Chemical Stoichiometry

Chemical Reaction Equations
A chemical equation is a shorthand description of a chemical reaction using symbols and formulas to represent the elements and compounds involved. It is a symbolic representation of a chemical reaction in terms of chemical formulas.
        Reactants -----> Products            or            Reactants = Products
        The symbols "------->" or "=", mean forms, yields or produces.

The reactants are the starting substances and the products are the substances formed.

Using the following abbreviations indicates the phases of the reacting participants.
(g) = gas;    (l) = liquid;    (s) = solid;    (aq) = aqueous

2Na(s) + 2H2O(l) ----> 2NaOH(aq) + H2(g)

The stoichometric coefficents (coefficients) are 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.

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.

The reaction of solid sodium metal with gaseous chlorine is indicated as follows:
    2 Na(s) + Cl2(g) ----> 2 NaCl(s)

Multiplying the equation by a factor gives the relative amounts at any scale.
    2[ 2 Na(s) + Cl2(g) ---> 2 NaCl(s) ] = 4 Na(s) + 2 Cl2(g) ---> 4 NaCl(s)

This balanced reaction can be interpreted on at least three scales or levels:
1. On the smallest scale possible (nanoscale level or atomic level), 2 atoms of sodium react with 1 molecule of chlorine to produce 2 formula units of sodium chloride.
2. On a molar scale, 2 moles of sodium atoms react with 1 mole of chlorine molecules to produce 2 moles of sodium chloride.
3. On a mass scale (practical scale) 45.98 g of sodium react with 70.90 g of chlorine to produce 116.88 g o sodium chloride.

Stoichiometry is the relationship that describes the relation between the masses of reactants and products.
Stoichiometric coefficients are the coefficients of the balanced equation.

Patterns of Chemical Reactions

In a combination reaction, two substances combine to form a third substance. {A + B ----> AB}
    2Na(s) + Cl2(g) ---> 2 NaCl(s)
    CaO(s) + SO2(g) ---> CaSO3(s)

In a decomposition reaction, a single compound reacts to give two or more substances. {AB ----> A + B}
    2 HgO(s) -----> 2 Hg(l) + O2(g)
    2 KClO3(s) -----> 2 KCl(s) + 3 O2(g)

In a displacement reaction (or single replacement reaction), an element reacts with a compound and takes the place of (displaces) one of the elements in the original compound. {A + BC -----> AB + C}
    Fe(s) + 2 HCl(aq) ---> FeCl2(aq) + H2(g)
    Cu(s) + 2 AgNO3(aq) ---> Cu(NO3)2(aq) + 2 Ag(s)

An exchange (metathesis reaction or double-replacement reaction), is a reaction between compounds that, when written as a molecular equation, appears to involve the exchange of parts between the two reactants. Such reactions often occur between ions in solution that can result in the formation of insoluble substances, weak electrolytes, or nonelectrolytes.
    BaCl2(aq) + ZnSO4(aq) ---> BaSO4(s) + ZnCl2(aq)
    Na2S(aq) + ZnCl2(aq) ---> NaCl(aq) + ZnS(s)

One other general reaction type is the combustion reaction.
A combustion is a reaction in which a substance reacts with oxygen, usually with the rapid release of heat to produce a flame.
    2 C4H10(g) + 13 O2(g) ---> 8 CO2(g) + 10 H2O(g)

Balancing Chemical Equations
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

Step 1. Balance any element that appears in only one compound on each side of the arrow. Carbon and hydrogen will be balanced in this step. There are two C atoms on the left, so a 2 is placed in front of CO2. There are eight hydrogen atoms on the left, which require 4 H2O molecules on the right:

C2H8N2 + N2O4 ----> N2 + 4 H2O + 2 CO2

Step 2. By inspection, balance any element that appears in more than one compound on either the right or the left. Only oxygen is balanced here, because nitrogen is balanced in step 3. There are four oxygen atoms on the left and eight on the right, so we place a 2 in front of N2O4 to get eight oxygen atoms on the left:

C2H8N2 + 2 N2O4 ----> N2 + 4 H2O + 2 CO2

Step 3. Balance any element that appears in elemental form on the right or left. Here, we balance the nitrogen atoms by placing a 3 in front of N2 on the right, so that there are six nitrogen atoms on each side:

C2H8N2 + 2 N2O4 ----> 3 N2 + 4 H2O + 2 CO2

Example
What is the sum of the coefficients when the following equation is balanced?

