HILL3

STOICHIOMETRY & CHEMICAL CALCULATIONS

MOLECULAR MASS AND FORMULA MASS
Molecular mass
- the sum of the atomic weights of the atoms in a molecule of the substance.
- sometimes referred to as molecular weight (MW)

Formula mass
- the sum of the atomic weights of the atoms in a formula unit of the compound.
- also referred to as formula weight (FW)

EXAMPLE
Calculate the molar mass for each of the following compounds.

H₃PO₄

        3 H @ 1.01 =     3.03
        1 P @ 30.97 = 30.97
        4 O @ 16.00 = 64.00
                                 98.00 u

K₂CO₃

        2 K @ 39.10 = 78.20
        1 C @ 12.01 = 12.01
        3 O @ 16.00 = 48.00
                               138.21 u

Al₂(SO₄)₃
    2 Al @ 26.98 =      53.96
    3 S @ 32.06   =      96.18
    12 O @ 16.00 =   192.00
                                 342.14 u

THE MOLE AND AVOGADRO'S NUMBER
Mole (mol)
- the quantity of substance in a sample that contains as many elementary entities as the
number of atoms in exactly 12 g of carbon-12.

- The elementary entities may be atoms, molecules, monatomic or polyatomic ions,
electrons, even formula units of salts such as NaCl.

- the mass of one mole in grams is numerically equal to the formula mass. The mole is
so defined that a sample of an element with a mass equal to its atomic mass in grams.

- the number of units is 6.022137 x 10²³ and is called Avogadro's number, N_A.

Molar mass
    -the mass in g of one mole of the substance.
        1 mol of Na = 22.99 g Na = 6.022 x 10²³ Na atoms
        1 mol of CO₂ = 44.01 g CO₂ = 6.022 x 10²³ CO₂ molecules
        1 mol of MgCl₂ = 95.21 g MgCl₂ = 6.022 x 10²³ MgCl₂ formula units

EXAMPLE
How many moles of Na are there in a 2.770 gram sample?
moles of Na = 2.770 grams/23.00 g per mol = 0.1204 moles of Na

EXAMPLE
How many moles of water are contained in a 355.0 gram sample?
moles of H₂O = grams of H₂O/ molecular mass
= 355.0 grams/18.02 g per mol = 19.70 moles of water.

EXAMPLE
How many grams of Ca(OH)₂ are required to provide 4.0102 x 10^-1 moles of Ca(OH)₂?
g Ca(OH)₂= [4.0102 x 10^-1 mol Ca(OH)₂][74.10 g/mol] = 29.716 grams Ca(OH)₂

MASS PERCENT COMPOSITION

Mass percent describes the proportions of the constituent
elements in a compound as the number of grams of each element per 100 g of the compound.

Percent composition = mass of atom A x 100
total mass of sample

EXAMPLE
What is the percent composition of sulfuric acid?
H₂SO₄ [MW = 98.08 g/mol]

%H = [2.02/98.08][100] = 2.06%

%S = [32.06/98.08][100] = 32.69%

%O = [64.00/98.08][100] = 65.25%

EXAMPLE
How many grams of oxygen are there in 36.45 g of hydrogen peroxide?
Hydrogen peroxide is 94.06% oxygen.

(36.45 g H₂O₂)(94.06 g O/100 g of H₂O₂) = 34.30 g O

EXAMPLE
What mass of oxygen is contained in a gallon of water?

%O = (16.00/18.02)(100) = 88.79%

(1 gallon)(3.7854 L)(1000 mL/1 L)(1.00 g/mL) = 3785.4 g

(3785.4 g H₂O)(88.79 g O/100 g H₂O)= 3361.06 g O

EMPIRICAL FORMULA
- the formula of a substance written with the smallest integer (whole number) subscripts

To convert percent compostion to an empirical formula:
    Step 1 Convert the percent of each element to a mass in grams.
    Step 2 Convert the mass of each element to an amount in moles.
    Step 3 Use the number of moles of the elements as subscripts in a tentative formula.
    Step 4 Attempt to get integers as subscripts by dividing each of the subscripts by
               the smallest subscript.
    Step 5 If any subscripts obtained after Step 4 are fractional quantities, multiply each
              of the subscripts by the smallest integer that will convert all the subscripts to integers.
              The result is an empirical formula.

