A 0.972g sample of a CaCl2*2H2O/K2C2O4*H2O solid salt mixture is dissolved in ~ 150ml of deionized water, previously adjusted to a pH that is basic. The precipitate, after having been filtered and air-dried, has a mass of 0.375 g. The limiting reactant in the salt mixture was later determined to be CaCl2*2H2O. What is the percent by mass of CaCl2*2H2O in the salt mixture?
I assume the ppt, if done properly, is CaC2O4.
Percent CaCl2*2H2O = (mass CaCl2*2H2O/mass sample)*100 =
(mass CaCl2*2H2O/0.972 g)*100 = see below.
The ppt is CaC2O4. Convert 0.375 g CaC2O4 to moles CaC2O4, then convert to moles CaCl2*2H2O, then to grams CaCl2*2H2O. Plug that number into the percent equation to calculate percent CaCl2*2H2O in the sample.
Check my thinking.
To find the percent by mass of CaCl2*2H2O in the salt mixture, we need to calculate the mass of CaCl2*2H2O present in the original sample.
First, let's calculate the moles of CaCl2*2H2O using its molar mass:
Molar mass of CaCl2*2H2O = (40.08 g/mol) + (2 * 35.45 g/mol) + (2 * 18.02 g/mol) = 147.02 g/mol
Moles of CaCl2*2H2O = (mass of CaCl2*2H2O) / (molar mass of CaCl2*2H2O) = 0.375 g / 147.02 g/mol
Next, let's calculate the moles of K2C2O4*H2O using the balanced chemical equation:
CaCl2*2H2O + K2C2O4*H2O -> CaC2O4*H2O + 2KCl
From the balanced chemical equation, we can see that 1 mole of CaCl2*2H2O reacts with 1 mole of K2C2O4*H2O to produce 1 mole of CaC2O4*H2O. Therefore, the moles of K2C2O4*H2O can be equated to the moles of CaCl2*2H2O.
Moles of K2C2O4*H2O = Moles of CaCl2*2H2O = 0.375 g / 147.02 g/mol
Now, let's calculate the mass of the K2C2O4*H2O in the sample:
Mass of K2C2O4*H2O = Moles of K2C2O4*H2O * Molar mass of K2C2O4*H2O
Molar mass of K2C2O4*H2O = (2 * 39.10 g/mol) + (2 * 12.01 g/mol) + (4 * 16.00 g/mol) + 18.02 g/mol = 214.20 g/mol
Mass of K2C2O4*H2O = (0.375 g / 147.02 g/mol) * 214.20 g/mol
Finally, to calculate the percent by mass of CaCl2*2H2O in the salt mixture, we can use the following formula:
Percent by mass = (Mass of CaCl2*2H2O / Total mass of the salt mixture) * 100
Total mass of the salt mixture = mass of CaCl2*2H2O + mass of K2C2O4*H2O
Percent by mass of CaCl2*2H2O = (Mass of CaCl2*2H2O / (mass of CaCl2*2H2O + mass of K2C2O4*H2O)) * 100
Now substitute the values and calculate the percent by mass of CaCl2*2H2O.
To determine the percent by mass of CaCl2*2H2O in the salt mixture, we need to calculate the mass of CaCl2*2H2O in the sample.
1. Calculate the number of moles of CaCl2*2H2O:
- The molar mass of CaCl2*2H2O = (1 mole of Ca) + (2 moles of Cl) + (2 moles of H2O)
- The molar mass of CaCl2*2H2O = (40.08 g/mol) + (2 * 35.45 g/mol) + (2 * 18.02 g/mol)
- The molar mass of CaCl2*2H2O = 147.02 g/mol
- Moles of CaCl2*2H2O = Mass of CaCl2*2H2O / Molar mass of CaCl2*2H2O
- Moles of CaCl2*2H2O = 0.375 g / 147.02 g/mol
2. Calculate the number of moles of K2C2O4*H2O:
- Since the limiting reactant is CaCl2*2H2O, we need to determine the moles of K2C2O4*H2O present.
- Moles of K2C2O4*H2O can be calculated using stoichiometry and the balanced equation.
3. Calculate the percent by mass of CaCl2*2H2O:
- Percent by mass = (Mass of CaCl2*2H2O / Total mass of mixture) * 100
- Total mass of mixture = Mass of CaCl2*2H2O + Mass of K2C2O4*H2O
By following these steps, you will be able to find the percent by mass of CaCl2*2H2O in the salt mixture.