2•Stoichiometry: Chemical Arithmetic Formula Conventions (1 of 24) Superscripts used to show the charges on ions Mg2+ the 2 means a 2+ charge (lost 2 electrons) Subscripts used to show numbers of atoms in a formula unit H2SO4 two H’s, one S, and 4 O’s Coefficients used to show the number of formula units 2Br– the 2 means two individual bromide ions Hydrates CuSO4 • 5 H2O some compounds have water molecules included 2•Stoichiometry: Chemical Arithmetic Stoichiometry Terms (2 of 24) stoichiometry study of the quantitative relationships in chemical formulas and equations. atomic mass weighted average mass of an atom, found on the periodic table formula mass sum of the atomic masses of the atoms in a formula molecular mass sum of the atomic masses of the atoms in a molecular formula gram molecular mass molecular mass written in grams molar mass same as gram molecular mass empirical formula formula reduced to lowest terms 2•Stoichiometry: Chemical Arithmetic Calculating Formula Mass (3 of 24) Formula or molecular mass is found by simply summing the atomic masses (on the periodic table) of each atom in a formula. H2SO4 1.01 + 1.01 + 32.06 + 16.0 + 16.0 + 16.0 + 16.0 = 98.08 u 2(1.01) + 32.06 + 4(16.0) = 98.06 u or 98.06 g/mole Generally, round off your answers to the hundredths or tenths place. Don’t round off too much (98.06 g/mol or 98.1 g/mol is OK, but don’t round off to 98 g/mol) Units Use u or amu if you are referring to one atom or molecule 2•Stoichiometry: Chemical Arithmetic Mole Facts (4 of 24) A mole (abbreviated mol) is a certain number of things. It is sometimes called the chemist’s dozen. A dozen is 12 things, a mole is 6.02 x 1023 things. Avogadro’s Number 1 mole of any substance contains 6.02 x 1023 molecules Molar Volume (measured at P = 760 mmHg and T = 0 °C) 1 mole of any gas has a volume of 22.4 Liters Molar Mass (see gram formula mass) 1 mole 6.02 x 1023 molecules 1 mole 22.4 L 1 mole molar mass 2•Stoichiometry: Chemical Arithmetic Line Equations (5 of 24) A Line Equation is the preferred way to show conversions between quantities (amount, mass, volume, and number) by canceling units (moles, grams, liters, and molecules) The line equation consists of the Given Value, the Desired Unit, and the line equation itself. Example: What is the mass of 135 Liters of CH4 (at STP)? Given: 135 L CH4 Desired: ? g CH4 135 L CH4 x 1 mol CH4 22.4 L CH4 x 16.0 g CH4 1 mol CH4 = 96.43 g CH4 2•Stoichiometry: Chemical Arithmetic Mole Relationships (6 of 24) The “Mole Map” shows the structure of mole problems Mass Volume at STP number of atoms or molecules Mass moles Volume at STP number of atoms or molecules ´ ´ ` ` 1) 1 mol molar mass 2) 1 mol 22.4 L 3) 1 mol 6.02 x 1023 molecules 2•Stoichiometry: Chemical Arithmetic Percentage Composition (by mass) (7 of 24) Percentage Composition quantifies what portion (by mass) of a substance is made up of each element. Set up a fraction: mass of element mass of molecule Change to percentage: 100 x mass of element mass of molecule Generally, round off your answers to the tenth’s place. The percentage compositions of each element should add up to 100% (or very close, like 99.9% or 100.1%) 2•Stoichiometry: Chemical Arithmetic Formula from % Composition (8 of 24) Given the Percentage Composition of a formula, you can calculate the empirical formula of the substance. Step 1 assume you have 100 g of substance so the percentages become grams Step 2 change grams of each element to moles of atoms of that element Step 3 set up a formula with the moles example: C2.4 H4.8 Step 4 simplify the formula by dividing moles by the smallest value C2.42.4 H 4.8 2.4 = CH2 Step 5 If ratio becomes… 1:1.5 multiply by 2 1:1.33 or 1:1.66 multiply by 3 2•Stoichiometry: Chemical Arithmetic Equation Terms (9 of 24) equation condensed statement of facts about a chemical reaction. reactants → substances that exist before a chemical rxn. Written left of arrow. → products substances that come into existence as a result of the reaction. Written to the right of the arrow. word equation an equation describing a chemical change using the names of the reactants and products. coefficients a number preceding atoms, ions, or molecules in balanced chemical equationns that showing relative #’s. 2•Stoichiometry: Chemical Arithmetic Other Mole Problems and Conversions (10 of 24) The gas density is often converted to molar mass: Example : The gas density of a gas is 3.165 g/Liter (at STP). What is the molar mass of the gas? Knowing that 22.4 L is 1 mole, you can set up the ratio: 3.165 g 1 Liter = molar mass 22.4 L Other metric conversions you should know: 1000 mL 1 Liter 1 kg 1000 grams 2•Stoichiometry: Chemical Arithmetic Writing Formula Equations Things To Remember (11 of 24) Example: Write the formula equation of... sodium metal + water → sodium hydroxide + hydrogen gas Na° + H2O → NaOH + H2 • metals often are written with the ° symbol to emphasize that the metal is in the neutral elemental state, not an ion. • some compounds have common names that you should just know... water, H2O; ammonia, NH3; methane, CH4 • remember the seven diatomic elements so they can be written as diatomic molecules when they appear in their elemental form. Other elemental substances are written as single atoms (e.g. sodium metal or helium gas, He) 2•Stoichiometry: Chemical Arithmetic Coefficients and Relative Volumes of Gases (12 of 24) Since every gas takes up the same amount of room (22.4 L for a mole of a gas at STP), the coefficients in an equation tell you about the volumes of gas involved. Example: N2(g) + 3 H2(g) → 2 NH3(g) + 2•Stoichiometry: Chemical Arithmetic Heart of the Problem (13 of 24) The “heart of the problem” conversion factor relates the Given and the Desired compounds using the coefficients from the balanced equation. Example: N2 + 3 H2 → 2 NH3 “ could be 3 moles H2 2 moles NH3 …which means that every time 2 moles of NH3 is formed, 3 moles of H2 must react. The format is always, moles of Desired moles of Given 2•Stoichiometry: Chemical Arithmetic Mass-Mass Problems Mass-Volume Problems (14 of 24) Mass-Mass problems are probably the most common type of problem. The Given and Desired are both masses (grams or kg). The pattern is: Given x molar mass of Given x “ x molar mass of Desired In Mass-Volume problems, one of the molar masses is replaced with 22.4 L1 mole depending on whether the Given or the Desired is Liters. 2•Stoichiometry: Chemical Arithmetic Mass-Volume-Particle Problems (15 of 24) If the Given or Desired is molecules, then the Avogadro’s Number conversion factor, 6.02 x 10 23 molecules 1 mole is used and the problem is a Mass-Particle or Volume-Particle problem. The units of the Given and Desired will guide you as to which conversion factor to use: Mass grams or kg Volume Liters or mL Particles molecules or atoms 2•Stoichiometry: Chemical Arithmetic Limiting Reactant Problems (16 of 24) In a problem with two Given values, one of the Given’s will limit how much product you can make. This is called the limiting reactant. The other reactant is said to be in excess. Solve the problem twice using each Given… the reactant that results in the smaller amount of product is the limiting reactant and the smaller answer is the true answer. Example: N2 + 3 H2 → 2 NH3 When 28.0 grams of N2 reacts with 8.00 grams of H2, what mass of NH3 is produced? (in this case, the N2 is the limiting reactant) 2•Stoichiometry: Chemical Arithmetic How Much Excess Reactant is Left Over (17 of 24) Example: N2 + 3 H2 → 2 NH3 When 28.0 grams of N2 reacts with 8.00 grams of H2, what mass of NH3 is produced? (in this case, the N2 is the limiting reactant) To find out how much H2 is left over, do another line