Chemistry B: Chemistry Calculations
Name: ____________________________ Hour: _____ Page 1
Chemistry B
Chemical
Calculations
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Chemistry B: Chemistry Calculations
Name: ____________________________ Hour: _____ Page 2
Worksheet #1: Limiting Reactants
When chemicals are mixed together to undergo a reaction, they are often mixed in stoichiometric quantities,
that is, in exactly the correct amounts so that all of the reactants “run out” and are used up at the same time.
A good analogy is baking chocolate chip cookies. Imagine you want to make as many cookies as you can. You
have lots of flour, eggs, sugar, and vanilla, but you only have 1 package of chocolate chips. Since the recipe
for a dozen cookies calls for 1 package of chocolate chip cookies, you can only make four dozen cookies. In
this case, you had an excess of all the other ingredients and what determined the amount of cookies (product)
you could make was the amount of chocolate chips. In this example the chocolate chips were the limiting
reactant.
Now let’s look at a chemical example. In the following reaction of making ammonia, two moles of ammonia
would be produced by reacting one mole of nitrogen and three moles of hydrogen.
N2(g) + 3H2(g)  2NH3(g)
1 mole 3 moles 2 moles
Now consider an example where the reactants are not in the correct stoichiometric quantities:
N2(g) + 3H2(g)  2NH3(g)
2 moles 3 moles ? moles
Since you now have twice as much nitrogen, you will use all of the hydrogen and some nitrogen will be left
over (1 mole) so hydrogen limits the amount of product produced. We will call hydrogen the limiting
reactant because the reaction will stop when you run out of hydrogen.
Consider another example where the reactants are not in the correct stoichiometric quantities:
N2(g) + 3H2(g)  2NH3(g)
1 mole 4 moles ? moles
Since you have more hydrogen than is needed, you will use up all of the nitrogen and some hydrogen will be
left over (1 mole) so nitrogen limits the amount of product produced and we call it the limiting reactant. The
reaction will stop when you run out of nitrogen.
How do you know that a problem is a limiting reactant problem and how do you solve it once you have
identified it? The answer to the first question is you know when you are given a limiting reactant problem
when you are given a stoichiometry problem and the masses of both reactants are given. In order to solve
the problem you must follow the following steps”
1. Determine the knowns (what you know/information given) and unknown (what you want to know) and
record them above the equation.
2. Using “what you know” (known masses), determine TWO possible masses produces. Use
stoichiometry and factor label.
3. Compare the two masses and see what reactant is limiting. The limiting reactant runs out first, SO it
limits how much product is produced, therefore the smaller amount of product is produced from the
limiting reactant.
EXAMPLE:
The reaction between solid white phosphorus and oxygen produces tetraphosphorus decaoxide. Determine the
mass in grams of P4O10 formed if 25.0 grams of phosphorus (P4) and 50.0 grams of oxygen are combined.
Balanced Equation = P4 + 5O2  P4O10
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Chemistry B: Chemistry Calculations
Name: ____________________________ Hour: _____ Page 3
1. Record 25 g above P4 and 50 g above O2 in the previous equation.
2. Calculate the number of moles of each reactant:
# moles of P4 = 25.0 g x 1 mole = 0.202 mol P4
123.9 g
# moles of O2 = 50.0 g x 1 mole
32.00 g
3.
Calculate the actual ratio of available moles of O2 and available moles of P4 (AKA: Make this a ration
over 1)
1.56 mol O2 =
0.202 mole P4
4.
= 1.56 mol O2
7.72 moles O2
1 mole P4
Determine the mole ratio of the two reactants from the balanced chemical equation:
5 moles O2
1 mole P4
5.
Since 5 moles of oxygen is needed to react and you have 7.72 moles, the oxygen is in excess and P4 is
the limiting reactant. Use the moles of P4 (0.202 moles) to finish the problem and calculate the
number of grams of P4O10 produced.
# grams of P4O10 = 0.202 mol P4 x 1 mole P4O10 x 283.9 g P4O10
1 mole P4
1 mole P4O10
= 57.3 grams of P4O10
Solve the following problems on a separate sheet of paper following all of the above steps.
1.
The reaction between solid sodium and iron (III) oxide is one in a series of reactions that inflate an
automotive air bag. The two products are sodium oxide and iron. If 100.0 grams of sodium reacts with
100.0 grams of Fe2O3, determine the mass of solid iron produced.
