Unit III The Mole

Lesson Date Topic WS

1. The Molar Mass 1

2. Mole Conversions 2

3. Calculating Atoms 3

4. Empirical Formula 4

5. Percentage Composition Empirical Formula 5

6. Mole Calculations to Molecular Formula 6

7. Identify that Gas 7

8. Practice Test 1 8

9. Practice Test 2 9

Worksheet # 1 Mole conversions

Describe each particle as an atom, molecule or a formula unit.

1. CO₂ m

2. KCl fu

3. C at

4. AgNO₃ fu

5. NH₄CH₃COO fu

6. O₂ m

7. Os at

8. SO₃ m

9. RbCl fu

10. CaCO₃ fu

11. Ag at

12. NH₃ m

13. Cl₂ m

14. Se at

15 The electrolysis of water using H₂SO₄ as a catalyst is used to generate hydrogen gas. For every 1 electron

consumed 1 atom of hydrogen (exact) is generated. An amp meter is used to measure the rate of electron

consumption.

1.00 coulomb = 6.24 x 10 ¹⁸ electrons.

If 1.00 g of hydrogen is generated at a current of 1.50 amps (1 amp = 1 coulomb per second) and over time

period of 17.865 hours, calculate Avogadro’s number (the number of atoms in 1.00g of hydrogen). Use unit

analysis! Start the unitanalysis with 17.865 h and end with atoms H.

17.865 h x 3600 s x 1.5 coulomb x 6.24 x 10¹⁸ e x 1 atom H = 6.02 x 10²³ atoms H

1 h 1s 1 coulomb 1 e

16. State Avogadro's # three different ways describing the number of atoms in a mole, formula units in a mole, and molecules in a mole.

6.02x 10²³ atoms = 1 mole

6.02x 10²³ formula units = 1 mole

6.02 x 10²³ molecules = 1 mole

Use Unit Analysis to Change from particles to moles or from moles to particles. Note that a particle is an atom, molecule, or formula unit.

17. Covert 5.44 x 10²⁶ at Co to moles.

5.44 x 10²⁶atoms Co x 1 mole = 904 moles

6.02 x 10²³ atoms

18. Convert 2.4 moles CO₂ to molecules.

2.4 moles CO₂ x 6.02 x 10²³ molecules = 1.4 x 10²⁴ molecules

1 mole

19. Covert 4.56 x 10²⁴ molecules CO₂ to moles.

4.56 x 10²⁴molecules CO₂ x 1 mole = 7.57 moles

6.02 x 10²³ molecules

20. Convert 10.9 moles CuSO₄ to FUs.

10.9 moles CuSO₄ x 6.02 x 10²³ FU’s = 6.56 x 10²⁴ FU’s

1 mole

21. Convert 5.33 x 10²⁵ FU to moles NaCl.

5.33 x 10²⁵FU’s NaCl x 1 mole = 88.5 moles

6.02 x 10²³ FU’s

22. Convert 2.4 moles C to atoms.

2.4 moles C x 6.02 x 10²³ atoms = 1.4 x 10²⁴atoms

1 mole

23. Covert 9.11 x 10²⁵ molecules SO₃ to moles.

9.11 x 10²⁵molecules SO₃ x 1 mole = 151 moles

6.02 x 10²³ FU’s

24. Convert 2.9 moles CaSO₄ to FUs.

2.9 moles CaSO₄ x 6.02 x 10²³ FU’s = 1.7 x 10²⁴ FU’s

1 mole

24. Convert 5.55 x 10²⁴ FU to moles KF.

5.55 x 10²⁴FU’s KF x 1 mole = 9.22 moles

6.02 x 10²³ FU’s

Draw Electron-dot Diagrams for each of the following. Remember, use brackets for cations or anions. Covalent compounds do not require brackets.

37. CCl₄

: Cl :

.. .. ..

: Cl ׃ C ׃ Cl :

.. .. ..

: Cl :

38. NaCl

[ Na ]⁺ [ : Cl : ]^-

39. CO₂

.. ..

: O : : C : : O :

40. Li₂SO₄

[ Li: ]⁺

: O :

.. .. ..

: O : S : O :

.. .. ..

: O :

[ Li ]⁺

41. HCN

H : C : : : N :

.. ..

: O : N :: O :

.. ..

: O :

Worksheet #2 The Mole

Calculate the molecular mass of the following compounds. Remember to determine the mass of each element and add the mass of the elements together.

1. KI 166.0 g/mole 2. Mg₃(PO₄)₂262.9 g/mole

3. NiO 74.7 g/mole 4. H₂O 18.02 g/mole

5. BaF₂ 175.3 g/mole 6. Ti(SO₄)₂^.6H₂O348.22 g/mole

7. CuS 95.6 g/mole 8. ZnHCO₃126.41 g/mole

9. Li₃N 34.7 g/mole 10. PdSO₄202.5 g/mole

11. NaI 149.9 g/mole 12. K₃PO₄212.3 g/mole

13. FeO^.6H₂O 179.92 g/mole 14. NaOH 40.01 g/mole

15. AlF₃^.8H₂O 228.16 g/mole 16. Ga₂(CO₃)₃319.4 g/mole

17. CuH₂ 65.52 g/mole 18. Zn(CH₃COO)₂183.46 g/mole

19. NiN 72.7 g/mole 20. BaSO₄^.5H₂O 323.5 g/mole

Define the following. The first one is done for you.

21. Atomic mass is the mass of 1 mole of atoms.

22. Molecular mass is the mass of 1 mole of molecules.

23. Formula mass is the mass of 1 mole of formula units.

24. Molar mass is the mass of 1 mole of particles.

25. Circle the formula units, square the molecules, and underline the atoms.

Oval: KCl Oval: NH4Cl

H₂O

CO₂

P₂O₅

K

Au

C₁₂H₂₄O₁₁

Oval: NaCl

Oval: K2SO4

NH₃

Cu

Br

C

Convert from grams to moles. Show all work using unit analysis

26. 100. g H₂O

100. g H₂O x 1 mole = 5.55 moles

18.02 g

27. 250. g MgCl₂

250. g MgCl₂x1 mole = 2.62 moles

95.3 g

28. 0.266 g C₆H₁₂

0.266 g C₆H₁₂x1 mole = 0.00316 moles

84.12 g

29. 1.2 x 10^-4 g Fe

1.2 x 10^-4 g Fe x1 mole = 2.2 x 10^-6 moles

55.8 g

Convert each quantity to moles.

