Your Name Lab #1 Date lab is done
LAB #1 : REACTIONS OF COPPER AND SEPARATION OF A MIXTURE
Data and Minor Calculations
Mass of Sodium Chloride + Copper(II) Sulfate _______________g 1.031
Mass of dried Copper (II) Oxide and filter paper _______________g 1.144
Mass of filter paper _______________g 1.032
Mass of dried Copper (II) Oxide _______________g 0.112
Calculations & Questions:
CuSO4(s) +2NaOH(aq) → Cu(OH)2(s) + Na2SO4(aq) REACTION 1
Cu(OH)2(s) → CuO(s) + H2S(l) REACTION 2
1. Mass of CuSO4:
1 mol CuO
0.112 g CuO x 79.5 g
= 0.0014088 mol CuO = 0.00141 mol CuO
Since the moles of CuO equals the moles of Cu(OH)2, it’s a 1:1 ratio due to the balanced chemical
equation.
159.6 g CuSO4
0.0014088 mol x 1 mol CuSO4
= 0.22484 g = 0.225 g CuSO4
2. Mass Percent :
calculated mass from experiment
Formula: Mass Percent = initial mass of mixture
Find NaCl mass:
Mass of mixture originally 1.031 g – Mass of CuSO4 calculated 0.2248 g = 0.806 g NaCl
0.806 𝑔
a) % of NaCl 1.031 𝑔
= 0.7817 g x 100% = 78.2%
0.2248 𝑔
b) % of CuSO4 = 0.2180 g x 100% = 21.8%
1.031 𝑔
LAB #1 : REACTIONS OF COPPER AND SEPARATION OF A MIXTURE
Data and Minor Calculations
Mass of Sodium Chloride + Copper(II) Sulfate _______________g 1.031
Mass of dried Copper (II) Oxide and filter paper _______________g 1.144
Mass of filter paper _______________g 1.032
Mass of dried Copper (II) Oxide _______________g 0.112
Calculations & Questions:
CuSO4(s) +2NaOH(aq) → Cu(OH)2(s) + Na2SO4(aq) REACTION 1
Cu(OH)2(s) → CuO(s) + H2S(l) REACTION 2
1. Mass of CuSO4:
1 mol CuO
0.112 g CuO x 79.5 g
= 0.0014088 mol CuO = 0.00141 mol CuO
Since the moles of CuO equals the moles of Cu(OH)2, it’s a 1:1 ratio due to the balanced chemical
equation.
159.6 g CuSO4
0.0014088 mol x 1 mol CuSO4
= 0.22484 g = 0.225 g CuSO4
2. Mass Percent :
calculated mass from experiment
Formula: Mass Percent = initial mass of mixture
Find NaCl mass:
Mass of mixture originally 1.031 g – Mass of CuSO4 calculated 0.2248 g = 0.806 g NaCl
0.806 𝑔
a) % of NaCl 1.031 𝑔
= 0.7817 g x 100% = 78.2%
0.2248 𝑔
b) % of CuSO4 = 0.2180 g x 100% = 21.8%
1.031 𝑔