Practice Problems on Working with Significant Figures


It is assumed that you already know the rules on working with significant figures.  If you don’t, you should review the “Tutorials on Significant Figures” accessed from my Homepage.


Quick Reminder:



Before you begin, remind yourself that these are not simple arithmetic problems.  The point of this exercise is for you to practice WATCHING YOUR SIGNIFICANT FIGURES!


Answers are provided at the bottom.  Be sure to WRITE DOWN YOUR ANSWERS before you look at the correct answers provided.










9.   The following are placed in a beaker weighing 39.457 g:

      2.689 g of NaCl, 1.26 g of sand and 5.0 g water

      What is the final mass of the beaker?


10. If the beaker containing a sample of alcohol weighs 49.8767 g and
the empty beaker weighs 49.214 g, what is the weight of the alcohol?












1.  40.9    (This is an addition problem. The limiting factor is 1.3, with one decimal place.)


2.   2       (This is a subtraction problem.  The limiting factor is 193, with NO decimal place.)


3.  1.8x103  (This is an addition problem.  The limiting factor is 1200, which has “ambiguous zeroes” which are assumed to be not significant.  We are therefore limited to the hundreds place.  The answer is 1800 but to eliminate the ambiguity of the tailing zeroes, it MUST be expressed in scientific notation, hence 1.8x103.)


4.   1.4 x 103  (900 and 500 have ambiguous zeroes that are assumed to be not significant.  The ans is limited to the hundreds place.  It is 1400 with ambiguous zeroes, so it must be expressed in scientific notation, hence 1.4x103.)


5.   5.4    (Both numbers in the question has one decimal place, so ans should have one decimal place.)


6.   18     (This involves only multiplication and division.  The limiting factor is 2.9 with 2 sig. fig., so the answer should have 2 sig. fig..  Note there is no reason to express it in scientific notation.)


7.   1.4    (This has a mix of subtraction and division, with different rules, so you must do them separately.  23.5 – 21.3 = 2.2 (with one decimal place) and then dividing 2.2 by 1.58, the answer is 1.4 with 2 significant figures.)


8.   0.396


9.   48.4 g


10. 0.663 g