# User:Elizabeth Ghias/Notebook/Experimental Chemistry/2012/10/05

(Difference between revisions)
 Revision as of 12:33, 8 October 2012 (view source)← Previous diff Revision as of 12:44, 8 October 2012 (view source)Next diff → Line 17: Line 17: # 0.0260 g CuSO4·5H2O was dissolved in 0.5 L water in a volumetric flask to create a 13.23 ppm solution of CuSO4·5H2O. # 0.0260 g CuSO4·5H2O was dissolved in 0.5 L water in a volumetric flask to create a 13.23 ppm solution of CuSO4·5H2O. #* Actual mass used was 0.0268 g CuSO4·5H2O #* Actual mass used was 0.0268 g CuSO4·5H2O - # + # 0.0148 g Na2SO4 was dissolved in 0.5 L water in a volumetric flask to create a 9.573 ppm solution co Na2SO4 + #* Actual mass used was 0.0153 g Na2SO4 + # The 13.23 ppm CuSO4·5H2O solution was diluted Line 29: Line 31: - CuSO4 Calculations + CuSO4·5H2O Calculations 20 ppm SO42- = 20 mg/L SO4 · 0.5 L = 10 mg SO42- · .1 g / 1000 mg = 0.01 g SO42- 20 ppm SO42- = 20 mg/L SO4 · 0.5 L = 10 mg SO42- · .1 g / 1000 mg = 0.01 g SO42- Line 54: Line 56: 2.082 × 10-4 moles Na+ · 22.99 g/mole Na+ = 0.00478 g Na+ · 1000 mg / 1 g = 4.786 mg Na+ ÷ 0.5 L = 9.573 mg/L Na+ = 9.573 ppm Na+ 2.082 × 10-4 moles Na+ · 22.99 g/mole Na+ = 0.00478 g Na+ · 1000 mg / 1 g = 4.786 mg Na+ ÷ 0.5 L = 9.573 mg/L Na+ = 9.573 ppm Na+ + + + CuSO4·5H2O Dilution Calculations + + # (13.23 mg/L)X = (10 mg/L)(0.01 L) → X = 0.00756 L = 7.558 mL CuSO4·5H2O in 2.442 mL H2O + # (13.23 mg/L)X = (7 mg/L)(0.01 L) → X = 0.005291 L = 5.291 mL CuSO4·5H2O in 4.079 mL H2O + # (13.23 mg/L)X = (3 mg/L)(0.01 L) → X = 0.002268 L = 2.268 mL CuSO4·5H2O in 7.732 mL H2O + # (13.23 mg/L)X = (1 mg/L)(0.01 L) → X = 0.000756 L = 0.0756 mL CuSO4·5H2O in 9.244 mL H2O + # (13.23 mg/L)X = (0.25 mg/L)(0.01 L) → X = 0.000189 L = 0.189 mL CuSO4·5H2O in 9.811 mL H2O + + + Na2SO4 Dilution Calculations + + #

## Revision as of 12:44, 8 October 2012

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## Objective

To remake the CuSO4·5H2O and Ha2SO4 standards

## Description

1. 0.0260 g CuSO4·5H2O was dissolved in 0.5 L water in a volumetric flask to create a 13.23 ppm solution of CuSO4·5H2O.
• Actual mass used was 0.0268 g CuSO4·5H2O
2. 0.0148 g Na2SO4 was dissolved in 0.5 L water in a volumetric flask to create a 9.573 ppm solution co Na2SO4
• Actual mass used was 0.0153 g Na2SO4
3. The 13.23 ppm CuSO4·5H2O solution was diluted

• See below for calculations

## Data

No data was collected today.

CuSO4·5H2O Calculations

20 ppm SO42- = 20 mg/L SO4 · 0.5 L = 10 mg SO42- · .1 g / 1000 mg = 0.01 g SO42-

0.01 g SO42- ÷ 96.06 g/mole = 1.041 × 10-4 moles SO42- · 1 mole CuSO4·5H2O / 1 mole SO42- = 1.041 × 10-4 moles CuSO4·5H2O

1.041 × 10-4 moles CuSO4·5H2O · 249.68 g/mole CuSO4·5H2O = 0.0260 g CuSO4·5H2O

1.041 × 10-4 moles CuSO4·5H2O · 1 mole Cu2+ / 1 mole CuSO4 = 1.041 × 10-4 moles Cu2+

1.041 × 10-4 moles Cu2+ · 63.546 g/mole Cu2+ = 0.0066 g Cu2+ · 1000 mg / 1 g = 6.615 g Cu2+ ÷0.5 L: = 13.23 mg/L = 13.23 ppm Cu2+

Na2SO4 Calculations

20 ppm SO42- = 20 mg/L SO4 · 0.5 L = 10 mg SO42- · .1 g / 1000 mg = 0.01 g SO42-

0.01 g SO42- ÷ 96.06 g/mole = 1.041 × 10-4 moles SO42- · 1 mole Na2SO4 / 1 mole SO42- = 1.041 × 10-4 moles Na2SO4

1.041 × 10-4 moles Na2SO4 · 142.04 g/mole Na2SO4 = 0.0148 g Na2SO4

1.041 × 10-4 moles Na2SO4 · 2 moles Na2+ / 1 mole Na2SO4 = 2.082 × 10-4 moles Na+

2.082 × 10-4 moles Na+ · 22.99 g/mole Na+ = 0.00478 g Na+ · 1000 mg / 1 g = 4.786 mg Na+ ÷ 0.5 L = 9.573 mg/L Na+ = 9.573 ppm Na+

CuSO4·5H2O Dilution Calculations

1. (13.23 mg/L)X = (10 mg/L)(0.01 L) → X = 0.00756 L = 7.558 mL CuSO4·5H2O in 2.442 mL H2O
2. (13.23 mg/L)X = (7 mg/L)(0.01 L) → X = 0.005291 L = 5.291 mL CuSO4·5H2O in 4.079 mL H2O
3. (13.23 mg/L)X = (3 mg/L)(0.01 L) → X = 0.002268 L = 2.268 mL CuSO4·5H2O in 7.732 mL H2O
4. (13.23 mg/L)X = (1 mg/L)(0.01 L) → X = 0.000756 L = 0.0756 mL CuSO4·5H2O in 9.244 mL H2O
5. (13.23 mg/L)X = (0.25 mg/L)(0.01 L) → X = 0.000189 L = 0.189 mL CuSO4·5H2O in 9.811 mL H2O

Na2SO4 Dilution Calculations

## Notes

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