Honors Chemistry I – Answer Key
Empirical / Molecular Formulas
Problem Set #1
1. A sample is analyzed as containing 24.09 grams of potassium, 0.308 moles of manganese,
23
and 7.42 x 10 atoms of oxygen. What is the empirical formula?
24.09 g K x (1 mol / 39.098 g)
= 0.61614 mol / .308 = 2
0.308 mol Mn
= 0.308 mol / .308 = 1
23
23
7.42 x 10 atoms O x (1 mol / 6.022 x 10 ) = 1.232 mol O / .308 = 4
K2MnO4 – potassium manganate
2. Samples of a compound are found to be 27.91% Fe, 24.08% S, and 48.01% O.
Can you figure out the empirical formula of the anion based on your knowledge of
nomenclature?
27.91 g Fe x (1 mol / 55.845 g) = .49978 / .49978 = 1
24.08 g S x (1 mol / 32.065 g) = .75097 / .49978 = 1.503
48.01 g O x (1 mol / 15.999 g) = 3.0008 / .49978 = 6.019
x2=2
x2=3
x 2 = 12
Fe2S3O12 → Fe2(SO4)3 – iron(III) sulfate or ferric sulfate
3. A certain compound is known to have a molecular weight slightly under 320 grams. If a
sample contains 0.226 moles of aluminum, 8.21 grams of carbon, and 21.6 grams of oxygen,
what is the molecular formula?
0.226 mol Al
= .226 mol / .226 = 1
8.21 g C x (1 mol / 12.011 g) = .6835 mol / .226 = 3.02
21.6 g O x (1 mol / 15.999 g) = 1.350 mol / .226 = 5.97
AlC3O6 EM = 159
320 / 159 = 2 Al2C6O12 → Al2(C2O4)3 – aluminum oxalate
22
23
4. A sample of a complex organic protein contains 8.60 x 10 atoms of carbon, 1.89 x 10
22
22
atoms of hydrogen, 5.16 x 10 atoms of oxygen, and 2.58 x 10 atoms of nitrogen. This
4.000-gram sample is known to be 0.007140 moles.
A) What is the empirical formula of this compound?
8.60 x 1022 atoms C x (1 mol / 6.022 x 1023) = .1428 mol / .04284 = 3.333
1.89 x 1023 atoms H x (1 mol / 6.022 x 1023) = .3138 mol / .04284 = 7.325
5.16 x 1022 atoms O x (1 mol / 6.022 x 1023) = .08569 mol / .04284 = 2.000
2.58 x 1022 atoms N x (1 mol / 6.022 x 1023) = .04284 mol /.04284 = 1
x 3 = 10
x 3 = 22
x3=6
x3=3
C10H22O6N3
B) What is the molecular weight of the empirical formula?
280.299 g/mol
C) What is the molecular weight of the molecular formula?
4.000 g / 0.007140 mol = 560.2 g/mol
D) What is the molecular formula?
560.2 / 280.299 = 2
C20H44O12N6
-2-
5. A compound with the empirical formula CH has a molar mass of 78 grams/mole. What is its
molecular formula?
78 / 13.02 = 6
C6H6 – benzene
6. 250.0 ml of a vapor weighs 1.035 grams. If the compound consists of 30.43% nitrogen and
69.57% oxygen, what is the empirical formula of this compound? What is the molecular
formula of this compound?
30.43 g N x (1 mol / 14.007 g) = 2.172 / 2.172 = 1
69.57 g O x (1 mol / 15.999 g) = 4.348 / 2.172 = 2.002
NO2 EM = 30.006 g/mol
MM = .2500 x (1 mol / 22.4) = 0.0112 mol
92.4 / 30.006 = 3
1.035 g / 0.0112 = 92.4 g/mol
N3O9
7. 50.0 grams of sulfur is mixed with 100.0 grams of metallic iron and the mixture is heated.
When the reaction is completed, 12.7 grams of iron remains unreacted. What is the empirical
formula of the compound formed?
50.0 g S x (1 mol / 32.065 g) = 1.559 / 1.559 = 1
87.3 g Fe x (1 mol / 55.845g) = 1.563 / 1.559 = 1.002
FeS – iron(II) sulfide or ferrous sulfide
8. When 40.0 grams of a sample of a pure chromium oxide is heated, 0.600 moles of oxygen gas
- O2 - is released. What is the empirical formula?
