These samples arrived in your lab on Friday October 23, 2015 with the following
ID: 514773 • Letter: T
Question
These samples arrived in your lab on Friday October 23, 2015 with the following label.
Unknown #1 Unknown # 2
The sample technician took the pictures of the samples after arival. He then filed the samples to shavings. The technician divided the unknowns into 15 samples of unknown 1 and 12 of unkown 2. He then further divided each sample size in half to increase his statistical set. There was only enough of unknown #2 for this many samples so the number of samples are not equal. The technician weighed each duplicate set of samples and recorded the mass of each metal as mass1 M and mass2 M on the spreadsheet.
Each experiment was conducted according to the following balanced equation.
M(s) + 2HCl(aq) à H2(g) + MCl2(s)
After separation of the samples the Technician recorded the room temperature which would be used as the temperature for the experiment. The atmospheric temperature was recorded using a barometer in mmHg. Once all experimental parameters were recorded the technician set up the experimental aparatus. He immersed a graduated cylinder upside down full of water in to a second tub of water. Using a hose he inserted into the graduated cylinder he connected it to his reaction vessle. In the reaction vessle he inserted a weighed amount of unknown and added 10 mL of HCl. The reaction proceeded immediately by producing gas bubbles which displaced the water in the graduated cylinder as seen in the following picture.
Upon completion of the experiment, when gas ceased to be produced the technician recorded the volume of water displaced by the produced hydrogen gas. He recorded each trial as V gas1 (mL) and V gas2 (mL). The technician performed this experiment in the same manner a total of 54 times. The technician did not do any further calculations of the raw data.
His report can be viewed in the excel sheet. You are the Lab chemist, please complete the calculations and interpret what you have been given and write a summary report.
Explanation / Answer
With the given excel sheet
For first series unknown#1
P(H2) = 745 - 18.7 = 726.3 mmHg
= 726.3/760 = 0.955 atm
V(H2) = 0.0225 L (gas 1)
R = gas constant
T = 21 + 273 = 294 K
uisng,
moles of H2 = PV/RT
= 0.955 x 0.0225/0.08206 x 294
= 0.00089 mol
1 mole of metal gives 1 mole of H2
moles of metal = 0.00089 mol
molar mass of metal = mass of metal taken/moles of metal
= 0.027/0.00089
= 30.34 g/mol
The metal in question here is Magnesium (Mg)
Similarly for unknown #2 series
P(H2) = 745 - 18.7 = 726.3 mmHg
= 726.3/760 = 0.955 atm
V(H2) = 0.0265 L (gas 1)
R = gas constant
T = 21 + 273 = 294 K
using,
moles of H2 = PV/RT
= 0.955 x 0.0265/0.08206 x 294
= 0.00105 mol
1 mole of metal gives 1 mole of H2
moles of metal = 0.00105 mol
molar mass of metal = mass of metal taken 1/moles of metal
= 0.071/0.00105
= 67.68 g/mol for mass 1
For mass 2
moles of H2 = PV/RT
= 0.955 x 0.0285/0.08206 x 294
= 0.00113 mol
1 mole of metal gives 1 mole of H2
moles of metal = 0.00113 mol
molar mass of metal = mass of metal taken 1/moles of metal
= 0.085/0.00113
= 75.33 g/mol for mass 1
average mass = 71.50 g/mol
the metal in question here matches with Zinc (Zn)
Similarly other data series can be used for the calculations,
The metal in question is closest to Mg metal for unknown #1 and Zn for unknown #2
Hence,
Mg + 2HCl --> MgCl2 + H2
Zn + 2HCl --> ZnCl2 + H2
1 mole of Mg or Zn upon reaction with HCl produces 1 mole of H2 gas
From the moles of H2 gas collected we can find moles of metal Mg or Zn reacted and from mass of original metal taken one may find the molar mass of the unknown metal which turns out to be magnesium and Zinc for unknown #1 and #2 metal in this case.
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