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Chemistry Problem Solving Strategies

Updated on July 21, 2015


Solving chemistry problems is not an easy task. To make matters worse, this maxim is repeated throughout school curricula and does more harm than good. The things are not that grim in truth, however, as most chemistry problems can be solved using a set of general strategies (which are generally not taught in schools), some of which may seem evident but often are overlooked. These strategies have use beyond chemistry as well, as they in general usually involve applying mathematical models to material, ‘touchable’ systems. Unsurprisingly, many chemistry graduates end up working in other fields where such skills are needed, such as finances. Here some of such strategies will be outlined (with solved examples at the bottom), discussed and examples provided, hoping that the reader will find such problems more understandable and easier afterwards (discussing theory in general is without the scope of this essay). Please do not hesitate to share your tips/insights or ask for problem solving help in the comment section below.

Determination of given and asked objects / quantities

This point seems obvious yet is often overlooked and not done in full. Typically all the given data has to be carefully interpreted and assigned various labels (and in some cases this is hard to do for experienced chemists as well), which most often include the name of the variable, the units and the system to which the variable belongs (may be a substance, a mixture of substances etc.), same considerations go for asked quantities. Some of the required data may not be given but instead determinable from the Periodic Table (such as molar masses), these can be determined during the solution of a problem. Probably the most often overlooked step is the assignment to a system to which the variable belongs so extra care has to be taken with that. See this put to practise in examples below.

Finding a path from given to asked quantities

Relating given quantities to asked quantities can also be a tricky business, hence is best done systematically. Many chemistry problems revolve around quantities in moles which are the convention for chemical reactions as well; hence usually it is indeed useful to convert various quantities to moles. For example, the relation line below may be used to solve a good deal of problems involving solutions:

V solution ↔ m solution ↔ m solute ↔ n solute ↔ c solute

Thus all five physical quantities are interconvertable through a set of specific constants (for example, Vsolution and msolution are interconverted via density of the solution). In general, if both given and asked quantities are within this interconversion line then the solution of the problem does indeed involve using formulas embedded in that line. Note that this is but one of such lines, most of them, however, do intersect through molar quantities (n solute in the line above). Example 1 below heavily utilises this line. After a path from given information to answer is found all that remains is transforming the steps into formulas by using the appropriate constants.

Checking the validity of equation and formulas

To minimise the chances of mistakes it is a good idea to check validities of equations and formulas used, both chemical and physical. A simple way to check the validity of a chemical equation is to count atoms on both the reagent and the product sides – both of them must have the same types and numbers of atoms (as they cannot appear nor disappear), if that is not the case then something is wrong with the chemical equation, implying either that something is missing or reaction coefficients have not been properly determined. If the problem involves determination of a chemical formula the validity of it may be checked by determining the oxidation states of atoms which must always sum to zero. Oxidation states also must follow the general rules of chemistry, thus it is not possible to have a nitrogen atom with state of +6 or a fluorine atom with state +1 as these are not quite normally possible (allowed oxidation states often can be found in periodic tables).

The validity of physical formulas can be checked by inspection of the dimensions which must be the same on both sides of an equation. For this purpose the dimensions may be treated the same as any mathematical quantity – if we multiply grams per mole and moles (g/mol X mol), moles cancel out and only grams remain. Keep in mind that only the dimensions matter here, not mathematical quantities. Examples 1 and 2 further elaborate on this method.

Checking the validity of an answer

In most cases it is quite easy to check whether the answer is right (though not all mistakes are detectable with this method) by simply going through the maths backwards, i.e. starting with the answer and obtaining the given data. Though if the initial numbers are not obtained it may mean that the mistake is either in forward or backward calculations. If the answer is right, it may mean that everything is done right or that the same mistake has been done in both forward and backward calculations. This sounds worse than it actually is – this method tends to work well when you know what you are doing.

