# Molecular Weight vs percentage



## cejohnsonsr (Jan 30, 2013)

I think I already know the answer to this (since I've done my reading), but I'd like to confirm what I think I know (since I'm not really a chemist). If the Molecular weight of a chemical is listed as 50.0, would that indicate a 50% concentration of that chemical in the solution? I ask because there are several chemicals I can buy from a company that I won't name (I don't want to be mistaken for advertising or spamming) where the description uses this instead of a percentage. The specific chemical I'm talking about is Nitric Acid. The relevant specs are listed below. I copied & pasted & removed any non-relevant information. For those who might want to know, I'm comparing prices. I've found a few sources that should be reliable, but not everything is listed the same way so I want to make sure I'm comparing apples to apples. If I'm correct, this should be a 63% concentration & would therefore be suitable for the AR process. 

Item Nitric Acid, ACS
Concentration 15.8N
Molecular Weight 63.01
Chemical Formula HNO3
Container Material Amber Glass

My standard disclaimer/caveat: I swear I'm reading as much as possible every day & I promise to continue to do so. I don't want anyone to do my work for me. I'm only trying to narrow down the list of things I need to read & sort them into the most useful & productive order. If I've asked a question, please believe that I've already searched & not found the answer I was looking for. I'm perfectly satisfied with being pointed in the right direction via a link to whatever it is I need to read in order to obtain my own answer. Thank you in advance for any help. Ed


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## FrugalRefiner (Jan 30, 2013)

ce,

The molecular weight of any substance is the sum of the atomic weights of the molecule's components. In the case of HNO3, that means 1 x the atomic weight of Hydrogen (1.008) plus 1 x the atomic weight of Nitrogen (14.007) plus 3 x the atomic weight of Oxygen (3 x 15.999 = 47.997) equals 63.012. The atomic weight remains the same regardless of the concentration / dilution. What does change is the density of the solution. The density of pure water is 1.000. The density of your example would be about 1.42.

The concentration is given as: "Concentration 15.8N". This is just under 70% concentrated acid, the standard we use. It is as concentrated as nitric acid can be made by simple distillation - its azeotrope.

Hope this helps,
Dave


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## butcher (Jan 30, 2013)

Ed 
cejohnsonsr, 

We do not mind you casually mentioning a company that can help members to find something they may need, but spamming the forum as you know are against what this forum is all about (we are not a source or website for everybody to advertise there wares for free), we are the gold refining forum.

Basically what it looks like what your asking, is this the nitric acid you need to recover and refine with, you posted a commercial concentrated nitric acid 68% acid and 32% water (azeotropic nitric acid ), this is concentrated enough for aqua regia, it would need to be diluted with equal volumes of water to use on silver or other base metals.

Below is 68% HNO3 nitric acid 32% H2O by weight
And the specifications for it:

HNO3
Molecular weight of HNO3 63.02g/mol
68% azeotropic nitric acid
Av. specific gravity of concentrated solution @ 68% HNO3 is about 1.42g/ml
Av. % percent in concentrated reagent (of 68% HNO3) is about 69.5%
Grams of active ingredient of HNO3 (in 68% HNO3) is 0.985g
Normality of 68% HNO3 is 15.6N
ml conc. reagent/liter N solution is 64

Molecular weight of HNO3 is how many grams per mol, this is gotten from the periodic table for each of the elements involved:
HNO3
Hydrogen gas (H) molecular weight = 1.00
Nitrogen gas (N) molecular weight = 14.00
Oxygen gas (O) molecular weight = 16
Note we have 3 oxygen so we multiply (3 x 16 = 48)
So here O3 = 48
Now adding these we get the molecular weight for HNO3
1 + 14 + 48 = 63 grams per mol for HNO3

Note 63 grams per mol has nothing to do with how dilute the solution is it gives the molecular weight for HNO3. Not the percentage of HNO3 to water content.

The percentage of acids can be expressed by many different means w/w, w/v, v/v and so on.
Read up about them in most any chemistry book (I find used chemistry books at second hand stores)
Here we are discussing azeotropic nitric acid so 68% HNO3 acid and 32% water.

As far as morality, molecular mass, percentages, Normality, specific gravity, density, concentration, parts per million, Molar solutions, osmolarity, and other ways to express how much of what is in what you will need to pick up that chemistry book.


