Calculate the mass percent of each mixture component. .

Question

sample of mixture containing only sodium chloride and potassium chloride has a mass of 4.000 g. When this sample is dissolved in water and excess silver nitrate is added, a white solide (silver chloride) forms. After filtration and drying, the solid silver chloride has a mass of 8.5904 g. Calcualte the mass percent of each mixture component.

Here is what I have so far:

NaCl (aq) KCl (aq) AgNO3 (aq) --> AgCl (s) (rest not important)

AgCl has an atomic mass of 143.35 g/mol so using molar ratios there are 0.059926 mol AgCl

I tried to use stoichiometry using the 1:1 molar ratio of NaCl to AgCl and the atomic mass of NaCl (58.44 g/mol) to get 3.5021 g NaCl

Using the same molar ratio of 1:1 of KCl to AgCl and the atomic mass of KCl (74.55 g/mol) to get 4.4675 g KCl and then...

I got stuck because the number of grams of KCl and NaCl did not add to what the problem said.

Answer 1

still trying to post an answer...

You know that the mass of the NaCl + the mass of the KCl = 4.000 g.

Let the mass of the NaCl = X

Let the mass of the KCl = Y

Then X + Y = 4.000 g.

All of the Chloride must come from either the NaCl or KCl.  The mass of  AgCl =  8.5904 g.

8.5904 g / 143.35 g/mole = 0.0599 moles of AgCl

Every mole of AgCl requires one mole of either NaCl or KCl

Moles of NaCl + moles KCl =  0.0599 moles

[X / 58.44 g/mol] + [Y / 74.55 g/mol]  =  0.0599 moles

but Y = 4.000 g – X

[X / 58.44 g/mol] + [(4.000 – X) / 74.55 g/mol]  =  0.0599 moles

[X *  74.55] + [(4.000 – X) *  58.44] =  0.0599 *  74.55 *  58.44

74.55 X – 58.44 X + (4 * 58.44) =  16.11 X + 233.76 = 260.97

16.11 X = 260.97 -  233.76 = 27.21

X = 27.21 / 16.11 =  1.6890 g

Y = 4.000 g – X = 4.000 – 1.6890 = 2.3110 g

Check:

(1.6890 g / 58.44 g/mol) + ( 2.3110 g /  74.55 g/mol) = 0.0289 + 0.0310 = 0.0599

for percents

NaCl =  (1.6890 g / 4.000 g) * 100 = 42.22 percent

KCl =  (2.3110 g / 4.000 g) * 100 = 57.78 percent

Answer 2

8.5904 g AgCl x 1 mol AgCl/143.3209 g AgCl

=0.059938 mol AgCl

=0.059938 mol Cl

(Y x 58.4424 g/mol)+(Z x 74.551 g/mol)=4.00 g

Y+Z=0.059938 mol

Solve the system of equation. For simplicity I will remove the units and just work with the numbers.

58.4424Y+74.551Z=4

Y+Z=0.059938

multiply the second equation by -58.4424 so we can eliminate Y.

58.4424Y+74.551Z=4

-58.4424Y-58.4424Z=-3.5029336

combine the two equations to eliminate Y

16.1086Z=0.497066

Z=0.0308572081 mol KCl

Y+Z=0.059938

Y+0.0308572081=0.059938

Y=0.0290810144 mol NaCl

So we had 0.0309 mol KCl and 0.0291 mol NaCl in the original mixture.

Convert that to grams. I like using nonrounded values so I will use that long decimal number.

0.0308572081 mol KCl x 74.551 g KCl/mol KCl=2.300 g KCl

NaCl=4.000 g-2.3000 g=1.700 g NaCl (or you could work it out like we did for KCl)

Mass% KCl=(2.300/4.000)*100=57.5%

Mass% NaCl=(1.700/4.000)*100=42.5%