Question

Chem 101 Study Guide for Chapter 3 Selected practice problems from the end of each chapter are recommended; these problems are listed in parentheses in the appropriate section. It's great to do more than those listed. Chapter 3: Stoichiometry (Questions: 25, 29, 33, 35, 39, 47, 51, 53, 55, 57, 59, 61, 65, 67, 73, 75, 79, 83, 87, 95, 99, 105) 3.1 Counting by Weighing 3.2 Atomic Masses What is an atomic mass, and how is it related to isotopes and natural abundance of isotopes? How are atomic masses determined? Calculate percent natural abundance. Calculate atomic mass from isotope abundance data 3.3 The Mole What is a mole? How many of something are found in a mole of that thing? Do calculations involving the number of moles, number of particles, and the mass of a substance. 3.4 Molar Mass What is a molar mass, and how is it calculated? 3.5 Learning to Solve Problems 3.6 Percent Composition of Compounds Calculate the percent composition by mass of an element in a compound from mass, formula, or mole data. 3.7 Determining the Formula of a Compound From mass, % composition, or mole data determine the formula of a compound. What is Empirical Formula? What is a Molecular Formula? 3.8 Chemical Equations What is a chemical equation? How is a chemical equation used? Define reactant and product Know how to specify states of matter in chemical equations: solid, liquid, gas, aqueous. Understand what it means to balance chemical equations. 3.9 Balancing Chemical Equations Balance chemical equations. 3.10 Stoichiometric Calculations: Amounts of Reactants and Products Understand relationship between reactants and products in terms of moles, masses, and number of molecules. 3.11 The Concept of Limiting Reactant What is a limiting reactant? Know how to determine the limiting reactant for a chemical reaction. Define actual yield, theoretical yield, and percent yield; be able to calculate all three.

Answer 1

3.1 Counting by Weighing:

It is an efficient way to calculate the weight of a particular compound of interest. Weight of a compound can be calculated by knowing molecular formula and using periodic table which provides the atomic weight of an individual atom.

Example: Finding molecular weight of H₂O.

Atomic weight of Hydrogen and Oxygen is 1 and 16 respectively.

M.W = weight of (2*Hydrogen atom) + (1* Oxygen atom) = 18 g/mole.

3.2 Atomic Masses:

Atomic mass is the sum of number of protons and neutrons in an atom. The number of protons in an element is always constant but number of neutrons may vary. These are known as isotopes of a particular element. The ‘average atomic mass’ is the sum of the masses of its isotopes, each multiplied by its % of natural abundance. % natural abundance is calculated as: (M1) (x) + (M2) (1-x) = Me. Me is the average atomic mass (from the periodic table). M1 is mass of the isotope for which you know abundance. M2 is mass of the isotope for which you don’t know the abundance. x is abundance to be known.

3.3 The Mole:

Mole is the counting unit used in chemistry to indicate the number of atoms or ions or molecules. In definition, mole of a substance is defined as the mass of substance containing the same number of fundamental units as there are atoms in exactly 12.000 g of 12C. The number of atoms in 12 gm of ¹²C is 6.022 * 10²³, this number is known as Avogadro’s number.

For example: if the mass of K is given to be 8.2 g. The atomic weight of K is 39.10 amu. Therefore, mole of K is = 8.2/39.10 = 0.2097 mol of K. Also number atoms in 8.2 g of K is 0.2097*6.022 * 10²³ = 1.26* 10²³.

3.4 Molar Mass:

Molar mass of a compound is the sum of the total mass of all the atoms that constitute a molecule per mole. See example above in 3.1.

3.6 Percent Composition of compounds:

The percent composition by mass of a compound can be calculated by dividing the mass of each element by the total mass of the compound. . In above example of H₂O, mass of H per mol of H₂O = (1/18)*100 = 5.56 % of H.

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3.7 Determining the formula of a compound:

It is reverse calculation of what we have done above. If you have given a compound of which composition is known. Say you have given chlorine (73.9%) and mercury (261%) by mass.

Solution:

                Known: M.W of Mercury = 200.59 g/mol, M.W of Chlorine = 35.45 g/mol

Basis: 100 gm of sample compound.

Number of moles of Mercury in given compound =

1 = 0.368 moles 73.9 20059

Number of moles of Chlorine in given compound =

1 26.1 = 0.736 moles a5 45.

The molar ratio of two elements =

0.736 mol of C= 2,0 0.368 mi of Ha

It means compound contains moles of Cl twice of Hg. Hence the molecular formula of a given compound is HgCl₂.

3.8 Chemical Equations:

It represents a chemical reaction, symbolically. In general, reactants represents on left-hand side and product on right-hand side.

Ex: .

CH4202 Co2 2H20

Here L.H.S compounds react known as a reactant and it produces product R.H.S.

States of compounds are represented as suffix. For Example A_(S) +B_(g) à C_(l)

3.9 Balancing Chemical Equation:

The number of atoms of an element taking part in a reaction (i.e. on reactant side) should be equal to number of atoms of an element on product side.

Example: Balance following chemical equation

___Mg + ___O₂ à ___MgO

Here, the number of atoms Mg and O₂ should be equal on both sides. So, by assuming 1 or 0.5 of O₂ we can balance.

1 Mg + 0.5 O₂ à 1 MgO

Or

2 Mg + 1 O₂ à 2 MgO

3.10 Stoichiometric Calculations: Amounts of reactants and products

4 NH_3(g) + 5 O_2(g)à 4 NO_(g) + 6 H₂O_(g)

Observing above reaction, we can say that 4 moles of NH₃ reacting with 5 moles of O₂ to give product of 4 moles of NO and 6 mol of H₂O. The explanation for masses and number of molecules is given above.

3.11 The concept of limiting reactant:

The reactant which react-out first from all of the reactant is limiting reactant.

Hence, the reactant in a chemical reaction limits the amount of product that can be formed.

Calculation for finding limiting reactant. Consider above reaction (3.10) of ammonia and oxygen. Let’s assume 2 gm of ammonia and 4 gm of oxygen of reactant.

Solution: Now, we have to calculate the amount of any particular product produced by 5 gm of ammonia and 3 gm of oxygen separately. Here we consider product NO

For ammonia:

2 gm*(1 mol of NH₃/17.0 g of NH₃) * ( 4 mol of NO/4 mol of NH₃) * ( 30 g NO/ 1 mol of NO) = 3.53 g of NO

For Oxygen:

4 gm*(1 mol of O₂/17.0 g of O₂) * ( 4 mol of NO/5 mol of O₂) * ( 30 g NO/ 1 mol of NO) = 3.00 g of NO

The oxygen produces less amount of No, hence it is limiting reactant. Whereas ammonia is in excess.

Amounts of products calculated from the complete reaction of the limiting reagent are called theoretical yields.

The amount actually produced of a product is the actual yield.

Actual Yield* 100 % yield = Theoreticalyisld