As a food chemist for a major potato chip company, you are responsible for deter

Question

As a food chemist for a major potato chip company, you are responsible for determining the salt content of new potato chips products for the packaging label. The potato chips are seasoned with table salt, NaCl. You weight out a handful of the chips, boil them in water to extract the salt, and then filter the boiled chips to remove the soggy chip pieces. You then analyze the chip filtrate for Cl- concentration using the Mohr method.

First you prepare a solution of silver nitrate, AgNO3, and titrate it against .500 g of KCl using the Mohr method. You find that it takes 62.2 mL of AgNO3 to reach the eqiuvalence point of the reaction.

You then use the same silver nitrate solution to analyze the chip filtrate in a Mohr reaction, finding that the solution yields a rusty brown precipitate when 46.1 mL of titrant is added.

Question: If the same sample of chips used to make the filtrate weighed 94.0g, how much NaCl is present in one serving (145g) of chips?

Explanation / Answer

As a food chemist for a major potato chip company, you are responsible for determining the salt content of new potato chip products for the packaging label. The company has just developed a new product, "Super Spud Chips", which are seasoned with table salt, NaCl. You weigh out a handful of the chips, boil them in water to extract the salt, and then pour the extract through a filter to remove the soggy chip pieces. You then analyze the chip filtrate for Cl^- concentration using the Mohr method.
You enter the lab to analyze the chip filtrate for Cl^-. First, you prepare a solution of silver nitrate, AgNO3 , and titrate it against 0.500 grams of KCl using the Mohr method. You find that it takes 63.9 mL of AgNO3 titrant to fully reach the equivalence point of the reaction.

You then use the same silver nitrate solution to analyze the chip filtrate in a Mohr reaction, finding that the solution yields a rusty brown precipitate when 46.1 mL of titrant is added.

If the sample of chips used to make the filtrate weighed 85.5 grams , how much NaCl is present in one serving (115 grams) of chips?

find the Molarity of your AgNO3 if it takes 63.9 mL of AgNO3 titrant against 0.500 grams of KCl:

first find moles of KCl , using its Molar mass:
0.500 g KCl @ 74.551 g/mol = 6.707e-3 moles KCl

since AgNO3 & KCl react 1 mole to 1 mole:
6.707e-3 moles KCl reacts with 6.707e-3 moles AgNO3

find Molarity of your 63.9 ml solution of AgNO3:
6.707e-3 mol AgNO3 / 0.0639 litres = 0.1050 Molar AgNO3
======================================...

You then analyze 85.5 grams of chips

find the moles in 46.1 mL of AgNO3 solution added:
0.0461 litres @ 0.1050 Molar AgNO3 = 4.8386e-3 moles of AgNO3

since NaCl reacts with AgNO3 in a 1 mole to 1 mole ratio:
4.8386e-3 mol AgNO3 reacted with 4.8386e-3 moles of NaCl

find grams of NaCl analyzed, using its molar mass:
4.8386e-3 mol NaCl @ 58.44 g/mol = 0.02828 grams of NaCl

find the salt in a 115 gram serving:
115 grams chips @ 0.0282 grams of NaCl / 85.5 grams analyzed= 0.03803 grams of salt

in this 3 sig fig problem your answer is 0.0380 grams of salt are in a 115 gram serving

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