Stoichiometry Introduction Rx between Hydrogen and Oxygen can be described as: Balanced equation: Or Or Avogadros Number: (number of Molecules)
Or Moles (amount of a substance containing avogadros number is a mole...so) Converting Moles to Grams Converting Once you have the molar mass, you can easily convert from grams to moles, and also from moles to grams. Number of moles = (# of grams) / (molar mass)
Number of grams = (# of moles) * (molar mass) Stoichiometry Stoichiometry is the accounting, or math, behind chemistry. Given enough information, one can use stoichiometry to calculate masses, moles, and percents within a chemical equation What You Should Expect The most common stoichiometric problem will present you with a certain amount of a reactant and then ask how much of a product can be formed. Here is a generic chemical equation: What You Should Expect Here is a typically-worded problem: Given 20.0 grams of A and sufficient B, how many grams of C can be produced? You will need to use molar ratios, molar masses, balancing and interpreting equations, and conversions between grams and moles. If you struggled with those in class, welcome to the club.
Go back and review them if you need to, because if you can't do that stuff, you can't do stoichiometry. Summary of molar ratios, molar masses, balancing and interpreting equations, and conversions between grams and moles Molar Ratio A comparison of the number of moles of one substance in a chemical equation. For example, The ratio of sodium carbonate to potassium chloride to sodium chloride to potassium carbonate is 1:2:2:1. Molar mass, symbol M, is the mass of one mole of a substance in grams (chemical element or chemical compound). The base SI unit for mass is the kilogram but, for both practical and historical reasons, molar masses are almost always quoted in grams per mole (g/mol or g mol 1 ), especially in chemistry. Balancing Chemical Equations A chemical equation is said to be balanced when there are equal numbers of each type of atom on each side.
Balancing is achieved by adjusting the stoichiometric coefficients not the subscripts in the chemical formulae. Chemical Equations A chemical equation describes a chemical reaction. Example: CH 4 + 2O 2 CO 2 + 2H 2 O Each reactant and product is described by its chemical formula reactants on the left and products on the right. A number in front of a chemical formula is a stoichiometric coefficient. To calculate the percent (%) of each element in a compound Example: Calc the % composition by mass of each element in CaO (calcium oxide) Example 1: Calc the % composition by mass of each element in CaO (calcium oxide)
Example 2: Calc. the % water of crystallisation in Copper II Sulfate crystals (CuSO 4.5H 2 O) An explanation: Water of crystallisation is water chemically combined but not bonded to a host molecule Substances containing these crystals are said to be Hydrated those with no crystals of this sort are said to be Anhydrous Calc. the % water of crystallisation in Copper II Sulfate crystals (CuSO 4.5H 2 O) Solution: M r CuSO 4.5H 2 O : H = 1; O = 16 ; S = 32 ; Cu = 63.5 Remember to count the number of atoms *Just water Empirical Formulas Each Chemical Element is composed of atom Each element is represented by a symbol Some elements are present as small molecules (H 2 ) Chemical formulas tell us how many atoms are present in the element
Element Formula Hydrogen H 2 Chlorine Cl 2 Nitrogen N 2 Bromine Br 2 Oxygen O 2 Iodine I 2 Fluorine F 2 A compound is represented by a chemical formula e.g. A molecule of water is represented by H 2 O Called the Molecular Formula Definition The empirical formula of a compound indicates what elements are present in the compound and the simplest whole number ratio in which the atoms of these elements are present
Example 1 What is the empirical formula of Ethyne (C 2 H 2 ) Ratio is 2carbon:2hydrogen Simplest WNR is 1:1 Therefore empirical formula is CH Example 2 What is the empirical formula of Glucose(C 6 H 12 O 6 ) Ratio is 6 carbon:12 hydrogen: 6 Oxygen Simplest WNR is 1:2:1 Therefore empirical formula is CH 2 O Example 3 What is the empirical formula of Water (H 2 O) Ratio is 2 hydrogen:1 Oxygen Simplest WNR is 2:1 Therefore empirical formula is H 2 O Using an empirical formula Chemists needing to have a chemical analysed determine the empirical formula using an elemental analyser then using the empirical formula they can determine the molecular formula of the unknown chemical
