Suppose that 0.205 g of a mixture containing Na2SO4 and K2SO4 is dissolved in water. An excess
of BaCl2 is added and a precipitate of 0.298 g of BaSO4 is produced. Calculate the percent by mass
of Na2SO4 and K2SO4 in the original mixture.
This is a two equation problem and the two are to be solved simultaneously. These are difficult to explain by writing so I'll do it in steps.
Let x = mass Na2SO4
and y = mass K2SO4
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equation 1 is x + y = 0.205
The second equation comes from the ppt of BaSO4. The idea is that you set up the mass BaSO4 from x (the Na2SO4) and the mass BaSO4 from y (the K2SO4), add them together and the sum will be 0.298g BaSO4. To avoid typing and the space used up let MM = molar mass and AM = atomic mass.
amount from Na2SO4 is x*(MM BaSO4/MM Na2SO4)
amount from K2SO4 is y(MM BaSO4/MM K2SO4)
Therefore, the second equation is
x*(MM BaSO4/MM Na2SO4) + y(MM BaSO4/MM K2SO4) = 0.298
Solve those two equations for x and y, then
%Na2SO4 = (x/mass sample)*100 = ?
%K2SO4 = (y/mass sample)*100 = ?
Post your work if you stuck.
To calculate the percent by mass of Na2SO4 and K2SO4 in the original mixture, follow these steps:
Step 1: Determine the molar mass of BaSO4
BaSO4 is composed of barium (Ba) and sulfate (SO4). The molar mass of Ba is 137.33 g/mol, and the molar mass of SO4 is 96.06 g/mol (32.07 g/mol for sulfur and 4 * 16.00 g/mol for oxygen). Therefore, the molar mass of BaSO4 is:
(1 * 137.33 g/mol) + (1 * 32.07 g/mol) + (4 * 16.00 g/mol) = 233.39 g/mol
Step 2: Calculate the number of moles of BaSO4
The number of moles of BaSO4 can be calculated using the formula:
moles = mass / molar mass
moles = 0.298 g / 233.39 g/mol = 0.001275 mol
Step 3: Determine the stoichiometry ratio
Looking at the balanced chemical equation between BaSO4 and Na2SO4/K2SO4:
BaCl2 + Na2SO4 -> BaSO4 + 2NaCl
BaCl2 + K2SO4 -> BaSO4 + 2KCl
You can see that one mole of BaSO4 is produced from one mole of Na2SO4 or one mole of K2SO4. Therefore, the 0.001275 mol of BaSO4 corresponds to the same number of moles of Na2SO4 and K2SO4 in the original mixture.
Step 4: Calculate the mass of Na2SO4 and K2SO4
The total mass of Na2SO4 and K2SO4 in the mixture can be calculated using the formula:
mass = moles * molar mass
For Na2SO4:
mass of Na2SO4 = 0.001275 mol * (2 * 22.99 g/mol + 32.07 g/mol + 4 * 16.00 g/mol) = 0.0975 g
For K2SO4:
mass of K2SO4 = 0.001275 mol * (2 * 39.10 g/mol + 32.07 g/mol + 4 * 16.00 g/mol) = 0.2010 g
Step 5: Calculate the percent by mass of Na2SO4 and K2SO4
The percent by mass can be calculated using the formula:
percent by mass = (mass component / total mass) * 100%
For Na2SO4:
percent by mass of Na2SO4 = (0.0975 g / 0.205 g) * 100% ≈ 47.56%
For K2SO4:
percent by mass of K2SO4 = (0.2010 g / 0.205 g) * 100% ≈ 98.05%
Therefore, the percent by mass of Na2SO4 in the original mixture is approximately 47.56%, and the percent by mass of K2SO4 is approximately 98.05%.
To calculate the percent by mass of Na2SO4 and K2SO4 in the original mixture, we'll need to use the concept of stoichiometry and the principles of mass and molar ratios.
Step 1: Calculate the number of moles of BaSO4 precipitated.
To do this, we'll use the molar mass of BaSO4, which is 233.39 g/mol:
Number of moles of BaSO4 = Mass of BaSO4 / Molar mass of BaSO4
= 0.298 g / 233.39 g/mol
Step 2: Determine the stoichiometric ratio between BaSO4 and Na2SO4/K2SO4.
From the balanced chemical equation, we know that 1 mole of BaSO4 is formed from 1 mole of Na2SO4 or K2SO4. Therefore, the ratio is 1:1.
Step 3: Calculate the number of moles of Na2SO4/K2SO4 in the mixture.
Since the stoichiometric ratio is 1:1, the number of moles of Na2SO4/K2SO4 is also equal to the number of moles of BaSO4:
Number of moles of Na2SO4/K2SO4 = Number of moles of BaSO4
Step 4: Calculate the mass of Na2SO4/K2SO4 in the mixture.
To do this, we'll assume that the mass of the mixture does not change during the reaction:
Mass of Na2SO4/K2SO4 = Mass of BaSO4 formed
Step 5: Calculate the percent by mass of Na2SO4 and K2SO4 in the original mixture.
To calculate the percent by mass of each compound, we'll use the following formula:
Percent by mass = (Mass of compound / Mass of mixture) x 100
Now, let's plug in the values:
Mass of Na2SO4/K2SO4 = Mass of BaSO4 formed = 0.298 g
Mass of mixture = 0.205 g
Percent Na2SO4 = (Mass of Na2SO4 / Mass of mixture) x 100
Percent K2SO4 = (Mass of K2SO4 / Mass of mixture) x 100
Since the stoichiometry is 1:1, we can safely assume that the mass of Na2SO4 = mass of K2SO4.
Percent Na2SO4 = Percent K2SO4 = (0.298 g / 0.205 g) x 100
Hence, the percent by mass of Na2SO4 and K2SO4 in the original mixture is approximately 145.4%.