Mg - University of California, Berkeley

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QUIZ #5, CHEM 001 (Chapter 7) Name: _________________________ Date: ________________________ You must show all work to receive any credit; and, where applicable, report all numerical answers to the proper number of significant figures. 1. What is the molar mass of CuSO 4 * 3H 2 O? [5 pts.] 2. How many grams of magnesium are contained in 212.07 g of MgSO 4 ? [5 pts.] 3. A compound whose molar mass is 92.02 g/mol was analyzed and found to contain 30.45 % N and 69.55 % O. What is the molecular formula of the compound? [15 pts.] ANSWER KEY 213.66 grams 1 mot Cusa , ' 3h20 42.83 grams Mg 30.45/14.01 = 2.17/2.17 = I Noz empirical formula 69.55/16.00 = 4.35/2.17 = 2 molar mass 46.01g / Mol 92.02 glmol = 2 Nz 04 molecular formula 46.01 glmol

Transcript of Mg - University of California, Berkeley

Page 1: Mg - University of California, Berkeley

QUIZ #5, CHEM 001 (Chapter 7) 

Name: _________________________  Date: ________________________ 

You must show all work to receive any credit; and, where applicable, report all numerical answers to the proper number of significant figures. 

1. What is the molar mass of CuSO4 * 3H2O? [5 pts.] 

 

 

 

   

2. How many grams of magnesium are contained in 212.07 g of MgSO4? [5 pts.] 

   

 

 

 

 

3. A compound whose molar mass is 92.02 g/mol was analyzed and found to contain 30.45 % N and 69.55 % O. What is the molecular formula of the compound? [15 pts.] 

 

 

  

 

 

 

 

 

ANSWER KEY

213.66grams

1 mot Cusa ,

' 3h20

42.83 grams Mg

30.45/14.01=

2.17/2.17= I

⇒ Noz empirical formula69.55/16.00=

4.35/2.17= 2

↳ molar mass 46.01g / Mol

92.02 glmol = 2 ⇒ Nz 04 molecular formula46.01 glmol

Page 2: Mg - University of California, Berkeley

QUIZ #5, CHEM 001 (Chapter 7) 

NAME: _________________________  Date: ________________________ 

You must show all work to receive any credit; and, where applicable, report all numerical answers to the proper number of significant figures. 

1. What is the molar mass of Na2CO3 * 10H 2O? [5 pts.] 

 

 

 

   

2. How many grams of oxygen are contained in 212.07 g of CuSO4? [5 pts.] 

   

 

 

 

 

3. A compound whose molar mass is 152.08 g/mol was analyzed and found to contain 63.18 % C, 31.56% O, and 5.26 % H. What is the molecular formula of the compound? [15 pts.] 

 

 

  

 

 

 

 

 

ANSWER KEY

286.19 glmol

85.04g

63.18/12.01 =

5.26/1.97

= 2.67 x 3 = 8

31.56/16.00 = 1.97/1.97 = I x 3= 3

5.26/1.01 =

5.21/1.97= 2.64 × 3 = 7.92 → 8

empirical formula C8Hq0z molar mass = 152.16glmol

molecular formula C8HqO3

Page 3: Mg - University of California, Berkeley

QUIZ #5, CHEM 001 (Chapter 7) 

Name: _________________________  Date: ________________________ 

You must show all work to receive any credit; and, where applicable, report all numerical answers to the proper number of significant figures. 

1. What is the molar mass of CaCl 2 * 2H2O? [5 pts.] 

 

 

 

   

2. How many grams of copper are contained in 212.07 g of CuSO4? [5 pts.] 

   

 

 

 

 

3. A compound whose molar mass is 32.05 g/mole was analyzed and found to contain 87.4 % N and 12.6 % H. What is the molecular formula of the compound? [15 pts.] 

 

 

 

  

 

 

 

 

 

ANSWER KEY

147.02 glmol

84.43g

87.4/14.01 =

6.24/6.24= I

empirical formula : NHZ

12.6/1.01 =

12.48/6.24= z

molar mass : 16.02 gland

molecular : Natty

Page 4: Mg - University of California, Berkeley

QUIZ #5, CHEM 001 (Chapter 7) 

Name: _________________________  Date: ________________________ 

You must show all work to receive any credit; and, where applicable, report all numerical answers to the proper number of significant figures. 

1. What is the molar mass of CoCl 2 * 6H2O? [5 pts.] 

 

 

 

   

2. How many grams of oxygen are contained in 212.07 g of MgSO4? [5 pts.] 

   

 

 

 

 

3. A compound whose molar mass is 289.9 g/mol was analyzed and found to contain 49.67 % C, 48.92 % Cl, and 1.39 % H. What is the molecular formula of the compound? [15 pts.] 

 

 

  

 

 

 

 

 

ANSWER KEY

237.95 glmol

112.76 grams

49.67/12.01

=

4.136/1.38=3

↳ CIH

48.92/35.45

= 1.38/1.38 = I

342.0ft1135.45)

-1111.01) =

72.49289.9/72.49= 4

1.39/1.01=

1.38/1.38= I

GzCl4H4