HS Chemistry - Essentials

Significant Figures & Measurements

Overview of The Page

This page will cover:

• What are significant figures?
• How do significant figures affect measurements and results?

When measurements are done in Chemistry, significant figures are often used. This is because in Chemistry, often the need arises to deal with very large or very small numbers. These numbers are often written in scientific notation. However, even when scientific notation isn't used, significant figures are still important.

Whenever we perform calculations in the lab, we convert the final answer to significant figures, keeping it as accurate as the least accurate number from the data set used to calculate it.

In simpler words, whenever we perform a calculation, chances are that not all the numbers used will have the same number of accurate significant figures/places after the decimal point. In these cases, we have to ensure that our final answer is only as accurate as the least accurate data figure $the number with the least significant figures or places after the decimal point$ used to calculate it. That way, we ensure that the final answer we present is as accurate as we can get it.

As an example, consider a situation where the average has to be taken of three data readings:

• 6.973 cm
• 8.127 cm
• 7.98 cm

All of these three data readings have been taken with a measurement tool, which means that they all have some margin of error, but more importantly, they are only accurate to a certain limit. The 6.973 cm reading, for example, could really be 6.9734 cm, but we don't know, as the measurement does not have that accuracy.

The 7.98 cm reading could be 7.989 cm, but we don't know, as the measurement does not have that accuracy. If it had, for example, said 7.980 cm, we would know that it is not 7.989 cm. However, since we have only been given accuracy to two decimal places, we cannot infer information about the value of the third decimal place.

The readings give us information about the value of the third decimal place for two of the data points $6.973 cm & 8.127 cm$, but not for the third $7.98 cm$. Since we can't infer information about the value of the third decimal place for 7.98 cm, it would be inaccurate to give our answer $the average$ to three decimal places. Doing so would assume that the 7.98 cm reading is equivalent to 7.980 cm. That's disingenuous - we don't know what the third decimal place is for that reading, so we can't have an answer that assumes the value of the third decimal place.

Thus, when we take the average, we can only give a final answer to two decimal places. Otherwise, we would be assuming the value of the third decimal place of the 7.98 cm reading, and we don't have that information.

The average of these readings is thus 7.69 cm when rounded to two decimal places. It's important to note that when calculating the answer, we still used the entirety of the readings we had $we didn't round 6.973 to 6.97 in our calculations$, but we rounded the final answer when reporting it.