EXAMPLE
What is the empirical formula of benzene? The elemental analysis of benzene is
92.2%C and 7.8%H.

STEP 1 92.2%C and 7.8%H can be thought of as 92.2 g C and 7.8 g H

STEP 2 (92.2g C)(1 mol C/12.0 g C) = 7.68 mol C
(7.8g H)(1 mol H/1.01 g ) =7.73 mol H

STEP 3 C_7.68H_27.73

STEP 4 3. 7.68/7.68 = 1 mol C
7.73/7.68 = 1.01 = 1 mol H

The empirical formula is: CH

EXAMPLE
A compound contains only nitrogen and oxygen. It is 30.4% N by mass.
Calculate the empirical formula.

    STEP 1    30.4% N means the the compound must contain (100 - 30.4 = 69.6% O).
                    These two values can be converted to mass by assuming there are 100 g
                    of the compound.

STEP 2 (30.4 g N)(1 mol N/14.01 g N) = 2.17 mol N
(69.6 g O)(1 mol O/16.00 g O) = 4.35 mol O

STEP 3 N_2.17O_4.35

STEP 4 2.17/2.17 = 1 mol N
4.35/2.17 = 2 mol O

The empirical formula is: NO₂

RELATING MOLECULAR FORMULAS TO EMPIRICAL FORMULAS

MW = n x empirical formula mass

n = number of empirical formula units in the molecular formula.

EXAMPLE
What is the molecular formula of benzene if the molecular weight is 78.12 and
the empirical formula is CH?
78.12/13.02 = 6 Therefore: C₆H₆

EXAMPLE
A compound with an empirical formula of NO₂ has a MW of 92.02.
What is the molecular formula?
92.02/46.01 = 2 Therefore: N₂O₄

EXAMPLE
Nicotine is 74.03% C, 8.70% H, and 17.27% N. If the molecular weight of nicotine
is found to be 162.26, what is the molecular formula?
    74.03 g of C = 6.164 mol C         6.164/1.233 = 4.9995
    8.70 g of H = 8.61 mol H             8.61/1.233 = 6.983 7
    17.27 g of N = 1.233 mol N        1.233/1.233 = 1

empirical formula = C₅H₇N₁ with EW = 81.13

162.26/81.13 = 2 Molecular formula C₁₀H₁₄N₂

Elemental Analysis: Experimental Determination of Mass Percent Composition
When a compound containing carbon and hydrogen is combusted, all of the carbon in the
compound is converted into carbon dioxide and all of the hydrogen is converted into water.

EXAMPLE
Benzene is a liquid compound composed of carbon and hydrogen. A sample of benzene
weighing 342 mg is burned in excess oxygen and forms 1156 mg of carbon dioxide.
What is the percentage composition of benzene?
C_XH_Y + O₂ ---> CO₂ + H₂O

(1.156gCO₂)(1molCO₂/44.0gCO₂)(1molC/1mol CO₂)(12.0gC/1mol C) = 0.3153g C

%C = (0.3153/0.342)(100) = 92.2 %C
%H = 100 - 92.2 = 7.8% H

EXAMPLE
When a 1.0000 g sample of a vitamin C was combusted, 1.4991 g of CO₂ and 0.4092 g
of H₂O were isolated. Calculate the percent composition and empirical formula of vitamin C.

