Yuria B.
asked • 10/19/18Finding the empirical formula of an unknown metal chloride using gravimetric analysis
Balanced Equation: AgNo3 (aq) + *MCl (s) —> AgCl (s) + MNO3 (aq)
*”M”stands for the unknown metal
Mass of filter paper: 1.140 grams
Mass of precipitate and filter paper: 1.2999 grams
What is the empirical formula of the metal chloride? What is the identity of the unknown metal?
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1 Expert Answer
Lauren H. answered • 10/19/18
Tutor
4.8
(24)
7 years experience teaching High School Chemistry and Honors Chemistry
1.2999 grams - 1.140 grams = .1599 g precipitate
.1599 g AgCl x 1 mole AgCl/143.32 g AgCl = .0011156852 mol AgCl
I'm going to call this 1.116 x 10^-3 mol
AgNo3 (aq) + *MCl (s) —> AgCl (s) + MNO3 (aq)
Can't go further...
Are you sure this is the entire problem?
M could be Li, or Na. or K... etc. But, the identity of M does not affect the amount of AgCl made..
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Ishwar S.
Yuria.... Is the starting mass of MCl given, but you forgot to include it in the question? Without this mass, the question cannot be answered past the point from Lauren’s answer. If you have it, kindly provide the value otherwise you can re-post the question.10/19/18