Question

SHOWING ALL WORK.. Calculate the percentcomposition of nitrogen in Methyl Orange(C14H14N3Na03S)

Answer 1

13%.

Step-by-step 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:

`\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)(g)/(mol), \\ Mol\text{ar mass of C}_(14)H_(14)N_3NaO_3S=(168+14+42+23+48+32)(g)/(mol), \\ Mol\text{ar mass of C}_(14)H_(14)N_3NaO_3S=327(g)/(mol). \end{gathered}`

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:

`Composition\text{ of X in compound=}\frac{mo\text{lar mass of X in compound}}{mo\text{lar mass of compound}}\cdot100\%`

X would be N (nitrogen) and the compound would be C14H14N3NaO3S, so replacing the values that we have, we obtain:

`\text{Composition of N=}\frac{42\text{ }(g)/(mol)}{327(g)/(mol)}\cdot100\%=12.84\%\approx13\%.`

The composition of nitrogen in methyl orange would be 13%.