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
Mole ratio of vanadium:sulfur................ 0.0401: 0.0611 => 1: 1.5 => 2 : 3
Empirical formula of compound............................................................. V
2
S
3
Number of atoms of vanadium used………. 0.0401 mol x Av# = 2.41 x 10 22
Number of atoms of sulfur used.................... 0.0611 mol x Av# = 3.68 x 10 22
Percentage of vanadium in the compound.. (2.041 g / 4.002 g) x 100 = 51.00%
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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 x 2 = 4
15.90 g / 50.9 g/mol = 0.312 mol / 0.312 = 1 x 2 = 2
Ag
4
V
2
O
7
17.30 g / 16.0 g/mol = 1.08 mol / 0.312 = 3.46 x 2 = 7

3. 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 0.273 g / 24.3 g/mol = 0.0112 mol Mg
0.0112 / 0.00750 = 1.50 : 1 = 3 : 2 => Mg
3
N
2
4.000 g / 0.01321 mol = 302.8 g/mol
M.Wt = 100.9 g/mol 302.8 / 100.9 = 3
Mg
9
N
6
4. Analysis of a compound contains 75.420% carbon, 3.9919 x 10
24
atoms of hydrogen, 8.380grams nitrogen, and 0.59806 moles oxygen. What is the empirical formula of this organic compound?
C: 75.420 g / 12.0 g/mol = 6.29 mol C / 0.598 = 10.5 x2 = 21
H: 3.9919 x 10
24
atoms / Av# = 6.63 mol H / 0.598 = 11 x2 = 22
N: 8.380 g / 14.0 g/mol = 0.599 mol N / 0.598 = 1 x2 = 2
O: 0.59806 moles = 0.598 mol O / 0.598 = 1 x2 = 2
C
21
H
22
N
2
O
2
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 H
2
O 54.603 g / 129.9 g/mol = 0.420 mol CoCl
2
2.52 mol / 0.420 mol = 6.00 : 1  CoCl
2
.
6 H
2
O
6. A 200.0-ml sample of a solution of sodium thiosulfate was analyzed to contain 3.22x10
22 atoms of sodium. What is the molarity of this solution? Na
2
S
2
O
3
3.22x10
22 atoms of Na ( 1 mol Na / Av#)( 1 mol cpd / 2 mol Na)
---------------------------------------------------------------------------- = [Na
2
S
2
O
3
] = 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. 1 mol cpd 1 mol Au 1000 g
8. 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(S
2
O
7
)
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?
M
1
V
1
= M
2
V
2
M
2
= (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 NO
3
-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 = [NO
3
-1 ] = 5.40 M