HSC Chemistry Guide:
How many significant figures I need?
In this HSC Chemistry Guide, we will take a look at the significant figures that you should have in your answer for different types of calculations used in the HSC Chemistry course.
Calculation Type #1: Final answer involving addition
When the last step in obtaining your final answer involves adding two or more values together, your answer should have identical decimal places as the value with the least decimal places that you added to obtain your answer.
Example: Suppose you have 7.27 moles of sodium chloride dissolved in a beaker of water and you just added another 4.2036 moles of sodium chloride into the same beaker. What is the total moles of sodium chloride is in the beaker?
Answer: Well, you will first add up 7.27 moles and 4.2036 moles using your calculator which will yield 11.4736 moles.
As stated earlier, if your final answer involves adding two or more values together, you answer should have identical decimal places as the value with the least decimal places that you added to obtain your answer.
Well, in our case, we added up 7.27 and 4.2036 to obtain our answer.
In terms of decimal places, 7.27 has two decimal places and 4.2036 has four decimal places.
Therefore, 7.27 has the fewest decimal places and therefore our final answer should have the same amount of decimal places as 7.27. That is, our answer should have two decimal places and NOT four or any other number.
Therefore our answer is 11.47 moles.
We DO NOT care about significant figures when the final step in obtaining our final answer involving the addition of two or more values. We only care about decimal places.
The reason why we use the value with fewest decimal place is to ensure we report our answer to as precise as possible without over-estimating precision in your reported final answer.
We should not report our answer to four decimal places as we are not sure what the exact mole of sodium chloride (7.27) is in four decimal places.
Calculation Type #2: Final answer requiring subtraction
If the last step of obtaining your final answer involves subtracting two or more values, you should report your final answer as the same rule as if your final answer involves adding two or more values.
That is, your final answer should have identical decimal places as the value with the least decimal places that you are subtracting to obtain your final answer.
Example: Suppose your teacher wants to you to weigh 2.83 grams of gummy bears for an experiment. You added gummy bears into a beaker and the reading shows 3.205 grams. What is the mass of gummy bears that over the mass which your teacher wants you to weigh for the experiment?
Answer: Well, to get the excess mass, we need to subtract 2.83 from 3.205 grams. If we calculate this using our calculator, we yield an excess mass of 0.375 grams.
Since the last step in obtaining our final answer involves subtraction, we are required to report our final answer equal to the value of the least decimal place which we used in our subtraction.
Therefore, our answer should be reported to two decimal places, i.e. 0.38 grams (after rounding up).
Again, similar to when the last step of your final answer involves addition, in the case of subtraction, we also don’t care or look at significant figures.
Calculation Type #3: Final answer requiring multiplication
When we are required to multiply two or more values to obtain our final answer, we need to report our final answer with significant figures that is identical to the value with the least significant figures which we multiplied to obtain our final answer.
Example: Suppose you want to determine the moles of sodium chloride present in a lab partner’s solution. You are given that your lab partner added 0.512 litres of sodium chloride at a concentration of 0.21M.
Answer: Well, to calculate the number of moles, we need to multiply the volume and concentration of sodium chloride added into the beaker together. If we put that into our calculator, we get 0.10941 moles.
However, we need to report our answer with significant figures that is identical to the value with the least number of significant figures which we multiplied with to obtain our final answer.
The volume of sodium chloride is 0.512 litres which has 3 significant figures
The concentration of sodium chloride is 0.21M which has 2 significant figures.
Therefore, our final answer should have 2 significant figures, i.e. 0.11 moles (after rounding up).
Calculation Type #4: Final answer requiring division
If the last step of obtaining your final answer involves dividing two or more values, your final answer should have significant figures that is identical to the value with the least significant figures which you divided to obtain your final answer.
Example: Suppose you want to determine the concentration of silver nitrate present in a lab partner’s solution. You are given that your lab partner added 0.0031 moles of silver nitrate with a volume of 0.002 litres into an empty beaker.
Answer: Well, to calculate the concentration, you need to divide moles by volume. If we put that into our calculator, we get 1.55M of silver nitrate.
Well, the moles of silver nitrate is 0.0031 moles which has 2 significant figures.
The volume of silver nitrate is 0.002 litres which has 1 significant figure.
Therefore, our final answer should have 1 significant figure, i.e. 2.0 x 101 M or 1M.
Question Type #5: Final answer requiring logarithm
When your final answer requires to use logarithms, you need to report your final answer with decimal places that is equal to the significant figures of value which you apply the logarithm to.
Example: Calculate the pH of sulfuric acid with a concentration of 0.101M.
Answer: Well, the applicable formula here is pH = -log10[H+] where [H+] is equal to 0.202M.
