1

Moles and Stoichiometry

I. Moles

A. Definition. 1. A mole (mol) = 6.022 x 1023 units. 6.022 x 1023= Avogadro's constant, with a dimension of particles mol–1 or mol–1 b. A mass in grams equal to the formula mass of a substance contains 6.02 x 1023 formula

units.

c. The practical unit of formula mass or atomic mass = grams/mol.

2. The following should be familiar to you. Type of Formula Composition of 10.0 g of substance Formula substance mass in moles _ in no. of units Na element 23.0 g/mol 10.0 /23.0= (0.435)(6.02 x1023) = 0.435 mol 2.62 x 1023 atoms

H2 molecular 2.0 g/mol 5.0 moles H2 3.01 x 1024 molecules of H2

10.0 moles H 6.02 x 1024 atoms of H

C2H4 molecular 28.0 g/mol 0.357 mol C2H4 2.15 x 1023 molecules C2H4 0.714 mol C 4.30 x 1023 atoms C 1.428 mol H 8.60 x 1023 atoms H

Al2O3 ionic 102 g/mol 0.098 mol Al2O3 5.90 x 1022 formula units 0.196 mol Al3+ 1.18 x 1023 Al3+ ions

0.294 mol O2- 1.77 x 1023 O2- ions

2

3. Additional Example. Consider a 48.60 g sample of the compound C4H10O2

a. How many moles of C4H10O2 are present?

MM of C4H10O2 = 4(12.0) + 10(1.0) + 2(16.0) = 90.0g/mol

Mol C4H10O2 = 48.60g

90.0g/mol= 0.540 mol

b. How many moles and how many grams of C are present in the sample?

The formula shows 4 C atoms for every molecule ∴ there are 4 moles of C for every mole of

C4H10O2.

Moles C = 0.540 mole C4H10O2 x

!

4 mol C

1 mol C4H

10O

2

= 2.16 mol C

grams C = (2.16 mol C)(12.0 g/mol C) = 25.92 g or

grams C =48.60g48.0 g C

90.0 g compound= 25.92g C

moles C = 25.92 g

12.0 g/mol= 2.16 mol C

c. How many O atoms are present in the sample? Mol O = 0.540 mole C4H10O2 x

2 mol O

1 mol C4H

1 0O

2

= 1.08 mol O

Atoms O = 1.08 mol O x 6.02x1023 atoms/mol = 6.50x1023

d. How many atoms of H and how many grams of H are present in the sample?

Moles of H = 0.540 mole C4H10O2x

!

10 mol H

1 mol C4H

10O

2

= 5.40 mol H

grams H = 5.40 molx1.0g/mol = 5.40g

Note: mol H could also be obtained by using the moles of C:

Mol H = 2.16 mol Cx

!

10mol H

4 mol C= 5.40 mol

3

B. Empirical formulas from analysis data.

1. A compound was analyzed and found to contain 29.1% Na, 40.6% S, and 30.3% O by mass. Calculate the empirical formula. moles of Na = 29.1g / 23.0 g/mol= 1.265 mol.----------> 1.000----------> 2 moles of S = 40.6g / 32.1g/mol = 1.265 mol -----------> 1.000 ---------> 2 moles of O = 30.3g / 16.0g/mol = 1.894 mol ------------>1.500--------- > 3 formula : Na2S2O3 Sodium thiosulfate

2. A 0.401 g sample of a compound was analyzed and found to contain 0.320 g of C and 0.081 g of H. In another experiment the molar mass was estimated to be equal to 30. Calculate the empirical and molecular formulas of the compound. moles C = 0.320g / 12.0g/mol = 0.0267 -------------> 1.00 ---------> 1 moles H = 0.081g / 1.0 g/mol = 0.081---------------> 3.04 ----------> 3 empirical formula = CH3 empirical formula mass (EFM) = 12.0 + 3(1.0) = 15

3015 = 2 = empirical formula units / molecule C2H6 = molecular formula.