30. 500. g H₂O x 1 mole = 27.7 mole

18.02 g

31. 0.269 g P₂O₅x 1 mole = 1.89 x10^-3 mole

142 g

32. 135.3 g CaSO₄x 1 mole = 0.9934 mole

136.2 g

33. 214.7 g CH₄x 1 mole = 13.39 mole

16.04 g

34. 25.3 g C₆H₁₂O₆x 1 mole = 0.141 mole

180.12 g

Convert from atoms, molecules, or formula units to grams.

35.

6.55 x 10²⁴ atoms Ag x 1 mole x 107.9 g = 1.17 x 10³ g

6.02 x 10²³ at 1 mole

36.

8.66 x 10²⁶ FU RbCl x 1 mole x 121 g = 1.174 x 10⁵ g

6.02 x 10²³ FU 1 mole

37.

5.00 x 10 ³² molecs N₂H₄ x 1 mole x 32.04 g = 2.66 x 10¹⁰ g

6.02 x 10²³ molecs 1 mole

38.

5.3 x 10²⁹ FU KBr x 1 mole x 119 g = 1.0 x 10⁸ g

6.02 x 10²³ FU 1 mole

39.

25.3 x 10²⁸ molecules C₄H₁₀x 1 mole x 58.1 g = 2.44 x 10⁷ g

6.02 x 10²³ molecules 1 mole

40.

1.33 x 10²⁵ atoms Ag x 1 mole x 107.9 g = 2.38 x 10³g Ag

6.02 x 10²³ at 1 mole

41.

1.55 x 10¹⁶ FU NaCl x 1 mole x 58.5 g = 1.51 x 10^-6 g NaCl

6.02 x 10²³ FU 1 mole

42.

2.55 x 10 ²⁷ molecules NH₃ x 1 mole x 17.03 g = 7.21 x 10⁴ g NH₃

6.02 x 10²³ molecules 1 mole

43.

5.3 x 10²⁹ FU KBr x 1 mole x 119 g = 1.05 x 10⁸ g

6.02 x 10²³ FU 1 mole

Convert from grams to atoms, molecules, or formula units.

44.

100. g H₂O₂ x 1 mole x 6.02 x 10²³ molecules = 1.77 x 10²⁴ molecules

34.02 g 1 mole

45.

2.6 g MgF₂x 1 mole x 6.02 x 10²³ FU = 2.5 x 10²² FU

62.3 g 1 mole

46.

0.211 g C₅H₁₂x 1 mole x 6.02 x 10²³ molecules = 1.76 x 10²¹ molecules

72.12g 1 mole

47.

3.33 x 10^-2 g Fe x 1 mole x 6.02 x 10²³ Atoms = 3.59 x 10²⁰ atoms

55.8 g 1 mole

48.

0.126 g Co x 1 mole x 6.02 x 10²³ at = 1.29 x 10²¹ atoms Co

58.9 g 1 mole

Worksheet #3 Calculating Atoms

1. Measure the mass some Calcium. Calculate the number of atoms. Add 50 mL water to a 250 mL fleaker. Add the calcium to the beaker and cover with a plastic funnel. Light the funnel after about 12 seconds. Write a balanced chemical equation for the reaction.

112 g Ca x 1 mole x 6.02 x 10²³ at = 1.68 x 10²⁴ at Ca

40.1 g 1 mole

2. Measure the mass of a Copper cylinder. Calculate the number of atoms in the cylinder.

112 g Cu x 1 mole x 6.02 x 10²³ at = 1.06 x 10²⁴ at Cu

63.5 g 1 mole

3. Measure the mass of an Iron cylinder. Calculate the number of atoms in the cylinder.

112 g Fe x 1 mole x 6.02 x 10²³ at = 1.21 x 10²⁴ at Fe

55.8 g 1 mole

4. Measure the mass of a Magnesium cylinder. Calculate the number of atoms in the cylinder.

112 g Mg x 1 mole x 6.02 x 10²³ at = 2.77 x 10²⁴ at Mg

24.3 g 1 mole

5. Measure the mass of an Aluminum cylinder. Calculate the number of atoms in the cylinder.

112 g Al x 1 mole x 6.02 x 10²³ at = 2.50 x 10²⁴ at Al

27.0 g 1 mole

6. Measure the mass of a small amount of dry ice (be quick- it won’t last long due to sublimation). Calculate the number of atoms of CO₂.

112 g CO₂ x 1 mole x 6.02 x 10²³ molecules = 1.53 x 10²⁴ molecules CO₂

44.0 g 1 mole

7. Calculate the mass of 4.56 x 10²⁵ atoms of Sr.

4.56 x 10²⁵ at Sr x 1 mole x 87.6 g = 6.64 x 10³ g Sr

6.02 x 10²³ at 1 mole

8. Calculate the mass of 6.33 x 10²⁰ molecules of CO₂.

6.33 x 10²⁰ molecule CO₂ x 1 mole x 44.0 g = 4.63 x 10^-2 g CO₂

6.02 x 10²³ molecs 1 mole

9. Calculate the mass of 8.66 x 10²⁶ FU of SrO.

8.66 x 10²⁶ FU SrO x 1 mole x 103.6 g = 1.49 x 10⁵ g SrO

6.02 x 10²³ FU 1 mole

10. Calculate the mass of 2.3 x 10²⁸ FU of SrCO₃^.H₂O.

2.3 x 10²⁸ FU SrCO₃^.H₂O x 1 mole x 165.62 g = 6.3 x 10⁶ g SrCO₃^.H₂O

6.02 x 10²³ FU 1 mole

11. Calculate the number of H atoms in 5.02g of CH₄. There are 4 atoms of H in one molecule of CH₄.

5.02 g CH₄ x 1 mole x 6.02 x 10²³ molecules CH₄ x 4 atoms H = 7.54 x 10²³ at H

16.04 g 1 mole molecule CH₄

12. Calculate the number of O atoms in 200. g Al₂(SO₄)₃. There are 12 atoms of O in one formula unit of Al₂(SO₄)₃.

200 g Al₂(SO₄)₃ x 1 mole x 6.02 x 10²³ FU Al₂(SO₄)₃ x 12 atoms O = 4.22 x 10²⁴ at O

