Stoichiometry: Calculations with Chemical Formulas and Equations Chapter 3 Chapter 3 1 1 1 1

Chemical Equations 2H2(g) + O2(g)  2H2O(g) The materials you start with are called Reactants. The materials you make are called Products. The numbers in front of the compounds (H2 and H2O) are called coefficients. Coefficients are multipliers, in this equation 2 in front of the H2 indicates that there are 2 molecules of H2 in the equation. Chapter 3

Chemical Equations 2H2(g) + O2(g)  2H2O(g) Notice that the number of hydrogen atoms and oxygen atoms on the reactant side and the product side is equal. Law of Conservation of Mass Matter cannot be created or lost in any chemical reaction. Chapter 3

___NH4NO3(s)  ___N2O(g) + ___H2O(g) Chemical Equations Balancing Chemical Reactions ___NH4NO3(s)  ___N2O(g) + ___H2O(g) Chapter 3

___NH4NO3(s)  ___N2O(g) + ___H2O(g) Chemical Equations Balancing Chemical Reactions ___NH4NO3(s)  ___N2O(g) + ___H2O(g) Reactants Products N 2 H 4 O 3 Chapter 3

___NH4NO3(s)  ___N2O(g) + _2_H2O(g) Chemical Equations Balancing Chemical Reactions ___NH4NO3(s)  ___N2O(g) + _2_H2O(g) Reactants Products N 2 H 4 2 4 O 3 2 3 Chapter 3

Chemical Equations Balancing Chemical Reactions ___Mg3N2(s) + ___H2O(l)  ___Mg(OH)2(s) + ___NH3(aq) Chapter 3

Chemical Equations Balancing Chemical Reactions ___Mg3N2(s) + ___H2O(l)  ___Mg(OH)2(s) + ___NH3(aq) Reactants Products Mg 3 1 N 2 H 5 O Chapter 3

Chemical Equations Balancing Chemical Reactions ___Mg3N2(s) + ___H2O(l)  _3_Mg(OH)2(s) + ___NH3(aq) Reactants Products Mg 3 1 3 N 2 1 H 5 9 O 2 6 Chapter 3

Chemical Equations Balancing Chemical Reactions ___Mg3N2(s) + ___H2O(l)  _3_Mg(OH)2(s) + _2_NH3(aq) Reactants Products Mg 3 1 3 N 2 1 2 H 5 9 12 O 1 2 6 Chapter 3

Chemical Equations Balancing Chemical Reactions ___Mg3N2(s) + _6_H2O(l)  _3_Mg(OH)2(s) + _2_NH3(aq) Reactants Products Mg 3 1 3 N 2 1 2 H 2 12 5 9 12 O 1 6 2 6 Chapter 3

C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) Patterns of Chemical Reactivity Combustion Reactions This is the combination of a substance with oxygen: C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) Combustion of hydrocarbons yield CO2 and H2O 4 Fe(s) + 3O2(g)  2 Fe2O3(s) Chapter 3

Patterns of Chemical Reactivity Combustion Reactions This is the combination of a substance with oxygen: C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) Less oxygen is required if the reactant has oxygen. C3H7OH(l) + 4 ½ O2(g)  3 CO2(g) + 4 H2O(l) Chapter 3

Patterns of Chemical Reactivity Combination and Decomposition Reactions Combination Reactions Two or more substances react to form one product. CaO(s) + H2O(l)  Ca(OH)2(s) Chapter 3

Patterns of Chemical Reactivity Combination and Decomposition Reactions Decomposition Reactions One substance reacts to form two or more substances. CaCO3(s)  CaO(s) + CO2(g) Chapter 3

Atomic and Molecular Weights The Atomic Mass Scale The mass of individual atoms is measured in atomic mass units (amu). 1 amu = 1.66054 x 10-24 g Chapter 3

Atomic and Molecular Weights The Atomic Mass Scale The amu is derived from carbon-12 (12C). The mass of one 12C is exactly 12 amu. Chapter 3

Atomic and Molecular Weights Average Atomic Mass The elements are actually a mixture of isotopes. The atomic mass shown in the periodic table is a weighted average of the various isotopes. Naturally occurring C: 98.892 % 12C, 1.108 % 13C. Average mass of C: (0.98892)(12 amu) + (0.01108)(13 amu) = 12.011 amu Chapter 3

