A solution of calcium nitrate, Ca(NO3)2, reacts with a solution of ammonium fluoride, NH4F, to form solid calcium fluoride, CaF2, and ammonium nitrate, NH4NO3(aq). When 45.00 mL of 4.8724 x 10-1 M Ca(NO3)2 solution was added to 60.00 mL of 9.9981 x 10-1 M NH4F solution, 1.524 g of CaF2 was isolated.
Calculate the percent yield of CaF2
This is a limiting reagent (LR) problem.
2NH4F + Ca(NO3)2 ==> CaF2 + 2NH4NO3
1. Write and balance the equation as above.
2. Convert numbers to mols. mols = M x L = ?
mols Ca(NO3)2 = M x L = ?
mols NH4F = M x L = ?
3a. Using the coefficients in the balanced equation, convert mols Ca9NO3)2 to mols CaF2.
3b. Do the same to convert mols NH4F to mols CaF2.
3c. It is likely that the two values for 3a and 3b will not agree which means one of them is not right. The correct value in LR problems is ALWAYS the smaller value.
4. Using the smaller value for mols CaF2 convert to grams. g = mols x molar mass. This is the theoretical yield (TY)
The actual yield (AY) is 1.524 in the problem.
5. %yield = (AY/TY)*100 = ?
To calculate the percent yield of CaF2, we need to compare the actual yield (the amount of CaF2 isolated) to the theoretical yield (the maximum possible amount of CaF2 that could have been obtained).
1. Calculate the moles of CaF2:
The molar mass of CaF2 is calculated by adding the atomic masses of calcium (Ca) and two fluorine (F) atoms.
Molar mass of CaF2 = 40.08 g/mol (Ca) + 2 * 18.99 g/mol (F) = 78.06 g/mol
Now, convert the mass of CaF2 isolated to moles using its molar mass.
Moles of CaF2 = 1.524 g / 78.06 g/mol = 0.0195 mol CaF2
2. Determine the limiting reactant:
To determine the limiting reactant, we need to compare the stoichiometry of the balanced chemical equation with the amount of reactants available.
The balanced chemical equation for the reaction is:
Ca(NO3)2 + 2NH4F -> CaF2 + 2NH4NO3
From the equation, we can see that 1 mole of Ca(NO3)2 reacts with 2 moles of NH4F to produce 1 mole of CaF2.
Moles of Ca(NO3)2 = volume (L) x molarity (mol/L) = 0.04500 L x 0.48724 mol/L = 0.0219252 mol Ca(NO3)2
Moles of NH4F = volume (L) x molarity (mol/L) = 0.06000 L x 0.99981 mol/L = 0.0599886 mol NH4F
From the stoichiometry, we can see that 1 mole of Ca(NO3)2 is equivalent to 1/2 moles of CaF2.
Therefore, the theoretical moles of CaF2 that can be produced are:
0.0219252 mol Ca(NO3)2 x (1/2) = 0.0109626 mol CaF2
Since Ca(NO3)2 is the limiting reactant (it is less in moles than NH4F), the moles of CaF2 formed are limited by the amount of Ca(NO3)2.
3. Calculate the theoretical yield:
The theoretical yield of CaF2 can be calculated using the moles of CaF2 formed and its molar mass.
Theoretical yield of CaF2 = moles of CaF2 x molar mass of CaF2 = 0.0109626 mol x 78.06 g/mol = 0.856 g
4. Calculate the percent yield:
Now, we can calculate the percent yield by dividing the actual yield (1.524 g) by the theoretical yield (0.856 g) and multiplying by 100.
Percent yield = (1.524 g / 0.856 g) x 100 = 178.08%
The percent yield of CaF2 is approximately 178.08%.
To calculate the percent yield of CaF2, you need to compare the actual yield (the amount of CaF2 obtained in the experiment) with the theoretical yield (the amount of CaF2 that should have been obtained according to stoichiometry).
First, let's start by determining the number of moles of CaF2 formed in the reaction. The balanced chemical equation for the reaction can be written as follows:
Ca(NO3)2 + 2 NH4F → CaF2 + 2 NH4NO3
From the equation, we can see that 1 mole of Ca(NO3)2 reacts with 2 moles of NH4F to produce 1 mole of CaF2. Therefore, the number of moles of CaF2 can be calculated using the given volume and molarity of the Ca(NO3)2 solution:
moles of Ca(NO3)2 = volume (in L) × molarity = 0.04500 L × 4.8724 × 10^-1 M = 2.192 × 10^-2 moles
Since the stoichiometric ratio between Ca(NO3)2 and CaF2 is 1:1, the number of moles of CaF2 formed in the reaction is also 2.192 × 10^-2 moles.
Next, we need to determine the theoretical yield of CaF2 in grams. The molar mass of CaF2 can be calculated as:
molar mass of CaF2 = (molar mass of Ca) + 2 × (molar mass of F)
= 40.08 g/mol + 2 × (18.99 g/mol)
= 78.06 g/mol
The theoretical yield of CaF2 can be calculated by multiplying the number of moles of CaF2 by its molar mass:
theoretical yield = moles of CaF2 × molar mass of CaF2
= 2.192 × 10^-2 moles × 78.06 g/mol
= 1.711 g
Now we can calculate the percent yield using the formula:
percent yield = (actual yield / theoretical yield) × 100
Given that the actual yield of CaF2 is 1.524 g, we can substitute these values into the formula:
percent yield = (1.524 g / 1.711 g) × 100
= 89.1%
Therefore, the percent yield of CaF2 is approximately 89.1%.