Have a dissolution of $\ce{HNO_3}$ in water. The density is $1.42\ce{g/mL}$ with a concentration of $\ce{16M}$. What is the concentration in % mass/mass?

Alright. So the % mass/mass is expressed as

$$\frac{\text{grams of } \ce{HNO_3}}{\text{grams of the dissolution}}\cdot100$$

So I need to find the grams of $\ce{HNO_3}$ first. They tell me that the Molarity is

$$16 = \frac{\text{moles of } \ce{HNO_3}}{\text{liters of dissolution}}$$

$\ce{HNO_3}$ has a molar mass of $63$. So to find out how many moles of $\ce{HNO_3}$ are in the dissolution I need to divide the grams of $\ce{HNO_3}$ by $63$... But how am I supposed to find out the grams of $\ce{HNO_3}$?

I was told that the numerator is $63$. But... $63$ is the molar mass of $\ce{HNO_3}$, not the number of moles of $\ce{HNO_3}$ in the dissolution.

Anyway, let's suppose that it is indeed $63$. Therefore

$$16 = \frac{63}{\text{liters of dissolution}} \implies \text{liters of dissolution} = 3.9$$

So there are $3.9$ liters of dissolution? I'm not sure - I feel like I got this all wrong. How do you know there are $63$ moles of $\ce{HNO_3}$?

  • $\begingroup$ You have your equation wrong. You were told you have 16 M (or 16 moles/liter). So, 1 liter has 16 moles (or $16\ moles\times 63\ g/mole$). $\endgroup$ – LDC3 Jul 15 '15 at 6:01
  • $\begingroup$ @LDC3 ah... I get it now. So there are $1008$ moles of $\ce{HNO_3}$ per liter in the dissolution. However, how do I know how many liters are in the dissolution? I am told that there is only one liter of dissolution but how do they even know that? $\endgroup$ – Voldemort Jul 15 '15 at 6:07
  • $\begingroup$ What happened to the units? It is not 1008 moles ! $\endgroup$ – LDC3 Jul 15 '15 at 6:08
  • $\begingroup$ @LDC3 oh, yeah you're right. It's $1008$ grams of $\ce{HNO_3}$ per liter. But I still don't get why is there only one liter of dissolution (according to the book). $\endgroup$ – Voldemort Jul 15 '15 at 6:11
  • $\begingroup$ I arbitrarily picked 1 liter to do the calculation with. The concentration would be the same no matter the volume. $\endgroup$ – LDC3 Jul 15 '15 at 6:15

Using 1L of solution, you have 16M of HNO3, each mole of which weighs 63g. Therefore, in 1L you'd have 63 X 16 = 1008g. Mass of 1L is given as 1420g from the density.

%mass/mass = 1008/1420 = 70% HNO3

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  • $\begingroup$ Ok, it all makes sense, but what I don't understand is why are you picking $1L$ of solution? Why one and not, say, five liters? Where/how do you get that number? In other words, how do you even know there is one liter. $\endgroup$ – Voldemort Jul 15 '15 at 6:23
  • $\begingroup$ 1L is convenient. Use 2.8L, for the same calculation, HNO3 = 16 X 2.8 X 63 = 2822.4, mass liquid = 2800 X 1.42 = 3976, 2822.4/3976 = 70%, same value. $\endgroup$ – roffster Jul 15 '15 at 6:28
  • $\begingroup$ Well I meant the Molarity part. The Molarity formula in your case is $16 = \frac{63}{1}$, but if you change that $1$ by a $2.8$ instead of getting $1008$ grams you would get 360 grams of $\ce{HNO_3}$. $\endgroup$ – Voldemort Jul 15 '15 at 6:35
  • $\begingroup$ Actually, the formula is: Total Moles = [concentration] X volume. So, 2.8L of 16M solution has 44.8 total moles, each with a mass of 63, which leads to a total mass of 2822.4. You have the concentration and volume, but you don't have the total moles, so that's why you are looking for. $\endgroup$ – roffster Jul 15 '15 at 6:40
  • $\begingroup$ Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the help center $\endgroup$ – Martin - マーチン Jul 15 '15 at 7:03

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