FORMULAS,
PERCENTAGE COMPOSITION, AND THE MOLE
FORMULAS: A chemical formula shows the elemental composition of a
substance: the
chemical symbols show what elements are present and the numerical
subscripts show how
many atoms of each element there are in a formula unit. Examples:
NaCl: one sodium atom, one chlorine atom in a formula unit
CaCl2: one calcium atom, two chlorine atoms in a formula unit
Mg3N2: three magnesium atoms, two nitrogen atoms in a formula unit
The presence of a metal in a chemical formula indicates an ionic
compound, which is composed
of positive ions (cations) and negative ions (anions).
A formula with
only nonmetals indicates a
molecular compound (unless it is an ammonium, NH4
+, compound). Only ionic compounds are
considered in this Tutorial.
There are tables of common ions in your lecture text, p 56 (cations)
and p 57 (anions). A
combined table of these same ions can be found on the inside back cover
of the lecture text. A
similar list is on the next page; all formulas needed in this and
subsequent Tutorial problems can
be written with ions from this list.
Writing formulas for ionic compounds is very straightforward: TOTAL
POSITIVE CHARGES
MUST BE THE SAME AS TOTAL NEGATIVE CHARGES. The formula must be
neutral. The
positive ion is written first in the formula and the name of the
compound is the two ion names.
EXAMPLE: Write the formula for potassium chloride.
The name tells you there are potassium, K+, and chloride,
Cl–, ions. Each potassium ion is +1
and each chloride ion is -1: one of each is needed, and the formula for
potassium chloride is
KCl. "1" is never written as a subscript.
EXAMPLE: Write the formula for magnesium hydroxide.
This contains magnesium, Mg2+, and hydroxide, OH–, ions. Each
magnesium ion is +2 and
each hydroxide ion is -1: two -1 ions are needed for one +2 ion, and
the formula for magnesium
hydroxide is Mg(OH)2. The (OH)2 indicates there are two OH–
ions. In a formula unit of
Mg(OH)2, there are one magnesium ion and two hydroxide ions; or one
magnesium, two
oxygen, and two hydrogen atoms. The subscript multiplies everything in
( ).
EXAMPLE: Write the formula for aluminum sulfate.
This contains aluminum, Al3+, and sulfate, SO4
2–, ions. The lowest common multiple of 3 and 2
is 6, so we will need six positive and six negative charges: two Al3+
and three SO4
2– ions, and
the formula for aluminum sulfate is Al2(SO4)3. Then, in a formula unit
of Al2(SO4)3 there are two
aluminum ions and three sulfate ions; or two aluminum, three sulfur,
and twelve oxygen atoms.
COMMON POSITIVE IONS COMMON NEGATIVE IONS COMMON NEGATIVE IONS
H+ hydrogen [Fe(CN)6]3– ferricyanide C2H3O2
– acetate
NH4
+ ammonium [Fe(CN)6]4– ferrocyanide CN– cyanide
Li+ lithium PO4
3– phosphate CNO– cyanate
Na+ sodium HPO4
2– hydrogen phosphate SCN– thiocyanate
K+ potassium H2PO4
– dihydrogen phosphate ClO– hypochlorite
Mg2+ magnesium CO3
2– carbonate ClO3
– chlorate
Ca2+ calcium HCO3
– hydrogen carbonate ClO4
– perchlorate
Sr2+ strontium SO3
2– sulfite IO3
– iodate
Ba2+ barium HSO3
– hydrogen sulfite MnO4
– permanganate
Al3+ aluminum SO4
2– sulfate NO2
– nitrite
Sn2+ tin(II) HSO4
– hydrogen sulfate NO3
– nitrate
Sn4+ tin(IV) S2O3
2– thiosulfate OH– hydroxide
Pb2+ lead(II)1 CrO4
2– chromate IO4
– periodate
Bi3+ bismuth Cr2O7
2– dichromate H– hydride
Cr3+ chromium(III)2 O2– oxide F– fluoride
Mn2+ manganese(II)3 O2
2– peroxide Cl– chloride
Fe2+ iron(II) S2– sulfide Br– bromide
Fe3+ iron(III) HS– hydrogen sulfide I– iodide
Co2+ cobalt(II)4
Ni2+ nickel(II)5
Cu+ copper(I)
Cu2+ copper(II) 1There is also a lead(IV)
Ag+ silver 2There is also a chromium(II)
Zn2+ zinc 3There is also a manganese(III)
Cd2+ cadmium 4There is also a cobalt(III)
Hg2
2+ mercury(I) 5There is also a nickel(III)
Hg2
2+ mercury(II)
T-8
PERCENTAGE COMPOSITION: Imagine a class of 40 boys and 60 girls: 100
students total.
