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Viewing:

Postlab for Percent Composition of a Hydrated Salt SD

front 1

Screen Reader Note: To calculate percent error, find the absolute value of the difference between experimental % and known percent. Take this and divide it the known %. After that, multiply by 100 %. End of note.

A student follows the experimental procedure in this lab finds the mass percent of water in a hydrate to be 24.94 %. The mass percent of this hydrate is known to be 26.64 %. What is the percent error for the student's result? Express the units as % in the units box.

back 1

Percent error = (26.64 - 24.94) / 24.94 x 100 = ? WRONG

ANSWER: 6.38%

front 2

A hydrate is known to have percentage of water of 25.67 %. The results obtained from the experiment were 18.72%, 52.73%, 32.74%. What conclusion would you most likely draw from these results?

back 2

The results are neither accurate nor precise.

front 3

Given the following data for the hydrate M(NO3)3 dot X H2O, where M is a metal with the atomic mass 64.30 g/mol,

Mass of Crucible and Lid 34.1431

Mass of Crucible, Lid and Hydrate 39.7279

Mass of Crucible, Lid and Anhydrous Salt (3rd Heating) 37.3706

What is the mass of hydrate, with correct significant figures?

back 3

39.7279 - 34.1431 =

5.5848 g

front 4

Given the following data for the hydrate M(NO3)3 dot X H2O, where M is a metal with the atomic mass 80.19 g/mol,

Mass of Crucible and Lid 34.6863

Mass of Crucible, Lid and Hydrate 39.9546

Mass of Crucible, Lid and Anhydrous Salt (3rd Heating) 37.0494

What was the mass of the anhydrous salt after the 3rd heating, with the correct significant figures?

back 4

37.0494 - 34.6863 =

2.0948 g WRONG

ANSWER: 2.3631 g

front 5

Given the following data for the hydrate M(NO3)3 dot X H2O, where M is a metal with the atomic mass 52.99 g/mol,

Mass of Crucible and Lid 34.4073

Mass of Crucible, Lid and Hydrate 39.9335

Mass of Crucible, Lid and Anhydrous Salt (3rd Heating) 37.3319

What was the mass of water lost, with the correct significant figures?

back 5

H - 1.01 x 2 = 2.02

O - 16.00 x 1 = 16

2.02 + 16 = 18.02

  • Mo - 95.950 x 1 = 95.950

N - 14.01 x 3 = 42.03
O - 16.00 x 9 = 144

  • 18.02 + 95.950 + 42.03 + 144 = 300

18.02 / 300 = 0.06 g WRONG
ANSWER: 2.6016 g

front 6

Given the following data for the hydrate M(NO3)3 dot X H2O, where M is a metal with the atomic mass 72.26 g/mol,

Mass of Crucible and Lid 34.5257

Mass of Crucible, Lid and Hydrate 39.5168

Mass of Crucible, Lid and Anhydrous Salt (3rd Heating) 37.0448

What was the mass percent of water in the hydrated salt, with the correct significant figures? Express units as % in the units box.

back 6

37.0448 - 34.5257 = 2.5191

H - 1.01 x 2 = 2.02

O - 16.00 x 1 = 16

2.02 + 16 = 18.02

18.02 / 2.5191 x 100% = 7.153% WRONG

ANSWER: 49.53%

front 7

Given the following data for the hydrate M(NO3)3 dot X H2O, where M is a metal with the atomic mass 33.01 g/mol,

Mass of Crucible and Lid 34.7334

Mass of Crucible, Lid and Hydrate 39.3280

Mass of Crucible, Lid and Anhydrous Salt (3rd Heating) 37.2814

How many molecules of water are in one formula unit of the hydrate? In other words what is the value of X? X should be a whole number, so choose the percentage of water for the hydrate that is closest to the percentage of water calculated for this experiment. HINT: X is between 6 and 18.

back 7

6 WRONG

ANSWER: 10