Chem lab

Data Collection and Processing
Aspect One
-2.00grams of Sr(NO3)2 used
-2.00grams of CuSO4
-Reactants: Sr(NO3)2 and CuSO4

Trial Mass of Beaker (g) Mass of Beaker with Sr(NO3)2 (g) Mass of Beaker (g) Mass of Beaker and CuSO4 (g) Mass of Filter Paper (g) Mass of Filter Paper and Contents (g)
1 111.08±0.01 113.08±0.01 111.1±0.01 113.1±0.01 1.28±0.01 2.93±0.01
2 111.1±0.01 113.1±0.01 111.23±0.01 113.23±0.01 1.27±0.01 2.98±0.01
3 111.26±0.01 113.26±0.01 111.09±0.01 113.09±0.01 1.27±0.01 2.95±0.01

Qualitative Data:
 Sr(NO3)2 (s) : white, baking soda like powder but rougher
 Sr(NO3)2 (s) and water: completely dissolved, clear solution
 CuSO4(s): blue, powdery
 CuSO4(s) and water: solution turned blue
 When the two solutions were mixed together a cloudy, light blue solution formed with white precipitate
 Light blue solution was dripping out of the funnel
 Contents on filter paper was a thin chalky powder

Aspect Two Sr(NO3)2 + CuSO4*5H2O -----> SrSO4 + CU(NO3)2+5H2O
Given: 2.00g 2.00g g=?
A.M: 1 mol 1 mol 1mol

Step 1
Sample Calculation for Mass of Precipitate in Trial 1
Mass of precipitate= mass of filter paper and contents – mass of filter paper
= 2.93g-1.28g
=1.65g

Uncertainties = absolute error of mass before + absolute error of mass after = 0.01g + 0.01g = ± 0.02g

Therefore the amount of precipitate formed in trial one was 1.65g± 0.02g.
Trial Mass of Precipitate Formed (g)
1 1.65g± 0.02g
2 1.71g± 0.02g
3 1.68g± 0.02g

Step 2
Average of precipitate formed
(mass of trial 1+mass of trial 2+mass of trial 3)/3
=(1.65g+1.71g+1.68g)/3
=1.68g

Error Average: (error for trial 1 + error for trial 2 +error for trial 3)/3 =(0.02g+0.02g+0.02g)/2 = ± 0.02g

Conversion to % error: (error/value) x100% =(0.02g/1.68g) x100% =1.2%