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(12) Weight of Elements Needed for Synthesis
Obj. 12. From the name of a compound, determine the
weight of each element needed to make a certain amount of it.
The specified amount of the compound to be made might be in moles or in grams. There
are a number of ways of going about solving problems like this. Two approaches are shown
in the examples that follow the exercises. One is to first come up with a balanced
equation and get the weight relationships from the balanced equation. The other
is to work with weight percentages. The balanced equation approach is
shown first and the weight percent approach is shown after that.
Exercises
How much of each element is needed to synthesize the indicated amount of each of the
following compounds?
a. 0.24 moles of magnesium bromide
b. 3.2 moles of sulfur trioxide
c. 17.0 g of sodium chloride
d. 14.2 g of calcium oxide
e. 4.33 g of potassium nitrate
f. 50.0 g of lithium oxide
g. 20.0 g of lead(IV) oxide
Worked-Out Examples (b,c) Using Balanced Equations
| (b) Synthesize 3.2 moles of sulfur trioxide. First we need an equation
that represents the formation of sulfur trioxide. Sulfur trioxide has the formula SO3.
So we write down SO3 on the right side of the equation. It is made from the
elements sulfur and oxygen. For sulfur we write down S and for oxygen, because it's one of
the diatomic elements, we write down the formula O2. |
| Synthesis of sulfur trioxide |
| S |
+ |
O2 |
 |
SO3 |
|
|
| In balancing this equation, the oxygen is what's going to cause the
problems because the oxygen occurs in pairs on the left and trios on the right. The lowest
number of oxygens that will allow us to do that will be six; so we need three O2's
on the left and two SO3's on the right. Two SO3 gives us six oxygens
and it also gives us two sulfurs. So we need a 2 in front of the S on the left. The
balanced equation is 2 S + 3 O2
2 SO3. |
| 2 S |
+ |
3 O2 |
|
2 SO3 |
|
|
| The mole interpretation of this balanced equation is that two moles of
sulfur react with three moles of oxygen to form two moles of SO3. Since we need
to find the weights of sulfur and oxygen that are required to make a
certain number of moles of SO3, we need to know how much two moles of sulfur
weigh and how much three moles of oxygen weigh. Two moles of sulfur, each mole of sulfur
weighs 32.1 grams so two moles of sulfur would weigh 64.2. Three moles of O2
weigh 96.0 grams. The weight for two moles of sulfur trioxide is shown but will not be
needed. |
| 2 S |
+ |
3 O2 |
|
2 SO3 |
| 64.2 g |
|
96.0 g |
|
160.2 g |
|
|
| That gives us the relationships that we need. 64.2 grams of sulfur reacts
with 96.0 grams of oxygen to make two moles of SO3. The question asks,
"How much of these two elements will be needed to make 3.2 moles of SO3?"
Start the calculation with 3.2 moles of SO3. Then multiply by a conversion
factor that will determine how many grams of sulfur are related to that, which is 64.2
grams of sulfur over two moles of SO3. Carrying out that calculation, we get
103 grams of sulfur. Since 3.2 moles only has two significant digits, we need to round
this off to get 1.0 x 102 grams of sulfur. By switching over to scientific
notation we can show significant digits without having to worry about 0's holding the
decimal place. |
| 3.2 moles SO3 |
x |
64.2 g S
2 moles SO3 |
= |
103 g S |
|
|
|
= |
1.0 x 102 g S |
|
|
| We also need to know how many grams of oxygen are needed. So 3.2 moles of
SO3 times a conversion factor that has 96.0 grams of oxygen on the top and 2
moles of SO3 on the bottom gives us 154 grams of O2. Again that
should be written as a two significant digit number. To avoid having a zero as a decimal
place holder, change it to scientific notation and round it to two digits. The answer is
1.5 x 102 grams of O2. |
| 3.2 moles SO3 |
x |
96.0 g O2
2 moles SO3 |
= |
154 g of O2 |
|
|
|
= |
1.5 x 102 g O2 |
|
|
(c) Synthesize 17.0 grams of sodium chloride. We want to know how much of each element
is needed. The elements, of course, are sodium and chlorine. First let's figure out a
balanced equation for this reaction. Sodium (Na) and chlorine (Cl2) are going
to react to form sodium chloride(NaCl).