342.3 g 1 mole FU Al₂(SO₄)₃

13. Calculate the mass of CaCO₃ that contains 2.00 x 10²⁸ atoms of O. There are 3 atoms of O per one FU of CaCO₃.

2.00 x 10²⁸ at O x 1 FU CaCO₃x 1mole x 100.1 g = 1.11 x 10⁶ g

3 at O 6.02 x 10²³ 1 mole

14. Calculate the mass of Al₂(SO₄)₃ that contains 2.00 x 10²⁸ atoms of O.

2.00 x 10²⁸ at O x 1 FU Al₂(SO₄)₃x 1mole x 342.3 g = 9.48 x 10⁵ g

12 at O 6.02 x 10²³ FU 1 mole

15. Calculate the mass of Al₂(SO₄)₃ that contains 2.00 x 10²⁰ atoms of Al.

2.00 x 10²⁰ at Al x 1 FU Al₂(SO₄)₃ x 1mole x 342.3 g = 0.0569 g

2 at Al 6.02 x 10²³ FU 1 mole

16. Calculate the number of Al atoms in 500.g Al₂(SO₄)₃.

500. g Al₂(SO₄)₃ x 1 mole x 6.02 x 10²³ FU Al₂(SO₄)₃ x 2 atoms Al = 1.76 x 10²⁴ at Al

342.3 g 1 mole FU Al₂(SO)₃

Draw Electron-dot Diagrams for each of the following. Remember, use brackets for cations or anions. Covalent compounds do not require brackets.

17. C₂F₆

.. ..

: F : : F :

.. .. .. ..

: F : C : C : F :

.. .. .. ..

: F: : F :

.. ..

18. KClO₄

[ K ] ⁺

: O :

.. .. ..

: O : Cl : O :

.. .. ..

: O :

19. NH₃

H : N : H

20. Li₂CO₃

21. N₂

: N ׃׃׃ N :

22. PO₃^3-

.. .. ..

: O : P : O :

.. .. ..

: O :

Worksheet #4 Empirical Formula

Use a calculation to determine the empirical formula of each of the following compounds. Show all of your work.

1. A compound is found to be 6.353 g Ag, 0.823 g N, and 2.824 g of O. Calculate the empirical formula of the compound.

6.353 g Ag x 1 mole = 0.05888 moles = 1

107.9 g 0.05879 moles ← smallest number

0.823 g N x 1 mole = 0.05879 moles = 1 AgNO₃

14.0 g 0.05879 moles

2.824 g O x 1 mole = 0.1765 moles = 3

16.0 g 0.05879 moles

2. A compound is found to be 6.25g Pb, 0.846g N, and 2.90g of O. Calculate the empirical formula of the above compound.

6.25 g Pb x 1 mole = 0.03016 moles = 1

207.2 g 0.03016 moles ← smallest number

0.846 g N x 1 mole = 0.06049 moles = 2 PbN₂O₆ or Pb(NO₃)₂

14.0 g 0.03016 moles 3 marks 4 marks only do this for ionic!

2.90 g O x 1 mole = 0.1813 moles = 6

16.0 g 0.03016 moles

3. A compound is found to be 1.00g Ca, 0.700g N, and 2.40 g O. Calculate the empirical formula of the above compound.

Ca(NO₃)₂

4. A compound is found to be 27.91% Fe, 24.08% S, and 48.0% O. Calculate the empirical formula.

27.91 g Fe x 1 mole = 0.5002 moles = 1 x 2 = 2

55.8 g 0.5002 moles

24.08 g S x 1 mole = 0.7502 moles = 1.5 x 2 = 3

32.1 g 0.5002 moles

48.0 g O x 1 mole = 3.000 moles = 6 x 2 = 12

16.0 g 0.5002 moles

Fe₂S₃O₁₂ Fe₂(SO₄)₃

5. A compound is found to be 15.38% Co, 40.74% Cr, and 43.88% O. Calculate the empirical formula.

Co₂(Cr₂O₇)₃

6. A compound is found to be 63.65% C, 10.71% H, 18.56% N, and 7.072% O. Calculate the empirical formula.

C₁₂H₂₄N₃O

7. Change 5.0 x 10²⁶ formula units CoCl₂^.6H₂O to grams.

5.0 x 10²⁶ FU CoCl₂^.6H₂O x 1 mole x 238.02 g = 2.0 x 10⁵ g CoCl₂^.6H₂O

6.02 x 10²³ FU 1 mole

8. Convert 2.36 g FeSO₄·5H₂O to formula units.

2.36 g FeSO₄·5H₂O x 1 mole x 6.02 x 10²³ FU FeSO₄·5H₂O = 5.87 x 10²¹ FU FeSO₄·5H₂O

242g 1 mole

9. Change 3.65 x 10²² molecules CO₂ to grams.

2.67 g

10. Convert 2.36 g grams P₂O₅ to molecules.

1.00 x 10²² molecules P₂O₅

Worksheet # 5 Percentage Composition, Molecular and Empirical Formula

Complete the following chart.

Empirical Formula Molar Mass Molecular Formula Molecular Mass

1. CH 13.01 g/mole C₆H₆ 78 g/mole

2. C₃H₂O 54.02 g/mole C₉H₆O₃ 162 g/mole

3. C₄H₇NO 85 g/mole C₁₂H₂₁N₃O₃ 255 g/mole

4. C₄H₈O 72.08 g/mole C₁₂H₂₄O₃ 216 g/mole

5. C₂H₄NO 58.04 g/mole C₆H₁₂N₃O₃ 174 g/mole

6. C₅H₁₁NO 101.11 g/mole C₁₅H₃₃N₃O₃ 303.33 g/mole

7. C₂H₂O 42.02 g/mole C₁₀H₁₀O₅ 210 g/mole

8. C₃H₇NO 73.07 g/mole C₁₂H₂₈N₄O₄ 292 g/mole

9. C₂H₄NO 58.04 g/mole C₄H₈N₂O₂ 116.08 g/mole

10. CCl₃ 118.5 g/mole C₂Cl₆ 237 g/mole

11. If the empirical formula for a compound is C₂H₃O and its molecular mass is 129 g/mole, what is the molecular formula.

