Question

Find the standard deviation of %H2O and relative standard deviation of %H2O in hydrated salt (%RSD) using the following data. Show work.

Trial 1 Trial 2 Trial 3

Mass of crucible and lid (g): 19.437 20.687 18.431

Mass of crucible, lid and

hydrated salt (g): 21.626 25.111 22.167

Mass of crucible, lid and

anhydrous salt (g): 21.441 24.702 21.762

Trial 1 Trial 2 Trial 3

Mass of Hydrated Salt (g): 2.189g 4.424g 4.186g

Mass of Anhydrous Salt (g): 2.004g 4.105g 3.331g

Mass of Water lost (g): 0.185g 0.409g 0.855g

Percent by mass of volatile

water in hydrated salt (%): 8.45% 9.25% 20.43%

Average percent H2O in hydrated salt (%H2O): 12.71%

Answer 1

Concepts and reason

The basic concept in the problem is the standard deviation of a data. It is a concise method of determining mean deviation from actual values of entries of a data.

Fundamentals

The standard deviation $\left( \sigma \right)$ of a data is

$\sigma = \sqrt {\frac{1}{{n - 1}}{{\sum {\left( {{x_i} - x} \right)} }^2}}$

Relative standard deviation $\left( {{\sigma _r}} \right)$ is

${\sigma _r} = \frac{\sigma }{{{x_i}}} \times 100$

Where n is the total number of entries, x is the average of the entries and ${x_i}$ is an entry.

Average value of the percent of water $\left( {{w_{av}}} \right)$ in hydrated salt is

${w_{av}} = \frac{{{w_1} + {w_2} + {w_3}}}{3}$

Substitute 8.45% for ${w_1}$ 9.25% for ${w_2}$ 20.43% for ${w_3}$

$\begin{array}{c}\\{w_{av}} = \frac{{8.45\% + 9.25\% \; + 20.43\% }}{3}\\\\ = \frac{{38.13\% }}{3}\\\\ = 12.71\% \\\end{array}$

Standard Deviation of value of percent of water in hydrated salt is

$\begin{array}{c}\\\sigma = \sqrt {\frac{1}{{n - 1}}{{\sum {\left( {{w_{av}} - w} \right)} }^2}} \\\\ = \sqrt {\frac{1}{{n - 1}}\left[ {{{\left( {{w_{av}} - {w_1}} \right)}^2} + {{\left( {{w_{av}} - {w_2}} \right)}^2} + {{\left( {{w_{av}} - {w_3}} \right)}^2}} \right]} \\\end{array}$

Substitute 12.71% for ${w_{av}}$ ,8.45% for ${w_1}$ , 9.25% for ${w_2}$ , 20.43% for ${w_3}$ and 3 for n

$\begin{array}{c}\\\sigma = \sqrt {\frac{1}{{3 - 1}}\left[ {{{\left( {12.71\% - 8.45\% } \right)}^2} + {{\left( {12.71\% - 9.25\% } \right)}^2} + {{\left( {12.71\% - 20.43\% } \right)}^2}} \right]} \\\\ = \sqrt {\frac{1}{2}\left[ {{{\left( {4.26\% } \right)}^2} + {{\left( {3.46\% } \right)}^2} + {{\left( {7.72\% } \right)}^2}} \right]} \\\\ = \sqrt {\frac{1}{2}\left[ {18.14\% + 11.97\% + 59.59\% } \right]} \\\\ = \sqrt {\frac{1}{2}\left[ {89.7\% } \right]} \\\end{array}$

Now

$\begin{array}{c}\\\sigma = \sqrt {\frac{1}{2}\left( {89.7} \right)\% } \\\\ = \sqrt {44.85\% } \\\\ = 6.69\% \\\end{array}$

Relative standard deviation $\left( {{\sigma _r}} \right)$ is

${\sigma _r} = \frac{\sigma }{{{w_{av}}}} \times 100$

Substitute 6.69% for $\sigma$ and 12.71% for ${w_{av}}$

$\begin{array}{c}\\{\sigma _r} = \frac{{6.69\% }}{{12.71\% }} \times 100\\\\ = 52.63\% \\\end{array}$

Ans:

Standard Deviation of value of percent of water in hydrated salt is 6.69%