Atomic and Molecular Weights Average Atomic Mass Atomic weight (AW) The average atomic mass (atomic weight) of an element. Atomic weights are listed on the periodic table. Chapter 3

Atomic and Molecular Weights Formula and Molecular Weights Formula weight (FW) The sum of the atomic weights of each atom in the chemical formula. Example: CO2 Chapter 3

Atomic and Molecular Weights Formula and Molecular Weights Formula weight (FW) The sum of the atomic weights of each atom in the chemical formula. Example: CO2 Formula Weight = 1(AW, carbon) + 2(AW, oxygen) Formula Weight = 1(12.011amu) + 2(16.0amu) Formula Weight = 44.0 amu Chapter 3

Atomic and Molecular Weights Formula and Molecular Weights Molecular weight The sum of the atomic weights of each atom in the molecular formula. Formula weight is the general term, molecule weight refers specifically to molecular compounds. Chapter 3

Atomic and Molecular Weights Percentage Composition from Formulas Chapter 3

Atomic and Molecular Weights Percentage Composition from Formulas Example: Calculate the percent oxygen in CH3CH2OH. Formula weight ethanol: 2(12.01amu) + 6(1.01amu) + 1(16.00amu) = 46.08amu Mass of oxygen: 1(16.00amu) = 16.00amu Chapter 3

Atomic and Molecular Weights Percentage Composition from Formulas Example: Calculate the percent oxygen in CH3CH2OH. Formula weight ethanol: 2(12.01amu) + 6(1.01amu) + 1(16.00amu) = 46.08amu Mass of oxygen: 1(16.00amu) = 16.00amu % oxygen Chapter 3

Atomic and Molecular Weights Percentage Composition from Formulas Example: Calculate the percent oxygen in CH3CH2OH. Formula weight ethanol: 2(12.01amu) + 6(1.01amu) + 1(16.00amu) = 46.08amu Mass of oxygen: 1(16.00amu) = 16.00amu % oxygen Chapter 3

The Mole That amount of matter that contains as many objects as exactly 12g of 12C. 12g 12C gives 6.02 x 1023 atoms of 12C 6.02 x 1023 is Avogadro’s Number Chapter 3

Molar Mass Mass in grams of 1 mole of substance (g/mol). This is determined by the sum of atomic weights (in grams) of the atoms in the formula of the molecule. Example: CO2 1(12.011 g/mol)+2(16.00g/mol) = Chapter 3

Molar Mass Mass in grams of 1 mole of substance (g/mol). This is determined by the sum of atomic weights (in grams) of the atoms in the formula of the molecule. Example: CO2 1(12.011 g/mol)+2(16.00g/mol) = 44.01g/mol Chapter 3

The Mole Interconverting Masses, Moles, and Numbers of Particles Chapter 3

The Mole Moles « Numbers of Particles Chapter 3

The Mole Mass  Moles Chapter 3

The Mole Moles  Mass Chapter 3

The Mole A sample of female sex hormone, estradiol, C18H24O2, contains 3.0 x 1020 atoms of hydrogen. How many atoms of carbon does it contain? Chapter 3

The Mole A sample of female sex hormone, estradiol, C18H24O2, contains 3.0 x 1020 atoms of hydrogen. How many atoms of carbon does it contain? Chapter 3

The Mole A sample of female sex hormone, estradiol, C18H24O2, contains 3.0 x 1020 atoms of hydrogen. How many atoms of carbon does it contain? Chapter 3

The Mole A sample of female sex hormone, estradiol, C18H24O2, contains 3.0 x 1020 atoms of hydrogen. How many atoms of carbon does it contain? Chapter 3

The Mole A sample of female sex hormone, estradiol, C18H24O2, contains 3.0 x 1020 atoms of hydrogen. How many atoms of carbon does it contain? Chapter 3

The Mole A sample of female sex hormone, estradiol, C18H24O2, contains 3.0 x 1020 atoms of hydrogen. How many molecules of estradiol does it contain? Chapter 3

The Mole A sample of female sex hormone, estradiol, C18H24O2, contains 3.0 x 1020 atoms of hydrogen. How many molecules of estradiol does it contain? Chapter 3