40/100 of the students are boys and 60/100 girls; or, 40% boys and 60%
girls. The "%" sign
means percent, or parts per 100.
Suppose a class has 10 boys and 15 girls, for a total of 25 students.
If you want to find out how
many boys there would be per 100 students, keeping the same ratio of
boys to girls, the
following proportion can be set up:
10 boys X boys
───────── = ──────────
25 students 100 students
This means: "if there are 10 boys per 25 students, then there would be
X boys per 100
students." Solving this equation,
10 boys
X = ────────── x 100 students = 40 boys
25 students
There would be 40 boys per 100 students, or 40% boys. Percent
calculation is based on this
kind of proportion. However, you do not have to set up the proportion
each time you want to
find percent composition. Simply divide the number of parts you are
interested in (number of
boys) by the total number of parts (number of students), and multiply
by 100. This is exactly
what we did to find X in the above example.
DIVIDE THE PART BY THE WHOLE AND MULTIPLY BY 100
Suppose a mixture contains 12.4 g salt, 15.3 g sugar and 50.3 g sand,
for a total 78.0 g of
mixture. The percentage of each component in the mixture can be found
by dividing the amount
of each (PART) by the total amount (WHOLE) and multiplying by 100:
12.4 g 15.3 g
────── x 100 = 15.9% salt ────── x 100 = 19.6% sugar
78.0 g 78.0 g
50.3 g
────── x 100 = 64.5%
78.0 g
The sum of these percentages is 100%. Everything must total 100%.
T-9
Percentage is written without units but the units are understood to be
"parts of whatever per 100
total parts." The "parts" and "whole" may be in any units as long as
they are both in the same
units. In the case of the above mixture, units could be put on the
numbers and the percentages
then used as conversion factors. For example, the 15.9% salt could
become
15.9 g salt 100.0 pounds mixture 15.9 tons salt
──────────── or ──────────────── or ──────────────
100.0 g mixture 15.9 pounds salt 100.0 tons mixture
Et cetera. We do not usually care about composition of salt/sugar/sand
mixtures, but we do
care about percentage composition of pure chemical substances. Unless
otherwise stated we
always mean percent by mass (weight).
The formula mass of a substance is the sum of the atomic masses of all
the atoms in the
formula. For example, the formula mass of sodium chloride, NaCl, 58.44,
is the sum of the
atomic mass of sodium, 22.99, and the atomic mass of chlorine, 35.45.
The formula mass of
sodium sulfate, Na2SO4, is found as follows:
2 Na: 2 x 22.99 = 45.98 → 45.98 g Na or 45.98 lbs Na
1 S: 1 x 32.06 = 32.06 → 32.06 g S or 32.06 lbs S
4 O: 4 x 16.00 = 64.00 → 64.00 g O or 64.00 lbs O
───── ───────── ──────────
Formula mass = 142.04 142.04 g Na2SO4 142.04 lbs Na2SO4
Atomic mass and formula mass are in atomic mass units, but are usually
written without units.
However, since these are relative masses of the atoms any mass unit can
be used, as shown
by the two columns to the right of the arrows, above.
In terms of atomic masses: in a total of 142.02 parts by mass of sodium
sulfate, 45.98 parts are
sodium, 32.06 parts are sulfur, and 64.00 parts are oxygen.
In terms of a gram mass unit (molar mass): in a total of 142.04 g of
sodium sulfate, 45.98 g are
sodium, 32.06 g are sulfur, and 64.00 g are oxygen.
