A sample of an iron chloride compound contained 69.97g of Fe and 111.11g of Cl ......

Question

Suppose a second sample was found to contain 21.66g of Fe. Assuming the compound follows the law of definite proportions, then what mass of Cl would be present in this sample?

Answer 1

To answer this question, the Law of Definite Proportions can be used simply as a ratio situation. We do need to be able to determine Empirical Formulas and Molecular Formulas from the masses of the elements provided, but this question is not really asking for that. The Law states that if it is the same compound, the relative ratios of moles (and therefore mass) of each element will be the same.

So for this situation we can say that in the original sample, the Cl to Fe mass ratio was

111.11 g Cl : 69.97 g Fe

The new sample contains only 21.66 g Fe but it is the same compound, so the ratio is

x g Cl : 21.66 g Fe

We can use a simple proportion 111.11 / 69.97 = x / 21.66 and solve for x

x = 34.39534944 on my calculator, round to 4 sig figs = 34.40 g of Cl in the new sample

To check this, you can see that 111.11 g Cl / 69.97 g Fe is 1.588 g Cl per g of Fe (when you divide the numbers). Which means your Cl mass should always be 1.588 times as big as your Fe mass in the same compound. 21.66 g Fe in the new sample should have 1.588 x 21.66 g = 34.40 g Cl with it.

Let me know if you need help with Percent Composition, Empirical Formula, and Molecular Formula determination, and their relationships to each other - good luck!

Answer 2

First thing to understand here is what is meant by the law of definite proportions. The law states that no matter where what portion of a specific compound you're working with you will always have the same ratio of elements. For instance if a compound had 30& of Hydrogen and 70% of C that would mean that that same proportion of atoms can be found in any other sample of that same compound. ( note those percentages refer to mols of those atoms not mass)

Next, we need to determine the atom proportion within sample #1 in order to be able to find the Cl proportion in sample #2.

Note: molar relationships are important here since they are how we properly determine atom proponents within molecules.

1st: Convert the measured mass (g) for both Fe and Cl into mols by using the molecular weights of the two elements.

Fe: 69.97 g * 1 mol/ 55.93 g = 1.251 mol Fe

Cl: 111.11 g * 1 mol/ 35.45 g = 3.1343 mol Cl

Now determine the molar relationship between the two elements in order to acquire an accurate chemical formula:

Fe: 1.251 mol / 1.251 mol = 1

Cl: 3.1343 mol / 1.251 mol = 2.5

We can't have a fraction of an atom so multiply both numbers by two:

Fe: 1 *2 = 2

Cl: 2.5 * 2 = 5

Now we have our chemical formula: Fe2Cl5 (in other words now we know that our unknown compound actually has the chemical formula Fe2Cl5, 2 Fe atoms for every 5 Cl atoms)

Our sample #2 here is the same compound as sample #1 and we now know the proper chemical formula thus we know the mass ratio for Fe and Cl within the compound.

Note: mol is just another way of saying atom or molecule in the chemistry sense at least, since we know that there are 2 Fe atoms for every 5 Cl atoms we can also say that there are 2 mols Fe for every 5 mols Cl.

First find out how many mols of Fe in sample #2 with the given information:

Fe: 21.66 g * 1 mol/ 55.93 g = 0.38727 mol Fe

Cl: mass unknown

There are now a couple ways we can go about this but I believe the simplest approach comes from understanding the atom ratios of the chemical formula Fe2Cl5.

Since there are 7 atoms total in the formula Fe2Cl5, we can say that 2/7 of the compound is composed of Fe and 5/7 is composed of Cl.

0.38727 mol * (7/2) = 1.3554 mol (total mols in Fe2Cl5)

1.3554 mol * (5/7) = 0.96818 mol Cl

Now multiply mol Cl by its molecular weight to acquire mass of Cl in sample.

0.96818 mol * 35.45 g/ 1 mol = 34.32 g Cl

Answer: Cl mass = 34.32 g

To prove that this answer is correct you can determine the Fe/Cl ratio for sample #2 just like with sample #1 in order to find a chemical formula which you will find is the same as like from sample #1.