3. Analysis by combustion.

A 1.000 g sample of a compound containing only C, H, and O was burned to give 2.197 g of CO2 and 1.199 g of H2O. Calculate the empirical formula. + O2 ∆ (CxHyOz)---------------> CO2 + H2O 1.000 g 2.197 g 1.199 g all the C goes to form CO2 mol C = mol CO2 = 2.197 g

44.0 g/mol = 0.04992 mol

grams C in the sample = (.04922 mol )(12.0 g/mol) = 0.600 g all the H ends up in H2O

mol H = 2xmol H2O = 2(1.199 g)

18.0 g/mol = 0.1331 mol

grams H in sample = (0.1331 mol)(1.0 g/mol) = 0.133 g total mass of sample = 1.000 g = g of C + g of H + g of O therefore g of O = 1.000 g - 0.600 g - 0.133 g = 0.266 g

mol O = 0.266 g

16.0 g/mol = 0.0166 mol

mol C = .04992--------------> 3.00

4

mol H = 0.1331 -------------> 8.01 empirical formula = C3H8O mol O = 0.0166 -------------> 1.00 II. Stoichiometry: Calculations from balanced chemical equations. A. Information in a balanced chemical equation: 4Al(s) + 3O2(g) ------> 2Al2O3(s) 1. In terms of atoms and molecules.

4 Al atoms + 3 O2 molecules -----> 2 Al2O3 formula units. 2. In terms of atomic mass units.

(4x27) = 108 u of Al + (3x32) = 96 u of O2 -----> (2x102) = 204 u of Al2O3 Note the Law of Conservation of mass (mass of products = mass of reactants) 3. In terms of moles.

4 moles of Al + 3 moles of O2 -----> 2 moles of Al2O3

4. In terms of grams. 108 g of Al + 96 g of O2 -----> 204 g Al2O3

5. In terms of any mass unit. 108 lb. of Al + 96 lb. of O2 ----> 204 lb. of Al2O3

108 slugs of Al + 96 slugs of O2 ----> 204 slugs of Al2O3 108 tons of Al + 96 tons of O2 ----> 204 tons of Al2O3

B. Calculations from balanced equations. 1. General. a. A balanced chemical equation gives stiochiometeric information directly in terms of moles of reactants and products.

Example, in the above equation, each mole of Al that reacts requires 34 mole of O2

and produces 12 mole of Al2O3.

b. In the laboratory, we do not obtain information directly in terms of moles. The normal laboratory measurements involve obtaining 1) the masses of pure solids and some pure liquids. 2) the volumes of solutions of known concentration. 3) the volumes of gases at known temperatures and pressures. 4) the volume of liquids of known densities. c. In stoichiometric calculations information is given, and asked for, in terms of these laboratory measurements. Therefore, the sequence to be followed is:

5

1) CONVERT LABORATORY INFORMATION TO MOLES 2) USE THE BALANCED CHEMICAL EQUATION TO GET INFORMATION IN TERMS OF MOLES.

3) RECONVERT MOLES BACK INTO LABORATORY UNITS.

2. Conversions to moles-these relationships should be familiar to you.

a. Direct mass measurements of pure solids and liquids. Moles = Mass in grams

formula mass

b. Pure liquids of known densities.

Mass = (density)x(volume) The densities are usually in g/mL and the volumes are in mL

Moles = Mass in grams

formula mass

c. Volumes of solutions of known Molarity:

Molarity (M) = Moles of Solute

Liters of Solution = mmol Solute

mL of Solution

moles of solute = (M)(no. of liters of solution)

mmoles of solute = (M)(no. of mL of solution)

d. Volumes of gases: Use the Ideal gas Law (see Chapter 5.1-5.6)

1) for a pure gases

PV = nRT = g

MM RT.

Where V = volume of the gas in L, P = pressure of the gas, MM= molar mass n = number of moles of the gas, T = temperature in K (= °C + 273.15), R = gas constant.

