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

Question

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Answer 1

Question

Chapter 12, Problem 7E

(a)

To determine

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

(a)

Expert Solution

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

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,

$Formula mass of C12H22O11=[12(atomic mass of C)+22 (atomic mass of H)+11(atomic mass of O)]$ (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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Answer 2

Question

Chapter 12, Problem 7E

(a)

To determine

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

(a)

Expert Solution

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

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,

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

Question

Chapter 12, Problem 7E

(a)

To determine

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

(a)

Expert Solution

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

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,

$Formula mass of C12H22O11=[12(atomic mass of C)+22 (atomic mass of H)+11(atomic mass of O)]$ (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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Answer 4

Question

Chapter 12, Problem 7E

(a)

To determine

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

(a)

Expert Solution

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

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,

$Formula mass of C12H22O11=[12(atomic mass of C)+22 (atomic mass of H)+11(atomic mass of O)]$ (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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Answer 5

Chapter 12, Problem 7E

(a)

To determine

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

(a)

Expert Solution

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

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,

$Formula mass of C12H22O11=[12(atomic mass of C)+22 (atomic mass of H)+11(atomic mass of O)]$ (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, $C12H22O11$ .

Answer 6

Chapter 12, Problem 7E

(a)

To determine

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

(a)

Expert Solution

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, $NaCl$ .

Answer 7

Chapter 12, Problem 7E

(a)

To determine

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

Answer 8

(a)

Expert Solution

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,

$NaCl$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

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, $NaCl$ .

Answer 13

(b)

To determine

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

(b)

Expert Solution

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,

$Formula mass of C12H22O11=[12(atomic mass of C)+22 (atomic mass of H)+11(atomic mass of O)]$ (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, $C12H22O11$ .

Answer 14

(b)

To determine

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

Answer 15

(b)

Expert Solution

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,

$C12H22O11$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

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, $C12H22O11$ .