In terms of pound mass unit: in a total of 142.06 lbs of sodium
sulfate, 45.98 lbs are sodium,
32.06 lbs are sulfur, and 64.00 lbs are oxygen.
T-10
Percent composition can be determined with the mass ratios from the
formula mass:
45.98
Sodium: ───── x 100 = 32.37% Na → 32.37 g Na or 32.37 lbs Na
142.04
32.06
Sulfur: ───── x 100 = 22.57% S → 22.57 g S or 22.57 lbs S
142.04
64.00
Oxygen: ───── x 100 = 45.06% O → 45.06 g O or 45.06 lbs O
142.04
Total = 100.00% 100.00 g total 100.00 lbs total
As mentioned before, percentage is written without any units but the
units are understood to be
"parts of whatever per 100 total parts". Since these are relative
masses any mass unit can be
used, as shown by the two columns to the right of the arrows, above.
In terms of percentage: in a total of 100.00 parts by mass of sodium
sulfate, 32.37 parts are
sodium, 22.57 parts are sulfur, and 45.06 parts are oxygen.
In terms of a gram mass unit (molar mass): in a total of 100.00 g of
sodium sulfate, 32.36 g are
sodium, 22.57 g are sulfur, and 45.06 g are oxygen.
In terms of a pound mass unit: in a total of 100.00 lbs of sodium
sulfate, 32.36 lbs are sodium,
22.57 lbs are sulfur, and 45.06 lbs are oxygen.
In the preceding calculations, the percent of each element in the
substance was found. The
percent of groups of elements may also be calculated. Suppose you want
percent sulfate,
SO4
2–, in sodium sulfate. The formula mass of sodium sulfate is
142.04 and of that total
32.06 + 64.00 = 96.06 is sulfate. Thus,
96.06
Sulfate: ───── x 100 = 67.63% SO4
2–
142.04
Notice: this is the sum of the percents sulfur and oxygen: 22.57% S +
45.06% 0 = 67.63%
SO4
2–.
We may say sodium sulfate is 32.37% sodium, 22.57% sulfur, and 45.06%
oxygen; or, 32.37%
sodium and 67.63% sulfate. The total is 100% in either case.
T-11
We now have two sets of numbers which show mass relationships among the
elements in
sodium sulfate: (1) the formula mass and (2) the percentage
composition.
From the formula mass on page T-9 various mass ratios, that is,
conversion factors, may be
obtained:
45.98 g Na 142.04 lbs Na2SO4 32.06 g S
──────────── or ────────────── or ───────── and others
142.04 g Na2SO4 32.06 lbs S 45.98 g Na
From the percentages on page T-10 various mass ratios may also be
obtained:
32.37 g Na 100.00 lbs Na2SO4 22.57 g S
──────────── or ────────────── or ───────── and others
100.00 g Na2SO4 22.57 lbs S 32.37 g Na
We can now solve problems like the following:
EXAMPLE: What mass of sulfur is contained in 17.2 g of sodium sulfate?
Use formula mass ratios on page T-9 for the appropriate conversion
factor:
32.06 g S
17.2 g Na2SO4 x ───────────── = 3.88 g S
142.04 g Na2SO4
OR, use percentage composition ratios on page T-10:
22.57 g S
17.2 g Na2SO4 x ───────────── = 3.88 g S
100.00 g Na2SO4
T-12
EXAMPLE: What mass of sodium is combined with 10.0 g of sulfur in
sodium sulfate?
You may use a conversion factor from the formula masses OR percentage
composition:
45.98 g Na
10.0 g S x ───────── = 14.3 g Na
32.06 g S
OR
32.37 g Na
10.0 g S x ────────── = 14.3 g Na
22.57 g S
Obviously, you do not have to calculate percent composition every time
you do a problem like
this: use formula mass ratios; if the percentage composition is given,
then use it.
THE MOLE: The most important concept you encounter this semester in
Chemistry is the mole.
The mole may be defined:
The mole is the amount (mass) of a substance which contains the same
number
of units (atoms, molecules, ions) as there are atoms in exactly 12 mass
units of
the carbon-12, 12C, isotope.
