Note: The calculations shown on this page sometimes show numbers that have been rounded off to converve space. However, the full unrounded values have been used to calculate the final answer, and then the final answer has been rounded back to the appropriate number of significant figures.

Problem 1

Al_{2}O_{3} FWT
= 2(26.981539)
+ 3(15.9994)
=

53.963178 +
47.9982 = 101.9613780

100% x __53.963178__
= 52.93%

101.9613780

Problem 2

Compound of Zn and O with 80.34 % Zn. Assume you have 100.00 g of the compound so there will be 80.34 g Zn.

100.00 g cmpd

- 80.34 g Zn

____________

19.66 g O

80.34 g Zn x __1 mol Zn __
= 1.229 mol Zn

65.39 g Zn

19.66 g O x __1 mol O __
= 1.229 mol O

15.9994 g O

Divide both numbers in the ratio by the smaller of the two numbers. In this case the numbers are the same, so we divide both numbers by themselves:

For Zn: __1.229__
= 1
For O __1.229__
= 1

1.229
1.229

Therefore, the empirical formula is ZnO

Problem 3

5.000 g of a compound containing only C and H, of which 3.994 g is C.

5.000 g cmpd

- 3.994 g C

____________

1.006 g H

3.994 g C x __1 mol C __
= 0.3325 mol C

12.011 g C

1.066 g H x __1 mol H __
= 0.9981 mol H

1.00794 g H

Divide both numbers in the ratio by the smaller of the two numbers:

__0.3325__ =
1 __
0.9981__ =
3

0.3325
0.3325

Therefore the empirical formula is CH_{3}

Problem 4

CO(g) +
3H_{2}(g) ---------->
CH_{4}(g) +
H_{2}O(g)

9.912 g
Excess
? g

9.912 g CO x __1 mol CO __
x __1 mol CH___{4} x
__16.04 g CH___{4}
= 5.677 g CH_{4}

28.01 g CO
1 mol CO
1 mol CH_{4}

Problem 5

CO(g) +
3H_{2}(g) ---------->
CH_{4}(g) +
H_{2}O(g)

9.912 g
Excess
? g

9.912 g CO x __1 mol CO __
x __1 mol H___{2}__O__
x __18.02 g H___{2}__O__
= 6.375 g H_{2}O

28.01 g CO
1 mol CO 1
mol H_{2}O

Problem 6

CO(g) +
3H_{2}(g) ---------->
CH_{4}(g) +
H_{2}O(g)

9.912 g
? g used

9.912 g CO x __1 mol CO __
x __3 mol H___{2}__ __
x __2.016 g H___{2}__ __
= 2.140 g H_{2}

28.01 g CO
1 mol CO 1 mol H_{2}

Problem 7

The limiting reactant always determines the amount of product that can be produced. That's because the limiting reactant is the reactant whoose supply runs out first. Once a reactant is missing, the chemical reaction can no longer proceed. Therefore, you can make no more product than the amount that has been produced as of the moment the limiting reactant runs out.

Problem 8

The reactant with the smallest mass is not necessarily the limiting reactant. Substances react on a molar basis rather than a mass basis. The more important question is not which reactant has the smallest mass but which reactant has the smallest number of moles. If all reactants in the balanced chemical equation react in a one to one mole ratio, then the reactant present in the smallest number of moles will be the limiting reactant.

However, if the stoichiometry is not one to one, then you can't even be sure that the reactant present in the smallest number of moles is limiting. As an example, consider the reaction

2H_{2}(g) +
O_{2}(g) ---------->
2H_{2}O(g)

Suppose you have 1.5 moles of H_{2} and 1 mole of O_{2}.
You have fewer moles of O_{2} but O_{2} is not the limiting
reactant, it is the excess reactant. Looking at the stoichiometry,
you see that 1 mole of O_{2} requires 2 moles of H_{2}
to be completely used up. Since we have only 1.5 moles of H_{2}
we will run out of H_{2} before we use up all the O_{2}.
So H_{2} is the limiting reactant, even though there are more moles
of H_{2} (1.5) than O_{2} (1).