The value of R depends on the dimensions of pressure.

R = 62.4 Torr Lmol K = 0.08206

atm Lmol K = 3.184

Pa m3mol K

2) For mixtures of gases use partial pressures and Dalton’s Law.

PxV = nxRT for a component x of a gaseous mixture (Px = Partial pressure of x) Ptot = ∑ Px

3) This is used when collecting a gas over water. A standard laboratory technique for collecting a gaseous product from a reaction is to allow the gas to displace water in a

6

container. The gas collected is saturated with water vapor. Therefore the total pressure in the container is due to a mixture of the gas (X) and H2O, so that

PTot

= PX+ P

H2O

PH2O

is the vapor pressure of water and is a function of temperature only, this can be looked

up in standard tables and subtracted from Ptot (usually atmospheric pressure) to obtain the partial pressure of the gas. C. Stoichiometric calculations. 1. Suppose that 80.0 g of Al was reacted according to the equation

4Al(s) + 3O2(g) ----> 2Al2O3(s) a. How many moles of O2 was consumed?

moles Al = 80.0 g

27.0 g/mol = 2.96 mol of Al

moles O2 = 2.96 mol Al x 3 mol O24 mol Al = 2.22 mol O2

b. What volume of O2 measured at 27°C and 750 Torr pressure would be used?

V = n O2RT

P = (2.22 mole)(62.4

Torr Lmol K )(300 K)

750 Torr = 55.3 L

c. How many grams of Al2O3 would be formed?

moles Al2O3 = 2.96 mol Al x ( 2 mol Al2O3

4 mol Al ) = 1.48 mol Al2O3

g Al2O3 = (1.48 mol)(102 g/mol) = 151 g Al2O3

7

2. Consider the reaction: 4KO2(s) + 2H2O(l) ----> 4KOH(aq) + 3O2(g) .

Suppose that 7.50 g of KO2 was placed in 50.0 mL of water.

a. What volume of O2 would be collected H2O at 27°C and 757 Torr atmospheric pressure?

moles KO2 = 7.50 g

71.1 g/mol = 0.105 mol KO2

moles of O2 produced = (0.105 mol KO2) x (3 mol O2

4 mol KO2 ) = 7.875x10-2 = nO 2

Since the gas is collected over water, the vapor pressure of water must be subtracted from the atmospheres pressure to get the partial pressure of O2, therefore PO

2 = Patm -P

H2O, where P

H2O= vapor pressure of water.

At 27°C, PH2O

= 27 Torr. ∴ PO 2 = 757 Torr - 27 Torr = 730 Torr

V = no

2RT

PO 2

= (7.875x10-2)(62.4)(300)

730 = 2.02 L

b. How many grams of KOH are formed?

moles of KOH formed = 0.105 mol KO2 x ( 4 mol KOH4 mol KO2 ) = 0.105 mol KOH

g KOH formed = 0.105 mol (56.1 g/mol) = 5.89 g

c. How many mL of water remains?

moles H2O required = 0.105 mol KO2 x ( 2 mol H2O4 mol KO2 ) = 0.0525 mol H2O

g H2O =( 0.0525 mol) x (18.0 g/mol) = 0.945 g H2O.

This is equivalent to 1 mL of H2O (the density of H2O = 1.0 g/mL).

∴ 49.0 mL of H2O will remain.

d. What is the final concentration of KOH?

MKOH = 0.105 mol

49.0x10-3 L = 2.14M

8

3. Limiting reagents. a. It is not necessary, and many times not desirable, to mix stoichiometric amounts of reactants. Under these conditions, the reagent used up first (the limiting reagent) determines the amount of product; the other reactant is the excess reactant. In such cases the first step is to decide which reactant is limiting. b. Consider the reaction 3Mg + N2 ----> Mg3N2 Suppose that 2.50 g Mg and 1.00 g N2 are mixed and the reaction takes place. 1) What reactant will be in excess and how many grams of this reactant remains unreacted?

moles Mg = 2.50 g

24.3 g/mol = 0.1029 mol moles N2 = 1.00 g

28.0 g/mol = 0.0357 mol

The stoichiometric

!

