-2
$\begingroup$

In chemistry class, I have been given a set of dimensional analysis questions to answer. I'm not sure if the units factor into whether or not significant figures will need to be applied. I know that when multiplying and dividing, the number in the equation with the least significant figures determines the number of significant figures in the problem. However, I'm confused how to use significant figures to answer these questions.


Question 1: At $\pu{4{^\circ}C}$, water has a density of exactly $\pu{1 g/mL}$. What mass would one gallon of water have at $\pu{4{^\circ}C}$? (There are roughly 3.78541 liters in a gallon.)

Here is my work:

$$ \begin{align} &1 \frac{\pu{g}}{\pu{mL}} * \frac{\pu{1000 mL}}{\pu{1 L}} * \frac{\pu{3.78541 L}}{\pu{1 gal}} * \pu{1 gallon} \\ &= \frac{\pu{1000 g}}{\pu{1 L}} * \frac{\pu{3.78541 L}}{\pu{1 gal}} * \pu{1 gallon} \\ &= \pu{3785.41 g} \end{align} $$

I'm not sure if I'm supposed to round 3785.41 to the one sig fig (making it $\pu{4000 g}$), five sig figs (making it 3785.4), or just leave my answer as it is.


Question 2: The currency used in much of Europe is called the "euro." If one euro is worth 1.2951 U.S. dollars, how much would one million U.S. dollars be in euros?

Here is my work:

$$ \begin{align} &\$1,000,000 * \frac{1 €}{1.2951} \\ &= 772141.1474 \\ &= \pu{700000 €} \end{align} $$

In this question, I assumed that either 1 or 1,000,000 determines the significant figures which is why I rounded the number to the first significant figure. I don't think the 1.2951 determines the significant figures because it contains the most significant figures.


Question 3: How many seconds are in 1.0000 weeks?

Here is my work:

$$ \begin{align} &\pu{1.000 weeks} * \frac{\pu{7 days}}{\pu{1 week}} * \frac{\pu{24 hours}}{\pu{day}} * \frac{\pu{60 min}}{\pu{1 hour}} * \frac{\pu{60 seconds}}{\pu{1 minute}} \\ &= \pu{604,800} \\ &= \pu{600,000 seconds} \end{align} $$

In this question, I believed that there would be only one significant figure due to the unit of days (7) having only one significant figure. Thus, the answer would be rounded to the first significant figure making the final answer 600,000.

$\endgroup$

closed as too broad by Mithoron, Tyberius, Jon Custer, Geoff Hutchison, Todd Minehardt Feb 1 '18 at 17:25

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

1
$\begingroup$

Question 1

It is reasonable to assume the 4°C, and the density of water are exact. The problem specifies that there 3.78541 liters in a gallon, which is six significant figures. So the answer should have six significant figures.

Question 2

The million dollars is an exact number of dollars, but the conversion rate is only specified to five significant figures. So the number of euros should be expressed to 5 significant figures.

Question 3

The number of weeks is given as 1.0000, so the value has 5 significant figures. There are exactly 7 days in a week, exactly 24 hours in a day, exactly 60 minutes in an hour, and exactly 60 seconds in a minute. So the number of seconds ought to be expressed to 5 significant figures.

$\endgroup$
0
$\begingroup$

Adding to what @MaxW has written, as a check it might be helpful to think about the errors in each of your cases.

In (1) the uncertainty in the number of litres is $\pm 0.000005 $ (just less than the number of places quoted assuming that the number given is subject to random error) and use this to determine the uncertainty in your final answer.

In (2) there can be no uncertainty in the exchange rate to euros as it is not subject to experimental error so the number could be argued to be 1.2951000$\cdots$. And so it could also be argued that any number of decimal places is ok, except that the currency is only divided into 1/100 of a dollar so this ultimately limits number of decimal places. (If you consider this to be too contrived deal with it as in (1).)

In case (3) this is essentially the same as (1) but is somewhat artificial in that we must assume that the week has an error of 0.00005 which gives an uncertainty of about 30 secs in the answer.

$\endgroup$
  • $\begingroup$ I agree that using significant figures is a rudimentary method of error propagation. // I think you've over analyzed (2). It is a toy problem. // You're just wrong that the week has an error in the number of days of 0.00005. The error of 0.00005 is in the number of weeks. $\endgroup$ – MaxW Jan 18 '18 at 4:48
  • $\begingroup$ @MaxW, possibly over egged it and yes my error, a typo .I have corrected it. $\endgroup$ – porphyrin Jan 18 '18 at 9:10
  • $\begingroup$ In (3) the error is 0.005%. For 604,800 seconds that is about 30 seconds, not 3. $\endgroup$ – MaxW Jan 18 '18 at 15:03

Not the answer you're looking for? Browse other questions tagged or ask your own question.