Chemistry 1

Volume 2

Worksheet 12
Percent Yield in Chemical Reactions

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1. Calculate the percent yield of a reaction that had a theoretical yield of 3.76 g and an actual
yield of 1.45 g.

2. If 2.5 g of K3PO4 was produced in the reaction below and the percent yield was 45%, what
was the theoretical yield?
3 KOH + H3PO4 à K3PO4 + 3 H2O

3. If the actual yield of a reaction was 0.567 g, and the theoretical yield was 0.750 g, what was
the percent yield of this reaction?

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4. If the percent yield of a reaction was 65.0% and its theoretical yield was 0.652 g of CO2 what
was the actual yield?

5. Calculate the theoretical yield of a reaction that produced 4.5 g of a product with a 38%
yield.

6. Calculate the percent yield of a reaction that produced 5.78 g NH4, if its theoretical yield
was 6.78 g.

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7. Calculate the percent yield of a reaction that produced 0.350 mol HCl if the theoretical yield
was 15.36 g.

8. Calculate the percent yield of H2 of a reaction that produced 0.050 moles H2 from 4.5 g HCl
and excess Al in the reaction below:
2 Al + 6 HCl à 2 AlCl3 + 3 H2

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9. Use the following equation to answer the questions that follow.
Fe + H2SO4 à Fe2(SO4)3 + H2
a. Balance the equation.

b. Calculate the theoretical yield of Fe2(SO4)3 if 1.4 g Fe and 3.4 g H2SO4 are reacted.

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 c. If the percent yield of Fe2(SO4)3 in the reaction was 67%, what was the actual yield?

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 Answer Key
1. Calculate the percent yield of a reaction that had a theoretical yield of 3.76 g and an actual
yield of 1.45 g.

To answer this, we just use the following equation:

!"#$%& ()*&+
Percent yield = ,-*./*#)"%& ()*&+ x 100

Substitute what we know (theoretical yield and actual yield).

0.23 4
Percent yield = 5.67 4 x 100

Don’t forget to multiply the decimal by 100 to convert it to a percentage.

Percent yield: 0.386 x 100 = 38.6%

Correct answer 38.6%

2. If 2.5 g of K3PO4 was produced in the reaction below and the percent yield was 45%, what
was the theoretical yield?
3 KOH + H3PO4 à K3PO4 + 3 H2O
!"#$%& ()*&+
Percent yield = ,-*./*#)"%& ()*&+ x 100

We just need to substitute what we know (percent yield and actual yield) into the
equation and solve for theoretical yield. Convert percent yield to a decimal so that we
can work with it.
8.3 4
45% = ,-*./*#)"%& ()*&+ x 100

8.3 4
0.45 = ,-*./*#)"%& ()*&+

Rearrange and solve for theoretical yield.

8.3 4
Theoretical yield = 9.23

Theoretical yield = 5.6 g

Correct answer: 5.6 g

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3. If the actual yield of a reaction was 0.567 g, and the theoretical yield was 0.750 g, what was
the percent yield of this reaction?

!"#$%& ()*&+
Percent yield = x 100
,-*./*#)"%& ()*&+

9.376 4
Percent yield = 9.639 4 x 100

Percent yield = 0.756 X 100 = 75.6 %

Correct answer: 75.6%

4. If the percent yield of a reaction was 65.0% and its theoretical yield was 0.652 g of CO2 what
was the actual yield?

!"#$%& ()*&+
Percent yield = ,-*./*#)"%& ()*&+ x 100

!"#$%& ()*&+
65% = 9.738 4
x 100
Actual yield = (0.65)(0.652 g)
Actual yield = 0.424 g

Correct answer: 0.424 g

5. Calculate the theoretical yield of a reaction that produced 4.5 g of a product with a 38%
yield.

!"#$%& ()*&+
Percent yield = ,-*./*#)"%& ()*&+ x 100

2.3 4
38% = ,-*./*#)"%& ()*&+ x 100

2.3 4
Theoretical yield = 9.5:

Theoretical yield = 12 g
Correct answer: 12 g

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6. Calculate the percent yield of a reaction that produced 5.78 g NH4, if its theoretical yield
was 6.78 g.
5.78 g NH4 is the actual yield. Any time you see “yield” or “produced,” assume that is
the actual yield.
3.6: 4
Percent yield = 7.6: 4 x 100

Percent yield = 0.853 x 100 = 85.3%

Correct answer: 85.3%

7. Calculate the percent yield of a reaction that produced 0.350 mol HCl if the theoretical yield
was 15.36 g.
To answer this, we need to convert the mol HCl into grams of HCl so that the units are
the same and cancel in the percent yield calculation.
0.350 mol HCl 36.46 g HCl = 12.8 g HCl
1 mol HCl

08.: 4
Percent yield = 03.57 4 x 100

Percent yield = 0.833 x 100 = 83.3 %

Correct answer: 83.3%

8. Calculate the percent yield of H2 of a reaction that produced 0.050 moles H2 from 4.5 g HCl
and excess Al in the reaction below:
2 Al + 6 HCl à 2 AlCl3 + 3 H2
First, we need to calculate our theoretical yield. Since Al is in excess, HCl is the limiting
reagent. Since the actual yield of H2 is given in moles in the problem, we can save
ourselves time and just convert 4.5 g HCl into moles H2.
4.5 g HCl 1 mol HCl 3 mol H2 = 0.062 mol H2
36.46 g HCl 6 mol HCl

9.939 ;.& <

Percent yield = 9.978 ;.& <!
!

Correct answer: 81%

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9. Use the following equation to answer the questions that follow.
Fe + H2SO4 à Fe2(SO4)3 + H2
a. Balance the equation.

Fe + H2SO4 à Fe2(SO4)3 + H2

Reactants Products
Fe 1 Fe 2
H 2 H 2
SO4 1 SO4 3

Step 1:
2 Fe + H2SO4 à Fe2(SO4)3 + H2
Reactants Products
Fe 2 Fe 2
H 2 H 2
SO4 1 SO4 3

Step 2:
2 Fe + 3 H2SO4 à Fe2(SO4)3 + H2
Reactants Products
Fe 2 Fe 2
H 6 H 2
SO4 3 SO4 3

Step 3:
2 Fe + 3 H2SO4 à Fe2(SO4)3 + 3 H2
Reactants Products
Fe 2 Fe 2
H 6 H 6
SO4 3 SO4 3

Correct answer: 2 Fe + 3 H2SO4 à Fe2(SO4)3 + 3 H2

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 b. Calculate the theoretical yield of Fe2(SO4)3 if 1.4 g Fe and 3.4 g H2SO4 are reacted.

1.4 g Fe 1 mol Fe 1 mol Fe2(SO4)3 399.88 g Fe2(SO4)3 = 5.0 g Fe2(SO4)3

55.845 g Fe 2 mol Fe 1 mol Fe2(SO4)3

3.4 g H2SO4 1 mol H2SO4 1 mol Fe2(SO4)3 399.88 g Fe2(SO4)3 = 4.6 g Fe2(SO4)3
98.079 g H2SO4 3 mol H2SO4 1 mol Fe2(SO4)3

Correct answer: 4.6 g Fe2(SO4)3

c. If the percent yield of Fe2(SO4)3 in the reaction was 67%, what was the actual yield?

!"#$%& =)*&+
67% = 2.78 4
x 100
(0.67)(4.6 g) = 3.1 g

Correct answer: 3.1 g

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Theoretical Yield & Percent Yield 
1. Given the following equation:

Al2(SO3)3 + 6 NaOH Æ 3 Na2SO3 + 2 Al(OH)3

a) If you start with 389.4 g of Al2(SO3)3 and you isolate 212.4 g of Na2SO3, what is your percent yield for this reaction?

