Chapter 3, Problem 3.AE

(a)

Interpretation Introduction

Interpretation:

The mass of precipitate contained in the crucible has to be calculated and the number of significant digits in that mass has to be given

Concept Introduction:

Significant Figures:

While reporting a reading of a measurement, the digits that are considered to be significant irrespective of the place of location of decimal point are called significant figures.

All the digits in a measurement except zero are considered significant, whereas zero is considered as significant only when it lies in the middle of the digit or at the end of number on the right-hand side of a decimal point.

Answer to Problem 3.AE

The mass of the precipitate contained in the crucible is $0.0924\text{ g}$

The number of significant figures present in the given mass is three

Explanation of Solution

Given data:

The weight of the empty crucible is $12.4372\text{ g}$

The weight of the same crucible containing a precipitate after the gravimetric analysis is $12.5296\text{ g}$

The mass of both the crucible has six significant digits.

Mass of the crucible:

The mass of the precipitate is obtained by finding the difference between two crucibles and is calculated as follows,

$\begin{array}{l}\text{ 12}\text{.5296}\\ \text{-12}\text{.4372}\\ \overline{\text{ 0}\text{.0924 g }}\end{array}$

Therefore, the mass of the precipitate is $0.0924\text{ g}$

The number of significant figures in the obtained mass of precipitate is three

Conclusion

The mass of the precipitate contained in the crucible is $0.0924\text{ g}$

The number of significant figures present in the given mass is three

(b)

Interpretation Introduction

Interpretation:

The absolute and relative uncertainty in mass of the precipitate has to be calculated and the mass with a reasonable number of digits has to be written

Concept Introduction:

Uncertainty:

Uncertainty means state of not certain in predicting a value.  In a measured value, the last digit will have associated uncertainty. Uncertainty has two values, absolute uncertainty and relative uncertainty.

Absolute uncertainty:

Expressed the marginal value associated with a measurement

Relative uncertainty:

Compares the size of absolute uncertainty with the size of its associated measurement

$\text{Relative uncertainty = }\frac{\text{absolute uncertainty}}{\text{magnitude of measurement}}$

For a set measurements having uncertainty values as $e_1,e_2\text{ and }{e}_3$, the uncertainty $(e_4)$ in addition and subtraction can be calculated as follows,

$e_4=\sqrt{e_1^2+e_2^2+e_3^2}$

Answer to Problem 3.AE

The absolute uncertainty of the mass of the precipitate is $0.0924(\pm 0.0004)\text{ g}$

The relative uncertainty of the mass of the precipitate is $0.0924(\pm 0.5\%)\text{ g}$

Explanation of Solution

Given data:

The weight of the empty crucible is $12.4372\text{ g}$

The weight of the same crucible containing a precipitate after the gravimetric analysis is $12.5296\text{ g}$

The mass of both the crucible has six significant digits.

The manufacturer states that the balance has an uncertainty of $\pm 0.3\text{ mg}$

Absolute Uncertainty in mass:

Assume,

The initial reading as $12.4372\text{ g}(\pm 0.3)\text{ mg}$

The final reading as $12.5296\text{ g}(\pm 0.3)\text{ mg}$

Convert grams to milligrams.

The mass of the precipitate is obtained from the difference between these two crucible values.

$\begin{array}{l}\text{ 12}\text{.5296 }(\pm 0.0003)\\ \text{-12}\text{.4372 }(\pm 0.0003)\\ \overline{\text{ 0}\text{.0924 }(\pm e)}\end{array}$

The uncertainty (e) is calculated as follows,

$\begin{array}{l}e=\sqrt{{(0.0003)}^2\text{ +(0}{\text{.0003)}}^{\text{2}}}\\ =\sqrt{0.00000018}\\ =0.00042\end{array}$

On writing the obtained value in significant figure, the uncertainty in mass is calculated as   $0.0004\text{ g}$

Therefore, the absolute uncertainty of the mass of the precipitate is $0.0924(\pm 0.0004)\text{ g}$

Relative uncertainty:

$\begin{array}{l}\text{Relative uncertainty = }\frac{\text{absolute uncetainty}}{\text{magnitude of measurement}}\\ \text{Percent relative uncertainty = 100}\times \text{relative uncertainty}\end{array}$

The relative uncertainty is calculated as follows,

$\begin{array}{l}\text{Relative uncertainty }\text{= }\frac{0.00042}{0.0924}\\ =0.0045\end{array}$

The percent relative uncertainty is calculated as follows,

$\begin{array}{l}\text{Percent relative uncertainty }\text{= 100}\times \text{relative uncertainty}\\ \text{=100}\times \pm \text{0}\text{.0045}\\ \text{=}\pm \text{0}\text{.5\%}\end{array}$

Therefore, the relative uncertainty of the mass of the precipitate is $0.0924(\pm 0.5\%)\text{ g}$

Conclusion

The absolute uncertainty of the mass of the precipitate is $0.0924(\pm 0.0004)\text{ g}$

The relative uncertainty of the mass of the precipitate is $0.0924(\pm 0.5\%)\text{ g}$