
Concept explainers
(a)
The percentage by mass of each element in salt, NaCl.
(a)

Answer to Problem 7E
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 %
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)
The percentage by mass of each element in sucrose, C12H22O11.
(b)

Answer to Problem 7E
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 mass of 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
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