%C
(1.4991 g CO₂)(1 mol CO₂/44.01 g CO₂)(1 mol C/1 mol CO₂)(12.01 g C/ 1 mol C) = 0.40909 g C

(0.40909 g C/1.0000 g Vitamin C)(100) = 40.909% C

%H
(0.4092 H₂O)( 1 mol H₂O/18.02 g H₂O)(2 mol H/1 mol H₂O)(1.01 g H/ 1 mol H) = 0.04587 g H

(0.04587 g H/1.0000 g Vitamin C)(100) = 4.587% H

%O = 100 - 40.909 - 4.587 = 54.504%

What is the empirical formula?
(40.909 g C)(1 mol C/12.01 g C) = 3.4062 mol C

(4.587 g H)( 1 mol H/ 1.01 g H) = 4.542 mol H

(54.504 g O)( 1 mol O/ 16 g O) = 3.4065 mol O

C_3.4062H_4.542O_3.4065

    3.4062/3.4062 = 1 mol C
   4.542/3.4062 = 1.333 mol H
   3.4065/3.4062 = 1.00008 mol O

C₁H_1.333O₁ = C₃H₄O₃

In another analysis the molecular mass of vitamin C was found to be 176.14.
What is the molecular formula of vitamin C?

molecular weight/empirical weight = 176.14/88.07 = 2

C₆H₈O₆

BALANCING CHEMICAL EQUATIONS

Chemical equation: a shorthand description of a chemical reaction using symbols
and formulas to represent the elements and compounds involved.

Reactants the starting substances

Products the substances formed

Stoichometric coefficents the numbers placed in front of formulas in a chemical equation to
balance the equation, thereby indicating the combining ratios of the reactants and the products.

2 Na + Cl₂ ----> 2 NaCl

Balanced equations obey the law of conservation of mass. That is the total mass before
a reaction takes place will equal the total mass after the reaction is complete. This also
holds true for number of atoms involved in the reaction. The number of atoms which take
part in a chemical reaction will remain constant regardless of whether the reaction goes to
completion or not.

Balancing an equation can sometimes involve trial and error. The following rules and
suggestions can go a long way in keeping down the errors.

1. The equation will be understood to proceed from left to right. The reactants are
on the left side of the arrow and the products are on the right.

2 H₂ + O₂ ----> 2 H₂O
Reactants -----> Products

2. Equations are balanced by adjusting COEFFICIENTS in front of formulas, never
by changing subscripts within formulas. Remember that a 1 is understood when a
coefficient is not present.

3. It is best to start with an element that appears in only one compound on each
side of the arrow.

4. Next balance any element that appears in more than one compound on either
the right or left.

5. Balance free elements last. That is balance any element that appears in elemental
form on the right or left.

6. When polyatomic species (NH₄⁺, SO₄^-2, OH^-, etc.) appear on both sides of the
arrow in compounds or as ions, balance them as units rather than individual elements.

7. Leave the coefficients in the lowest whole number ratio.

EXAMPLE
Balance the following.
C₂H₈N₂ + N₂O₄ ----> N₂ + H₂O + CO₂

[Answer: C₂H₈N₂ + 2 N₂O₄ ----> 3 N₂ + 4 H₂O + 2 CO₂]

EXAMPLE
Balance the following equation. (Note: balance the SO₄^-2 as a unit rather than individual atoms.)
Fe₂(SO₄)₃ + BaCl₂ ----> BaSO₄ + FeCl₃

[Answer: Fe₂(SO₄)₃ + 3 BaCl₂ ----> 3 BaSO₄ + 2 FeCl₃]

EXAMPLE
What is the sum of the coeffiecients when the following equations are balanced?
a. C₂H₆ + O₂ ------> CO₂ + H₂O [Answer: 17]

b. C₄H₁₀ + O₂ -----> CO₂ + H₂O [Answer: 33]

STOICHIOMETRY

EXAMPLE 1
All alkali metals react with water to produce hydrogen gas and the corresponding alkali
metal hydroxide. A typical reaction is that between lithium and water:

2Li(s) + 2H₂O(l) ---> 2LiOH(aq) + H₂(g)

How many moles of H₂ can be formed by the complete reaction of 6.23 moles of
Li with water?

mol H₂ = 6.23 mol Li x 1 mol H₂ = 3.12 mol H₂
2 mol Li

EXAMPLE 2
How many grams of H₂ can be formed by the complete reaction of 80.57 g of Li with water?

