3 Stoichiometry: Calculations with Chemical Formulas and Equations

Transcription

1 3 Stoichiometry: Calculations with Chemical Formulas and Equations 3.1 Chemical Equations Balance chemical equations. 3. Simple Patterns of Reactivity Predict products of a chemical reaction in a combination reaction between a metal and nonmetal. Predict products of a chemical reaction in a decomposition reaction involving metal carbonates, metal chlorates, and metal hydroxides. Predict products of complete combustion reactions of hydrocarbons and simple compounds containing C, H, and. 3.3 Formula Weights Calculate the formula weight of a substance given its chemical formula. Calculate the molecular weight of a molecular substance. Calculate the percent composition of the elements in a compound from its formula. 3.4 Avogadro s Number and the Mole Use Avogadro s number to calculate the number of particles in a given number of moles and vice versa. Calculate the molar mass of a substance from its formula. Interconvert between the number of moles of a substance and its mass. Use Avogadro s number and molar mass to calculate the number of particles making up a substance and vice versa. bjectives 3.5 Empirical Formulas Calculate the empirical formula of a compound, having been given appropriate analytical data such as elemental percentages or the quantity of C and H produced by combustion. Calculate the molecular formula, having been given the empirical formula and molecular weight. 3.6 Qualitative Informational from Balanced Equations Calculate the mass of a particular substance produced or used in a chemical reaction (massmass problem). Calculate the number of particles or moles of a particular substance produced or used in a chemical reaction. 3.7 Limiting Reactants Determine the limiting reagent in a reaction. Calculate the theoretical and actual yields of chemical reactions given the appropriate data. Calculate the percent yield given appropriate data.

2 Chapter 3 Notes Stoichiometry: Calculations with Chemical Formulas and Equations 3.1 Chemical Equations The quantitative nature of chemical formulas and reactions is called stoichiometry. Lavoisier observed that mass is conserved in a chemical reaction. o This observation is known as the law of conservation of mass. Chemical equations give a description of a chemical reaction. o There are two parts to any equation: Reactants (written to the left of the arrow) and Products (written to the right of the arrow): H + d H There are two sets of numbers in a chemical equation: 1. Numbers in front of the chemical formulas (called stoichiometric coefficients) Stoichiometric coefficients give the ratio in which the reactants and products exist o H means that there are two water molecules present. o Note: in H there are four hydrogen atoms present (two for each water molecule).. Numbers in the formulas (they appear as subscripts). The subscripts give the ratio in which the atoms are found in the molecule. o H means there are two H atoms for each one molecule of water. Matter cannot be lost in any chemical reactions. o Therefore, the products of a chemical reaction have to account for all the atoms present in the reactants--we must balance the chemical equation. o When balancing a chemical equation we adjust the stoichiometric coefficients in front of chemical formulas. o Subscripts in a formula are never changed when balancing an equation. Example: The reaction of methane with oxygen: CH 4 + dc + H Counting atoms in the reactants: In the products: 1 C 1 C 4 H H 3 It appears as though H has been lost and has been created. To balance the equation, we adjust the stoichiometric coefficients: CH 4 + d C + H The physical state of each reactant and product may be added to the equation: CH 4(g) + (g) d C (g) + H (g) Reaction conditions occasionally appear above or below the reaction arrow (e.g., " " often is used to indicate the addition of heat). Equation Writing on the AP Test 1. All equations must be balanced. You do not have to include states 3. Write as net ionic equation Substances that dissociate should be written as ions (All reactions are assumed to take place in water unless otherwise stated!) Leave out any molecule/ion that is not changed in the reaction 4. Scoring 1 point correct reactants points correct products 1 point balanced equation 1 point answering question about reaction 1

