Question

For Trials 2 and 3, the percent CH_3COOH in vinegar was 5.01% and 4.66% respectively. What is the average percent of CH_3COOH in the vinegar sample? Data Analysis, B. What are the standard deviation and the relative standard deviation (%RSD) for the percent of CH_3COOH in the vinegar sample? Data Analysis, C and D.

Answer 1

Concepts and reason

The given problem is based on the concept of average percent and relative standard deviation (RSD) of a data.

Average of a data is equal to the sum of the data values divided by the total number of values in the data.

Relative standard deviation is a measure of precision with respect to the average value. It is expressed in percentage.

Fundamentals

The formula to calculate the average $\left( {{\rm{\bar x}}} \right)$ is:

$\bar x = \frac{{{x_1} + {x_2} + {x_3}.....{x_n}}}{n}$

Here,

$n$ is the total number of value.

The formula to calculate standard deviation $\left( \sigma \right)$ is:

$\sigma = \sqrt {\frac{1}{{n - 1}}\Sigma {{\left( {{x_{\rm{i}}} - \bar x} \right)}^2}}$

The formula to calculate relative standard deviation $\left( {{\rm{\% RSD}}} \right)$ is:

${\rm{\% RSD}} = \left( {\frac{\sigma }{{\bar x}}} \right)100{\rm{ \% }}$

(5.a)

The formula to calculate the average percent is:

$\bar x = \frac{{{x_1} + {x_2}}}{2}$ …… (1)

Substitute $5.01{\rm{ \% }}$ for ${x_1}$ and ${\rm{4}}{\rm{.66 \% }}$ for ${x_2}$ in equation (1).

$\begin{array}{c}\\\bar x = \frac{{\left( {5.01{\rm{ \% }}} \right) + \left( {{\rm{4}}{\rm{.66 \% }}} \right)}}{2}\\\\ = 4.84{\rm{ \% }}\\\end{array}$

(5.b)

The formula to calculate standard deviation $\left( \sigma \right)$ is:

$\sigma = \sqrt {\frac{{{{\left( {{x_{\rm{1}}} - \bar x} \right)}^2} + {{\left( {{x_{\rm{2}}} - \bar x} \right)}^2}}}{{n - 1}}}$ …… (2)

Substitute $5.01{\rm{ \% }}$ for ${x_1}$ , ${\rm{4}}{\rm{.66 \% }}$ for ${x_2}$ , $4.84{\rm{ \% }}$ for $\bar x$ and 2 for $n$ in equation (2).

$\begin{array}{c}\\\sigma = \sqrt {\frac{{{{\left( {5.01{\rm{ \% }} - 4.84{\rm{ \% }}} \right)}^2} + {{\left( {{\rm{4}}{\rm{.66 \% }} - 4.84{\rm{ \% }}} \right)}^2}}}{{2 - 1}}} \\\\ = \sqrt {\frac{{\left( {0.0289{\rm{ \% }}} \right) + \left( {0.0324{\rm{ \% }}} \right)}}{1}} \\\\ = 0.248{\rm{ \% }}\\\end{array}$

The formula to calculate relative standard deviation $\left( {{\rm{\% RSD}}} \right)$ is:

${\rm{\% RSD}} = \left( {\frac{\sigma }{{\bar x}}} \right)100{\rm{ \% }}$ …… (3)

Substitute $0.248{\rm{ \% }}$ for $\sigma$ and $4.84{\rm{ \% }}$ for $\bar x$ in equation (3).

$\begin{array}{c}\\{\rm{\% RSD}} = \left( {\frac{{0.248{\rm{ \% }}}}{{4.84{\rm{ \% }}}}} \right)100{\rm{ \% }}\\\\ = {\rm{5}}{\rm{.12 \% }}\\\end{array}$

Ans: Part 5.a

The average percent of acetic acid $\left( {{\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{COOH}}} \right)$ in the vinegar sample is ${\bf{4}}.{\bf{84 \% }}$ .