# How many significant figures should I report calculating volume concentrations? [closed]

Let's say a $$\pu{300mL}$$ solution of $$\ce{CuCl_2}$$ was diluted to $$\pu{900mL}$$. I am then asked to calculate the concentration of the new solution. My solution to the problem is as follows:

\begin{align}M_1 \times V_1 &= M_2 \times V_2\\[0.5em] \pu{300mL} \times \pu{4M} &= M_2 \times \pu{900mL}\\[0.5em] M_2 &= \pu{1.33M}\end{align}

Now I'm wondering how many significant figures my final answer should have. From what I know, I have to refer to the least significant measurement, which happens to have one significant figure. Should my answer be 1 instead?

• I've tried to clarify your question, but I don't see what has to do with Stoichiometry... – G M Apr 10 at 11:35
• The data are not clearly defined. For example, M1 is suddenly taken as 4 mol/L. Where does it come from ? Then M2 is 1.33 without any unit . What does it mean ? Further more : What are the errors on the given volumes (300 mL and 900 mL) Are they ± 1 mL ? We cannot guess them ... – Maurice Apr 10 at 16:48
• While you cannot ever assume the precision or accuracy of measurements when it is not given directly, most of the time measurements between 250 mL and 1000 mL have a precision of 5 mL and an accuracy of ±5 mL; however, you should always check your graduated cylinder (or volumetric flask) to determine the exact accuracy and precision of your measurements. – Eli Jones Apr 13 at 13:47
• RANT - I'll go off on a bit of a tangent here. It seems that it is common to give answers for problems in most books to 3 significant figures regardless of how the problem is setup. To me "4 molar" has one significant figure. If the book's author wanted three significant figures then the problem should specify "4.00 molar." – MaxW Apr 13 at 20:41
• I don't think that this question should be closed, the user attempted to answer and explained its considerations – G M Apr 14 at 6:59

We cannot say how many significant figures you should be reporting because:

• We cannot be sure about the precision of your volume measurements; and
• We cannot be sure about the precision of the original solution’s concentration.

Concerning volumes: most of the time when working with such volumes in the lab, I would use measuring cylinders. These come with a rather poor precision but are adequate for most purposes. However, considering we need $$\pu{300ml}$$ and $$\pu{900ml}$$, we would probably want to work with very large cylinders so the error would be in the range of $$5{-}\pu{10ml}$$. It is possible to use much more precise equipment: in analytical lab courses, we would use pipettes that measure exactly $$\pu{25.0 \pm 0.1 ml}$$ and dilute using measuring flasks at a similar level of precision.

If, as it seems from the values you plugged, the initial solution was $$\pu{4M}$$ (1 significant figure), I would argue the result should be given to 1 significant figure too – introducing a large error. Arguably, a proper error propagation calculation would probably give you a better precision in the final result.

• The point about volumes is that "$\pu{3.00\times 10^2 ml}$" and "$300. \ \pu{ml}$" have three significant figures, but "$\pu{300 ml}$" is ambiguous since there could be 1, 2, or 3 significant figures. – MaxW Apr 13 at 19:42

The concept of significant figures is not always clear, and the rule that you find in many textbooks can be misleading. If you read the international standard e.g. Evaluation of measurement data — Guide to the expression of uncertainty in measurement you will hardly find a comparable definition of a significant figure.

In fact, the answer here, in my opinion, depends on the context. If you are adding solution with a precision of one millilitre. Would be correct to report $$\pu{1.333M}$$ because if you added, for instance, $$\pu{0.301 mL}$$ you would register $$\pu{1.337M}$$. So I would say that the correct answer is $$\pu{1.333 M}$$.

Eventually, it depends on what are you going to do with that solution and you might not need such precision.

• -1 ignoring the molarity of the solution, the volume contains at most 3 significant figures. "1.333 molar" would have four. Inventing extra precision is exactly what significant figures is trying to avoid. – MaxW Apr 13 at 19:53
• @MaxW thanks a lot for your feedback, I have to ponder more about this subject but given the precision of one millilitre, I am still not sure that in practice is not better to preserve also the third digit after the comma because excessive truncation I think it eventually leads to biases. – G M Apr 14 at 6:57
• In my days in high school and college a table of 4 digit log tables or a slide rule was the norm. 3 digits was all you got. Now with a calculator I carry two extra digits in intermediate calculations and round off the final result to the desired number of digits. This will reduce round off errors in the overall calculation but round-off errors can't be prevented entirely. Using the 20 digits that a calculator might have is just to tedious. – MaxW Apr 14 at 7:10
• @MaxW you are right, thanks for your comments! – G M Apr 14 at 7:30
• One more thing. The guide you pointed to uses more sophisticated methods for error propagation. However the Wiki article on Significant Figures gives a good explanation. – MaxW Apr 14 at 7:41

Short answer: Without knowledge of the partial errors, expressing the final concentration as 1.33 M or 1.333 M is fine.