Fe2(SO4)3 + BaCl2 ----> BaSO4 + FeCl3
[Answer: The sum of the coefficients is 9.]

The Mole and Chemical Reactions
A balanced equation can be used to derive molar ratios of the participants in the reaction.
The balanced equation for the combustion of the hydrocarbon butane, C4H10 is as follows.
        2 C4H10(g) + 13 O2(g) ---> 8 CO2(g) + 10 H2O(g)

Using this equation, the following relationships can be derived.
13 mol O2/2 mol C4H10     or         8 mol CO2/2 mol C4H10 or        10 mol H2O/13 mol O2

These ratios can be used to calculate the amounts of one participant from an amount of another participant.

Example
How many moles of oxygen are required to react with 0.50 mol of butane?
        2 C4H10(g) + 13 O2(g) ---> 8 CO2(g) + 10 H2O(g)
        (0.50 mol C4H10)(13 mol O2/2 mol C4H10) = 3.25 mol O2

Example
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?
        (6.23 mol Li)(1 mol H2/2 mol Li ) = 3.12 mol H2

Using molar masses and molar ratios, the mass of one participant can be used to calculate the mass of any other participant in the chemical equation.
Step 1.  Convert grams of one substance to moles.
Step 2.  Using the appropriate molar ratio to convert the moles of that reactant or product to moles of another participant.
Step 3.  Convert the moles of the second participant back into grams.

Example
How many grams of H2 can be formed by the complete reaction of 80.57 g of Li with water?
    2Li(s) + 2H2O(l) ---> 2LiOH(aq) + H2(g)

        Step 1.    (80.57 g Li)(1 mol Li/6.941 g Li) = 11.6078 mol Li
        Step 2.    (11.6078 mol Li)(1 mol H2/2 mol Li) = 5.8039 mol H2
        Step 3.    (5.8039 mol H2)(2.02 g H2 /1 mol H2) = 11.72 g H2

Example
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?

        Step 1.   (856g C6H12O6)(1 mol C6H12O6/180.2g C6H12O6) = 4.7503 mol C6H12O6
        Step 2.   (4.7503 mol C6H12O6)(6 mol CO2/1 molC6H12O6) = 28.502 mol CO2
        Step 3.   (28.502 mol CO2)(44.01 g CO2/1 mol CO2) = 1.25 x 103 g CO2

Example
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)
What mass of ammonia is need to react with 100 g of K2PtCl4?

        Step 1.    (100 g K2PtCl4)(1 mol K2PtCl4/415.08 g K2PtCl4) = 0.2409 mol K2PtCl4
        Step 2.    (0.2409 mol K2PtCl4)(2 mol NH3/1 mol K2PtCl4) = 0.4818 mol NH3
        Step 3.    (0.4818 mol NH3)(17.04 g NH3/1 mol NH3) = 8.210 g NH3

Limiting Reactants
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 means 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.

The following steps can be used to calculate the mass produced of a given substance when one or more of the reactants is present in a limiting mass.