Why 0.202M? Well, the moles of hydrogen ions is twice the moles of sulfuric acid due to mole ratio. So, 0.101M x 2 = 0.202M. The volume of the acid also did not change so if we are given the concentration of the sulfuric acid, we can determine the concentration of hydrogen ions.
So, pH = -log10[0.202] = 0.695.
Note that since [H+] = 0.202 which has 3 significant figures.
Therefore, our final answer needs to have decimal places that is equal to the amount of significant figures as the value which we applied the logarithm to.
Therefore, our final answer should be reported to three significant figures, i.e. 0.695.
Side Note: The second ionisation of sulfuric acid, i.e. HSO42- -> H+ (aq) + SO42-(aq), is not complete but rather the HSO42- ion has approximately 30% degree of ionisation. However, for HSC purposes, we assume that sulfuric acid is a strong acid and thus it has complete ionisation. Therefore, we multiply 0.101M by two to obtain 0.202M of hydrogen ions.
OH NO, AN EXCEPTION!
If we are going to take the average of two or more values, it means that we are going to divide. As mentioned in calculation type #4, the final answer therefore should have the same number of significant figures as the value with the least significant figures which you divided to obtain the final answer.
However, we will illustrate an exception to previous calculation types here.
Example: What is the average of 71.32, 87.1, 91.8?
Answer: Well, to take the average of the three values, we need to add up all three values before dividing by three.
If we add up all three values, we get 250.22 and we can leave it to two decimal places on paper (or you can write it down on paper as 250. It does not really matter as it is not our final answer to the question yet).
Next, to get our final answer, we divide 250.22 (exact value stored in calculator) by 3 which gives us 83.406666..
Recall that from calculation type #4, we say that if the final answer involves division, the final answer should report its significant figures equal to the value of lowest significant figures which you divided to obtain the final answer.
Since ‘3’ is an exact value here, we ignore it in our consideration of significant figures.
Since 250.22 has five significant figures, you should therefore expect your final answer to have five significant figures. However, notice that from our raw data values, the data value with the LEAST significant figures is three.
Therefore, our final answer CANNOT have more than three significant figures!
So, our final answer should be reported to three significant figures which will be 83.4.
Practical Example involving multi-step calculations
Random calculation question: Suppose that we have a beaker of aqueous sulfuric acid (with water) sitting on an electronic balance that has a reading of 300.46 grams. For fun, you took a spoon and scooped out some sulfuric acid from the beaker. The new electronic balance reading is now 297.24 grams.
Later, you transfer the sulfuric solution that you scooped up into a conical flask.
Part (a): Calculate the concentration of hydrogen ions present in the conical flask? You are given that the molar mass of sulfuric acid is 98.079 grams. Assume that the density of H2SO4 is 1 g/mL.
Answer to part (a): m(H2SO4) scooped = 300.46g – 297.24g = 3.22g
n(H2SO4) in conical flask = m(H2SO4) / MM(H2SO4) = 3.22g / (98.079g/mol) = 0.0328 moles (store exact value of the moles of sulfuric acid in calculator)
[H2SO4] = n(H2SO4) / V(H2SO4) = 0.0328 moles / (3.22/1000) = 10.2M
NOTE: 3.22mL has 3 significant figures and, in litres, it still has 3 significant figures, i.e 3.22 x 10-3 litres.
This is because the conversion factor between mL and Litres (1000mL) is an exact value (fixed) in all situations. Therefore, we do not need to factor it in our significant figures calculations.
Similarly, the hypothetical density used for sulfuric acid here is also an exact value or is fixed and hence we do not consider it in our significant figures calculations.
All in all, this gives, [H+] = 2x [H2SO4] = 2 x 10.2M = 20.4M
NOTE: The ‘2‘ used to compare the mole ratio between sulfuric acid and hydrogen ion is an exact value or is fixed. Therefore, we do not need to factor it in our significant figures calculations.
NOTE: Notice that although the lowest significant figures provided in the data values is 5, our answer has significant figures of 3. This is because prioritise our calculated value’s significant figures over raw data’s significant figures. As long as our calculated value’s significant figures does not exceed a raw data value with the less than three significant figures, we are okay!
Part (b): What is the pH of the acid solution in the conical flask?
Answer to part (b): pH = -log10[H+] = -log10[20.4] = -1.31
*NOTE: We stored “20.4” from part (a) as an exact value on our calculator, i.e. 20.39172504.
Yes, if the concentration of a strong acid is high enough, it is possible to have negative pH as shown above. Similarly, if the concentration of a strong base is high enough, it is possible to have a pH greater than 14!