It does not matter what mass units are used as long as the same mass
units are used for the
substance and the carbon-12. This is a strict definition of the mole,
but it is operationally
useless. A mole of a monoatomic element is the atomic mass taken with
mass units; a mole of
anything else is the formula mass taken with mass units. The kind of
mole you get depends on
the mass units you use. For example:
22.99 grams Na = 1 gram-mole Na has the same number of Na atoms as the
number of atoms in 12 grams of 12C.
63.55 milligrams Cu = 1 milligram-mole Cu has the same number of Cu
atoms
as the number of atoms in 12 milligrams of 12C.
70.90 grams Cl2 = 1 gram-mole Cl2 has the same number of Cl2 molecules
as
the number of atoms in 12 grams of 12C.
253.80 lbs I2 = 1 lb-mole I2 has the same number of I2 molecules as the
number of atoms in 12 lbs of 12C.
58.44 grams NaCl = 1 gram-mole NaCl has the same number of NaCl formula
units as the number of atoms in 12 grams of 12C.
T-13
142.04 tons Na2SO4 = 1 ton-mole Na2SO4 has the same number of Na2SO4
formula units as the number of atoms in 12 tons of 12C.
18.02 g H2O = 1 gram-mole H2O has the same number of H2O molecules as
the number of atoms in 12 g of 12C.
137.32 kg PCl3 = 1 kilogram-mole PCl3 has the same number of PCl3
molecules as the number of atoms in 12 kg of 12C.
Chemists generally use the gram-mole, written mole, abbreviated mol.
This is the mole your
textbook uses and is the only one for which Avogadro's Number is
applicable. In one grammole
of a substance there is Avogadro's Number of units: 6.022 x 1023 of
them. Unless further
qualification is noted the term "mole" is understood to mean the
gram-mole.
Each of the equalities above becomes a conversion factor:
22.99 g Na 1 mol Na
22.99 g Na = 1 mol Na becomes ────────── or ─────────
1 mol Na 22.99 g Na
And so on.
EXAMPLE: How many moles is 31.65 g of sodium metal?
1 mol Na
31.65 g Na x ────────── = 1.377 mol Na
22.99 g Na
EXAMPLE: What is the mass of 0.362 mol of sodium sulfate?
142 g Na2SO4
0.362 mol Na2SO4 x ──────────── = 51.4 g Na2SO4
1 mol Na2SO4
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The number of significant figures used in the atomic mass or formula
mass should be at least as
many as in the given (measured) data.
EXAMPLE: How many atoms of copper are in a 125 mg sample of copper
metal?
When a problem asks "how many atoms or molecules" or "what is the mass
of an atom or
molecule," you must use Avogadro's Number. The units associated with
that Number then
depend on the problem. In this problem you are looking for atoms of
copper so the units will be:
T-14
6.022 x 1023 atoms Cu
──────────────────
1 mol Cu
The number of atoms of copper is related to moles of copper; moles of
copper is related to
grams of copper; the mass of copper given in the problem is in
milligrams. The conversion
sequence will be:
mg Cu → g Cu → mol Cu → atoms Cu
1 g Cu 1 mol Cu 6.022 x 1023 atoms Cu
125 mg Cu x ───────── x ──────── x ───────────────── =
1000 mg Cu 63.5 g Cu 1 mol Cu
1.19 x 1021 atoms Cu
EXAMPLE: What is the mass of one water molecule?
This time the units associated with Avogadro's Number are
6.022 x 1023 H2O molecules
─────────────────────
1 mol H2O
We are converting from a number of molecules (1, an exact number) to
mass:
1 mol H2O 18.0 g H2O
1 H2O molecule x ─────────────────────── x ────────── =
6.022 x 1023 H2O molecules 1 mol H2O
2.99 x 10–23 g H2O
If four significant figures had been used for the formula mass of
water, 18.02, the answer would
have been 2.992 x 10–23 g H2O. In this case the precision of
the answer depends on the
conversion factors, since the given data is an exact number.
