“1.25 g of crushed limestone was added to 50.0 cm3 of 1.00 mol dm-3 hydrochloric acid (an excess). The mixture was left until all bubbling stopped and was then made up carefully to 250 cm3 with pure water. A 25.0 cm3 sample of this was pipetted into a conical flask and some methyl orange indicator was added. Sodium hydroxide solution of concentration 0.100 mol dm-3 was added from a burette. 30.0 cm3 were needed to reach the end-point of the indicator. We can then calculate the percentage of calcium carbonate in the sample.” What is the mass percentage of carbonate in the sample? Assume all carbonate is calcium carbonate, what is the mass percentage of the remainder ( or is this assumption not tenable) ?
The idea here is that the CaCO3 reacts with a known excess of HCl, the excess HCl is then titrated in a regular acid/base titn to determine how much of an excess was there.
CaCO3 + 2HCl ==> CaCl2 + H2O + CO2
1.25 g sample-->added 50 x 1 mol/dm^3 HCl. = 50 mmoles HCl-->diluted to 250 cc-->25 cc aliquot taken and titrated with 30 cc of 0.1 mole/dm^3 HCl.
I would start with the titration step.
30 cc x 0.1 mol/dm^3 = 3 mmoles NaOH.
3 mmoles NaOH = 3 mmoles excess HCl (in the 25 cc aliquot). Since that was a 25 cc aliquote from a 250 cc flask, then 3 mmoles x (250/25) = 30 mmoles in the entire flask. We started with 50 mmoles HCl; therefore 50-30=20 mmoles HCl were used by the CaCO3. Looking at the coefficients in the equation, 20 mmoles HCl means 10 mmoles CaCO3 reacted. I would then convert 10 mmoles to moles, grams CaCO3 = moles CaCO3 x molar mass CaCO3 and %CaCO3 = (mass CaCO3/mass sample)*100 = ??
mass of sample, of course, is 1.25 g.
I get an answer about 80% CaCO3 and it makes no difference which part we talk about. It's 80% in the original sample and 80% in the sample taken for titration.
Where did you get this question from
To calculate the mass percentage of carbonate in the sample, we need to determine the amount of calcium carbonate that reacted with the hydrochloric acid.
First, let's calculate the amount (in moles) of hydrochloric acid used in the reaction:
Amount of HCl = concentration * volume
= 1.00 mol dm^-3 * 0.0500 dm^3
= 0.0500 moles
Since the molar ratio between calcium carbonate and hydrochloric acid is 1:2, the amount of calcium carbonate that reacted can be calculated as:
Amount of CaCO3 = 0.0500 moles * (1/2)
= 0.0250 moles
Next, let's calculate the molar mass of calcium carbonate:
Molar mass of CaCO3 = (40.08 g/mol) + (12.01 g/mol) + (3 * 16.00 g/mol)
= 100.09 g/mol
Now, we can calculate the mass of calcium carbonate:
Mass of CaCO3 = amount * molar mass
= 0.0250 moles * 100.09 g/mol
= 2.5025 g
So, the mass percentage of carbonate in the sample is:
Mass percentage of carbonate = (mass of CaCO3 / mass of sample) * 100
= (2.5025 g / 2.5025 g) * 100
= 100%
Since we assumed all carbonate is calcium carbonate, the mass percentage of the remainder (non-carbonate) is 0%.
To calculate the mass percentage of carbonate in the sample, we need to find the amount of calcium carbonate that reacted with the hydrochloric acid.
First, let's calculate the number of moles of hydrochloric acid reacted with calcium carbonate:
- The concentration of the hydrochloric acid is given as 1.00 mol dm^-3.
- The volume of hydrochloric acid used is 50.0 cm^3, which is equal to 50.0 × 10^-3 dm^3.
- So, the number of moles of hydrochloric acid reacted is: 1.00 mol dm^-3 × 50.0 × 10^-3 dm^3 = 0.0500 moles.
Since the hydrochloric acid is present in excess, we can assume that all the calcium carbonate reacts completely.
Next, let's calculate the number of moles of calcium carbonate that reacted:
- According to the balanced chemical equation for the reaction between calcium carbonate (CaCO3) and hydrochloric acid (HCl), one mole of calcium carbonate reacts with two moles of hydrochloric acid.
- Therefore, the number of moles of calcium carbonate is 0.0500 moles ÷ 2 = 0.0250 moles.
Now we can calculate the mass of calcium carbonate that reacted using its molar mass:
- The molar mass of calcium carbonate (CaCO3) is the sum of the atomic masses of calcium (Ca), carbon (C), and three oxygen (O) atoms.
- The atomic masses of Ca, C, and O are approximately 40.08 g/mol, 12.01 g/mol, and 16.00 g/mol, respectively.
- So, the molar mass of calcium carbonate is 40.08 g/mol + 12.01 g/mol + (16.00 g/mol × 3) = 100.09 g/mol.
- The mass of calcium carbonate reacted is therefore 0.0250 moles × 100.09 g/mol = 2.5025 g.
Finally, let's calculate the mass percentage of carbonate in the sample:
- The given mass of crushed limestone added to the hydrochloric acid was 1.25 g.
- So, the percentage of carbonate in the sample is (mass of calcium carbonate reacted ÷ mass of limestone sample) × 100.
- Plugging in the values, we get (2.5025 g ÷ 1.25 g) × 100 ≈ 200.2%.
Based on the given information, the mass percentage of carbonate in the sample is approximately 200.2%.
As for the mass percentage of the remainder, we cannot calculate it without additional information about the other components of the limestone sample. Assuming that all the carbonate is calcium carbonate, the remainder would be composed of other components such as impurities or non-carbonate minerals present in the limestone.