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## Sucho (Jan 30, 2013)

your nitric is 15.8 N ( Normal) in this case, you can count as 15.8 M ( Molar) // sorry i will not answer the difference between Normal and Molar cos i think its useless for you

15.8 Molar means that in 1 litre you have 15.8 Moles of Acid

1 Mole of nitric acid has a weight of 63,01 g ( that comes from Molar Mass - 63,01 g per mole)

15,8 Moles X 63,01 gram per mole gives 995,558 grams of acid in one litre

density is 1,51 gram per cubic centimetre

995,558 grams / 1,51 gram per cubic centimetre gives 659, 309 cubic centimetres in one litre

1000 cubic centimetres - 659,309 cubic centimetres of HNO3 gives 340, 691 cubic centimetres of water

percentage of acid is stated as a weight to weight ratio - w/w 

995,588 grams of HNO3 + 340,691 grams of water gives 1336,249 grams 

995,588 grams pure HNO3 / 1336,249 gives 0,74,5 - that is 74,5 %

those are basic calculations.the inaccuration is caused, because when you mix two different fluids, theres always a reaction between them so retardation or dilatation of volume occurs ( same when you are mixing pure ethanol with water )

the best way to calculate this more accurate is to use density of 15,8 M nitric, which is 1,42 g / mole

so the best calculation is - 995,558 grams of pure HNO3 / 1420 grams of 1 litre of solution gives 0,701 - that is 70,1 %

you are able to buy 70 % HNO3


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## cejohnsonsr (Jan 30, 2013)

Thank you so much for the best answers I could have hoped for. I truly appreciate the help.


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## Richard NL (Feb 25, 2017)

After much searching/googling/reading still can not find a satisfying answer(2 weeks all my free time).

The government wants the know the "concentration of the substance in percent by weight" of my nitric acid 60%, 40%Baumé technical grade.
In order to request a license for Regulation(EU)No98/2013 of the European Parliament and of the Council of 15 January 2013 on the marketing and use of explosives precursor. 
I assume that the percentage of which is indicated on the label is weight/weight 60%.
After a telephone contact with inspection environment and transport, I was informed clearly that I really had to fill it in correctly.
After thinking about it, i got confused!!!

1# 4HNO₃=2H₂O+4NO₂+O₂.

2# 100+10%=110, 110-10%=101
100-10%=99, 99+10%= 99
A difference of 2 with a calculation of 10%.

3# by butcher » January 30th, 2013, 11:12 pm


butcher said:


> Grams of active ingredient of HNO3 (in 68% HNO3) is 0.985g



4# Density of 60%HNO₃ 1,3630g/mL, density of water 0,9982 g/mL = 1,3630-0,9982=0,3648≙364.8 gram/liter

I have in almost 5 years not 1 question asked (May 31st, 2012).
Do i have to write down 60% concentration of the substance in percent by weight, or is there a flaw in the system?
If there is an error in the way of calculating the concentration of the the substance in percent by weight with Nitric acid, it cut have effect on the not regulated 3% nitric acid which is available to members of the general public in Europe.

I would really appreciate if you could help.


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## g_axelsson (Feb 25, 2017)

I can't follow your calculations, but the 60% acid is percent weight.

Göran


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## justinhcase (Feb 25, 2017)

Richard NL said:


> 3% nitric acid which is available to members of the general public in Europe.
> 
> I would really appreciate if you could help.


I do not think 3% will be of much use to you.
I have never used such a dilute acid but I would think it would be too slow to be of any real practical use to you. Think about the amount of H2O you will have to evaporate if you intend to use it for A.R.
You will have to apply for a licence for non-commercial use.


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## goldsilverpro (Feb 25, 2017)

Richard NL,

3% Nitric is almost worthless to the refiner.

Your 1# and 2# are meaningless.

3# At 20C, the density of 68% w/w Nitric acid 1.408. Therefore, one ml will weigh 1.408g. Therefore, the nitric acid content in one ml = 1.408 X .68 = .957g. The difference in this and butcher's is probably due to him using the density at a different temperature than I did. I always do this at 20C.

4# Your math is faulty. Density of 60% w/w Nitric acid at 20C is 1.3667 g/ml. Therefore, one ml will contain 1.3667 X .60 = .820g of nitric. If this is 60% nitric, it must be 40% water. Therefore, the amount of water in one ml of 60% nitric = 1.3667 X .40 = .547g. Interestingly enough, 60% nitric by weight is 54.7% water by volume. This is because HNO3 is so much denser than H2O.

However, I would forget about the water - it just makes things more confusing. If you focus on my method, you will see it is simple and makes sense. 

I used this chart for the densities:
http://www.handymath.com/cgi-bin/nitrictble2.cgi?submit=Entry


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## Richard NL (Feb 26, 2017)

Thanks so much for your time and help, much appreciated!
I just had to make sure, without a doubt.
I will continue to apply for a license for non-commercial use.

Thumbs up!


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## Ravn (Mar 9, 2017)

Hi

Here are the formulars for calculating concentration or molarity. 
I hope it will usefull. 


Conversion from Molarity to % concentration (W/W)

*Cm = Cp x d / (100 % x Mw)*


Conversion from % concentration to Molarity

*Cp = Cm x 100 % x Mw / d*


Cm = Molarity 
Cp = Concentration in % W/W 
d= Density in grams / litre
Mw = Molecula weight (of the dissolved substance)


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