CHEM 101 MODULE 2 EXAM

CHEM 101 MODULE 2 EXAM


Question 1
pts
Click this link to access the Periodic Table. This may be helpful throughout the exam.


Show the calculation of the molecular weight for the following compounds, reporting
your answer to 2 places after the decimal.


1. (NH4)2CO3


2. C8H6NO4Br
Your Answer:
1. (NH4)2CO3 =96.09grams


N = 14.01 x 2 2(14.01) + 8(1.008)+3(16.0) = 96.09grams
H = 1.008 x 8
C = 12.01 x 1
O = 16 x 3 = 96.09grams


2. C8H6NO4Br = 260.04grams
C= 12.01 x 8 8(12.01) + 6(1.008)+(14.01) + 4(16) = 260.04grams
H=1.008 x 6
N = 14.01
O = 16 x 4
Br = 79.90 Added = 260.04
1. 2N + 8H + C + 3O = 96.09


2. 8C + 6H + N + 4O + Br = 260.04


Question 2
pts


This study source was downloaded by 100000842568006 from CourseHero.com on 04-15-2022 14:06:00 GMT -05:00


https://www.coursehero.com/file/61070130/CHEM-103-Module-2-Examdocx/

, Click this link to access the Periodic Table. This may be helpful throughout the exam.


Show the calculation of the number of moles in the given amount of the following
substances. Report your answerto 3 significant figures.


1. 13.0 grams of Al2(CO3)3


2. 16.0 grams of C8H6NO4Cl
Your Answer:
1. 13.0 grams of Al2(CO3)3
Mole = Grams/ MW
13..99 = .0555 moles


2. 16.0 grams of C8H6NO4Cl
Mole = Grams/ MW
16/215.59 = .0742moles


1. Moles = grams / molecular weight = 13..99 = 0.0556 mole


2. Moles = grams / molecular weight = 16..59 = 0.0742 mole


Question 3
pts
Click this link to access the Periodic Table. This may be helpful throughout the exam.


Show the calculation of the number of grams in the given amount of the following
substances. Report your answer to 1 place after the decimal.


1. 1.05 moles of (NH4)2SO4


2. 1.18 moles of C9H9NO3
Your Answer:


This study source was downloaded by 100000842568006 from CourseHero.com on 04-15-2022 14:06:00 GMT -05:00


https://www.coursehero.com/file/61070130/CHEM-103-Module-2-Examdocx/