCl is 1.10 g mL⁻¹. Calculate the molality of the solution. (Molar mass of NaCl = 58.5 g mol⁻¹)
A fixed mass of gas at constant pressure occupies 400 mL at 320 K. Using the relation \(V \propto T\) read from the average curve, calculate the volume the gas would occupy at 400 K.
How many significant figures are in each of the following quantities?
- 45.26 ft
- 0.109 in
- 0.00025 kg
- \(2.3659 \times 10^{-8}\) cm
- 52.0 cm³
- 0.00020 kg
- \(8.50 \times 10^4\) mm
- 300.0 cg
Round off each of the following quantities to two significant figures:
- 25.55 mL
- 0.00254 m
- \(1.491 \times 10^5\) mg
- 199 g
Round off each of the following quantities to three significant figures:
- 1.43 cm³
- \(458 \times 10^2\) cm
- 643 cm²
- 0.039 m
- \(6.398 \times 10^{-3}\) km
- 0.0179 g
- 79,000 m
- 42,150
- 649.85
- 23,642,000 mm
- 0.0041962 kg
Express the following sums to the appropriate number of significant figures:
- \(2.3 \times 10^3\) mL \(+\ 4.22 \times 10^4\) mL \(+\ 9.04 \times 10^3\) mL \(+\ 8.71 \times 10^5\) mL
- 319.5 g \(-\) 20460 g \(-\) 0.0639 g \(-\) 45.642 g \(-\) 4.173 g
Express the following quantities in exponential terms.
- 0.0003498
- 235.4678
- 70000.0
- 1569.00
Give the number of significant figures in each of the following:
- \(1.230 \times 10^4\)
- 0.002030
- \(1.23 \times 10^4\)
- \(1.89 \times 10^{-4}\)
Express the quantities in (b) above with or without exponents, as the case may be.
Find out the molar masses of the following compounds:
- Copper sulphate crystal (\(\mathrm{CuSO_4 \cdot 5H_2O}\))
- Sodium carbonate, decahydrate (\(\mathrm{Na_2CO_3 \cdot 10H_2O}\))
- Mohr’s salt \([\mathrm{FeSO_4(NH_4)_2SO_4 \cdot 6H_2O}]\) (At. mass: Cu = 63.5; S = 32; O = 16; H = 1; Na = 23; C = 12; Fe = 56; N = 14)
Work out the percentage composition of the constituent elements in the following compounds:
- Lead phosphate \([\mathrm{Pb_3(PO_4)_2}]\)
- Potassium dichromate (\(\mathrm{K_2Cr_2O_7}\))
- Macrocosmic salt — sodium ammonium hydrogen phosphate, tetrahydrate (\(\mathrm{NaNH_4HPO_4 \cdot 4H_2O}\)) (At. mass: Pb = 207; P = 31; O = 16; K = 39; Cr = 52; Na = 23; N = 14)
Find the percentage composition of the constituents of green vitriol crystals (\(\mathrm{FeSO_4 \cdot 7H_2O}\)). Also find out the mass of iron and the water of crystallisation in 4.54 kg of the crystals. (At. mass: Fe = 56; S = 32; O = 16)
The red colour of blood is due to a compound called “haemoglobin”. It contains 0.335 % of iron. Four atoms of iron are present in one molecule of haemoglobin. What is its molecular weight? (At. mass: Fe = 55.84)
A substance, on analysis, gave the following percent composition: Na = 43.4 %, C = 11.3 % and O = 45.3 %. Calculate the empirical formula. (At. mass: Na = 23 u, C = 12 u, O = 16 u)
Assuming the atomic weight of a metal M to be 56, find the empirical formula of its oxide containing 70.0 % of M.
1.00 g of a hydrated salt contains 0.2014 g of iron, 0.1153 g of sulfur, 0.2301 g of oxygen and 0.4532 g of water of crystallisation. Find the empirical formula. (At. wt.: Fe = 56; S = 32; O = 16)
An organic compound containing oxygen, carbon, hydrogen and nitrogen contains 20 % carbon, 6.7 % hydrogen and 46.67 % nitrogen. Its molecular mass was found to be 60. Find the molecular formula of the compound.
A compound on analysis gave the following percentage composition by mass: H = 9.09; O = 36.36; C = 54.55. The molar mass of the compound is 88. Find its molecular formula.
Carbohydrates are compounds containing only carbon, hydrogen and oxygen. When heated in the absence of air, these compounds decompose to form carbon and water. If 310 g of a carbohydrate leaves a residue of 124 g of carbon on heating in the absence of air, what is the empirical formula of the carbohydrate?
