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Determination of Toluene By Gas Chromatographic Method

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Experiment 5 Lab Report Determination of Toluene By Gas Chromatographic Method Module Name: Fundamental Analytical Chemistry (Practical Section) Date of experiment: 16/10/2007 Introduction: Gas chromatography is a specifically gas-liquid chromatography that involves a sample being vaporized and injected onto the head of the chromatographic column. The sample is transported through the column by the flow of inert, gaseous mobile phase, which is usually an inert gas such as helium or an unreactive gas such as nitrogen. The column itself contains a liquid stationary phase, which is a microscopic layer of liquid or polymer on an inert solid support, inside glass or metal tubing that is adsorbed onto the surface of an inert solid. The instrument used to perform gas chromatographic separations is called a gas chromatograph. A gas chromatograph is a chemical analysis instrument for separating chemicals in a complex sample. A gas chromatograph uses a flow-through narrow tube known as the column, through which different chemical constituents of a sample pass in a gas stream (carrier gas, mobile phase) at different rates depending on their various chemical and physical properties and their interaction with a specific column filling, called the stationary phase. As the chemicals exit the end of the column, they are detected and identified electronically. ...read more.


Injection 2 37091520 9386726 Unknown 1 Sample 1 Injection 1 19637872 17390980 14001504 12632516 Injection 2 15144088 11263528 Unknown 1 Injection 1 14390448 13539784 13046920 12395596 Sample 2 Injection 2 12689120 11744272 Unknown 2 Injection 1 13885472 14136844 7647261 8076117 Sample 1 Injection 2 14388216 8504973 Unknown 2 Injection 1 14865376 15788384 8390445 8671136 Sample 2 Injection 2 16711392 8951827 Table 2: Peak area ratio (Toluene to Cumene) Peak Area Ratio (corr. to. 4 d.p.) Peak Area Ratio Average (corr. to. 4 d.p.) Standard 1 Injection 1 0.9020 0.8871 (1.0 ml) Injection 2 0.8722 Standard 2 Injection 1 1.9327 1.9115 (2.0 ml) Injection 2 1.8903 Standard 3 Injection 1 3.7689 3.8602 (4.0 ml) Injection 2 3.9515 Unknown 1 Injection 1 1.4026 1.3736 Sample 1 Injection 2 1.3445 Unknown 1 Injection 1 1.1030 1.0918 Sample 2 Injection 2 1.0805 Unknown 2 Injection 1 1.8157 1.7537 Sample 1 Injection 2 1.6917 Unknown 2 Injection 1 1.7717 1.8193 Sample 2 Injection 2 1.8668 Q4.) Graph 1: Q5.) From Table 2: 1. The mean of unknown 1 peak area ratio is (1.3736 + 1.0918)/2= 1.2327 Put y = 1.2327 into y = 0.0002x - 0.0349 1.2327 = 0.0002x - 0.0349 x = (1.2327+0.0349)/0.0002 x = 6338 The toluene concentration is 6338 microgram/ml in unknown 1. ...read more.


The %RSD is quite close to 0 so that the result is acceptable. Q11.) From Q5, the concentration of toluene is 6338 microgram/ml, it is assumed all toluene in 6L air which adsorbed on the activated charcoal is all dissolved in the unknown solution. Since 1 L = 1*10-3 m3, 6 L = 6*(1* 10-3) m3, so the concentration of toluene (mg/ m3) in the laboratory environment = (6338/1000) mg / (6*10-3) m3 = 1056.333 mg/m3 Q12.) From Q11, the concentration of toluene in the laboratory environment is 1056.333 mg/m3. Number of moles of toluene = (1056.333/1000) / 92.14 = 0.0114644 At NTP, all gases follow the following equation: Mole = Volume of the gas / Molar Volume of air (Molar Volume of air = 24.46 L/mol) Thus, the Volume of air toluene in 6L = Mole of toluene x Molar Volume of air = 0.0114644 x 24.46 =0.280419 dm3 =0.280419 L =0.280419 *10-3 m3 Since 1 ppm = 1 * 10-6 m3/m3, so the concentration of air toluene = (0.280419 * 10-3) / 1*10-6 =280.419 ppm Conclusions: 1. The concentration of the toluene in the unknown sample solution by using internal standard method is 31.69 mg/ml. 2. The concentration of the toluene in the unknown sample solution by using external standard method is 33.38435 mg/ml. 3. Using internal standard method is more accurate than using external standard method. 4. ...read more.

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