Stoichiometry: Understanding Chemical Formulas and Equations, Summaries of Stoichiometry

Download Stoichiometry: Understanding Chemical Formulas and Equations and more Summaries Stoichiometry in PDF only on Docsity! 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 ➂ ➂ ➁ ➁ ➀ ➀ 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 C 2.4 2.4 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 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 Yield Theoretical 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 solute Liters 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 HCl 1 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.