Mathematical Skills: Algebra, Standard Form & Accuracy

Why very large numbers appear in chemistry

Atoms, ions and molecules are extremely small, so an ordinary laboratory sample contains an enormous number of particles. Chemists group particles using the mole. One mole contains the Avogadro constant number of specified particles:

$N_A=6.02\times10^{23}\ \text{mol}^{-1}$

The particles must be identified: they might be atoms, molecules, ions or formula units. Standard form makes numbers of this size easier to write, compare and calculate with.

Writing numbers in standard form

A number in standard form has the structure

$A\times10^n$

where $1\leq A<10$ and $n$ is a whole-number power, which may be positive, zero or negative. The first part must therefore contain exactly one non-zero digit before the decimal point.

For a large number, move the decimal point to the left until the first part lies between 1 and 10. The number of places moved becomes a positive power: $602\,000\,000\,000\,000\,000\,000\,000$ becomes $6.02\times10^{23}$.

For a small number, moving the decimal point to the right gives a negative power. For example, $0.000701=7.01\times10^{-4}$. A negative power does not mean that the number itself is negative; it means the number is smaller than 1.

Always check that the coefficient is in the permitted range. For example, $60.2\times10^{22}$ has the correct value but is not correctly presented in standard form; it should be $6.02\times10^{23}$.

Calculating with powers of ten

When multiplying numbers in standard form, multiply the coefficients and add the powers. When dividing, divide the coefficients and subtract the powers:

$$(a\times10^m)(b\times10^n)=ab\times10^{m+n}$$ $$\frac{a\times10^m}{b\times10^n}=\frac{a}{b}\times10^{m-n}$$

The resulting coefficient may need to be adjusted so that it lies between 1 and 10. For instance, $15.05\times10^{22}$ is rewritten as $1.505\times10^{23}$.

On a calculator, use its exponent-entry key, often labelled EXP, EE or $\times10^x$. Enter $6.02\times10^{23}$ as one number and use brackets where necessary. Reading the display carefully matters: a calculator might show $1.505\text{E}23$, meaning .

Linking particles, moles and the Avogadro constant

If $N$ is the number of particles and $n$ is the amount in moles, then

$N=nN_A$

Multiplying by $N_A$ converts moles into particles because every mole contains $6.02\times10^{23}$ particles. Dividing by $N_A$ reverses the conversion:

$n=\frac{N}{N_A}$

For example, $5.42\times10^{23}$ particles correspond to

$n=\frac{5.42\times10^{23}}{6.02\times10^{23}}=0.900\ \text{mol}$

to three significant figures. The powers of ten cancel because $10^{23}/10^{23}=10^0=1$.

Changing the subject of an equation

The subject is the quantity isolated on one side of an equation. In $N=nN_A$, the subject is $N$. A question may instead require $n$, so the equation must be rearranged.

An equation is like a balanced pair of scales: whatever operation is performed on one side must also be performed on the other. To make $n$ the subject, divide both sides by $N_A$:

$$N=nN_A$$ $$\frac{N}{N_A}=\frac{nN_A}{N_A}$$

The $N_A$ factors on the right cancel, leaving $n=N/N_A$.

The same reasoning applies to other chemical relationships. Starting with

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

multiplying both sides by $M_r$ gives

$m=nM_r$

while dividing $m=nM_r$ by $n$ gives

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

This is safer than relying only on a formula triangle because it still works with more complicated equations. A useful check is to substitute the rearranged expression back into the original equation and see whether it simplifies correctly.

What significant figures show

Significant figures communicate the precision of a value. Begin counting at the first non-zero digit. Zeros before that digit only locate the decimal point and are not significant, so $0.000701$ has three significant figures.

Zeros between non-zero digits are significant: $9.005$ has four significant figures. Trailing zeros after a decimal point are also significant, so $0.900$ has three significant figures. They show that the value has been stated more precisely than $0.9$.

Trailing zeros in a whole number can be ambiguous. Standard form removes that ambiguity: has three significant figures, whereas has four.

To round, locate the final digit that must be retained and inspect the next digit. If the next digit is 5 or more, increase the retained digit by 1; otherwise leave it unchanged. Thus $1.505\times10^{23}$ becomes $1.51\times10^{23}$ to three significant figures.

Accuracy through a complete calculation

Use the precision requested in the question or an appropriate precision for the data given. Do not round intermediate results: store the calculator value and round only the final answer. Early rounding can accumulate and change the result of a multistep calculation.

A clear chemical calculation therefore follows a dependable sequence:

1. Write the relevant equation.
2. Rearrange it before inserting numbers.
3. Convert quantities into the required units.
4. Substitute values, keeping powers of ten in brackets where helpful.
5. Calculate without premature rounding.
6. Give the final answer to an appropriate number of significant figures and include its unit.
7. Check the scale: a fraction of a mole should still represent a very large number of particles.