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Unit 7A- The Mole
Accelerated Chemistry I
The Avogadro Constant and the Mole
The number of atoms of an element that have a total mass in grams equivalent to the mass of one atom in amu is
called the Avogadro constant, and is equal to 6.02 × 1023. One mole (mol) of things equals 6.02 × 1023 things.
The figure below shows how to convert from one unit (grams, moles, or particles, i.e. atoms, molecules, formula
units) to another.
1. Write the missing conversion factors
a.
b.
c.
c
.
a
.
b
.
2. How many moles are in 25.0 g of sodium? Estimate before you solve: is 25 g more or less than one
mole? Answ: 1.09 mol Na
3. Calculate the moles of sodium a sample of 2.41 × 1048 atoms of sodium. Answ: 4.00x1024 mol Na
4. Calculate the moles of fluorine in a sample containing 8.6 × 1018 atoms of fluorine. Answ: 1.4x10-5 mol
5. How many moles of magnesium are in 4.32 × 10-4 g magnesium? Answ: 1.78x10-5 mol Mg
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Complete the following 2-step conversions. SHOW YOUR WORK!
6. 3.68 × 1040 atoms C = ? g C Answ: 7.34x1017 g C
7.
284 g Mg = ? atoms Mg Answ: 7.03x1024 atoms Mg
8.
5.22 × 108 atoms B = ? g B Answ: 9.37x10-15 g B
Molar Mass of a compound = mass of one mole of that compound Find the molar mass of the following
compounds:
9. zinc chloride Answ: 136.27 g/mol
10. 2-methylpropane Answ: 58.14 g/mol
11. 1-bromo, 3,4-dimethylpentane Answ: 179.12 g/mol
12. Sulfurous acid Answ: 82.09 g/mol
13. sodium carbonate Answ: 105.98 g/mol
Solve the following problems using the factor-label method; SHOW YOUR WORK!
14. 25.0 g zinc chloride = ? moles zinc chloride Answ: 0.183 mol
2
15. 2.5×1024 molecules 2-methylpropane = ? moles 2-methylpropane Answ: 4.2mol
16. 6.42 µg 1-bromo, 3,4-dimethylpentane = ? moles 1-bromo, 3,4-dimethylpentane Answ: 3.58x10-8 mol
17. 5.32 mg sulfurous acid = ? atoms of H? Answ: 7.81x1019 atoms H
18. 4.38 × 1026 formula units sodium carbonate = ? g sodium ions Answ: 3.35x104 g Na+
19. Calculate the numbers of individual C, H, and O atoms in 1.50 g of glucose (C6H12O6).
Answ: 3.01x1022 atoms of O and C
6.02x1022 atoms H
20. Pheromones are a special type of compound secreted by the females of many insect species to attract the
males for mating. One pheromone has the molecular formula C19H38O. Normally, the amount of this
pheromone secreted by a female insect is about 1.0 × 10-12 g. How many molecules are there in this
quantity? Answ: 2.1x109 molecules
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21. How many water molecules are present in 2.56 cm3 of water at standard temperature and pressure?
Answ: 8.55x1022 molecule
Percentage Composition
Example: Find the percentage composition of arsenic(III) oxide.
