Experiment 7 Lab Report (final) PDF

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Experiment 7: Empirical Formulas Hypothesis When oxygen combines with magnesium during heating, the mole ratio will be 1:1, creating the empirical formula to be MgO. If KCL O 3

decomposes then KCl and O 2 will

separate, which will yield a 1.5:1 ratio.

Data Measurement s Mass of crucible and lid (g) Mass of crucible, lid, and sample (g) Mass of sample (g) Mass of crucible, lid, and product (g). 1st mass measurement 2nd Mass Measurement (g) 3rd Mass Measurement (g) Final Mass of crucible, lid, and product (g) Mass of product (g)

Trial 1 Part B 49.209 g

Trial 2 Part B 38.773 g

Trial 1 Part C 22.901g

Trial 2 Part C 24.696g

49.401 g

38.962 g

23.902 g

25.608 g

0.192 g

0.189 g

1.001 g

1.400 g

49.512 g

39.069 g

23.878 g

25.584 g

49.513 g

39.068 g

23.878 g

25.577 g

X

X

X

X

49.513 g

39.068 g

23.875 g

25.572 g

0.304 g

0.297 g

0.974 g

0.877 g

Part B. Combination Reaction of Magnesium and Oxygen Measurements Trial 1 Trial 2 Mass of O (g) 0.112 g 0.108 g Mass ratio of Mg to O 1.7g: 1.00g 1.75g: 1:00g Moles of Mg (mol) 0.0079 mol 0.0078 mol Moles of O (mol) 0.007 mol 0.0068 mol Mole ratio of Mg to O 1.13: 1.00 mol 1.15 :1.00 mol MgO Consensus empirical formula of magnesium oxide Percent by mass (%) 60.304% Mg and 39.696% O Part C. Decomposition Reaction of a Pure Compound Measurements Trial 1 Trial 2 Mass of product = mass KCl 0.974 g 0.877 g 0.027 g 0.035 g Mass of O 2 (g) 36.5: 1.00 g 27.8 g: 1.00 g Mass ratio of KCl to O 2 Mole of KCl 0.013 mol 0.012 mol 8.34 x 10−4 mol 1.12 x 10−3 Moles of O 2 15.7: 1.00 mol 10.5: 1 mol Mole ratio of KCl to O 2 Consensus mole ratio of KCl 13.2:1 to O 2 Percent by mass (%) 60.8 % KCl and 26.1% O 2

Results Part B: Trail One a. Mass of O = mass of product – mass of sample a. 0.304 – 0.192 = 0.112 grams b. Mass of Mg in mass ratio =

a.

0.192 0.112

mass of Mg mass of O

= 1.71g of Mg

c. Moles of Mg = (mass of Mg) x

0.112 0.112

Mass of O in mass ratio =

mass of O mass of O

= 1.00g of O ratio= 1.71g Mg: 1.00g O

1 mole Mg Molar mass of Mg

a. (0.192 Mg) x

1 mole Mg Molar mass of Mg

d. Moles of O = (mass of O) x

1 mole O Molar mass of O

1 mole O 15.999 g O

a. (0.112g O) x

= 0.00790 mol Mg

= 0.00700

Moles of Mg Moles of O

e. Moles of Mg in mole ratio =

Moles of O in mole ratio =

Moles of O Moles of O a.

0.00790 0.00700

= 1.13 Moles of Mg

0.00700 0.00700

= 1.00 Moles of O

b. 1.13 mol Mg: 1.00 mol O Part B: Trial Two a. Mass of O = mass of product – mass of sample a. 0.297 – 0.189 = 0.108 grams b. Mass of Mg in mass ratio =

a.

0.189 0.108

mass of Mg mass of O

0.108 = 1.00g 1.75g Mg: 1.00g O 0.108

= 1.75g

c. Moles of Mg = (mass of Mg) x

a. (0.189g Mg) x

a. (0.108) x

1 mole Mg Molar mass of Mg

1 mole Mg 24.305

d. Moles of O = (mass of O) x 1 mole O 15.999

Mass of O in mass ration =

= 0.00778 mol Mg

1 mole O molar mass of O = 0.00675 mol O

mass of O mass of O

e. Moles of Mg in mole ratio =

moles of Mg moles of O

a.

