Calculate the percentage by weight of atoms in given compound:
(a) C in ${ CO }_{ 2 }$
(b) Na in ${ Na }_{ 2 }{ CO }_{ 3 }$
(c) Al in aluminium nitride.
[C = 12, O = 16, H = 1, Na = 23, Al = 27, N = 14]
Answer
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Hint: Solutions are sometimes represented in terms of relative percent concentration of solute in a solution. To determine the weight percent of a solution, divide the mass of solute by mass of the solution and multiply by 100 to obtain percent.
Complete step by step answer:
Let us try to solve the given question.
(a) Weight of C = 12
Weight of ${CO} _ {2}$ = 12 + 2(16) = 44
% age of C in ${CO} _ {2}$
= $\dfrac {weight\quad of\quad carbon} {weight\quad of\quad {CO} _ {2}} \times 100$
$\dfrac {12} {44} \times 100\quad =\quad \dfrac {300} {11}$
= 27.27%
Therefore, the percentage of C in ${CO} _ {2}$ is 27.3%
(b) Weight of Na in ${ Na }_{ 2 }{ CO }_{ 3 }$
= 2 (Na) + C + 3 (O)
= 2 (23) + 12 + 3 (16)
= 46 + 12 + 48 = 106
Percentage of sodium in ${ Na }_{ 2 }{ CO }_{ 3 }$
Weight $\dfrac { Weight\quad of\quad Na }{ Weight\quad of\quad { Na }_{ 2 }{ CO }_{ 3 } \times 100}$
= $\dfrac {46} {106} \times 100$=43.4%
Therefore, the percentage of Na in ${ Na }_{ 2 }{ CO }_{ 3 }$ is 43.3%.
(c) Weight of aluminium in aluminium nitride.
Molecular weight of aluminium nitride
= 27 + 14 = 41g
Percentage of aluminium
= $\dfrac {Weight\quad of\quad aluminium} {Weight\quad of\quad aluminium\quad nitride} \times 100$
= $\dfrac { 27 }{ 41 } \times 100$=65.85%
Therefore, percentage of Al in aluminium nitride is 65.85%
Note: If a raw material in your formula is a liquid and measured by the volume, you must know the mass of this, which requires a density value and remember the total mass of the solution is always equal to the sum of the mass of each of the solutes.
Complete step by step answer:
Let us try to solve the given question.
(a) Weight of C = 12
Weight of ${CO} _ {2}$ = 12 + 2(16) = 44
% age of C in ${CO} _ {2}$
= $\dfrac {weight\quad of\quad carbon} {weight\quad of\quad {CO} _ {2}} \times 100$
$\dfrac {12} {44} \times 100\quad =\quad \dfrac {300} {11}$
= 27.27%
Therefore, the percentage of C in ${CO} _ {2}$ is 27.3%
(b) Weight of Na in ${ Na }_{ 2 }{ CO }_{ 3 }$
= 2 (Na) + C + 3 (O)
= 2 (23) + 12 + 3 (16)
= 46 + 12 + 48 = 106
Percentage of sodium in ${ Na }_{ 2 }{ CO }_{ 3 }$
Weight $\dfrac { Weight\quad of\quad Na }{ Weight\quad of\quad { Na }_{ 2 }{ CO }_{ 3 } \times 100}$
= $\dfrac {46} {106} \times 100$=43.4%
Therefore, the percentage of Na in ${ Na }_{ 2 }{ CO }_{ 3 }$ is 43.3%.
(c) Weight of aluminium in aluminium nitride.
Molecular weight of aluminium nitride
= 27 + 14 = 41g
Percentage of aluminium
= $\dfrac {Weight\quad of\quad aluminium} {Weight\quad of\quad aluminium\quad nitride} \times 100$
= $\dfrac { 27 }{ 41 } \times 100$=65.85%
Therefore, percentage of Al in aluminium nitride is 65.85%
Note: If a raw material in your formula is a liquid and measured by the volume, you must know the mass of this, which requires a density value and remember the total mass of the solution is always equal to the sum of the mass of each of the solutes.
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