Empirical formula C₂H₃O Empirical Mass 43.03 g/mole x 3

Molecular Formula C₆H₉O₃ Molecular Mass 129 g/mole

129/43.03 = 3x

12. A compound is 49.2% P and 50.8% O, calculate the empirical formula. If the molecular mass is 126 g/mole, calculate the molecular formula.

49.2 g P x 1 mole = 1.587 moles = 1

31.0 g 1.587 moles

50.8 g O x 1 mole = 3.175 moles = 2 PO₂

16.0 g 1.587 moles

Empirical formula PO₂ Empirical Mass 63.0 g/mole x 2

Molecular Formula P₂O₄ Molecular Mass 126 g/mole

126/63.0 = 2x

13. A compound is 62.54% Pb, 8.46% N, and 29.0% O, calculate the empirical formula.

Pb(NO₃)₂

14. A compound is 46.08 % C, 27.0% N, and 27.0 % O. Calculate the empirical formula. If the molecular mass is 832 g/mole, calculate the molecular formula.

46.08 g C x 1 mole = 3.84 moles = 2.2749 x 7 = 16

12.0 g 1.688 moles

27.0 g N x 1 mole = 1.929 moles = 1.1428 x 7 = 8

14.0 g 1.688 moles

27.0 g O x 1 mole = 1.688 moles = 1 x 7 = 7 C₁₆N₈O₇

16.0 g 1.688 moles

Empirical formula C₁₆N₈O₇ Empirical Mass 416 g/mole x 2

Molecular Formula C₃₂N₁₆O₁₄ Molecular Mass 832 g/mole

832/416 = 2x

15. Oil of citronella is a mosquito repellent is 87.8% C and 12.2% H. Calculate the empirical formula. If the molecular mass is 205 g/mole, calculate the molecular formula.

87.8 g C x 1 mole = 7.3167 moles = 1 x 3 = 3

12.0 g 7.3167 moles

12.2 g H x 1 mole = 12.079 moles = 1.651 x 3 = 5

1.01 g 7.3167 moles

C₃H₅

Empirical formula C₃H₅ Empirical Mass 41.05 g/mole x 5

Molecular Formula C₁₅H₂₅ Molecular Mass 205 g/mole

205/41.05 = 5x

Calculate the percentage composition for:

16. ZnSO₄

1 Zn = 1 x 65.4 = 65.4 g

1 S = 1 x 32.1 = 32.1 g

4 O = 4 x 16 = 64.0 g

161.5 g/mole

% Zn = 65.4 g x 100% = 40.5 %

161.5 g

% S = 32.1 g x 100% = 19.9 %

161.5 g

% O = 64.0 g x 100% = 39.6 %

161.5 g

100%

17. Al₂(CO₃)₃

2 Al = 2 x 27.0 = 54.0 g

3 C = 3 x 12.0 = 36.0 g

9 O = 9 x 16.0 = 144.0 g

234.0 g/mole

% Al = 54.0 g x 100% = 23.1 %

234.0 g

% C = 36.0 g x 100% = 15.4 %

234.0 g

% O = 144 g x 100% = 61.5 %

234.0 g

100%

18. Ca₃(PO₄)₂

3 Ca = 3 x 40.1 = 120.3 g

2 P = 2 x 31.0 = 62.0 g

8 O = 8 x 16.0 = 128.0 g

310.3 g/mole

% Ca = 120.3 g x 100% = 38.8 %

310.3 g

% P = 62.0 g x 100% = 20.0 %

310.3 g

% O = 128 g x 100% = 41.3 %

310.3 g

100.1%

19. Convert 3.66 Kg CO₂to molecules.

3.66 Kg CO₂ x 1000 g x 1 mole x 6.02 x 10²³ molecules = 5.01 x 10²⁵ molecules CO₂

1 Kg 44.0 g 1 mole

20. Covert 4.0 x 10²⁶ FU of MgCl₂ to Kg.

4.0 x 10²⁶ FU MgCl₂ x 1mole x 95.3 g x 1 Kg = 63 Kg

6.03 x 10²³ FU 1 mole 1000 g

21. In a propane tank there are 9.0 Kg of C₃H₈, calculate the number of H atoms. (First calculate molecules and then H atoms).

9.0 Kg C₃H₈x 1000 g x 1 mole x 6.02 x 10²³ molecules x 8 at H = 9.8 x 10²⁶ H at C₃H₈

1 Kg 44.08 g 1 mole 1 molecule C₃H₈

22. A certain mass of complex Co(NH₃)₆Cl₃ was found to contain 2.65 x 10²¹ atoms of H, calculate the mass of cobalt III chloride hexammine. (Change the number of atoms to FU’s first).

2.65 x 10²¹ at H x 1 FU Co(NH₃)₆Cl₃ x 1 mole x 267.58 g = 6.54 x 10^-2 g

18 at H 6.02 x 10²³ at 1 mole

Worksheet # 6 Mole Calculations to Molecular Formula

1 The electrolysis of water using H₂SO₄ as a catalyst is used to generate hydrogen gas. For every 1 electron consumed

1 atom of hydrogen (exact) is generated. An amp meter is used to measure the rate of electron consumption.

1.00 coulomb = 6.24 x 10 ¹⁸ electrons.

If 1.00 g of hydrogen is generated at a current of 1.50 amps (1 amp = 1 coulomb per second) and over time period

of 17.865 hours, calculate Avogadro’s number (the number of atoms in 1.00g of hydrogen). Use unit analysis!

Start the unit analysis with 17.865 h and end with atoms H.