equation: Given: 28.0 g N2 Desired: ? g H2 subtract the answer of this problem from 8.00 g H2 2•Stoichiometry: Chemical Arithmetic Limiting Reactants (18 of 24) It is difficult to simply guess which reactant is the limiting reactant because it depends on two things: (1) the molar mass of the reactant and (2) the coefficients in the balanced equation The smaller mass is not always the limiting reactant. Example: N2 + 3 H2 → 2 NH3 1 mole (28 g N2) will just react with 3 moles (6.06 g H2) so, if we react 28.0 g N2 with 8.0 g H2, only 6.06 g H2 will be used up and 1.94 g of H2 will be left over. In this case, N2 is the L.R. and H2 is in X.S. 2•Stoichiometry: Chemical Arithmetic Theoretical Yield and Percentage Yield (19 of 24) The answer you calculate from a stoichiometry problem can be called the Theoretical Yield. Theoretically , you should get this amount of product. In reality , you often get less than the theoretical amount due to products turning back to reactants or side reactions. The amount you actually get is called the Actual Yield. Percentage Yield = Actual YieldTheoretical Yield x 100 2•Stoichiometry: Chemical Arithmetic Balancing Chemical Equations (20 of 24) The balanced equation represents what actually occurs during a chemical reaction. Since atoms are not created or destroyed during a normal chemical reaction, the number and kinds of atoms must agree on the left and right sides of the arrows. __Na2CO3 + __HCl → __ NaCl + __H2O + __CO2 To balance the equation, you are only allowed to change the coefficients in front of the substances... not change the formulas of the substances themselves. Reduce the coefficients to the lowest terms. Fractions may be used in front of diatomic elements. 2•Stoichiometry: Chemical Arithmetic Combustion Equations (21 of 24) The burning of fuels made of C, H, and O is called combustion. You need to memorize O2, CO2 and H2O Example: The combustion of propane, C3H8, is written: C3H8 + 5O2 → 3CO2 + 4H2O Be careful when writing equations for alcohols, such as butanol, C4H9OH • don’t forget to add the H’s (a total of 10 of them) • don’t forget to take account of the O atom in the alcohol C4H9OH + 6O2 → 4CO2 + 5H2O 2•Stoichiometry: Chemical Arithmetic Solutions -- Molar Concentration (22 of 24) Many reactions are carried out in solution. Solutions are convenient and speed up many reactions. Concentration is often expressed as Molarity ( M) = moles of soluteLiters of solution You can calculate the molarity of a solution when given moles (or grams) of a substance and its volume. You can use the molarity of a solution as a conversion factor 0.150 M HCl ≈ 0.150 moles HCl1 Liter HCl or 1 Liter HCl 0.150 moles HCl to convert moles to Liters and vice versa. Volumetric flasks are used to make solutions. 2•Stoichiometry: Chemical Arithmetic Dilution Problems (23 of 24) You can calculate the moles of a solute using the volume and molarity of the substance. Since diluting a solution adds water and no solute, the moles of solute before and after the dilution remains constant. So... Vi · Mi = Vf · Mf where “I” means “initial” and “f” means “final” The units of volume or concentration do not really matter as long as they match on the two sides of the equation. 2•Stoichiometry: Chemical Arithmetic Acid-Base Titrations (24 of 24) Acids form the H+ ion. Bases form the OH– ion. Acids + bases mix to form H2O (HOH) and a salt. The moles of H+ = the moles of OH– in a neutralization. An acid-base titration is the technique of carefully neutralizing an acid with a base and measuring the volumes used. An indicator (we used phenolphthalein) allows us to observe when the endpoint is reached. If a monoprotic acid is neutralized with a base that only has one OH– ion per formula unit, the simple formula: Va · Ma = Vb · Mb allows you to determine the molarity of the unknown. Paul Groves