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Chapter 3. Formulas, Equations, and Moles. 3.1 Balancing Equation. Balancing Equations. Chemical Reaction Reactants  Products Steps: Write the unbalance equation Find an appropriate coefficient and place in front of the formula unit to balance the equation

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Chapter 3

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Chapter 3
Chapter 3

Formulas, Equations,

and Moles

Balancing equations
Balancing Equations

  • Chemical Reaction

    Reactants  Products

  • Steps:

    • Write the unbalance equation

    • Find an appropriate coefficient and place in front of the formula unit to balance the equation

    • Reduce the coefficients to the smallest ratio

    • Check your answer

Balancing equation
Balancing Equation

  • Propane, C3H8 is a colorless gas often used as a heating and cooking fuel in campers and rural homes. Write a balance equation for the combustion reaction of propane with oxygen to yield carbon dioxide and water.

    • Step 1: Write out the unbalanced equation

    • Step 2: Balance C atoms

    • Step 3: Balance C and H

    • Step 4: Balance C, H and O

    • Reduce coefficients to the smallest ratio

    • Check answer


  • When solid potassium reacts with liquid water, the products are hydrogen gas and potassium hydroxide, the later remains dissolved in the water. From these information, write the balanced equation for the reaction

  • Write the balanced equation for the following reaction

    • Solid carbon reacts with gaseous oxygen to form gaseous carbon dioxide

    • Glass is sometimes decorated by etching patterns on its surface. Etching occurs when hydrofluoric acid (in a aqueous solution of HF) reacts with the silicon dioxide in the glass to form gaseous silicon tetrafluoride and the liquid water.

Chemical symbols on different levels
Chemical symbols on Different Levels

  • Chemical symbols represent both a microscopic and a macroscopic level

  • Microscopic level – chemical symbols represent the behavior of individual atoms and molecules

  • E.g Atoms and molecules are much too small to be seen  use microscopic behavior to describe

  • 2 H2 + O2 2 H2O

  • Two molecules of hydrogen react with one molecule of oxygen to yield two molecule of water.

    • Help to understand how reaction occur

Chemical symbols on different levels1
Chemical symbols on Different Levels

  • Macroscopic level – formula and equations represent the large-scale behavior of atoms and molecules that give rise to observable properties

    • Deal with macroscopic behavior in the laboratory

  • E.gWeighing amount of reactants, place them in a flask and observe visible changes.

3 3 avogadro number and the mole
3.3 Avogadro Number and The Mole

  • Described the number of objects present in a dozen

    • Based on C-12

  • 1 mol of anything = 6.02 x 1023 units of that substance (atoms, molecules, particles, dollars etc..)

    • Avogadro’s number or nA


  • or Formula unit

Chapter 3

  • Molecular mass (molecular weight) – sum of the atomic masses of all atoms in a molecule (covalent molecule)

  • Formula mass – sum of atomic masses all atoms in a formula unit of any substances (ionic salts)

    • Units = amu

Molecular weight
Molecular Weight

  • What is the molecular weight of CH4?

  • Calculate molecular weight of sulfur dioxide, a gas produced when sulfur containing fuels are burned.

  • Calculate the formula weight of the substance below (record to two decimal places)

    • Ba3(PO4)2

Molar mass
Molar Mass

  • Like density, it’s a ratio of two numbers.

    • Mass per one mole of substance

      • Unit => g/mol

  • Replaced with molecular weight because molar mass described the concept more accurately

    E.gMolecular weightMolar Mass

    18.02 amu18.02 g/mol

Molar mass1
Molar Mass

  • Calculate the molar mass for each of the substance below (record to two decimal places)

    • C2H4

    • NH3

    • C2H5Cl

3 4 stoichiometry chemical arithmetic
3.4 Stoichiometry Chemical Arithmetic

  • Recall

    • Molar mass is the mass per one mole of a substance

    • 1 mol of substance = # g

    • Use as a conversion factor

    • E.gif the molar mass of NaCl = 58.44 g/mol

  • Moles  Gramsor Grams  Moles


  • Calcium carbonate (also called as calcite, CaCO3), is the principal mineral found in limestone, marble, chalk, pearls and the shells of marine animals such as clams

    • Calculate the molar mass of calcium carbonate

    • A certain sample of calcium carbonate contains 4.86 mol. What is the mass in grams of this sample?


