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×04th May 2019 @ 11 min read

In chemistry, we may deal with different varieties of calculations. Sometimes, we may come across very large numbers like Avogadro’s constant (*N*_{A} = 6.022 × 10^{23} mol^{−1}), or sometimes, with very small numbers like the diameter of hydrogen (120 pm). Many times, chemistry experiments involve the handling of large experimental data like vapour pressure data. So, there is a need for the right technique to manage numbers conveniently and maintain the accuracy of numbers. This is where the concept of significant figures arises.

Significant figures are digits which tell certainty in a number. Significant figures are always a whole number.

Let say we are measuring the weight of a calcium carbonate sample in the chemistry laboratory. The measuring instrument is of traditional type say a beam balance, we get reading of 150 g. If the same sample is measured in an analytical balance, we get reading of 153.471 2 g. The weight of the sample in the latter instrument is higher than the former. This shows the former reading is accurate up to first two digits which are 1 and 5. Significant digits are the total numbers with certainty plus one uncertainty digit. Therefore, in the above reading of the beam balance, the significant figure is 3: two certainty digits (1 and 5) plus one uncertainty digit (0).

Note: The last digit of a measurement or calculation is always subjected to uncertainty. If we say the volume of sodium hydroxide solution in a beaker is 305 mL, the last digit 5 has uncertainty while the first two digits are certain. Unless specified, the uncertainty of ± 1 is understood in the last digit.

There are six rules for estimating the significant figures in which are states below.

All non-zero digits are significant whether the number is without the decimal point or not.

- 94 122 has five significant figures.
- 123.232 has six significant figures.

Zeros between two non-zero digits are significant. In other words, entrapped zeros are significant.

- 22 300 093 has eight significant figures.
- 909 has three significant figures.
- 203.004 has six significant figures.
- 40.04 has four significant figures.

All zeroes preceding the first non-zero digit are insignificant.

- 0 232 has three significant figures.
- 0.12 has two significant figures.
- 0.000 000 000 230 2 has four significant figures.

If a number has the decimal point, all the zeros at the end (the right side of the decimal point) are significant.

- 83.720 0 has six significant figures.
- 0.130 0 has four significant figures.
- 0.001 630 000 00 has nine significant figures.
- 10.010 has five significant figures

When a number is without the decimal point, all zeros at the end (after the last non-zero digit) may or may not be significant. Consider a number 1 400, It may have 2, 3, or 4 significant figures. It is not possible to estimate whether the number is certain up to ± 1, ± 10, or ± 100.

For example, the weight of sodium chloride is 1 400 g. It is clear the first digit 1 is certain, but the certainty of the rest three digits is difficult to know. It might be the real weight of salt is 1 400.1 g or 1 403 g or 1 311 g. If the real weight of salt is 1 400.1, then uncertainty exists in the last of 1 400, and it has four significant figures. Likewise, if the real weight is 1 403 g, uncertainty is in the last two digits of 1 400, and significant figures are 3. Similarly, for 1 311, significant figures are 1.

This ambiguity can be overcome by writing a number in scientific notations (exponential forms). All digits in scientific notations are significant (excluding the exponential part). The above example of sodium chloride in scientific notations is as follows.

- 1.400 × 10
^{3}g has 4 significant figures. - 1.40 × 10
^{3}g has 3 significant figures. - 1.4 × 10
^{3}g has 2 significant figures.

The whole numbers have infinite significant figures. The number of quantities is always whole numbers; they cannot be expressed in the decimal or in fraction. For example, 2 dollars, 1 aeroplane, 50 molecules of water. These numbers can be expressed in decimals if there are infinite zero after the decimal point as shown below.

- 2.000 000 000…∞
- 1.000 000 000…∞
- 50.000 000 000…∞

Before we move to addition, subtraction, multiplication, and division, it is necessary to understand the rounding off numbers. There are three rules which are explained below.

