Question

solve each of the following problems. show all your work and remember to include units! percent error = |experimental value - accepted value| / accepted value × 100% 1. as the result of experimental work, a student finds the density of a liquid to be 0.1369 g/cm³. the known density of that liquid is 0.1478 g/cm³. what is the percent error of this students work? 2. after heating a 10.00 g sample of potassium chlorate, a student obtains an amount of oxygen calculated to be 3.90 g. theoretically, there should be 3.92 g of oxygen in this amount of potassium chlorate. what is the percent error in this experiment? 3. the melting point of potassium thiocyanate determined by a student in the laboratory turned to be 174.5 °c. the accepted value of this melting point is 173.2 °c. what is the percent error in this reading? 4. a person attempting to lose weight on a diet weighed 175 lb. on a bathroom scale at home. an hour later at the doctors office, on a more accurate scale, this persons weight is recorded 178 lb. assuming that there was no real weight change in that hour, what is the percent error for these reading? 5. in an exercise to teach students how to use and analytical balance, the instructor gives student a quarter which has been pre - weighed as 5.6026 g. the weight that the student reads for the same quarter is 5.6013 g. what is the percent error in the students reading?

Explanation:

Step1: Recall percent - error formula

Percent Error=$\frac{\vert\text{experimental value}-\text{accepted value}\vert}{\text{accepted value}}\times100\%$

Step2: Solve problem 1

Experimental density = $0.1369\ g/cm^{3}$, Accepted density = $0.1478\ g/cm^{3}$
Percent Error=$\frac{\vert0.1369 - 0.1478\vert}{0.1478}\times100\%=\frac{0.0109}{0.1478}\times100\%\approx7.37\%$

Step3: Solve problem 2

Experimental amount of oxygen = $3.90\ g$, Accepted amount of oxygen = $3.92\ g$
Percent Error=$\frac{\vert3.90 - 3.92\vert}{3.92}\times100\%=\frac{0.02}{3.92}\times100\%\approx0.51\%$

Step4: Solve problem 3

Experimental melting - point = $174.5^{\circ}C$, Accepted melting - point = $173.2^{\circ}C$
Percent Error=$\frac{\vert174.5 - 173.2\vert}{173.2}\times100\%=\frac{1.3}{173.2}\times100\%\approx0.75\%$

Step5: Solve problem 4

Experimental weight = $175\ lb$, Accepted weight = $178\ lb$
Percent Error=$\frac{\vert175 - 178\vert}{178}\times100\%=\frac{3}{178}\times100\%\approx1.69\%$

Step6: Solve problem 5

Experimental weight of quarter = $5.6013\ g$, Accepted weight of quarter = $5.6026\ g$
Percent Error=$\frac{\vert5.6013 - 5.6026\vert}{5.6026}\times100\%=\frac{0.0013}{5.6026}\times100\%\approx0.023\%$

Answer:

1. Approximately $7.37\%$
2. Approximately $0.51\%$
3. Approximately $0.75\%$
4. Approximately $1.69\%$
5. Approximately $0.023\%$