2. Given the following equation:

Al(OH)3 (s) + 3 HCl (aq) Æ AlCl3 (aq) + 3 H2O (l)

a) If you start with 50.3 g of Al(OH)3 and you isolate 39.5 g of AlCl3, what is the percent yield?

3. Given the following equation:

K2CO3 + HCl Æ H2O + CO2 + KCl

a) Balance the equation.

b) Determine the theoretical yield of KCl if you start with 34.5 g of K2CO3.
c) Starting with 34.5 g of K2CO3, and you isolate 3.4 g of H2O, what is the percent yield?

4. Given the following equation:

H2SO4 + Ba(OH)2 Æ BaSO4 + H2O

a) If 98.0 g of H2SO4 is reacted with excess Ba(OH)2, determine the percent yield of BaSO4 if you isolate 213.7 g of
BaSO4.

5. Given the following equation:

CaCl2 + Li3PO4 Æ LiCl + Ca3(PO4)2

a) If you start with 82.4 g of CaCl2 and you isolate 82.4 g of Ca3(PO4)2, what is your percent yield for this reaction?

6. Given the following equation:

Cr(OH)3 + HI Æ CrI3 + HOH

a) If you start with 50.3 g of Cr(OH)3 and you isolate 39.5 g of CrI3, what is the percent yield?
Limiting Reagent, then % Yeild calculations 

Reaction 1: NaCl + H2SO4 Æ NaHSO4 + HCl

a) If a reaction vessel contains 10.0 g of sodium chloride and 12.0 g of sulfuric acid, what is the
limiting reactant?
b) What is the theoretical yield of hydrochloric acid?
c) What is the percent yield if 4.45g of HCl is produced?

Reaction 2: C7H6O3 + C4H6O3 Æ C9H8O4 + C2H4O2

a) When 2.00 g of C7H6O3 is heated with 4.00 g of C4H6O3, what is the limiting reagent?
b) What is the theoretical yield (in grams) of aspirin, C9H8O4
c) If the actual yield of aspirin is 2.21g, what is the percentage yield?

Reaction 3: CuCl2 + NaNO3 Æ Cu(NO3)2 + NaCl

a) If 15 grams of copper (II) chloride react with 20 grams of sodium nitrate, What is the limiting
reagent?
b) How many grams of sodium chloride are formed?
c) If 11.3 grams of sodium chloride are formed, what is the percent yield of this reaction?
 Chapter 12 Worksheet 2 (ws12.2)
Concentration Units

molarity (M)

moles of solute / liter of solution (most commonly used unit of concentration)

molality (m)

moles of solute / kg of solvent

Compare to molarity and notice differences in the denominator. (L vs kg; solution vs. solvent)
Molality is independent of temperature so it is used for calculating boiling point elevation
and freezing point depression.

mole fraction (X) and mole %

moles of component / total moles of all components

A component is any substance in the solution so it can be a solute or the solvent.

mole % = mole fraction x 100

mass fraction and mass (or weight) percent (%)

mass fraction = mass of component / total mass of soln

mass % = mass fraction x 100

parts per million/billion/trillion (ppm/ppb/ppt)

ppm = (mass fraction) x 106

ppb = (mass fraction) x 109
ppt = (mass fraction) x 1012

Used for very dilute solutions (for example pollutants in drinking water). Notice that mass percent could
also be called parts per hundred.

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When an ionic compound dissolves in water, it dissociates into its component ions. So how do we
specify the concentration of such a solution? Is it the concentration of the cation? The concentration of
the anion? The sum of these two concentrations? IT IS NONE OF THESE. Unless stated otherwise,
the concentration of a solution of an ionic compound is the concentration of the original undissociated
solute.

1. In a 2 M solution of magnesium phosphate:

a. What is the molarity of magnesium ions?

b. What is the molarity of phosphate ions?

c. In this chapter you will learn about properties of solutions that depend on the total
concentration of solute particles. What is the total molarity of particles in this solution (the
sum of the two concentrations above)?

2. What is the molarity of an ammonium carbonate solution if the concentration of ammonium ions
is 2 M? What is the concentration of carbonate ions and what is the total concentration of solute
particles?

3. A solution was made by dissolving 800.0 g of NaOH in 2.00 L of water. Calculate the molality,

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 mole fraction, mass % and ppm of NaOH in this solution. (You will need the density of water.)

4. Consider a 0.300 M solution of NaOH?

a. Calculate the molality of NaOH. (Assume the density of the solution is the same as the density of
pure water.)

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b. Notice that the molality is a little bigger than molarity. Why are these two numbers similar and
why is molality always greater than molarity? Why can’t one convert molality to molarity
without additional information?

5. What is the molarity of pure water at 4oC? (Hint: Use the density and molar mass of water.)
What happens to the concentration of water when a solute is dissolved in it?

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 Worksheet #14
Limiting Reagents
1. Potassium superoxide, KO2, is used in rebreathing masks to generate oxygen according to the
reaction below. If the mask contains 0.150 mol KO2 and 0.100 mol water, how many moles
of oxygen can be produced? What is the limiting reagent?
4KO2(s) + 2H2O(ℓ) → 4KOH(s) + 3O2(g)

2. Suppose 13.7 g of C2H2 reacts with 18.5 g O2 according to the reaction below. What
is the mass of CO2 produced? What is the limiting reagent?
2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(ℓ)

3. Nitrogen gas can react with hydrogen gas to form gaseous ammonia. If 4.7 g of nitrogen
reacts with 9.8 g of hydrogen, how much ammonia is formed? What is the limiting reagent?

4. One of the most common acids found in acid rain is sulfuric acid. Sulfuric acid is formed
when gaseous sulfur dioxide reacts with ozone (O3) in the atmosphere to form gaseous sulfur
trioxide and oxygen. The sulfur trioxide forms sulfuric acid when it comes in contact with
water. If 5.13 g of sulfur dioxide reacts with 6.18 g of ozone, how much sulfur trioxide is
formed? What is the limiting reagent?

5. Another way that sulfuric acid is formed in the atmosphere is when sulfur dioxide reacts with
oxygen in a reaction catalyzed by dust in the atmosphere to form sulfur trioxide. If 7.99 g of
sulfur dioxide reacts with 2.18 g of oxygen, how much sulfur trioxide can form? What is the
limiting reagent?

Determining Excess Reactants

6. In the reaction in problem #5 above, how much of the excess reactant remains after all of the
limiting reactant has reacted?

7. Heating together the solids NH4Cl and Ca(OH)2 can generate ammonia. Aqueous CaCl2 and
liquid H2O are also formed. If a mixture of 33.0 g each of NH4Cl and Ca(OH)2 is heated,
how many grams of NH3 will form? What is the limiting reagent? Which reactant remains in
excess, and in what mass?

8. Nitrogen monoxide is formed primarily in car engines, and it can react with oxygen to form
gaseous nitrogen dioxide. Nitrogen dioxide forms nitric acid when it comes in contact with
water, another component of acid rain. If 3.13 g of nitrogen monoxide reacts with 4.16 g of
oxygen, how much nitrogen dioxide will form? What is the limiting reagent? Which reactant
remains in excess, and in what mass?

Percent Yield
9. Liquid nitroglycerine (C3H5(NO3)3) is a powerful explosive. When it detonates, it produces a
gaseous mixture of nitrogen, water, carbon dioxide, and oxygen. What is the theoretical
yield of nitrogen 5.55 g of nitroglycerine explodes? If the actual amount of nitrogen obtained
is 0.991 g, what is the percent yield of nitrogen?

