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








**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
 
 






---








 **Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)




Closed 3 years ago.








[Improve this question](/posts/131513/edit) 



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? 




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.


 2

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](https://www.bipm.org/utils/common/documents/jcgm/JCGM_100_2008_E.pdf) 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. 


 0

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


**Not so short answer**


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 )


**Longer answer**:


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.


$$\Delta c\_2 = \sqrt {{\left(\Delta c\_1 \cdot \frac{ \pu{0.3 L} }{ \pu{0.9 L}}\right)}^2 + {\left(\pu{ 4 mol/L} \cdot \frac{\Delta V\_1 }{ \pu{0.9 L } }\right)}^2 + {\left(\pu{ 4 mol/L} \cdot \frac{\pu{0.3 L} }{({ \pu{0.9 L})}^2} \cdot \Delta V\_2 \right)}^2} \tag{6}$$


 0