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Chapter 3: Problem 104

The iodide ion in a \(1.552 \mathrm{g}\) sample of the ionic compound MI is removed through precipitation. The precipitate is found to contain \(1.186 \mathrm{g}\) I. What is the element M?

Short Answer

Expert verified
The element 'M' in the ionic compound MI is 'Lithium'.

Step by step solution

01

Calculate the mass of M

First, calculate the mass of the unknown element (M) in the ionic compound by subtracting the mass of iodide from the total mass of the compound. Use the formula \( M_{compound}= M_{M}+ M_{I}\). Therefore, \(M_{M}= M_{compound} - M_{I}\). The mass of iodide (I) is given as 1.186g and the mass of the compound is 1.552g. Therefore, \(M_{M}= 1.552g - 1.186g = 0.366g \).
02

Calculate the mole of I and M

To assign the element M, calculate the number of moles for iodide (I) and the unknown element M respectively. The number of moles is typically calculated using the formula \( \text{n} = \frac{M}{M_r} \), where \(M\) is the mass and \(M_r\) is the molar mass. However, in this case, they are in a 1:1 ratio as the formula of the compound is MI. Therefore, the moles of both, M and I will be the same.
03

Calculate the molar mass of M

The molar mass of M can be calculated using the formula: \( M_{r} = \frac{M}{n} \). Since the moles of I and M are the same, they cancel out, and the molar mass of M equals the mass of M, which is 0.366g.
04

Identify the element M

Using a periodic table, identify the element with the molar mass closest to 0.366g/mol. The atomic mass is usually given in atomic mass units (amu), but when considering a full mole of atoms, it's equivalent to g/mol. Here, the element M corresponds to 'Lithium (Li)' since the atomic mass of Lithium is approximately 0.366g/mol (7 amu).

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