Worksheet 14 1
10. Solid copper(I) oxide reacts with oxygen to form copper(II) oxide. If 4.18 g of copper(I)
oxide reacts with 5.77 g of oxygen, what is the theoretical yield of copper(II) oxide? If the
actual amount of copper(II) oxide obtained is 4.28 g, what is the percent yield?

11. What is the percent yield of a reaction in which 41.5 g of solid tungsten(VI) oxide reacts with
excess hydrogen to produce metallic tungsten and 9.50 mL of water? The density of water is
1.00 g/mL

12. What is the percent yield of a reaction in which 201 g of solid phosphorous trichloride reacts
with excess water to form 128 g of aqueous hydrogen chloride and aqueous phosphorous
acid, H3PO3?

13. When 18.5 g of gaseous methane and 43.0 g of chlorine gas undergo a reaction that has an
80.0% yield, what mass of liquid chloromethane, CH3Cl, forms? Gaseous hydrogen chloride
also forms.

14. When 56.6 g of calcium and 30.5 g of nitrogen undergo a reaction that has a 93.0% yield,
what mass of solid calcium nitride forms?

15. How many moles of MnCl can be produced by the reaction of 5.0 mol KMnO4, 3.0 mol
H2C2O4, and 22 mol HCl?
2KMnO4 + 5H2C2O4 + 6HCl = 2MnCl2 + 10 CO2 + 2KCl + 8H2O

16. How many grams of Fe are produced by reacting 2.00 kg Al with 300 g Fe2O3?
Fe2O3 + 2Al = Al2O3 + 2Fe

17. How many grams of which reactant are left over in Problem 16?

18. Gaseous H2S dissociates into H2 and S gases at very high temperatures: H2S = H2 + S.
When 0.620 g of H2S was held at 2000º C, it was found that 13 mg of H2 were produced.
What is the percent yield?

19. The first step in the Ostwald process for manufacturing nitric acid is the reaction of
ammonia, NH3, with oxygen, O2, to produce nitric oxide, NO, and water. The reaction
consumes 595 g of ammonia. How many grams of water are produced? Write the
balanced equation.

20. Sodium reacts violently with water to produce hydrogen and sodium hydroxide. How many
grams of hydrogen are produced by the reaction of 400 mg of sodium with water?

Worksheet 14 2
 Answers to Worksheet #14

Limiting Reagents
A Limiting Reagent is the reactant that is completely used up in a reaction. This reagent is the
one that determines the amount of product formed. Limiting reagent calculations are performed
in the same manner as the stoichiometric equations on Worksheet #11. However, with a limiting
reagent, you must calculate the amount of product obtained from each reactant (that means doing
math/stoichiometry at least twice!). Note that the limiting reagent is not always the lowest
number of grams, so you absolutely must do the math twice! The actual amount of product
obtained will be the lowest answer from stoichiometry (do not add, average, multiply, etc. – just
take the lowest one). Remember to balance the equations! This also might be a good time to
review stoichiometry if you are still struggling.

1. 4KO2(s) + 2H2O(ℓ) → 4KOH(s) + 3O2(g)

3molO2
molO2 = 0.150molKO2 = 0.113molO2
4molKO2
3molO2
molO2 = 0.100molH 2O = 0.150molO2
4molH 2O
The lowest amount of O2 obtained by calculation is 0.113 mol. Therefore, only 0.113 mol O2
can be obtained. KO2 is the reagent that is totally consumed in the reaction, and so KO2 is the
limiting reagent (this is the reagent that led to the lowest number of moles of O2).

2. 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(ℓ)

1molC2 H 2 4molCO2 44.01g
mass ( g )CO2 = 13.7 gC2 H 2 = 46.3 gCO2
26.036 g 2molC2 H 2 1molCO2
1molO2 4molCO2 44.01g
mass ( g )CO2 = 18.5 gO2 = 20.4 gCO2
32.00 g 5molO2 1molCO2
The amount of CO2 obtained is 20.4 g and oxygen is the limiting reagent (note that there was
a higher number of grams of oxygen, but it is still the limiting reagent!).

3. _N2(g) + 3H2(g) → 2NH3(g)

1molN 2 2molNH 3 17.034 g
mass ( g ) NH 3 = 4.7 gN 2 = 5.7 gNH 3
28.02 g 1molN 2 1molNH 3
1molH 2 2molNH 3 17.034 g
mass ( g ) NH 3 = 9.8 gN 2 = 5.5 gNH 3
2.016 g 3molH 2 1molNH 3
The amount of NH3 obtained is 5.7 g, and N2 is the limiting reagent.

Worksheet 14 3
4. _SO2(g) + _O3(g) → _SO3(g) + _O2(g)
1molSO2 1molSO3 80.07 g
mass ( g ) SO3 = 5.13 gSO2 = 6.41gSO3
64.07 g 1molSO2 1molSO3
1molO3 1molSO3 80.07 g
mass ( g ) SO3 = 6.18 gO3 = 10.3 gSO3
48.00 g 1molO3 1molSO3
The amount of SO3 obtained is 6.41 g, and SO2 is the limiting reagent.
5. 2SO2(g) + _O2(g) → 2SO3(g)
1molSO2 2molSO3 80.07 g
mass ( g ) SO3 = 7.99 gSO2 = 9.99 gSO3
64.07 g 2molSO2 1molSO3
1molO2 2molSO3 80.07 g
mass ( g ) SO3 = 2.18 gO3 = 10.9 gSO3
32.00 g 1molO2 1molSO3
The amount of SO3 obtained is 9.99 g, and SO2 is the limiting reagent.

Determining Excess Reagents

The reagent that is not the limiting reagent is the reagent in excess. In other words, we have
plenty of it left over when the reaction is completed.

6. The excess reagent is O2. First, we must determine how much of it was used in the
reaction:
1molSO3 1molO2 32.00 g
mass ( g )O2 = 9.99 gSO3 = 2.00 g
80.07 g 2molSO3 1molO2
Now, subtract the amount used from the amount of oxygen we started with to get the amount
left over: 2.18 g O2 – 2.00 g O2 = 0.18 g O2 left over.

7. 2NH4Cl(s) + _Ca(OH)2(s) → 2NH3(g) + _CaCl2(aq) + 2H2O(ℓ)

First, we must determine what the limiting reagent is, as was done above:
1molNH 4Cl 2molNH 3 17.034 g
mass ( g ) NH 3 = 33.0 gNH 4Cl = 10.5 gNH 3
53.492 g 2molNH 4Cl 1molNH 3
1molCa(OH ) 2 2molNH 3 17.034 g
mass ( g ) NH 3 = 33.0 gCa(OH ) 2 = 15.2 gNH 3
74.096 g 1molCa (OH ) 2 1molNH 3
The amount of NH3 obtained is 10.5 g, and NH4Cl is the limiting reagent. The reagent in
excess is Ca(OH)2. Before we can determine how much is left over, we have to determine
how much we used through stoichiometry.
1molNH 3 1molCa (OH ) 2 74.096 g
mass ( g )Ca (OH ) 2 = 10.5 gNH 3 = 22.8Ca (OH ) 2
17.034 g 2molNH 3 1molCa (OH ) 2

So, we used 22.8 g Ca(OH)2 in the reaction. The amount of Ca(OH)2 left over is how much
we started with minus how much we used:
mass ( g )Ca (OH ) 2 = 33.0 g − 22.8 g = 10.2 gCa (OH ) 2
So, we have 10.2 g Ca(OH)2 left over at the end of the reaction.

