An Introduction to Physical Science
An Introduction to Physical Science
14th Edition
ISBN: 9781305079137
Author: James Shipman, Jerry D. Wilson, Charles A. Higgins, Omar Torres
Publisher: Cengage Learning
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Question
Chapter 12, Problem 7E

(a)

To determine

The percentage by mass of each element in salt, NaCl .

(a)

Expert Solution
Check Mark

Answer to Problem 7E

The percentage by mass of sodium (Na) is 39.3 % and the percentage by mass of chlorine (Cl) is 60.7 % in salt, NaCl .

Explanation of Solution

The percentage by mass of an element in a compound is defined as the percentage of total mass of that element present in the total mass or formula mass of the compound.  Mass percent of any element B in compound AB is given as,

% B component = (total mass of component Bformula mass of compound AB)100 %

Atomic mass of sodium (Na) is 23.0 u and atomic mass of chlorine (Cl) is 35.5 u .

The formula mass of NaCl is calculated as,

Formula mass of NaCl=atomic mass of Na+atomic mass of Cl

Substitute 23.0 u for atomic mass of Na and 35.5 u for atomic mass of Cl in the above equation.

Formula mass of NaCl=23.0 u+35.5 u=58.5 u

The percentage by mass of sodium (Na) is given as,

% Na = (total mass of Naformula mass of NaCl)100 % (1)

Substitute 23.0 u for total mass of Na and 58.5 u for formula mass of NaCl in equation (1).

% Na=(23.0 u58.5 u)100%= 39.3 %

The percentage by mass of chlorine (Cl) is given as,

% Cl = (total mass of Clformula mass of NaCl)100 % (2)

Substitute 35.5 u for total mass of Cl and 58.5 u for formula mass of NaCl in equation (2).

% Cl=(35.5 u58.5 u)100%60.7 % 

Conclusion:

Therefore, the percentage by mass of sodium (Na) is 39.3 % and percentage by mass of chlorine (Cl) is 60.7 % in salt, NaCl .

(b)

To determine

The percentage by mass of each element in sucrose, C12H22O11 .

(b)

Expert Solution
Check Mark

Answer to Problem 7E

The percentage by mass of carbon (C) is 42.1 % , the percentage by mass of hydrogen (H) is 6.4% and percentage by mass of oxygen (O) is 51.5 % in sucrose, C12H22O11 .

Explanation of Solution

The percentage by mass of an element in a compound is defined as the percentage of total mass of that element present in the total mass or formula mass of the compound.  Mass percent of any element B in compound AB is given as,

% B component = (total mass of component Bformula mass of compound AB)100 %

Atomic mass of carbon (C) is 12.0 u .

Atomic mass of hydrogen (H) is 1.0 u .

Atomic mass of oxygen (O) is 16.0 u .

The chemical formula of sucrose is C12H22O11 , it consists of 12 atoms of carbon, 22 atoms of hydrogen and 11 atoms of oxygen.

The formula mass of sucrose, C12H22O11 is calculated as,

Formula mass of C12H22O11=[12(atomic massof C)+22(atomic mass of H)+11(atomic mass of O)] (3)

Substitute 12.0 u for atomic mass of C , 1.0 u for atomic mass of H and 16.0 u for atomic mass of O in equation (3).

Formula mass of C12H22O11=12(12.0 u)+22(1.0 u)+11(16.0 u)=342.0 u .

The total mass of carbon (C) atoms is,

total mass of C = 12(atomic mass of carbon)

Substitute 12.0 u for atomic mass of carbon .

total mass of C = 12(12.0 u)=144.0 u

The total mass of hydrogen (H) atoms is,

total mass of H = 22(atomic mass of hydrogen)

Substitute 1.0 u for atomic mass of hydrogen .

total mass of C = 22(1.0 u)=22.0 u

The total mass of oxygen (O) atoms is,

total mass of O = 12(atomic mass of oxygen)

Substitute 16.0 u for atomic mass of oxygen .

total mass of O = 11(16.0 u)=176.0 u

The percentage by mass of carbon (C) is given as,

% C = (total mass of Cformula mass of C12H22O11)100 % (4)

Substitute 144.0 u for total mass of C and 342.0 u for formula mass of C12H22O11 in equation (4).

% C = (144.0 u342.0 u)100 %=42.1%

The percentage by mass of hydrogen (H) is given as,

% H = (total mass of H formula mass of C12H22O11)100 % (5)

Substitute 22.0 u for total mass of H and 342.0 u for formula mass of C12H22O11 in equation (5).

% Cl=(22.0 u342.0 u)100%= 6.4 %

The percentage by mass of oxygen (O) is given as,

% O = (total mass of O formula mass of C12H22O11)100 % (6)

Substitute 176.0 u for total mass of O and 342.0 u for formula mass of C12H22O11 in equation (6).

% O=(176.0 u342.0 u)100%= 51.5 %

Conclusion:

Therefore, the percentage by mass of carbon (C) is 42.1 % , the percentage by mass of hydrogen (H) is 6.4% and percentage by mass of oxygen (O) is 51.5 % in sucrose, C12H22O11 .

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Chapter 12 Solutions

An Introduction to Physical Science

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