Respuesta :
The percentage composition by mass of nitrogen in [tex]{\left( {{\text{N}}{{\text{H}}_{\text{4}}}} \right)_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}[/tex] is [tex]\boxed{\left( 2 \right){\text{ 29}}{\text{.2 \% }}}[/tex].
Further explanation:
The most commonly applied concentration terms are as follows:
1. Molality
2. Mole fraction
3. Molarity
4. Parts per million
5. Mass percent
6. Volume percent
7. Percentage composition
The given compound is [tex]{\left( {{\text{N}}{{\text{H}}_{\text{4}}}} \right)_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}[/tex]. It contains two atoms of nitrogen, eight atoms of hydrogen, one atom of carbon and three atoms of oxygen.
The mass of nitrogen in [tex]{\left( {{\text{N}}{{\text{H}}_{\text{4}}}} \right)_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}[/tex] can be calculated as follows:
[tex]{\text{Mass of N}} =\left( {{\text{Number of nitrogen atoms}}} \right)\left( {{\text{Molar mass of nitrogen}}} \right)[/tex] …… (1)
The number of N atoms in [tex]{\left( {{\text{N}}{{\text{H}}_{\text{4}}}} \right)_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}[/tex] is 2.
The molar mass of nitrogen is 14.01 g/mol.
Substitute these values in equation (1).
[tex]\begin{aligned}{\text{Mass of nitrogen}} &= \left( {\text{2}} \right)\left( {{\text{14}}{\text{.01 g/mol}}} \right)\\&= 28.02{\text{ g/mol}}\\\end{aligned}[/tex]
The formula to calculate the percentage composition of N in [tex]{\left( {{\text{N}}{{\text{H}}_{\text{4}}}} \right)_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}[/tex] is as follows:
[tex]{\text{\% composition of N}} = \left( {\dfrac{{{\text{Mass of N}}}}{{{\text{Mass of }}{{\left( {{\text{N}}{{\text{H}}_{\text{4}}}} \right)}_{\text{2}}}{\text{C}{{\text{O}}_{\text{3}}}}}} \right){\text{100 \% }}[/tex] …… (2)
The mass of N is 28.02 g/mol.
The mass of [tex]{\left( {{\text{N}}{{\text{H}}_{\text{4}}}} \right)_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}[/tex] is 96.0 g/mol.
Substitute these values in equation (2).
[tex]\begin{aligned}{\text{\% composition of N}} &= \left( {\frac{{{\text{28}}{\text{.02 g/mol}}}}{{{\text{96}}{\text{.0 g/mol}}}}} \right){\text{100 \% }}\\&= {\text{29}}{\text{.1875 \% }} \\ &\approx {\text{2}}{\text{.92 \%}}\\\end{aligned}[/tex]
Therefore the percentage composition by mass of nitrogen in [tex]{\left( {{\text{N}}{{\text{H}}_{\text{4}}}} \right)_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}[/tex] is 29.2 %.
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Answer details:
Grade: Senior School
Chapter: Concentration terms
Subject: Chemistry
Keywords: (NH4)2CO3, N, nitrogen, mass, 2.92 %, 96.0 g/mol, 28.02 g/mol, nitrogen, hydrogen, carbon, oxygen, mass of N, mass of (NH4)2CO3.
The % composition of nitrogen is 29.166% ~ 29.2%. Thus option 2 is correct.
The percent composition of nitrogen in ammonium carbonate can be calculated as:
% Composition = [tex]\rm \dfrac{mass\;of\;element}{mass\;of\lcompound}\;\times\;100[/tex]
Mass of nitrogen in the compound has been:
Number of Nitrogen atoms = 2
Mass of 1 nitrogen atom = 14 g/mol
Mass of 2 nitrogen atom = 28 g/mol
Mass of compound = 96 g/mol
% Composition of nitrogen = [tex]\rm \dfrac{28}{96}\;\times\;100[/tex]
% Composition of nitrogen = 29.166 %
The % composition of nitrogen is 29.166% ~ 29.2%. Thus option 2 is correct.
For more information about the percent composition, refer to the link:
https://brainly.com/question/17505281
