Download Topic 3 MOLE Avodagro`s number = 6.02 x 1023 things = 1 mole 1

Topic 3
MOLE
 Avodagro’s number = 6.02 x 1023 things = 1 mole
 1 mole of any substance weighs its formula weight in grams (molar mass)
EMPIRICAL/MOLECULAR FORMULAS
 From % to empirical = % to mass, mass to mole, divide by small, times ’til whole.
 From molecular to empirical = (molecular mass) / (empirical mass) = how many
times to multiply the formula up
 Combustion analysis
REACTIONS
 Reactants, products, coefficients, subscripts
 Balancing reactions to comply with law of conservation of mass
 Combustion (complete & incomplete), combination (or synthesis), decomposition
STOICHIOMETRY
 Stoichiometry = ratio of reactants and products in a balanced chemical equation
Always use the ratio of moles (cannot compare grams)
 Limiting Reactants (the reactant you will run out of first)
 Theoretical vs. actual yield
_________________________________________________________________________
1. Balance the following reactions and write the type of reaction:
2KNO3 (s)  2 KNO2 (s) + O2 (g)
__decomposition___________
4 CH3NH2(g) + 9 O2 (g)  4 CO2 (g) + 10 H2O(g) + 2N2 (g) combustion__
CO (g) + O2 (g)  CO2 (g)
_____combination or synthesis_
2. Calculate the percent by mass of carbon in acetylene, C2H2, a gas used in welding.
Each carbon has a mass of 12, so total mass of carbon in compound = 24.
(24 / 26) x 100 = 92%
3. Calculate the percent by mass of hydrogen in ammonium sulfate.
Each H has a mass of 1, so total mass of H in (NH4)2(SO4) = 8
(8 / 132) x 100 = 6%
4. What is the mass of 8.11 x 1023 atoms of Sulfur?
8.11 x 1023 atoms x (1 mole / 6.02 x1023 atoms) X (32g / 1 mole) = 43.1 grams
5. How many molecules of propane, C3H8 are in 0.0666 mole?
0.0666 mole x (6.02 x1023 molecules / 1 mole) = 4.01 x1022 molecules
6. How many molecules of acetaminophen, C8H9O2N are in a 50.0 mg tablet?
50.0mg x (1g/1000mg) x (1 mole/151g) x (6.02 x1023 molecules /1mole) =
1.99 x1020 molecules
7. The characteristic odor of pineapple is due to ethyl butyrate, a compound containing
carbon, hydrogen, and oxygen. Combustion of 2.78mg of ethyl butyrate produces
6.32mg of CO2 and 2.58 mg of H20. What is the empirical formula of the compound?
6.32mg of CO2 x (1g/1000mg) x (1 mole/44g) = 0.000144mole CO2 = 0.000144mole C or 0.00173 grams C
2.58mg of H2O x (1g/1000mg) x (1 mole/18g) = 0.000143mole H2O = 0.000286mole H or 0.000286 grams H
2.78 mg total – 1.73 mg C - .286 mg H = 0.764 mg O = .000764 grams O = 0.0000478 moles O
0.000144 mole C /0.0000478 = 3.01 = 3
0.000286 mole H /0.0000478 = 5.98 = 6
0.0000478 mole O /0.0000478 = 1
C3H6O
8. Water is added to 4.267 grams of UF6. The only products are 3.730 grams of a solid
containing only uranium, oxygen and fluorine and 0.970 gram of a gas. The gas is 95.0%
fluorine, and the remainder is hydrogen.
(a) From these data, determine the empirical formula of the gas.
Assume 100 grams, F = 95.0 g (1mole / 19 g) = 5 moles F and H = 5.0 g (1mole / 1 g) = 5 moles H
Formula = HF
(b) What fraction of the fluorine of the original compound is in the solid and what fraction
in the gas after the reaction?
95% of the 0.970 g gas = 0.9215 grams F in gas
4.267 grams of UF6 (1mole UF6/352g) (6 moles F / 1 mole UF6) (19g/1 mole F) = 1.382g F in UF6
0.9215 / 1.382 = 0.667 x 100 = 66.7% or 2/3 of F in the gas, 1/3 in the solid
(c) What is the formula of the solid product?
1.382 grams of F in UF6 - 0.9215g F in gas = 0.4605 g F in solid product = 0.0242 mole F
4.267 grams of UF6 (1mole UF6/352g) (1 moles U / 1 mole UF6) = 0.0121 mole U (238g/mole)=
2.885g U
3.370 g of product – 2.885 g U – 0.4605 g F = 0.3845g O (1 mole / 16 g) = 0.02403 mole O
0.0121 mole U / 0.0121 =1
0.02403 mole O / 0.0121 = 1.986 = 2
UF2O2
0.0242 mole F / 0.0121 = 2
(d) Write a balanced equation for the reaction between UF6 and H2O. Assume that the
empirical formula of the gas is the true molecular formula.
UF6 (s) + 2H2O (l) --> UF2O2 (s) + 4HF (g)
9. What is the molecular formula for each of the following:
Empirical formula = HCO2, molar mass = 90.0 g
HCO2 = 45g/mol x2 = H2C2O4
Empirical formula = C2 H4O, molar mass = 88 g/mol
C2H4O = 44g/mol x2 = C4H8O2
94.1 % O and 5.9 % H; molar mass = 34 g
94.1 g O (1mol/16g)= 5.88 = 6 and 5.9g H (1mol/1g) = 5.9mole = 6
So formula is OH with empirical mass = 17, molecular mass = 34 (or 2x17) so molecular formula is
H2O2
10. In a reaction vessel, 0.600 mol of Ba(NO3)2(s) and 0.300 mol of H3PO4(aq) are combined
with deionized water to a final volume of 2.00 L. The reaction represented below
occurs.