| Balancing this equation is a fairly simple task and the balanced equation
is shown. The mole interpretation of this equation is two moles of sodium reacts with one
mole of Cl2 and forms two moles of NaCl. Using the atomic weights and the
formula weights we can figure that two moles of sodium weighs 46.0 grams, one mole of Cl2
weighs 71.0 grams, and two moles of NaCl weighs 117.0 grams. We now have the weight
relationship that 46.0 grams of sodium reacts with 71.0 grams of chlorine to form 117.0
grams of sodium chloride. |
| Synthesis of sodium chloride |
| 2 Na |
+ |
Cl2 |
|
2 NaCl |
|
|
|
The questions given are "How much sodium do we need to make 17.0 grams of sodium
chloride?" and "How much chlorine do we need to make 17.0 grams of sodium
chloride?"
| To figure the weight of sodium needed, start with 17.0 grams of NaCl and
multiply by a conversion factor that has 46.0 grams of sodium over 117.0 grams of sodium
chloride. Multiplying that through, gives 6.68 grams of sodium. |
| 17.0 g NaCl |
x |
46.0 g Na
117.0 g NaCl |
= |
6.68 g Na |
|
|
| We also want to find out how much chlorine is needed. To do that start
with 17.0 grams of NaCl times the conversion factor that has 71.0 grams of chlorine on the
top and 117.0 grams of sodium chloride on the bottom. Multiplying that through we get 10.3
grams of chlorine. |
| 17.0 g NaCl |
x |
71.0 g Cl2
117.0 g NaCl |
= |
10.2 g Cl2 |
|
|
Worked-Out Example (c) Using Weight Percentage
These problems can also be figured out by using percentages. The percentage approach
can be used here because there is only one product. Here's how you would use that
approach.
| Use the formula of sodium chloride to figure out the weight percentage of
sodium and chlorine in the compound. The formula, of course, is NaCl. The atomic weight of
sodium is 23.0; the atomic weight of chlorine is 35.5. We can use these to get the formula
weight of 58.5 grams in one mole of NaCl. So one mole of NaCl weighs 58.5 grams and it
contains 23.0 grams of sodium and 35.5 grams of chlorine. |
| Na |
23.0 |
| Cl |
35.5 |
| NaCl |
58.5 |
|
| From those figures we can come up with percentage of chlorine and sodium
in the compound. The percentage of sodium is 23.0 divided by 58.5 to give 39.3% sodium.
The percentage of chlorine is 35.5 out of 58.5 grams total, and so that comes out to be
60.7% chlorine in the compound. |
| |
Fraction |
Percent |
| Na |
23.0
58.5 |
39.3 % |
| Cl |
35.5
58.5 |
60.7 % |
|
| The amount of sodium that is needed to make 17.0 grams of sodium chlorine
will be 39.3% of 17.0 grams. That comes out to be 6.68 grams of sodium. The amount of
chlorine needed will be 60.7% of the 17.0 grams, which comes out to be 10.3 grams of
chlorine. |
| Na |
39.3 % of 17.0 g = 6.68 g |
| Cl |
60.7 % of 17.0 g = 10.3 g |
|
Keep in mind that this percentage approach can (only) be used when only one product is
formed.
Answers to Exercises
How much of each element is needed to synthesize the indicated amount of each of the
following compounds?
a. 0.24 moles of magnesium bromide requires 5.8 g of magnesium and 38 g of bromine
b. 3.2 moles of sulfur trioxide requires 1.0 x 102 g of sulfur and 1.5 x 102
g of oxygen
c. 17.0 g of sodium chloride requires 6.68 g of sodium and 10.3 g of chlorine
d. 14.2 g of calcium oxide requires 10.2 g of calcium and 4.05 g of oxygen
e. 4.33 g of potassium nitrate requires 1.67 g of potassium, 0.600 g of nitrogen and
2.06 g of oxygen
f. 50.0 g of lithium oxide requires 23.3 g of lithium and 26.8 g of oxygen
g. 20.0 g of lead(IV) oxide requires 17.3 g of lead and 2.68 g of oxygen
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E-mail instructor:
Eden Francis
Clackamas Community College
©1998, 1999 Clackamas Community College, Hal Bender
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