Module 5 Chemical Quantities Practice Problems Q&A PDF

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Module 5 Chemical Quantities Practice Problems Q&A...

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Chemical Quantities – Practice Problems Q&A 1. Calculate the molar mass for each of the following substances; a. Cr2O3 b.

Cu(NO3)2

c.

P4O6

d.

Bi2O3

e.

CS2

f.

H2SO3

2. Calculate the number of moles in a given mass a) 21.4 mg of nitrogen dioxide b) 1.56 g of copper(II) nitrate c) 2.47 g of carbon disulfide d) 5.04 g of aluminum sulfate 3. Calculate the mass in grams of each of the following a. 1.25 mol of aluminum chloride b.

1.31 x 10-3 mol uranium

c.

0.00104 mol of carbon dioxide

d.

1.49 x 102 mol of iron

4. Calculate the number of particles given the mass or number of moles a. 0.00475 mol PH 3 b.

4.75 g PH3

c.

1.25 x 10-2 g

d.

1.25  10–2 mol Pb(C 2H3O2)2

Pb(C2H3O2)2

5. Calculate mass given number of particles a. 10,000,000,000 nitrogen molecules b.

2.49  1020 carbon dioxide molecules

6. Calculate moles given number of particles a. 6.02 x 1023 molecules of C 2H6 b. 1.27 x 1024 molecules of acetylsalicylic acid, HC 9H7O4 7. Percent composition by mass a. mass % C in CH4

8.

b.

mass % Na in NaNO3

c.

mass % C in CO

d.

mass % N in NO2 Empirical Formula a) A compound was analyzed to contain the following percentages by mass: 28.03% calcium; 22.38% oxygen; 49.59% chlorine. Determine the empirical formula of the compound. b) When 1.25 g of aluminum metal is heated with fluorine gas, 3.89 g of aluminum fluoride results. Determine the empirical formula of the compound.

9. Molecular Formula a) A compound with the empirical formula CH was found to have a molar mass of approximately 78 g. What is the molecular formula of the compound? b) A compound has the percentage composition: 78.14% boron; 21.86% hydrogen. If the molar mass of the compound is determined to be between 27 and 28 g, what are the empirical and molecular formulas of the compound?

Answers Chemical Quantities – Practice Problems Solutions 1. Calculate the molar mass for each of the following substances; a. Cr2O3, chromium(III) oxide mass of 2 mol Cr = 2(52.00 g) = 104.00 g mass of 3 mol O = 3(16.00 g) = 48.00 g molar mass of Cr2O3 = (104.00 g + 48.00 g) = 152.00 g b.

Cu(NO3)2, copper(II) nitrate mass of 1 mol Cu = 63.55 g = 63.55 g mass of 4 mol N = 4(14.01 g) = 56.04 g mass of 6 mol O = 6(16.00 g) = 96.00 g molar mass of Cu(NO3)2 = (63.55 g + 56.04 g + 96.00 g) = 215.59 g

c.

P4O6, tetraphosphorus hex(a)oxide mass of 4 mol P = 4(30.97 g) = 123.9 g mass of 6 mol O = 6(16.00) g = 96.00 g molar mass of P 4O6 = (123.9 g + 96.00 g) = 219.9 g

d.

Bi2O3, bismuth(III) oxide mass of 2 mol Bi = 2(208.98 g) = 417.96 g mass of 3 mol O = 3(16.00 g) = 48.00 g molar mass of Bi2O3 = (417.96 g + 48.00 g) = 465.96 g

e.

CS2, carbon disulfide mass of 1 mol C = 12.01 g = 12.01 g mass of 2 mol S = 2(32.07 g) = 64.14 g molar mass of CS2 =

f.

(12.01 g + 64.14 g) = 76.15 g

H2SO3, sulfurous acid mass of 2 mol H = 2(1.01 g) = 2.02 g mass of 1 mol S = 32.07 g = 32.07 g mass of 3 mol O = 3(16.00 g) = 48.00 g molar mass of H2SO3 = (2.02 g + 32.07 g + 48.00 g) = 82.09 g

2. Calculate the number of moles in a given mass a. 21.4 mg of nitrogen dioxide molar mass of NO2 = 46.01 g;

21.4 mg = 0.0214 g

0.0214 g NO2  b.

1 mol = 4.65  10–4 mol NO2 46.01 g

1.56 g of copper(II) nitrate molar mass of Cu(NO3)2 = 187.57 g 1.56 g Cu(NO3)2  1 mol / 187.57 g = 8.32  10–3 mol Cu(NO3)2

c.

