CHE-121 Written Assignment 3 PDF

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Written assignment 3....

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Name: College ID: Thomas Edison State University General Chemistry I with Labs (CHE-121) Section no.: 1 Semester and year: MAR2018

Written Assignment 3: Mass Relationships for Compounds Answer all assigned questions and problems, and show all work. 1. How many atoms are there in 5.10 moles of sulfur (S)? (6 points) 5.10 x 6.02x1023=3.07x1024

(Reference: Chang 3.13)

2. How many moles of calcium (Ca) atoms are there in 77.4 g of Ca? (6 points) 77.4g Ca *1 mol/ 40.1|gCa=1.93 mol Ca (Reference: Chang 3.15)

3. What is the mass in grams of 1.00 × 1012 lead (Pb) atoms? (8 points) (Reference: Chang 3.19)

1.0x1012x207.2/6.02x10^23=3.44gx10^-10

4. Describe how you would determine the isotopic abundance of an element from its mass spectrum. (6 points) (Reference: Chang 3.32)

By using a mass spectrometer, and observing a spike associated with the appropriate atomic weight. This allows us to determine the quantity, or abundance, as well. 5. Cinnamic alcohol is used mainly in perfumery, particularly in soaps and cosmetics. Its molecular formula is C9H10O. a. Calculate the percent composition by mass of C, H, and O in cinnamic alcohol. (8 points) 1 Copyright © 2017 by Thomas Edison State University. All rights reserved.

C9H10O=(9x12)+(10x1)+16=134 C=((9x12)/134)x100=80.6% H=((10x1)/134)x100=7.5% O=(16/134)x100=11.9% b. How many molecules of cinnamic alcohol are contained in a sample of mass 0.469 g? (8 points) 1 mole=134 grams (calculated above)=6.02x1023 0.469g=(.0469g x 6.02X1023) / 134= 0.02107 x 1023 c. How many carbon atoms are in 0.469 g of cinnamic alcohol? (8 points) 0.469g x 1mol/134= 0.0035 0.0035 x 9 (because there are 9 carbons in each molecule)= 0.0315 0.0315 x 6.02x10^23= 1.896 Carbon Atoms (Reference: Chang 3.41)

6. Allicin is the compound responsible for the characteristic smell of garlic. An analysis of the compound gives the following percent composition by mass: C: 44.4 percent; H: 6.21 percent; S: 39.5 percent; O: 9.86 percent. Calculate its empirical formula. What is its molecular formula given that its molar mass is about 486g? (20 points) C: 486 x 0.444= 215.784 215.784/12=17.982= 18 Carbons H: 486 x 0.0621=30.1806 30.1806/1=20.1806= 30 Hydrogens S: 486 x 0.395= 191.97 191.97/32= 5.9990625= 6 Sulfurs O: 486 x 0.0986= 47.9196 47.9196/16=2.994975= 3 Oxygens Overall that gives us...C18H30S6O3>C6H10S2O 6(12)+10(1)+2(32)+1(16)=162g/mol (Reference: Chang 3.43)

7. Tin(II) fluoride is often added to toothpaste as an ingredient to prevent tooth decay. What mass of fluoride in grams is in 258 g of toothpaste that has 15% tin(II) fluoride? (10 points) SnF2=156.69amu 258g / 156.69=1.65 2 x 1.65=3.3 3.3x18.99(amu of Fluoride)= 62.667g 62.667 x 0.15=9.4g (I believe the question was asking for the 9.4g based on the toothpaste only containing 15% of the total 258g in question, if not the answer would be 62.667g) 2 Copyright © 2017 by Thomas Edison State University. All rights reserved.

8. What are the empirical formulas of the compounds with the following compositions? (a) 2.1 percent H, 65.3 percent O, 32.6 percent S, H: 2.1/1.007=2.085 2.085/1.017=2.05=2 Hydrogens O: 65.3/15.999=4.082 4.082/1.017=4.01=4 Oxygens S: 32.6/32.066=1.0167 1.0167/1.017=0.999=1 Sulfur H2O4S (b) 20.2 percent Al, 79.8 percent Cl (20 points) Al: 20.2/26.98= 0.75 Cl: 79.8/35.45=2.25 Cl3Al

0.75/0.75=1 Aluminum 2.25/0.75=3 Chlorines

(Reference: Chang 3.49)

3 Copyright © 2017 by Thomas Edison State University. All rights reserved....