E5 C60 chlorate 2022-1
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Experiment 5
Name Albert Hunanyan _ Oxygen in Potassium Chlorate Report Sheet Section Number______________ Table I. Data and *Calculations *For calculations, show setups with numerical values and units. Mass of “container” –
test tube, beaker, and catalyst
32.721 g Mass of “container”
and KClO
3
33. 745 g Mass of “container” and contents after the first heating
33. 314 g Mass of “container” and contents after the second heating 33.314 g Mass of “container” and contents after the third heating if needed 33.314 g *Calculate the mass of KClO
3
used. 33. 745 - 32.721 = 1.024 g *Calculate the mass of oxygen lost by heating (the total weight lost). 33.745 - 33.314 = 0.931 g *Calculate the experimental percentage of oxygen in KClO
3 % ?????? = ?𝑎?? ????
?𝑎?? ?? 𝐾𝐶?𝑂
3
? 100
(0.931/ 1.024) x 100 = 42.09 % *Calculate the theoretical percentage of oxygen in KClO
3
(from the chemical formula). (39.17 –
42.1)/ 39.17) x 100 = 2.43 % *Calculate the % error 2.43/ 39.14 x 100 = 6.2%
Post Laboratory Questions Name: Albert Hunanyan 1. Write equations similar to the decomposition of KClO
3, for the thermal decompositions of two chlorate compounds to form oxygen and chloride salts. a.
LiClO
3
→
LiCl + 3O2 b.
Ca(ClO
3
)
2 →
CaCl2 + 3O2 c.
Al(ClO
3
)
3 →
2AlCl3 + 9O2 2. Calculate the percentage of oxygen in Al(ClO
3
)
3 from the chemical formula. 1 mole Al(ClO3)3 contains 9 mol oxygen Mass of 9 mol oxygen = 144 g % of Oxygen = (144/277) x 100 = 0.519 x 100 = 52.0% 3. A sample of an unknown metal chlorate, weighing 1.725 g, is heated until all of the oxygen is driven off. The residue remaining in the container weighs 0.859 g. Calculate the percentage of oxygen in this metal chlorate. Weight of Oxygen= total weight –
residue weight = 1.725 g –
0.859 g = 0.866 g Weight percentage= (weight of oxygen / total weight of sample) x 100 = (0.866 g/ 1.725 g) x 100 = 0.502 x 100 = 50.2 % 4. A student records the following data in a laboratory experiment to determine the percentage of oxygen in Ca(ClO
3
)
2
. Mass of container 58.957 g Mass of container and Ca(ClO
3
)
2
60.734 g Mass of container and contents after: First heating 60.221 g Second heating 59.910 g Third heating 59.899 g Determine: a.
The experimental percentage of oxygen in Ca(ClO
3
)
2
. Mass of Ca(ClO
3
)
2 = 60.734 –
58.95 = 1.777 g Mass of Oxygen in Ca(ClO
3
)
2 after heating = 60.734 –
59.899 = 0.835 g Experiment % of Oxygen in of Ca(ClO
3
)
2 = (0.835/ 1.777) x 100 = 0. 469 x 100
= 47.0 % b. The theoretical percentage of oxygen in Ca(ClO
3
)
2. = (96/ 207) x 100 = 0.463 x 100 = 46.4% b.
The percentage error. = (46.4 –
47.0)/ 46.4) x 100 = (0.6/ 46.4) x 100 = 0.0129 x 100 = 1.29%
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