Honors Chemistry I



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, and 7.42 x 1023 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

7.42 x 1023 atoms O x (1 mol / 6.022 x 1023) = 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 x 2 = 2

24.08 g S x (1 mol / 32.065 g) = .75097 / .49978 = 1.503 x 2 = 3

48.01 g O x (1 mol / 15.999 g) = 3.0008 / .49978 = 6.019 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

 

 

4. A sample of a complex organic protein contains 8.60 x 1022 atoms of carbon, 1.89 x 1023 atoms of hydrogen, 5.16 x 1022 atoms of oxygen, and 2.58 x 1022 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 x 3 = 10

1.89 x 1023 atoms H x (1 mol / 6.022 x 1023) = .3138 mol / .04284 = 7.325 x 3 = 22

5.16 x 1022 atoms O x (1 mol / 6.022 x 1023) = .08569 mol / .04284 = 2.000 x 3 = 6

2.58 x 1022 atoms N x (1 mol / 6.022 x 1023) = .04284 mol /.04284 = 1 x 3 = 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 1.035 g / 0.0112 = 92.4 g/mol

92.4 / 30.006 = 3 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 x 2 = 2

12.60 g Mg x (1 mol / 24.305 g) = .5184 mol / .3455 = 1.500 x 2 = 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

Mole ratio of vanadium:sulfur................ 0.0401: 0.0611 => 1: 1.5 => 2 : 3

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%

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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 Ag4V2O7

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 => Mg3N2 4.000 g / 0.01321 mol = 302.8 g/mol

M.Wt = 100.9 g/mol 302.8 / 100.9 = 3

Mg9N6

4. Analysis of a compound contains 75.420% carbon, 3.9919 x 1024 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 1024 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

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 54.603 g / 129.9 g/mol = 0.420 mol CoCl2

2.52 mol / 0.420 mol = 6.00 : 1 ( CoCl2. 6 H2O

6. A 200.0-ml sample of a solution of sodium thiosulfate was analyzed to contain 3.22x1022 atoms of sodium. What is the molarity of this solution? Na2S2O3

3.22x1022 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. 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(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

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