6.02 x 10²³ at

2. Convert 500. g NaCl to formula units.

500 g x 1 mole x 6.02 x 10²³ FU = 5.15 x 10²⁴ Fu

58.5 g 1 mole

3. Convert 9.8 x 10²⁴ molecules of C₂H₆ to grams.

9.8 x 10²⁴molecules x 1 mole x 30.06 g = 4.9 x 10²g

6.02 x 10²³molecules 1 mole

4. Calculate the number of O atoms in 50.0 lb of dry ice CO₂.

(2.21 lb = 1.00Kg)

50.0 lb x 1.00 Kg x 1000 g x 1 mole x 6.02 x 10 ²³ molecules x 2 at O

2.21 lb 1.00 Kg 44.0 g 1 mole 1 molecules

= 6.19 x 10²⁶ atoms O

5. Calculate the percentage composition of Al₂(SO₄)₃. 3 significant figures!

2 Al 2 x 27.0 = 54.0 % Al = 15.8 %

3 S 3 x 32.1 = 96.3 % S = 28.1 %

12 O 12 x 192.0 = 192.0 % O = 56.1 %

342.3 g/mole

6. Calculate the molar mass of Co₂(SO₄)₃ ^. 6H2O

514.22 g/mole

7. The empirical formula for a compound is C₂H₅O and its molecular mass is 135g/mol. The olecular formula is:

Empirical formula C₂H₅Ox 3 Empirical Mass 45.05 g/mole x 3

Molecular Formula C₆H₁₅O₃ Molecular Mass 135 g/mole

135/45.05 = 3x

8. A compound is 24.4% Ca, 17.1% N and 58.5% 0. Calculate the empirical formula.

Ca(NO₃)₂

9. Hydroquinone, a chemical used for photographic developing, is 65.45% C, 5.51% H and 29.09% O. Calculate the empirical and molecular formula. The molecular mass is 110 g/mol.

65.45 g C x 1 mole = 5.454 moles = 3 = 3

12.0 g 1.818 moles

5.51 g H x 1 mole = 5.455 moles = 3 = 3

1.01 g 1.818 moles

29.09 g O x 1 mole = 1.818 moles = 1 = 1

16.0 g 1.818 moles

C₃H₃O

Empirical formula C₃H₃O Empirical Mass 55.03 g/mole x 2

Molecular Formula C₆H₆O₂ Molecular Mass 110 g/mole

110/55.03 = 2x

10. A compound is 50.5 % C, 5.26 % H, and 44.2% N. Calculate the empirical formula. If the molecular mass is 380.2 g/mole, calculate the molecular formula.

50.5 g C x 1 mole = 4.2083 = 1.33 x 3

12.0 g 3.1571

5.26 g H x 1 mole = 5.2079 = 1.65 x 3

1.01 g 3.1571

44.2 g N x 1 mole = 3.1571 = 1

14.0g 3.1571

Empirical Formula Empirical Mass

C₄H₅N₃95.05 g/mole

Molecular Formula Molecular Mass

C₁₆H₂₀N₁₂ 380.2 g/mole

11. A compound is 50.5% C, 5.26% H, and 44.2% N, calculate the empirical formula. If the molecular mass is 285.15 g/mole, calculate the molecular formula.

50.5 g C x 1 mole = 4.208 moles = 1.333 x 3 = 4

12.0 g 3.157 moles

5.26 g H x 1 mole = 5.208 moles = 1.650 x 3 = 5

1.01 g 3.157 moles

44.2 g N x 1 mole = 3.157 moles = 1 x 3 = 3 C₄H₅N₃

14.0 g 3.157 moles

Empirical formula C₄H₅N₃ Empirical Mass 95.05 g/mole x3

Molecular Formula C₁₂H₁₅N₉ Molecular Mass 285.15 g/mole

285.15/95.05 = 3x

Challenge Problems

12. When 15.0 g of a compound known to contain C, H, O, and S was burned, 16.2 g of CO₂, 6.63 g of H₂O, and 15.7 g of SO₂ were produced. What is the empirical formula of the compound?

16.2 g CO₂ x 1 mole x 1 mole C = 0.3682 mole C x 12.0 g = 4.418 g C

44.0 g 1 mole CO₂ 1 mole

6.63 g H₂O x 1 mole x 2 mole H = 0.7358 mole H x 1.01 g = 0.7432 g H

18.02 g 1 mole H₂O 1 mole

15.7 g SO₂ x 1 mole x 1 mole S = 0.2449 mole S x 32.1 g = 7.862 g S

64.1 g 1 mole SO₂ 1 mole

Mass of C H S = 13.023 g

Mass of O = Total C H S O – Mass of C H S

Mass of O = 15.0 g – 13.023

Mass of O = 1.977 g

1.977 g O x 1 mole = 0.1236 moles O

16.0 g

Mole Ratio 0.3682 mole C = 3

0.1236 moles

0.7358 mole H = 6

0.1236 moles

0.1236 moles O = 1 Empirical Formula C₃H₆OS₂

0.1236 moles

0.2449 mole S = 2

0.1236moles

13. When 51.3 g of a compound known to contain C, H, N, and O was burned, 84.2 g of CO₂, 20.7 g of H₂O, and 10.7 g of N₂ were produced. What is the empirical formula of the compound?

84.2 g CO₂ x 1 mole x 1 mole C = 1.9136 mole C x 12.0 g = 22.96 g C

44.0 g 1 mole CO₂ 1 mole

20.7 g H₂O x 1 mole x 2 mole H = 2.2974 mole H x 1.01 g = 2.320 g H

18.02 g 1 mole H₂O 1 mole

10.7 g N₂ x 1 mole x 2 mole N = 0.7643 mole N x 14.0 g = 10.70 g N

28.0 g 1 mole N₂ 1 mole

Mass of C H N = 35.98 g

Mass of O = Total C H N O – Mass of C H N

Mass of O = 51.3 g – 35.98 g

Mass of O = 15.32 g

15.32 g O x 1 mole = 0.9575 moles O

16.0 g

Mole Ratio 1.9136 mole C = 2.50 x 4 = 10

0.7643 moles

2.2974 mole H = 3.00 x 4 = 12

0.7643 moles

0.7643 mole N = 1 x 4 = 4

0.7643 moles

0.9575 moles O = 1.253 x 4 = 5

0.7643 moles

Empirical Formula C₁₀H₁₂N₄O₅ Wow, finally a challenge!!

Worksheet #7 Identify that Gas

1. A volume of a gas weighs 0.256 g. An equal volume of H₂ weighs 0.01163g.

Calculate the molecular mass of the gas. Assuming that this gas is a common gas,

what do you think the gas could be? Show some work!

Gas X = 0.256 g = 22.01 x 2.02 g = 44.5 g/mole

H₂ 0.01163 g 1 mole

Basically, we are just saying that the molar mass of the unknown gas is 22.01 x heavier

than a known gas H₂ which we know is 2.02 g/mole.