6Na + Fe2O3  3Na2O + 2Fe
2. Photosynthesis reactions in green plants use carbon dioxide and water to produce glucose (C6H12O6)
and oxygen. Calculate the mass of glucose produced if 88.0 grams of carbon dioxide and 64.0 grams of
water are available to a plant growing in sunlight.
6CO2 + 6H2O  C6H12O6 + 6O2
3. Disulfur dichloride is used to vulcanize rubber, a process invented by Charles Goodyear. The process
makes rubber harder, stronger, and less likely to become soft when hot or brittle when cold. In the
production of disulfur dichloride, molten sulfur reacts with chlorine gas to the following equation: S8
+ 4Cl2  4S2Cl2 If 200.0 grams of sulfur reacts with 100.0 grams of chlorine, what mass of disulfur
dichloride is produced?
4. Iron is obtained commercially in a blast furnace by the reaction of hematite (Fe2O3) with carbon
monoxide. The two products are iron and carbon dioxide. How many grams of iron are produced if
75.0 grams of hematite react with 40.0 grams of carbon monoxide?
Fe2O3 + 3CO  2Fe + 3 CO2
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Chemistry B: Chemistry Calculations
Worksheet#2: Calculation of Percent by Mass
Name: ____________________________ Hour: _____ Page 4
An extremely common calculation in chemistry is to find the percent by mass of a component in a compound
or a mixture. As you know, a percent is calculated by making a fraction of “what you are talking about” over
the “whole thing” and then multiplying the fraction by 100.
Example 1: Calculate the percent (%) oxygen by mass in aluminum oxide Al2O3
Step 1: Molar Mass of Al2O3
2 (Al) = 2(26.98) = 53.96
3 (O) = 3(16.00) = 48.00
101.96 g/mol
Step 2: Calculate Percent Mass
%O =
48.00 g O
x
1014.96 g Al2O3
100 = 47.077 = 47.08 % O
Example 2: Calculate the % phosphate ion in calcium phosphate Ca3(PO4)2
Step 1: MM of PO4-3
1 (P) = 1(30.97) = 30.97
4 (O) = 4(16.00) = 64.00
Step 3: Calculate Percent Mass
94.97g/mol % PO4-3 = 189.94 g PO4-3
x 100 = 61.2354 = 61.24% PO4-3
Step 2: MM of Ca3(PO4)2
310.18 g Ca3(PO4)2
3 (Ca) = 3(40.08) = 120.24
2 (PO4-3) = 2(97.97)= 189.94
310.18 g/mol
Solve the following problems showing all work just as in the examples above.
1. Calculate the percent by mass for the named element in the following compounds: (see example 1)
a. Sodium in sodium chloride NaCl
b. Silver in silver oxide Ag2O
c. Oxygen in hydrogen peroxide H2O2
d. Calcium in calcium borate Ca3(BO3)2
e. Nitrogen in aluminum nitrate Al(NO3)3
2. Calculate the percent by mass for the named polyatomic ion in each of the following compounds: (see
example 2)
f. carbonate CO3-3 in potassium carbonate K2CO3
g. ammonium NH4+1 in ammonium sulfate(NH4)2SO4
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Chemistry B: Chemistry Calculations
Name: ____________________________ Hour: _____ Page 5
Worksheet #3: Calculation of Simplest Formula or Empirical Formula
A formula is the ratio of atoms of each element present in a compound; since moles of elements contain equal
numbers of atoms it follows that formulas are also mole ratios. A simplest or empirical formula is the smallest
whole number ratio of the atoms of each elements present. For example: H2O says there are two atoms of
hydrogen for each atom of oxygen in water.
Example Problem: A compound is found to contain 36.11% calcium and 63.88% chlorine by mass. What is
the simplest formula of this compound?
Solution: Assume a 100.00 gram sample (that means you have 36.11 grams of calcium and 63.88 grams of
chlorine).
# moles Ca = 36.11 g Ca x 1 mole Ca = 0.90094 moles Ca (keep extra digits)
40.08 g Ca
# moles Cl = 63.887 g Cl x 1 mole Cl = 1.8019 moles Cl
35.45 g Cl
We then write the formula by dividing through by the smallest number.
Ca0.90094 Cl1.8019 = CaCl2
0.90094
0.90094
Calculate the simplest formula for the following compounds showing all work as done in the example:
1. 35.94% aluminum; 64.06% sulfur
2.
74.185% sodium; 25.815% oxygen
3.
77.73% iron; 22.27% oxygen
4.
26.6% potassium; 35.4% chromium; and 38.0% oxygen
5.