mol Li = 80.57 g Li x 1 mol Li = 11.61 mol Li
6.941 g Li

mol H₂ = 11.61 mol Li x 1 mol H₂ = 5.805 mol H₂
2 mol Li

g H₂ = 5.805 mol H₂ x 2.016 g H₂ = 11.70 g H₂
1 mol H₂

EXAMPLE 3
The food we eat is degraded, or broken down, in our bodies to provide energy for
growth and function. A general overall equation for this very complex process represents
the degradation of glucose (C₆H₁₂O₆) to carbon dioxide (CO₂) and water (H₂O):

C₆H₁₂O₆ + 6O₂ ---> 6CO₂ + 6H₂O

If 856 g of C₆H₁₂O₆ is consumed by the body over a certain period, what is the
mass of CO₂ produced?

mol of C₆H₁₂O₆ = 856 g C₆H₁₂O₆ x 1 mol C₆H₁₂O₆ = 4.75 mol C₆H₁₂O₆
180.2 g C₆H₁₂O₆

mol of CO₂ produced = 4.75 mol C₆H₁₂O₆ x 6 mol CO₂ = 28.5 mol CO₂
1 molC₆H₁₂O₆

mass of CO₂ produced = 28.5 mol CO₂ x 44.01 g CO₂ = 1.25 x 10³ g CO₂
1 mol CO₂

EXAMPLE 3
The compound cisplatin (Pt(NH₃)₂Cl₂) has been used as an antitumor agent. It is prepared by the reaction between potassium tetrachloroplatinate (K₂PtCl₄) and ammonia (NH₃):

K₂PtCl₄(aq) + 2NH₃(aq) Pt(NH₃)₂Cl₂(s) + 2KCl(aq)

How many grams of cisplatin can be obtained starting with 0.8862 g of K₂PtCl₄? You may assume that there is enough of NH₃ to react with all of the K₂PtCl₄.

0.8862 g K₂PtCl₄ x 1 mol K₂PtCl₄ x 1 mol Pt(NH₃)₂Cl₂ x 300.1 g Pt(NH₃)₂Cl₂
415.1gK₂PtCl₄ 1 mol K₂PtCl₄ 1 mol Pt(NH₃)₂Cl₂

= 0.6407 g Pt(NH₃)₂Cl₂

LIMITING REACTANTS (LIMITING REAGENTS)
A balanced equation shows the RELATIVE proportions between reactants and between
products. This relation is fixed but the actual amounts of reactants present can vary. This
can mean that sometimes there can be an excess of one or more of the reactants. In other
words the quantity of one reactant will control the amount of products that will be formed.

1. Calculate the amount of product (moles or grams, as needed) that can be
formed from each reactant.

    2.     Determine which reactant is limiting.(The reactant that gives the least amount
            of product is the limiting reactant and the other reactant is in excess. The limiting
            reactant will determine the amount of product formed in the reaction.)

EXAMPLE
How many moles of H₂O will be formed when 4.0 moles of H₂ are allowed to react
with 1.0 mole of O₂ according to the reaction:

2 H₂ + O₂ ----> 2 H₂O

[Answer: 2 mol H₂O]

EXAMPLE
How many moles of BaSO₄ will be produced from a mixture of 3.5 moles H₂SO₄
and 2.5 moles BaCl₂ using the reaction:

H₂SO₄ + BaCl₂ -----> BaSO₄ + 2 HCl

[Answer: 2.5 moles BaSO₄]

EXAMPLE
A mixture of 35.0 g of hydrogen and 270 grams of oxygen react to form water.
How many grams of water will form?

2 H₂ + O₂ ----> 2 H₂O

[Answer: 304 g H₂O]

EXAMPLE
How many grams of SiO₂ will form when 10.0 grams of silicon are mixed with
11.0 grams of oxygen gas?

Si + O₂ ----> SiO₂

[Answer: 20.7 g SiO₂]

PERCENT YIELD

Theoretical yield calculated maximum amount of product that can be obtained
from a given amount of reactant, according to the chemical equation.

Actual yield the measured amount of product that is obtained when the reaction
is carried out.