Another related technique is checking whether an answer makes physical sense. Molarity of 0.100 M for some solute usually makes sense, whereas molarity of 100 M does not. If half-life for some first order reaction is an hour, it make sense that in an hour and a half concentration fall to about thirty five percent of the original reagent concentration (as one and a half half-lifes have passed), it makes no sense if it falls to seventy percent or ten percent. If the answer makes no physical sense (it takes some skill to figure it out) then usually it is best to scrap the old solution and try again.

Conversion table for selected units

Volume measurement
1 ml = 10^-3 l
Mass measurement
1 mg = 10^-3 g
Chemical bond lengths
1 Å = 10^-10 m
ÅChemical bond lengths
1 pm = 10^-12 m
(Ambient) pressure measurement
1 atm = 101325 Pa
(Ambient) pressure measurement
1 bar = 100000 Pa
Torr / milimetre of Mercury
760 Torr = 1 atm
1 Torr = 133.3 Pa

Metric prefixes and non-SI units

Metric prefixes and non-SI units (milli-, micro-, angstroms etc.) are a source of common mistakes, sometimes even for experienced yet somewhat clumsy chemists. Probably the least error-prone yet somewhat laborious way is to carefully convert everything to prefix-neutral units (e.g. milligrams to grams, angstroms to metres), an useful conversion table is shown to the right.

A less laborious way to circumnavigate this problem is to check the magnitude of the answer – often it will be quite clear whether the result is thousand times higher or lower than it should be, though this method requires quite some experience.

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Example 1


The concentration of a potassium hydroxide solution was determined by titrating a solution of tosylic acid (TsOH) made from tosylic acid monohydrate. TsOH is an acid with one acidic hydrogen, molar mass of its monohydrate is 190.22 g/mol. During the titration 25.85 ml of KOH solution were used to neutralise a TsOH solution made from 0.1027 g of tosylic acid monohydrate. Calculate the molarity of potassium hydroxide solution.

Assigning the information:

The problem involves two reacting substances - potassium hydroxide and tosylic acid. For tosylic acid, we are given its monohydrate molar mass – 190.22 g/mol (molar mass of non-hydrate is irrelevant here as monohydrate was used to make the solution) , the mass used to make its solution – 0.103 g and the fact that it has just one acidic hydrogen. For potassium hydroxide, we are given the volume of the solution used to neutralise TsOH (25.85 ml) and we are required to find molarity of the solution.

Finding a path from given to asked quantities:

Luckily, the problem is mole-centred hence the relation line m solute ↔ n solute

c solute is sufficient to find the related formulas

Validity of equation and formulas:

Numbers of atoms / groups on both sides of the chemical equation are the same (one Tosyl group, two hydrogen and two oxygen atoms) so the neutralisation reaction equation should be correct. Dimensions in formulas agree so they should be alright. Note that millilitres have been converted to litres.

Validity of the answer:

Backwards solution has been done, the same answer is obtained.

Solution of the Problem in Example 1


Example 2


The first order reaction A → B has a characteristic half-life of 30 seconds. Find the fraction of A concentration remaining in the reaction mixture after 70 seconds have passed.

Assigning the information:

This time we have a first order kinetic problem – the half-life is characteristic to the reaction as is the time of the reaction (70 s). The problem is essentially asking to find a ratio [A]/[A]0 (fraction of A concentration), the product B is of no interest in this problem whatsoever.

Finding a path from given to asked quantities:

Kinetic problems in general are not mole-centric hence a different path has to be used. A good way to express it is t1/2 ↔ k ↔ [A]. All that remains is to find or derive the formulas involved in the path (integrated, not differential kinetic equations have to be used).

Validity of formulas:

The first order kinetic constant is supposed to have s-1 as its dimensions. Dividing a number by time does indeed yield a constant with dimensions s-1. Concentration ratio should be dimensionless and by multiplying s-1 with s we indeed obtain a dimensionless ratio.

Validity of the answer:

70 seconds correspond to two and a bit half-lifes so we can expect concentration at that time to be between 0.25 (corresponding to two half-lifes) and 0.125 (corresponding to three half-lifes) of the initial. The obtained answer 0.198 is indeed in between the two so should be right.

Solution of the Problem in Example 2



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