2.47 g of carbon disulfide molar mass of CS2 = 76.15 g 2.47 g CS2 

d.

1 mol = 0.0324 mol 76.15 g

5.04 g of aluminum sulfate molar mass of Al2(SO4)3 = 342.17 g 5.04 g Al2(SO4)3  1 mol / 342.17 g = 0.0147 mol Al2(SO4)3

3. Calculate the mass in grams of each of the following a. molar mass AlCl 3 = 133.33 g 1.25 mol × 133.33g/1mol= 167 g b.

molar mass U = 238.03 g 1.31 × 10–3 mol × 238.03g/1mol= 0.312 g

c.

molar mass CO2 = 44.01 g 0.00104 mol ×

d.

44.01 g = 0.0458 g 1 mol

molar mass Fe = 55.85 g 1.49 × 102 mol Fe ×

55.85 g = 8.32 × 103 g 1 mol

4. Calculate the number of particles given the mass or number of moles a. 0.00475 mol  6.02 x 1023 molecules / 1mol = 2.86  1021 molecules b.

molar mass of PH3 = 34.00 g 4.75 g  1mol / 34.00 g x 6.02 x 1023 molecules / 1mol

c.

= 8.41  1022 molecules

molar mass of Pb(C 2H3O2)2 = 325.30 g 1.25  10–2 g  1mol / 325.30 g x 6.02 x 1023 molecules / 1mol

d.

1.25  10–2 mol 

= 2.31  1019 molecules

6.02 x 1023 molecules / 1mol = 7.53  1021 molecules

5. Calculate mass given number of particles a. Assuming 10,000,000,000 = 1.0  1010

b.

1.0  1010 molecules 

28.02 g N 2 = 4.7  10–13 g N 2 6.022  1023 molecules

2.49  1020 molecules 

44.01 g CO2 = 0.0182 g CO2 6.022  1023 molecules

6. Calculate moles given number of particles a. 6.02 x 1023 molecules x 1mol / 6.02 x 1023 molecules = 1.00 mol of C2H6

b. 1.27 x 1024 molecules x 1mol / 6.02 x 1023 molecules = 2.11 mol of acetylsalicylic acid, HC9H7O4 7. Percent composition by mass a. molar mass of CH 4 = 16.05 g % C = 12.01g / 16.05 g  100 = 74.83% C b.

molar mass NaNO3 = 85.00 g % Na =

c.

molar mass of CO = 28.01 g %C=

d.

22.99 g Na  100 = 27.05% Na 85.00 g

12.01 g C  100 = 42.88% C 28.01 g

molar mass of NO2 = 46.01 g %N=

14.01 g N × 100 =30.45% N 46.01 g

8. Empirical Formula a) Assume we have 100.0 g of the compound, so that the percentages become masses. 28.03 g Ca ×

1 mol = 0.6994 mol Ca 40.08 g

22.38 g O ×

1 mol = 1.3988 mol O 16.00 g

49.59 g Cl ×

1 mol = 1.3989 mol Cl 35.45 g

Dividing each number of moles by the smallest number of moles (0.6994 mol Ca) gives

0.6994 mol Ca = 1.000 mol Ca 0.6994 mol

1.3988 mol O = 2.000 mol O 0.6994 mol 1.3939 mol Cl = 2.000 mol Cl 0.6994 mol The empirical formula is CaO2Cl2 b) The amount of fluorine that reacted with the aluminum sample must be (3.89 g – 1.25 g) = 2.64 g of fluorine 1.25 g Al ×

2.64 g F ×

1 mol = 0.04633 mol Al 26.98 g 1 mol = 0.1389 mol F 19.00 g

Dividing each number of moles by the smaller number of moles gives

0.04633 mol Al = 1.000 mol Al 0.04633 mol

0.1389 mol F = 2.999 mol F 0.04633 mol The empirical formula is AlF3. 9. Molecular Formula a) empirical formula mass of CH = 13 g n=

molar mass 78 g = =6 empirical formula mass 13 g

The molecular formula is (CH)6 or C6H6 b) Assume that we have 100.00 g of the compound; then 78.14 g will be boron, and 21.86 g will be hydrogen. 78.14 g B 

1 mol = 7.228 mol B 10.81 g

21.86 g H  1mol / 1.01g = 21.64 mol H Dividing each number of moles by the smaller number (7.228 mol B) gives the empirical formula as BH3. The empirical molar mass of BH3 would be [10.81 g + 3(1.01 g)] = 13.84 g. This is approximately half of the indicated actual molar mass (13.84/28 =2), and therefore the molecular formula must be B2H6....