chemistry handbook review 6.calculate the mass of 6.89 mol of antimony (sb). 6.89 mol sb x (121.757...

Chemistry Handbook Review

6. Calculate the mass of 6.89 mol of Antimony (Sb).

6.89 mol Sb x (121.757 g Sb / 1 mol Sb) =

839 g Sb

Chemistry Handbook Review

7. A chemisty needs 0.0700 mol selenium (Se) for a reaction. What mass of selenium should the chemist use?

0.0700 mol Se x (78.96 g Se / 1 mol Se) =

5.53 g Se

Chemistry Handbook Review

8. A sample of sulfur (S) has a mass of 223 g. How many moles of sulfur are in the sample?

223 g S x (1 mol S / 32.066 g S) = 6.95 mol S

Chemistry Handbook Review

9. A tank of compressed helium (He) contains 50.0 g helium. How many moles of helium are in the tank?

50.0 g He x (1 mol He / 4.003 g He) = 12.5 mol He

Chemistry Handbook Review

10.A nickel coin is 25.0% nickel and 75.0% copper. If a nickel coin has a mass of 5.00 g, how many moles of Ni are in the coin?

5.00 g x 0.25 = 1.25 g Ni

1.25 g Ni x (1 mol Ni / 58.693 g Ni) = 0.0213 mol Ni

Chemistry Handbook Review

16.Calculate the number of moles in 17.2 g of benzene (C6H6).

Molar mass of C6H6 is:

C: 6 x 12.01 = 72.06 g C

H: 6 x 1.008 = 6.048 g H

Total = 78.1 g C6H6

17.2 g C6H6 x (1 mol C6H6 / 78.1 g C6H6) = 0.220 mol C6H6

Chemistry Handbook Review

17. Calculate the number of moles in 350.0 g of potassium chlorate (KClO3).

Molar mass of KClO3 is:

K: 1 x 39.098 = 39.098 g K

Cl: 1 x 35.453 = 35.453 g Cl

O: 3 x 15.999 = 47.997 g O

Total = 122.548 g KClO3

350.0 g KClO3 x (1 mol KClO3 / 122.548 g KClO3) = 2.856 mol KClO3

Chemistry Handbook Review

18. Determine the mass of 0.187 mol of tin (II) sulfate (SnSO4).

Molar mass of SnSO4 is:

Sn: 1 x 118.710 = 118.710 g Sn

S: 1 x 32.066 = 32.066 g S

O: 4 x 15.999 = 63.996 g O

Total = 214.772 g SnSO4

0.187 mol SnSO4 x (214.772 g SnSO4 / 1 mol SnSO4 ) = 40.2 g SnSO4

Chemistry Handbook Review

19. A chemist needs 1.35 mol of ammonium dichromate for a reaction. The formula for this substance is (NH4)2Cr2O7. What mass of ammonium dichromate should the chemist measure out?

Molar Mass of (NH4)2Cr2O7 :N: 2 x 14.007 = 28.014 g NH: 8 x 1.008 = 8.06 g HCr: 2 x 51.996 = 103.99 g CrO: 7 x 15.999 = 111.99 g O

Total: 252.06 g (NH4)2Cr2O7

1.35 mol (NH4)2Cr2O7 x (252.00 g (NH4)2Cr2O7 / 1 mol (NH4)2Cr2O7) = 340. g (NH4)2Cr2O7

Chemistry Handbook Review

19.b. Find the number of H atoms found in 15.5 grams of methane (CH4).

Molar mass of CH4 : 1 mole C x 12.01 g = 12.01 g C4 moles H x 1.008 g = 4.032 g HTotal mass of 16.04 g/mol

15.5 g CH4 x (1 mol CH4 /16.04 g CH4) = 0.966 mol CH4

0.966 mol CH4 x (6.02 x 1023 molecules CH4 / 1 mol CH4) = 5.82 x 1023 molecules CH4

5.82 x 1023 molecules CH4 x (4 atoms H/ 1 molecule CH4) = 2.33 x 1024 atoms H

Chemistry Handbook Review

20.A student needs 0.200 mol each of zinc metal and copper (II) nitrate (Cu(NO3)2) for an experiment. What mass of each should the student obtain?

0.200 mol Zn x (65.39 g Zn / 1 mol Zn) = 13.1 g Zn

Chemistry Handbook Review

Molar Mass of Cu(NO3)2:

Cu: 1 x 63.546 = 63.546 g Cu

N: 2 x 14.007 = 28.014 g N

O: 6 x 15.999 = 95.994 g O

Total = 187.554 g Cu(NO3)2

0.200 mol Cu(NO3)2 x (187.554 g Cu(NO3)2 / 1 mol Cu(NO3)2) = 37.5 g Cu(NO3)2

Chemistry Handbook Review

21.Calculate the percent composition of aluminum oxide (Al2O3).