3 3. Some Simple Patterns of Chemical Reactivity Combination (Synthesis) Reactions In combination reactions two or more substances react to form one product. o Combination reactions have more reactants than products. o Consider the reaction: Mg (s) + (g) d Mg (s) When both reactants are elements, they must combine to form a single product! Decomposition Reactions In decomposition reactions one substance undergoes a reaction to produce two or more other substances. o Decomposition reactions have more products than reactants. o Consider the reaction that occurs in an automobile air bag: NaN 3(s) d Na (s) + 3N (g) o This is an example of one of five types of decomposition you should know for the AP Test. 1. Binary Compounds (except for peroxides) two elemental substances. Metal carbonates metal oxide + C 3. Metal hydrogen carbonates (bicarbonates) metal carbonate + C + H 4. Metal chlorates metal chloride and 5. Metal hydroxides metal oxide + H o If any of these products are seen as the reactants, then the reaction is a combination rewaction written as the reverse of the above reactions. Combustion in Air Combustion reactions are rapid reactions that require from air and produce a flame. o Hydrocarbon combustion (complete) produces C and H o xygenated hydrocarbons will combust in the same way C 3 H 8(g) + 5 (g) d 3C (g) + 4H (l) CH 3 H + 3 C + 4H 3.3 Formula Weights Formula and Molecular Weights Formula weight (FW) is the sum of atomic weights for the atoms shown in the chemical formula. o Example: FW (H S 4 ) = AW(H) + AW(S) + 4AW() = (1.01 amu) amu + 4(16.0 amu) = 98.0 amu. Molecular weight (MW) is the sum of the atomic weights of the atoms in a molecule as shown in the molecular formula. o Example: MW (C 6 H 1 6 ) = 6(1.0 amu) + 1 (1.01 amu) + 6 (16.0 amu) = amu. Formula weight of the repeating unit (formula unit) is used for ionic substances. o Example: FW (NaCl) = 3.0 amu amu = 58.5 amu. Percentage Composition from Formulas Percent composition is obtained by dividing the mass contributed by each element (number of atoms times AW) by the formula weight of the compound and multiplying by 100. % element = (number of atoms of that element)(atomic weight of element)(100) (formula weight of compound)

4 3.4 Avogadro s Number and The Mole The mole (abbreviated "mol") is a convenient measure of chemical quantities. o 1 mole of something = 6.0 x 10 3 of that thing (this is a rounded value). This number is called Avogadro s number. Thus, 1 mole of carbon atoms = 6.0 x 10 3 carbon atoms. o Experimentally, 1 mole of 1 C has a mass of 1.0 g. Molar Mass The mass in grams of 1 mole of substance is said to be the molar mass of that substance. Molar mass has units of g/mol. o The mass of 1 mole of 1 C = 1 g. o The molar mass of a molecule is the sum of the molar masses of the atoms: Example: The molar mass of N = x (molar mass of N). o Molar masses for elements are found on the periodic table. o The formula weight (in amu s) is numerically equal to the molar mass (in g/mol). Interconverting Masses, Moles, and Number of Particles Look at units: o Mass: g o Moles: mol o Molar mass: g/mol o Number of particles: 6.0 x 10 3 mol 1 (Avogadro s number per mole). Note: g/mol x mol = g (i.e. molar mass x moles = mass), and mol x mol 1 = a number (i.e. moles x Avogadro s number = molecules). To convert between grams and moles, we use the molar mass. 1mol molar mass mass = # mol moles = # grams molar mass 1mol To convert between moles and molecules we use Avogadro s number. 3 1mol particles particles # mol moles = # particles particles = 1mol To convert between atoms and molecules/formula units we use the number of atoms in the formula molecules C 1 H atoms 1 molecules C6H To convert between particles and mass, we use both. 6 H = atoms H 3 mass 1mol particles 1mol molar mass # particles particles = # grams molar mass 1mol = particles 1mol 3.5 Empirical Formulas from Analyses Recall that the empirical formula gives the relative number of atoms in the molecule. Finding empirical formula from mass percent data: o We start with the mass percent of elements (i.e. empirical data) and calculate a formula. Assume we start with 100 g of sample. The mass percent then translates as the number of grams of each element in 100 g of sample. From these masses, the number of moles can be calculated (using the atomic weights from the periodic table). The lowest whole-number ratio of moles is the empirical formula. o Divide each by the smallest mole amount o If not all whole numbers, multiply by,3,4 3