2. A volume of a gas weighs 2.12 g. An equal volume of He weighs 0.265 g. Calculate the molecular mass of the gas. Assuming that this gas is a common gas, what do you think the gas could be? Show some work!

Gas X = 2.12 g = 8.000 x 4.0 g = 32.0 g/mole

He 0.265 g 1 mole

Basically, we are just saying that the molar mass of the unknown gas is 8.000 x heavier

than a known gas He which we know is 4.0 g/mole.

3. A volume of a gas weighs 0.235g. An equal volume of He weighs 0.02043g. Calculate the molecular mass of the gas. This gas is a gas produced by automobile pollution, what do you think the gas could be? Show some work!

Gas X = 0.235 g = 11.50 x 4.0 g = 46.0 g/mole

He 0.02043 g 1 mole

NO₂ nitrogen dioxide

4. A volume of a gas weighs 0.37216 g. An equal volume of H₂ weighs 0.01163 g. Calculate the molecular mass of the gas. Assuming that this gas is a on that is produced by burning matches, what do you think the gas could be? Show some work!

64.6 g/mole SO₂

5. A volume of a gas weighs 0.02051g. An equal volume of CO₂ weighs .02654g. Calculate the molecular mass of the gas. Assuming that this gas is a smelly one that is produced by rotten eggs, what do you think the gas could be? Show some work! This gas is also added to natural gas to make it smell so that leaks can be detected.

34.0 g/mole H₂S

6. Convert 1.65 g CO₂ to molecules.

1.65 g CO₂ x 1 mole x 6.02 x 10²³ molecules = 2.26 x 10²² molecules

44.0 g 1 mole

7. Convert 12.5 Kg of SO₂ to molecules.

12.5 Kg SO₂ 1000 g x 1 mole x 6.02 x 10²³ molecules = 1.17 x 10²⁶ molecules

1 Kg 64.1 g 1 mole

8. Covert 2.0 x 10²⁶FU of MgCl₂ to g.

2.0 x 10²⁶ FU MgCl₂ x 1 mole x 95.3 g = 3.2 x 10⁴ g

6.02 x 10²³ FU 1 mole

9. In a propane tank there are 8.0 Kg of C₃H₈. Calculate the number of H atoms. (First calculate molecules and then H atoms).

8.0 Kg C₃H₈ x 1000 g x 1 mole x 6.02 x 10²³ molecules C₃H₈ x 8 atoms H = 8.7 x 10²⁶ at H

1 Kg 44.08 g 1 mole molecule C₃H₈

10. A certain mass of complex Co(NH₃)₆Cl₃ was found to contain 2.65 x 10²⁵ FU, calculate the mass of cobalt III chloride hexammine.

2.65 x 10²⁵ FU x 1 mole x 267.58 g = 1.18 x 10⁴ g

6.02 x 10²³ FU 1 mole

11. The empirical formula of a compound is C₃H₁₁O₃ and its molecular formula is 380.44 g/mole. Determine the molecular formula.

C₃H₁₁O₃ 95.11 g/mole

C₁₂H₄₄O₁₂ 380.44 g/mole

12. Calculate the percentage composition of Co(NH₃)₆Cl₃ to three significant figures.

1 Co = 1 x 58.9 = 58.9 22.0 %

6 N = 6 x 14.0 = 84.0 31.4 %

18 H = 18 x 1.01 = 18.18 6.79 %

3 Cl = 3 x 35.5 = 106.5 39.8 %

267.58

13. A compound is 27.73% Mg, 23.58% P, and 48.69% O, calculate the empirical formula.

Mg₃(PO₄)₂

14. 9, 10-dihydro-6-methylergoline-8-carboxylic acid (LSD) a drug with psychomimetric properties is 71.6 % C, 6.03 % H, 10.4 % N, and 11.9 % O. If the molecular mass of the compound is 268.16 g/mol, calculate the empirical and the molecular formula.

71.6 g C x 1 mole = 5.967 moles = 8

12.0 g 0.7438 moles

6.03 g H x 1 mole = 5.970 moles = 8

1.01 g 0.7438 moles

10.4 g N x 1 mole = 0.7429 moles = 1 C₈H₈NO

14.0 g 0.7438 moles

11.9 g O x 1 mole = 0.7438 moles = 1 C₄H₅N₃

16.0 g 0.7438 moles

Empirical formula C₈H₈NO Empirical Mass 134.08 g/mole x 2

Molecular Formula C₁₆H₁₆N₂O₂ Molecular Mass 268.16 g/mole

268.16/134.08 = 2x

15. A volume of a gas containing sulphur and oxygen that is produced by burning coal has a mass of 2.088g. An equal volume of H₂ has a mass of 0.0658 g. Calculate the molar mass of the gas and determine its formula.

Gas X = 2.088 g = 31.73 x 2.02 g = 64.1 g/mole

H₂ 0.0658 g 1 mole

SO₂ sulphur dioxide

16. Here is the tough one. You can do it- just take it one step at a time! A compound was known to contain C, H, N, O, and S. When a 5.43 g sample was burned the products were 8.43 g CO₂, 1.15 g H₂O, 0.450 g N₂, and 3.07 g of SO₂. Determine the empirical formula of the compound. I left you a whole page- I hope that’s enough. Good luck!

8.43 g CO₂ x 1 mole x 1 mole C = 0.1916 mole C x 12.0 g = 2.299 g C

44.0 g 1 mole CO₂ 1 mole

1.15 g H₂O x 1 mole x 2 mole H = 0.1276 mole H x 1.01 g = 0.1289 g H

18.02 g 1 mole H₂O 1 mole

0.450 g N₂ x 1 mole x 2 mole N = 0.03214 mole N x 14.0 g = 0.4500 g N

28.0 g 1 mole N₂ 1 mole

3.07 g SO₂ x 1 mole x 1 mole S = 0.04779 mole S x 32.1 g = 1.537 g S

64.1 g 1 mole SO₂ 1 mole

Mass of C H N S = 4.4149 g

Mass of O = Total C H N S O – Mass of C H N S

Mass of O = 5.43 g – 4.4149

Mass of O = 1.0151 g

1.0151 g O x 1 mole = 0.06344 moles O

16.0 g

Mole Ratio 0.1916 mole C = 6 x 2 = 12

0.03214 moles

0.1276 mole H = 4 x 2 = 8

0.03214 moles

0.03214 mole N = 1 x 2 = 2

0.03214 moles

0.06344 moles O = 2 x 2 = 4

0.03214 moles

0.04779 mole S = 1.5 x 2 = 3

0.03214 moles

Empirical Formula C₁₂H₈N₂O₄S₃

Thank goodness that one is finished!!!!