Molar Mass of Al2O3 :Al: 2 x 26.982 = 53.964 g AlO: 3 x 15.999 = 47.997 g O

Total = 101.96 g Al2O3

% Al = 53.964 / 101.96 = 52.93%% O = 47.997 / 101.96 = 47.07%

Chemistry Handbook Review

22. Determine the percent composition of magnesium nitrate, which has the formula Mg(NO3)2.

Molar Mass of Mg(NO3)2 :Mg: 1 x 24.305 = 24.305 g MgN: 2 x 14.007 = 28.014 g NO: 6 x 15.999 = 95.994 g O

Total = 148.313 g Mg(NO3)2

% Mg = 24.305 / 148.313 = 16.39%% N = 28.014 / 148.313 = 18.89%% O = 95.994 / 148.313 = 64.72%

Chemistry Handbook Review

23. Calculate the percent oxygen in potassium chlorate KClO3.

Molar Mass of KClO3 :K: 1 x 39.098 = 39.098 g KCl: 1 x 35.453 = 35.453 g ClO: 3 x 15.999 = 47.997 g O

Total = 122.548 g KClO3

% O = 47.997 / 122.548 = 39.17%

Chemistry Handbook Review

24. Calculate the percent nitrogen in ammoinium hexacyanoiron (II), which has the formula (NH4)4Fe(CN)6.

Molar Mass of (NH4)4Fe(CN)6:N: 10 x 14.007 = 140.07 g NH: 16 x 1.008 = 16.13 g HFe: 1 x 55.847 = 55.847 g FeC: 6 x 12.011 = 72.066 g C

Total = 284.11 g (NH4)4Fe(CN)6

% N = 140.07 / 284.11 = 49.30%

Chemistry Handbook Review

25.Acetylene gas has the molecular formula C2H2. How does the percent composition of acetylene compare with that of benzene (C6H6)?

The percent compositions will be the same because the ratio of carbon atoms to hydrogen atoms is the same, 1:1

Chemistry Handbook Review

26. The composition of acetic acid is 40.00% carbon, 6.71% hydrogen, and 53.29% oxygen. Calculate the empirical formula for acetic acid.

40.00 g C x (1 mol C / 12.011 g C) = 3.330 mol C6.71 g H x (1 mol H / 1.008 g H) = 6.66 mol H53.29 g O x (1 mol O / 15.999 g O) = 3.331 mol O

3.330 mol C / 3.330 = 1.000 mol C6.66 mol H / 3.330 = 2.00 mol H3.331 mol O / 3.330 = 1.000 mol O

Empirical formula = CH2O

Chemistry Handbook Review

27. An unknown compound is analyzed and found to be composed of 14.79% nitrogen, 50.68% oxygen, and 34.53% zinc. Calculate the empirical formula for the compound.

14.79 g N x (1 mol N / 14.007 g N ) = 1.056 mol N 50.68 g O x (1 mol O / 15.999 g O ) = 3.168 mol O 34.53 g Zn x(1 mol Zn/65.39 g Zn) =0.5281 mol Zn

1.056 mol N / 0.5281 = 2.000 mol N3.168 mol O / 0.5281 = 5.999 mol O0.5281 mol Zn / 0.5281 = 1.000 mol ZnEmpirical formula = N2O6Zn = Zn(NO3) 2

Chemistry Handbook Review

28. A lab investigation is done to determine the empirical formula for a compound that combines aluminum with carbon. An empty crucible and lid is measured at 35.656 grams, and then again measured after aluminum was added at 39.098 g. After reacting only with available carbon, the final mass was 40.247 g. Determine the empirical formula of the compound.

Chemistry Handbook Review

• Mass of Aluminum:

39.098 g – 35.656 g = 3.442 g Al

• Mass of Carbon:

40.247 g – 39.098 g = 1.149 g C

Chemistry Handbook Review

3.442 g Al x (1 mol Al / 26.982 g Al) = 0.1276 mol Al

1.149 g C x (1 mol C / 12.011 g C) = 0.09566 mol C

0.1276 mol Al / 0.09566 = 1.334 mol Al

0.09566 mol C / 0.09566 = 1.000 mol C

Ratio of (1.334 : 1.000) x 3 = (4.002 : 3.000)

Empirical formula = Al4C3

Chemistry Handbook Review

29. The composition of ascorbic acid (vitimin C) is 40.92% carbon, 4.58% hydrogen, and 54.50% oxygen. What is the empirical formula for vitimin C?

40.92 g C x (1 mol C / 12.011 g C ) = 3.407 mol C 4.58 g H x (1 mol H/ 1.008 g H) = 4.54 mol H 54.50 g O x (1 mol O / 15.999 g O ) = 3.406 mol O

3.407 mol C / 3.406 = 1.000 mol C4.54 mol H / 3.406 = 1.333 mol H3.406 mol O / 3.406 = 1.000 mol ORatio of (1.000: 1.333 : 1.000) x 3 = (3.000: 3.999 : 3.000)

Empirical formula = C3H4O3

Chemistry Handbook Review

30. Ricinine is one of the poisonous compunds found in the castor plant. The composition of ricinine is 58.54% carbon, 4.91% hydrogen, 17.06% nitrogen, and 19.49 % oxygen. Ricinine’s molar mass is 164.16 g/mol. Determine its molecular formula.