The Mole A sample of female sex hormone, estradiol, C18H24O2, contains 3.0 x 1020 atoms of hydrogen. How many molecules of estradiol does it contain? Chapter 3

The Mole A sample of female sex hormone, estradiol, C18H24O2, contains 3.0 x 1020 atoms of hydrogen. How many molecules of estradiol does it contain? Chapter 3

The Mole A sample of female sex hormone, estradiol, C18H24O2, contains 3.0 x 1020 atoms of hydrogen. How many molecules of estradiol does it contain? Chapter 3

The Mole A sample of female sex hormone, estradiol, C18H24O2, contains 3.0 x 1020 atoms of hydrogen. How many moles of estradiol does it contain? Since we know the number of molecules, we can calculate the number of moles Chapter 3

The Mole A sample of female sex hormone, estradiol, C18H24O2, contains 3.0 x 1020 atoms of hydrogen. How many moles of estradiol does it contain? Since we know the number of molecules, we can calculate the number of moles Chapter 3

The Mole A sample of female sex hormone, estradiol, C18H24O2, contains 3.0 x 1020 atoms of hydrogen. How many moles of estradiol does it contain? Since we know the number of molecules, we can calculate the number of moles Chapter 3

The Mole A sample of female sex hormone, estradiol, C18H24O2, contains 3.0 x 1020 atoms of hydrogen. How many moles of estradiol does it contain? Since we know the number of molecules, we can calculate the number of moles Chapter 3

The Mole A sample of female sex hormone, estradiol, C18H24O2, contains 3.0 x 1020 atoms of hydrogen. What is the mass of this sample in grams? Now we will convert the moles to grams Chapter 3

The Mole A sample of female sex hormone, estradiol, C18H24O2, contains 3.0 x 1020 atoms of hydrogen. What is the mass of this sample in grams? Now we will convert the moles to grams Chapter 3

The Mole A sample of female sex hormone, estradiol, C18H24O2, contains 3.0 x 1020 atoms of hydrogen. What is the mass of this sample in grams? Now we will convert the moles to grams Chapter 3

The Mole A sample of female sex hormone, estradiol, C18H24O2, contains 3.0 x 1020 atoms of hydrogen. What is the mass of this sample in grams? Now we will convert the moles to grams Chapter 3

The Mole A sample of female sex hormone, estradiol, C18H24O2, contains 3.0 x 1020 atoms of hydrogen. What is the mass of this sample in grams? Now we will convert the moles to grams Chapter 3

The Mole A sample of female sex hormone, estradiol, C18H24O2, contains 3.0 x 1020 atoms of hydrogen. What is the mass of this sample in grams? Now we will convert the moles to grams Chapter 3

Empirical Formulas from Analyses Chapter 3

Empirical Formulas from Analyses Analysis Hg  73.9% Cl  26.1% assume 100g sample Hg  73.9 g Cl  26.1g - convert grams to moles Hg  73.9g / 200.59g/mol = 0.367 mol Cl  26.1g/ 35.45g/mol = 0.736 mol determine the empirical formula by using the moles of the elements to get the smallest whole number ratio of the elements. Chapter 3

Empirical Formulas from Analyses Analysis Hg  73.9% Cl  26.1% assume 100g sample Hg  73.9 g Cl  26.1g - convert grams to moles Hg  73.9g / 200.59g/mol = 0.367 mol Cl  26.1g/ 35.45g/mol = 0.736 mol determine the empirical formula by using the moles of the elements to get the smallest whole number ratio of the elements. Chapter 3

Empirical Formulas from Analyses Analysis Hg  73.9% Cl  26.1% assume 100g sample Hg  73.9 g Cl  26.1g convert grams to moles Hg  73.9g / 200.59g/mol = 0.367 mol Cl  26.1g/ 35.45g/mol = 0.736 mol determine the empirical formula by using the moles of the elements to get the smallest whole number ratio of the elements. Chapter 3

Empirical Formulas from Analyses Analysis Hg  73.9% Cl  26.1% assume 100g sample Hg  73.9 g Cl  26.1g convert grams to moles Hg  73.9g / 200.59g/mol = 0.367 mol Cl  26.1g/ 35.45g/mol = 0.736 mol determine the empirical formula by using the moles of the elements to get the smallest whole number ratio of the elements. Chapter 3

