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Example 2.1 A Conceptual Example
Jan Baptista van Helmont (1579–1644) first measured the mass of a young willow tree
and, separately, the mass of a bucket of soil and then planted the tree in the bucket. After
five years, he found that the tree had gained 75 kg in mass even though the soil had lost
only 0.057 kg. He had added only water to the bucket, and so he concluded that all the
mass gained by the tree had come from the water. Explain and criticize his conclusion.
Analysis and Conclusions
Van Helmont’s explanation anticipated the law of conservation of mass by
dismissing the possibility that the increased mass of the tree could have been
created from nothing. His main failure, however, was in not identifying all the
substances involved. He did not know about the role of carbon dioxide gas in the
growth of plants. In applying the law of conservation of mass, we must focus on all
the substances involved in a chemical reaction.
Exercise 2.1A
A sealed photographic flashbulb containing magnesium and oxygen has a mass of
45.07 g. On firing, a brilliant flash of white light is emitted and a white powder is formed
inside the bulb. What should the mass of the bulb be after firing? Explain.
Exercise 2.1B
Is the mass of the burned match pictured in the photograph the same as, less than, or
more than the mass of the burning match? Explain your answer.
Example 2.2
The mass ratio of oxygen to magnesium in the compound magnesium oxide is 0.6583:1.
What mass of magnesium oxide will form when 2.000 g of magnesium is completely
converted to magnesium oxide by burning in pure oxygen gas?
Strategy
We usually write mass ratios in a form such as “the ratio of O to Mg is 0.6583:1.”
The first number represents the mass of the first element named—in this case, a
mass of oxygen, say 0.6583 g oxygen—and the second number represents the mass
of the second element named—here a mass of magnesium. Although written only as
“1,” we assume that the second number is as precisely known as the first, that is,
1.0000 g magnesium. According to the law of constant composition, the mass of
oxygen that combines with the 2.000 g of magnesium must be just the right amount
that makes the mass ratio of oxygen to magnesium in the product 0.6583:1.
According to the law of conservation of mass, the mass of magnesium oxide must
equal the sum of the masses of the magnesium and oxygen that react.
Solution
We can state the mass ratio in the form of a conversion factor and then determine
the required mass of oxygen.
The mass of the sole product, magnesium oxide, equals the total of the masses of
the substances entering into the reaction.
Example 2.2 continued
Assessment
As a simple check, compare the calculated mass with the starting mass of
magnesium. From the law of conservation of mass, the product of the reaction
of magnesium with oxygen must have a mass greater than the starting mass of
magnesium; and 3.317 g magnesium oxide is indeed greater than 2.000 g
magnesium.
Exercise 2.2A
What mass of magnesium oxide is formed when 1.500 g of oxygen combines with
magnesium?
Exercise 2.2B
When a strip of magnesium metal was burned in pure oxygen gas, 1.554 g of oxygen was
consumed and the only product formed was magnesium oxide. What must have been the
masses of magnesium metal burned and magnesium oxide formed?
Example 2.3
How many protons, neutrons, and electrons are present in a 81Br atom?
Strategy
We can use the identity of the element and several simple relationships between the
subatomic particles described above to determine the numbers of these particles.
Solution
The atomic number of bromine
is not given here, but we can
obtain it from the list of elements
on the inside front cover.
Z = 35 = number of protons
In the bromine atom, the number
of positively charged protons
equals the number of negatively
charged electrons.
Number of protons = number of electrons = 35
From mass number 81 and
atomic number 35, we calculate
the number of neutrons, using
Equation (2.1).
Number of neutrons = A – Z = 81 – 35 = 46
Example 2.3 continued
Exercise 2.3A
Use the notation
to represent the isotope of tin having 66 neutrons.
Exercise 2.3B
Isobars are atoms with the same mass number but different atomic numbers. Indicate
the numbers of protons, neutrons, and electrons and the
notation for a cadmium atom
that is an isobar of tin-116.
Example 2.4
Use the data cited above to determine the weighted average atomic mass of carbon.
Strategy
The contribution each isotope makes to the weighted average atomic mass is given
by Equation (2.2). The weighted average atomic mass is the sum of the two
contributions.
Solution
The contributions are
Contribution of carbon-12 = 0.98892 x 12.00000 u = 11.867 u
Contribution of carbon-13 = 0.01108 x 13.00335 u = 0.1441 u
The weighted average mass is
Atomic mass of carbon = 11.867 u + 0.1441 u = 12.011 u
Assessment
This is the value listed in a table of atomic masses. As expected, the atomic mass of
carbon is much closer to 12 u than to 13 u.
