State of Matter Sleuth
EXPERIMENT #4: STATE OF MATTER SLEUTH
Calculations
To find the molecular weight of the unknown salt: The unknown salt that was assigned is unknown salt in vial #2, which was weighed to 0.12 g mass. Initial Volume of NaOH | Final Volume of NaOH | Volume of NaOH needed | 46.6 mL | 34.8 mL | 46.6-34.8=11.8 mL
46.6-34.8=0.0118 L |
1) Number of moles of NaOH used=0.0118 L of NaOH0.1 M of NaOH
1) Number of moles of NaOH used=0.00118 mol 2) The number of moles of NaOH used is equal to the number of moles of H+ions present in the solution, which is equal to the number of moles of the unknown salt.
Moles of NaOH=Moles of H+=Moles of Unknown Salt=0.00118 mol 3) Molecular weight is equal to grams per mole. The mass of the salt was measured before the titration was carried out.
Molecular weight of the salt=mass in grams of saltnumber of moles of salt
Molecular weight of the salt=0.12 g0.00118 mol
Molecular wei hsalt=102 g mol-1
The molecular weight was compared to all possible salts. Possible Salts | Molecular Weight (g/mol) | NaCl | 58.44 | NaNO3 | 85.00 | NaI | 149.89 | KBr | 119.00 | KCl | 74.55 | KI | 166.00 | NaBr | 102.89 |
Table 1: List of possible salts and their respective molecular weights
Discussion and Conclusion
Part 1: Identifying the ionic compound
By visual inspection, the compound in bottle A was pure element, and the compound in bottle D was heterogeneous mixture because the mixture does not mix well and the proportions of substances were uneven visually.
To determine the ionic compound from compound B and compound C, a conductivity test and solubility test were carried out. Both compounds were mixed with water. Both compounds were soluble in water. Therefore, the type of compounds could not yet be determined. Both solutions were then tested with a conductivity meter. The larger value of conductance of solution of compound C significantly assured that compound C is the ionic compound whereas compound
Calculations
To find the molecular weight of the unknown salt: The unknown salt that was assigned is unknown salt in vial #2, which was weighed to 0.12 g mass. Initial Volume of NaOH | Final Volume of NaOH | Volume of NaOH needed | 46.6 mL | 34.8 mL | 46.6-34.8=11.8 mL
46.6-34.8=0.0118 L |
1) Number of moles of NaOH used=0.0118 L of NaOH0.1 M of NaOH
1) Number of moles of NaOH used=0.00118 mol 2) The number of moles of NaOH used is equal to the number of moles of H+ions present in the solution, which is equal to the number of moles of the unknown salt.
Moles of NaOH=Moles of H+=Moles of Unknown Salt=0.00118 mol 3) Molecular weight is equal to grams per mole. The mass of the salt was measured before the titration was carried out.
Molecular weight of the salt=mass in grams of saltnumber of moles of salt
Molecular weight of the salt=0.12 g0.00118 mol
Molecular wei hsalt=102 g mol-1
The molecular weight was compared to all possible salts. Possible Salts | Molecular Weight (g/mol) | NaCl | 58.44 | NaNO3 | 85.00 | NaI | 149.89 | KBr | 119.00 | KCl | 74.55 | KI | 166.00 | NaBr | 102.89 |
Table 1: List of possible salts and their respective molecular weights
Discussion and Conclusion
Part 1: Identifying the ionic compound
By visual inspection, the compound in bottle A was pure element, and the compound in bottle D was heterogeneous mixture because the mixture does not mix well and the proportions of substances were uneven visually.
To determine the ionic compound from compound B and compound C, a conductivity test and solubility test were carried out. Both compounds were mixed with water. Both compounds were soluble in water. Therefore, the type of compounds could not yet be determined. Both solutions were then tested with a conductivity meter. The larger value of conductance of solution of compound C significantly assured that compound C is the ionic compound whereas compound