Chapter 1, Problem 62RGQ

You set out to determine the density of lead in the laboratory. Using a top loading balance to determine the mass and the water displacement method (Study Question 41) to determine the volume of a variety of pieces of lead, you calculate the following densities: 11.6 g/cm³, 11.8 g/cm³, 11.5 g/cm³, and 12.0 g/cm³. You consult a reference book and find that the accepted value for the density of lead is 11.3 g/cm³. Calculate your average value, percent error, and standard deviation of your results.

Interpretation Introduction

Interpretation: The average, percent error and standard deviation has to be calculated.

Concept introduction:

Precision: Is a measurement indicates how well several determination of the same quantity agree.

Accuracy: Is the agreement of a measurement with the accepted value of the quantity.

Percent error: The difference between measured result and the accepted value.

$\text{Percent}\text{error}\text{=}\frac{\text{Error}\text{in}\text{measurement}}{\text{Accepted}\text{value}}\text{×}\text{100\%}$

$\text{Error in measurement = Experimentally determined value - Accepted value}\text{.}$

Answer to Problem 62RGQ

The average value is $\text{11}{\text{.725 g/cm}}^3\text{.}$

The percent error obtained is $3.7\%$

The obtained Standard deviation is $0.19g/c{m}^3$

The average value, percent error and standard deviation is calculated as,

Explanation of Solution

Average density Value:

  $\begin{array}{l}\text{Average density : }\frac{11.6+\text{ 11}\text{.8 + 11}\text{.5 + 12}\text{.0}}{4}\\ \text{ = 11}{\text{.725 g/cm}}^3\text{.}\end{array}$

Hence, the average density value is obtained as shown above. Hence, the average value is $\text{11}{\text{.725 g/cm}}^3\text{.}$

Percent error:

  $\begin{array}{l}\text{Error}\text{in}\text{measurement =}\text{11}{\text{.725 g/cm}}^3\text{ - 11}{\text{.3 g/cm}}^3\\ \text{Percent}\text{error}\text{=}\frac{\text{11}{\text{.725 g/cm}}^3\text{ - 11}{\text{.3 g/cm}}^3}{\text{11}{\text{.3 g/cm}}^3}\text{×}\text{100\%}\\ \text{ = 3}\text{.7\%}\end{array}$

From the above calculation, the percent error is obtained as shown above.

Standard deviation:

  $\begin{array}{l}\text{SD =}\sqrt{\frac{\Sigma {\text{|x-μ|}}^{\text{2}}}{\text{N}}}\\ \text{where,}\text{x}\text{is values in the data set,}\\ \text{ μ}\text{is}\text{average of the data set (mean value),}\\ \text{ N is no}\text{.of data points}\text{.}\\ \\ \text{Given : }\\ \text{Find mean value (μ):}\\ \text{μ = }\frac{\text{11}\text{.6+ 11}\text{.8 + 11}\text{.5 + 12}\text{.0}}{\text{4}}\\ \text{ = 11}{\text{.725 g/cm}}^{\text{3}}\text{.}\\ \begin{array}{cccc}\text{Determination}& \text{Measured values}& \begin{array}{l}\text{Difference}\text{between}\\ \text{Measurement}\text{and}\text{Average}\\ \text{(x-μ)}\end{array}& \begin{array}{l}\text{Square of Difference}\\ {\text{|x-μ|}}^{\text{2}}\end{array}\\ \text{1}& \text{11}\text{.6}& \text{-0}\text{.125}& \text{0}\text{.016}\\ \text{2}& \text{11}\text{.8}& \text{0}\text{.075}& \text{0}\text{.006}\\ \text{3}& \text{11}\text{.5}& \text{-0}\text{.225}& \text{0}\text{.051}\\ \text{4}& \text{12}\text{.0}& \text{0}\text{.275}& \text{0}\text{.076}\end{array}\\ \Sigma {\text{|x-μ|}}^{\text{2}}\text{=0}\text{.016 + 0}\text{.006 + 0}\text{.051 + 0}\text{.076}\\ \text{ = 0}\text{.149}\\ \text{SD=}\sqrt{\frac{\text{0}\text{.149}}{\text{4}}}\text{=}\text{0}{\text{.19g/cm}}^{\text{3}}\end{array}$

The obtained Standard deviation is $0.19g/c{m}^3$

Conclusion

The average value was calculated as $\text{11}{\text{.725 g/cm}}^3\text{.}$

The percent error obtained was calculated as $3.7\%$

The obtained Standard deviation was calculated as $0.19g/c{m}^3$