Chemical formulas should be considered in terms of moles of atoms
combined: the formula
NaCl means there is one mole of sodium combined with one mole of
chlorine; the formula H2O
means there are two moles of hydrogen combined with one mole of oxygen
to form one mole of
water.
T-15
The formula mass for sodium sulfate was calculated on page T-9.
Referring to those numbers,
the formula Na2SO4 means: two moles sodium (45.98 g), one mole sulfur
(32.06 g), and four
moles oxygen (64.00 g) combine to form one mole of sodium sulfate
(142.04 g). Let us use this
concept and do the first problem on page T-11 a different way:
EXAMPLE: What mass of sulfur is contained in 17.2 g of sodium sulfate?
1 mol Na2SO4 1 mol S 32.1 g S
17.2 g Na2SO4 x ──────────── x ──────────── x ──────── = 3.89 g S
142 g Na2SO4 1 mol Na2SO4 1 mol S
Please notice: The same numbers are used in the same places, so the
overall arithmetic is the
same. The thought process is different: on page T-11 we thought in
terms of mass ratios, now
we are thinking in terms of mole ratios.
This answer is slightly different from the one obtained on page T-11.
Can you see why?
EMPIRICAL (SIMPLEST) FORMULA: The empirical formula is the formula
expressed as the
lowest whole number ratio of atoms; it may or may not be the same as
the formula. The
empirical formula and formula of water, H2O, are the same; the
empirical formula of
phosphorous pentoxide is P2O5, but the formula is P4O10. The empirical
formula is derived
solely from experimental, percentage composition data. On page T-10 we
calculated
percentage composition from a formula; let us now calculate an
empirical formula from
percentage composition data. Fundamental to this calculation is the
fact: chemical formulas
show mole ratios of elements.
EXAMPLE: Calculate the empirical formula for a compound with the
analysis:
32.37% Na 22.57% S 45.06% O
Remember percentage means "parts of whatever per 100 total parts," so
let us translate these
percentages into grams of each element out of a total 100 grams of
compound. Then convert
these masses to moles:
1 mol Na
32.37 g Na x ────────── = 1.408 mol Na
22.99 g Na
1 mol S
22.57 g S x ───────── = 0.7040 mol S
32.06 g S
1 mol O
45.06 g O x ───────── = 2.816 mol O
16.00 g O
T-16
Please note: these are the number of moles of each element contained in
100.00 g of
compound.
Now find the ratio of moles of each element to the smallest number of
moles, by dividing each
number of moles by the smallest:
1.408 mol Na
─────────── = 2.000 mol Na / mol S
0.7040 mol S
0.7040 mol S
─────────── = 1.000
0.7040 mol S
2.816 mol O
─────────── = 4.000 mol O / mol S
0.7040 mol S
These ratios say there are 2 mol Na per 1 mol S; and 4 mol O per 1 mol
S. The lowest wholenumber
ratio of moles is: Na2S1O4; and since we do not write "1" as a
subscript, the formula is:
Na2SO4.
Sometimes these two steps do not give whole number ratios. In that case
an additional step is
needed, as shown below.
EXAMPLE: Calculate the empirical formula for a compound with the
analysis:
29.1% Na 40.5% S 30.4% O
Let us do as with the previous EXAMPLE: (1) translate the percents into
grams of the element
per 100 g of compound and find moles of each element; then (2) find the
ratio of moles of each
element to the smallest number of moles:
1 mol Na 1.27 mol Na
29.1 g Na x ──────── = 1.27 mol Na ────────── = 1.01 mol Na/mol S
23.0 g Na 1.26 mol S
1 mol S 1.26 mol S
40.5 g S x ──────── = 1.26 mol S ────────── = 1.00
32.1 g S 1.26 mol S
1 mol O 1.90 mol O
30.4 g O x ──────── = 1.90 mol O ────────── = 1.51 mol O/mol S
16.0 g O 1.26 mol S
T-17
These ratios say there is 1 mol Na per 1 mol S, and 1.5 mol O per 1 mol
S, which are not whole
number ratios. To make them whole number ratios we must multiply each
by 2, giving ratios of
2 mol Na per 2 mol S and 3 mol O per 2 mol S; and the formula is
Na2S2O3.