0.00778 0.00675

b.

1.15 moles Mg :1.00 moles O

= 1.15 moles of Mg

Moles of O in mole ratio = 0.00675 0.00675

moles of O Moles of O

= 1.00 moles of O

f. Consensus Empirical Formula = mole ratio = MgO a.

1.13 + 1.17 2

= 1.15:1

g. Calculate molar mass of MgO = molar mass of Mg + molar mass of O a. 24.305g + 15.999g = 40.304 grams Calculate percent by mass of Mg and O: Percent by mass of Mg =

molar mass of Mg x 100 Molar mass of compound

24.305 x 100 = 60.304% of Mg 40.304 Percent by mass of O =

molar mass of O x 100 molar mass of compund

15.999 x 100 = 39.696% of O 40.304 Part C: Trial 1 a. Mass of product (KCl) = final mass of crucible, lid, and product – mass od crucible and lid a. 23.8750 – 22.9007 = 0.9743 grams b. Mass of O 2 = mass of crucible, lid, and sample – final mass of crucible, lid, and product a. 23.9017 – 23.8750 = 0.0267 grams

c. Mass of KCl in mass ratio =

mass of Mg Mass of O 2

Mass of

O 2 in mass ratio =

mass of O 2 mass of O 2

a.

0.9743 0.0267

d. Mass of KCl = (mass of KCl) x

a. (0.9743) x

1 mol KCl 74.551

e. Mole of O 2 = (mass of

a. (0.0267) x

0.0267 0.0267

= 36.5g KCl

f. Moles of KCl in mole ratio =

36.5g KCl: 1.00g O 2

1 mol KCl molar mass of KCl = 0.01307 mol KCl

O2 ) x

1 mole O2 31.998

= 1.00g O 2

1 mol O2 molar mass O 2

= 8.34 x

10−4

moles of Mg moles of O 2

mol O 2

Moles of

O 2 in mole ratio =

moles of O 2 moles of O 2

a.

0.01307 0.000834

= 15.7 mol KCl

0.000834 = 1.00 mol O 2 0.000834

b. 15.7 mol KCl: 1.00 mol O 2 Part C: Trial Two a. Mass of product (KCl) = final mass of crucible, lid, and product – mass od crucible and lid a. 25.5723 – 24.6955 = 0.8768 grams

b. Mass of O 2 = mass of crucible, lid, and sample – final mass of crucible, lid, and product a. 25.6077 – 25.5723 = 0.0354 grams c. Mass of KCl in mass ratio =

mass of Mg Mass of O 2

Mass of

O 2 in mass ratio =

mass of O 2 mass of O 2 a.

0.8768 0.0354

b. 27.8g KCl :1.00g

a. (0.0354) x

1 mol KCl molar mass of KCl

1 mole KCl 74.551

e. Mole of O 2 = (mass of

O2

O2

d. Mass of KCl = (mass of KCl) x

a. (0.8768) x

0.0354 = 1.00g 0.0354

= 27.8g KCl

= 0.01176 mole KCl

O2 ) x

1 mol O2 31.998

f. Moles of KCl in mole ratio =

1 mol O2 molar mass O 2

= 1.12 x 10−3 moles of Mg moles of O 2

mol O 2

Moles of

O 2 in mole ratio =

moles of O 2 moles of O 2

a.

0.01176 0.00112

= 10.5 mol KCl

b. 10.5 mol KCl: 1.00 mol O 2 g. Consensus mole ratio KCl to O 2

0.00112 = 1.00 mol 0.00112

O2

a.

