(14) Weight of Compound Made

Lesson 10

(14) Weight of Compound Made

Obj. 14. From the name of a compound, determine the weight of the compound that can be made from specified amounts of each element.

Again we are dealing with synthesis. But now we need to find out how much product can be made from a certain amount of reactants - the elements. The tricky part here is that we are given two amounts of elements that are reacting with one another and those may or may not be the proper amounts to react with one another to make the product. In other words, we are dealing with limiting reagent problems.

Exercises

Although they don't say so, the following questions are limiting reagent problems.

a. How much copper(II) sulfide can be made from 4.20 g of copper and 2.40 g of sulfur?

b. How much sulfur dioxide can be made by burning 0.14 g of sulfur in a flask that contains 0.24 g of oxygen?

c. How much potassium chloride can be made from 2.50 g of potassium and 2.50 g of chlorine?

d. How much potassium bromide can be made from 2.50 g of potassium and 2.50 g of bromine?

e. Starting with 5.0 g of iron and 8.0 g of chlorine, how much ferrous chloride could be made?

f. Starting with 5.0 g of iron and 8.0 g of chlorine, how much ferric chloride could be made?

g. How much magnesium iodide can be made from 1.20 g each of magnesium and iodine?

Worked-Out Examples (f,g)

(f) "Starting with 5.0 grams of iron and 8.0 grams of chlorine, how much ferric chloride could be made?" In order to do this, we need the formulas for the chemicals involved and a balanced equation based on those formulas.

In this case we have a reaction in which iron reacts with chlorine to make ferric chloride. Start by writing the unbalanced skeleton equation. Iron, Fe, reacts with chlorine (a diatomic element), Cl₂, to form ferric chloride. Ferric chloride is an ionic compound. Iron is a transition metal which can form more than one oxidation state. In ferric the "ic" ending indicates that we are dealing with the higher charge. Because the two possible charges on an iron ion are +2 and +3, ferric is the +3. So we have iron in a +3 state. Chlorine is a nonmetal which likes to gain one electron, because there is room for just one more electron in its last energy level. So iron with a +3 charge and chlorine with a -1 charge allows us to figure out that the formula for ferric chloride is FeCl₃. All of that gives us the equation Fe + Cl₂ FeCl₃. Next comes the task of balancing this equation. When that is done we get 2 Fe + 3 Cl₂ 2 FeCl₃.

iron + chlorine

ferric chloride

Fe + Cl₂

FeCl₃

Fe + 3 Cl₂

2 FeCl₃

The next thing we need are some weight relationships. We can get those from the formula weights and the number of moles. The mole interpretation of this equation is that two moles of Fe reacts with three moles of Cl₂ to give two moles of FeCl₃. Two moles of iron weighs 111.6 grams. Three moles of Cl₂ weighs 213.0 grams. Two moles of FeCl₃ weighs 324.6 grams. So the weight relationships are that 111.6 grams of iron reacts with 213.0 grams of chlorine to make 324.6 grams of FeCl₃.

3 Cl₂

2 FeCl₃

111.6 g

213.0 g

324.6 g

The question here is "How much FeCl₃ can be made from 5.0 grams of iron and 8.0 grams of chlorine?" One approach would be to figure out what the limiting reagent is. In this case it is not obvious (to me) what the limiting reagent is because the weights of chlorine and iron are more or less in the right proportions. (It looks like chlorine might be the limiting reagent. If you were very confident about that, you could simply do the calculation based on chlorine.) One approach that can be used in such a situation is this: calculate the amount of FeCl₃ that can be made from 5.0 grams of iron and then calculate the amount of FeCl₃ that can be made from 8.0 grams of chlorine. The lesser of these two amounts will be the amount of FeCl₃ (ferric chloride) that can actually be made.

To do this, start with 5.0 grams of Fe and multiply by a conversion factor that has 324.6 grams of FeCl₃ on the topand 111.6 grams of Fe on the bottom. That comes out to 14.543 grams of FeCl₃. Since we started with two significant digits, round the answer off to 15 grams of FeCl₃. That's the amount of ferric chloride that can be made from 5.0 grams of iron.

5.0 g Fe

324.6 g FeCl₃
111.6 g Fe

15 g FeCl₃

Next, figure how much ferric chloride could be made from 8.0 grams of chlorine. Start with 8.0 grams of Cl₂ and multiply that by a conversion factor that has 324.6 grams of FeCl₃ on the top and 213.0 grams of Cl₂ on the bottom. Rounding off to two significant digits, that comes to 12 grams of FeCl₃.

8.0 g Cl₂

324.6 g FeCl₃
213.0 g Cl₂

12 g FeCl₃

Compare the amounts that can be made: 5.0 grams of iron is enough to make 15 grams of ferric chloride; 8.0 grams of chlorine is enough to make 12 grams of ferric chloride. So the amount that will actually be made would be 12 grams of ferric chloride.

(g) "How much magnesium iodide can be made from 1.20 grams each of magnesium and iodine?" Again, we need to have a balanced equation so that we can get the proper weight relationships between the chemicals. In order to get the balanced equation we need to know the correct formulas.

We start with magnesium and iodine to make magnesium iodide. For magnesium we use Mg. Iodine is one of the diatomic elements so we use I₂. The formula for magnesium iodide is based on the fact that magnesium is a metal which likes to lose two electrons and iodine is a nonmetal that likes to gain one electron. Because we will need to have two iodines for every magnesium, the formula for magnesium iodide is MgI₂. So our skeleton equation is Mg + I₂

MgI₂. It is also the balanced equation for this reaction. The formula weights give the weight relationships for the reaction: 24.3 grams of magnesium reacts with 253.8 grams of iodine to make 278.1 grams of magnesium iodide.

magnesium + iodine magnesium iodide

I₂

MgI₂

24.3 g

253.8 g

278.1 g

In this case it is fairly easy to figure out the limiting reagent. We have the same quantity (1.20 grams each) of magnesium and iodine available. But the weight relationships show that it takes more than 10 times as much iodine, by weight, as magnesium. We are obviously going to run out of iodine and iodine will be the limiting reagent. In this case it is only necessary to do the one calculation. The amount of magnesium iodide formed will be dependent on the limiting reagent iodine. So the calculation is this: 1.20 grams of I₂ multiplied by the conversion factor that has 278.1 grams of MgI₂ over 253.8 grams of I₂. Carrying out that calculation, we get 1.31 grams of MgI₂.

1.20 g I₂

278.1 g MgI₂
253.8 g I₂

1.31 g MgI₂

Answers to Exercises

Although they don't say so, the following questions are limiting reagent problems.
The limiting reagent in each question has been highlited.

a. How much copper(II) sulfide can be made from 4.20 g of copper and 2.40 g of sulfur? 6.32 g CuS

b. How much sulfur dioxide can be made by burning 0.14 g of sulfur in a flask that contains 0.24 g of oxygen? 0.28 g SO₂

c. How much potassium chloride can be made from 2.50 g of potassium and 2.50 g of chlorine? 4.76 g KCl

d. How much potassium bromide can be made from 2.50 g of potassium and 2.50 g of bromine? 3.72 g KBr

e. Starting with 5.0 g of iron and 8.0 g of chlorine, how much ferrous chloride could be made? 11 g FeCl₂

f. Starting with 5.0 g of iron and 8.0 g of chlorine, how much ferric chloride could be made? 12 g FeCl₃

g. How much magnesium iodide can be made from 1.20 g each of magnesium and iodine? 1.31 g MgI₂

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