Worksheet 14 4
8. 2NO(g) + _O2(g) → 2NO2(g)
1molNO 2molNO2 46.01g
mass( g ) NO2 = 3.13 gNO = 4.80 gNO2
30.01g 2molNO 1molNO2
1molO2 2molNO2 46.01g
mass ( g ) NO2 = 4.16 gO2 = 12.0 gNO2
32.00 g 1molO2 1molNO2
The amount of NO2 obtained is 4.80 g, and the limiting reagent is NO. The reagent in
excess is O2. The amount of O2 used is:
1molNO2 1molO2 32.00 g
mass ( g )O2 = 4.80 gNO2 = 1.67 gO2
46.01g 2molNO2 1molO2
The amount of O2 left over is:
mass ( g )O2 = 4.16 g − 1.67 g = 2.49 gO2

Percent Yield
No reaction, when performed in the lab, gives as much product as stoichiometry says it should.
When reporting yields in literature, in addition to stating a gram amount of product obtained, the
percent yield is also reported. The percent yield is the actual yield of a reaction expressed as a
percent of the theoretical yield. In order to do these equations, you must first do stoichiometry to
determine the amount of product you should obtain.
⎛ actual ⎞
% yield = ⎜ ⎟100 Where actual means the yield obtained in the lab
⎝ theoretical ⎠
and theoretical means the amount that stoichiometry said you should have obtained.

9. 4C3H5(NO3)3(ℓ) → 6N2(g) + 10H2O(g) + 12CO2(g) + _O2(g)

1molC 3 H 5 ( NO3 ) 3 6molN 2 28.02 g

mass ( g ) N 2 = 5.55 gC 3 H 5 ( NO3 ) 3 = 1.03 gN 2
227.1g 4molC 3 H 5 ( NO3 ) 3 1molN 2
⎛ 0.991g ⎞
% yield = ⎜ ⎟100 = 96.2%
⎝ 1.03g ⎠

10. 2Cu2O(s) + _O2(g) → 4CuO(s)

1molCu 2 O 4molCuO 79.55 g
mass ( g )CuO = 4.18 gCu 2 O = 4.65 gCuO
143.1g 2molCu 2 O 1molCuO
1molO2 4molCuO 79.55 g
mass ( g )CuO = 5.77 gO2 = 57.4 gCuO
32.00 g 1molO2 1molCuO
The theoretical yield of CuO is 4.65 g, and Cu2O is the limiting reagent.
⎛ 4.28 g ⎞
% yield = ⎜ ⎟100 = 92.0%
⎝ 4.65 g ⎠

11. __WO3(s) + 3H2(g) → __W(s) + 3H2O(ℓ)

Worksheet 14 5
 1molWO3 3molH 2O 18.02 g 1mL
vol (mL ) H 2O = 41.5 gWO3 = 9.67mLH 2O
231.9 g 1molWO3 1molH 2O 1.00 g
⎛ 9.50mL ⎞
% yield = ⎜ ⎟100 = 98.2%
⎝ 9.67 mL ⎠

12. __PCl3(aq) + 3H2O(ℓ) → 3HCl(aq) + __H3PO3(aq)

1molPCl3 3molHCl 36.46 g
mass ( g ) HCl = 201gPCl3 = 160.gHCl
137.32 g 1molPCl3 1molHCl
⎛ 128 g ⎞
% yield = ⎜⎜ ⎟⎟100 = 80.0%
⎝ 160.g ⎠
13. __CH4(g) + __Cl2(g) → CH3Cl(ℓ) + __HCl(g)
First, find the limiting reagent, and thus the theoretical yield of CH3Cl:
1molCH 4 1molCH 3Cl 50.48 g
mass ( g )CH 3Cl = 18.5 gCH 4 = 58.2 gCH 3Cl
16.04 g 1molCH 4 1molCH 3Cl
1molCl2 1molCH 3Cl 50.48 g
mass ( g )CH 3Cl = 43.0 gCl2 = 30.6 gCH 3Cl
70.90 g 1molCl2 1molCH 3Cl
So, Cl2 is the limiting reagent, and 30.6 g CH3Cl is the theoretical yield.
⎛ actual ⎞
Remembering that % yield = ⎜ ⎟100 , we can solve for actual yield:
⎝ theoretical ⎠
% yield (theoretical ) (80.0)(30.6 g )
actual = = = 24.5 g
100 100
So 24.5 g of CH3Cl was obtained in the lab from this experiment.

14. 3Ca(s) + __N2(g) → __Ca3N2(s)

1molCa 1molCa 3 N 2 148.26 g
mass( g )Ca 3 N 2 = 56.6 gCa = 69.8 gCa3 N 2
40.08 g 3molCa 1molCa3 N 2
1molN 2 1molCa 3 N 2 148.26 g
mass( g )Ca 3 N 2 = 30.5 gN 2 = 161gCa 3 N 2
28.02 g 1molN 2 1molCa 3 N 2
So, Ca is the limiting reagent, and 69.8 g is the theoretical yield of Ca3N2.
⎛ actual ⎞
Remembering that % yield = ⎜ ⎟100, we can solve for actual yield:
⎝ theoretical ⎠
% yield (theoretical ) (93.0)(69.8 g )
actual = = = 64.9 g
100 100

15. 1.2 mol MnCl2 (First determine which is limiting:

5.0 mol KMnO 4 3.0 mol H 2C2 O 4
= 2.5, = 0.66 (lim iting)
2 5

Worksheet 14 6
 22 mol HCl
= 3.7
6
2 mol MnCl2 ⎞
Then solve ? mol MnCl2 = 3.0 mol H 2 C2 O 4 ⋅ ⎟
5 mol H 2 C2 O 4 ⎟
⎠
16. 210 g Fe (First determine which is limiting:
103 g Al mol Al 74.0 mol Al
mol Al = 2.00 kg Al ⋅ ⋅ = 74.0 mol Al; = 37.0
kg Al 27.0 g Al 2
mol Fe2 O3 1.89 mol Fe 2 O3
mol Fe 2 O3 = 3.00 g Fe 2O3 ⋅ ⋅ (lim iting)
159.6 g Fe 2 O3 1
mol Fe 2 O3
Then solve ? g Fe = 300 g Fe 2 O3 ⋅
159.6 g Fe 2 O3
2 mol Fe 55.8 g Fe ⎞
⋅ ⋅ ⎟
mol Fe 2O3 mol Fe ⎟⎠
17. 1.90 kg Al (Since you found in Problem 16 that Fe2O3 is limiting and is therefore completely
used up, all you need do is find out how much Al is used by the Fe2O3 and subtract this
amount from the amount of Al you started with.
mol Fe2 O3 2 mol Al 27.0 g Al
? g Al = 300 g Fe2 O3 ⋅ ⋅ ⋅
158.6 g Fe 2 O3 mol Fe2 O3 mol Al
= 102 g Al are used.
Since 2000 g Al were present, then 1898 g Al must remain. Rounded off to the correct
number of significant figures, this is 1.90 kg.)
mol H 2S mol H 2 2.02 g H 2
18. 35.4% ( ? g H 2 = 0.620 g H 2S ⋅ ⋅ ⋅
34.1 g H 2S mol H 2S mol H 2
= 0.0367 g H2 theoretical
10−3 g H 2
13 mg H 2 ⋅
mg H 2
Percent yield = ⋅ 100 = 35.4% yield)
0.0367 g H 2