Worksheet #8 Review of Mole Calculations

1. Convert 2.59 g of SO₃ to molecules.

2.59 g of SO₃ x 1 mole x 6.02 x 10 ²³ = 1.95 x 10²² molecules

80.1 g 1 mole

2. Convert 3.56 x 10²⁵ FU of CoCl₄ to Kg.

11.9 Kg

3. In a propane tank there are 26.5 Kg. of C₂H₆. Calculate the number of H atoms.

26.5 Kg C₂H₆ x 1000 g x 1 mole x 6.02 x 10²³ molecs C₂H₆ x 6 atoms H = 3.18 x 10²⁷ at H

1 Kg 30.06 g 1 mole molecule C₂H₆

4. Describe each atom as an atom, molecule, anion, cation, or a formula unit:

CO₂ molecule

Co atom

AgNO₃FU

KCl FU

Cr₂O₇^2-anion

NH₄⁺ cation

5. Convert 568 g of H₃PO₄to moles.

568 g x 1 mole = 5.79 moles

98.03 g

6. Convert 3.25 x 10⁵g Rh into atoms.

3.25 x 10⁵ g x 1 mole x 6.02 x 10²³ at = 1.90 x 10²⁷ at

102.9 g 1 mole

7. Calculate the molar mass of the following compounds:

Hg₃(PO₄)₂791.8 g/moleCuI₂ 317.3 g/mole

Pb₂SO₄510.5 g/moleLi₂SO₃ 93.9 g/mole

8. What is the percent composition of C₂H₆

% C = 79.8 %

% H = 20.2 %

9. What is the percent composition of CaCl₂^. 2H₂O

1 x 40.1 = 40.1 g % Ca = 27.3 %

2 x 35.5 = 71.0 g % Cl = 48.3 %

4 x 1.01 g = 4.04 g % H = 2.75 %

2 x 16.0 = 32.0 g % O = 21.7 %

147.14

10. The empirical formula of a compound is SiH₃. If 0.0275 mol of the compound has a mass of 1.71g, what is the compounds molecular formula?

Empirical Formula SiH₃Empirical Mass = 31.13 g/mole

Molecular Formula Si₂H₆ Molar Mass = 1.71 g = 62.18 g/mole

0.0275 mole

11. A compound is 27.73% Mg, 23.58% P, and 48.69% O, calculate the empirical formula.

27.73 g Mg x 1 mole = 1.141 moles = 1.5 x 2 = 3

24.3 g 0.7606 moles

23.58 g P x 1 mole = 0.7606 moles = 1 x 2 = 2

31.0 g 0.7606 moles

48.69 g O x 1 mole = 3.043 moles = 4 x 2 = 8

16.0 g 0.7606 moles

Mg₃P₂O₈Mg₃(PO₄)₂

12. Find the empirical formula for a compound containing 46.3% Li and 53.7% O.

46.3 g Li x 1 mole = 6.710 moles = 2

6.9 g 3.356 moles

53.7 g O x 1 mole = 3.356 moles = 1

16.0 g 3.356 moles

Li₂O

13. Convert 5.65 x 10¹⁷ atoms of Fe into grams.

5.24 x 10^-5 g

14. A compound contains C, H and O. A 5.90 g sample is burned to yield 11.18 g of CO₂ and 3.66 g H₂O. What is the empirical formula of the compound?

11.18 g CO₂ x 1 mole x 1 mole C = 0.2541 mole C x 12.0 g = 3.0491 g C

44.0 g 1 mole CO₂ 1 mole

3.66 g H₂O x 1 mole x 2 mole H = 0.4062 mole H x 1.01 g = 0.4103 g H

18.02 g 1 mole H₂O 1 mole

Mass of C H = 3.4594 g

Mass of O = Total C H O – Mass of C H

Mass of O = 5.90 g – 3.4594

Mass of O = 2.441 g

2.441 g O x 1 mole = 0.1525 moles O

16.0 g

Mole Ratio 0.2541 mole C = 1.666 x 3 = 5

0.1525 moles

0.4062 H = 2.664 x 3 = 8

0.1525 moles

0.1525 moles O = 1 x 3 = 3

0.1525 moles

Empirical Formula C₅H₈O₃

15. A volume of a gas weighs 2.12 g. An equal volume of He weighs 0.265g. Calculate the molecular mass of the gas. Assuming that this gas is a common diatomic gas, what do you think the gas could be? Show your work.

Gas X = 2.12 g = 8 x 4.0 g = 32 g/mole

He 0.265 g 1 mole

O₂ oxygen

Worksheet # 9 Moles Practice Test # 1

1. Convert 5.65 g of SO₂ to molecules.

5.65 g x 1 mole x 6.02 x 10²³ molecules = 5.31 x 10²² molecules

64.1 g 1 mole

2. Convert 3.56 x 10²⁵ FU of CuCl₂ to Kg.

3.56 x 10²⁵ FU of CuCl₂ x 1 mole x 134.5 g x 1 Kg = 7.95 Kg

6.02 x 10²³ FU 1 mole 1000 g

3. In a lighter there are 0.52 g. of C₄H₁₀. Calculate the number of H atoms.

0.52 g x 1 mole x 6.02 x 10²³ molecules x 10 at H = 5.4 x 10²² at H

58.1 g 1 mole 1 molecule C₄H₁₀

4. Describe each atom as an atom, molecule, or a formula unit:

CO₂ molecule

Zn atom

Ca(NO₃)₂ FU

NaCl FU

CH₃COOH molecule

5. Convert 572 g of HNO₂to moles.

572 g HNO₂ x 1 mole = 12.2 mole

47.01 g

6. Convert 3.25 x 10⁵g Ru into atoms.

3.25 x 10⁵ g x 1 mole x 6.02 x 10²³ at = 1.94 x 10²⁷ at

101.1 g 1 mole

7. Calculate the molar mass of the following compounds:

Au₃(PO₄)₃876 g/moleCuBr₂ 223.3 g/mole

Pb₃(PO₄) ₂811.6 g/moleLi₂CO₃ 73.8 g/mole

8. What is the percent composition of C₃H₁₀

% C = 78.1 %

% H = 21.9 %

9. What is the percent composition of SrCl₂^.3H₂O

% Sr = 41.2 %

% Cl = 33.4 %

% H = 2.85 %

% O = 22.6 %

10. The empirical formula of a compound is C₃H₈. If 2.00 mol of the compound has a mass of 264.48 g, what is the compounds molecular formula? Hint: first calculate the molar mass by dividing grams by moles.