Example of an Empirical formula from analytical data On analysis a compound is found to contain 68.85% carbon ; 4.92% hydrogen ; 26.23% Oxygen What is the empirical formula? Answer: Imagine we have 100g of the compound then the ratios in grams is 68.85: 4.92: 26.23 We divide by the molar mass (periodic table) of each element Carbon = 12 Hydrogen = 1 Oxygen = 16 68.85: 4.92: 26.23 We divide by the molar mass (Mr) 68.85 12: 4.92 1: 26.23 16 This gives us a relative atomic mass (Ar) ratio 5.74:4.92:1.64 Divide each by the lowest Ar which in this case is 1.64 5.74:4.92:1.64 Divide these by the lowest Ar
5.74 1.64 ; 4.92 1.64; 1.64 1.64 And we get the basis for an empirical formula for the unknown 3.5:3:1 The rule is it must be a whole number so! The empirical formula is 7:6:2 or C 7 H 6 O 2 Another Example A compound containing only C (52.17%); H (13.04%); O (34.79%) You are told The Mr is 92 What is the empirical formula? What is the Molecular formula? Remember C = 12, O = 16 and H = 1 Solution To answer this problem we need to under stand...if we have 100 grams of the compound then Carbon will make up 52.17 grams; Hydrogen, 13.04g and Oxygen, 34.79g Convert each of these masses to moles by dividing by the relative atomic mass to get a mole ratio If we then divide by the smallest mole ratio we can get a simplest whole number mole ratio
Remember Sometimes when we divide by the smallest mole ratio we do not get a whole number and it is necessary to multiply by a factor to get a whole number Compound contains C (65.11%);H (8.83%);O (26.06%) Find the empirical formula
Empirical formulas from combination data Example 0.72 g of Magnesium is heated in excess oxygen, 1.2g of Magnesium Oxide is formed what is the empirical formula of Magnesium Oxide Mass of Magnesium Oxide formed = 1.2g Mass of Oxygen consumed = 1.2 0.72 = 0.48g Moles of Magnesium atoms consumed = 0.72 24 = 0.03 Moles of Oxygen atoms consumed = 0.48 16 = 0.03
Ratio of Magnesium to Oxygen atoms = 0.03 : 0.03 = 1:1 Empirical Formula = MgO Empirical formulas from decomposition data 5.8g of an oxide of Iron was heated with carbon and 4.2g of iron was formed What is the empirical formula of the oxide? Mass of Iron in the compound = 4.2g Mass of Oxygen in the compound = 5.8-4.2g = 1.6 Moles of iron atoms in the compound = 4.2 56 = 0.075 Moles of oxygen atoms in the compound = 1.6 16 = 0.1 Ratio of Iron atoms to Oxygen atoms in the compound = 0.075 : 0.1 Whole numbers = 3:4 Empirical formula = Fe 3 O 4 Molecular Formulas Empirical formulas are used only for ionic compounds Covalent compounds are more complex (molecules) and so we use molecular formulae
Definition The Molecular formula of a compound indicates the actual number of atoms of an each element present in the molecule of the compound To find the molecular formula of a compound we need to know the empirical formula and the relative molecular mass Example Empirical formula for benzene is CH & Mr = 78. what s the molecular formula? Answer Formula mass = 12+1 = 13 Mr = 78 Number of CH units in a benzene molecule is = 78/13 = 6 Molecular formula is = C 6 H 6 Another Example Mr of propene = 42. The solution contains 85.7% C ; 14.3% H by mass. Molecular Formula? Answer Carbon = 85.7 12 = 7.14 Hydrogen = 14.3 1 = 14.3 Simplest ratio = 1:2 Therefore empirical formula = CH 2
Therefore empirical formula = CH 2 Formula Mass of CH 2 = 14 Relative Molecular Mass of propene = 42 Number of CH 2 units in a propene molecule = 42 / 14 = 3 Molecular formula of Propene = C 3 H 6 % composition by Mass If the empirical formula of a compound is known the % by mass of each element present can be calculated To calculate the percent (%) of each element in a compound Example: Calc the % composition by mass of each element in CaO (calcium oxide)
Example Calc the % composition by mass of Nitrogen in ([NH 4 ] 2 SO 4 ) Example Calc. the % water of crystallisation in Copper II Sulfate crystals (CuSO 4.5H 2 O) An explanation: Water of crystallisation is water chemically combined but not bonded to a host molecule. Substances containing these crystals are said to be Hydrated those with no crystals of this sort are said to be Anhydrous
Structural Formulas Molecular formulas of covalent compounds give the chemist a lot of detail about the compound and the molecules in it The structural formula shows the arrangement of the atoms within a molecule of the compound
IF THE STRUCTURAL FORMULA IS KNOWN THEN THE EMPIRICAL AND MOLECULAR FORMULAS ARE EASY Example Structural formula of ethene is
To get the Molecular formula, count the atoms C 2 H 4 To get the empirical formula just assess the simplest ratio CH 2 Chemical Equations Tells us what substances are in and what is produced during a chemical reaction The correct formula for each of the reactants and the products must be used and the equation must be balanced To balance an equation the formulas cannot be altered just multiplied Example Carbon reacts with oxygen to form carbon monoxide Balance the equation from this unbalanced one There is one carbon atom on the left There is one carbon atom on the right There are two oxygen atoms on the left There is one oxygen atom on the right The equation is unbalanced
