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%
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%