58.54 g C x (1 mol C / 12.011 g C ) = 4.874 mol C 4.91 g H x (1 mol H/ 1.008 g H) = 4.87 mol H 17.06 g N x (1 mol N / 14.007 g N ) = 1.218 mol N 19.49 g O x (1 mol O / 15.999 g O ) = 1.218 mol O

4.874 mol C / 1.218 = 4.002 mol C4.87 mol H / 1.218 = 3.999 mol H1.218 mol N / 1.218 = 1.000 mol N1.218 mol O / 1.218 = 1.000 mol OEmpirical formula = C4H4NO

Chemistry Handbook Review

Molar Mass of C4H4NO:C: 4 x 12.011 = 48.044 g CH: 4 x 1.008 = 4.032 g HN: 1 x 14.007 = 14.007 g NO: 1 x 15.999 = 15.999 g O

Total = 82.082 g C4H4NO164.16 g/mol / 82.082 g/mol = 2.0000

Molecular formula = C8H8N2O2

Chemistry Handbook Review

31. The compound borazine consists of 40.29% boron (B), 7.51% hydrogen, and 52.20% nitrogen, and its molar mass is 80.50 g/mol. Calculate the molecular formula for borazine.

40.29 g B x (1 mol C / 10.811 g B) = 3.727 mol B 7.51 g H x (1 mol H/ 1.008 g H) = 7.45 mol H 52.20 g N x (1 mol N / 14.007 g N) = 3.727 mol N

3.727 mol B / 3.727 = 1.000 mol B7.45 mol H / 3.727 = 2.00 mol H3.727 mol N / 3.727 = 1.000 mol NEmpirical formula = BH2N

Chemistry Handbook Review

Molar Mass of BH2N :

B: 1 x 10.811 = 10.811 g C

H: 2 x 1.008 = 2.016 g H

N: 1 x 14.007 = 14.007 g N

Total = 26.834 g BH2N

80.50 g/mol / 26.834 g/mol = 3.000

Molecular formula = B3H6N3

Chemistry Handbook Review

33. Find the percent of water and the percent of anhydrate in the compound Cr2(SO4)3•18H2O

Molar Mass of Cr2(SO4)3: 2 moles Cr x 52.0 g/mol = 104.0g3 moles S x 32.1 g/mol = 96.3g12 moles O x 16.0 g/mol = 192.0gTotal of 392.3 g/mol

Molar Mass of 18 moles H2O: 18 x 18.02 g/mol = 324.0 g/mol

Chemistry Handbook Review

392.3 g/mol of anhydrate + 324.0 g/mol water = 716.3 g hydrate

% anhydrate (Cr2(SO4)3): 392.3g / 716.3 g = 54.8%

% water (H2O): 324.0g / 716.3 g = 45.2%

Chemistry Handbook Review

35. Cerium (III) iodide (CeI3) occurs as a hydrate. This was evaulated during a lab by measuring an empty crucible and lid at 47.511 grams, then adding the hydrate and re-measuring the mass at 49.505g. After heating over a bunsen burner, the final mass was 49.031 grams. Calculate the formula for the hydrate.

Mass of water= 49.505g – 49.031g = 0.474 g H2O

Mass of anhydrate= 49.031g – 47.511 = 1.520 g CeI3

Molar Mass of CeI3:Ce: 1 mol Ce x (140.115 g Ce / 1 mol Ce) = 140.115 g CeI: 3 mol I x (126.904 g I / 1 mol I) = 380.712 g I

Total = 520.827 g/molMolar Mass of H2O :H: 2 mol H x (1.008 g H / 1 mol H) = 2.016 g HO: 1 mol O x (15.999 g O / 1 mol O) = 15.999 g O

Total = 18.015 g/mol

Chemistry Handbook Review

1.520g CeI3 x (1 mol CeI3 / 520.827 g CeI3) = 0.002918 mol CeI3

0.474g H2O x (1 mol H2O / 18.015 g H2O ) = 0.0263 mol H2O

mol H2O / mol CeI3 = 0.0263 / 0.002918 = 9.01

Formula of hydrate = CeI3•9H2O

Chemistry Handbook Review

36. Cobalt (II) nitrate (Co(NO3)2) is used in ceramic glazes and produces a blue color when the ceramic is fired. This compound exists as a hydrate whose composition is 37.1% water and 62.9% Co(NO3)2 by mass. Write the molecular formula for this hydrate.

Molar Mass of Co(NO3)2 :Co: 1 x 58.933 = 58.933 g CoN: 2 x 14.007 = 28.014 g NO: 6 x 15.999 = 95.994 g O

Total = 182.94 g Co(NO3)2

Chemistry Handbook Review

62.9 g Co(NO3)2 x (1 mol Co(NO3)2 / 182.94 g Co(NO3)2) = 0.344 mol Co(NO3)2

37.1 g H2O x (1 mol H2O / 18.015 g H2O ) = 2.06 mol H2O

mol H2O / mol Co(NO3)2 = 2.06 / 0.344 = 5.99

Formula of hydrate = Co(NO3)2 •6H2O