0.600 mol O2 = 1.200 mol O = 19.1988 g O
40.0 – 19.2 g O = 20.8 g Cr x (1 mol / 51.996) = .4000 mol Cr / .4000 = 1
= 1.200 mol O / .4000 = 3
CrO3 – chromium(VI) oxide
9. When 12.60 grams of magnesium metal is ignited in air, it reacts with nitrogen to produce
17.44 grams of a compound that has a molecular weight of 100.9 g/mole. What is the
formula of this compound?
17.44 g – 12.60 g = 4.84 g N x (1 mol / 14.007 g) = .3455 mol / .3455 = 1
12.60 g Mg x (1 mol / 24.305 g) = .5184 mol / .3455 = 1.500
x2=2
x2=3
Mg3N2 – magnesium nitride
10.
When 4.000 grams of the hydrated salt of sodium tetraborate is heated, the anhydrous
residue is found to make up only 52.78% of the original mass (the remainder being water).
What is the correct formula for this hydrate?
Na2B4O7 – sodium tetraborate
52.78 g Na2B4O7 x (1 mol / 201.217 g) = .2623 / .2623 = 1
47.22 g H2O x (1 mol / 18.015 g) = 2.621 / .2623 = 9.993
Na2B4O7·10H2O - sodium tetraborate decahydrate
Honors Chemistry I
Molarity / Empirical Formula / Molecular Formula Worksheet
Solution Set
1. An experimenter heated a finely divided sample of vanadium metal with flowers of sulfur
(powdered sulfur), recording the data as indicated below. Using these data, complete the data
table. Be sure to include the proper units.
DATA TABLE:
--------------------------------------------------------------------------------------------------------------------------Mass of crucible, cover, and vanadium sample........................................... 28.316 g
Mass of crucible, and cover......................................................................... 26.275 g
Mass of vanadium sample........................................................................ 2.041 g
Mass of crucible, cover, and residue......................................................... 30.277 g
Mass of sulfur used.................................................................................
1.961 g
Mass of vanadium sulfide residue............................................................. 4.002 g
Moles of vanadium reacted..................................................................... 0.0401 mol
Moles of sulfur reacted........................................................................... 0.0611 mol
2:3
Mole ratio of vanadium:sulfur................ 0.0401: 0.0611 => 1: 1.5 =>
Empirical formula of compound............................................................. V2S3
Number of atoms of vanadium used………. 0.0401 mol x Av# =
2.41 x 1022
Number of atoms of sulfur used.................... 0.0611 mol x Av# =
3.68 x 1022
Percentage of vanadium in the compound.. (2.041 g / 4.002 g) x 100 =
51.00%
=====================================================================
=====
2.
A student analyzes a compound and found it to contain 66.80% silver, 15.90% vanadium,
with the remainder being oxygen. Determine the empirical formula of the compound.
66.80 g / 107.9 g/mol = 0.619 mol
/ 0.312 = 2
x2 =4
3.
15.90 g / 50.9 g/mol
= 0.312 mol
/ 0.312 = 1
x2 =2
17.30 g / 16.0 g/mol
= 1.08 mol
/ 0.312 = 3.46 x 2 = 7
Ag4V2O7
A student reacted 0.273 grams of magnesium metal with atomic gaseous nitrogen,
producing a residue having a mass of 0.378 grams. If 4.000 grams of this compound is
analyzed to be 0.01321 moles of compound, what is the true formula?
0.378 g - 0.273 g = 0.105 g = mass of nitrogen
0.105 g / 14.0 g/mol = 0.00750 mol N
mol Mg
0.273 g / 24.3 g/mol = 0.0112
0.0112 / 0.00750 = 1.50 : 1 = 3 : 2 => Mg3N2
g/mol
4.000 g / 0.01321 mol = 302.8
M.Wt = 100.9 g/mol
302.8 / 100.9 = 3
Mg9N6
4.
24
Analysis of a compound contains 75.420% carbon, 3.9919 x 10 atoms of hydrogen,
8.380grams nitrogen, and 0.59806 moles oxygen. What is the empirical formula of this
organic compound?