How many moles of sucrose are in a table spoon of sugar that contains 2.85 g? The molar mass of sucrose is 342.0 g/mol

  • Calculate the mass of 4.85 mol of acetic acid, HC2H3O2. Vinegar is a dilute solution of acetic acid

  • How many water molecules are in a 10.0g sample of water?

Mass calculation
Mass Calculation

  • Steps for Calculating the Masses of Reactants and Product in Chemical Reactions

    • Step 1: Balance the equation for the reaction

    • Step 2: convert the masses of reactants or product to moles

    • Step 3: use the balanced equation to set up the appropriate mole ratio(s)

    • Step 4: Use the mole ratio(s) to calculate the number of moles of the desired reactant or product

    • Step 5: Convert from moles back to masses (of the desired reactant or product)


  • Aqueous solution of sodium hypochlorite (NaOCl), best known as household bleach, are prepared by reaction of sodium hydroxide with chlorine:

    …..NaOH(aq) + ….Cl2(g)  …NaOCl(aq) + NaCl(aq) + ….H2O(l)

    How many gram of NaOH are needed to react with 2.5g Cl2?

Limiting reactant
Limiting Reactant

  • Limiting Reactant (limiting reagent): is the reactant that is completely consumed in a chemical reaction and limits the amount of product.

  • Excess Reactant: Any of the other reactants still present after determination of the limiting reactant.

Steps in determination limiting reagent
Steps in Determination Limiting Reagent

  • i.Check to be sure you have a balanced equation

    • aA + bB  c C

  • ii. Convert the amount of reagent one that was given into the number molesproduct that you could form if that reagent was completely consumed.

    • gA  moles A  moles C

  • iii.Convert the amount of reagent two that was given into the number moles of product that you could form if that reagent was completely consumed.

    • gB  moles B  moles C

  • Steps in determination limiting reagent1
    Steps in Determination Limiting Reagent

    • iv. The reactant that produced the LEAST amount of product in step 2 or 3 will the limiting reagent.

    • v. Convert the LEAST moles into the number grams product and that will be your theoretical yield.

      • Theoretical yield is the amount of product that can be made in a chemical reaction based on the amount of limiting reagent

    Excess reactant
    Excess Reactant

    • To determine how much of reactants are left over from the reaction:

      • Larger moles of product – smaller moles of product

      • Convert this moles of product to the g of the excess reactant


    • If we have 42.5 g Mg and 33.8 g O2, what is the limiting reactant?

      2 Mg(s) + O2(g)  2 MgO(s)


    • Ammonia, NH3, can be synthesis by the following reaction:

      2 NO(g) + 5 H2(g)  2 NH3(g) + 2H2O(g)

      • Starting with 86.8 g NO and 25.6 g H2, find the theoretical yield of ammonia in grams

      • Calculate the number grams of excess reactant that are unused from the reaction

    3 5 percent yield
    3.5 Percent Yield

    • Recall

      • Theoretical yield is the amount of product that can be made in a chemical reaction based on the amount of limiting reagent

        actual yield

    • Percent yield = --------------------- x 100

      theoretical yield


    • Ethyl alcohol is prepared industrially by the reaction of ethylene, C2H4 with water. What is the percent yield of the reaction if 4.6 g of ethylene gives 4.7g of ethyl alcohol?

      • C2H4(g) + H2O(l)  C2H6(l)


    • Titanium (IV) oxide is a white compound used as a coloring pigment. Solid titanium (IV) oxide can be prepared by reacting gaseous titanium (IV) chloride with oxygen gas. A second product of this reaction is chlorine gas.

      • Suppose 6.71 x 103 g of titanium (IV) chloride is reacted with 2.45 x103 g of oxygen. Calculate the mass of titanium (IV) oxide that can form

      • If the percent yield of TiO2 is 75%, what mass is actually formed?

    3 11 percent composition and empirical formula
    3.11 Percent Composition and Empirical Formula

    • Formula of a compound represents the relative numbers of various types of atoms present.

      • E.g CO2, H2O etc..

