MASS RELATIONSHIP IN CHEMICAL REACTIONS

MASS RELATIONSHIP IN CHEMICAL REACTIONS

MOLAR MASS AND MOLE CONCEPT

            Mole refers to the number of individual particles present in any chemical species. The mole concept was introduced using Carbon-12 as standard. Scientist found out that the number of particles in a substance is equal to that present in 12grams of Carbon-12. This number is known as Avogadro’s number. The value of Avogadro’s number has been determined to be 6.022 x 10²³.

            It has been mentioned that the term formula weight is used for atoms, molecules or ions that exist as molecules or compound. The formula of a compound gives the relative number of atoms in a formula unit. Therefore, 1 mole of formula unit contains 6.022 x 10²³ of these units, whether they are atoms, molecules or ions. The mass of 6.022 x 10²³ formula units whether atoms, molecules or ions is the formula weight expressed in grams. The atomic/molecular mass in amu is numerically equal to the molar mass in grams (molar mass – mass of 1 mole of substance).

            Mole concept also gives us another way of expressing atomic, molecular or formula weights as gram per mole. For example, the formula weight of hydrogen atom
(H) is 1.0g, but the formula weight of hydrogen molecule (H₂) is 2.0g. Therefore:

                        1 mole of H atom = 1.0g

                        1 mole of H₂ molecule = 2.0g

Another example to illustrate mole concept is as follows:

            The formula weight of water, H₂O is 18 amu. We can say also that 1 mole of water (that contains 6.022 x 10²³ molecules of water) has a mass of 18 grams.

            Let us now consider several types of calculations that can be done using the mole concept. Here the mole represents the amount of substances. This can be expressed as number of particles whether it is an atom, molecule or ion.

Illustrative Example 1

Calculate the number of moles of 100 grams of KMnO₄.

Solution:

            Determine the molar mass of KMnO₄

 

Solve for the number of moles.

  

Practice Exercises

1.      Calculate the number of moles in each of the following:

a.      47.8 grams of C₂H₆O

b.      95.0 grams of phosphoric acid

c.      20.7 grams of KCl

2.      Determine the mass in grams of each of the following:

a.      6.5 mole of carbonic acid

b.      0.32 mole of CO₂

c.      1.5 x 10²⁰ molecules of H₂

3.      Calculate the number of atoms/ molecules in each of the following

a.      0.025 grams of S

b.      0.040 mole of NaOH

PERCENTAGE COMPOSITION OF COMPOUNDS

We have learned that a compound consists of two or more substances. Here, we will calculate the amount of substances that compose a compound. We are going to express this in percentage. Percent is defined as part per hundred, or fraction of one hundred. To determine the percentage composition of a compound first, calculate the relative formula weight then divide the relative atomic mass of each element by the formula weight and multiply by 100.

Consider the following examples:

 

Practice Exercises:

1. Calculate the percent composition of each of the following:

a.      Na₃PO₄

b.      Ca(HCO₃)₂

c.      Al(NO₃)₃

d.      NH₄NO₃

    2.      Calculate the percent of sulfur in each of the following:

a.      Iron (III) sulfide

b.      Sulfur dioxide

c.      Sulfurous acid

d.      Barium sulfide

     3.  A magnesium ribbon weighing 1.25 grams formed 3.45 grams of oxide when made to react with oxygen. What is the percent of oxygen in this compound?

