Calbraton Lecture Notes Calbraton Wth Standards Sgnal from Standards: N 11 1, 2.. N x 0 0.05 0.7 1.1 2.2 6.5 11.7 15.2 20.1 24.2 29.4 y 2.2152. 10 4 6.785. 10 4 0.00642 0.01057 0.02204 0.06488 0.11676 0.15202 0.20115 0.24206 0.29400 0.3 Calbraton Data Ponts Sgnal (Absorbance) y 0.2 0.1 0 0 5 10 15 20 25 30 x Concentraton
Regresson Analyss: x avg x N x avg = 10.10455 y avg y N y avg = 0.10098 x. y s xy. x y N s xy = 11.48360 y. y s yy y 2 N s yy = 0.11491 x. x s xx x 2 N s xx = 1.14761 10 3 Calculaton of lne: Slope: m s xy m = 0.01001 s xx Intercept: b y avg m. x avg b = 1.29441 10 4 y calc. m x b
Uncertanty Calculatons: In the regresson s yy m 2. s xx s r N 2 s r = 2.60153 10 4 In the slope 2 s r s m s = s m 7.67947 10 6 xx In the ntercept x 2 s b. s r N. x 2 x 2 s b = 1.10336 10 4 Analyss of an Unknown: j 1.. 5 M 5 sgnal j 0.04247 0.04251 0.04242 0.04262 0.04258 sgnal avg j sgnal j M sgnal avg = 0.04252
Calculaton of unknown unknown sgnal avg m b unknown = 4.26217 Calculaton of uncertanty n unknown s r s. unknown m 1 M 1 N sgnal avg y avg 2. m 2 s xx s unknown = 0.01472 RSD s unknown unknown RSD = 0.34541 % y 0.3 0.2 unknown Data and Calbraton y calc 0.1 0 sgnal avg 0.1 0 5 10 15 20 25 30 trace 1 calbraton x
Calbraton by Standard Addton (M&M's example): Mass of the M&M's n the bag: Mass ntal. 50 gm Next you add 5 M&M's to the bag and fnd the new mass: MM added 5 Mass fnal. 55 gm mass of the 5 M&M's that you added: Mass added Mass fnal Mass ntal Mass added = 5.00000 gm From number of M&M's added, determne mass of each M&M Mass MM Mass added MM added Mass MM = 1.00000 gm The number of M&M's ntally n the bag MM ntal Mass ntal Mass MM MM ntal = 50.00000 And now t s tme to eat!!!
CAL_LE~1.MCD And now t s tme to eat!!! 6/12/97
Standard Addton (Chemstry Problem) Analyss an sample of drt to determne the concentraton of lead Sgnal sample 0.5879 ppm 10 6 C standard. 500 ppm V standard. 0.1 ml Amount of lead added n the spke. mass.. 1. gm spke C standard V standard 1. ml mass spke = 5.00000 10 5 gm Calculate the ncrese n the concentraton of the sample. V total. 100.0 ml C spked mass spke V. 1. gm total 1. ml C spked = 5.00000 10 7 C spked = 0.50000 ppm Measure the sgnal of the spked sample Sgnal spked 0.7816
The sgnal of the spke: Sgnal spke Sgnal spked Sgnal sample Sgnal spke = 0.19370 Calculate nstrument response sgnal concentraton Response Sgnal spke C spked Response = 0.38740 ppm 1 Now, recall the sgnal of the orgnal sample: Sgnal sample = 0.58790 Ths corresponds to a sample concentraton of: C sample Sgnal sample Response C sample = 1.51755 ppm
But we assumed that the sgnal from the "spked sample" s equal to the sgnal from the sample plus the sgnal from the spke. Or that: Sgnal spked Sgnal sample Sgnal spke However, the sample s slghtly dluted by the volume of the spkng soluton (0.1 ml n ths example). So "scale" the sgnal to account for dluton: V Sgnal. total V standard sample_adjusted Sgnal sample V total Sgnal sample_adjusted = 0.58731 Repeat the above calculatons wth adjusted sgnal The sgnal of the spke: Sgnal spke_adjusted Sgnal spked Sgnal sample_adjusted Sgnal spke_adjusted = 0.19429 The response Response_adjusted Sgnal spke_adjusted C spked Response_adjusted = 0.38858 ppm 1 The adjusted response corresponds to a sample concentraton of: C sample_true C sample_true Sgnal sample Response_adjusted = 1.51296 ppm
C sample_true = 1.51296 ppm CAL_LE~1.MCD 6/12/97
Calbraton wth an Internal Standard: Run #1, standard mxture wth known concentraton of Alar and decane. Alar Concentraton C alar_std. 2.871 ppm Peak Area A. alar_std 6.824 10 4 Decane Concentraton C decane_std. 5.55 ppm Peak Area A. decane_std 7.31 10 10 Calculate Response of Detector Alar Response alar A alar_std C alar_std Response alar = 2.37687 10 4 ppm 1 Decane Response decane A decane_std C decane_std Response decane = 1.31712 10 10 ppm 1 Relatve Response Response relatve Response alar Response decane Response relatve = 1.80460 10 6
Run #2, Unknown Alar concentraton spked wth a known concentraton of decane. Alar Concentraton C alar_unk Peak Area A. alar_unk 8.367 10 4 Decane Concentraton C decane_unk. 2.46 ppm Peak Area A. decane_unk 3.23 10 12 Response to standard (decane) Response decane A decane_unk C decane_unk Response decane = 1.31301 10 12 ppm 1 Response to unknown (from relatve response) Response relatve Response alar Response decane Response alar. Response relatve Response decane Response alar = 2.36946 10 6 ppm 1
Concentraton of unknown Response alar A alar_unk C alar_unk C alar_unk A alar_unk Response alar C alar_unk = 0.03531 ppm