2.2: Scientific Notation - Writing Large and Small Numbers
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Learning Objectives
- Express a large number or a small number in scientific notation.
- Carry out arithmetical operations and express the final answer in scientific notation
Chemists often work with numbers that are exceedingly large or small. For example, entering the mass in grams of a hydrogen atom into a calculator would require a display with at least 24 decimal places. A system called scientific notation avoids much of the tedium and awkwardness of manipulating numbers with large or small magnitudes. In scientific notation, these numbers are expressed in the form
\[ N \times 10^n\]
where N is greater than or equal to 1 and less than 10 (1 ≤ N < 10), and n is a positive or negative integer (10^{0} = 1). The number 10 is called the base because it is this number that is raised to the power \(n\). Although a base number may have values other than 10, the base number in scientific notation is always 10.
A simple way to convert numbers to scientific notation is to move the decimal point as many places to the left or right as needed to give a number from 1 to 10 (N). The magnitude of n is then determined as follows:
- If the decimal point is moved to the left n places, n is positive.
- If the decimal point is moved to the right n places, n is negative.
Another way to remember this is to recognize that as the number N decreases in magnitude, the exponent increases and vice versa. The application of this rule is illustrated in Example \(\PageIndex{1}\).
Example \(\PageIndex{1}\): Expressing Numbers in Scientific Notation
Convert each number to scientific notation.
- 637.8
- 0.0479
- 7.86
- 12,378
- 0.00032
- 61.06700
- 2002.080
- 0.01020
Solution
Explanation | Answer | |
---|---|---|
a |
To convert 637.8 to a number from 1 to 10, we move the decimal point two places to the left: 637.8 Because the decimal point was moved two places to the left, n = 2. |
\(6.378 \times 10^2\) |
b |
To convert 0.0479 to a number from 1 to 10, we move the decimal point two places to the right: 0.0479 Because the decimal point was moved two places to the right, n = −2. |
\(4.79 \times 10^{−2}\) |
c | This is usually expressed simply as 7.86. (Recall that 10^{0} = 1.) | \(7.86 \times 10^0\) |
d | Because the decimal point was moved four places to the left, n = 4. | \(1.2378 \times 10^4\) |
e | Because the decimal point was moved four places to the right, n = −4. | \(3.2 \times 10^{−4}\) |
f | Because the decimal point was moved one place to the left, n = 1. |
\(6.106700 \times 10^1\) |
g | Because the decimal point was moved three places to the left, n = 3. | \(2.002080 \times 10^3\) |
h | Because the decimal point was moved two places to the right, n = -2. | \(1.020 \times 10^{−2}\) |
Addition and Subtraction
Before numbers expressed in scientific notation can be added or subtracted, they must be converted to a form in which all the exponents have the same value. The appropriate operation is then carried out on the values of N. Example \(\PageIndex{2}\) illustrates how to do this.
Example \(\PageIndex{2}\): Expressing Sums and Differences in Scientific Notation
Carry out the appropriate operation and then express the answer in scientific notation.
- \( (1.36 \times 10^2) + (4.73 \times 10^3) \nonumber\)
- \((6.923 \times 10^{−3}) − (8.756 \times 10^{−4}) \nonumber\)
Solution
Explanation | Answer | |
---|---|---|
a |
Both exponents must have the same value, so these numbers are converted to either \((1.36 \times 10^2) + (47.3 \times 10^2) = (1.36 + 47.3) \times 10^2 = 48.66 × 10^2\) or \((0.136 \times 10^3) + (4.73 \times 10^3) = (0.136 + 4.73) \times 10^3) = 4.87 \times 10^3\). Choosing either alternative gives the same answer, reported to two decimal places. In converting 48.66 × 10^{2} to scientific notation, \(n\) has become more positive by 1 because the value of \(N\) has decreased. |
\(4.87 \times 10^3\) |
b |
Converting the exponents to the same value gives either \((6.923 \times 10^{-3}) − (0.8756 \times 10^{-3}) = (6.923 − 0.8756) \times 10^{−3}\) or \((69.23 \times 10^{-4}) − (8.756 \times 10^{-4}) = (69.23 − 8.756) \times 10^{−4} = 60.474 \times 10^{−4}\). In converting 60.474 × 10^{-4} to scientific notation, \(n\) has become more positive by 1 because the value of \(N\) has decreased. |
\(6.047 \times 10^{−3}\) |
Multiplication and Division
When multiplying numbers expressed in scientific notation, we multiply the values of \(N\) and add together the values of \(n\). Conversely, when dividing, we divide \(N\) in the dividend (the number being divided) by \(N\) in the divisor (the number by which we are dividing) and then subtract n in the divisor from n in the dividend. In contrast to addition and subtraction, the exponents do not have to be the same in multiplication and division. Examples of problems involving multiplication and division are shown in Example \(\PageIndex{3}\).
Example \(\PageIndex{3}\): Expressing Products and Quotients in Scientific Notation
Perform the appropriate operation and express your answer in scientific notation.
- \( (6.022 \times 10^{23})(6.42 \times 10^{−2}) \nonumber\)
- \( \dfrac{ 1.67 \times 10^{-24} }{ 9.12 \times 10 ^{-28} } \nonumber \)
- \( \dfrac{ (6.63 \times 10^{−34})(6.0 \times 10) }{ 8.52 \times 10^{−2}} \nonumber \)
Solution
Explanation | Answer | |
---|---|---|
a |
In multiplication, we add the exponents: \[(6.022 \times 10^{23})(6.42 \times 10^{−2})= (6.022)(6.42) \times 10^{[23 + (−2)]} = 38.7 \times 10^{21} \nonumber\] In converting \(38.7 \times 10^{21}\) to scientific notation, \(n\) has become more positive by 1 because the value of \(N\) has decreased. |
\(3.87 \times 10^{22}\) |
b |
In division, we subtract the exponents: \[{1.67 \times 10^{−24} \over 9.12 \times 10^{−28}} = {1.67 \over 9.12} \times 10^{[−24 − (−28)]} = 0.183 \times 10^4 \nonumber\] In converting \(0.183 \times 10^4\) to scientific notation, \(n\) has become more negative by 1 because the value of \(N\) has increased. |
\( 1.83 \times 10^3\) |
c |
This problem has both multiplication and division: \[ {(6.63 \times 10^{−34})(6.0 \times 10) \over (8.52 \times 10^{−2})} = {39.78 \over 8.52} \times 10^{[−34 + 1 − (−2)]} \nonumber \] |
\( 4.7\times 10^{-31}\) |
Contributions & Attributions
This page was constructed from content via the following contributor(s) and edited (topically or extensively) by the LibreTexts development team to meet platform style, presentation, and quality:
Henry Agnew (UC Davis)