Mg

N2

molar ratio = 31 , the experimental molar ratio is

0.10290.0357 = 2.88 which is less than 3

∴ N2 is in excess and Mg is the limiting reagent and will determine the amount of product.

moles of N2 required = 0.1029 mol Mg x ( 1 mol N23 mol Mg ) = 0.0343 mol

moles N2 left = 0.0357 mol - 0.0343 mol = 0.0014 mol g in excess = (0.0014 mol) x (28.0 g/mol) = 3.92x10-2 g. b. How many grams of Mg3N2 will be formed.

moles Mg3N2 = 0.1029 mol Mg x ( 1 mol Mg3N2

3 mol Mg ) = 3.43x10-2 mol

grams of Mg3N2 = (3.43x10-2 mol) x (100.9 g/mol) = 3.46 g 4. Yields. a. The above calculations give the maximum amount of product that could be obtained in a reaction. In practice, this is never obtained, we do not get 100% of what we expect. b. The percent yield, or yield, is:

Yield = experimental amount of product

theoretical amount of productx100

The theoretical amount of product is that calculated in the above examples. c. There are a number of reasons for low yields.

1) Poor technique. 2) An unfavorable equilibrium. In such cases the yield can be improved by having one of the reactants in excess. 3) Competing reaction or consecutive reactions.

9

d. When new reactions are reported in the literature, they are usually described under conditions which maximize the yield.

10

Problems

1. On analysis a 2.75 g sample of a compound was found to contain 1.55 g of phosphorus

and 1.20 g of sulfur. Calculate the empirical formula of this compound. (P4S3)

2. A sample of a compound contains 3.96 g of carbon, 0.66 g of hydrogen, and 3.52 g of oxygen.

In another experiment it was found that a 2.05 g sample of the gaseous compound occupied a volume

of 584 mL at 125 °C and 400 Torr pressure. a. What is the empirical formula of this compound? b.

What is the molar mass of this compound? c. What is the molecular formula of this compound?

(C3H6O2; 218; C9H18O6)

3. A sample of a compound containing only nitrogen and sulfur was burned in oxygen completely

converting the nitrogen to N2O3 and the sulfur to SO2. The oxides formed were trapped and

weighed giving 0.198 g of N2O3 and 0.512 g of SO2. What is the empirical formula of the

compound? (N2O3)

4. Calculate the empirical formulas from the following percent compositions.

a. 34.3% Na; 17.9% C; 47.8% O b. 39.3% C; 8.2% H; 52.5% O

c. 72.3% Fe; 27.7% O d. 69.9% Fe; 30.1% O

e. 77.7% Fe; 22.3% O f. 9.7% Al; 38.4% Cl; 51.9% O

(Answer: a.NaCO2 b. C2H5O2 c. Fe3O4 d. Fe2O3 e. FeO f. AlCl3O9)

5. A 5.00 g sample of a compound containing C, H, and N was burned in oxygen to give 9.778 g

of CO2 and 7.000 g of H2O. Calculate the empirical formula of the compound. (C2H7N)

6. A 3.500 g sample of a mixture of NaCl and KCl was dissolved in water and the chloride

precipitated as AgCl. If the mass of the AgCl precipitate is 7.522 g, calculate the mass percent

NaCl in the sample. (42.9% NaCl)

7. Calcium hydroxide reacts with H3PO4 to give Ca3(PO4)2. Suppose the 0.850 g of Calcium

hydroxide was placed in 200 mL of a 0.045M H3PO4 solution. What reactant would be in excess

and how many grams of Ca3(PO4)2 would be formed. (H3PO4, 1.19 g Ca3(PO4)2)

8. What volume of H2 could be collected over water at 27°C and 774 Torr atmospheric pressure by

the reaction of 0.256 g of Na with excess water to give H2 and NaOH? (139 mL)

