Miscelaneous information about mathematics in chemistry.
Basically, dimensions can be treated as algebraic quantities. So dimensions can be added and subtracted only if they have the same base. Simple! In chemistry you can use this technique to do problems you don't understand on multiple choice tests. Look at the answers and see what units they are in, then look at the question and see what units are given. After that use conversion factors to try and cancel them out.
Every measurement entails a degree of uncertainty; this can vary depending on the measure device and skill with which the device is operated. It is accepted that there is doubt of at least one figure in the last digit. For example, a miniscus occurs when water is stored in glass, it is the thin line that curves upward. Someone might read the top of the miniscus and another person the bottom, this would cause an inaccuracy in the measurement.
Significant figures are the digits in any measurement that are known with a certainty plus a digit that is uncertain on the end. For example, a graduated cylinder is labeled in one millimeter increments, you can be sure of the number to the last line and then guess if the liquid is halfway between the next solid measurement.
The rules for finding significant figures are as follows:
Rule 1: In numbers that do not contain zeros, all the digits are significant. For example, the number 3.1228 has 5 significant figures, 3.14 has 3 and 514 has 3 significant figures.
Rule 2: All zeros between significant digits are significant. For example, the number 7.05 has 3 significant figures, 6002 has 4 significant figures and 3.0041 has 5 significant figures.
Rule 3: Zeros to the left of the first nonzero digit serve only to fix the position of the decimal point and are not significant. For example, 0.00058 has 2 significant figures, 0.0003094 has 4 significant figures, and 0.000001 has 1 significant figure.
Rule 4: In a number with digits to the right of a decimal point, zeros to the right of the last nonzero digit are significant. 43 has 2 significant figures, 43.0 has 3 significant figures, 43.00 has 4 significant figures, and 0.40050 has 5 significant figures.
Rule 5 is harder to explain and needs an example, suppose a 100 gram brick was measured, in this notation it is impossible to know how precise the measurement was. It could have been measured to the nearest gram (100 plus or minus 1 g) if so it has 3 significant figures, or it could have only been weighted to the nearest 10 grams (100 plus or minus 10) in which case it has 2 significant figures.
The mass of the brick should be reported as either 1.00 x 10^2 g (in which case there are 3 significant figures
1.0 x 10^2 grams (2 significant figures)
or finally it could be reported as 1 x 10^2 g (only one significant figure)
It is important to remember that when multiplying or dividing, the number of significant figures in the answer should be the same as the number that had the fewest significant figures when working out the problem.
Example: A bus leaves a location and arrives in another 8.05 hours later. Let’s pretend the distance between them was 486.9 miles. What is the average speed of the bus in miles per hour?
To solve for speed we need to formulate the following equation, speed = distance traveled/time so filling in the equation it would be s=486.9 miles/8.05 hours. First calculate the number of significant figures in both numbers, 486.9 has 4 significant figures, and 8.05 has 3 significant figures, so that means our answer must have 3 significant figures. Now solve 486.9/8.05 which is equal to 60.48447205, however we are only allowed 3 significant figures so the answer is changed to 60.5 miles per hour.
For addition and subtraction the process is a little different, the number of decimal places in the result will be the same as the number with the smallest number of decimal places. For example, 10.21 + 0.2 + 256 would be equal to 266.
The official way to arrive at that number is to add the numbers normally and then change the answer
so that it has the corresponding number of significant figures. However I find it just as simple to
ignore the excess decimals during the equation, so instead of working the entire equation out I
would simplify it into 10 + 0 + 256, which is equal to 266. The normal degree of inaccuracy caused
by this faster method is generally considered acceptable.
This can be important because it simplifies chemistry by shaving off irrelevant information.