knowns: As2O3 = 2×As + 3×O = 2×74.9 amu + 3×16.0 amu = 198 amu or 198 g/mol
Percentage of As =
2 × 74.9 amu
× 100% = 75.8%
198 amu
Percentage of O =
3 × 16.0 amu
× 100% = 24.2%
198 amu
22. What is the percentage composition of calcium hydroxide? Answ: 54.10% Ca, 2.73% H, 43.17% O
23. It is determined that in a 500.0-g sample of pitchblende, a uranium ore, there are 2.5 g of uranium, U.
What percent of the ore is uranium? Answ: 0.50% U
24. What is the percent of copper and chlorine in
a. Copper (I) chloride? Answ: 64.19% Cu, 35.81% Cl
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b. Copper (II) chloride? Answ: 47.27% Cu, 52.73% Cl
25. A sample of an unknown compound with a mass of 2.876 g has 66.07 percent carbon, 6.71 percent
hydrogen, 4.06 percent nitrogen and 23.16 percent oxygen. What is the mass of each element in this
compound? Answ: 1.900 g C, 0.195 g H, 0.117 g N, 0.6661 g O
26. Find the percent composition of a compound that contains 2.7369 g of chlorine, 0.4116 g of oxygen and
0.7971 g of phosphorus. Answ: 69.366% Cl, 10.43% O, 20.20 % P
27. A sample of a compound with a mass of 0.432 g is analyzed. The sample is found to contain only
fluorine and oxygen. If the compound contains 0.128 g oxygen, calculate the percent composition of
the compound. Answ: 29.6% O, 70.4% F
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Empirical and Molecular Formulas
Example: Determine the empirical formula for the compound with a composition of 46.56% Fe and 53.44% S.
Assume you have 100 g of compound; there are 46.56 g Fe and 53.44 g S
Fe:
46.56 g Fe
Ratio of moles:
1 mol Fe
= 0.834 mol Fe
55.8 g Fe
S:
53.44 g s
1 mol S
= 1.66 mol S
32.1 g S
0.834 / 0.834
1.00
Fe
0.834
=
=
=
so FeS2 (divide through by the smaller number)
1.66 / 0.834
1.99
S
1.66
or Fe0.834S1.66 = Fe(0.834/0.834)S1.66/0.834 = FeS2
28. Determine the empirical formula of a compound containing 2.644 g of gold and 0.476 g chlorine.
Answ: AuCl
29. Determine the empirical formula of a compound containing 0.928 g of gallium and 0.412 g phosphorus.
Answ: GaP
30. What is the empirical formula for a compound containing zinc and oxygen if a 5.09 g sample contains
4.09 g of zinc? Answ: ZnO
31. Determine the empirical formula of a compound containing 2.16 g of aluminum, 3.85 g of sulfur and
7.68 g oxygen. Answ: Al2S3O12 = Al2(SO4)3
Unit 7 Student Handout – The Mole
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32. Determine the molecular formula of a compound that is 30.45% nitrogen and 69.55% oxygen. The
molar mass of the compound is 92 g/mol. Answ: N2O4
33. Determine the molecular formula of a compound containing 56.36 g of oxygen and 43.64 g of
phosphorus. The molar mass of the compound is 283.9 g/mol. Answ: P4O10
34. Ascorbic acid, aka vitamin C, has 40.9% carbon, 4.58% hydrogen, and 54.5% oxygen. It’s molar mass
is 176.1 g/mol. What is its molecular formula? Answ: C6H8O6
Hydrates
Example: A 10.0 g sample of hydrated copper(II) sulfate is heated to drive off the water. The dry sample has a
mass of 6.4 g. What is the mass of the water driven off? What is the formula of the hydrate?
10.0 g CuSO4 hydrate – 6.4 g CuSO4 dry = 3.6 g water driven off
Convert the masses of dry CuSO4 and water to moles
6.4g CuSO4 1 mol CuSO4
160 g CuSO4
= 0.040 mol CuSO4
3.6 g H2O 1 mol H2O
= 0.20 mol H2O
18.0 g H2O
The ratio between moles of CuSO4 and H2O is 1 to 5, so CuSO45H2O.
Unit 7 Student Handout – The Mole
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Find the formulas for the hydrates containing the following masses of components. Name the hydrates.
35. 3.71 g sodium carbonate, 6.29 g water Answ: Na2CO3•10H2O
36. Two hydrates of cobalt(II) chloride exist. One is 21.68% water the other is 45.43% water. What are
their empirical formulas? Answ: CoCl2•2H2O and CoCl2•6H2O
Molarity
Molarity is a measure of a solution’s concentration. It is equal to the ratio between the moles of dissolved
substance (solute) and the volume of a solution in L.
M=
mol of solute
L of solution
37. Complete the following table by filling in the correct values.
Concentration
a.
10.0M
b.
0.062M
c.
d.
Moles of solute
3.00 L
0.651 mol
6.3 mol
6.0M
e.