19. 945 g H2O (4NH3 + 5O2 = 4NO + 6H2O

mol NH3 6 mol H 2 O 18.0 g H 2O
g H2O = 595 g NH3 ⋅ ⋅ ⋅
17.0 g NH3 4 mol NH3 mol H 2 O
= 945 g H2O)

20. 0.0176 g H2 (2Na + 2H2O = 2NaOH + H2

10−3 g Na mol Na mol H 2 2.02 g H 2
? g H2 = 400 mg Na ⋅ ⋅ ⋅ ⋅
mg Na 23.0 g Na 2 mol Na mol H 2
= 0.0176 g H2)

Worksheet 14 7
 Chapter 12 Worksheet 2 (ws12.2)
Concentration Units

molarity (M)

moles of solute / liter of solution (most commonly used unit of concentration)

molality (m)

moles of solute / kg of solvent

Compare to molarity and notice differences in the denominator. (L vs kg; solution vs. solvent)
Molality is independent of temperature so it is used for calculating boiling point elevation
and freezing point depression.

mole fraction (X) and mole %

moles of component / total moles of all components

A component is any substance in the solution so it can be a solute or the solvent.

mole % = mole fraction x 100

mass fraction and mass (or weight) percent (%)

mass fraction = mass of component / total mass of soln

mass % = mass fraction x 100

parts per million/billion/trillion (ppm/ppb/ppt)

ppm = (mass fraction) x 106

ppb = (mass fraction) x 109
ppt = (mass fraction) x 1012

Used for very dilute solutions (for example pollutants in drinking water). Notice that mass percent could
also be called parts per hundred.

1
When an ionic compound dissolves in water, it dissociates into its component ions. So how do we
specify the concentration of such a solution? Is it the concentration of the cation? The concentration of
the anion? The sum of these two concentrations? IT IS NONE OF THESE. Unless stated otherwise,
the concentration of a solution of an ionic compound is the concentration of the original undissociated
solute.

1. In a 2 M solution of magnesium phosphate:

a. What is the molarity of magnesium ions?

b. What is the molarity of phosphate ions?

c. In this chapter you will learn about properties of solutions that depend on the total
concentration of solute particles. What is the total molarity of particles in this solution (the
sum of the two concentrations above)?

2. What is the molarity of an ammonium carbonate solution if the concentration of ammonium ions
is 2 M? What is the concentration of carbonate ions and what is the total concentration of solute
particles?

3. A solution was made by dissolving 800.0 g of NaOH in 2.00 L of water. Calculate the molality,

2
 mole fraction, mass % and ppm of NaOH in this solution. (You will need the density of water.)

4. Consider a 0.300 M solution of NaOH?

a. Calculate the molality of NaOH. (Assume the density of the solution is the same as the density of
pure water.)

3
b. Notice that the molality is a little bigger than molarity. Why are these two numbers similar and
why is molality always greater than molarity? Why can’t one convert molality to molarity
without additional information?

5. What is the molarity of pure water at 4oC? (Hint: Use the density and molar mass of water.)
What happens to the concentration of water when a solute is dissolved in it?

4
 Chemistry 1
Volume 2

Worksheet 13
Molar Concentration of Solutions – Part 1

1
© MathTutorDVD.com
1. Which of the following is the correct mathematical representation of molarity?
!"#$% %"#'($
a. Molarity = #)($*% "+ %"#'()",

-*.!% %"#'($
b. Molarity = #)($*% "+ %"#'()",

!"#$% %"#'($
c. Molarity = -*.!% "+ %"#'()",

!"#$% %"#'($
d. Molarity = !"#$% "+ %"#'()",

2. Which of the following are valid units for molar concentration?

a. g/mol
b. mol/L
c. kg/mol
d. mol/kg

3. Triethylamine (1.5 mL) is added to 20 mL of chloroform. Which is the solute and which is the
solvent in this solution?

2
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4. LiBr (0.63 mol) was added to exactly 1 L of water. What is the molarity of this solution?

5. Calculate the molarity of a solution that has 0.75 mol dissolved in 0.87 L of water.

6. Calculate the molarity of the following solutions.

a. 0.15 mol NaCl in 1.4 L H2O

b. 1.4 mol LiCl in 3.5 L H2O

c. 0.843 mol CO2 in 25 L H2O

3
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7. Which solution is more concentrated: A solution made of 4.5 mol NaOH in 150 mL water or
a solution made of 10.0 mol NaOH in 500 mL water?

8. What volume of water was used to create a 4.5 M solution that contains 3.4 mol LiBr?

9. If 4.0 g of Ba(OH)2 were dissolved in 4.5 L of water, what is the molarity of the final
solution?

4
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10. What is the molarity of a solution that has 2.1 g SrCl2 dissolved in exactly 500 mL water?

5
© MathTutorDVD.com
 Answer Key
1. Which of the following is the correct mathematical representation of molarity?
!"#$% %"#'($
a. Molarity = #)($*% "+ %"#'()",

-*.!% %"#'($
b. Molarity = #)($*% "+ %"#'()",

!"#$% %"#'($
c. Molarity = -*.!% "+ %"#'()",

!"#$% %"#'($
d. Molarity = !"#$% "+ %"#'()",

Molarity represents the number of moles of a solute in the total volume of a

solution in liters.

𝒎𝒐𝒍𝒆𝒔 𝒔𝒐𝒍𝒖𝒕𝒆
Correct answer: a. Molarity = 𝒍𝒊𝒕𝒆𝒓𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏

2. Which of the following are valid units for molar concentration?

a. g/mol
b. mol/L
c. kg/mol
d. mol/kg
Molarity is defined using the following equation:
!"#$% %"#'($
Molarity = #)($*% "+ %"#'()",

As you can see, moles are the units in the top of the equation and liters (L) are the
units in the bottom. Therefore, molarity has units of mol/L, but it is typically written
as M.
Correct answer: b. mol/L

6
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3. Triethylamine (1.5 mL) is added to 20 mL of chloroform. Which is the solute and which is the
solvent in this solution?
The solute is the compound that is present in the smaller amount, while the solvent is
present in the larger amount. Thus, triethylamine is the solute and chloroform is the
solvent.
Correct answer: Solute: triethylamine; solvent: chloroform

4. LiBr (0.63 mol) was added to exactly 1 L of water. What is the molarity of this solution?

!"#$% %"#'($
Molarity = #)($*% "+ %"#'()",

:.<= !"#
Molarity = > ?

Molarity = 0.63 M
Correct answer: 0.63 M

5. Calculate the molarity of a solution that has 0.75 mol dissolved in 0.87 L of water.

!"#$% %"#'($
Molarity = #)($*% "+ %"#'()",

:.@A !"#
Molarity = :.B@ ?

Molarity = 0.86 M
Correct answer: 0.86 M

6. Calculate the molarity of the following solutions.

a. 0.15 mol NaCl in 1.4 L H2O

:.>A !"#
Molarity = >.C ?

Molarity = 0.11 M
Correct answer: 0.11 M

7
© MathTutorDVD.com
 b. 1.4 mol LiCl in 3.5 L H2O
>.C !"#
Molarity = =.A ?

Molarity = 0.40 M
Correct answer: 0.40 M

c. 0.843 mol CO2 in 25 L H2O

:.BC= !"#
Molarity = DA ?

Molarity = 0.033 M

Correct answer: 0.033 M

7. Which solution is more concentrated? A solution made of 4.5 mol NaOH in 150 mL water or
a solution made of 10.0 mol NaOH in 500 mL water?

