Chapter 3:

Calculations with Chemical Formulas and

Equations

Molecular Mass and Formula Mass◆ for molecular compounds:

the molecular mass is the mass (in amu) of one molecule of the compound

molecular mass = ∑atomic masses of elements present

ex. P2O5 molecular mass = 2(30.97 amu) + 5(16.00 amu) = 141.94 amu

◆ for ionic compounds:the formula mass is the mass (in amu) of one formula unit of the compound

formula mass = ∑atomic masses of elements present

ex. Ca(ClO3)2 formula mass = 40.08 amu + 2(35.45 amu) + 6(16.00 amu) = 206.98 amu

The Mole

◆ SI base unit for amount of substance◆ the mole is a counting unit

1 mole contains the same number of particles as there are atoms in exactly 12 g of 12C.

by definition: 1mol = 6.022 x 1023 particlesAvogadro’s number, NA = 6.022 x 1023 mol!1

so: 1 mol Li = 6.022 x 1023 Li atoms

1 mol H2O = 6.022 x 1023 H2O molecules

1 mol NaNO3 = 6.022 x 1023 NaNO3 formula units

Using NA as a Conversion Factor

◆ NA relates amount of substance in mol to number of particles*

* particles may be atoms, molecules, formula units or ions depending on sample under consideration

examples:

Calculate the amount (in mol) of molybdenum in a sample containing 6.295 x 1025 atoms Mo.

Calculate the number of SO3 molecules in 1.25 mol SO3.

What Does 1 Mole Look Like? Molar Mass

◆ Molar Mass of an Element - mass (in g) per mole of an element.

units g/mol

◆ Molar Mass of a Compound - mass (in g) per mole of a compound.

units g/mol

molar mass compound = ∑molar mass of elements present

Molar Mass of an Element

◆ calculate the molar mass of an element:

convert average atomic mass (in amu/atom) to molar mass (in g/mol)

ex: Determine the molar mass of rhodium.

◆ use molar mass as a conversion factor relating sample mass (in g) to amount of substance (in mol)

ex: Determine the mass (in g) of 0.775 mol Sn.

◆ combine molar mass and NA as conversion factors ...

ex: Determine the number of atoms in 105.7 g Sn.

ex: Determine the mass (in g) of a single atom of Sn.

Molar Mass of an Element

Molar Mass of a Compound

◆ Molar Mass of a Compound - mass (in g) per mole of a compound.

units g/mol

molar mass compound = ∑molar mass of elements present

ex: Calculate the molar mass of phosphorus pentachloride.

ex: Calculate the number of molecules in 10.0 mg PCl5.

ex: Calculate the mass (in g) of a single PCl5 molecule.

Compound Stoichiometry◆ the subscripts in a chemical formula tell us the ratio of how elements combine

◆ use these ratios as conversion factors in calculations

ex: Consider calcium nitrate. Write the chemical formula.

What relationships exist between ...mass in g mass in amu

amount in mol # of formula units# of calcium ions # of nitrate ions

mol of calcium ion mol of nitrate ion#’s of Ca, N, and O atom

mol of Ca, N, and O

Compound Stoichiometry Calculations

Consider a 25.0 g sample of Ca(NO3)2. Determine the following:

a. molar massb. mol Ca(NO3)2

c. mol Ca2+

d. mol NO3!

e. mass (in g) of Of. number of NO3! ionsg. number of formula unitsh. number of O atomsi. mass (in g) of 1 formula unit

Mass Percent Composition◆ composition of compound with respect to percent mass of each element present

consider AxBy: mass% A = ———————— x 100

mass% B = ———————— x 100

x(molar mass A)molar mass AxBy

y(molar mass B)molar mass AxBy

ex: Determine the mass % composition of barium nitrate.

and: mass % A + mass % B = 100%

Chemical Formulas:Empirical Formula vs. Molecular Formula

empirical formula: relative numbers of atoms present

subscripts in chemical formula are smallest possible whole number ratio

molecular formula:exact numbers of atoms present

subscripts in chemical formula may or may not be smallest possible whole number ratio

Determination of Empirical and Molecular Formulas: Using Mass Percent Composition Data

ex: Vitamin C (molar mass = 176.1 g/mol) is composed of 40.9% C, 4.57% H, and 54.5% O by mass. Determine the empirical and molecular formula of Vitamin C.

1. Determine the mol of each element present.◆ assume a 100 g sample and use mass % composition to interpret g of each element present

◆ convert g element ! mol element

2. Determine the mol ratio of elements present in the smallest whole numbers.

Determination of Empirical and Molecular Formulas: Using Mass Percent Composition Data (Cont’d)

ex: Vitamin C (molar mass = 176.1 g/mol) is composed of 40.9% C, 4.57% H, and 54.5% O by mass. Determine the empirical and molecular formula of Vitamin C.

3. Divide the actual molar mass of the compound by the molar mass of the empirical formula. This will equal a small whole number, n.

4. Multiply the subscripts of the empirical formula by n ! molecular formula

Another Formula Determination Example:

A 500.0 mg sample of a compound composed of only cadmium and phosphorus is analyzed and found to contain 422.5 mg Cd.

Determine the empirical and molecular formulas of this compound. The actual molar mass of the compound is 399.1 g/mol.

Stoichiometry of Chemical Reactions

Answer questions like:

◆ How many g of reactant B are required to react completely with x g of reactant A?

◆ How many g of product C can be formed from the complete reaction of x g of reactant A?

Consider the reaction:CaO (s) + 3 C (s) " CaC2 (s) + CO (g)

56.08 g/mol 12.01 g/mol 64.01 g/mol 28.01 g/mol

Calculate the mass (in g) of C required to react completely with 155.2 g CaO.

