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

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(Autocreate 2012/10/05 Entry for User:Elizabeth_Ghias/Notebook/Experimental_Chemistry)
Current revision (14:29, 8 October 2012) (view source)
 
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==Objective==
==Objective==
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To remake the CuSO<sub>4</sub>·5H<sub>2</sub>O and Ha<sub>2</sub>SO<sub>4</sub> standards
==Description==
==Description==
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# 0.0260 g CuSO<sub>4</sub>·5H<sub>2</sub>O was dissolved in 0.5 L water in a volumetric flask to create a 13.23 ppm solution of CuSO<sub>4</sub>·5H<sub>2</sub>O.
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#* Actual mass used was 0.0268 g CuSO<sub>4</sub>·5H<sub>2</sub>O
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# 0.0148 g Na<sub>2</sub>SO<sub>4</sub> was dissolved in 0.5 L water in a volumetric flask to create a 9.573 ppm solution co Na<sub>2</sub>SO<sub>4</sub>
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#* Actual mass used was 0.0153 g Na<sub>2</sub>SO<sub>4</sub>
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# The 13.23 ppm CuSO<sub>4</sub>·5H<sub>2</sub>O solution was diluted
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*See below for calculations
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==Data==
==Data==
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* Add data and results here...
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No data was collected today.
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<u>CuSO<sub>4</sub>·5H<sub>2</sub>O Calculations</u>
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20 ppm SO<sub>4</sub><sup>2-</sup> = 20 mg/L SO<sub>4</sub> · 0.5 L = 10 mg SO<sub>4</sub><sup>2-</sup> · .1 g / 1000 mg = 0.01 g SO<sub>4</sub><sup>2-</sup>
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0.01 g SO<sub>4</sub><sup>2-</sup> ÷ 96.06 g/mole = 1.041 × 10<sup>-4</sup> moles SO<sub>4</sub><sup>2-</sup> · 1 mole CuSO<sub>4</sub>·5H<sub>2</sub>O / 1 mole SO<sub>4</sub><sup>2-</sup> = 1.041 × 10<sup>-4</sup> moles CuSO<sub>4</sub>·5H<sub>2</sub>O
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1.041 × 10<sup>-4</sup> moles CuSO<sub>4</sub>·5H<sub>2</sub>O · 249.68 g/mole CuSO<sub>4</sub>·5H<sub>2</sub>O = 0.0260 g CuSO<sub>4</sub>·5H<sub>2</sub>O
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1.041 × 10<sup>-4</sup> moles CuSO<sub>4</sub>·5H<sub>2</sub>O · 1 mole Cu<sup>2+</sup> / 1 mole CuSO<sub>4</sub> = 1.041 × 10<sup>-4</sup> moles Cu<sup>2+</sup>
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1.041 × 10<sup>-4</sup> moles Cu<sup>2+</sup> · 63.546 g/mole Cu<sup>2+</sup> = 0.0066 g Cu<sup>2+</sup> · 1000 mg / 1 g
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= 6.615 g Cu<sup>2+</sup> ÷0.5 L: = 13.23 mg/L = 13.23 ppm Cu<sup>2+</sup>
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<u>Na<sub>2</sub>SO<sub>4</sub> Calculations</u>
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20 ppm SO<sub>4</sub><sup>2-</sup> = 20 mg/L SO<sub>4</sub> · 0.5 L = 10 mg SO<sub>4</sub><sup>2-</sup> · .1 g / 1000 mg = 0.01 g SO<sub>4</sub><sup>2-</sup>
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0.01 g SO<sub>4</sub><sup>2-</sup> ÷ 96.06 g/mole = 1.041 × 10<sup>-4</sup> moles SO<sub>4</sub><sup>2-</sup> · 1 mole Na<sub>2</sub>SO<sub>4</sub> / 1 mole SO<sub>4</sub><sup>2-</sup> = 1.041 × 10<sup>-4</sup> moles Na<sub>2</sub>SO<sub>4</sub>
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1.041 × 10<sup>-4</sup> moles Na<sub>2</sub>SO<sub>4</sub> · 142.04 g/mole Na<sub>2</sub>SO<sub>4</sub> = 0.0148 g Na<sub>2</sub>SO<sub>4</sub>
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1.041 × 10<sup>-4</sup> moles Na<sub>2</sub>SO<sub>4</sub> · 2 moles Na<sup>2+</sup> / 1 mole Na<sub>2</sub>SO<sub>4</sub> = 2.082 × 10<sup>-4</sup> moles Na<sup>+</sup>
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2.082 × 10<sup>-4</sup> moles Na<sup>+</sup> · 22.99 g/mole Na<sup>+</sup> = 0.00478 g Na<sup>+</sup> · 1000 mg / 1 g = 4.786 mg Na<sup>+</sup> ÷ 0.5 L = 9.573 mg/L Na<sup>+</sup> = 9.573 ppm Na<sup>+</sup>
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<u>CuSO<sub>4</sub>·5H<sub>2</sub>O Dilution Calculations</u>
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# (13.23 mg/L)X = (10 mg/L)(0.01 L) → X = 0.00756 L = 7.558 mL CuSO<sub>4</sub>·5H<sub>2</sub>O in 2.442 mL H<sub>2</sub>O
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# (13.23 mg/L)X = (7 mg/L)(0.01 L) → X = 0.005291 L = 5.291 mL CuSO<sub>4</sub>·5H<sub>2</sub>O in 4.079 mL H<sub>2</sub>O
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# (13.23 mg/L)X = (3 mg/L)(0.01 L) → X = 0.002268 L = 2.268 mL CuSO<sub>4</sub>·5H<sub>2</sub>O in 7.732 mL H<sub>2</sub>O
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# (13.23 mg/L)X = (1 mg/L)(0.01 L) → X = 0.000756 L = 0.0756 mL CuSO<sub>4</sub>·5H<sub>2</sub>O in 9.244 mL H<sub>2</sub>O
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# (13.23 mg/L)X = (0.25 mg/L)(0.01 L) → X = 0.000189 L = 0.189 mL CuSO<sub>4</sub>·5H<sub>2</sub>O in 9.811 mL H<sub>2</sub>O
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<u>Na<sub>2</sub>SO<sub>4</sub> Dilution Calculations</u>
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# (9.573 ppm)
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# (9.573 ppm)X = (7 ppm)(
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==Notes==
==Notes==

Current revision

Experimental Chemistry Main project page
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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

  1. (9.573 ppm)
  2. (9.573 ppm)X = (7 ppm)(



Notes

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