Empirical Formulas from Analyses Molecular Formula from Empirical Formula To determine the molecular formula from an empirical formula, you must have the molecular weight of the substance. Chapter 3

Empirical Formulas from Analyses Molecular Formula from Empirical Formula To determine the molecular formula from an empirical formula, you must have the molecular weight of the substance. Chapter 3

Empirical Formulas from Analyses Molecular Formula from Empirical Formula To determine the molecular formula from an empirical formula, you must have the molecular weight of the substance. Chapter 3

Empirical Formulas from Analyses Molecular Formula from Empirical Formula Empirical formula: CH Empirical formula weight: 13.019 g/mol Molecular weight: 78.114g/mol Chapter 3

Empirical Formulas from Analyses Molecular Formula from Empirical Formula Empirical formula: CH Empirical formula weight: 13.019 g/mol Molecular weight: 78.114g/mol Chapter 3

Empirical Formulas from Analyses Molecular Formula from Empirical Formula Empirical formula: CH Empirical formula weight: 13.019 g/mol Molecular weight: 78.114g/mol Chapter 3

Empirical Formulas from Analyses Molecular Formula from Empirical Formula Empirical formula: CH Empirical formula weight: 13.019 g/mol Molecular weight: 78.114g/mol Chapter 3

Empirical Formulas from Analyses Molecular Formula from Empirical Formula Empirical formula: CH Empirical formula weight: 13.019 g/mol Molecular weight: 78.114g/mol Chapter 3

Empirical Formulas from Analyses Determine the empirical formula of the compound with the following compositions by mass: C, 10.4%; S, 27.8%, Cl, 61.7%. Assume a 100g sample, so C  10.4% 10.4g S  27.8%  27.8g Cl  61.7%  61.7g Moles of each element C  10.4g/12.011g/mol = 0.866 mol S  27.8g/32.066g/mol = 0.867 mol Cl  61.7g/35.453g/mol = 1.74 mol Chapter 3

Empirical Formulas from Analyses Determine the empirical formula of the compound with the following compositions by mass: C, 10.4%; S, 27.8%, Cl, 61.7%. Moles of each element C  10.4g/12.011g/mol = 0.866 mol S  27.8g/32.066g/mol = 0.867 mol Cl  61.7g/35.453g/mol = 1.74 mol Chapter 3

Empirical Formulas from Analyses Determine the empirical formula of the compound with the following compositions by mass: C, 10.4%; S, 27.8%, Cl, 61.7%. Moles of each element C  10.4g/12.011g/mol = 0.866 mol S  27.8g/32.066g/mol = 0.867 mol Cl  61.7g/35.453g/mol = 1.74 mol Chapter 3

Empirical Formulas from Analyses Determine the empirical formula of the compound with the following compositions by mass: C, 10.4%; S, 27.8%, Cl, 61.7%. Moles of each element C  10.4g/12.011g/mol = 0.866 mol S  27.8g/32.066g/mol = 0.867 mol Cl  61.7g/35.453g/mol = 1.74 mol Chapter 3

Empirical Formulas from Analyses Combustion Analysis Chapter 3

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What is the empirical formula for menthol? Mass of Carbon mass CO2  moles CO2  moles C  grams C Chapter 3

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What is the empirical formula for menthol? Mass of Carbon mass CO2  moles CO2  moles C  grams C Chapter 3

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What is the empirical formula for menthol? Mass of Carbon mass CO2  moles CO2  moles C  grams C Chapter 3

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What is the empirical formula for menthol? Mass of Carbon mass CO2  moles CO2  moles C  grams C Chapter 3

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What is the empirical formula for menthol? Mass of Carbon mass CO2  moles CO2  moles C  grams C Chapter 3

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What is the empirical formula for menthol? Mass of Hydrogen mass H2O  moles H2O  moles H  grams H Chapter 3

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What is the empirical formula for menthol? Mass of Hydrogen mass H2O  moles H2O  moles H  grams H Chapter 3

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What is the empirical formula for menthol? Mass of Hydrogen mass H2O  moles H2O  moles H  grams H Chapter 3

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What is the empirical formula for menthol? Mass of Hydrogen mass H2O  moles H2O  moles H  grams H Chapter 3

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What is the empirical formula for menthol? Mass of Hydrogen mass H2O  moles H2O  moles H  grams H Chapter 3