43.64% phosphorus; 56.36% oxygen
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Chemistry B: Chemistry Calculations
Name: ____________________________ Hour: _____ Page 6
Worksheet #4: Calculation of Molecular Formulas
For the vast majority of ionic compounds the simplest or empirical formula is the only formula that is possible.
However, with covalent compounds, many times the formula for the molecule is a whole number multiple of
the simplest formula.
For example:
Simplest formula
CH
CH
CH2
Acetylene
Benzene
Propene
Molecular formula
C2H2
C6H6
C3H6
So we need to learn to calculate the molecular formula.
EXAMPLE: Calculate the simplest and molecular formula for a gaseous hydrocarbon that is 85.60% carbon
and has a molecular weight of 70.1 grams per mole.
A. First we calculate the simplest formula and find out how many grams are in a simplest formula
unit.
100.00% CH – 85.60% C = 14.40% H
Assume 100.00 gram sample of the compound. This means you have 85.60 g C and 14.40 g H.
# moles of C = 85.60 g C x 1 mole C
= 7.12739 moles C
12.01 g C
# moles of H = 14.40 g H x 1 mole H
= 14.2574 moles H
1.01 g H
Don’t worry about significant figures here when you are finding moles of each element because when you are
done with a formula calculation you always end up with whole number ratios. So do not round off the number
of moles at this last step.
B. Divide by the smallest number of moles:
C7.12739
H14.2574
7.12739
7.12739
CH2 so we know the simplest formula us CH2
Add up the mass of one simplest formula, this is the grams in a simplest formula unit or SFU
Molar Mass of CH2
1 (C) = 1(12.01) = 12.01
2 (H) = 2(1.01) = 2.02
14.03 grams/SFU
C. Now calculate the actual molecular formula, you are trying to find out the number of times the
simplest formula divides into the molecular weight, and you will show a neat factor label step for
this.
#SFU/mol = 70.1 grams
Mole
x 1 SFU = 4.996 = 5 SFU/mol
14.03 grams
So the molecular formula is (CH2) x 5 = C5H10
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Chemistry B: Chemistry Calculations
Name: ____________________________ Hour: _____ Page 7
EXAMPLE PROBLEM #1
Calculate the simplest formula and the molecular formula for a hydrocarbon that is 84.09% carbon and has a
molecular weight of 114.24 grams per mole.
A. Calculate the simplest formula
B. Calculate the number of grams in a simple formula unit for this compound.
C. Calculate the molecular formula.
D. Write the balanced chemical equation for the complete combustion of this hydrocarbon.
On a separate sheet of paper, for each of the following compounds calculate: the simplest formula and
the molecular formula.
1.
An oxide of nitrogen is 46.68% nitrogen and 53.32% oxygen by mass. Its molecular weight is 60.03
grams per mole.
2.
A gaseous hydrocarbon is found to be 92.24% carbon. It has a molecular weight of 26.01 grams per
mole.
3. An oxide of phosphorus is 43.64% phosphorus and has a molecular weight of 283.78 grams per mole.
4.
A gaseous hydrocarbon is 82.63% carbon and has a molecular weight of 58.08 grams per mole.
5.
A compound is found to have 40.00% carbon, 6.72% hydrogen, and 53.28% oxygen by mass. Its
molecular weight is 60.06 grams per mole.
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Chemistry B: Chemistry Calculations
Worksheet #5: Introduction to Hydrates
Name: ____________________________ Hour: _____ Page 8
Hydrates are compounds which contain molecules of water within their crystal structures. In these
compounds, the positive and negative ions of the compound are surrounded by a definite number of water
molecules when the compound is in its solid crystalline state. The compounds are called hydrates and the
water is called water of hydration or water of crystallization. When we write the formulas for these
compounds we indicate the number of water molecules in the crystal by following the formula with a dot and
then the appropriate number of water. Examples:
copper (II) sulfate pentahydrate
CuSO4 * 5H2O (s)
magnesium sulfate heptahydrate
MgSO4 * 7H2O (s)
calcium chloride dihydrate
CaCl2 * 2H2O (s)
1 = mono
2 = di
3 = tri
4 = tetra
5 = penta
Prefixes
6 = hexa
7 = hepta
8 = octa
9 = nona
10 = deca
In the writing of formulas for hydrated compounds the dot is NOT a multiplication sign – it is simply the
method used to indicate the water of hydration. The (s) is called a state symbol and means the compound is a
solid. NOTICE also, the (s) comes at the end of the formula – the whole formula including the molecules of
water represent one substance.
