Chapter 27, Problem 27.30QAP

Interpretation Introduction

(a)

Interpretation:

Retention time for cinnamaldehyde should be determined using Figure b.

Concept introduction:

The retention time of a substance can be determined using the chromatogram. The distance between the starting position (0) and the peak of the interested substance. From the x-axis of the chromatogram, retention time of the interested substance can be determined.

Answer to Problem 27.30QAP

Retention time = 19 min

Explanation of Solution

Figure b peak is obtained on the time of 19 mins, as shown using blue arrow. Therefore, retention time for cinnamaldehyde is 19 mins.

Interpretation Introduction

(b)

Interpretation:

Number of theoretical plates should be calculated using Figure b.

Concept introduction:

$\begin{array}{l}N=16{\left(\frac{t_R}{W}\right)}^2\\ where:\\ N=Numberofplates\\ t_R=\mathrm{Re}tentiontime\\ W=Widthofpeakbase\end{array}$

Answer to Problem 27.30QAP

N= 36,100

Explanation of Solution

Peak width, (W) should be determined first, in order to calculate the number of theoretical plates. Peak width can be determined using Figure b, indicated in black arrow.

$W=19.2-18.8=0.4$

$\begin{array}{l}N=16{\left(\frac{t_R}{W}\right)}^2\\ N=16{\left(\frac{19}{0.4}\right)}^2=36,100\\ \end{array}$

Interpretation Introduction

(c)

Interpretation:

Plate height should be calculated using the information in part a and b.

Concept introduction:

Plate height can be calculated using the following equation.

$\begin{array}{l}H=\frac{L}{N}\\ where:\\ H=Plateheight\\ L=Lengthofcolumnpacking\\ N=Numberofplates\end{array}$

Answer to Problem 27.30QAP

Plate height = 0.83 mm

Explanation of Solution

Given information:

L= 30 cm

$\begin{array}{l}H=\frac{L}{N}\\ H=\frac{30cm}{36100}=8.31\times {10}^{-4}cm=0.83mm\end{array}$

Interpretation Introduction

(d)

Interpretation:

Calibration curves should be plotted for cinnamaldehyde, eugenol, and thymol. The R² values for each compound also should be determined.

Concept introduction:

Standard addition method is an analytical method used in quantitative analysis of unknown samples. Calibration curves can be plotted, concentration vs relative peak area. That plot can be used to determine the unknown sample concentration.

Answer to Problem 27.30QAP

$\begin{array}{l}{R}_{Cinnamaldehyde}^2=0.9964\\ R_{Eugenol}^2=0.9885\\ R_{Thymol}^2=0.9952\end{array}$

Explanation of Solution

Given information:

Interpretation Introduction

(e)

Interpretation:

Sensitivity levels of the calibration curves should be determined, as highest and lowest.

Concept introduction:

Linear regression R² can be used to determine the linearity between two variables, with regard to this question the sensitivity of the calibration curve can be determined by the R² value.

Answer to Problem 27.30QAP

Highest calibration curve sensitivity= Cinnamaldehyde

Lowest calibration curve sensitivity= Eugenol

Explanation of Solution

Higher the linear regression, higher the sensitivity of calibration curve. Linear regression R² can be used to determine the linearity between two variables. Statistically, higher linear regression represents higher linearity/fitting of two variables to a linear equation.

Therefore, in this case cinnamaldehyde has the highest sensitivity, while eugenol has the lowest sensitivity.

Interpretation Introduction

(f)

Interpretation:

Concentrations of each component in the sample should be calculated. Then the standard deviation also should be calculated.

Concept introduction:

The equation obtained from the calibration curve in part d can be used in the calculation of concentrations of unknown samples.