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

Step 2. The reactant that gives the least amount of product is the limiting reactant. 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

        (4.0 mol H2)(2 mol H2O/2 mol H2) = 4.0 mol H2O
        (1.0 mol O2)(2 mol H2O/1 mol O2) = 2.0 mol H2O

        Because only 2.0 mol of H2O can be formed from this mixture

Limiting Reagent Sample Problem #1


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


(2.5 moles BaSO4)

H2SO4 + BaCl2 -----> BaSO4 + 2 HCl



Limiting Reagent Sample Problem #2


A mixture of 35.0 g of hydrogen and 270 grams of oxygen react to form water. How many grams of water will form? (304 g H2O)


2 H2 + O2 ----> 2 H2O


Limiting Reagent Sample Problem #3


How many grams of silicon will form when 15.0 grams of boron are reacted with 50.0 grams of silicon dioxide according to the following equation? (29.2 g Si)



4 B + 3 SiO2 -----> 3 Si + 2 B2O3
Percent Yield
Theoretical yield is the calculated maximum amount of product that can be obtained from a given amount of reactant, according to the chemical equation.

Actual yield is the measured amount of product that we finally obtain.

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

If the theoretical yield calculated for a reaction is 14.8 g, and the amount of product obtained is 9.25 g,
the percent yield is:   (9.25 g / 14.8) x 100 = 62.5%.

Percent Yield Sample Problem #1
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. [89.9%]

CS2 + 3 Cl2 ---> CCl4 + S2Cl2




Percent Yield Sample Problem #2
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. [91.9%]

MgBr2 + 2 AgNO3 ---> Mg(NO3)3 + 2 AgBr



Percent Yield Sample Problem #3
When 100 g of oxalic acid (H2C2O4) and 100 g of methanol (CH3OH) were allowed to react, 38 g of methyl oxalate (C4H6O4) were isolated. What is the percent yield? [29%]

H2C2O4 + 2 CH3OH <---> C4H6O4 + 2 H2O
Percent Composition and Empirical Formula
In a combustion analysis of a compound containing carbon and hydrogen, the compound reacts with with oxygen and all of the carbon in the compound is converted to carbon dioxide and the hydrogen in the compound is converted to water.
    2 C4H10(g) + 13 O2(g) ---> 8 CO2(g) + 10 H2O(g) + Heat

The law of conservation of mass and the concept of following the mass of elements from one single compound to several compounds provides a basis for determining percent composition, empirical formula and molecular formula.

Combustion Analysis Sample Problem #1

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 percent composition of benzene?  [92.25% C & 7.75% H]

If a compound contains oxygen as well as carbon and hydrogen, the mass of the oxygen can be determined by subtracting the masses of the carbon and the hydrogen from the total mass of the compound.

mass of O = mass of compound – [mass of C + mass of H]

Combustion Analysis Sample Problem #2

n-Butyl phthalate is used as an insect repellant and is composed of carbon, hydrogen, and oxygen. When a 0.3413 g sample was subjected to combustion analysis, 0.2430 g of water and 0.8633 g of carbon dioxide were produced. In another analysis, the molecular weight was determined to be 278.38 g/mol. Calculate the empirical formula and the molecular formula.

[empirical formula = C8H11O2 & molecular formula = C16H22O4]

Combustion Analysis Sample Problem #3

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.

[40.909% C, 4.587% H, 54.504% O & empirical formula = C3H4O3]

In another analysis the molecular mass of vitamin C was found to be 176.14. What is the molecular formula of vitamin C? [C6H8O6]

Combustion Analysis Sample Problem #4

A 1.500 g sample of hydrocarbon undergoes complete combustion to produce 4.400 g of CO2 and 2.700 g of H2O. What is the empirical formula of this compound? [CH3]

Combustion Analysis Sample Problem #5

A 0.250 g sample of hydrocarbon undergoes complete combustion to produce 0.845 g of CO2 and 0.173 g of H2O. What is the empirical formula of this compound? [CH]

Combustion Analysis Sample Problem #6

A 0.1034 g sample of a compound composed of carbon, hydrogen and oxygen is subjected to combustion analysis, producing 0.2351g CO2 and 0.0962 g H2O. What is the empirical formula for the compound? [C3H6O]

If the molecular weight is 116.18 g/mol, what is the molecular formula?

[C6H12O2]