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Experiment 7 Lab Report (final)
General Chemistry I/Lab (CHEM 1300)
Nova Southeastern University
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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 KCLO 3 decomposes then KCl and O 2 will separate, which will yield a 1:1 ratio.
Data
Measurement s Trial 1 Part B Trial 2 Part B Trial 1 Part C Trial 2 Part C Mass of crucible and lid (g)
49 g 38 g 22 24 Mass of crucible, lid, and sample (g)
49 g 38 g 23 g 25 g
Mass of sample (g) 0 g 0 g 1 g 1 g Mass of crucible, lid, and product (g). 1st mass measurement
49 g 39 g 23 g 25 g
2 nd Mass Measurement (g)
49 g 39 g 23 g 25 g 3 rd Mass Measurement (g)
X X X X
Final Mass of crucible, lid, and product (g)
49 g 39 g 23 g 25 g
Mass of product (g) 0 g 0 g 0 g 0 g
Part B. Combination Reaction of Magnesium and Oxygen Measurements Trial 1 Trial 2 Mass of O (g) 0 g 0 g Mass ratio of Mg to O 1: 1 1: 1:00g Moles of Mg (mol) 0 mol 0 mol Moles of O (mol) 0 mol 0 mol Mole ratio of Mg to O 1: 1 mol 1 :1 mol Consensus empirical formula of magnesium oxide MgO Percent by mass (%) 60% Mg and 39% O Part C. Decomposition Reaction of a Pure Compound Measurements Trial 1 Trial 2 Mass of product = mass KCl 0 g 0 g Mass of O 2 (g) 0 g 0 g Mass ratio of KCl to O 2 36: 1 g 27 g: 1 g Mole of KCl 0 mol 0 mol Moles of O 2 8 x 10 − 4 mol 1 x 10 − 3 Mole ratio of KCl to O 2 15: 1 mol 10: 1 mol Consensus mole ratio of KCl to O 2
13:
Percent by mass (%) 60 % KCl and 26% O 2
Results
Part B: Trail One a. Mass of O = mass of product – mass of sample a. 0 – 0 = 0 grams b. Mass of Mg in mass ratio = massof Mgmassof O Mass of O in mass ratio = massof Omassof O a. 0.1920 = 1 of Mg 0.1120 = 1 of O ratio= 1 Mg: 1 O c. Moles of Mg = (mass of Mg) x Molarmassof Mg 1 moleMg
b. 1:1 f. Consensus Empirical Formula = mole ratio = MgO a. 1+ 2 1 = 1: g. Calculate molar mass of MgO = molar mass of Mg + molar mass of O a. 24 + 15 = 40 grams Calculate percent by mass of Mg and O:
Percent by mass of Mg = Molarmassof compoundmolarmassof Mg x 100
- 40 100 = 60% of Mg
Percent by mass of O = molarmassof compundmolarmassofO x 100
15.99940 100 = 39% of O
Part C: Trial 1 a. Mass of product (KCl) = final mass of crucible, lid, and product – mass od crucible and lid a. 23 – 22 = 0 grams b. Mass of O 2 = mass of crucible, lid, and sample – final mass of crucible, lid, and product a. 23 – 23 = 0 grams c. Mass of KCl in mass ratio = massof MgMassof O 2 Mass of O 2 in mass ratio = massof Omassof O 22
a. 0.97430 = 36 KCl 0.02670 = 1 O 2 36 KCl: 1 O 2
d. Mass of KCl = (mass of KCl) xmolarmassof KCl 1 molKCl
a. (0) x 1 molKCl74 = 0 mol KCl
e. Mole of O 2 = (mass of O 2 ) x molarmassO 1 molO 22
a. (0) x 1 31 2 = 8 x 10 − 4 mol O 2
f. Moles of KCl in mole ratio = molesof Mgmolesof O 2 Moles of O 2 in mole ratio = molesof Omolesof O 22
a. 0.0008340 = 15 mol KCl 0.0008340 = 1 mol O 2 b. 15 mol KCl: 1 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 – 24 = 0 grams b. Mass of O 2 = mass of crucible, lid, and sample – final mass of crucible, lid, and product a. 25 – 25 = 0 grams c. Mass of KCl in mass ratio = massof MgMassof O 2 Mass of O 2 in mass ratio = massof Omassof O 22
a. 0.87680 = 27 KCl 0.03540 = 1 O 2 b. 27 KCl :1 O 2 d. Mass of KCl = (mass of KCl) xmolarmassof KCl 1 molKCl a. (0) x 1 moleKCl74 = 0 mole KCl
e. Mole of O 2 = (mass of O 2 ) x molarmassO 1 molO 22
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: and 1:1 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 KClO 3 was supposed to be 1:1, but our data shows the mole ratio was 15:1. 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 grams. We decided to heat it once more under a highly intense flame but after measuring it, the difference decreased by 0 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 KCLO 3 should have yielded a ratio of 1:1, but according to our data the ratio was 15:1. 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 1:1. 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 KCLO 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 1:1 but our results were 15:1 which did not prove our expected outcome. Due to the large error in the experiment, our hypothesis was not accepted.
Post Lab Questions
- 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.
Experiment 7 Lab Report (final)
Course: General Chemistry I/Lab (CHEM 1300)
University: Nova Southeastern University
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