3 Ba(NO3)2(aq) + 2 H3PO4(aq)  Ba3(PO4)2(s) + 6 HNO3(aq)
Calculate the mass of Ba3(PO4)2(s) formed.
.600 mole Ba(NO3)2 (1 Ba3(PO4)2mole /3 mol Ba(NO3)2) = 0.200 mole Ba3(PO4)2
0.200 mole Ba3(PO4)2 (601.93g/mol) = 120 grams
11. Write the balanced equation for the complete combustion of octane, C8H18.
2C8H18 + 25O2 16CO2 + 18H2O
a) How many grams of O2 are needed to burn 1.50 mole of octane?
1.5 mole octane (25mole O2 /2mole octane) (32 g /mole O2) = 600 g
12. One of the steps in the commercial process for converting ammonia to nitric acid
involves the conversion of N H3 to NO:
4NH3 (g) + 5 O2 (g)  4NO (g) + 6 H2O (g)
In a certain experiment 2.50 grams of NH3 reacts with 2.85 grams of O2.
a) Which is the limiting reactant?
2.50 grams of NH3 (1 mol NH3 /17g) (4 mol NO/4 mol NH3) = 0.147 mol NO
2.85 grams of O2 (1 mol O2 /32 g) (4 mole NO/5 mol O2) = 0.0713 mol NO O2 is limiting
b) How many grams of NO form?
0.0713 mol NO (30g/mol) = 2.14 grams NO
c) How many grams of the excess reactant remain after the limiting reactant is all used up?
0.0713 mol NO (4 mol NH3 /4 mole NO) (17g /1 mol NH3 ) = 1.21 g used
2.50 grams – 1.21 grams = 1.29 grams left over
13. The molecular formula of a hydrocarbon is to be determined by analyzing its combustion
products. The hydrocarbon burns completely, producing 7.2 grams of water and 7.2 liters
of CO2 at standard conditions. What is the empirical formula of the hydrocarbon?
7.2 grams of water = .80 mol H2O / .32 = 2.5 x 2 = 5
14.08 grams of CO2 = 0.32 mol CO2 / .32 =1 x 2 = 2
C2H5
14. Answer the following questions about acetylsalicylic acid, the active ingredient in aspirin.
a) The amount of acetylsalicylic acid in a single aspirin tablet is 325 mg, yet the tablet has a
mass of 2.00 g. Calculate the mass percent of acetylsalicylic acid in the tablet.
(0.325g / 2.00 g ) x 100 = 16.25%
b) The elements contained in acetylsalicylic acid are hydrogen, carbon, and oxygen. The
combustion of 3.000 g of the pure compound yields 1.200 g of water and 3.72 L of dry
carbon dioxide, measured at STP. Determine the mass, in g, of each element in the 3.000 g
sample of the compound.
1.200 g water (1 mol water/ 18 g) (2 mol H / 1 mol water)= 0.133 mol H = .133g H
3.72 L CO2 (1 mole CO2/22.4L) (1 mol C / 1 mole CO2) = 0.17 mol C = 2 g C
3.000 g compound - 2 g C - .133g H = 0.867 g O = 0.054 moles O
15. A compound containing silver and chlorine contains 75.3% Ag. What is the empirical
formula?
Ag = 75.3 g (1 mol / 107.9 g) = 0.698 mol
AgCl
Cl = 24.7 g (1 mol / 35.4 g) = 0.698 mol
16. The reaction of 25.0 g benzene, C6H6, with excess HNO3 resulted in 21.4 g C6H5NO2.
C6H6 + HNO3  C6H5NO2 + H2O
a) What is the theoretical yield?
25.0 g benzene (1 mol/78 g) (1 mol C6H5NO2 /1 mol benzene) (123g/1 mol C6H5NO2)= 39.4g
(21.4 / 39.4 ) 100 = 54.3 %
b) If 30.0 g C6H5NO2 was obtained, what is the percentage yield?
(30.0 / 39.4 ) 100 = 76.1 %
17. If 70.9 g of chlorine reacts with sufficient cesium to produce cesium chloride, what is the
theoretical yield?
2 Cs + Cl2  2CsCl
a. 70.9 g
b. 132.9 g
c. 146 g
d. 337 g
e. 285 g
70.9 g Cl2 (1 mol/70g)(2 mol CsCl/1 mol Cl) (168.5 g/1 mol)
18. In the reaction below, 8.0 g of H2 react with 9.0 g of O2. Which of the following statements
is true?
2 H2 + O2  2H2O
a. The equation is not balanced. 4 H atoms and 2 O atoms on each side
b. The H2 is the limiting reactant.
c. The O2 is the limiting reactant.
d. 2.0 moles of H2O would be produced.
e. 36 grams of H2O would be produced.
19. Aluminum reacts with oxygen to form aluminum oxide. Which of the reactions below is
correct and properly balanced?
a. Al(s) + O(g)  AlO(s)
b. 2Al(s) + O2 (g)  2AlO(s)
c. 3Al(s) + O2 (g)  Al3O2 (s)
d. 4Al(s) + O2 (g)  2Al2O(s)
e. 4Al(s) + 3O2 (g)  2Al2O3 (s)
20. 3.0 g of N2 react with 3.0 g of H2 according to the reaction below. What will the final
mixture contain?
N2 + 3H2  2NH3
a. H2 only
b. NH3 only
c. N2 & H2 & NH3
d. H2 & NH3
e. N2 & 2NH3