Write each of the following in exponential notation:
- 3,672,199
- 0.000098
- 0.00461
- 198.75
Write each of the following numbers in ordinary decimal form:
- \(3.49 \times 10^{-11}\)
- \(3.75 \times 10^{-1}\)
- \(5.16 \times 10^4\)
- \(43.71 \times 10^{-4}\)
- \(0.011 \times 10^{-3}\)
- \(14.3 \times 10^{-2}\)
- \(0.00477 \times 10^5\)
- 5.00858585
Perform each of the following calculations. Round off your answers to two digits.
- \(\dfrac{1}{3.40 \times 10^{24}}\)
- \(\dfrac{33}{9.00 \times 10^{-4}}\)
- \(\dfrac{1.4 \times 10^9}{(2.77 \times 10^3)(3.76 \times 10^5)}\)
- \(\dfrac{(4 \times 10^{-3})(9.9 \times 10^{-7})}{(789)(1.002 \times 10^{-10})(0.3 \times 10^2)}\)
Perform each of the following calculations. Round off your answers to three digits.
- \((3.26 \times 10^4)(1.54 \times 10^6)\)
- \((8.39 \times 10^7)(4.53 \times 10^9)\)
- \(\dfrac{8.94 \times 10^6}{4.35 \times 10^4}\)
- \(\dfrac{(9.28 \times 10^9)(9.9 \times 10^{-7})}{(511)(2.98 \times 10^{-6})}\)
Perform the following operations:
- \(3.971 \times 10^7 + 1.98 \times 10^4\)
- \(1.05 \times 10^{-4} - 9.7 \times 10^{-5}\)
- \(4.11 \times 10^{-3} + 8.1 \times 10^{-4}\)
- \(2.12 \times 10^6 - 3.5 \times 10^5\)
A 1.000 mL sample of acetone, a common solvent used as a paint remover, was placed in a small bottle whose mass was known to be 38.0015 g. The following values were obtained when the acetone-filled bottle was weighed: 38.7798 g, 38.7795 g and 38.7801 g. How would you characterise the precision and accuracy of these measurements if the actual mass of the acetone was 0.7791 g?
Your laboratory partner was given the task of measuring the length of a box (approx. 5 in) as accurately as possible, using a metre stick graduated in millimetres. He supplied you with the following measurements: 12.65 cm, 12.6 cm, 12.65 cm, 12.655 cm, 126.55 mm, 12 cm.
- State which of the measurements you would accept, giving the reason.
- Give your reason for rejecting each of the others.
What weight of calcium oxide will be formed on heating 19.3 g of calcium carbonate? (At. wt.: Ca = 40; C = 12; O = 16)
The hourly energy requirements of an astronaut can be satisfied by the energy released when 34 grams of sucrose are “burnt” in his body. How many grams of oxygen would be needed to be carried in a space capsule to meet his requirement for one day?
2.12 Objective Questions
- The branch of chemistry which deals with the study of separation, identification, and quantitative determination of the composition of different substances is called …………
- Physical chemistry
- Inorganic chemistry
- Organic chemistry
- Analytical chemistry
- Which one of the following properties of matter is NOT quantitative in nature?
- Mass
- Length
- Colour
- Volume
- SI unit of mass is ……..
- kg
- mol
- pound
- m³
- The number of significant figures in \(1.50 \times 10^4\) g is ………..
- 2
- 3
- 4
- 6
- In Avogadro’s constant \(6.022 \times 10^{23}\) mol⁻¹, the number of significant figures is ……….
- 3
- 4
- 5
- 6
- By decomposition of 25 g of \(\mathrm{CaCO_3}\), the amount of CaO produced will be …………
- 2.8 g
- 8.4 g
- 14.0 g
- 28.0 g
- How many grams of water will be produced by complete combustion of 12 g of methane gas?
- 16
- 27
- 36
- 56
- Two elements A (at. mass 75) and B (at. mass 16) combine to give a compound having 75.8 % of A. The formula of the compound is
- AB
- \(\mathrm{A_2B}\)
- \(\mathrm{AB_2}\)
- \(\mathrm{A_2B_3}\)
- A hydrocarbon contains 79.87 % carbon and 20.13 % of hydrogen. What is its empirical formula?
- CH
- \(\mathrm{CH_2}\)
- \(\mathrm{CH_3}\)
- \(\mathrm{C_2H_5}\)
- How many grams of oxygen will be required to react completely with 27 g of Al? (Atomic mass: Al = 27, O = 16)
- 8
- 16
- 24
- 32
- In \(\mathrm{CuSO_4 \cdot 5H_2O}\) the percentage of water is …… (Cu = 63.5, S = 32, O = 16, H = 1)
- 10 %
- 36 %
- 60 %
- 72 %
- When two properties of a system are mathematically related to each other, the relation can be deduced by
- Working out mean deviation
- Plotting a graph
- Calculating relative error
- All the above three