Solution 1:
Convert 150 mL to L
150 mL 1L = 0.15 L
1,000 mL

C.A !"#
Molarity = :.>A ?
= 30 M

Solution 2:
500 mL 1L = 0.5 L
1,000 mL

>:.: !"#
Molarity = :.A ?
= 20 M

Since the molarity of Solution 1 is higher, it is more concentrated.

Correct answer: Solution 1 is more concentrated.

8
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8. What volume of water was used to create a 4.5 M solution that used 3.4 mol LiBr?

=.C !"#
4.5 M = ?)($*% "+ %"#'()",

Liters of solution = 0.76 L

Correct answer: 0.76 L

9. If 4.0 g of Ba(OH)2 were dissolved in 4.5 L of water, what is the molarity of the final
solution?

Use the molar mass of Ba(OH)2 to convert g to moles.

4.0 g Ba(OH)2 1 mol Ba(OH)2 = 0.023 mol Ba(OH)2

171.34 g Ba(OH)2

:.:D= !"#
Molarity = C.A ?
= 0.0052 M

Correct answer: 0.0052 M

10. What is the molarity of a solution that has 2.1 g SrCl2 dissolved in exactly 500 mL water?

Step 1:
Convert the grams of SrCl2 to moles using the molar mass of SrCl2 (158.52 g/mol).
2.1 g SrCl2 1 mol SrCl2 = 0.013 mol SrCl2
158.52 g SrCl2

Step 2:
Then, convert 500 mL to L:

500 mL 1L = 0.5 L
1,000 mL

Step 3:
Use the molarity formula to calculate the molarity
:.:>= !"#
Molarity = :.A ?
= 0.026 M

Correct answer: 0.026 M

9
© MathTutorDVD.com
 Chemistry 1
Volume 2

Worksheet 14
Molar Concentration of Solutions – Part 2

1
© MathTutorDVD.com
1. Calculate the molarity of the following solutions.
a. 0.67 mol Fe(NO3)2 in 3.5 L H2O

b. 5.6 mol fructose in 75 mL water

2. Calculate the molarity of a solution of 6.5 g NaCl in 3.4 L H2O

3. Calculate the molarity of a solution of 4.5 g KI in 50.0 mL H2O

2
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4. How many mL of solvent were used to create a 0.45 M solution that contains 3.4 moles of
sugar?

5. How many moles of H2SO4 are needed to prepare 100.0 mL of a 2.5 M solution?

6. A chemist has 40 mL of a 0.5 M solution of NaOH. How many grams of NaOH are in this
solution?

3
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7. How many liters of water are needed to create a 4.5 M solution from 3.5 g NaCl?

8. A chemist needs to create 250.0 mL of a 5.00 M solution of NaHCO3. How many grams of
NaHCO3 do they need?

4
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9. A solution of LiNO3 is 45% by mass and has a density of 1.05 g/mL. What is the molarity of
this solution?

5
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10. A solution of 4.5 g triethylamine (TEA) and 25.0 mL benzene was created during a reaction.
The volume of the solution increased after TEA was added. If the density of TEA is 0.726
g/mL and its molar mass is 101.19 g/mol, what is the molarity of the final solution?

6
© MathTutorDVD.com
 Answer Key
1. Calculate the molarity of the following solutions.
a. 0.67 mol Fe(NO3)2 in 3.5 L H2O

!"#$% %"#'($
Molarity = )*($+% %"#'(*",

-./0 !"#
Molarity = 1.2 )

Correct answer: 0.19 M

b. 5.6 mol fructose in 75 mL water

Convert 75 mL to L.
75 mL 1L = 0.075 L
1,000 mL

2./ !"#
Molarity = -.-02 )

Molarity = 75 M

Correct answer: 75 M

2. Calculate the molarity of a solution of 6.5 g NaCl in 3.4 L H2O

Since we’re not given number of moles, we need to convert the grams of NaCl to moles
using its molar mass (58.44 g/mol).

6.5 g NaCl 1 mol NaCl = 0.11 mol NaCl

58.44 g NaCl

!"#$% %"#'($
Molarity = )*($+% %"#'(*",
-.33 !"#
Molarity = 1.4 )

Correct answer: 0.032 M

7
© MathTutorDVD.com
3. Calculate the molarity of a solution of 4.5 g KI in 50.0 mL H2O

4.5 g KI 1 mol KI = 0.027 mol KI

166.0 g KI
Since molarity has units of mol/L, we need to convert 50 mL to L.

50.0 mL 1L = 0.0500 L
1,000 mL

-.-50 !"#
Molarity = -.-2-- )

Correct answer: 0.54 M

4. How many mL of solvent were used to create a 0.45 M solution that contains 3.4 moles of
sugar?

We need to rearrange the molarity formula to solve for liters of solution.

!"#$% %"#'($
Molarity = )*($+% %"#'(*",

1.4 !"#
0.45 M = )*($+% %"#'*",

Liters of solution = 3.4 mol/0.45 M

Liters of solution = 7.6 L

Then, convert the liters of solution to mL, since that is what the question asks for.
7.6 L 1,000 mL = 7,600 mL
1L

Correct answer: 7,600 mL

8
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5. How many moles of H2SO4 are needed to prepare 100.0 mL of a 2.5 M solution?

!"#$% %"#'($
Molarity = )*($+% %"#'(*",

Since M has units of mol/L, we need to convert 100 mL to L.

100.0 mL 1L = 0.1000 L
1,000 mL

!"#$% %"#'($
2.5 M = -.3--- )

Moles solute = 0.25 mol

Correct answer: 0.25 mol

6. A chemist has 40 mL of a 0.5 M solution of NaOH. How many grams of NaOH are in this
solution?

Convert 40 mL to L first.

40 mL 1L = 0.04 L
1,000 mL

!"#$% %"#'($
Molarity = )*($+% %"#'(*",

!"#$% %"#'($
0.5 M = -.-4 )

Moles NaOH = 0.02 mol NaOH

Use the molar mass of NaOH (39.99 g/mol) to convert moles to grams NaOH.
0.02 mol NaOH 39.99 g NaOH = 0.8 g NaOH
1 mol NaOH

Correct answer: 0.8 g NaOH

9
© MathTutorDVD.com
7. How many liters of water are needed to create a 4.5 M solution from 3.5 g NaCl?

First, convert the g NaCl to moles using its molar mass (58.44 g/mol).

3.5 g NaCl 1 mol NaCl = 0.060 mol NaCl

58.44 g NaCl

-.-/- !"#
4.5 M = )*($+% %"#'(*",

Liters solution = 0.013 L

Correct answer: 0.013 L

8. A chemist needs to create 250.0 mL of a 5.00 M solution of NaHCO3. How many grams of
NaHCO3 do they need?

Step 1:
250.0 mL 1L = 0.2500 L
1,000 mL

Step 2:
Then, use the molarity formula to convert the number of moles NaHCO3 present in this
volume of solution.
!"#$% 6789:!
5.00 M = -.52-- )

Moles NaHCO3 = 1.25 mol NaHCO3

Step 3:
Use the moles of solute and the molar mass of NaHCO3 (84.007 g/mol) to calculate the
grams of NaHCO3
1.25 mol NaHCO3 84.007 g NaHCO3 = 105 g NaHCO3
1 mol NaHCO3

Correct answer: 105 g NaHCO3

10
© MathTutorDVD.com
9. A solution of LiNO3 is 45% by mass and has a density of 1.05 g/mL. What is the molarity of
this solution?

Assume 1 L of solution to simplify the calculations.

