Lab 7 Empirical Formula Lab Report-3 PDF

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Lab 7 Empirical Formula Name: Tashfia Hasan Lab Partner: Tina Takla, Dev Gohel, Som Keshav, Makaela King, Leslie Course: E01, Tuesdays 6 PM - 8:45 PM Instructor Name: Obiajulu Nwanze Laboratory Assistant Name: Camila Gonzalez Date Experiment was Performed: September 15, 2020

Data: Combination Reaction of Magnesium and Oxygen Question

Trial 1

1

Mass of Crucible and lid

50.174

2

Mass of Mg used

0.02

3

Mass of crucible, lid, and Mg (g )

50.194

5

Crucible, lid, and MgO

50.198

Combination Reaction (Question 8 Continued) Question 8

Parts

Trial 1

a

Mass of O (g)

0.004

b

Mass ratio of Mg and O

0.02 g to 0.004

c

Moles of Mg (mol)

0.0008

d

Moles of O (mol)

0.0002

e

Mole ratio of Mg to O

0.0008 mol Mg to 0.0002 mol O

8f. Consensus Empirical Formula Of Magnesium Oxide: MgO 8g. Percent by Mass: 43.16% Mg, 56.83% O Decomposition Reaction Question 9 (Replace

Trial 1

Trial 2

Mass of Crucible and lid

50.166

50.201

Mass of KCl used

1.043

1.041

Mass of crucible, lid,

51.209

51.242

50.513

50.550

CaO with KCl)

and KCl (g)

Crucible, lid, KCl, O2

Decomposition Reaction Question 9 Continued Question 9

Parts

Trial 1

Trial 2

a

Mass of product = mass product KCl (g)

1.043

1.041

b

Mass of O2 (g)

0.696

0.657

c

Mass ratio of KCl to O2

0.347 to 0.696

0.384 to 0.657

d

Mole of KCl

0.00465

0.00515

e

Mole of O2

0.0218

0.0205

f

Mole ratio of KCl to

0.00465 to 0.0218

0.00515 to 0.0205

O2 9g. Consensus Empirical Formula Of Potassium Chlorite: KClO2 9h. Percent by Mass: 69.9 % KCl 30.0% O2 Calculations: Synthesis/Combustion Reaction Equation: 2Mg + O2 ->2MgO Mass of Sample (Mg): Mass of crucible, lid, and Mg (g ) - Mass of Crucible and lid = 50.194 g - 50.174 g = 0.02 g Mass of Compound (MgO): Mass of Crucible, lid, and MgO - Mass of Crucible and lid = 50.198 g - 50.174 g = 0.024 g Mass of O2 (g): Mass of Compound - Mass of Mg = 0.024 g - 0.02 g = 0.004 g Moles of Mg(mol): 0.02 g * (1 mol/ 24.30 g Mg) = 0.0008 mol Mg Moles of O2(mol): 0.004 g * (1 mol/32 g O2) = 0.0001mol O2 Empirical Formula: Percent Mass of Mg: 43.16% -> 43.16 g Mg/24.30 g Mg = 1.776 mol Mg Percent Mass of O2: 56.83% -> 56.83 g/32.0 g O2 = 1.776 mol O2 1.776 mol Mg/1.776 = 1 mol Mg 1.776 mol O/1.776 = 1 mol O Percent By Mass: Molar Mass of MgO: 24.30 g Mg (1 mol) + 32 g O2 (1 mol) = 56.30 g MgO Percent Mass of Mg: (24.30 g/56.30 g) * 100 = 43.16% Mg Percent Mass of O2: (32 g/56.30 g) * 100 = 56.83% O2 Decomposition Reaction Formula: 2KClO3 -> 2KCl + 3O2 Mass of Product: Trial 1: Mass of crucible and lid - Mass of crucible, lid, and KCl (g) = 51.209 g - 50.166 g = 1.043 g KClO3 Trial 2: Mass of crucible and lid - Mass of crucible, lid, and KCl (g) (mass of compound)= 51.242 g - 50.201 g = 1.0410 g KClO3