C:
H:
N:
O:
75.420 g / 12.0 g/mol
24
3.9919 x 10 atoms / Av#
8.380 g / 14.0 g/mol
0.59806 moles
= 6.29 mol C / 0.598
= 6.63 mol H / 0.598
= 0.599 mol N / 0.598
= 0.598 mol O / 0.598
= 10.5 x2 = 21
= 11
x2 = 22
= 1
x2 = 2
= 1
x2 = 2
C21H22N2O2
5.
If a hydrate of cobalt (II) chloride was analyzed to be 45.397% water. What is the formula
of this hydrate?
45.397 g / 18.0 g/mol = 2.52 mol H2O
CoCl2
2.52 mol / 0.420 mol = 6.00 : 1 
54.603 g / 129.9 g/mol = 0.420 mol
CoCl2. 6 H2O
6.
A 200.0-ml sample of a solution of sodium thiosulfate was analyzed to contain 3.22x10
atoms of sodium. What is the molarity of this solution? Na2S2O3
22
22
3.22x10 atoms of Na ( 1 mol Na / Av#)( 1 mol cpd / 2 mol Na)
---------------------------------------------------------------------------- = [Na2S2O3] =
0.134 M
0.2000 L
7.
How many kilograms of metallic gold can be recovered from 15.0 liters of a 5.00 M
solution of auric oxalate?
15.0 L
5.00 mol cpd
2 mol Au
197.0 g Au
1 kg
x --------------- x --------------- x -------------- x ----------
=
2.96 kg
Au
1 L soln.
8.
1 mol cpd
1 mol Au
1000 g
How many milliliters of solution can be prepared by dissolving 4.05 grams of stannic
pyrosulfate in enough water to make a 2.00 M solution?
Sn(S2O7)2 M.Wt = 471.1
g/mol
4.05 g ( 1 mol cpd / 471.1 g) ( 1000 mL soln / 2.00 mol cpd) = 4.30 mL
9.
From the problem above (#8), what would be the concentration of tin (+4) ion?
[Sn+4] = 2.00 M
10.
If you took 5.00 ml of the original solution in problem #8 and added 35.0 ml of water to it,
what would be the molarity of the final, diluted solution?
M1V1 = M2V2
M2 = (2.00 M)(5.00 mL) / 40.00 mL) = 0.250 M
11.
-
What is the [Br ] if 56.0 grams of chromium (III) bromide is dissolved into 300.0ml of a
1.00 M solution of cupric bromide?
(56.0 g cpd / 291.7 g/mol )x 3 (/ 0.3000 L x (3 mol Br-1 / 1 mol cpd) = [Br-1] =
1.92 M
12.
How many grams of mercuric nitrate must be dissolved in 500.0 ml of water if you wish to
make a 0.650 M solution of the compound?
0.5000 L soln. ( 0.650 mol cpd / 1 L soln) ( 324.6 g / 1 mol cpd) = 105 g cpd.
13.
If 2.50 liters of nitric oxide gas is bubbled through (dissolved) 5.00 liters of water, what is
the molarity of the resulting solution? This solution is prepared at STP.
[2.50 L NO / (22.4 L/ 1 mol)] / 5.00 L = 0.0223 M
14. Three solutions are mixed:
250.0 ml of a 3.00 M ferric nitrate solution
500.0 ml of a 1.90 M ferric chloride solution
100.0 ml of a 5.85 M plumbic nitrate solution
What is the concentration (molarity) of each of the four ions?
moles Fe+3 = (0.2500 L)(3.00 M) = 0.750 moles + (0.5000 L)(1.90 M) = 0.950
moles
= 1.700 moles Fe / 0.850 L = [Fe+3] = 2.00 M
moles Pb+4 = (0.1000 L)(5.85 M) = 0.585 moles
= 0.585 moles Fe / 0.850 L = [Pb+4] = 0.688 M
moles Cl-1 = (0.5000 L)(1.90 M) x 3 = 2.850 moles
= 2.850 moles Cl / 0.850 L = [Cl-1] = 3.35 M
moles NO3-1 = (0.2500 L)(3.00 M) x 3 = 2.25 moles + (0.1000 L)(5.85 M) x 4 =
2.34 moles
= 4.59 moles Fe / 0.850 L = [NO3-1] = 5.40 M