If the digit to be dropped is less than 5 (i.e. 0, 1, 2, 3, and 4), drop it and all the digits right to it. The preceding digit to dropped digit remains unchanged. The numbers which are dropped before the decimal point are replaced with zeros.

Round off number 864.731 92 to 4 significant figures. To make the number to four significant number, we must drop number 3 and all the numbers right to it (i.e., 1, 9, and 2). Since number 3 is less than 5, the preceding number (which is 7) remains unchanged as per rule. Therefore, the round off figure is 864.7.

If the same number is round off to 2 significant figures, we must drop 4 and all the numbers following it (7, 3, 1, 9, and 2). Since number 4 is less than 5, the preceding number (which is 6) remains unchanged. Also, 4 is replaced by 0, because it is before the decimal. The new round off number is 860.

If the digit to be dropped is more than 5 (i.e., 6, 7, 8, and 9), drop it and all the digits right to it. And importantly, the preceding number is increased by one. The numbers which are dropped before the decimal point are replaced with zeros.

Consider rounding off the same number, 864.731 92, to 6 significant figures. We have to drop number 9 and all the numbers following it. Since 9 is more than 5, preceding number 1 is up by one. Thus, the round off number is 864.732.

For 1 significant figure, the round off number will be 900.

When the number to be dropped is 5, and the number preceding 5 is an odd number, we increase the preceding number by one. But if the preceding number is even, it remains unchanged. The numbers which are dropped before the decimal point are replaced with zeros.

Consider number 0.081 585. Round off the number to four significant figures. We must drop the last number (which is 5). Since the preceding number is 8 which is even. It is unaffected. The round off number is 0.081 58.

For 2 significant figures, the round off number will be 0.082.

In addition and/or subtraction of significant figures, the resultant number should not have more digits right to the decimal point than either of the precedent numbers. In simple words, the output should have the same decimal places as the least decimal places in the original numbers. Consider an example below.

12.23 + 9 329.568 + 0.019 33 = 9 341.817 33 = 9 341.82

In the above numbers, the least number of decimal places is 2. So, the result should be limited to 2 decimal places. The addition of three numbers is 9 341.817 33 which is round off to two decimal places. The same is true in subtraction.

7 182.119 − 1 121.232 1 0.119 31 = 6 060.767 59 = 6 060.768

2.21 − 1.565 + 2.951 = 3.596 = 3.60

In multiplication and division, the result should not have more significant figures than the original numbers. Consider an example below.

98.23 × 5.3402 = 524.567 846 = 524.6

The least significant figures in the original numbers are 4. So, the result must be limited to 4 significant figures. The same is true in the division.

23.34 ÷ 2.01 = 11.611 940 3… = 11.6

885.45 ÷ 2. = 442.725 = 400

Note: When we are working on a long series of calculation, only the final result needed to be round off to desired significant figures, not intermediate ones. To maintain the accuracy, we retain an extra significant digit in intermediate calculations.

2 mol of hydrochloric acid (HCl) is dissolved into 220.0 g of water. Find the total weight of hydrochloric acid solution?

The molecular weight of HCl is 36.46 g mol^{−1}.

Thus, the weight of 2 mol of HCl is 2 mol × 36.46 g mol^{−1} = 72.92 g

(here, number 2 is the whole number which has infinite significant figures and molar mass of HCl has 4 significant figures; in multiplication, we must consider the least significant figures.)

The total weight of the solution is 220.0 + 72.92 = 292.92 = 292.9 g.

3.23 g of copper sulphate is dissolved into 0.150 L of water. Calculate the molar concentration of copper sulphate?

The molar mass of copper sulphate is 159.609 g mol^{−1}.

Thus, the moles of copper sulphate is 3.23 g ÷ (159.609 g mol^{−1}) = 0.020 236 9 mol = 0.020 23 mol.

(An extra significant digit is considered in the above calculation since it is the intermediate result. But the number is certain only up to 3 significant figures.)

The concentration is 0.020 23 mol ÷ 0.150 L = 0.134 866 66 mol L^{−1} = 0.135 mol L^{−1}.

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