Mass of KCl: Trial 1: Mass of crucible, lid, KCl and O2 - Mass of crucible and lid = 50.513 g - 50.166 g = 0.347 g mol KCl Trial 2: Mass of crucible, lid, KCl and O2 - Mass of crucible and lid = 50.550 g - 50.166 g = 0.384 g mol KCl Mass of O2: Trial 1: Mass of compound - Mass of KCl = 1.043 g - 0.347 g = .696 g O2 Trial 2: Mass of compound - Mass of KCl = 1.0410 g - 0.384 g = .657 g 02 Mole of KCl: Trial 1: 0.347 *(1 mole/74.5513 g KCl) = 0.00465 mol KCl Trial 2: 0.384*(1 mole/74.5513 g KCl) = 0.00515 mol KCl Mole of O2: Trial 1: 0.696 * (1 mole/32 g O2) = 0.0218 mol O2 Trial 2: 0.657*(1 mole/32 g O2) = 0.0205 mol O2 Empirical Formula: Percent Mass of KCl: 69.9% KCl -> 69.9 g O2/74.5513 g KCl = 0.938 mol KCl Percent Mass of O2: 30.0% O2 -> 30.0 g O2/32 g O2 = 0.938 mol O2 0.938 mol KCl/0.938 = 1 mol KCl 0.938 mol O2/0.938 = 1 mol O2 Percent By Mass: Molar Mass of KClO3: 74.5513 g/mol KCl + 32.0 g/mol O2 = 106.6 g/mol KClO3 Percent Mass of O2: (32.0 g O2/106.6 g/mol KClO3) * 100 = 30.0% O2 Percent Mass of KCl: (74.5513 g KCl/106.6 g/mol KClO3) * 100 = 69.9% KCl Results: Based on the data and calculations of this experiment there is an observation that was apparent. When a substance is heated up in a combination reaction, the mass of the element increases when it comes in contact with heat. For instance, in Trial 1 of the combination reaction with Mg and O, the mass of the including the crucible, lid, and Mg is 50.194 g. But after the

crucible has been heated the mass of the crucible, lid and compound increases to 50.198 g. However, in a decomposition reaction the opposite effect occurs where the mass of the substance including the crucible and lid is less than it was prior to heating. Heating the compound causes bonds in the compound to break and release energy and ultimately mass as a result. This observation is demonstrated during the decomposition reaction where the mass of the crucible, lid and KCl decreases after the crucible has been heated. Specifically, the mass decreases from 51.209 g to 50.513 g in Trial 1. Brief Conclusion: In this experiment, the objective was to determine the empirical formula of a compound based on the moles and grams of the respective elements within the compound. In creating a combination reaction of magnesium (Mg) and oxygen(O), magnesium oxide was formed. To find the empirical formula of this compound, conversion factors were formed using the molar masses of the elements and forming conversion factors based on the ratios and the mass of the compound and specific element measured during the lab. Similarly, the same process was used to find the empirical formula for a decomposition reaction. As discussed in the results section, increased heat applied to the crucible led to an increase in mass in the combination reaction but had the opposite effect on a decomposition reaction where mass was lost during the reaction. Post Lab Questions: 3) When magnesium is burning and becomes uncontrolled, the moles of oxygen will be too high because more oxygen is being added to the magnesium causing an increase in the moles of oxygen since the mass of O2 in the reaction will be larger after heating excessively.

5) During a decomposition reaction the compound is broken down so if it isn’t fully decomposed then the mole ratio of CaO to CO2 will be too high because there will be much more CaO measured than there actually is in the compound. 6) If the compound is contaminated then the mole ratio of CaO to CO2 will be too high because the decomposition reaction would thermally decompose the CO2 but will have a difficult time decomposing the contaminated substance. References Beran, J. A. (2014). Pages 115-122 Laboratory Manual for Principles of General Chemistry Tenth Edition...