Respuesta :
Answer:
13%.
Explanation:
What is given?
Molar mass of C = 12 g/mol.
Molar mass of H = 1 g/mol.
Molar mass of N = 14 g/mol.
Molar mass of Na = 23 g/mol.
Molar mass of O = 16 g/mol.
Molar mass of S = 32 g/mol.
Step-by-step solution:
First, we have to calculate the molar mass of C14H14N3NaO3S by doing an algebraic sum with the given molar masses of each element. We have 14 carbons (C), 14 hydrogens (H), 3 nitrogens (N), 1 sodium (Na), 3 oxygens (O), and 1 sulfur (S). The calculation of this molar mass will look like this:
[tex]\begin{gathered} Mol\text{ar mass of C}_{14}H_{14}N_3NaO_3S=(14\cdot12+14\cdot1+3\cdot14+1\cdot23+3\cdot16+1\cdot32)\frac{g}{mol}, \\ Mol\text{ar mass of C}_{14}H_{14}N_3NaO_3S=(168+14+42+23+48+32)\frac{g}{mol}, \\ Mol\text{ar mass of C}_{14}H_{14}N_3NaO_3S=327\frac{g}{mol}. \end{gathered}[/tex]Now that we have the molar mass of methyl orange (C14H14N3NaO3S), we have to calculate the composition of nitrogen. As we have 3 nitrogens and the molar mass of nitrogen is 14 g/mol, the molar mass of this only element in methyl orange would be 3 x 14 = 42 g/mol, so the composition formula is given by this:
[tex]Composition\text{ of X in compound=}\frac{mo\text{lar mass of X in compound}}{mo\text{lar mass of compound}}\cdot100\%[/tex]X would be N (nitrogen) and the compound would be C14H14N3NaO3S, so replacing the values that we have, we obtain:
[tex]\text{Composition of N=}\frac{42\text{ }\frac{g}{mol}}{327\frac{g}{mol}}\cdot100\%=12.84\%\approx13\%.[/tex]The composition of nitrogen in methyl orange would be 13%.
