Anonymous

A sample of 0.600 mol of a metal M reacts completely with excess fluorine to form 46.8 g of MF2.

(a) How many moles of F are in the sample of MF2 that forms?

(b) How many grams of M are in this sample of MF2?

(c) What element is represented by the symbol M?

Relevance

Because your question states that the 0.600 mol of metal M reacts COMPLETELY and because there is one equivalent of M in your molecule MF2, you know that the 46.8g of MF2 product is 0.600 mol.

a. Because there are two equivalents of fluorine for every MF2, there are twice as many moles of F as there are of MF2:

2(0.600 mol MF2) = 1.20 mol F in the sample product formed.

b. In order to find how many grams of M are in the sample product we need to find how much of the mass is F first. We know that we have 1.20 mol F, so how many grams F is that? To find this use the atomic mass of fluorine is which approximately 18.9984 g/mol. This is from your periodic table. Then: (1.20 mol F)(18.9984 g/mol F) = 22.80g F

Subtract the total mass of the fluorine in the sample from the mass of the sample and you will get the mass of M: 46.8g MF2 - 22.80 g F = 24g M

c. You know you have 24 g M in 0.600 mol. This gives an atomic mass of 24g/0.600mol = 40g/mol. Referencing the periodic table the element with the atomic mass of 40 is Calcium. So M = Calcium (and your compound is CaF2).

To find the

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• (a) 1.200 mol MF2 means a 1:2 ratio of M to F so .600 x 2 is 1.200

(b) 46.8g/mol/.600mol MF2 = 78g/mole F is 19.0 g/moles and there are 2 so 2x 19.0g = 38.0g

78.0mole/g [MF2] - 38.0mole/g [F2] = 40g/mole [M]

(c) M hass a molar mass of 40.0 grams per mole which is the molar mass of calcium so M is Ca

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