Calculate the mass (in g) of CaC2 that can form from the complete reaction of 155.2 g CaO.

Reaction Yields

◆ theoretical yield: maximum amount of product that can form in a reaction; typically reported in g or mol

◆ 1 actual or experimental yield: actual amount of product collected from a reaction

◆ percent yield: percentage of the theoretical yield that is actually collected

% yield = ——————— " 100actual yieldtheoretical yield

Identification of the Limiting Reactant

The limiting reactant is the reactant that limits the theoretical yield of product.

◆ it is the reactant that is completely consumed during the course of the chemical reaction

◆ as soon as one of the reactants is gone, the reaction must stop!

TO CALCULATE A THEORETICAL YIELD, YOU MUST ALWAYS START WITH THE LIMITING REACTANT!!

Identification of the Limiting Reactant: (a not very scientific example)

to build a perfect s’more you need:

2 graham crackers2 marshmallows

1 square of chocolate

Consider 2 sets of ingredients:A

50 graham crackers50 marshmallows

25 squares chocolate

B60 graham crackers75 marshmallows

22 squares chocolate

Identification of the Limiting Reactant: (a not very scientific example)

set up s’more reaction tables:

2 graham crackers

+ 2 marshmallows +1 square chocolate " 1 s’more

before 50 50 25 0

change -50 -50 -25 +25

after 0 0 0 25

2 graham crackers

+ 2 marshmallows +1 square chocolate " 1 s’more

before 60 75 22 0

change -44 -44 -22 +22

after 16 31 0 22

Identification of Limiting Reactant:Calculations and Reaction Tables

100 g CaC2 and 100 g H2O are combined and allowed to react.

◆ identify the limiting reactant

◆ calculate the theoretical yield of products

◆ determine the mass of the excess reactant that remains unconsumed when the reaction is complete

◆ complete the reaction table

CaC2 (s) + 2 H2O (l) " Ca(OH)2 (s) + C2H2 (g)

CaC2 (s) + 2 H2O (l) " Ca(OH)2 (s) + C2H2 (g)

Calculations to fill in the reaction table:

◆ determine mol of each reactant

100 g CaC2 " ––––––––––– = 1.56 mol CaC2

100 g H2O " ––––––––––– = 5.55 mol H2O

1 mol CaC2

64.01 g

1 mol H2O18.02 g

Calculations to fill in the reaction table:

◆ pick one reactant; calculate mol of 2nd reactant required to react completely with the 1st reactant

1.56 mol CaC2 " ––––––––––– = 3.12 mol H2O

or conversely

5.55 mol H2O " ––––––––––– = 2.78 mol CaC2

CaC2 (s) + 2 H2O (l) " Ca(OH)2 (s) + C2H2 (g)

2 mol H2O1 mol CaC2

1 mol CaC2

2 mol H2O

Calculations to fill in the reaction table:

◆ compare mol of 2nd reactant required for complete reaction to mol present in the system

here - more H2O is present (5.55 mol) than is required (3.12 mol) to completely consume CaC2

∴ H2O is the excess reactant

OR

there is less CaC2 present (1.56 mol) than is required (2.78 mol) to completely consume H2O present

∴ CaC2 is the limiting reactant

CaC2 (s) + 2 H2O (l) " Ca(OH)2 (s) + C2H2 (g) CaC2 (s) + 2 H2O (l) " Ca(OH)2 (s) + C2H2 (g)

before 1.56 mol 5.55 mol 0 0

change !1.56 mol !3.12 mol +1.56 mol +1.56 mol

after 0 2.43 mol 1.56 mol 1.56 mol

CaC2 (s) + 2 H2O (l) " Ca(OH)2 (s) + C2H2 (g)

before 1.56 mol 5.55 mol 0 0

change !1.56 mol !(2∕1)1.56 mol +(1∕1)1.56 mol +(1∕1)1.56 mol

after 0 2.43 mol 1.56 mol 1.56 mol

CaC2 (s) + 2 H2O (l) " Ca(OH)2 (s) + C2H2 (g)

Calculations based on the reaction table:

◆ determine the theoretical yield of each product

1.56 mol Ca(OH)2 " –––––––––––– = 116 g Ca(OH)2

1.56 mol C2H2 " ––––––––––– = 40.6 g C2H2

◆ determine the mass of excess reactant unconsumed when the reaction is complete

2.43 mol H2O " ––––––––––– = 43.8 g H2O

74.10 g1 mol Ca(OH)2

26.04 g1 mol C2H2

18.02 g1 mol H2O

Homework Problem: 3.95

3 TiO2 (s) + 4 C (s) + 6 Cl2 (g) ! 3 TiCl4 (g) + 2 CO2 (g) + 2 CO (g)

3 TiO2 + 4 C + 6 Cl2 ! 3 TiCl4 + 2 CO2 + 2 CO

before 0.0519 mol 0.472 mol 0.0956 mol 0 0 0

change

after

4.15 g 5.67 g 6.78 g79.88 g/mol 12.01 g/mol 70.90 g/mol

A Final Reaction Stoichiometry Example:Gravimetric Analysis

A 27.2569 g sample containing some unknown amount of potassium superoxide (KO2) and other substances is treated with excess H2O; the following reaction occurs*:

4 KO2 (s) + 2 H2O (l) " 4 KOH (s) + 3 O2 (g)

*note: of the substances present in the original sample, only the KO2 will react with water

Assuming that the you achieve complete conversion of KO2 to KOH, determine the mass percent KO2 in the original sample if the experiment yields13.6247 g KOH.