We must check the equation and multiply the atoms to ensure that they balance out There are two carbon atoms on the left There are two carbon atoms on the right There are two oxygen atoms on the left There are two oxygen atoms on the right The equation is balanced Example Unbalanced equation There is one carbon atom on the left There is one carbon atom on the right There are four hydrogen atoms on the left There are two hydrogen atoms on the right There are two oxygen atoms on the left There are three oxygen atoms on the right The equation is unbalanced
Multiply the atoms until balanced The equation is short 2 oxygen atoms on the left and 2 hydrogen atoms on the right Calculations based on balanced equations A balanced equation gives the relative amounts of each of the reactants consumed and each of the products produced in a chemical reaction If the amount of any one of the reactants consumed is known then it is possible to calculate the amounts of the other reactants and products If the amount of methane consumed is known we can calculate the amounts of carbon dioxide and water produced. We can also calculate the amount of oxygen consumed The reverse is also true...if we know the amount of a product then we can calculate the others example Methane burns in air
If 2.5 moles of methane are reacted fully with oxygen calculate the number of moles of carbon dioxide and oxygen produced and the number of moles of oxygen consumed We know that 2.5 moles of methane were consumed Calculating Masses of Reactants or products from Balanced Chemical Equations Example Magnesium reacts with Oxygen to produce MgO The first step is to change the given quantity (gas or volume of gas at s.t.p.)to moles
To achieve this divide the given mass by the molar mass If we burn 9g of Mg in excess O 2 (enough to react all the Mg) what mass of MgO will be formed Solution First write down the equation and circle what we want to know Which is the same as :
Another way of achieving the same thing 9 g of Mg = 9 24 = 0.375 From the equation
USE YOUR MOLE MAP Calculations of volumes of gaseous reactants or products from balanced chemical equations example Propane burns in air according to the equation What volume of O 2 (measured at s.t.p.) is needed for complete combustion of 11g of propane
Answer: 11g of C 3 H 8 = 11/44 moles = 0.25 Another type of calculation A solution of NaOH is reacted with enough H 2 SO 4 Solution to easily neutralise it. The equation for the reaction is On evaporation of the water 284g of Sodium Sulfate are obtained. Calculate : A. The number of moles of H 2 SO 4 acid consumed B. The number of water molecules formed
C. The mass of the NaOH used to make up the solution Answer: A. The number of moles of H 2 SO 4 acid consumed 284g NaSO4 = 284 142 moles = 2 From the equation Answer: B. The number of water molecules formed B. Avagadros number = 6 x 10 23 From the equation
Answer: c. The mass of the NaOH used to make up the solution c. Relative Molecular mass (Mr) From the equation Calculations involving excess of one reactant Sometimes when a chemical reaction occurs there is excess of one of the reactants. If the balanced equation is known and the initial quantities in the reaction are known then it is possible to identify the chemical that is in excess. The substance that is not present in excess is called the limiting reactant as it is the amount of this substance that will dictate how much product will be produced Example Zinc reacts with sulphuric acid according to the equation
A 250 cm 3 aqueous solution containing 9.8g sulfuric acid is added to 13g Zinc (a) Show that zinc is present in excess (b) Calcuate the mass of Zinc Sulfate formed (c) Calculate the volume of hydrogen gas (measured at s.t.p) formed Answer: a : Show that zinc is present in excess Moles of zinc present initially = 13 65 = 0.2 Moles of Sulfuric acid present initially = 9.8 98 = 0.1 Answer: b : Calcuate the mass of Zinc Sulfate formed Sulfuric acid is the limiting reactant
Answer: c : Calculate the volume of hydrogen gas (measured at s.t.p) formed Percentage Yields When a real reaction occurs the amounts of product isolated are often less than those calculated. This may be due to the reaction been reversible, or to some of the products reacting further to form yet other products or to loss of product during purification
Percentage yield accounts for these factors Percentage Yields Example 10.2 g of ethanol (C 2 H 5 OH)were heated with aluminium Oxide and 1.7g ethene (C 2 H 4 ) were formed Calculate the percentage yield of ethene 10.2 g of C 2 H 5 OH = 10.2 46 moles = 0.22