Thus, in cases where steps (1) and (2) do not lead to whole numbers,
(3) multiply each ratio by
some integer (2, 3, etc.) to make all the ratios whole numbers.
1) Write formulas for the following:
a) bismuth nitrate j) zinc phosphate s) copper(II) oxide
b) calcium hydride k) magnesium carbonate t) lithium nitrate
c) mercury(II) cyanide l) tin(II) chloride u) aluminum sulfide
d) strontium iodide m) lead(II) bromide v) iron(II) phosphate
e) tin(IV) hydroxide n) ammonium dichromate w) barium sulfate
f) potassium sulfite o) nickel(II) chromate x) silver nitrate
g) sodium carbonate p) cadmium cyanate y) copper(I) sulfide
h) cobalt(II) nitrite q) mercury(I) chloride z) iron(III) sulfate
i) chromium(III) iodide r) manganese(II) periodate
2) Calculate the percentage composition, by mass, of the following. Use
four significant
figures.
a) potassium phosphate e) zinc chloride
b) ammonium cyanide f) aluminum nitrate
c) sodium hydrogen sulfate g) chromium(III) hydroxide
d) potassium perchlorate h) copper(II) carbonate
3) Using the percentages from problem (2) calculate the following:
a) grams of phosphate in 65.5 g of potassium phosphate
b) milligrams of nitrogen in 250.0 mg of ammonium cyanide
c) pounds of sodium in 10.0 lbs of sodium hydrogen sulfate
d) grams of chlorine in 37.5 g of potassium perchlorate
e) tons of zinc in 15.25 tons of zinc chloride
f) tons of nitrate in 125 tons of aluminum nitrate
g) milligrams of oxygen in 1.500 g chromium(III) hydroxide
h) pounds carbon in 138 lbs copper(II) carbonate
T-18
4) Convert each of the following to moles:
a) 25.00 g of iron(III) sulfate
b) 3.67 x 10–2 g of lead(II) nitrate
c) 5.990 x 103 g of tin(II) bromide
d) 0.07785 kg of ammonium chromate
e) 175 g of manganese(II) carbonate
f) 8.5 x 10–6 g of zinc oxide
g) 0.4670 mg of barium sulfate
h) 3.55 x 105 mg of aluminum chloride
5) Calculate the mass (in grams) of each of the following:
a) 1.00 mol of barium sulfate
b) 2.550 mol of zinc nitrate
c) 0.155 mol of sodium hydrogen carbonate
d) 1.255 mol of potassium permanganate
e) 1.67 x 10–2 mol of sodium hydroxide
f) 4.990 x 10–4 mol of copper(I) chloride
g) 0.00585 mol of magnesium chloride
h) 1.45 x 102 mol of calcium bromide
6) Calculate the following:
a) grams of manganese in 1.55 g of manganese(II) nitrate
b) tons of phosphorous in 5.73 tons of potassium phosphate
c) moles of sulfur in 7.33 g of aluminum sulfate
d) pounds of iron in 12.6 lbs of iron(II) phosphate
e) moles of nitrogen in 5.55 g of lead(II) nitrate
f) grams of nitrogen in 3.95 g of ammonium phosphate
g) grams of iron in 1.37 kg of iron(III) sulfide
h) moles of oxygen in 100.0 g of iron(III) sulfate
7) Calculate the following:
a) number of sodium atoms in 6.79 g of sodium metal
b) number of sodium ions in 7.77 g of sodium phosphate
c) number of water molecules in 15.5 mg of water
d) number of cobalt atoms in 111.1 mg of cobalt metal
e) number of ammonia molecules, NH3, in 375 kg of ammonia
f) number of gold atoms in 1.00 x 10–6 mg of gold
g) number of ozone molecules, O3, in 1.73 x 10–8 g of ozone
h) mass, in grams, of ten water molecules
i) mass, in grams, of one mercury atom
j) mass, in milligrams, of three sulfuric acid, H2SO4, molecules
k) mass, in kilograms, of 1.25 x 1015 sulfate ions
l) mass, in kilograms, of 2.57 x 1025 ammonium ions
T-19
8) Calculate the empirical formulas for the compounds with the
following analyses:
a) 34.4% Fe 65.6% Cl
b) 29.4% Ca 23.6% S 47.0% O
c) 28.7% K 1.5% H 22.8% P 47.0% O
d) 28.0% Fe 24.1% S 48.0% O
e) 88.8% Cu 11.2% O
f) 72.3% Fe 27.7% O
g) 69.9% Fe 30.1% O
h) 11.1% N 3.2% H 41.3% Cr 44.4% O
i) 68.4% Cr 31.6% O
Answers to Problems
1) a) Bi(NO3)3 h) Co(NO2)2 o) NiCrO4 u) Al2S3
b) CaH2 i) CrI3 p) Cd(CNO)2 v) Fe3(PO4)2
c) Hg(CN)2 j) Zn3(PO4)2 q) Hg2Cl2 w) BaSO4
d) SrI2 k) MgCO3 r) Mn(IO4)2 x) AgNO3
e) Sn(OH)4 l) SnCl2 s) CuO y) Cu2S
f) K2SO3 m) PbBr2 t) LiNO3 z) Fe2(SO4)3
g) Na2CO3 n) (NH4)2Cr2O7
2) a) 55.26% K; 14.59% P; 30.15% O for K3PO4
b) 63.59% N; 9.152% H; 27.26% C for NH4CN
c) 19.15% Na; 0.8396% H; 26.71% S; 53.30% O for NaHSO4
d) 28.22% K; 25.59% Cl; 46.19% O for KCl04
e) 47.98% Zn; 52.02% Cl for ZnCl2
f) 12.67% Al; 19.73% N; 67.60% O for Al(NO3)3
g) 50.47% Cr; 46.59% O; 2.935% H for Cr(OH)3
h) 51.43% Cu; 9.720% C; 38.85% O for CuCO3
3) a) 29.3 g PO4
3– e) 7.317 tons Zn
b) 159.0 mg N f) 109 tons NO3
1–
c) 1.92 lbs Na g) 698.8 mg O
d) 9.60 g Cl h) 13.4 lbs C
T-20
4) a) 0.06252 mol Fe2(SO4)3 e) 1.52 mol MnCO3
b) 1.11 x 10–4 mol Pb(NO3)2 f) 1.0 x 10–7 mol ZnO
c) 21.51 mol SnBr2 g) 2.001 x 10–6 mol BaSO4
d) 0.5122 mol (NH4)2CrO4 h) 2.66 mol AlCl3
5) a) 233 g BaSO4 e) 0.668 g NaOH
b) 483.0 g Zn(NO3)2 f) 0.04940 g CuCl
c) 13.0 g NaHCO3 g) 0.558 g MgCl2
d) 198.3 g KMnO4 h) 2.90 x 104 g CaBr2
6) a) 0.476 g Mn in Mn(NO3)2 e) 0.0335 mol N in Pb(NO3)2
b) 0.837 ton P in K3PO4 f) 1.11 g N in (NH4)3PO4
c) 0.0642 mol S in Al2(SO4)3 g) 735 g Fe in Fe2S3
d) 5.90 lbs Fe in Fe3(PO4)2 h) 3.001 mol O in Fe2(SO4)3
7) a) 1.78 x 1023 Na atoms g) 2.17 x 1014 O3 molecules
b) 8.56 x 1022 Na1+ ions in Na3PO4 h) 2.992 x 10–22 g H2O
c) 5.19 x 1020 H2O molecules i) 3.331 x 10–22 g Hg
d) 1.135 x 1021 Co atoms j) 4.887 x 10–19 mg H2SO4
e) 1.33 x 1028 NH3 molecules k) 1.99 x 10–10 kg SO4
2–
f) 3.06 x 1012 Au atoms l) 0.768 kg NH4
1–
8) a) FeCl3
b) CaSO4
c) KH2PO4
d) Fe2(SO4)3
e) Cu2O
f) Fe3O4
g) Fe2O3
h) (NH4)2Cr2O7
i) Cr2O3
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