15.5 + 10.8 2

= 13.2:1

Discussion 1. The objective of this experiment was to determine the empirical formula of magnesium and oxygen through combination reactions, and determining the mole ratio of the decomposition products of potassium chlorate. For the experiment Part B we measured the mass of the fired crucible and lid first so we would know the mass of the sample after heating the crucible, lid, and sample (0.15-0.20g of magnesium). We heated the magnesium until it was a powdery ash and periodically opened the lid to ensure that the oxygen was fully reacting with the magnesium. After recording our data for the combination reaction trial two, we started experiment Part C. We recorded the mass of potassium chlorate and heated it with the lid off for 20-25 minutes so that the oxygen fully reacted. During the decomposition, the KCl and the oxygen should have separated. After recording trial two we then calculated Part B and Part C of the lab. 2. The results of our experiment proved our hypothesis that the empirical formula would be MgO when oxygen combines with magnesium during heating. We took the mass of MgO and the mass of the original sample and subtracted them, so we would know the amount of oxygen that was added. We found the mole ratio of MgO by taking the mass of magnesium and oxygen, and dividing them by their molar masses. The mole ratios from trial one and two were; 1.13:1 and 1.17:1 moles of MgO which were close to a 1:1 mole ratio. Since our experiment created a mole ratio greater than one, there was an error made during the experiment. The error could have been created by not heating the magnesium long enough in order to fully react with the oxygen, or by not opening the lid enough.

The results of Part C did not approve our hypothesis because the mole ratio of KCl O 3 was supposed to be 1.5:1, but our data shows the mole ratio was 15.7:1. The chemical equation shows that there should be 2KCl’s for every 3 moles of

O 2 . Due to our results, the sample

was not heated under an intense enough flame in order to separate the oxygen completely from KCl. We measured the sample two different times after heating it after each measurement to separate more oxygen, but the difference was less than 0.01 grams. We decided to heat it once more under a highly intense flame but after measuring it, the difference decreased by 0.027 grams. Our results were precise but inaccurate and resulted in a large experimental error. 3. The results recorded in Part B were expected because the empirical formula should have been MgO. Knowing that magnesium has a +1 charge and oxygen has a -1 charge that when they react they would create MgO. Our mole ratio was similar to 1:1 due to a slight error during the experiment, but our hypothesis was supported. Our results in Part C, however were not. The decomposition of KCL O 3 should have yielded a ratio of 1.5:1, but according to our data the ratio was 15.7:1. Our results were not expected, and our hypothesis was not supported due to the large amount of error during the experiment.

Conclusion This experiment was to determine the final empirical formula of a combination reaction and to find the mole ratio of a decomposition reaction. The procedure involved taking precise measurements of the heated crucible and lid first, in order to have a control. We then measured a strip of magnesium and heated it inside the crucible, and repeatedly lifted the lid allowing oxygen to fully react. The experiment resulted in MgO being the empirical formula and the ratio being 1.13:1. The ratio was close to being 1:1 so there was a small error during the procedure,

such as not heating the crucible long enough which would allow the oxygen to fully react with the magnesium. Our expected outcome for MgO were produced and our data supports our hypothesis. When forming the decomposition experiment, we heated the KCL O 3

in the

crucible for 20-25 minutes with the lid off ensuring the release of oxygen. We weighed the crucible twice and each time we saw a small difference in the amount of oxygen being released. We then heated the sample under a highly intense flame but when we measured it again, we still saw little difference in mass. The mole ratio should have been 1.5:1 but our results were 15.7:1 which did not prove our expected outcome. Due to the large error in the experiment, our hypothesis was not accepted.

Post Lab Questions 2. Since Javier forgot to polish the magnesium metal, the mole ratio of magnesium to oxygen would be higher than expected as a result of his error. The mole ratio will be higher because scrubbing the strip of magnesium removes any surface chemicals that it has been exposed to. Since he didn’t scrub it, the ratio will be higher because it can add mass to the sample which will make the mole ratio greater than 1:1. 4. Since Sylvia was in a hurry to finish the experiment, she did not allow all of the magnesium to react so her magnesium to oxygen ratio will be higher than expected. This is due to the fact that there will be a smaller difference in the mass that was measured before and after. The mass ratio will be smaller which will create a higher mole ratio than 1:1. 5. Since the sample is not completely decomposed then the mole ratio will be much higher than a 1.5:1 mole ratio. As seen in the results of our experiment, the sample did not completely decompose, and it resulted in an exceedingly high mole ratio.

Reference Beran, J. A. Laboratory Manual for Principles of General Chemistry; 10th ed.; Wiley: Hoboken, 2014...