I am trying to solve a question in my study and I don't get the answer required.

A mixture of $\ce{NaNO3}$ and $\ce{Na2SO4}$ weight $5.37$ grams, the $\ce{Na}$ = $1.61$ grams. What is the percentage mass of the $\ce{NaNO3}$ in the mixture?

The correct answer according to the book is $45.1\%$, but I get $37\%$.

What is did is:

The weight of the Na is 1.61 total, which means the rest is 3.76 grams total mass/Molar mass of the rest is 3.76/158 = 0.0238 mol
m(N) = 0.0238*14 = 0.33 grams
m(O3) = 0.0238*48 = 1.09 grams
m(Na) = 1.61/3 = 0.54 grams
total mass = 1.96 grams

1.96/5.37 = 37%

Where and what am I missing or mistaken?


1 Answer 1


You need to do a bit more math than that. In order to solve the majority of problems in chemistry you need to switch to the amounts of substances.

So, let's use the following notations: $\omega$ – mass fraction; $m$ – mass; $n$ – amount; $M$ – molecular mass. I refer to $\ce{NaNO3}$ as compound $1$ and to $\ce{Na2SO4}$ as compound $2$. Let's solve this algebraically first. from the definition:

$$\omega_1 = \frac{m_1}{m} = \frac{n_1M_1}{m}$$

In order to find $n_1$, we use the sodium content in both salts

$$n(\ce{Na}) = \frac{m(\ce{Na})}{M(\ce{Na})} = n_1 + 2n_2 \implies n_1 = \frac{m(\ce{Na})}{M(\ce{Na})} - 2n_2$$

and total mass:

$$m = n_1M_1 + n_2M_2$$

Combining both equations (I let you do the math), you find $n_1$:

$$n_1 = \frac{m(\ce{Na})}{M(\ce{Na})} - \frac{2\left(m - \frac{m(\ce{Na})}{M(\ce{Na})}M_1\right)}{M_2 - 2M_1}$$

Now you can find the mass fraction of sodium nitrate:

$$ \begin{align} \omega_1 &= \frac{\left(\frac{m(\ce{Na})}{M(\ce{Na})} - \frac{2\left(m - \frac{m(\ce{Na})}{M(\ce{Na})}M_1\right)}{M_2 - 2M_1}\right)M_1}{m} \\ &= \frac{\left(\frac{\pu{1.61 g}}{\pu{23.00 g mol-1}} - \frac{2\left(\pu{5.37 g} - \frac{\pu{1.61 g}}{\pu{23.00 g mol-1}}\cdot\pu{85.00 g mol-1}\right)}{\pu{142.04 g mol-1} - 2\cdot\pu{85.00 g mol-1}}\right)\cdot\pu{85.00 g mol-1}}{\pu{5.37 g}} \\ &= 0.4513~\text{or}~45.13\% \end{align} $$


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