Top Posters
Since Sunday
a
5
k
5
c
5
B
5
l
5
C
4
s
4
a
4
t
4
i
4
r
4
r
4
A free membership is required to access uploaded content. Login or Register.

CHY203 - UV-Visible Spectrophotometry lab report

Ryerson University
Uploaded: 6 years ago
Contributor: cloveb
Category: Chemical Engineering
Type: Report
Rating: N/A
Helpful
Unhelpful
Filename:   CHY 203 -Instrumental Method of Analysis-UV-Visible Spectrophotometry Lab Report.docx (1017.77 kB)
Page Count: 1
Credit Cost: 2
Views: 227
Downloads: 2
Last Download: 6 years ago
Description
CHY 203 Instrumental Methods of Analysis
Transcript
Department of Chemical Engineering UV-Visible Spectrophotometry Student name: Student number: CHY203, Section 02 Group number: 2 April 3rd, 2013 Instructor: David Naranjit TA: Saif Data & Result Sheets for UV-Visible Spectrophotometry Calculations: Part 1: Note: the absorbance value of solution A was ignored due to the difficulty in reading the value of the absorbance from graph A in part B. Table1. Showing the absorbance obtained at 625nm for each solution at different volume of 5g mL-1 fluoride added Solutions Volume of 5g mL-1 fluoride Standard added (mL) Absorbance at 625nm RB 0 ---- A 0 ---- B 1 0.1613 C 2 0.2096 D 3 0.2432 E 4 0.2750 Part 2: The concentration of fluoride in the sample unknown #8 (g /mL) was calculated using graph1 as shown below: y = 0.0375x + 0.1286 The slope m= 0.0375 Let x be the volume of fluoride added Let y-intercept b= 0.1286 be the absorbance The volume of fluoride added to the solution sample can be calculated by using solution A since it has absorbance of 0. 0= 0.0375x + 0.1286 x = - 0.1286/0.0375 = -3.4293 mL of the fluoride in the sample cx = -cs Vs,0 / Vx = b cs/ mVx Where cx is the concentration of the unknown sample Vs or Vs,0 is the volume of fluoride in the unknown #8 sample Cx = -cs Vs,0 / Vx = - (5) * (-3.4293)/ (5) = 3.43 g mL-1 To Check: Cx = b cs/ mVx = (0.1286)* (5) / (0.0375)*(5) = 3.43 g mL-1 Table2. Showing the Amount of Concentration of unknown #8 in g mL-1 Added for each Solution Solutions Concentration of standards solutions (unknown #8) added Volume 5g mL-1 F- Solutions A 5 0 B 5 1 C 5 2 D 5 3 E 5 4 Graph1. The absorbance of each solution at 625nm against the volume of standard Fluoride added in mL-1 Discussion Questions: 1. Explain the differences in the spectra obtained in section A.1. Spectrum in UV-visible spectrometers can effect by a number of parameters such as scan speed. In section A.1; however, spectrum of holmium oxide had been scanned under two different scan speed conditions one was under scan speed of (30 nm min-1) and the other was under scan speed of (1000nm min-1 ) in order to investigate the effect of the scan speed parameter on the absorbance of spectrum. By comparing the peaks value of the two scanned speed , the absorbance of the highest peaks obtained in the graph of holm1 of low speed of 30nm/min was about 0.115, while the absorbance for holm2 of high speed of 1000nm/min for the second highest peak was 0.114 relatively close to the value of the highest peaks of holm1 which indicate that the spectrum in the holm2 had a less time when it analyzed due to the value of scan speed used comparing to the holm1 that took longer time thus holm1 had a better analyzed of absorbance. The longer the time the analysis sample can takes, the more accuracy the result can be obtained. Consequently, by looking and examining the spectrum in two graphs of holmium oxide sample, one can conclude that the scan speed of 30nm/min had a better absorbance and resolution since the graph appears smoother comparing to that of 1000nm/min which in return can contribute to a better result of spectrum analysis. 2. Explain the trend obtained in section A.2. Why is a wide slit more often used for quantitative analysis and a narrow slit for qualitative analysis? The trend obtained in section A.2, was using a different slit width value in order to determine and observe the effect of slit width parameter on the absorbance of the analysed sample. By examining the graph in part A.2 of holm 3 and 4 slit width study, the slit of 0.2 had a higher absorbance value of about 0.115 and thus shaper peaks than those of 0.4 slit and 0.5 slit. Therefore, the narrow the slit width, the higher absorbance obtained and thus a better result of spectrum analysis. However, the narrow slit used for qualitative analysis because it allows only a small portion of the UV- visible light to pass through the sample which result in higher absorbance, sharp peaks and thus a good analysis and identify of the sample. Whereas, a wide slit is often used for quantitative analysis since the slit size allows the more dispersive light to pass through the sample being analyzed which cause more brightener and hence a low absorbance value and less sharp peaks. Graphs for Analysis Spectra of Solutions Part A: Parameter Adjustment Using a Holmium Oxide Standard 1. Effect of Scan Speed Holmium Oxide1-Absorbance at 30 nm min-1 scan speed study Holm 2-Absorbance at 1000 nm min-1 scan speed study 2. Effect of Slit Width Holmium Oxide-Absorbance at 4.0 nm , 2.0 nm and 5.0 nm Slit width Study Part B: Fluoride Analysis Absorbance Analysis of Different Solutions with different volume of F- solution added References: Instrumental Methods of Analysis Laboratory Manual, UV-Visible Spectrophotometry, Ryerson University 2013, page 45-51

Related Downloads
Explore
Post your homework questions and get free online help from our incredible volunteers
  1160 People Browsing
Your Opinion
Do you believe in global warming?
Votes: 370