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CHM130 Lab 7 Determining the Formula of a Compound Name: A. Data Table (show completed data table here) (12 points) Measurements Mass of empty 100 mL beaker Mass of zinc metal Mass of beaker containing zinc chloride

Mass (in grams) 55.45 g 1.80 g 58.16 g

B. Follow-Up Questions (Show all work for calculations.) 1. Calculate the mass of zinc chloride produced. You know the original mass of the beaker. You also know the mass of the beaker plus the product you produced, which is zinc chloride. The difference between these two masses is the mass of zinc chlorine produced. Subtract the original mass of the beaker from the mass of the beaker plus the zinc chloride produced. Show your work. (20 points) Formula: (mass beaker + zinc chloride) - mass of beaker = mass of zinc chloride produced 58.16 g – 55.45 g = 2.71 g mass of zinc chloride produced. 2. Calculate mass of chlorine reacted. To do this, you know that the zinc and the hydrochloric acid combined to create zinc chloride. You know the mass of zinc used, and the mass of zinc chloride produced; so, the difference (subtract) between these values is the mass of chlorine reacted. (20 points) Formula: Mass of zinc chloride produced - mass of zinc used = mass of chloride reacted. 2.71 g – 1.80 g = 0.91 g mass of chloride reacted 3. Calculate moles of zinc based on original mass of zinc. Show your calculations. (20 points) Formula:

Mass Zn used x 1 mol Zn = moles of Zn used Molar mass of Zn 1.80 Zn used x 1 mol Zn = 0.027527 moles of Zn used 65.39 g of Zn Answer: 0.0275 moles of Zn used (3 significant figures)

4. Calculate moles of chlorine reacted based on your answer for #2. Show your work. (20 points) Formula: Mass Cl used x 1 mol Cl = moles of Cl used Molar mass of Cl 0.91 g Cl used x 1 mol Cl = 0.02566 moles of Cl used 35.45 g of Cl Answer= 0.0257 moles of Cl used (3 significant figures) 5. What is the experimental empirical formula of zinc chloride? Determine the simplest whole number ratio of moles of zinc used, to moles of chlorine reacted in the zinc chloride that was produced. (Take your answers to #3 and #4, and divide both values by whatever value is smaller.) (20 points) Zn = 0.0257mol = 1 0.0257mol Cl = 0.0275 mol = 1.07 0.0257 mol Ratio: 1:1

6. Determine the actual empirical formula of zinc chloride, based on the ionic charges of zinc and chloride ions, using your rules for ionic nomenclature. (20 points) Since zinc has 2 positive ions and chloride has 1 negative ion, the actual empirical formula of zin chloride is ZnCl2 with a molar mass of 136.286 g/mol. Molar mass: 65.38 g of Zn + 35.453(2) g of Cl =136.286 7. Does your experimental empirical formula (answer to #5) agree

with the known empirical formula (answer to #6) of zinc chloride? List and describe some sources of error that may have affected your results. (20 points) 8. Is zinc chloride an ionic compound or a molecular compound? Explain. (8 points) Zinc chloride is an ionic compound. I concluded this because Zn is a metal and Cl is a nonmetal, therefore, there is an ionic compound between the two. Also, Zn has a positive charge and Cl has a negative charge. 9. List at least three reasons why this lab could not be completed in the home lab environment. (8 points) Three reasons why this lab couldn’t be completed at home are as follows: 1) Lack of equipment, such as the scale. 2) The compound being created in this experiment can be hazardous. 3) Lack of materials, such as Zinc. 10. In the reaction between phosphoric acid and ammonia, a compound forms containing 28.2 grams nitrogen, 8.1 grams hydrogen, 20.8 grams phosphorus, and 42.9 grams oxygen by mass. a. What is the empirical formula for this compound? (20 points) N= 28.2g / 14.007g/mol = 2.013 moles of nitrogen H= 8.1 g / 1.008g/mol = 8.035 moles of hydrogen P= 20.8g / 30.974g/mol = 0.6715 moles of Phosphorus O= 42.9g / 15.999g/mol = 2.6814 moles of oxygen N= 2mol / 1mol = 2 moles of Nitrogen H= 8mol / 1mol = 8 moles of Hydrogen P= 1mol / 1mol = 1 mole of Phosphorus O= 3mol / 1mol = 3 moles of Oxygen N2H8PO3 b. If the molecular mass for the compound is 149 g/mol, what is the molecular formula for this compound? (12 points) Nitrogen = 14.007g x 2mol = 28.014g/mol Hydrogen = 1.008g x 8mol = 8.064g/mol Phosphorus = 30.974g x 1mol = 30.974g/mol Oxygen = 15.999g x 3mol = 47.997g/mol

Empirical Mass = 115.05g/mol 149g/mol / 115.05g/mol = 1.2950 = 1 Molecular Formula = N2H8PO3...