Lab 1 - Answers to the Lab PDF

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Summary

The first analytical chemistry lab is tricky, I found it helpful to go over other's calculations and work to understand the theory ...

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1) Volume of H2O Delivered by Pipet (6 sig figs): T °=25 ℃ ± 0.5 ℃

ρH2O = 0.9972995g/ml Samples

Mass of Water (3 values selected by TA) 10.0096g 10.0041g 10.0055g 10.0064g

1 2 3 Average

10.0096 g+10.0041 g+10.0055 g X´ = =10.0064 g 3

Volume of pipet =

m of water 10.0064 g = =10.0335 ml ρ of water 0.9972995 g/ml

2) Density of unknown solution (4 sig figs):

Samples 1 2 3 Average

Mass of Unknown Densities Solution 10.8800g 1.0844g/ml 10.8732g 1.0837g/ml 10.8889g 1.0853g/ml --1.0857g/ml

ρUnk =

m Solution V solution

ρUnk =

10.8800 g =1.0844 g/ml 10.0335ml

1

2 ρUnk =

10.8732 g =1.0837 g/ml 10.0335ml

ρUnk =

10.8889 g =1.0853 g /ml 10.0335ml

3

1.0844 g /ml +1.0837 g /ml+1.0853 g/ml X´ = =1.0857 g 3

√

SD=

(1.0844−1.0857)2 +(1.0837−1.0857 )2 +( 1.0853−1.0857 )2 =0.008 g /ml 3−1

Deviation from Average: g 1) 1.0857−1.0844 =1.3E-3 ml g 2) 1.0857−1.0837 =2.0E-3 ml g 3) 1.0857−1.0853 =0.4E-3

ml

Average Of Deviation: 1.3E-3 + 2.0E-3 + 0.4E-3 =1.2E-3 g/ml 3

Precision in PPT: 1.2E-3 g /ml ∗1000=1.1053 1.0857 g /ml...