Exercise 2.4A
There are three naturally occurring isotopes of neon. Their percent abundances and
atomic masses are neon-20, 90.51%, 19.99244 u; neon-21, 0.27%, 20.99395 u; neon-22,
9.22%, 21.99138 u. Calculate the weighted average atomic mass of neon.
Exercise 2.4B
The two naturally occurring isotopes of copper are copper-63, mass 62.9298 u, and
copper-65, mass 64.9278 u. What must be the percent natural abundances of the two
isotopes if the atomic mass of copper listed in a table of atomic masses is 63.546 u?
Example 2.5 An Estimation Example
Indium has two naturally occurring isotopes and a weighted average atomic mass of
114.82 u. One of the isotopes has a mass of 112.9043 u. Which is likely to be the second
isotope: 111In, 112In, 114In, or 115In?
Analysis and Conclusions
The masses of isotopes differ only slightly from whole numbers, which tells us that
the isotope with mass 112.9043 u is 113In. To account for the observed weighted
average atomic mass of 114.82 u, the second isotope must have a mass number
greater than 114. It can be only 115In.
Exercise 2.5A
The masses of the three naturally occurring isotopes of magnesium are 24Mg,
23.98504 u; 25Mg, 24.98584 u; 26Mg, 25.98259 u. Use the atomic mass given inside the
front cover to determine which of the three is the most abundant. Can you determine
which is the second most abundant? Explain.
Exercise 2.5B
For the three magnesium isotopes described in Exercise 2.5A, (a) could the percent
natural abundance of 24Mg be 60.00%? (b) What is the smallest possible value for the
percent natural abundance of 24Mg?
Example 2.6
Write the molecular formula and name of a compound for which each molecule contains
six oxygen atoms and four phosphorus atoms.
Strategy
We need to write the chemical symbols of the two elements and use the stated
numbers of atoms as subscripts following the symbols. We then determine which
element to place first in the molecular formula.
Solution
We represent the six atoms of oxygen as O6 and the four atoms of phosphorus as P4.
According to the scheme in Figure 2.9, the element O is followed only by F.
Phosphorus therefore comes first in the formula; we write P 4O6.
The name of a compound with four (tetra-) P atoms and six (hexa-) oxygen
atoms in its molecules is tetraphosphorus hexoxide.
Exercise 2.6A
Write the molecular formula and name of a compound for which each molecule contains
four fluorine atoms and two nitrogen atoms.
Exercise 2.6B
Write the molecular formula and name of a compound the molecules of which each
contain one oxygen atom and eight sulfur atoms.
Example 2.7
Write (a) the molecular formula of phosphorus pentachloride and (b) the name of S2F10.
Solution
(a) Choosing which element symbol goes first. The order of elements shown in the
molecular formula must be the same as the order in the name.
Writing subscripts. The lack of a prefix on phosphorus signifies one P atom per
molecule. The prefix penta- indicates five chlorine atoms. The molecular formula
is PCl5.
(b) The subscripts indicate two (di-) sulfur atoms and ten (deca-) fluorine atoms. The
compound is disulfur decafluoride.
Exercise 2.7A
Write (a) the molecular formula of tetraphosphorus decoxide and (b) the name of S7O2.
Exercise 2.7B
Write a plausible molecular formula for a compound that has one sulfur atom, two
oxygen atoms, and two fluorine atoms in each of its molecules. Comment on any
ambiguity that exists in this case.
Example 2.8
Determine the formula for (a) calcium chloride and (b) magnesium oxide.
Solution
(a) First we write the symbols for the ions, with the cation first: Ca 2+ and Cl–. The
simplest combination of these ions that gives an electrically neutral formula unit is
one Ca2+ ion for every two Cl– ions. The formula is CaCl2 .
Ca2+ + 2 Cl– = CaCl2
(b) Figure 2.10 tells us that the ions are Mg2+ and O2–. The simplest ratio for an
electrically neutral formula unit is 1:1. The formula of this binary ionic compound
is MgO.
Mg2+ + O2– = MgO
Exercise 2.8A
Give the formula for each of the following ionic compounds:
(a) potassium sulfide
(b) lithium oxide
(c) aluminum fluoride
Exercise 2.8B
Give the formula for each of the following ionic compounds:
(a) chromium(III) oxide
(b) iron(II) sulfide
(c) lithium nitride
Example 2.9
What are the names of (a) MgS and (b) CrCl3?
Solution
(a) MgS is made up of Mg2+ and S2– ions. Its name is magnesium sulfide.
