Presentation on theme: "C H E M I S T R Y Chapter 3 Mass Relationships in Chemical Reactions."— Presentation transcript:
C H E M I S T R Y Chapter 3 Mass Relationships in Chemical Reactions
Chemical Reaction In a chemical reaction a chemical change produces one or more new substances. there is a change in the composition of one or more substances.
Chemical Reaction In a chemical reaction old bonds are broken and new bonds are formed. atoms in the reactants are rearranged to form one or more different substances.
Chemical Reaction In a chemical reaction the reactants are Fe and O 2. the new product Fe 2 O 3 is called rust.
5 Chemical Equations A chemical equation gives the formulas of the reactants on the left of the arrow. the formulas of the products on the right of the arrow. ReactantsProduct C(s) O 2 (g) CO 2 (g)
6 Symbols Used in Equations Symbols in chemical equations show the states of the reactants. the states of the products. the reaction conditions. TABLE
7 Chemical Equations are Balanced In a balanced chemical reaction no atoms are lost or gained. the number of reacting atoms is equal to the number of product atoms.
Balancing Chemical Equations A balanced chemical equation shows that the law of conservation of mass is adhered to. In a balanced chemical equation, the numbers and kinds of atoms on both sides of the reaction arrow are identical. 2NaCl(s)2Na(s) + Cl 2 (g) right side: 2 Na 2 Cl left side: 2 Na 2 Cl
Balancing Chemical Equations 2.Find suitable coefficients—the numbers placed before formulas to indicate how many formula units of each substance are required to balance the equation. 2H 2 O(l)2H 2 (g) + O 2 (g) 1.Write the unbalanced equation using the correct chemical formula for each reactant and product. H 2 O(l)H 2 (g) + O 2 (g)
Balancing Chemical Equations Chapter 3/10 3.Reduce the coefficients to their smallest whole-number values, if necessary, by dividing them all by a common denominator. 2H 2 O(l)2H 2 (g) + O 2 (g) 4H 2 O(l)4H 2 (g) + 2O 2 (g) divide all by 2
Balancing Chemical Equations Do not change subscripts when you balance a chemical equation. You are only allowed to change the coefficients. H 2 O(l)H 2 (g) + O 2 (g) unbalanced 2H 2 O(l)2H 2 (g) + O 2 (g) Chemical equation changed! H2O2(l)H2O2(l)H 2 (g) + O 2 (g) Balanced properly
13 Balance the coefficients from reactants to products. A. __N 2 (g) + __H 2 (g) __ NH 3 (g) B. B. __Co 2 O 3 (s) + __ C(s) __Co(s) + __CO 2 (g) Write a balanced equation for the reaction between carbon dioxide and potassium hydroxide to form potassium carbonate and water. Examples
Chemical Symbols on Different Levels 2H 2 O(l)2H 2 (g) + O 2 (g) 0.56 kg of hydrogen gas react with 4.44 kg of oxygen gas to yield 5.00 kg of liquid water. macroscopic: 2 molecules of hydrogen gas react with 1 molecule of oxygen gas to yield 2 molecules of liquid water. microscopic: How can we relate the two to each other?
Avogadro’s Number and the Mole 2(12.0 amu) + 4(1.0 amu) = 28.0 amuC2H4:C2H4: HCl: 1.0 amu + 35.5 amu = 36.5 amu Molecular Mass: Sum of atomic masses of all atoms in a molecule. Formula Mass: Sum of atomic masses of all atoms in a formula unit of any compound, molecular or ionic.
Molar Mass of Na 2 SO 4 Calculate the molar mass of Na 2 SO 4. 16 ElementNumber of Moles Atomic MassTotal Mass in Na 2 SO 4 Na S O
Stoichiometry: Chemical Arithmetic Stoichiometry: The relative proportions in which elements form compounds or in which substances react. aA + bB cC + dD Moles of A Grams of A Moles of B Grams of B Mole Ratio Between A and B (Coefficients) Molar Mass of BMolar Mass of A
Stoichiometry: Chemical Arithmetic How many grams of NaOH are needed to react with 25.0 g Cl 2 ? 2NaOH(aq) + Cl 2 (g) NaOCl(aq) + NaCl(aq) + H 2 O(l) Aqueous solutions of sodium hypochlorite (NaOCl), best known as household bleach, are prepared by reaction of sodium hydroxide with chlorine gas: Moles of Cl 2 Grams of Cl 2 Moles of NaOH Grams of NaOH Mole Ratio Molar Mass
Stoichiometry: Chemical Arithmetic The commercial production of iron from iron ore involves the reaction of Fe 2 O 3 with CO to yield iron metal plus carbon dioxide: Fe 2 O 3 (s) + 3CO(g) 2Fe(s) + 3CO 2 (g) Predict how many grams of CO will react with 0.500 moles Fe 2 O 3 ?
