A certain metal M forms a bromide containing 53.79 percent Br by mass. What is the chemical form

A certain metal M forms a bromide containing 53.79 percent...

Key Concepts

Percent Composition

Percent composition tells us what fraction of a compound’s total mass is contributed by each element. This concept is used to deduce the masses of individual elements in a given mass of a compound, serving as the starting point for determining its formula.

Atomic Mass and Molar Conversions

Using the known atomic masses of the elements in a compound, one can convert the mass of each element (obtained from the percent composition) to moles. This conversion is critical to compare the amounts of different elements on a mole basis rather than by mass.

Empirical Formula Determination

Once the masses are converted to moles, the next step involves finding the simplest whole number ratio of the elements present. This ratio gives the empirical formula of the compound, which is a simplified representation of its composition.

Stoichiometric Ratios

The stoichiometric ratios derived from the mole values indicate how many atoms of one element combine with atoms of another in a fixed, whole-number ratio. These ratios are essential for constructing the correct chemical formula for the compound.

Transcript

00:01 So for this problem, we are given some certain metal m, and we know that this m, when it forms a bromide, that it contains 53 .79 % br by mass.

00:18 And we want to find the molecular formula for this compound.

00:22 Well, if we just assume that the mass of our entire compound is 100 grams, then we know that the mass of the bromine that we are dealing with.

00:36 With is 53 .79 grams and that the mass of our certain unknown metal m is just the mass of the entire compound subtracted by the mass of the bromine.

00:52 And when you do this, you get that 46 .2 grams of our compound is our unknown metal m.

01:01 So for this problem, we're going to go about guessing and checking.

01:05 Essentially, we're going to assume, or guess the chemical formula for our compound and we're going to see if it fits our description and we're going to go one by one well because we know that the charge on br is minus 1 when it's in an ionic compound we know that the combinations can include m br if m is plus 1 m br 2 if m is plus 2 m br 3 if m is plus 2 and br 3 if m is plus 3 and so on and so forth.

01:40 So let's first assume that the chemical formula is mbr, or in this case, that the m is a plus one charge.

01:51 So if that's the case, then we can do the following.

01:57 Well, we know that we have 53 .79 grams of bromine, and you want to convert this to moles of bromine, so we just divide by the atomic mass, which is 74 .90.

02:14 Now for the multiple ratio, well if we assume that this is our compound, then the multiple ratio is 1 to 1.

02:23 And when we multiply this, you'll get 0 .673 moles of m.

02:31 So that's how many moles of m there would be if our compound was 100 grams.

02:38 So we know that in this compound, that 46 .21 grams of that will be m.

02:44 So we can just take the mass of m and divide it by the moles of m in our compound.

02:55 So we can see what the molar mass of m is and then we can identify the metal.

02:59 Well, if we do this calculation, we'll get 68 .64 grams per mole.

03:05 So that's a smaller mass of a metal m.

03:08 But to verify whether or not this m is a real metal, we have to look at the periodic table and see if there is an element that has a smaller mass...

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