Standard Form for Very Large and Very Small Numbers

Introduction

Mole calculations are full of extreme numbers: 6.022 × 10²³ particles in a mole, 1.99 × 10⁻²³ g for a carbon atom. Writing these out with all their zeros is slow and invites mistakes. Standard form keeps them short and makes their size obvious at a glance. This page revises how to write numbers in standard form and, more importantly for this unit, how to multiply and divide them quickly and correctly.

Core explanation

The format. A number in standard form is written

A × 10ⁿ, where 1 ≤ A < 10 and n is a whole number (positive, negative or zero).

- 602 000 = 6.02 × 10⁵ (decimal point moves 5 places left, so n = +5) - 0.000 35 = 3.5 × 10⁻⁴ (decimal point moves 4 places right, so n = −4) - 7.1 = 7.1 × 10⁰

A positive index means a large number; a negative index means a number smaller than 1.

Multiplying. Multiply the coefficients and add the indices:

(A × 10ᵃ) × (B × 10ᵇ) = (A × B) × 10ᵃ⁺ᵇ

Example: (2.0 × 10³) × (3.0 × 10⁴) = 6.0 × 10⁷.

Dividing. Divide the coefficients and subtract the indices:

(A × 10ᵃ) ÷ (B × 10ᵇ) = (A ÷ B) × 10ᵃ⁻ᵇ

Example: (8.0 × 10⁶) ÷ (2.0 × 10²) = 4.0 × 10⁴.

Tidying up. If the new coefficient falls outside 1 to 10, adjust it and change the index to compensate:

- 15 × 10²³ = 1.5 × 10²⁴ (coefficient ÷ 10, index + 1) - 0.25 × 10²⁴ = 2.5 × 10²³ (coefficient × 10, index − 1)

Negative indices. Subtracting a negative index is the same as adding: 10⁵ ÷ 10⁻³ = 10⁵⁺³ = 10⁸. This happens often when dividing a sample mass by the mass of one atom.

Using a calculator. Enter 6.022 × 10²³ as 6.022 then the EXP (or ×10ˣ) key then 23. Do not type "× 10 EXP 23", which gives ten times too much. For negative indices, use the (−) key: 1.99 EXP (−) 23. When dividing by a standard-form number, put it in brackets or use EXP, so the calculator divides by the whole number.

Formulae

(A × 10ᵃ) × (B × 10ᵇ) = AB × 10ᵃ⁺ᵇ and (A × 10ᵃ) ÷ (B × 10ᵇ) = (A/B) × 10ᵃ⁻ᵇ, with the final coefficient adjusted to lie between 1 and 10.

Step-by-step reasoning

To multiply or divide in standard form:

1. Separate the coefficients from the powers of ten. 2. Multiply (or divide) the coefficients. 3. Add (for ×) or subtract (for ÷) the indices. 4. Adjust the coefficient to lie between 1 and 10, changing the index to match. 5. Round to the appropriate number of significant figures.

Visual explanation

Picture a number line marked in powers of ten rather than in ones: 10⁻²⁴, 10⁻¹², 10⁰, 10¹², 10²⁴. Each step to the right multiplies by ten. An atom's mass sits near the far left; the Avogadro constant near the far right; a laboratory sample of a few grams sits in the middle, where the two meet.

Real-world analogy

Standard form is like giving a postcode instead of a full address. "6.022 × 10²³" tells you the region (10²³) at once and the exact location (6.022) within it, just as a postcode tells a courier first the area and then the street.

Real-world example

Scientists at CERN quote collision energies, particle masses and event rates spanning more than thirty powers of ten. Astronomers do the same with distances. Every scientific field, from medicine (virus sizes around 1 × 10⁻⁷ m) to economics (national debts in the 10¹² range), depends on standard form to keep numbers readable.

Why?

Why do we add indices when multiplying? Because 10³ × 10⁴ means (10 × 10 × 10) × (10 × 10 × 10 × 10), which is seven tens multiplied together: 10⁷. Dividing cancels tens from top and bottom, which is why indices are subtracted.

Common misconception

"A bigger negative index means a bigger number." The opposite is true: 10⁻²³ is far smaller than 10⁻³. A number with index −23 has 22 zeros after the decimal point before the first significant digit.

Worked example

Question: A sample contains 1.2 × 10²² atoms. How many moles is this? Use NA = 6.0 × 10²³ mol⁻¹.

Reasoning: n = N ÷ NA = (1.2 × 10²²) ÷ (6.0 × 10²³). Coefficients: 1.2 ÷ 6.0 = 0.20. Indices: 22 − 23 = −1. So n = 0.20 × 10⁻¹ = 2.0 × 10⁻² mol.

Answer: 2.0 × 10⁻² mol, or 0.020 mol.

Quick check

1. Calculate (3.0 × 10⁵) × (4.0 × 10⁻²) in standard form. Answer: 12 × 10³ = 1.2 × 10⁴.

Exam focus

Show the coefficient and index working separately so that method marks are available even if the arithmetic slips. Always finish with a correctly adjusted coefficient between 1 and 10, and double-check calculator entries that use EXP with negative indices.

Advanced insight

Taking the logarithm of a standard-form number gives its order of magnitude directly: log₁₀(6.022 × 10²³) ≈ 23.78. Chemists use logarithmic scales, such as pH, precisely because concentrations span so many powers of ten. Standard form is the everyday doorway to that way of thinking.

Summary

Standard form writes numbers as A × 10ⁿ with 1 ≤ A < 10. To multiply, multiply the coefficients and add the indices; to divide, divide the coefficients and subtract the indices. Adjust the coefficient afterwards. Use the EXP key on a calculator rather than typing "× 10". These skills underpin every calculation involving the Avogadro constant.

Practice questions

1. Write 0.000 000 52 in standard form. Answer: 5.2 × 10⁻⁷. 2. Calculate (6.0 × 10²³) × 2.5 in standard form. Answer: 15 × 10²³ = 1.5 × 10²⁴. 3. Calculate (9.0 × 10⁴) ÷ (3.0 × 10⁻²). Answer: 3.0 × 10⁴⁺² = 3.0 × 10⁶. 4. A sample of 12 g of carbon contains 6.0 × 10²³ atoms. Find the mass of one atom. Answer: 12 ÷ (6.0 × 10²³) = 2.0 × 10⁻²³ g.