    • Empirical formula: simplest formula

    • Molecular formula: the actual formula of a compound


    • In each case below, the molecular formula for a compound is given. Determine the empirical formula for each compound

      • C6H6. This is the molecular for benzene; a liquid commonly used in industry as a starting material for many important products

      • H2O2. This is hydrogen peroxide, a substance commonly diluted with water and used as a disinfectant

      • CCl4. This is carbon tetrachloride, an organic solvent

    Percent composition
    Percent Composition

    • Percent composition – the mass present of each element present in a compound

      Atomic mass of an element

    • % composition =……………………………….. X 100

      Molar mass of the compound

    Percent composition1
    Percent Composition

    • To calculate the percent composition (percentage composition) of a compound

      • Calculate the molar mass of the compound

      • Calculate the total mass of each element present in the formula of the compound

      • Calculate the percent composition (percentage composition):% by weight (mass) of element = (total mass of element present ÷ molecular mass) x 100


    • Glucose or blood sugar, has the molecular formula C6H12O6. What is the empirical formula, and what is the percent composition?

    Determination of an empirical formula
    Determination of an empirical formula

    • Step 1: Obtain the mass of each element present (in grams)

    • Step 2: Determine the number of moles of each type of atom present

    • Step 3: Divide the number of mole of each element by the smallest number of moles to convert the smallest number to 1. If all of the numbers so obtained are integers, these are the subscripts in the empirical formula. If one or more of these number are not integers, go on to step 4

    Determination of an empirical formula1
    Determination of an empirical formula

    • Step 4: Multiply the numbers you derived in step 3 by the smallest integer that will convert all of them to whole numbers. This set of whole numbers represents the subscripts in the empirical formula


    • In a lab experiment it was observed that 0.6884 g of lead combines with 0.2356 g of chlorine to form a binary compound. Calculate the empirical formula of this compound

    • When a 2.00 g sample of iron metal is heated in air, it reacts with oxygen to achieve a final mass of 2.573 g. Determine the empirical formula for this iron oxide


    • What is the empirical formula for sodium thiosulfate with the percent composition of 30.36% O, 29.08% Na and 40.56%S


    • Cisplatin, the common name for a platinum compound that is used to treat cancerous tumors, has the composition (mass percent) 65.02% platinum, 9.34 % nitrogen, 2.02% hydrogen, and 23.63% chlorine. Calculate the empirical formula for csiplatin

    Molecular formula
    Molecular Formula

    • Molecular formula – gives the actual numbers of atoms in a molecule

      • Molecular formula = n x empirical formula

        • may be the same as the empirical formula

    Molecular formula1
    Molecular Formula

    • a multiple of the empirical formula (or # of empirical unit) is defined:

      Emperical mass

    • Multiple or n = --------------------------

      ( # of empirical unit) Molecular mass


    • Ribose, a sugar present in the cells of all living organism, has a molecular mass of 150 amu and the empirical formula CH2O. What is the molecular formula?


    • A compound used as an additive for gasoline to help prevent engine knock shows the following percent composition

      71.65 % Cl24.27 % C4.07 % H

      The molar mass is known to be 98.96 g/mol. Determine the empirical formula and the molecular formula for this compound

    Combustion analysis
    Combustion Analysis

    CxHyOz (O can be replaced with any other element)

    • gCO2 moles CO2  moles C

    • gH2O  moles H2O  moles H

    • g of O or unknown element = grams of sample – (g of H + g of C)

    • Follow the steps in determination of an empirical formula

    Combustion analysis1
    Combustion Analysis

    • Caproic acid, the substance responsible for the aroma of dirty gym socks and running shoes, contains carbon, hydrogen and oxygen. On combustion analysis, a 0.450 g sample of caproic acid gives 0.418 g of H2O and 1.023 g of CO2. what is the empirical formula of caproic acid? If the molecular mass of caproic acid is 116.2 amu, what is the molecular formula?

      • Determine the percent composition of caproic acid?


    • Coniine, a toxic substance isolated from poison hemlock, contains only carbon, hydrogen and nitrogen. Combustion analysis of a 5.024 mg sample yields 13.90 mg of CO2 and 6.048 mg of H2O. What is the empirical formula of coniine?

      EXAM 1

      (Chapter 1-3, except 3.7-3.10)

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