When we heat hydrated salts we drive off their water of hydration as water vapor and we are left with the
dehydrated salt (or it may be referred to as the anhydrous salt). Dehydrated is a term that means “water has
been removed”. Anhydrous is a term that means “no water is present”. Dehydrated and anhydrous are not
exactly synonymous; both mean no water present but the dehydrated says that water was once there and has
been removed, anhydrous says there is no water present but it doesn’t tell anything about the history of the
material.
Many times we will do laboratory experiments where we heat a hydrate to drive off its water of hydration. We
can represent this process by an equation with state symbols.
Example: copper (II) sulfate pentahydrate is heated until the anhydrous salt is formed.
CuSO4 * 5H2O (s)  CuSO4 (s) + 5H2O (g)
Write the equations including state symbols for dehydrating the following substances.
1. magnesium sulfate heptahydrate
4. sodium carbonate decahydrate
2. calcium chloride dihydrate
5. calcium sulfate dihydrate
3. barium chloride dihydrate
6. sodium thiosulfate pentahydrate
1.
___________________________________________________________________
2.
___________________________________________________________________
3. ____________________________________________________________________
4. ____________________________________________________________________
5. ____________________________________________________________________
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Chemistry B: Chemistry Calculations
Worksheet #6: Percent Mass of Hydrates
Name: ____________________________ Hour: _____ Page 9
In worksheet #5 we learned that many salts form crystals with water trapped in their crystal structure and that
we call salts hydrated and the water is called water of hydration of water of hydration. It is frequently valuable
to know what percent by mass of a crystal is water and what percent is actually the anhydrous salt. We can
calculate these values from the formula.
Example: Calculate the percent of water by mass in copper (II) sulfate pentahydrate CuSO4 * 5H2O (s)
Step #1: MM of CuSO4 * 5H2O (s)
Step #2: Calculate Percent Mass
Cu = 63.54
% H2O =
%H2O
x 100
S = 32.06
CuSO4 * 5H2O
4 O = 64.00
5H2O = 90.10
= 90.10 g x 100 = 36.083% H2O
Total = 249.70 g
249.70 g
USE THIS EXACT SET UP TO CALCULATE THE % OF WATER BY MASS FOR EACH OF THE
SIX HYDRATES ON WORKSHEET #5
1.
2.
3.
4.
5.
6.
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Chemistry B: Chemistry Calculations
Name: ____________________________ Hour: _____ Page 10
Chemistry Calculations Review Sheet
This review covers: % Composition by mass, simplest formula, molecular formula and limiting reactant s
Solve the problems on another sheet of paper. Show all work for credit.
1. A compound was found to contain 41.33% arsenic by mass and 58.67% chlorine by mass. Calculate
the simplest formula for this compound. (arsenic trichloride, AsCl3)
2. An oxide of iron was analyzed and a 12.00 gram sample of the oxide was found to contain 8.39 grams
of iron. Calculate the simplest formula for this compound. (iron(III) oxide, Fe2O3)
3. Calculate the percent by mass for each element in aluminum hydroxide Al(OH)3.
% O and 3.88% H)
(34.59 % Al, 61.53
4. A student was given a sample of scandium oxide Sc2O3 to analyze for the percent composition by mass.
She found that a 23.00 gram sample of the scandium oxide contained 15.00 grams of scandium.
Calculate the percent by mass for each element in this compound. (65.22% Sc and 34.8% O)
5. A hydrocarbon was found to contain 84.09% carbon by mass and 15.91% hydrogen by mass. This
compound has a molecular weight of 114 grams per mole. Calculate the simplest formula for the
compound and the molecular formula for the compound. (C4H9 and C8H18)
6. Nitrogen and hydrogen react to form ammonia NH3. If 300.0 grams of nitrogen react with 50.0 grams
of hydrogen and they were heated under pressure in the presence of a catalyst, how many grams of
ammonia are produced? (281 grams of ammonia)
7. If 50.0 grams of aluminum is mixed with 70.0 liters of chlorine gas Cl2 at STP, what mass of solid
product would be formed? (247 grams)
Extra Challenge:
8. The old fashioned soda-acid fire extinguishers contained a pound of baking soda NaHCO3 dissolved in
water and a container of sulfuric acid H2SO4 which when inverted mixed with the baking soda and
produced carbon dioxide CO2 which put out the fire. If the acid bottle contained 250.0 grams of
sulfuric acid, what volume of carbon dioxide would be produced at STP? (114 L CO2)
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