Finding Average Molarity: SF in Data

  • Context: High School 
  • Thread starter Thread starter frozonecom
  • Start date Start date
  • Tags Tags
    Data
Click For Summary
SUMMARY

The discussion focuses on calculating the average molarity of a solution with values of 0.092 M and 0.095 M, both having two significant figures. The correct method involves first summing the values to obtain 0.187 M, which has three decimal places, and then dividing by 2 to find the average. The debate centers on whether to report the average as 0.0935 M or 0.094 M, with the consensus leaning towards 0.094 M due to the significant figures of the original data. Additionally, uncertainty in the measurements is discussed, suggesting a preferred notation of 0.094 M with an uncertainty of ±0.0004 M.

PREREQUISITES
  • Understanding of significant figures in scientific measurements
  • Familiarity with basic arithmetic operations (addition and division)
  • Knowledge of error propagation techniques
  • Awareness of the PEMDAS order of operations
NEXT STEPS
  • Research the rules of significant figures in calculations
  • Learn about error propagation methods in experimental data
  • Explore the concept of uncertainty in measurements
  • Study the implications of rounding in scientific reporting
USEFUL FOR

Students in chemistry, laboratory technicians, and anyone involved in scientific data analysis who needs to understand the principles of significant figures and uncertainty in measurements.

frozonecom
Messages
63
Reaction score
0
Significant Figures in Data

Suppose I have two recorded Molarity of a certain solution.
And the data I have are:

0.092 M and 0.095 M --- which both has 2 significant figures.

If I were asked to get the AVERAGE MOLARITY of my substance, then I'd do:

(0.092 M + 0.095 M)/2 = My AVG. MOLARITY.

Since 2 is an exact number, I will not base the SF of my final answer to that number.
Instead, I'd base it to my Molarities, but I know I should do the addition first, since they're enclosed in a parenthesis. (following PEMDAS)

0.092 M + 0.095 M = 0.187 M ----- (It should have 3 decimal places, since that is my least decimal place in my data)

Then, dividing my answer by 2, should I report it as:

0.0935 M ----- since 0.187 has 3 SF
OR
0.094 M ------- since my initial working data only has 2 SF?

Which one is more correct? Which one should I follow and why?
 
Last edited:
Physics news on Phys.org
Anyone??
 
I would quote (0.0935 +- 0.0004) M, where the 4 comes from the assumption that both initial values have an uncertainty of 0.0005 each and the remaining part is error propagation.
If you cannot give an explicit uncertainty for some reason, I think I would prefer 0.094 M.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
8K
  • · Replies 25 ·
Replies
25
Views
2K
Replies
2
Views
4K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
7K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 32 ·
2
Replies
32
Views
5K