1.984 g of a hydrate of K2CO3 is dissolved in 250. mLs of H2O. 10.0 mLs of this solution is tirade against 0.100 M HCl, producing the following results.
Flask 1 - Initial 0, final 9.2
Flask 2 - Initial 9.2, final 18.3
Flask 3 - initial 18.3, final 27.8
Flask 4 - initial 27.8, final 36.9
Determine the formula of the hydrate.
For some reason my instructor doesn't like to teach us anything until the day after it's due. I really have no idea how to even start this question. Help would be much appreciated!! Thanks!
What indicator did you use?
I'm assuming the indicator was something in the acid range, say change pH about 4 to 5 which would mean CO3^2- was titrated all the way to H2O and CO2.
K2CO3.xH2O + 2HCl ==> H2O + CO2 + xH2O _ 2KCl
mols HCl. Subtract from each other and take the average. That's 9.2 for #1 and 9.1 for #2 etc.
mols HCl = M x L = ?
mols K2CO3.xH2O = 1/2 that (from the coefficients in the balanced equation)
Then since mol = grams/molar mass you can rearrange to molar mass = grams/mol
g of sample = 1.984 x (10/250) = 0.07936
So molar mass = about 0.0936/0.000461 = about 172 or so.
K2CO3 weighs 138.2 so 172 - 138.2 = about 34 which tells me there must be 2 H2O for 2*18 = 36.
Why is it not closer than that. Probably for a couple of reasons. First, you have a sample weighed to 4 places but you read the buret to only 2 places. Second is you can read another place on the buret if you had used above 10 and we like to use more like 35 or 40 mL. It improves the precision. Third, I don't know how pure the K2CO3.2H2O was.
Thank you very much! this makes much more sense now! I really appreciate your help.
To determine the formula of the hydrate, we need to use the given data and perform a series of calculations. Here's how you can approach this problem step by step:
Step 1: Calculate the moles of HCl reacted.
To calculate the moles of HCl reacted, you can use the equation:
moles HCl = volume of HCl (in L) × molarity of HCl
Given:
volume of HCl = 10.0 mL = 0.010 L
molarity of HCl = 0.100 M
Substitute the values into the equation:
moles HCl = 0.010 L × 0.100 M
moles HCl = 0.001 mol
Step 2: Calculate the moles of K2CO3 reacted.
From the balanced chemical equation between K2CO3 and HCl, we know that 2 moles of HCl react with 1 mole of K2CO3.
Since the moles of HCl reacted are 0.001 mol (from Step 1), the moles of K2CO3 reacted will be half of that: 0.001 mol ÷ 2 = 0.0005 mol.
Step 3: Calculate the mass of K2CO3.
To calculate the mass of K2CO3, you can use the equation:
mass = moles × molar mass
The molar mass of K2CO3 can be obtained from the periodic table:
K2CO3 = (2 × atomic mass of K) + atomic mass of C + (3 × atomic mass of O)
Given:
atomic mass of K = 39.10 g/mol
atomic mass of C = 12.01 g/mol
atomic mass of O = 16.00 g/mol
Substitute the values into the equation:
molar mass of K2CO3 = (2 × 39.10 g/mol) + 12.01 g/mol + (3 × 16.00 g/mol)
molar mass of K2CO3 = 138.21 g/mol
Now calculate the mass of K2CO3:
mass of K2CO3 = 0.0005 mol × 138.21 g/mol
mass of K2CO3 = 0.06911 g (rounded to 5 decimal places)
Step 4: Calculate the mass of water lost during heating.
To find the mass of water lost, we need to subtract the mass of K2CO3 from the original mass of the hydrate.
Given:
mass of the hydrate = 1.984 g
mass of K2CO3 = 0.06911 g
mass of water lost = mass of the hydrate - mass of K2CO3
mass of water lost = 1.984 g - 0.06911 g
mass of water lost = 1.91489 g (rounded to 5 decimal places)
Step 5: Calculate the moles of water lost.
To calculate the moles of water lost, you can use the equation:
moles = mass ÷ molar mass
Given:
mass of water lost = 1.91489 g
The molar mass of water (H2O) is:
molar mass of H2O = (2 × atomic mass of H) + atomic mass of O
Given:
atomic mass of H = 1.01 g/mol
atomic mass of O = 16.00 g/mol
Substitute the values into the equation:
molar mass of H2O = (2 × 1.01 g/mol) + 16.00 g/mol
molar mass of H2O = 18.02 g/mol
Now calculate the moles of water lost:
moles of water lost = 1.91489 g ÷ 18.02 g/mol
moles of water lost = 0.10631 mol (rounded to 5 decimal places)
Step 6: Calculate the ratio of moles of water to moles of K2CO3.
To find the ratio of moles of water to moles of K2CO3, divide the moles of water lost by the moles of K2CO3:
ratio = moles of water lost ÷ moles of K2CO3
ratio = 0.10631 mol ÷ 0.0005 mol
ratio = 212.62 (rounded to 2 decimal places)
The ratio of moles of water to moles of K2CO3 is approximately 212.62.
Step 7: Determine the empirical formula of the hydrate.
The empirical formula gives the simplest whole number ratio of elements in a compound. To find the empirical formula, divide the ratio obtained in Step 6 by the greatest common divisor (GCD) to simplify it.
In this case, the GCD of 212.62 is 2. So, divide the ratio by 2:
simplified ratio = 212.62 ÷ 2
simplified ratio = 106.31
The empirical formula for the hydrate of K2CO3 is K2CO3·106.31H2O.