11

9. A hydrocarbon was analyzed and found to contain 84.1% C and 15.9% H by mass. In an

experiment it was found that a 0.488 g sample of the gaseous compound occupied a volume of

215 mL at 50°C and 400 Torr. Calculate the molecular formula of the compound. (C8H18)

10. Aluminum sulfide reacts with oxygen to give aluminum sulfate. How much oxygen would be

consumed when 6.60 g of aluminum sulfide reacts? (8.45 g)

11. Propene, C3H6, burns in oxygen to form carbon dioxide and water.

a) Write the balanced equation for this reaction.

b) Explain what the equation states in a quantitative way.

c) Per mole of propene, how many moles of oxygen would be required?

d) Suppose 0.84 g of propene is burned. How many moles of Propene is present? How

many moles of oxygen would be required for the complete combustion? How many

moles of water and carbon dioxide would be formed? Calculate the grams of oxygen and

the grams of carbon dioxide and water formed.

e) Show that the masses in (d) are in accordance with the law of conservation of mass.

(d. 0.02 mol C3H6; O2, 0.09 mol, 2.88g; CO2, 0.06 mol, 2.64 g; H2O, 0.06 mol, 1.08g)

12. Calcium carbonate decomposes on heating to give calcium oxide and CO2 (g). What

volume of CO2, measured at 100° C and 757 torr, would be generated by the

decomposition of 5.0 g of calcium carbonate? (1.54 L)

13. Naphthalene, C10H8, reacts with O2 to give CO and H2O. What volume of CO could be

collected over water at 24° C and 750 torr total pressure by the complete reaction of 1.50 g

of naphthalene? (Vapor pressure of water at 24° C = 22 torr.) (2.98 L)

14. a. A compound containing only carbon and hydrogen when reacted with O2 produced 1.62

g of H2O and 2.40 liter of CO2 gas when measured at 27 ° C and 700 torr pressure.

Assuming that all of the carbon in the compound was converted to CO2 and all the

hydrogen was converted to H2O, calculate the empiricalformula of the compound. (CH2)

b. In another experiment, a 0.30 g sample of this gaseous compound was found to occupy

a volume of 137 mL at 27° C and 730 torr pressure. Calculate the molar mass of the

compound. (56.1 g/mol)

c) what is the molecular formula of the compound? (C4H8)

12

d) Write the balanced equation for the reaction of this compound with O2.

(C4H8 + 6 O2 ----> 4 CO2 + 4 H2O)

15. Aluminum reacts with HCl to give AlCl3 and H2 (g). What volume of H2 would be

collected over water at 28° C and 748 torr pressure when 5.0 g of Al is placed in 250 mL of

a 2.0 M HCl solution? (Vapor pressure of water at 28° C = 28 m torr.) (0.500 mol HCl

(LIMITING REAGENT); 0.185 mol Al; 13.0 L of H2)

16. Zinc sulfide reacts with O2 (g) to give zinc (II) oxide and SO2 (g). What volume of O2 (g),

measured at 25° C and 740 torr pressure, would be required to react with 0.25 g of zinc

sulfide? What volume of SO2 (g), measured under the same conditions, would be produced

in this reaction?

(volume of O2 = 96.7 mL volume of SO2 = 64.5 mL)

17. Phosphorus burns in O2 to give P2O5. Suppose that 10.0 g of P is ignited in a 30.0 L

container of O2 at a temperature of 100 °C and a pressure of 400 torr.

a. What reactant is in excess, and how many moles of that reactant will be left unreacted?