EMPIRICAL AND MOLECULAR FORMULAS

            Empirical formula gives the simplest ratio of atoms that make up the compound. It can be derived from chemical analysis or from the percentage composition. The molecular formula shows the actual number of atoms of each element present in one molecule of the compound. To illustrate how the empirical formula can be determined from percentage composition, consider the case of calcium carbonate. Analysis shows that this compound has the following composition: Ca = 40%, C = 12%, and O = 48%. Here you are going to convert the given composition to the relative number of grams. Therefore, the given composition would mean that out of 10 grams of calcium carbonate, 40g is Ca, 12g is C, and 48g is O. then using the relative atomic weight of each element, these figures can be converted to moles of atoms. Take note that: 1 mole of Ca atom = 40g, 1 mole of C atom = 12g, 1 mole of O atom = 16g. You can find here the simplest ratio of the number of atoms of calcium, carbon and oxygen. Therefore you have,

 

Illustrative Problem 1. Determining Empirical Formula

Calculate the empirical formula for a compound that contains 27.48% magnesium, 23.67% phosphorus and 48.85% oxygen.

Solution:

Basis: 100 g sample

Therefore the sample contains:                     27.48g Mg

                                                                           23.67g P

                                                                           48.85g O

To find the ratio of atoms in the compound

Again the quotient for Mg is not a whole-number. Here, you have to multiply each value by the smallest possible number that will give a whole number. Hence, the number is 2. Therefore,

Mg = 1.5 x 2           = 3

P    = 1.0 x 2           = 2

O   = 4.0 x 2           = 8

The empirical formula is Mg₃P₂O₈ or Mg₃ (PO₄)₂

Illustrative Problem 2. Determining Molecular Formula

          To determine the molecular formula from percentage composition, the molecular weight must be given also.

          A compound upon analysis is found to contain 60% C, 13.3% H and 26.7% O. the molecular weight of the compound is 120. Determine the molecular formula of the compound.

Practice Exercises:

1.      Calculate the empirical formula of each of the following:

a.      A compound that contains 38.71% Ca, 20% P and 41.28% O.

b.      A compound that contains 55.85g Fe combined with 32.07g S.

c.      A compound that contains 16.2g Na, 11.26g S and 22.53g O.

d.      A compound that contains 17.98mg Al and 31.968mg of S.

2.      Calculate the molecular formula of each of the following:

a.      Empirical formula = CH

Molecular weight = 26

b.      Empirical formula = CH₂

Molecular weight = 56

c.      A compound that contains 30.44% nitrogen and 69.56% oxygen; the molecular weight is 92.

d.      A compound consist of 10g of C, 1.66g H and 13.33g O. The molecular weight of the compound is 180.

MASS RELATIONSHIP FROM BALANCED CHEMICAL EQUATION

              

             The balanced equation shows the relative masses of the substances that participate as a reactant or as products. To understand this clearly, first consider the significance of a chemical equation. For example, the balanced equation in a combination reaction of aluminum oxide and water is Al₂O₃ + 3H₂O → 2Al(OH)₃

              Quantitatively, this means that 1 mole of aluminum oxide reacts with 3 moles of water yields 2 moles of aluminum hydroxide. Also from the coefficients in the balanced equation you can derive the following relationships:

1 mole Al₂O₃ = 2 moles Al(OH)₃

1 mole Al₂O₃ = 3 moles H₂O

               Then you are going to consider the mole concept to find the mass of a mole of each of the substances involve. Recall that the formula weight in amu can be expressed in g/mol.

Now consider several stoichiometric problems from balanced equation. In doing such calculation observe the following steps:

1. Write the balanced equation.
2. Convert from the given units to moles.
3. Convert the moles of a given quantity to moles of the desired quantity using the relationship from the balanced equation.
4. Convert the moles of the new quantity to the desired units using formula weights, density, Avogadro’s number and so on.

Practice Exercises:

1. Given the equation:

3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO + 4H₂O

a.      Calculate the number of grams of copper (II) nitrate that could be produced from 5.25 moles of copper.

b.      Calculate the number of grams of copper that would be required to produce 100g of copper (II nitrate).

c.      Calculate the number of grams of nitric acid that would react with 2.5 g of copper.

d.      Calculate the number of moles of nitric acid that would be required to produce 7.45g of copper (II) nitrate.