$\begin{array}{l}{y}_{Cinnamaldehyde}=1.7077x-0.2074\\ y_{Eugenol}=1.0906x-0.0741\\ y_{Thymol}=2.2094x+0.2925\\ \\ S\tan darddevation(s)=\sqrt{\frac{{\displaystyle \sum _{i=1}^N{\left(x_i-x\right)}^2}}{N-1}}\\ Where:\\ x_i=concentrationof\mathrm{int}erestedsubs\tan ce\\ x=Meanconcentrationof\mathrm{int}erestedsubs\tan ce\\ N=Numberofsamples\end{array}$

Answer to Problem 27.30QAP

$\begin{array}{l}{x}_{Cinnamaldehyde}=1.64mg/200\mu l\\ x_{Eugenol}=0.893mg/200\mu l\\ x_{{}_{Thymol}}=1.58mg/200\mu l\end{array}$

Standard deviations of:

Cinnamaldehyde=1.130

Eugenol=0.564

Thymol= 1.523

Explanation of Solution

Given information:

Information provided in the question that is used to solve this particular category

Relative peak areas:

Cinnamaldehyde=2.6

Eugenol=0.9

Thymol=3.8

Detailed explanation/work out of the complete problem.

$\begin{array}{l}{y}_{Cinnamaldehyde}=1.7077x-0.2074\\ 2.6=1.7077x-0.2074\\ 2.6+0.2074=1.7077x\\ x_{Cinnamaldehyde}=1.64mg/200\mu l\\ \\ y_{Eugenol}=1.0906x-0.0741\\ 0.9=1.0906x-0.0741\\ 0.9+0.0741=1.0906x\\ x_{Eugenol}=0.893mg/200\mu l\\ \\ y_{Thymol}=2.2094x+0.2925\\ 3.8-0.2925=2.2094x\\ x_{{}_{Thymol}}=1.58mg/200\mu l\end{array}$

Standard deviations were calculated using Excel sheets.

Interpretation Introduction

(g)

Interpretation:

The statistical effect should be calculated for the decomposition of cinnamaldehyde with the temperature.

Concept introduction:

Analysis of variance (ANOVA) test can be done in order to determine whether there is a statistical effect on temperature for the decomposition of cinnamaldehyde.

Answer to Problem 27.30QAP

There is a statistical effect on the decomposition of cinnamaldehyde.

Explanation of Solution

Given information:

Detailed explanation/work out of the complete problem.

The test was carried out using Excel.

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Column 1	7	930	132.8571	4623.81
Column 2	7	490.7	70.1	82.59
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	13784.61	1	13784.61	5.857814	0.0323	4.747225
Within Groups	28238.4	12	2353.2
Total	42023	13

Anova: Two-Factor Without Replication
SUMMARY	Count	Sum	Average	Variance
20	7	490.7	70.1	82.59
40	7	463.6	66.22857	131.359
60	7	470.1	67.15714	149.1929
40	3	263.8	87.93333	0.063333
60	3	204.4	68.13333	21.40333
100	3	186.8	62.26667	18.58333
140	3	175.2	58.4	29.89
180	3	177.6	59.2	36.91
200	3	190.9	63.63333	0.573333
210	3	225.7	75.23333	4.083333
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Rows	57.19143	2	28.59571	2.06938	0.168992	3.885294
Columns	2013.03	6	335.5049	24.27942	4.75E-06	2.99612
Error	165.8219	12	13.81849
Total	2236.043	20

Interpretation Introduction

(h)

Interpretation:

The test should be carried out to the hypothesis that, there is no effect of temperature or time on the decomposition of the sample.

Concept introduction:

Statistical test ANOVA should be carried out in order to check the hypothesis.

Answer to Problem 27.30QAP

Check the explanation part.

Explanation of Solution

Anova: Two-Factor Without Replication
SUMMARY	Count	Sum	Average	Variance
20	6	403	67.16667	26.83067
40	6	375.4	62.56667	44.99067
60	6	382.2	63.7	78.636
60	3	204.4	68.13333	21.40333
100	3	186.8	62.26667	18.58333
140	3	175.2	58.4	29.89
180	3	177.6	59.2	36.91
200	3	190.9	63.63333	0.573333
210	3	225.7	75.23333	4.083333
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Rows	68.92444	2	34.46222	2.238356	0.157272	4.102821
Columns	598.3244	5	119.6649	7.772354	0.003183	3.325835
Error	153.9622	10	15.39622
Total	821.2111	17