We first need to calculate the number of moles of LiNO3 in this solution. We do that by
following these steps, which are combined in the calculation below.
1) Convert the assumed 1 L of solution into mL since density has the unit of mL.
2) Then, multiply this by the density of the solution to obtain g of solution.
3) Multiply this by the mass percent of the solution (45 g LiNO3/100 g solution) to get g
of LiNO3.
4) Finally, convert the grams of LiNO3 into moles LiNO3 using its molar mass from the
periodic table.

1L 1,000 mL 1.05 g solution 45 g LiNO3 1 mol LiNO3 = 6.85 mol LiNO3

1L 1 mL 100 g solution 68.946 g LiNO3

Now that we have moles, we just divide this number of moles by the assumed 1 L of
solution to get the molarity of the solution.
!"#$% %"#'($
Molarity = ;"#'!$ "< %"#'(*",

/.=2 !"#
Molarity = 3 )

Molarity = 6.85 M
Correct answer: 6.85 M

11
© MathTutorDVD.com
10. A solution of 4.5 g triethylamine (TEA) and 25.0 mL benzene was created during a reaction.
The volume of the solution increased after TEA was added. If the density of TEA is 0.726
g/mL and its molar mass is 101.19 g/mol, what is the molarity of the final solution?

Remember, molarity is based on the total volume of solution. So, since the problem
states that the volume of the solution increased after addition of triethylamine, we need
to calculate the volume of triethylamine that was added. Use the density to calculate
this.

Step 1:
4.5 g TEA 1 mL TEA = 6.2 mL TEA
0.726 g TEA

Step 2:
Next, determine the volume of the solution. To do this, add 6.2 mL TEA to 25.0 mL
benzene.

Total solution volume: 6.2 mL + 25.0 mL = 31.2 mL

Convert this to liters by dividing by 1,000.

31.2 mL/1,000 = 0.0312 L solution

Step 3:
Finally, we need to calculate how many moles of triethylamine we have. Use the molar
mass provided (101.19 g/mol).

4.5 g TEA 1 mol TEA = 0.044 mol TEA

101.19 g TEA

Step 4:
Use the mol TEA and total volume of solution to calculate molarity.

-.-44 !"# >?@

Molarity = -.-135 )

Molarity = 1.4 M

Correct answer: 1.4 M

12
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Name: _________________________________ #: ____ Class: ____ Date: ________________
Concentrations of Solutions Practice
Calculate the concentration in each of the following solutions:
1. 10 moles of potassium hydroxide in 5.16 L of solution

2. 0.5 moles of calcium chloride in 485 mL of solution

3. 0.079 moles of magnesium sulfate in 25 mL of solution

Find the number of moles of solute in each of the following solutions:

4. 2L of 2.3 M nitric acid

5. 200 mL of 0.8 M sodium carbonate solution

6. 300 mL of 4 M sodium hydroxide solution

7. How many grams of NaOH must be used to make 250 mL of a 1.842 M solution of NaOH?
a. Start with what you know, the final volume and concentration of the solution. Arrange these
factors to calculate the moles of NaOH needed.

b. Change moles of NaOH to grams of NaOH.

8. Calculate the concentration of a solution of CO2 in water, which contains 20.25 g of CO2 per 675 mL
of solution.
a. Start with what you know, change grams of CO2 to moles of CO2.

b. With the moles value found in part a, use the concentration formula to calculate the
concentration of CO2.

9. The initial concentration of the solution H2SO4 is 18 M. What would the volume of the initial stock
solution (V1) need to be if the concentration was being changed to 0.4 M of a 1 L solution?

10. The initial concentration of the solution HCl is 12 M. What would the volume of the initial stock
solution (V1) need to be if the concentration was being changed to 1.1 M of a 500-mL solution?

11. The initial concentration of the solution HNO3 is 16 M. What would the volume of the initial stock
solution (V1) need to be if the concentration was being changed to 5.5 M of a 325-mL solution?
 Solution Concentrations Worksheet (Section 12.3)

Name ________________________ Period:

Measuring Concentration: There are several different ways to measure and express the concentration of a
solution. Molarity (Section 12.3) the term we learned earlier, refers to the concentration of a solution
expressed in moles of solute per liter of solution. We also use several other units, including the following:

mass of solute (g)

1. Percent by Mass x100%
mass of solution (g)
volume of solute (mL)
2. Percent by Volume x100%
volume of solution (mL)
mass of solute (mg)
3. Mass/Volume Percent x100% (units are used in medicine)
volume of solution (dL)
1 g of solute
4. Parts per million
1 x 106 g of solution
For this type
! of unit, these equivalents work for water solutions:
1 ppm = 1 mg/L 1 ppb = 1µg/L 1 ppt = 1 ng/L
ppm = parts per million ppb = billion ppt = trillion
amount of solute (mol)
5. Molality this unit is independent of temperature
mass of solvent (kg)

Problems – Do work on Separate Paper. Show Dimensional Analysis.

1. Glucose is a sugar that is found abundantly in nature. What is the percent by mass of a solution made by
dissolving 163 g of glucose in 755 g of water? Do you need to know the formula of glucose? Why or why
not?
2. What is the mass percent sucrose in a solution obtained by mixing 225 g of an aqueous solution that is
6.25% sucrose by mass with 135 g of an aqueous solution that is 8.20% sucrose by mass?
3. Determine the volume percent of toluene in a solution made by mixing 40.0 mL toluene with 75.0 mL of
benzene.
4. What is the concentration of Na+, in parts per million by mass, in 0.00152 M Na2SO4?
5. What is the molality of a solution prepared by dissolving 225 mg of glucose (C6H12O6) in 5.00 mL of
ethanol (density = 0.789 g/mL)
6. How many milliliters of water (density = 0.998 g/mL) are required to dissolve 25.0 g of urea (CO(NH2)2)
in order to produce a solution that is 1.65 m? (m is the abbreviation for molality)
7. Describe the process you would use in order to prepare 5.00 kg of an aqueous solution that is 8.00% NaCl
by mass.
8. What is the mass percent of solute when 4.12 g is dissolved in 100.0 g of water?
9. What is the volume percent of 10.00 g of acetone (d = 0.789 g/mL) in 1.55 L of an acetone-water
solution?
10. On average, glucose makes up about 0.10% of human blood, by mass. What is the approximation
concentration of glucose in blood in milligrams per deciliter? (You have to assume something about blood).
11. Convert 0.0035% NaCl by mass into parts per million of NaCl.
12. Convert 2.4 ppm F- into molarity of fluoride ion.
13. Calculate the molality of a solution prepared by dissolving 125 mL of pure methanol (density = 0.791
g/mL) into 275 g of ethanol.
 Concentrations Worksheet KEY
1. 163 g glucose/918 g soln x 100% = 17.8% glucose

2.

