University:
Arizona State University
Course:
CHM 101 | Introductory Chemistry
Academic year:

2022
Views:

388

Pages:

2
Author:

kundgetanufv

Laboratory Skills Part 1 1. Calculate and record the following data in the table below. Mixture Label: A. Mass of the empty evaporating dish (g) 50.000 g B. Mass of the evaporating dish plus sample (g) 60 g 2. Calculate and record the following data in the table below. Mixture Label: A. Mass of the sample (g) 10 g B. Mass of the contents after heating (g) 59.662 g C. Mass of NH4Cl in the sample (g) 9.662 g Part 2 1. Calculate and record the following data in the table below. Mixture Label: A. Mass of beaker #1 (g) 32 g B.Mass of the beaker and solid after decanting (g) 71.662 g C. Mass of beaker #2 (g) 2. Calculate the mass of the SiO2 (sand) in the mixture. 6.645 g 38.645 g Part 3 1. What did you observe during the evaporation technique? 2. Calculate and record the following quantities. Mixture Label: A.Mass of beaker with NaCl after boiling (g) 35.018 g B.Mass of NaCl obtained (g) 3.018 g 3. The amount of sodium chloride in the original mixture can be determined indirectly by subtracting the masses of ammonium chloride and sand from the initial mass of the powdered mixture. Calculate the percent difference between the results from the two methods using the following equation: NH4Cl ((10 g- 9.662 g)/(9.662 g)) *100 = 3.498 % Sand ((10 g - 6.645g)/(6.645 g)) *100 = 50.489 % Conclusions 1. How is the difference in the physical properties of the powdered mixture components being used to separate them in this lab? I believe its due to solubility of the powdered mixture. 2. Suppose a student is given 5.475 g of a powdered mixture to separate. Because the mixture contains iron shavings, he decides to separate them using a magnet. He obtains 1.824 g of iron. What is the mass percent of the iron in the mixture? %m/m = m of a / m of sample * 100 = 1.824 g/ 5.475 g *100 = 33.315 %