     2. If sodium sulfate reacts with aluminum nitrate, calculate

a.       Number of moles sodium sulfate that would react 15 grams of aluminum          nitrate

b.      Number of moles of aluminum nitrate that could be formed 2.75 moles of sodium sulfate.

c.       Number of grams of aluminum sulfate that could be formed 6.25 grams of sodium sulfate.

d.      Number of grams sodium sulfate that would be required to produce 120 grams of aluminum sulfate.

LIMITING REACTANTS

            In general, when a chemical reaction is carried out, one of the reactants will be used in excess of the amount needed. The reactant that is not present in excess is the one that will determine how much of the product can be obtained and I thus referred to as the limiting reactant. For example in the reaction of hydrogen gas and oxygen gas to form water, we have seen that 2 moles of hydrogen will combine with 1 mole of oxygen to form 2 moles of water. If there is only one mole of oxygen present and an excess of hydrogen, only 2 moles of water can be obtained. Ni matter how much hydrogen is present, the oxygen will limit the amount of water that can be produced. Thus oxygen is the limiting reactant.

The following steps can be used in doing calculations of this type:

1.      Calculate the number of moles of product that can be obtained for each reactant given.

2.      The reactant that gives the least number of moles of product is the limiting reactant and is the one that will determine the theoretical yield in the reaction; that is no matter how much of the excess reactant is present, no more product can be obtained than that calculated from the limiting reactant.

3.      Next, the moles of the theoretical yield are converted to any desired units, such as grams.

4.      The amount of excess reactant that is used can be calculated either from the theoretical amount of product obtainable or from the amount of the limiting reactant present.

Illustrative Problem 1

            A 50g calcium carbonate is reacted with 35g of phosphoric acid. Calculate (a) the number of grams of calcium phosphate that could be produced. (b) The number of grams of excess reactant that will remain.

Factors that Influence Reaction Rate

1.      Concentration: molecules must collide to react. The more molecules are present in the container, the more frequently they collide, and more often a reaction occurs. Thus, reaction rate is proportional to the concentration of reactants.

2.      Physical State: molecules must mix to collide. When the reactants are in the same phase, as in an aqueous solution, random thermal motion brings them into contact. When they are in different phases, contact occurs only at the interface, so vigorous stirring and grinding may be needed. In these cases, the more finely divided a solid or liquid reactant, the greater its surface area per unit volume, the more contact it makes with the other reactant, and the faster the reaction occurs.

3.      Temperature: molecules must collide with enough energy to react. At a higher temperature, more collisions occur in a given time. Thus, raising the temperature increases the reaction rate by increasing the number and, especially, the energy of the collisions.

Increasing the temperature of a reaction increases the average speed of particles and therefore their collision frequency. But collision frequency cannot be the only factor affecting rate, in the vast majority of collision, the molecules rebound without reacting. Arrhenius propose that every reaction has an energy threshold that the colliding molecules must exceed in order to react. This minimum collision energy is the activation energy (E_a), the energy required to activate the molecules into a state from which a reactant bonds can change into product bond. Only those collisions with enough energy to exceed E_a can lead to reaction.

4.      Catalyst. There are many situations in which the rate of a reaction must be increased for it to be useful. To increase the rate of reaction catalyst can be added.a catalyst is a substance that increases the rate of chemical reaction without itself being consumed.

RECORDED LECTURES

Percent Composition By Mass

https://www.youtube.com/watch?v=bGmbyztdCHE

Empirical Formula & Molecular Formula Determination From Percent Composition

https://www.youtube.com/watch?v=JeSSucG-CVw

ONLINE PUBLISHED RESEARCH

Molecules for devices: The inorganic chemistry behind modern technologies

https://doi.org/10.1016/j.poly.2018.01.007.

(http://www.sciencedirect.com/science/article/pii/S0277538718300093)

Molecular Inorganic Chemistry

https://doi.org/10.1016/B978-0-12-409547-2.05397-X.

(http://www.sciencedirect.com/science/article/pii/B978012409547205397X