Problem in finding relation between molarity and weight by weight percentage

While finding the relation I did the following steps:-

$$\frac WW$$% $$= \frac{W_{solute}}{W_{solution}}\times100$$ $$.......(eq^n 1)$$

(where both masses are in grams)

Since weight of solution can be written as $$volume_{solution}\times density_{solution}$$

(where volume is in ml and density in g/ml)

and weight of solute can be written as $$Molar mass_{solute}\times Moles_{solute}$$

(where molar mass is in g/mol)

Plugging these values we found after our first equation in the first equation we get

$$\frac WW$$% $$=$$ $$M_{solute}\times n_{solute}\times100\over {V_{solution}}\times D_{solution}$$

where M is molar mass of solute, D is density of solution, V is volume of solution(in ml ) and n is moles of solute

Since V is ml it can be written as $$V\over 1000$$ liters.

Now putting this in equation above we get

$$\frac WW$$% $$=$$ $$M_{solute}\times n_{solute}\times100\over{V_{solution}}\times D_{solution}\times 0.001$$ (now V is in liters)

$$\frac WW$$% $$=$$ $$M_{solute}\times n_{solute}\times100000\over {V_{solution}}\times D_{solution}$$

Since $$Molarity$$ $$= n_{solute}\over V_{solution}(in...liters)$$

$$\frac WW$$% $$=$$ $$M_{solute}\times M \times100000\over D_{solution}$$

(where M is molarity of solute). But the relation is not true so I was confused what was the mistake that i did while solving the problem?

• Guides for formatting of chemical/mathematical formulas/expressions/equations: Basics / Detailed / Upright vs Italics / Math SE Mathjax tutorial Commented Jul 21 at 6:57
• The mole and mol are the unit name and symbol, like the kilogram and kg, while the quantity name is the (molar) amount, like the mass. Commented Jul 21 at 7:02

$$C_\mathrm{A}=\frac{n_\mathrm{A}}{V}=\frac{m_\mathrm{A}}{M_\mathrm{A}\;V}=\frac{\rho_\mathrm{A}}{M_\mathrm{A}}=\frac{y_\mathrm{A}\;\rho}{M_\mathrm{A}}$$

$$\frac{w}{w}\%=y_\mathrm{A}\cdot100\%$$

$$\boxed{\frac{w}{w}\%=\frac{C_\mathrm{A}\;M_\mathrm{A}}{\rho}\cdot100\%}$$

Where:

$$C_A$$ = molar concentration of species A $$(\pu{mol/L})$$

$$M_A$$ = molar mass of species A $$(\pu{g/mol})$$

$$\rho$$ = density of solution $$(\pu{g/L})$$

• You take mass concentration formally as fractional density. Interesting. Commented Jul 21 at 7:13
• @Poutnik Indeed. It is analogous to its mole fraction counterpart in terms of molar concentrations: $x_A=\frac{C_A}{C}$ Commented Jul 21 at 7:32
• What I mean is that, IMHO, terms fractional mass/molar densities are very rarely used, in favour of respective concentrations. Commented Jul 21 at 8:00
• I was able to prove the relation but when I did it in my class for the first time I used this approach. My question is what is the mistake here. Commented Jul 21 at 8:03
• @Ur-Friend you forgot to convert the density units from $\pu{g/mL}$ to $\pu{g/L}$. In other words, your final expression needs to be divided by 1000. Commented Jul 21 at 8:12

\begin{align} \left(\frac{w}{W}\%\right) &= \frac{\text{wt. of solute}\ (w_\ce{A})}{\text{wt. of solution}\ (w)}\times100 \\ C_\ce{A} &= \frac{n_\ce{A}}{V_\text{solution}\ \ce{(in m\ell)}}\times1000=\frac{w_\ce{A}}{M_\ce{A}}\times\frac{1000}{V_\text{solution}} \tag{1} \\ V_\text{solution} &= \frac{\text{wt. of solution}}{\ce{density of solution (in g/m\ell)}}= \frac{w}{\rho} \tag{2} \\ C_\ce{A} &=\frac{w_\ce{A}}{M_\ce{A}}\cdot\frac{\rho}{w}\times1000 \\ C_\ce{A} &=\left(\frac{w_A}{w}\times100\right)\cdot\left(\frac{10\rho}{M_\ce{A}}\right) \\ C_\ce{A} &= \left(\frac{w}{W}\%\right)\cdot\left(\frac{10\rho}{M_\ce{A}}\right) \tag{3}\label{3} \end{align} Rearranging equation $$\eqref{3}$$ $$\left(\frac{w}{W}\%\right) =\frac{C_\ce{A}M_\ce{A}}{10\rho}$$