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What is the empirical formula for menthol? Mass of Oxygen mass O = mass of sample – (mass C +mass H) Chapter 3

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What is the empirical formula for menthol? Mass of Oxygen mass O = mass of sample – (mass C +mass H) Chapter 3

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What is the empirical formula for menthol? Now we can determine the empirical formula Mass of elements: C  0.07721g H  0.01299g O  0.01030g Chapter 3

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What is the empirical formula for menthol? Now we can determine the empirical formula Moles of elements: C  0.07721g/12.011g/mol = H  0.01299g/1.01g/mol = O  0.01030g/16.00g/mol = Chapter 3

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What is the empirical formula for menthol? Now we can determine the empirical formula Moles of elements: C  0.07721g/12.011g/mol = 0.006428 mol H  0.01299g/1.01g/mol = 0.01286 mol O  0.01030g/16.00g/mol = 0.0006438 mol Chapter 3

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What is the empirical formula for menthol? Now we can determine the empirical formula Moles of elements: C  0.07721g/12.011g/mol = 0.006428 mol H  0.01299g/1.01g/mol = 0.01286 mol O  0.01030g/16.00g/mol = 0.0006438mol Chapter 3

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What is the empirical formula for menthol? Now we can determine the empirical formula Moles of elements: C  0.07721g/12.011g/mol = 0.006428 mol H  0.01299g/1.01g/mol = 0.01286 mol O  0.01030g/16.00g/mol = 0.0006438 mol Chapter 3

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What is the empirical formula for menthol? Now we can determine the empirical formula Moles of elements: C  0.07721g/12.011g/mol = 0.006428 mol H  0.01299g/1.01g/mol = 0.01286 mol O  0.01030g/16.00g/mol = 0.0006438 mol Chapter 3

Quantitative Information The coefficients in a balanced equation represent both the number of molecules and the number of moles in a reaction. The coefficients can also be used to derive ratios between any two substances in the chemical reaction. 2 H2 : 1 O2 2 H2 : 2 H2O 1 O2 : 2 H2O The ratios can be used to predict - The amount of product formed - The amount of reactant needed Chapter 3

Quantitative Information Chapter 3

2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g) Quantitative Information 2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g) How many grams of CO2 are formed if 1.00g of butane (C4H10) is allowed to react with excess oxygen? Chapter 3

2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g) Quantitative Information 2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g) How many grams of CO2 are formed if 1.00g of butane (C4H10) is allowed to react with excess oxygen? Moles of C4H10 F.W. 58.124g/mol Chapter 3

2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g) Quantitative Information 2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g) How many grams of CO2 are formed if 1.00g of butane (C4H10) is allowed to react with excess oxygen? Moles of C4H10 F.W. 58.124g/mol Chapter 3

2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g) Quantitative Information 2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g) How many grams of CO2 are formed if 1.00g of butane (C4H10) is allowed to react with excess oxygen? Ratio of C4H10:CO2 2 C4H10 : 8 CO2 or Chapter 3

2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g) Quantitative Information 2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g) How many grams of CO2 are formed if 1.00g of butane (C4H10) is allowed to react with excess oxygen? Set-up ratio and proportion between known and unknown quantities Chapter 3

2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g) Quantitative Information 2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g) How many grams of CO2 are formed if 1.00g of butane (C4H10) is allowed to react with excess oxygen? Set-up ratio and proportion between known and unknown quantities Chapter 3

2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g) Quantitative Information 2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g) How many grams of CO2 are formed if 1.00g of butane (C4H10) is allowed to react with excess oxygen? Set-up ratio and proportion between known and unknown quantities Chapter 3

2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g) Quantitative Information 2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g) How many grams of CO2 are formed if 1.00g of butane (C4H10) is allowed to react with excess oxygen? Convert the moles of unknown substance into the desired units Chapter 3

2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g) Quantitative Information 2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g) How many grams of CO2 are formed if 1.00g of butane (C4H10) is allowed to react with excess oxygen? Convert the moles of unknown substance into the desired units FW of CO2: 44.011g/mol Chapter 3

2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g) Quantitative Information 2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g) How many grams of CO2 are formed if 1.00g of butane (C4H10) is allowed to react with excess oxygen? Convert the moles of unknown substance into the desired units FW of CO2: 44.011g/mol Chapter 3

