HW#10 Chapter 7 Problem 7.55 - Clinical Application ResourcesConstantsPeriodic T

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HW#10 Chapter 7

Problem 7.55 - Clinical Application

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Problem 7.55 - Clinical Application

Calculate the molar mass for each of the following:

Part A

Citalopram, an antidepressant, C20H21FN2O

Express your answer to four significant figures and include the appropriate units.

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Part B

calcium gluconate, a calcium supplement, CaC12H22O14

Express your answer to four significant figures and include the appropriate units.

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Part C

rubbing alcohol, C3H8O

Express your answer to four significant figures and include the appropriate units.

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Part D

Levofloxacin, used to treat a range of bacterial infections , C18H20FN3O4

Express your answer to four significant figures and include the appropriate units.

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Problem 7.55 - Clinical Application

Calculate the molar mass for each of the following:

Part A

Citalopram, an antidepressant, C20H21FN2O

Express your answer to four significant figures and include the appropriate units.

Molar mass of C20H21FN2O =

SubmitMy AnswersGive Up

Part B

calcium gluconate, a calcium supplement, CaC12H22O14

Express your answer to four significant figures and include the appropriate units.

Molar mass of CaC12H22O14 =

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Part C

rubbing alcohol, C3H8O

Express your answer to four significant figures and include the appropriate units.

Molar mass of C3H8O =

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Part D

Levofloxacin, used to treat a range of bacterial infections , C18H20FN3O4

Express your answer to four significant figures and include the appropriate units.

Molar mass of C18H20FN3O4 =

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Explanation / Answer

A)

Molar mass of C20H21FN2O,

MM = 20*MM(C) + 21*MM(H) + 1*MM(F) + 2*MM(N) + 1*MM(O)

= 20*12.01 + 21*1.008 + 1*19.0 + 2*14.01 + 1*16.0

= 324.4 g/mol

Answer: 324.4 g/mol

B)

Molar mass of CaC12H22O14,

MM = 1*MM(Ca) + 12*MM(C) + 22*MM(H) + 14*MM(O)

= 1*40.08 + 12*12.01 + 22*1.008 + 14*16.0

= 430.4 g/mol

Answer: 430.4 g/mol

C)

Molar mass of C3H8O,

MM = 3*MM(C) + 8*MM(H) + 1*MM(O)

= 3*12.01 + 8*1.008 + 1*16.0

= 60.09 g/mol

Answer: 60.09 g/mol

D)

Molar mass of C18H20FN3O4,

MM = 18*MM(C) + 20*MM(H) + 1*MM(F) + 3*MM(N) + 4*MM(O)

= 18*12.01 + 20*1.008 + 1*19.0 + 3*14.01 + 4*16.0

= 361.4 g/mol

Answer: 361.4 g/mol

Feel free to comment below if you have any doubts or if this answer do not work

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