Volume of solution
5.9 L
1.00 L
0.0555 mol
500 mL
38. What is the molarity of a solution in which 90.0 g of C6H12O6, glucose, is dissolved in enough water to
make 250.0 mL of solution? Answ: 2.00 M C6H12O6
Unit 7 Student Handout – The Mole
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39. If 0.0100 kg of sodium hydroxide is dissolved in enough water to make 250.0 mL of solution, what is
the molarity? Answ: 1.00 M NaOH
40. Describe the preparation of 250.0 mL of 0.2 M calcium nitrate.
41. What mass of potassium phosphate is needed to prepare 4.00 liters of 1.50 M solution? Answ: 1.27 kg
42. What is the molarity of a solution that contains 85.6 grams of phosphorous acid in 0.385 L of solution?
Answ: 2.70 M
Unit 7 Student Handout – The Mole
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43. What volume of 0.025M magnesium nitrate would contain 4.50 g of dissloved nitrate ions? Answ: 1.5
L
THE MOLE VOCABULARY REVIEW
Match the term with the correct description.
a. molecular mass
b. Avogadro constant
c. molarity
d. empirical formula
e.
f.
g.
h.
formula mass
molar mass
percentage composition
molecular formula
______
44. the sum of the atomic masses of all atoms in the formula unit of an ionic compound
______
45. the simplest ratio of the elements in a compound
______
46. the sum of all the atomic masses in a molecule
______
47. the mass of 6.02 ×1023 molecules, atoms, ions, or formula units of a species
______
48. the ratio between the moles of dissolved substance and the volume of solution in liters
______
49. shows the actual number of atoms in a molecule
______
50. 6.02 × 1023
______
51. a statement of the relative mass each element contributes to the mass of a compound as a
whole
Challenge Problems
52. A typical virus particle is 5.00 × 10-6 cm in diameter. If Avogadro’s number of these virus particles
were laid in a row, how many kilometers long would the line be? Answ: 3.01x1013 km
Unit 7 Student Handout – The Mole
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53. How many molecules are in a sample of water that requires 8.40 kcal of heat energy to raise the
temperature by 34.0°C? Answ: 8.25x1024 molecules
54. Nitroglycerine contains 60% as many carbon atoms as hydrogen atoms; three times as many oxygen
atoms as nitrogen atoms; and the same number of carbon and nitrogen atoms. The number of moles in a
gram of nitroglycerine is 0.00441. What is the molecular formula of nitroglycerin? Answ: C3H5N3O9
55. A compound of hydrogen and carbon has a molar mass 114 g/mol. If one mole of the compound
contains 18.17 g of hydrogen, what is the compound’s molecular formula? Answ: C8H18
Unit 7 Student Handout – The Mole
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56. The element gold has properties that have made it much sought after through the ages. A cubic meter of
ocean water contains 6.00 × 10-6 g gold. If the total mass of the water in the oceans of the world is 4.00
× 1020 kg, how many kilograms of gold are distributed throughout the oceans? (Assume the density of
sea water is 1.00 g/mL). How many liters of seawater would have to be processed to recover a kilogram
of gold (which had a value of about $11,000 at 1989 prices)? Do you think this recovery operation is
feasible? Answ: 2.40x1012 g Au distributed throughout the oceans, 1.67x1011 L seawater to process
57. What volume of 2.00M aluminum nitrate solution contains 0.250 moles of nitrate ions? Answ: 41.7 mL
of solution
58. A solution of ammonium sulfate is made that is 2.25 M for ammonium ions. What volume of this
solution can be produced using only 50.0 g of ammonium sulfate? Answ: 336 mL of solution
Unit 7 Student Handout – The Mole
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59. You have 250. mL of 0.136 M HCl. Using a volumetric pipet, you take 25.00 mL of that solution and
dilute it to 100.00 mL in a volumetric flask. Now you take 10.00 mL of the second solution, using a
volumetric pipet, and dilute it to 100.00 mL in a volumetric flask. What is the concentration of HCl in
the final solution? (This process is called a serial dilution.) Answ: 0.0034 M HCl
60. Iron metal has density of 7.874 g/cm3. You have a piece of metal in the shape of a rectangular
prism that is 5.500 x 108 pm by 1.250 x 10–8 Mm by 1.345 x 109 nm. How many atoms of iron
do you have? (7 points) Answ: 7.848x1023 atoms
Unit 7 Student Handout – The Mole
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