Quiz 5 Name: _______________________________________ Chemistry 102, Lecture 401 – Fall 2016 DS Section: (1 point) ____________________________ 1. What is the empirical and molecular formula of a hydrocarbon that contains 93.7% carbon by mass (the remaining mass is hydrogen) and has a molar mass of about 128.2 g∙mol–1? (3 points)  93.7 g C   mol C 
128.2 g total  

  10 mol C
 100 g total   12.01 g C 
 6.3 g H   mol H 

  8 mol H
 100 g total   1.01 g H 
128.2 g total  
molecular formula = C10 H8 empirical formula  C5 H 4
2. In lab this week, you precipitated a number of salts. Using the balanced equation and given information, complete the table. (6 points) Work: Mg(NO3)2(aq) + 2NaOH(aq)  Mg(OH)2(s) + 2NaNO3(aq) molar mass (Mg(NO3)2) 148.33 g∙mol–1 molar mass (NaOH) 40.00 g∙mol–1 molar mass (Mg(OH)2) 58.33 g∙mol–1 mass of Mg(NO3)2 10.0 g mass of NaOH 6.00 g limiting reactant Mg(NO3)2 theoretical yield (of solid) 3.93 g percent yield 80.2 % actual yield 3.15 g 

10.0 g Mg  NO    148.33 g Mg  NO 


mol Mg NO3
  2 mol NaOH   40.00 g NaOH 
 
 
  5.39 g Na 3 PO 4
mol NaOH 
3 2   1 mol Mg  NO 3  2  
Because 5.39 g NaOH  need  < 6.00 g NaOH  have  , Mg  NO3 2 is the limiting reactant.

mol Mg  NO3 2
2
3 2
10.0 g Mg  NO    148.33 g Mg  NO 

3 2
3 2
  1 mol Mg  OH 2
 
  1 mol Mg  NO3 2
 58.33 g Mg  OH 2

 mol Mg  OH 

2

  3.93 g Ag 3 PO 4

 3.15 g Mg  OH 2 exp 

  100%  80.2%
 3.93 g Mg  OH 2 theo 
3. Give an anion and the correct formula for a silver salt that would result in the mixtures shown below. (2 points) Anion: halides, SO42–, CO32–, PO43–, CrO42–, OH–, S2– Formula of salt (as an example): AgCl Anion: NO3–, HCO3–, C2H3O2– Formula of salt (as an example): AgNO3