[(6.25% x 225mL) + (8.20% x 135mL)] g sucrose

x 100% = 6.98%
(225 + 135)g solution

3. 40.0 mL/115 mL x 100 % = 34.8%

!
0.00152 mol Na 2SO 4 2 mol Na +
x x
4. L 1 mol Na 2SO 4
22.9898 g 1000 mg
x = 69.9 mg/L or ppm
mol Na + g

1g 1 mol glucose
225 mg x
1000 mg 180.158 g
5. = 0.317 m
0.789 g 1 kg
5.00 mL x
mL 1000 g

1 kg water 1 mol urea

x x 25.0 g urea x
1.65 mol urea 60.055 g
6.
1000 g 1 mL
x = 253 mL water
kg 0.998 g
 7. 5.00 kg x 8.00% = 400. g NaCl, so dissolve 400. g NaCl in
4.60 kg of water to make solution

8. 4.12 g solute/104.12 g solution x 100% = 3.96 %

1 mL
10.00 g x
0.789 g
x 100% = 0.818%
9. 1.55 L x 1000 mL
L

0.10 g glucose 1g 1000 mg 100 mL

10. 100 g blood x x x = 1.0 x 10 2 mg/dL
! 1 mL g dL

0.0035 g NaCl 1g 1000 mL 1000 mg

x
11. 100 g solution 1 mL x x = 35 mg/L or ppm
! L g

2.4 mg F - 1g 1 mol F - -4 -
! 12. x x = 1.3 x 10 M F
L 1000 mg 18.9984 g

0.791 g 1 mol CH 3OH

! 125 mL x
mL 32.042 g
= 11.2 m
1 kg
275 g x
1000 g
Naming Chemical Compounds Worksheet
Name the following ionic compounds:
1) NaBr __________________________________________
2) CaO __________________________________________
3) Li2S __________________________________________
4) MgBr2 __________________________________________
5) Pu(OH)3 __________________________________________
6) Hg(CN)2 __________________________________________
7) Mo(C2H3O2) __________________________________________
8) Cr(CrO4)3 __________________________________________
9) WO2 __________________________________________
10) Mg(ClO)2 __________________________________________

Write the formulas for the following ionic compounds:

11. potassium dichromate __________________________________________
12. gold(III) oxide __________________________________________
13. aluminum periodate __________________________________________
14. barium nitrite __________________________________________
15. silver carbonate __________________________________________
16. lithium sulfite __________________________________________
17. zinc hydrogen carbonate __________________________________________
18. iron(III) hydroxide __________________________________________
19. ammonium phosphate __________________________________________
20. copper(II) bromite __________________________________________

21) SO3 __________________________________________

22) N2S __________________________________________
23) BF3 __________________________________________
24) P2Br4 __________________________________________
25) SiO2 __________________________________________
26) SF6 __________________________________________
27) NO2 __________________________________________
28) nitrogen trichloride __________________________________________
29) dinitrogen trioxide __________________________________________
30) phosphorus pentafluoride ______________________________________
31) diboron tetrahydride __________________________________________
 Naming Chemical Compounds - Answers
Name the following ionic compounds:
1) NaBr sodium bromide

2) CaO calcium oxide

3) Li2S lithium sulfide

4) MgBr2 magnesium bromide

5) Be(OH)2 beryllium hydroxide

Write the formulas for the following ionic compounds:

6) potassium iodide KI

7) magnesium oxide MgO

8) aluminum chloride AlCl3

9) sodium nitrate NaNO3

10) calcium carbonate CaCO3

11) lithium sulfate Li2SO4

12) beryllium phosphide Be3P2

13) magnesium hydroxide Mg(OH)2

14) sodium phosphate Na3PO4

15) aluminum carbonate Al2(CO3)3

16) calcium chloride CaCl2

17) sodium cyanide NaCN

18) aluminum oxide Al2O3

19) magnesium acetate Mg(C2H3O2)2

20) ammonium chloride NH4Cl

Write the names of the following covalent compounds:
21) SO3 sulfur trioxide

22) N2S dinitrogen sulfide

23) PH3 phosphorus trihydride

24) BF3 boron trifluoride

25) P2Br4 diphosphorus tetrabromide

26) CO carbon monoxide

27) SiO2 silicon dioxide

28) SF6 sulfur hexafluoride

29) NH3 ammonia

30) NO2 nitrogen dioxide

Write the formulas of the following covalent compounds:

31) nitrogen trichloride NCl3

32) boron carbide BC

33) dinitrogen trioxide N2O3

34) phosphorus pentafluoride PF5

35) methane CH4

36) sulfur dibromide SBr2

37) diboron tetrahydride B2H4

38) oxygen difluoride OF2

39) carbon disulfide CS2

40) nitrogen monoxide NO

 Molecular Geometry Chart
# of Electron Number of Electron Pair Molecular Geometry Approximate
Groups Lone Pairs Arrangement Bond Angles
2 0 linear 180o 180°

0 trigonal planar 120°

120o

1 bent <120°
3
<120o

0 tetrahedral 109.5°
109.5o

1 trigonal pyramid <109.5° (~107°)

<109.5o
4
2 bent <109.5°(~105°)

<109.5o

0 trigonal 90o
90°, 120°
bipyramidal
120o

1 see-saw <90o <90°, <120°

<120o

2 T-structure <90°
<90o
5

3 linear 180°
180o

0 octahehral 90o 90°, 90°

90o

6 1 square pyramidal 90°, <90°

90o

<90o

2 square planar 90°

90o
 Pauling
electronegativity
values
H He
2.1 N/A
Li Be B C N O F Ne
1.0 1.5 1.5 2.5 3.0 3.5 4.0 N/A
Na Mg Al Si P S Cl Ar
0.9 1.2 1.5 1.8 2.1 2.5 3.0 N/A

K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
0.8 1.0 1.3 1.5 1.6 1.6 1.5 1.8 1.9 1.8 1.9 1.6 1.6 1.8 2.0 2.4 2.8 N/A*
Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe
0.8 1.0 1.2 1.4 1.6 1.8 1.9 2.2 2.2 2.2 1.9 1.7 1.7 1.8 1.9 2.1 2.5 N/A*
Cs Ba Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn
0.7 0.9 1.3 1.5 1.7 1.9 2.2 2.2 2.2 2.4 1.9 1.8 1.9 1.9 2.0 2.2 N/A
Fr Ra Rf Db Sg Bh Hs
0.7 0.9
 Lewis Structures, Shapes, and Polarity W 319

Draw Lewis structures, name shapes and indicate polar or non-polar for the
following molecules:

a. CH4

b. NCl3

c. CCl2F2

d. CF2H2

e. CH2O

f. CHN

g. PI3

h. N2O

i. SO2

j. CS2

k. CO

l. H2O

m. COF2

n. N2

o. O2

p. H2

q. Cl2

r. HF

s. O3

t. NI3

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a. CH4 tetrahedral, non-polar

b. NCl3 trigonal pyramidal, polar

c. CCl2F2 tetrahedral, polar

d. CF2H2 tetrahedral, polar

e. CH2O trigonal planar, polar

f. CHN linear, polar

g. PI3 trigonal pyramidal, polar

h. N2O linear, polar

i. SO2 bent, polar

j. CS2 linear, non-polar

k. CO linear, polar

l. H2O bent, polar

m. COF2 trigonal planar, polar

n. N2 linear, non-polar

o. O2 linear, non-polar

p. H2 linear, non-polar

q. Cl2 linear, non-polar

r. HF linear, polar

s. O3 bent, non-polar

t. NI3 trigonal pyramidal, polar

 Lewis Structures, Shapes, and Polarity W 319

(a) (b) (c) (d)

:
H :
Cl
:
H

:
: Cl N Cl :

:
:
:F Cl : :F H

:
H C H C C

:
: Cl :
:F: :F:

:
H

:
: :
:
(e) (f) H C N (g) (h) :
N N O
:

:
:

O H : l P l :
C
:

:
H : l :

:
:

(i) (j) (k) :

C S
: (l)
S C S O

:
:
:

S
:

H H
O O
:

:
:

:
:

(m) :O: (n) (o) (p)

:

O O
: :
N N H H
:

C
:

:
:F F
:

(q) (r) (s) (t)

:

:
:

: Cl H F : O
Cl : :
: l N l :
:

O O
:
:

:
:

:
:
:
:

: l :
:

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