(b) From Figure 2.10 we see that there are two simple ions of chromium, Cr3+ and
Cr2+. Because there are three Cl– ions in the formula unit CrCl3, the cation must
have a charge of 3+, that is, it must be Cr3+. Because there are two chromium
cations, there are two chlorides, CrCl2 and CrCl3, and we must assign a different
name to each. Therefore, the name of our compound cannot be simply chromium
chloride; instead, to indicate the 3+ cation, we say the name is chromium(III)
chloride.
Exercise 2.9A
Name the following compounds:
(a) CaBr2
(b) Li2S
(c) FeBr2
(d) CuI
Exercise 2.9B
Write the name and formula for each of the following compounds:
(a) the sulfide of copper(I) (b) the oxide of cobalt(III) (c) the nitride of magnesium
Example 2.10
Write the formula for (a) sodium sulfite and (b) ammonium sulfate.
Solution
(a) It is possible to identify the sulfite ion without memorizing all the ions in Table 2.4.
If you remember the name and formula of one of the sulfur–oxygen polyatomic
anions, you should be able to deduce the names of others. Suppose you remember
that sulfate is SO42–. The -ite anion has one fewer oxygen atom, 3 instead of 4, and
so it is SO32–. The charges of the two ions in a formula unit must balance, which
means the formula unit of sodium sulfite must have Na + and SO32– in the ratio 2:1.
The formula is therefore Na2SO3.
(b) The ammonium ion is NH4+, and the sulfate ion is SO42–. A formula unit of
ammonium sulfate has two NH4+ ions and one SO42– ion. To represent the two
NH4+ ions, we place parentheses around the NH4, followed by a subscript 2,
(NH4)2, and thus arrive at the formula (NH4)2SO4.
Exercise 2.10A
What is the formula for (a) ammonium carbonate, (b) calcium hypochlorite, and
(c) chromium(III) sulfate?
Exercise 2.10B
Write a plausible formula for
(a) potassium aluminum sulfate
(b) magnesium ammonium phosphate
Example 2.11
What is the name of (a) NaCN and (b) Mg(ClO4)2?
Solution
(a) The ions in this compound are Na+, sodium ion, and CN–, cyanide ion
(see Table 2.4). The name of the compound is sodium cyanide.
(b) The ions present are Mg2+, magnesium ion, and ClO4–, perchlorate ion. The name
of the compound is magnesium perchlorate.
Exercise 2.11A
Name each of the following compounds:
(a) KHCO3
(b) FePO4
(c) Mg(H2PO4)2
Exercise 2.11B
Give a plausible name for the following:
(a) Na2SeO4
(b) FeAs
(c) Na2HPO3
Cumulative Example
Show that the following experiment is consistent with the law of conservation of mass
(within the limits of experimental error): A 10.00-g sample of calcium carbonate was
dissolved in 100.0 mL of hydrochloric acid solution (d = 1.148 g/mL). The products were
120.40 g of solution (a mixture of hydrochloric acid and calcium chloride) and 2.22 L of
carbon dioxide gas (d = 0.0019769 g/mL).
Strategy
This problem may give the initial impression of being quite formidable because
several data are given. Upon examination, it is much less challenging than you
might think. We must show that mass is conserved in the experiment. That means
that we must compare the mass of the starting materials to the mass of the end
products of the chemical change. If mass is conserved, the two masses should be
identical, within the limits of experimental error. Our job then is to find the masses
of the starting materials and of the end products.
Solution
Let’s begin by identifying the starting materials and end products. The context of
the problem makes it clear that calcium carbonate reacts with a hydrochloric acid
solution, and so calcium carbonate and the HCl solution are the starting materials.
The end products are another solution and carbon dioxide gas.
Starting mass: The mass of calcium carbonate is given. We can use the density and
the volume of the HCl solution to find its mass.
Cumulative Example continued
Solution continued
Then we can add the masses of the two starting materials.
Mass of products: This time, the mass of the solution is given. We must use volume
and density of carbon dioxide gas to find its mass. However, we must first convert
the volume in liters to milliliters because the density is given in grams per milliliter.
Then we can add the masses of the two products.
Assessment
We note that the masses of reactants and products are not exactly the same.
However, the 124.8 g of reactants is reported to four significant figures, and so it is
precise only to 0.1 g. The difference between the masses of the starting materials
and end products is less than 0.1 g. Clearly, the difference in masses is smaller than
the uncertainty in the mass of the starting materials. We can therefore conclude that
this experiment is consistent with the law of conservation of mass.