5 Molecular Formula from Empirical Formula The empirical formula (relative ratio of elements in the molecule) may not be the molecular formula (actual ratio of elements in the molecule). o Example: ascorbic acid (vitamin C) has empirical formula C 3 H 4 3 and a molecular formula is C 6 H 8 6. To get the molecular formula from the empirical formula, we need to know the molecular weight, MW. o The molecular weight (MW) must be some whole number multiple of the formula weight (FW) of the empirical formula. o Find this multiple using the following formula: molecular weight multiple = empirical formula weight o The the molecular formula is the empirical formula times this multiple. Combustion Analysis Empirical formulas are routinely determined by combustion analysis. o A sample containing C, H and is combusted in excess oxygen to produce C and H. The amount of C gives the amount of C originally present in the sample. 1.0 g C ( ) mass carbon = mass C 44.0 g C The amount of H gives the amount of H originally present in the sample..0 g H ( ) mass hydrogen = mass H 18.0 g H The amount of originally present in the sample is given by the difference between the amount of sample and the amount of C and H accounted for. mass oxygen = mass sample mass carbon mass hydrogen 3.6 Quantitative Information from Balanced Equations (Stoichiometry) The coefficients in a balanced chemical equation give the relative numbers of particles (atoms, molecules, or formula units) involved in the reaction. H (g) + (g) d H (l) o The stoichiometric coefficients in the balanced equation may be interpreted in ways: 1. The relative numbers of atoms, molecules, or formula units involved in the reaction o moles of H used = 1 mole of used = moles of H produced. r the relative numbers of moles involved in the reaction. o molecules of H used = 1 molecule of used = molecules of H produced o The molar quantities indicated by the coefficients in a balanced equation are called stoichiometrically equivalent quantities. o Stoichiometric relations or ratios may be used to convert between quantities of reactants and products in a reaction. mol H mol H mol H mol H mol H o ALL STICHIMETRY PRBLEMS MUST USE THE MLE RATI!!! o The number of grams of reactant cannot be directly related to the number of grams of product. To get grams of product from grams of reactant you must do a 3-step dimensional analysis problem: o Convert grams of reactant to moles of reactant (use molar mass), o Convert moles of one reactant to moles of other reactants and products (use the stoichiometric ratio from the balanced chemical equation), o Convert moles back into grams for desired product (use molar mass). mol H mol H mol H 18.0 g H 4.8 g = 7.9 g H 3.0 g 1mol H mol H 4

6 3.7 Limiting Reactants It is not necessary to have all reactants present in stoichiometric amounts. ften, one or more reactants is present in excess. Therefore, at the end of reaction, those reactants present in excess will still be in the reaction mixture. The one or more reactants which are completely consumed are called the limiting reactants or limiting reagents. o ne of the easiest ways to find the limiting reactant is to remember the LIMITING MAKES LESS! Reactants present in excess are called excess reactants or excess reagents. Limiting Reactant Problem If.50 g of H reacts with 3.40 g of, what is the limiting react, excess reactant, and how much water can be made? H + H 1mol H mol H 18.0 g H.50 g H =.3 g H.0 g H mol H 1mol H mol H 18.0 g H 3.40 g = 3.83 g H 3.0 g 1mol H Limiting Reactant = Water produced = 3.83 g H What is the excess reactant and how much is left over? mol H.0 g H 3.40 g =.49 g H 1mol H 3.0 g.50 g g.07 g is used Excess Reactant = H Excess left =.07 g Theoretical Yields The amount of product predicted from stoichiometry taking into account limiting reagents is called the theoretical yield. This is often different from the actual yield - the amount of product actually obtained in the reaction. The percent yield relates the actual yield (amount of material recovered in the laboratory) to the theoretical yield: actual yield Percent yield = 100 theoretical yield 5