# Concentration uncertainty due to volume measurement with cylinder

(All volumes were measured using a measuring cylinder which has a $$\pm\pu{0.5 cm3}$$ uncertainty)

• You have a $$\pu{50 cm3}$$ of $$\pu{0.5 M}$$ solution to begin with
• You then make a $$\pu{0.4 M}$$ solution by mixing $$\text{40 cm}^3$$ of the above solution with $$\pu{10 cm3}$$ water.

If I'm not mistaken, the uncertainty for the volume should be $$\pm\pu{1.0 cm3}$$ because the second solution is the result of two measurements (and also uncertainties).

My question is: what is the uncertainty for the concentration?

• It depends whether the 0.5 uncertainty represent a property of the cylinder (systematic error) or if the cylinder is perfect and 0.5 represent random error in reading the cylinder. In really life it would be some of each. Jan 29, 2015 at 21:36
• Thank you for the reply. In this case, I'm talking about the uncertainty as a property of the cylinder. How would you work out the random error in reading?
– Nol
Jan 30, 2015 at 9:13

1. Let's assume that you're talking about random error. If you specify the error in the volume with $\pm$, this is probably the case---systematic error in this case would have a single sign; for example, if the scale marks on the side of cylinder were shifted by $\rm 0.5\ cm^3$ down from what their true values should have been, you'd have a systematic error of $-0.5\ \rm cm^3$ (the negative sign tells you that your readings are $0.5\ \rm cm^3$ lower than they should have been).

You can say that the error in a sum will be less than or equal to the sum of errors. It's better to say that the uncertainty in a sum is the square root of the sum of squared uncertainties:

$$\delta(V_1+V_2) = \sqrt{\delta(V_1)^2 + \delta(V_2)^2} = \sqrt{\rm(0.5\ cm^3)^2 + (0.5\ cm^3)^2} = 0.70_7\ \rm cm^3$$

For a product or quotient, you can say that the relative error will be less than or equal to the sum of the relative errors in the measurements. It's better to say that the relative error in a product or quotient is the square root of the sum of squared relative errors:

$$\begin{array}{rcl} \delta(C_{dilute}) &=& \delta\left(\frac{C_{stock}V_1}{V_1+V_2}\right)\\ \frac{\delta(C_{dilute})}{C_{dilute}} &=& \sqrt{\left(\frac{\delta(C_{stock})}{C_{stock}}\right)^2 + \left(\frac{\delta(V_{1})}{V_{1}}\right)^2 + \left(\frac{\delta(V_{1}+V_2)}{V_{1}+V_2}\right)^2 } \end{array}$$

...can you take it from there?

2. Let's assume you have systematic error, that is, you actually meant something like "The cylinder always reads $\rm 0.5\ cm^3$ too high". You will need a definite sign on a systematic error, not a $\pm$.

There are several ways to proceed here. Let's take the easiest route, which is to estimate the systematic error in the concentration as the difference between the concentration and its perturbed value (the value with all of the volumes having the systematic error added in). If your systematic error in volume was $\delta V = +0.5\ \rm cm^3$, you can estimate the systematic error in the concentration as

$$\begin{array}{rcl} \delta{C_{dilute}} &\approx& C_{dilute}^{perturbed} - C_{dilute} \\ &\approx& \frac{C_{stock}(V_1+\delta(V)}{(V_1+\delta(V))(V_2+\delta(V))} - C_{dilute} \end{array}$$

The sign on this error will be important; it tells you whether the systematic error in volume makes the concentration higher or lower than its "true" value.

Note: The above calculation is just an estimate of the error. You can do a better job using derivatives here. I'll show you that if you want to see it.

• Thank you very much for the reply. So for the volume uncertainty, we cannot simply add up two ±0.5's? I added them up because I thought that ±1.0 would be the extreme minimum and maximum boundaries. As for the concentration, I have no idea at all before asking. Is the equation Cstock(V1) = Cdilute(V1+V2) derived from nstock = ndilute?
– Nol
Jan 30, 2015 at 9:19
• Some people do just add them up to get a MAXIMUM error; you can say $\delta(V_1+V_2) \le \delta(V_1) + \delta(V_2)$ so $\delta(V_1+V_2) \le 1.0\ \rm cm^3$ in this case. Notice the inequality; the way I do it in the answer above gives a better error estimate. And yes, that's where that equation came from. Jan 30, 2015 at 12:28
• My value of ±0.5 cm^3 came from the fact that 1 cm^3 was the smallest division of the measuring cylinder. Thus I believe the error is half of the smallest division for analog instruments. I am not entirely sure which method to use but am more leaning towards method 2 of systematic errors. In this case, should the final error be written with signs or ±?
– Nol
Jan 30, 2015 at 15:25
• Actually, no, it's random error then because you're estimating the error from the measuring scale; it would be systematic error if the scale on the side of the cylinder was stamped incorrectly by $0.5\ \rm cm^3$. Usually people say you can read a scale to 1/10th of the smallest scale division, so $\pm 0.5\ \rm cm^3$ is pretty conservative. Jan 30, 2015 at 15:31
• I see, it seems my understanding on errors was quite off. Thank you for explaining in detail this far, I'll definitely upvote your answer when I can (currently I only have 14 reputation). Sorry for the hassle but I have two more questions. 1) Can you show me the maths behind the random error? 2) What are the values for V1 and V2 in this example?
– Nol
Jan 30, 2015 at 15:46