Lecture 3- Stoichiometry

General Chemistry Course # 111, two credits Second Semester 2009

King Saud bin Abdulaziz University for Health Science Textbook: Principles of Modern Chemistry by David W. Oxtoby, H. Pat Gillis, and Alan Campion (6 edition; 2007)

Dr. Rabih O. Al-Kaysi Ext: 47247 Email: [email protected]

Lecture 3

Stoichiometry

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1 - Chemical Changes • Chemical properties describes the reaction a substance undergo to form new substances. • The study of chemical changes is at the heart of chemistry. – Some chemical changes are simple and some are complex. • For example, changes that occur in your brain and eyes allow you to see and think. These are complex chemical changes

• We will explore the quantity of substances consumed and produced in a chemical reaction. • Chemical Equations are used to describe chemical reactions.

2 - Defining Chemistry • Lavoisier: mass is conserved in a chemical reaction. • His careful measurements turned chemistry into a science (Father of Chemistry)…Not true! Jaber Bin Hayan is the father of chemistry • Investigate the reaction of hydrogen with oxygen to produce water (2H2 + O2 → 2H2O).

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3 - Chemical Equations • The chemical equation for the formation of water can be visualized as two hydrogen molecules reacting with one oxygen molecule to form two water molecules: 2H2 + O2 → 2H2O

Products

Reactants

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Reading Chemical Equations • The plus sign (+) means “react” and the arrow points towards the substance produce in the reaction. • The chemical formulas on the right side of the equation are called reactants and after the arrow are called product. • The numbers in front of the formulas are called stoichiometric coefficients. 2Na + 2H2O → 2NaOH + H2 Co ef fi

ci

en t

Reactants

Products Products

Understanding Chemical Equations

Coefficients and subscripts included in the chemical formula have different effects on the composition.

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4 - Balancing Chemical Equations • Law of conservation of mass: matter cannot be lost in any chemical reactions.

O

Class Practice Problem Balance the following equations: •

(a) Na(s) + H2O(l)

NaOH(aq) + H2(g)

(b) Al(s) + HCl(aq)

AlCl3(aq) + H2(g)

(c) C2H4(g) + O2(g)

CO2(g) + H2O(l)

–

–

–

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5 - Combination and Decomposition Reactions

• Combination reactions: or Synthesis Reaction; two or more substances react to form products: 2Mg(s) + O2(g) → 2MgO(s)

• The Mg has combined with O2 to form MgO (ionic compounds). • Decomposition reactions: is when one substance undergoes a reaction to produce two or more simpler substances: 2NaN3(s) → 2Na(s) + 3N2(g) (the reaction that occurs in an air bag) • The NaN3 has decomposed into Na and N2 gas.

• Single Replacement reaction: A single uncombined element reacts to replace another element in a compound: Zn + CuSO4  ZnSO4 + Cu

The Zn has replaced the Cu in CuSO4 • • Double Replacement reaction: parts of two compounds switch places to form two new compounds:

AgNO3 + NaCl  AgCl (solid) + NaNO3 • Transformation: A single compound transforms to another simple compound under the influence of heat, light or catalyst light

Trans-Cinnamic acid  Cis-Cinnamic acid

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6 - Patterns in Chemical Reactivity

• The periodic table can be used to predict how elements will react in a combination reaction: 2K(s) + 2H2O(l) → 2KOH(aq) + H2(g)

• All alkali metals will react with water to form the hydroxide compound and hydrogen. • let M represent the alkali metal, we are able to write: 2M(s) + 2H2O(l) → 2MOH(aq) + H2(g) Alkali metal + water → Metal hydroxide + hydrogen

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7 - Formula Weights Formula and Molecular Weights

• Formula weights (FW) is the sum of the atomic weights of each atom in the chemical formula. FW (H2SO4) = 2AW(H) + AW(S) + 4AW(O) = 2(1.0 amu) + (32.0 amu) + 4(16.0) = 98.0 amu • If the chemical formula is also its molecular formula then the weight is called the molecular weight (MW). MW(C6H12 O6) = 6(12.0 amu) + 12(1.0 amu) + 6(16.0 amu) * More complicated when isotopes are present

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Formula Weights Percentage Composition from Formulas • Percent composition is the atomic weight for each element divided by the formula weight of the compound multiplied by 100:

( No. of Atoms of Element )( AW ) % Element = ×100 FW of Compound

Class Practice Problem • Calculate the FW of C12 H22 O11 . 12C = 12 x 12.01 = 144.12 22H = 22 x 1.01 = 22.22 11O = 11 x 16.00 = 176.0 C12 H22 O11 = 342.34 amu

• Calculate the percent composition of H2O. •% H = 11.21% •% O = 88.79%

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8 - Molar Mass • Molar mass: mass in grams of 1 mole of substance (units g/mol, g.mol-1 ). • Experimentally, 1 mole of 12 C = 12 g, which can be written as 12g/mol.

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9 - The Mole • The unit we use to express the quantity of atoms, ions, and molecules that an object contains is called mole. • Mole: convenient measure chemical quantities. • The actual number of atoms, ions, or molecules in 1 mole of something = 6.0221367 × 1023 (Advogadro’s number) of that thing. • Thus, • 1 mole of 12 C atoms = 6.02 x 1023 12 C atoms • 1 mole of H2O molecules = 6.02 x 1023 molecules • 1 mole of NO3- ions = 6.02 x 1023 ions

Visualizing The Mole Concept

Different Units

Class Practice Problem • How many C atoms are in 0.350 mol of C6H12 O6? •C atoms = [0.350 mol C6H12 O6 (6.02 x 1023 molecules/1 mole C6H12 O6)(6 C atoms/1 molecule)] = 1.26 x 1024 C atoms

Class Practice Problem Converting moles to mass • Calculate the number of moles of glucose C6H12 O6 in 5.380 g of C6H12 O6. •Moles of C6H12 O6 = [5.380 g C6H12 O6 (1 mole C6H12 O6/ 180 g C6H12 O6)] = 0.02989 mol C6H12 O6

Class Practice Problem Converting mass to particles • Calculate the number of atoms of Cu in 3 g of Cu? •Atoms of Cu = [3 g Cu (1 mole Cu/180 g Cu)(6.02 x 1023 atoms Cu/1 mole Cu)] = 3 x 1023 Cu atoms

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8 - Empirical Formulas from Analyses • Start with mass % of elements (i.e. empirical data) and calculate a formula, or • Start with the formula and calculate the mass % elements. • For example: • Ascorbic acid contains 40.92 percent C, 4.58 percent H, and 54.50 percent O by mass. What is the empirical Formula? The experimentally determined molecular weight is 176 amu.

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Empirical Formulas from Analyses Molecular Formula from Empirical Formula • Once we know the empirical formula, we need the MW to find the molecular formula. • Subscripts in the molecular formula are always wholenumber multiples of subscripts in the empirical formula •

Eg: we have 11.11% 12H and 88.88% 816 O. The formula weight = 36 amu. What is the empirical formula and molecular formula? # moles Deuterium = 5.55, # moles Oxygen = 5.55  ratio is 1:1 Empirical formula = 12HO formula weight of empirical formula ( 12HO) = 18 Formula weight/ Empirical formula weith = 32/18 = 2  multiply empirical formula by 2 2 1 H2 O2

Quantitative Information from Balanced Equations • Balanced chemical equation gives number of molecules that react to form products. • Interpretation: ratio of number of moles of reactant required to give the ratio of number of moles of product. • These ratios are called stoichiometric ratios. NB: Stoichiometric ratios are ideal proportions • Real ratios of reactants and products in the laboratory need to be measured (in grams and converted to moles).

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9 - Limiting Reactants • If the reactants are not present in stoichiometric amounts, at end of reaction some reactants are still present (in excess). • Limiting Reactant: one reactant that is consumed • Used 23 g Na and 32 g Oxygen. Which is the limiting reagent • 4 x 23 1 x 32 4Na + O2  4Na2O • • 23 32 • Oxygen needed = 23 x 32/ (4 x23) = 8 g • Sodium needed = 32 x 4 x 23/ (32) = 4 x 23 = 72 g

Limiting Reactants

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Limiting Reactants Theoretical Yields • The amount of product predicted from stoichiometry taking into account limiting reagents is called the theoretical yield. • The percent yield relates the actual yield (amount of material recovered in the laboratory) to the theoretical yield:

Actual yield % Yield = × 100 Theoretical yield