The accepted value was either 3.10 g or 3.20 g
Why?
There are two possibilities:
- The accepted value is lower than the experimental value by 1.61%
- The accepted value is higher than the experimental value by 1.61%
The formula for the percent error is the following:
[tex]\% error = \frac{|Experimental Value-AcceptedValue|}{|Accepted Value|}*100[/tex]
Clearing for Accepted Value and inputting the values we get the following:
[tex]Accepted Value=\frac{ExperimentalValue}{\frac{\% Error}{100}+1 } =\frac{3.15g}{\frac{1.61}{100}+1}=3.10 g[/tex]
Another possibility is that the the Accepted value is higher. In that case we'd get:
[tex]Accepted Value=\frac{ExperimentalValue}{1- \frac{\% Error}{100} } =\frac{3.15g}{1-\frac{1.61}{100}}=3.20 g[/tex].
The most likely value for the accepted value is 3.20 g, as most experiments in which mass is recovered yield a lower experimental value than the initial one.
Have a nice day!
