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Postlab for Percent Composition of a Hydrated Salt SD ATTEMPT 2

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 29.53 %. The mass percent of this hydrate is known to be 24.64 %. What is the percent error for the student's result? Express the units as % in the units box.

back 1

Percent error = (29.53 - 24.64) / 24.64 x 100 =

19.85 %

front 2

A hydrate is known to have percentage of water of 18.67 %. The results obtained from the experiment were 25.72%, 25.73%, 25.74%. What conclusion would you most likely draw from these results?

back 2

The results are precise but not accurate.

front 3

Given the following data for the hydrate M(NO3)3 dot X H2O, where M is a metal with the atomic mass 77.44 g/mol,

Mass of Crucible and Lid 34.2064

Mass of Crucible, Lid and Hydrate 39.2665

Mass of Crucible, Lid and Anhydrous Salt (3rd Heating) 37.2103

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

back 3

39.2665 - 34.2064 = 5.0601 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 87.70 g/mol,

Mass of Crucible and Lid 34.5525

Mass of Crucible, Lid and Hydrate 39.8826

Mass of Crucible, Lid and Anhydrous Salt (3rd Heating) 37.2183

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

back 4

37.2183 - 34.5525 = 2.6658 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 76.06 g/mol,

Mass of Crucible and Lid 34.5333

Mass of Crucible, Lid and Hydrate 39.0518

Mass of Crucible, Lid and Anhydrous Salt (3rd Heating) 37.0842

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

back 5

mass of hydrate:

39.0518 - 34.5333 = 4.5185 g

mass of the anhydrous salt after the 3rd heating:

37.0842 - 34.5333 = 2.5509 g

mass of water lost:

4.5185 g - 2.5509 g = 1.9676 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 80.20 g/mol,

Mass of Crucible and Lid 34.1507

Mass of Crucible, Lid and Hydrate 39.6121

Mass of Crucible, Lid and Anhydrous Salt (3rd Heating) 37.1157

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

mass of hydrate:

39.6121 - 34.1507 = 5.4614 g

mass of the anhydrous salt after the 3rd heating:

37.1157 - 34.1507 = 2.9650 g

mass of water lost:

5.4614 g - 2.9650 g = 2.4964 g

mass percent of water in the hydrated salt:

(2.4964 g / 5.4614 g) x 100 = 45.71 %

front 7

Given the following data for the hydrate M(NO3)3 dot X H2O, where M is a metal with the atomic mass 28.65 g/mol,

Mass of Crucible and Lid 34.3417

Mass of Crucible, Lid and Hydrate 39.3530

Mass of Crucible, Lid and Anhydrous Salt (3rd Heating) 37.0140

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.

mass of hydrate:

39.3530 - 34.3417 = 5.0113 g

mass of the anhydrous salt after the 3rd heating:

5.0113 g - 1.3390 g = 3.6723 g

mass of water lost:

5.0113 g - 3.6723 g = 1.3390 g

mass percent of water in the hydrated salt:

(1.3390 g / 5.0113 g) x 100 = 26.72 %

back 7

Answer: 10

IGNORE THE REST

mass of water:

39.3530 - 37.0140 = 1.3390 g

molar mass of water:

H - 1.01 x 2 = 2.02

O - 16.00 x 1 = 16.00

2.02 + 16.00 = 18.02

moles of water:

1.3390 g / 18.02 g/mol = 0.0743 mol

molar mass of anhydrous salt:

N - 14.01 x 3 = 42.03

O - 16.00 x 9 = 144

28.65 g/mol + 42.03 + 144 = 214.68 g/mol

mass of hydrate:

39.3530 - 34.3417 = 5.0113 g

mass of the anhydrous salt:

5.0113 g - 1.3390 g = 3.6723 g

moles of anhydrous salt:

3.6723 g / 214.68 g/mol = 0.01711 mol

anhydrous salt : water -> 0.01711 mol : 0.0743 mol

^Divide by smallest number:

0.01711 mol / 0.01711 mol = 1 mol

0.0743 mol / 0.01711 mol = 4.34 -> 4 mol

anhydrous salt : water -> 1 : 4