# Finding percent sulfuric acid by mass

1. Jul 1, 2013

### leroyjenkens

1. The problem statement, all variables and given/known data
Determine the percent sulfuric acid by mass of a 1.49m (molality) aqueous solution of H2SO4.

H2SO4 has a molar mass of about 98 g/mol

2. Relevant equations

Percent by mass = mass solute / mass solution
Molarity = mol solute / liter of solution
Molality = mol solute / mass solvent (in kg)

3. The attempt at a solution

It doesn't seem like there's enough information. This is practice problem B, so I'm not sure if it wants me to use information from practice problem A or not. On the last page, I used information from A to solve B and got the wrong answer, so I don't think we need any information from A. It just seems like any equation arrangement I make, I have 2 unknowns.

1.49 = mol solute / kg solv
2 unknowns

% by mass = mass solute / mass solution * 100
There's 3 unknowns in this one. What other equations are there that I could use?

And do I have to assume a volume of the solution for this? Like 1 liter?

Thanks.

2. Jul 1, 2013

### Saitama

Even if I had to assume, I would assume that mass of solvent is 1kg.

You can even do it by considering the mass of solvent as x and find the mass of the solute in terms of x.

3. Jul 2, 2013

### leroyjenkens

I assumed the mass of solvent was 1kg and got an answer of 14.6%. The answer is 12.8%. I did everything right.
If I had assumed the mass of the solvent was 2kg, I would have gotten a different answer. So how can I get the correct answer by assuming values?

4. Jul 2, 2013

### Saitama

I can't point out the error until you show your calculations. :)

5. Jul 2, 2013

### Staff: Mentor

That definitely means you are doing something wrong. But as Pranav-Arora said, it is impossible to tell what not seeing your approach.

I got 12.7% (12.741) when using 98 for molar mass, and 12.8% (12.751) when using 98.08 for molar mass.

6. Jul 2, 2013

### leroyjenkens

Good point.

Alright, I'll start from scratch with the information I have:

Molality = 1.49 = mol solute / kg solvent
% H2SO4 by mass = mass solute / mass solution *100%
Molar mass of H2SO4 = 98g/mol
Molarity = mol solute / L solution (not even sure if I need this)

Let's say I have 1 liter of solution. The solution is a combo of H2SO4 and H2O.

Molar mass of H2O is 18.

And this time I'm using the table that my teacher taught us to use. We fill in the table with all the information we know, and then we figure out using the information we know, to fill in the rest.

Ok so the stuff I know is above, and here's another thing I know.
1g of water = 1 mL

So I convert that to 1000g water per liter, multiply that by 1 mol / 18 g to give me the mols of water, because I'm assuming I'm using just 1 liter of water. I use the equation 1.49 = mol solute / kg solv, plugging in 1kg of my solvent to get 1.49 as my moles of H2SO4. Using that an the molar mass of H2SO4 by multiplying 1.49moles * 98g/mole = 146.02g. So I have mass of both solute and solvent, so I add those together to find mass of solution, which is 1146.02 and use equation %mass of H2SO4 = 146.02 / 1146.02 * 100% to get the answer of 12.74148793%, which is pretty close to the answer in the back of the book, which is 12.8%. I guess they made a mistake by not keeping enough sig figs throughout the calculations? Or did I make a tiny mistake?

Thanks. (I just figured all this out. I didn't make this thread because of a 0.1 difference between my answer and the book.)

Oh ok thanks. I guess I got it right. I forgot the molar masses of those chemicals wasn't exactly 98 and 18.

7. Jul 2, 2013

### Staff: Mentor

Will you ever use it?

Will you ever use it?

Will you ever use it?

Sounds like a homework template. Not without a reason.

Will you ever use it?

Will you ever use it?

Have you used any of the information posted earlier? :tongue2:

My bet is they calculated using exact data and forgot to check for rounding errors. In this particular case last digit of the final answer happens to be very sensitive to data, because of the rounding. Completely accidental.

Have you used 18 in your calculations?

8. Jul 2, 2013

### leroyjenkens

I guess I did a roundabout way of doing this problem. Oh well, it's good practice, I guess. We have a quiz tomorrow where we have to fill in the tables with all the information, so even if I don't need the information for the problem, we're expected to fill in all the information into the tables.

Thanks for the help.

9. Jul 3, 2013

### symbolipoint

The density of the solution would help. You want to convert 1000 milliliters of the solution to grams of solution.

10. Jul 3, 2013

### Staff: Mentor

No - you are given molality, not molarity.

11. Jul 3, 2013

### chemisttree

And... poster perhaps still has an imperfect understanding of the pertinent differences between the two. That is the purpose of exercises like this one.

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