Yields of Chemical Reactions The amount actually formed in a reaction. The amount predicted by calculations from the limiting reactant. Actual Yield: Theoretical Yield: actual yield theoretical yield x 100% Percent Yield =
Reactions with Limiting Amounts of Reactants Limiting Reactant: The reactant that is present in limiting amount. The extent to which a chemical reaction takes place depends on the limiting reactant. Excess Reactant: Any of the other reactants still present after determination of the limiting reactant.
Reactions with Limiting Amounts of Reactants At a high temperature, ethylene oxide reacts with water to form ethylene glycol which is an automobile antifreeze and a starting material in the preparation of polyester polymers: C 2 H 4 O(aq) + H 2 O(l) C2H6O2(l)C2H6O2(l)
Examples How many sandwiches can be made from 10 slices of bread and 8 slices of cheese? Which one is the limiting reactant? The balanced chemical equation is 2Bd + Ch Bd 2 Ch
Example Ammonia, NH 3, can be synthesized by the following reaction: 2 NO(g) + 5 H 2 (g) 2 NH 3 (g) + 2 H 2 O(g) Starting with 86.3 g NO and 25.6 g H 2 a. Determine the limiting reactant b. Find the theoretical yield of NH 3 in grams
Reactions with Limiting Amounts of Reactants Li 2 O(s) + H 2 O(g) 2LiOH(s) Lithium oxide is used aboard the space shuttle to remove water from the air supply according to the equation: If 80.0 g of water are to be removed and 65.0 g of Li 2 O are available, which reactant is limiting? How many grams of excess reactant remain? How many grams of LiOH are produced?
Percent Composition and Empirical Formulas Percent Composition: Expressed by identifying the elements present and giving the mass percent of each. Empirical Formula: It tells only the ratios of the atoms in a compound. Molecular Formula: It tells the actual numbers of atoms in a compound. It can be either the empirical formula or a multiple of it. multiple = molecular mass empirical formula mass
Steps in determine the Empirical formula Step 1: Obtain the mass of each element (in grams) Step 2: Determine the numbers of atom present Step 3: Divide the smallest moles by numbers of each atoms to obtain the closet integer as possible. Step 4: If the result ended with 0.5, 0.33, 1.125, 1.50 etc… then multiply with a factor to get the nearest integer as possible.
Example A compound containing nitrogen and oxygen is decomposed in the laboratory and produces 24.5 g nitrogen and 70.0 g oxygen. Calculate the empirical formula of the compound.
Example An unknown sample gives the following mass percent: 17.5% Na, 39.7% Cr and 42.8% O. What is the empirical formula?
Example Butanedione – a main component in the smell and taste of butter and cheese – contains the elements carbon, hydrogen, and oxygen. The empirical formula of butanedione is C 2 H 3 O and its molar mass is 86.09 g/mol. Find its molecular formula
Examples Ibuprofen, an aspirin substitute, has the following mass percent composition: C 75.69%, H 8.80% an O 15.51%. What is the empirical formula of ibuprofen?
Determining Empirical Formulas: Elemental Analysis Combustion Analysis: A compound of unknown composition (containing a combination of carbon, hydrogen, and possibly oxygen) is burned with oxygen to produce the volatile combustion products CO 2 and H 2 O, which are separated and weighed by an automated instrument called a gas chromatograph. hydrocarbon + O 2 (g) xCO 2 (g) + yH 2 O(g) carbon hydrogen
Percent Composition and Empirical Formulas A colorless liquid has a composition of 84.1 % carbon and 15.9 % hydrogen by mass. Determine the empirical formula. Also, assuming the molar mass of this compound is 114.2 g/mol, determine the molecular formula of this compound.
Example Upon combustion, a 0.8233 g sample of a compound containing only carbon, hydrogen, and oxygen produced 2.445 g CO 2 and 0.6003 g H 2 O. Find the empirical formula of the compound
Example A compound contains only carbon, hydrogen, and oxygen. Combustion of 10.68 mg of compound yields 16.01 mg CO 2 and 4.37 mg H 2 O. The molar mass of the compound is 176.1 g/mol.