Actual is always less than theoretical.

Percent yield the ratio of the actual yield to the theoretical yield multiplied by 100.

actual yield x 100 = percent yield
theoretical yield

EXAMPLE
If the theoretical yield calculated for a reaction is 14.8 g, and the amount of product
obtained is 9.25 g, calculate the percent yield.

percent yield = 9.25 g x 100 = 62.5%
14.8 g

EXAMPLE
Carbon tetrachloride was prepared by reacting 100. g of carbon disulfide and 100. g
of chlorine. Calculate the percent yield if 65.0 g of CCl₄ was obtained from the reaction.

CS₂ + 3 Cl₂ ---> CCl₄ + S₂Cl₂

100. g CS₂ x 1 mol CS₂ x 1 mol CCl₄ x 154 g CCl₄ = 202 g CCl₄
76.2 g CS₂ 1 mol CS₂ 1 mol CCl₄

100. g Cl₂ x 1 mol Cl₂ x 1 mol CCl₄ x 154 g CCl₄ = 72.3 g CCl₄
71.0 g Cl₂ 3 mol Cl₂ 1 mol CCl₄

Percent Yield = 65.0 g x 100 = 89.9%
72.3 g

EXAMPLE
Silver bromide was prepared by reacting 200.0 g of magnesium bromide and an adequate amount of silver nitrate. Calculate the percent yield if 375.0 g of silver bromide was obtained from the reaction.

MgBr₂ + 2 AgNO₃ Mg(NO₃)₃ + 2 AgBr

200.0 g MgBr₂ x 1 mol Mgbr₂ x 2 mol AgBr x 187.8 g AgBr = 408.0 g AgBr
184.1 g MgBr₂ 1 mol MgBr₂ 1 mol AgBr

The theoretical yield is 408.0 g AgBr.

Percent yield = 375.0 g AgBr x 100 = 91.9%
408.0 g AgBr

MOLARITY

Molarity (M) = mole of solute per liter of solution = mol solute/L solution

EXAMPLE
If 400.0 mL of a solution contains 5.00 x 10^-3 moles of AgNO₃, what is the molarity of this solution?

M = 5.00 x 10^-3 mol AgNO₃ / 0.4000 L = 0.0125 M AgNO₃

EXAMPLE
Calculate the molarity of an HCl solution which contains 18.23 g of HCl in 355.0 mL of solution.
A. Calculate the number of moles of HCl.
(18.23 g HCl) (1 mol HCl/36.46 g HCl) = 0.5000 mol HCl

B. Divide the number of moles of HCl by the total volume in liters.
0.5000 mol HCl/0.3550 L solution = 1.408 M HCl

EXAMPLE
    How many moles of NaCl are contained in a 27.49 mL sample of a
     0.350 M solution of NaCl?
    (0.350 mol/L)(0.02749 L) = 9.62 x 10^-3 mol NaCl

Dilution
When a solution is diluted the concentration will be less. This can be stated mathematically
by the following formula.
M₁V₁ = M₂V₂

    M₁ = The original molarity. It will be the concentrated solution which will be diluted.
    V₁ =   Is the volume of concetrated solution which will be diluted.
    M₂ = the final molarity.
    V₂ =   the final volume.

EXAMPLE
What volume of 18.0 M H₂SO₄ must is required to make 100 mL of a 5.0 M solution
of H₂SO₄? [Answer: 28 mL]

EXAMPLE
How many milliliters of water are required to dilute 6 M H₂SO₄ to 175 mL
of 0.3 M H₂SO₄? [Answer: 166 mL]

SOLUTION STOICHIOMETRY

EXAMPLE
How many mL of 6.0 M HCl are needed to react with 3.78 g of Mg?

Mg(s) + 2 HCl(aq) ---> MgCl₂(aq) + H₂(g)
[Answer: 52 mL]

EXAMPLE
How many mL of 0.0487 M Ba(OH)₂ are needed to react with 35.67 mL of 0.0748 M HCl?
[Answer: 27.3 mL Ba(OH)₂]