Molar Mass = 264.48 g = 132.24 g/mole 132.24 = 3 x

2.00 mole 44.08

C₉H₂₄

11. Find the empirical formula for a compound containing 24.45 % Ca, 17.07 % N and 58.53 % O.

Ca(NO₃)₂

12. Find the empirical formula for a compound containing 74.40 % Ga and 25.60 % O.

Ga₂O₃

13. Convert 5.55 x 10¹⁷ atoms of Co into grams.

5.55 x 10¹⁷ at Co x 1 mole x 58.9 g = 5.43 x 10^-5 g Co

6.02 x 10²³at 1 mole

14. A volume of a gas weighs 3.0475 g. An equal volume of He weighs 0.265g. Calculate the molecular mass of the gas. Assuming that this gas causes acid rain, what do you think the gas could be? Show your work.

Gas x = 3.04475 = 11.489 x 4.0 g/mole = 46 g/mole NO₂

He 0.265

15. Methamphetamine, MDMA, or commonly called ecstasy is an illegal drug from the family called “entactogens” which literally means in Greek “touching within”. It is considered to be a mood elevator that is 59.506 % C, 8.0135 % H, 6.9424 % N, 7.934 % O, and 17.604 % Cl. Calculate the empirical formula for MDMA.

             C₁₀H₁₆NOCl

16.     Brodifacoum, a Coumarin derivative used to poison rats, is also used a blood thinner in small doses. An antidote for this poison is vitamin K, a blood-clotting agent. This compound contains C, H, Br and O.  A 10.00 g sample is burned to yield 25.633 g of CO₂, 3.894 g H₂O, and 1.502 g of Br₂. What is the empirical formula of the compound? The structural formula is shown below.

25.633 g CO₂ x 1 mole x 1 mole C = 0.5826 mole C x 12.0 g = 6.991 g C

44.0 g 1 mole CO₂ 1 mole

3.894 g H₂O x 1 mole x 2 mole H = 0.4322 mole H x 1.01 g = 0.4365 g H

18.02 g 1 mole H₂O 1 mole

1.502 g Br₂ x 1 mole x 2 mole Br = 0.01880 mole Br x 79.9 g = 1.502 g Br

159.8 g 1 mole Br₂ 1 mole

Mass of C H Br = 8.930 g

Mass of O = Total C H Br O – Mass of C H Br

Mass of O = 10.00 g – 8.930

Mass of O = 1.071 g

1.071 g O x 1 mole = 0.06691 moles O

16.0 g

Mole Ratio 0.5826 moles C = 31 x 2 = 62

0.01880 moles

0.4321 mole H = 23 x 2 = 46

0.01880 moles

0.01880 mole Br = 1 x 2 = 2

0.01880 moles

0.06691 moles O = 3.56 x 2 = 7

0.01880 moles

Empirical Formula C₆₂H₄₆Br₂O₇

That’s all folks!!!!-

Worksheet # 10 Moles Practice Test 2

Convert 3.567 x 10²³molecules of C₄H₁₀to g.

34.4 g

Convert 1.547 kg of Fe₂(CO₃)₃ to FU’s.

3.19 x 10²⁴ FU

What is the percent composition of Cr(C₂H₃O₂)₂ ? Three significant figures.

Cr = 30.6%

C = 28.2%

H = 3.56%

O = 37.6%

The empirical formula of a compound is SiH₃. If 1.5 moles of the compound have a mass of 139.95g, what is the compound's molecular formula? (Hint: calculate molecular mass first.)

Si₃H₉

A compound has an empirical formula of ClCH₂ and a molecular weight of 99.0 g/mol. What is its molecular formula?

Cl₂C₂H₄

Aspirin has a chemical formula of C₉H₈O₄. A sample of aspirin has 9.70 x 10¹² atoms of carbon. How many atoms of hydrogen does this sample contain?

8.62 x 10¹² atoms H

A volume of a gas weighs 2.647 g. An equal volume of O₂ weighs 3.025 g. Calculate the molecular mass of the gas. Assuming that this gas is an element, what is the gas?

N₂ (atomic mass 28.0 g/mole)

Find the empirical formula for the oxide that contains 42.05 g of nitrogen and 95.95 g of oxygen.

NO₂

A compound is 75.46% carbon, 4.43% hydrogen, and 20.10% oxygen by mass. It has a molecular weight of 318.31 g/mol. What is the molecular formula for this compound?

C₂₀H₁₄O₄

0.487 grams of quinine (a compound made C, H, N and O with a molar mass = 324 g/mol) is combusted and found to produce 1.321 g CO₂, 0.325 g H₂O and 0.0421 g of N₂. Determine the empirical and molecular formulas.

Emperical C₁₀H₁₂NO Molecular C₂₀H₂₄N₂O₂

If 0.500 g of hydrogen was generated at a current of 0.750 amps and over time period of 17.865 hours, calculate Avogadro’s number (the number of atoms in 1.00 g of hydrogen). Use unit analysis! Start the unit analysis with 17.865 h and end with atoms H.

1.00 coulomb = 6.24 x 10 ¹⁸ electrons 1 electron = 1 H atom 1 amp = 1 coulomb per

17.865 h x 60 min x 60 s x 0.750 coul x 6.24x10¹⁸ el x 1H atom = 3.01x10²³ H atoms

h min s coul 1 el

3.01 x 10²³ H atoms = 6.02x10²³ H atoms/g

0.5 g