Question

A chemistry student carried out an experiment; the mass she recovered was 3.15 g; her percent error was 1.61%. What was the accepted value?

Answer 1

The accepted value was either 3.10 g or 3.20 g

Why?

There are two possibilities:

1. The accepted value is lower than the experimental value by 1.61%
2. The accepted value is higher than the experimental value by 1.61%

The formula for the percent error is the following:

`\% error = (|Experimental Value-AcceptedValue|)/(|Accepted Value|)*100`

Clearing for Accepted Value and inputting the values we get the following:

`Accepted Value=(ExperimentalValue)/((\% Error)/(100)+1 ) =(3.15g)/((1.61)/(100)+1)=3.10 g`

Another possibility is that the the Accepted value is higher. In that case we'd get:

`Accepted Value=(ExperimentalValue)/(1- (\% Error)/(100) ) =(3.15g)/(1-(1.61)/(100))=3.20 g`.

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!