Mathematical Techniques and Unit Conversions

Why quantitative chemistry needs several mathematical skills

Quantitative chemistry connects a chemical description with a numerical result. A question may require several linked operations: interpret a formula or balanced equation, convert units, rearrange an equation, use a ratio, calculate a percentage and finally round sensibly. Treating these as a sequence of small decisions is more reliable than trying to complete the whole calculation in one step.

A useful routine is:

1. identify what the question gives and what it asks for;
2. write down the relevant chemical equation or mathematical relationship;
3. convert quantities into compatible units;
4. rearrange the relationship if necessary;
5. substitute values, keeping units alongside them;
6. carry unrounded values into later stages;
7. check the size and unit of the answer, then round appropriately.

Converting units before using an equation

A number without its unit is incomplete. Conversions must therefore be made before substitution. Common mass relationships are

$1\ \text{kg}=1000\ \text{g}$

and

$1\ \text{g}=1000\ \text{mg}.$

To convert milligrams to grams, divide by 1000; to convert grams to milligrams, multiply by 1000. For example, $600\ \text{mg}=0.600\ \text{g}$. Writing the conversion explicitly helps prevent a factor-of-1000 error.

The conversion between mass and amount in moles is not simply a change of unit: it also depends on the substance's relative formula mass, $M_r$. When mass $m$ is in grams,

$n=\frac{m}{M_r},$

where $n$ is the amount in moles. Numerically, one mole has a mass of $M_r$ grams. Thus a substance with $M_r=40$ has a mass of per mole, whereas a different substance with has a mass of per mole.

For example, a $2.40\ \text{g}$ sample of a substance with $M_r=40.0$ contains

$n=\frac{2.40}{40.0}=0.0600\ \text{mol}.$

The unit conversion and the chemical conversion should not be confused: first express the mass in grams, then use $M_r$ to find moles.

Volume conversions are also important in solution calculations:

$1\ \text{dm}^3=1000\ \text{cm}^3.$

Therefore, divide a volume in $\text{cm}^3$ by 1000 to express it in $\text{dm}^3$.

Changing the subject of an equation

The subject is the quantity isolated on one side of an equation. Rearranging changes the form of the relationship, not the relationship itself. Whatever operation is performed on one side must also be performed on the other.

Starting with

$n=\frac{m}{M_r},$

multiply both sides by $M_r$ to make mass the subject:

$m=nM_r.$

To make relative formula mass the subject, divide this result by $n$:

$M_r=\frac{m}{n}.$

The same algebra applies to concentration in $\text{g dm}^{-3}$:

$c=\frac{m}{V},\qquad m=cV,\qquad V=\frac{m}{c}.$

Here $m$ is the mass of dissolved substance in grams and $V$ is the solution volume in cubic decimetres. Checking the units provides a useful test: multiplying $\text{g dm}^{-3}$ by $\text{dm}^3$ leaves grams, as required for mass.

Using ratios in chemical calculations

A balanced equation gives the ratio in which particles react and therefore the ratio of amounts in moles. For example,

$2A+B\rightarrow 2C$

shows a mole ratio $A:B:C$ of $2:1:2$. It does not state that the substances react in a mass ratio of $2:1:2$, because different substances usually have different masses per mole.

Suppose the $0.0600\ \text{mol}$ calculated above is substance $A$. The equation shows that two moles of $A$ correspond to one mole of $B$, so

$n(B)=0.0600\times\frac{1}{2}=0.0300\ \text{mol}.$

The conversion factor is written with the wanted coefficient on top and the known coefficient underneath. If the mass of $B$ were required, the next step would be $m=nM_r$ using the $M_r$ of . This creates the common multistep pathway

$\text{mass of known substance}\rightarrow\text{moles of known substance}\rightarrow\text{moles of wanted substance}\rightarrow\text{mass of wanted substance}.$

Calculating percentages

A percentage compares a part with its whole:

$\text{percentage}=\frac{\text{part}}{\text{whole}}\times100.$

The part and whole must be measured in the same unit so that the units cancel. For instance, if a $1.30\ \text{g}$ sample contains $0.988\ \text{g}$ of a particular substance, its percentage by mass is

$\frac{0.988}{1.30}\times100=76.0\%.$

Read the wording carefully to decide which quantity is the part and which is the whole. Reversing them can produce an impossible percentage above 100%. In longer problems, the required part may first have to be found using a mole ratio or another equation.

Managing multistep calculations

Do not round after every stage. Store the calculator value or write several extra digits, because early rounding can alter the final result. A clear calculation chain should also label quantities rather than present a line of unexplained numbers. For a reacting-mass problem, for example, show the separate mass-to-moles, mole-ratio and moles-to-mass stages.

A quick estimate can reveal major mistakes. Dividing by 1000 should make a positive number smaller; multiplying a number of moles by a mass per mole should give a mass; and a constituent's percentage of a sample should normally be between 0% and 100%.

Choosing significant figures

Significant figures communicate the precision of a value. Count from the first non-zero digit: $0.00330$ has three significant figures, while $12.06$ has four. Zeros between non-zero digits are significant, and trailing zeros after a decimal point can also be significant.

Unless the question gives a different instruction, give the final answer to a precision appropriate to the data supplied—usually no more significant figures than the least precise measured value used. Exact counting numbers and exact conversion factors such as $1000\ \text{mg}=1\ \text{g}$ do not limit the precision.

To round, inspect the next digit: 0–4 leaves the final retained digit unchanged, while 5–9 increases it by one. Include trailing zeros when they are needed to show precision: $2.40\ \text{g}$ states three significant figures, whereas $2.4\ \text{g}$ states two.