If you are wondering just how many moles of O_{2} would be left
over, we can easily reason that out. We needed 2 moles of H_{2}
but have only 1.5 moles or 75% of what we need. Therefore, we can
use up only 75% of the O_{2}. We have 1 mole of O_{2}
so we can use up 0.75 moles and 0.25 moles will be left over.

You can determine the limiting reactant by dividing the number of moles of each reactant by the coefficient of that reactant in the balanced equation. The reactant that gives the smallest quotient is the limiting reactant. This method is most convenient when you know the number of moles of the reactnats.

In most of the problems you will solve in general chemistry, you will be given reactant masses rather than number of moles. That's why in class, I taught you to just calculate for each reactant, the mass of one of the products that can can be produced by that reactant. The reactant that calculates the smallest mass of product is the limiting reactant. Of course, you have to calculate the mass of the SAME PRODUCT for each of the reactants in order to have a meaningful comparison. If there is more than one product in the reaction, it does not matter which product you use, as long as you use the same one for each reactant. Sometimes, the problem will ask you for the mass of one of the products, and it then makes sense to use THAT product as the one you calculate for all reactants. That way, you accomplish two things at the same time -- determining which reactant is the limiting reactant and determining what mass of product can be produced.

Problem 9

The excess reactant calculates the larger amount of product. Of course, you can not actually obtain that much product, because the limiting reactant will have run out before you get that much product. Always remember that you can never obtain a larger mass of any product than the limiting reactant will allow you to produce.

Problem 10

N_{2}(g) +
O_{2}(g) ---------->
2NO(g)

5.113 g
8.976 g
? g

5.113 g N_{2}(g) x __1 mol
N___{2}__ __ x
__2 mol NO __ x __30.01
g NO__ =
10.953 g NO

28.01 g N_{2} 1 mol N_{2}
1 mol NO

8.976 g O_{2}(g) x __1 mol
O___{2}__ __ x
__2 mol NO__ x __30.01
g NO__ =
19.229 g NO

28.01 g N_{2} 1 mol O_{2}
1 mol NO

The above calculations show that N_{2} is the limiting reactant,
since it restricts us to making a smaller mass of NO than does the O_{2}.
Therefore, 10.953 g NO is the maximum mass of NO that can be produced in
this chemical system, and producing this mass of NO will require completely
using up our supply of N_{2}.

Problem 11

As stated above, N_{2} is the limiting reactant, because it
calculates the smaller mass of NO.

Problem 12

This is a trivial problem if you understand problem 10. We identified
N_{2} as the limiting reactant -- this means that it is completely
used up. Therefore, there will be NONE of it left when the reaction
is complete. It's mass will be 0.000 g.

Problem 13

N_{2}(g) +
O_{2}(g) ---------->
2NO(g)

5.113 g
? g used

5.113 g N_{2} x __1 mol N___{2}__ __
x __1 mol O___{2}__ __
x __32.00 g O___{2}__ __
= 5.840 g O_{2}

28.01 g N_{2} 1 mol N_{2}
1 mol O_{2}_{}

It is important to note that STOICHIOMETRY CAN ONLY CALCULATE REACTIVE
MASSES. Therefore, the 5.840 g O_{2} is the amount USED,
not the amount left over. We can, however, easily find the amount
left over by subtraction:

8.976 g O_{2}

- 5.840 g O_{2}

____________

3.136 g O_{2}_{}

So, assuming the reaction goes to completion (completely exhausting
the available supply of N_{2}) there will be 3.136 g O_{2}
left over that can not be used because no more N_{2} is available.

Problem 14

percent yield = 100%
x __actual yield__

theoretical yield

percent yield = 100%
x __2.087 g __ =
19.05%

10.953 g

Therefore, we only obtained 19.05 % of the maximum possible amount we could have obtained. A 100% yield would have given 10.953 g NO, as calculated in problem 10.

*This page was last modified Tuesday April 13, 1999*