# Mass percentage

1. Jun 30, 2016

### mmmeraki

Hi! I got a question. I have been given the reaction of the equation and no other data. I have to calculate the mass percentage of H2CO3. The gas dissolves in water.

1. The problem statement, all variables and given/known data

CaCO3 + H2SO4 = CaSO4 + H2O(l) + CO2(g)

2. Relevant equations

3. The attempt at a solution
CaCO3 + H2SO4 = CaSO4 + H2O(l) + CO2(g)
1 mol 1 mol 1 mol 1 mol 1 mol
m (CO2) = 1 mol * 44g/mol = 44 g
m (H2O) = 1 mol * 18g/mol = 18 g
m (solution) = 44 g + 18 g = 62 g
% = (44 g / 66 g) * 100 = 70,96%
or
CaCO3 + H2SO4 = CaSO4 + H2O(l) + CO2(g)
n (CO2) = 1 dm3 / 22,4 dm3/mol = 0,045 mol
m (CO2) = 0,045 mol * 44 g/mol = 1,98 g
n (H2O) = 0,045 mol (1:1)
m (H2O) = 0,045 mol * 18 g/mol = 0,81 g
m (solution) 1,98 g + 0,81 g = 2,79 g
% = (1,98 g / 2,79 g) * 100 = 70,96%

Does it even make sense?

Last edited: Jun 30, 2016
2. Jun 30, 2016

### .Scott

I would interpret "the mass percentage of the H2CO3" as: the ratio of the mass of the H2CO3 over the total mass of the products, expressed as a percentage.

If that's what you want, then you will need to know the mass of one mol of CaSO4.

3. Jun 30, 2016

### Staff: Mentor

What is the exact statement of your problem?

4. Jun 30, 2016

### mmmeraki

Assume that it's 136 g. How do I calculate the ratio?

5. Jun 30, 2016

### .Scott

The mass of the H2CO3 (62) over the total mass of the products (__ + __).

But Chestermiller has a point. I am not convinced you have paraphrased the problem correctly and completely.

6. Jun 30, 2016

### mmmeraki

It's possible that I forgot something. It was on the test.

7. Jun 30, 2016

### .Scott

Your original computation was the percentage of mass of the CO2 in the solution to the solution (H2CO3).
I interpreted what you reported as the percentage of mass of the solution (H2CO3) to the total mass of the product (CaSO4+H2CO3).

8. Jun 30, 2016

### mmmeraki

Thanks! Now I got it. :)