(O2 in excess, 0.113 mol left)

b. How many grams of P2O5 will be formed? (22.9 g)

18. CO2 can be removed from a gaseous mixture by reacting it with Na2O (s) to give Na2CO3

(s). A mixture of CO2 (g) and an inert gas in a 5.0 liter container originally exerted a

pressure of 500 torr at 25° C. After the gas mixture was exposed to Na2O, the pressure in

the container decreased to 200 torr.

a) What was the partial pressure of CO2 in the gas mixture? (300 Torr)

b) How many grams of CO2 was in the gas mixture? (3.55 g)

c) How many grams of Na2CO3 was formed? (8.55 g)

19. Consider the following reaction:

Al + H2SO4 -----> Al2(SO4)3 + H2

a. Balance the equation.

b. For a 8.1 g samples of Al,

1) Calculate the moles of Al present.

13

2) Calculate the moles and grams of H2SO4 required for complete reaction.

3) Calculate the moles and grams of Al2(SO4)3 and of H2 that would be formed.

4) Show that mass is conserved in this reaction.

20. For the following reaction:

HCl + O2 -----> H2O + Cl2

a. Balance the equation.

b. For a 0.16 g sample of O2,

1) Calculate the moles of O2 present

2) Calculate the moles and grams of HCl required for complete reaction.

3) Calculate the moles and grams of H2O and of Cl2 formed.

4) Show that mass is conserved in this reaction.

21. Butane, C4H10, reacts with O2 to give CO2 and H2O.

a. Write the balanced equation for the reaction.

b. Suppose 17.4 g of C4H10 and 64.0 g of O2 are mixed and the reaction allowed to

take place.

1) Which reactant will be in excess? How many moles of this reactant will be left?

How may grams?

2) How many grams of CO2 is formed?

3) How many grams of H2O is formed?

22. Zinc sulfide, ZnS, reacts with O2 to give ZnO and SO2.

a. Write the balanced equation for the reaction.

b. Suppose 20.0 g of ZnS and 15.0 g of O2 are mixed and the reaction allowed to

take place.

1) What reactant will be in excess? How many moles of this reactant will be left?

How many grams will be left?

2) How many grams of SO2 will be formed?

c. SO2 reacts with CaO according to the equation: CaO + SO2 -----> CaSO3

14

How many grams of ZnS would be required to produce enough SO2 to react with 8.4 g

of CaO?

23. Aluminum carbonate reacts with HBr according to the equation:

Al2(CO3)3(s) + 6 HBr(aq) 2 AlBr3(aq) + 3 H2O(l) + 3 CO2(g)

Suppose that 5.84 g of Al2(CO3)3 reacts., a. how many grams of AlBr3 (FM = 267) is formed? b.

what volume of CO2(g) would be collected over H2O at 24 °C and 754.4 torr total pressure? (The vapor

pressure of H2O at 24 °C = 22.4 torr) c. how many mL of a 3.0 M HBr solution would be needed in this

reaction?

24. Suppose that iron reacted with H2S to give Fe2S3 and H2 according to the equation:

2 Fe(s) + 3 H2S(aq) -----> Fe2S3(s) + 3 H2(g)

When 8.40 g of Fe reacts, a. what volume of H2(g) could be collected over water at 25 °C and

758.8 Torr atmospheric pressure from the reaction? (vapor pressure of water at 25 °C,

PH2O

= 23.8Torr ) b. how many Liters of a 0.20 M H2S solution would be required in the reaction?

c. How many grams of Fe2S3 would be produced in the reaction?

Answers 19 - 24 19. b. 1) 0.30; 2)0.45 mol, 44.1 g; 3) Al2(SO4)3, 0.15 mol, 51.3 g; H2, 0.45 mol, 0.90 g 20. b. 1) 5.0x10–3 mol; 2) 0.02 mol, 0.73 g; 3) Cl2, 0.01 mol, 0.71 g; H2O, 0.01 mol, 0.18 g 21. b. 1) O2 in excess, 0.05 mol or 1.6 g remain; 2) 52.8 g; 3) 27.0 g 22. b. 1) O2 in excess, 0.16 mol or 5.2 g remain; 2) 13.1 g; c. 14.6 g 23. a. 13.33 g; b. 1.9 L; c. 50 mL 24. a. 5.71 L; b. 1.13 mL; c. 15.6

15

16