2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g) Quantitative Information 2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g) How many grams of CO2 are formed if 1.00g of butane (C4H10) is allowed to react with excess oxygen? Convert the moles of unknown substance into the desired units FW of CO2: 44.011g/mol Chapter 3

Limiting Reactants “What runs out first” 2 C8H18 + 25 O2  16 CO2 + 18 H2O Chapter 3 4 4 4 4

Limiting Reactants “What runs out first” 2 C8H18 + 25 O2  16 CO2 + 18 H2O If you have 2 moles of C8H18 and 20 moles of O2 all the O2 will be used and the reaction will stop Chapter 3 4 4 4 4

Limiting Reactants “What runs out first” 2 C8H18 + 25 O2  16 CO2 + 18 H2O If you have 2 moles of C8H18 and 20 moles of O2 all the O2 will be used and the reaction will stop O2 is call the limiting reagent (reactant) Chapter 3 4 4 4 4

Limiting Reactants “What runs out first” 2 C8H18 + 25 O2  16 CO2 + 18 H2O If you have 2 moles of C8H18 and 20 moles of O2 all the O2 will be used and the reaction will stop O2 is call the limiting reagent (reactant) Limiting Reagent – The reagent present in the smallest stoichiometric quantity in a mixture of reactants. Chapter 3 4 4 4 4

Limiting Reactants Example 2 C8H18 + 25 O2  16 CO2 + 18 H2O Determine the limiting reagent of this reaction if 10.0 grams of C8H18 and 25.0 grams of O2 are allowed to react. Chapter 3 4 4 4 4

Limiting Reactants Example 2 C8H18 + 25 O2  16 CO2 + 18 H2O Determine the limiting reagent of this reaction if 10.0 grams of C8H18 and 25.0 grams of O2 are allowed to react. Convert grams to moles Chapter 3 4 4 4 4

Limiting Reactants Example 2 C8H18 + 25 O2  16 CO2 + 18 H2O Determine the limiting reagent of this reaction if 10.0 grams of C8H18 and 25.0 grams of O2 are allowed to react. Convert grams to moles FW(C8H18) 114.268g/mol FW(O2) = 32.00g/mol Chapter 3 4 4 4 4

Limiting Reactants Example 2 C8H18 + 25 O2  16 CO2 + 18 H2O Determine the limiting reagent of this reaction if 10.0 grams of C8H18 and 25.0 grams of O2 are allowed to react. Convert grams to moles FW(C8H18) 114.268g/mol FW(O2) = 32.00g/mol Chapter 3 4 4 4 4

Limiting Reactants Example 2 C8H18 + 25 O2  16 CO2 + 18 H2O Determine the limiting reagent of this reaction if 10.0 grams of C8H18 and 25.0 grams of O2 are allowed to react. Convert grams to moles FW(C8H18) 114.268g/mol FW(O2) = 32.00g/mol Chapter 3 4 4 4 4

Limiting Reactants Example 2 C8H18 + 25 O2  16 CO2 + 18 H2O Determine the limiting reagent of this reaction if 10.0 grams of C8H18 and 25.0 grams of O2 are allowed to react. Divide each reagent by its own coefficient Chapter 3 4 4 4 4

Limiting Reactants Example 2 C8H18 + 25 O2  16 CO2 + 18 H2O Determine the limiting reagent of this reaction if 10.0 grams of C8H18 and 25.0 grams of O2 are allowed to react. Divide each reagent by its own coefficient Chapter 3 4 4 4 4

Limiting Reactants Example 2 C8H18 + 25 O2  16 CO2 + 18 H2O Determine the limiting reagent of this reaction if 10.0 grams of C8H18 and 25.0 grams of O2 are allowed to react. The substance with the smallest calculated value will be the limiting reagent. In this case, O2 is the limiting reagent. Chapter 3 4 4 4 4

Limiting Reactants Theoretical Yields - The calculated amount of product based on the limiting reagent (Theoretical yield). Chapter 3

Limiting Reactants Theoretical Yields - The calculated amount of product based on the limiting reagent (Theoretical yield). Chapter 3

Practice Problems 3.6, 3.16, 3.18, 3.32, 3.40, 3.48, 3.60, 3.72 Chapter 3