Stoichiometry: Calculations with Chemical Formulas and Equations

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

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

___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

___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

___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) ½ O2(g)  3 CO2(g) H2O(l) Chapter 3

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

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 = x g Chapter 3

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

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: % 12C, % 13C. Average mass of C: ( )(12 amu) + ( )(13 amu) = amu Chapter 3

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

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

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

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

Percentage Composition from Formulas Chapter 3

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

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( 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( 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 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. 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

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

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

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

Molecular Formula from Empirical Formula Empirical formula: CH Empirical formula weight: g/mol Molecular weight: g/mol Chapter 3

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 = mol S  27.8g/32.066g/mol = mol Cl  61.7g/35.453g/mol = 1.74 mol Chapter 3

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 = mol S  27.8g/32.066g/mol = mol Cl  61.7g/35.453g/mol = 1.74 mol Chapter 3

Combustion Analysis Chapter 3

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

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

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

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

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

Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A g sample of menthol is combusted, producing g of CO2 and g of H2O. What is the empirical formula for menthol? Now we can determine the empirical formula Moles of elements: C  g/12.011g/mol = mol H  g/1.01g/mol = mol O  g/16.00g/mol = 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) O2(g)  8 CO2(g) 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) O2(g)  8 CO2(g) 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 g/mol Chapter 3

2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g)
Quantitative Information 2 C4H10(l) O2(g)  8 CO2(g) 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) O2(g)  8 CO2(g) 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) O2(g)  8 CO2(g) 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) O2(g)  8 CO2(g) 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: g/mol Chapter 3

Limiting Reactants “What runs out first”
2 C8H O2  16 CO H2O Chapter 3 4 4 4 4

2 C8H O2  16 CO 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

2 C8H O2  16 CO 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

2 C8H O2  16 CO 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

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

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) g/mol FW(O2) = 32.00g/mol Chapter 3 4 4 4 4

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

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

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

Stoichiometry: Calculations with Chemical Formulas and Equations

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