The initial point is to know the significant digits of your input values. E.g. what are significant digits of the 4 M solution ? 300 mL ? 900 ml ?

The are 2 major rules for 4 arithmetic operations and significant digits:

1/ For addition/subtraction, the result has the last significant digit as the number with less absolute accuracy:

$$4.005 + 3.2 \simeq 7.2$$ ( takes the last significant digit place from the latter )

1/ For multiplication/division, the result has the count of significant digit as the number with less relative accuracy:

$$2.0 * 3.211 \simeq 6.2$$ ( takes the significant digit count from the former )

As a general rule, if there is given a function:

$$y=f(x_1,x_x,..) \tag{1}$$

calculate its differential form:

$$\mathrm{d}y=\dfrac {\partial f}{\partial x_1} \cdot \mathrm{d}x_1 + \dfrac {\partial f}{\partial x_2} \cdot \mathrm{d}x_2 + ... \tag{2}$$

Take the given or estimated values of $$\mathrm{d}x_\mathrm{i}$$ as estimation of the current or maximal error of independent variables $$x_\mathrm{i}$$ , calculate $$\mathrm{d}y$$ and then decide the valid digits.

If you evaluate rather standard deviation errors or confidency intervals, you have to evaluates the square, implying there is no correlation between the input variable errors.

$$\mathrm{d}y=\sqrt{ {\left(\dfrac {\partial f}{\partial x_1} \cdot \mathrm{d}x_1 \right)}^2 + {\left(\dfrac {\partial f}{\partial x_2} \cdot \mathrm{d}x_2\right)}^2 + ...} \tag{3}$$

For the example in the question, we have:

$$c_2 = c_1 * \frac{V_1 }{ V_2} \tag{4}$$

and by differentiation:

$$\mathrm{d}c_2 = \mathrm{d}c_1 * \frac{V_1 }{ V_2} + c_1 * \frac{\mathrm{d}V_1 }{ V_2} - c_1 * \frac{V_1 }{ V_2^2} \cdot \mathrm{d}V_2 \tag{5}$$

If we plug in the numbers for $$c_1, V_1, V_2$$, replace $$\mathrm{d}$$ by $$\Delta$$ and apply rather the absolute value for error cumulation:

$$\Delta c_2 = \left|\Delta c_1 \cdot \frac{ \pu{0.3 L} }{ \pu{0.9 L}}\right| + \left|\pu{ 4 mol/L} \cdot \frac{\Delta V_1 }{ \pu{0.9 L } }\right| + \left|\pu{ 4 mol/L} \cdot \frac{\pu{0.3 L} }{({ \pu{0.9 L})}^2} \cdot \Delta V_2 \right| \tag{6}$$

Estimate the error of initial concentration $$c_1$$, of the initial and final volumes $$V_1$$ resp. $$V_2$$.

The $$\mathrm{d}c_2$$ is the error of the final concentration. Plug in the numbers. The order of the final error should be the last valid/significant digit.

Be aware the concentration info 4 M says nothing about its accuracy, in contrary to e.g. 4.00 M.

In case of the SD or confidency intervals, do the same but with the squares as above.

$$\Delta c_2 = \sqrt {{\left(\Delta c_1 \cdot \frac{ \pu{0.3 L} }{ \pu{0.9 L}}\right)}^2 + {\left(\pu{ 4 mol/L} \cdot \frac{\Delta V_1 }{ \pu{0.9 L } }\right)}^2 + {\left(\pu{ 4 mol/L} \cdot \frac{\pu{0.3 L} }{({ \pu{0.9 L})}^2} \cdot \Delta V_2 \right)}^2} \tag{6}$$

• Think you went down the rabbit hole. The OP specifically asked about significant figures, not error prorogation in general. – MaxW Apr 13 at 19:34
• "The order of the final error should be the last significant digit" – Poutnik Apr 13 at 20:30