Chapter 9 – Quantitative Relationships in Chemistry Stoichiometry - is a branch of chemistry that deals with the relative quantities of reactants and products in chemical reactions. 9.1 Mole and Molar Mass of Elements Mole (abr: mol) – The amount of matter that contains as many objects (atoms, molecules or ions) as the number of atoms in exactly 12 g of 12 C. Conversion Factor: 1 mol = 6.022 x 10 23 (Avogadro’s number) Example: 1 mol 12 C atoms = 6.022 x 10 23 12 C atoms 1 mol H 2 O molecules = 6.022 x 10 23 H 2 O molecules 1 mol NO 3 – ions = 6.022 x 10 23 NO 3 – ions 1 mol NaCl units = 6.022 x 10 23 NaCl units 1. How many moles are present in 5.23 x 10 24 F – ions? 2. Determine the number of moles in 3.02 x 10 22 CaCl 2 . 3. Calculate the number of particles in 0.0463 mol of NH 3 .
Transcript
Chapter 9 – Quantitative Relationships in Chemistry
Stoichiometry - is a branch of chemistry that deals with the relative quantities of reactants and products in chemical reactions.
9.1 Mole and Molar Mass of Elements
Mole (abr: mol) – The amount of matter that contains as many objects (atoms, molecules or ions) as the number of atoms in exactly 12 g of 12C.
One NaCl unit weighs 58 amu ----------> 1 mol NaCl weighs 58 g
Formula Weight (FW) or Formula Mass (FM) – The sum of the atomic weights of each atom in a chemical formula (ionic formula)
1. NaCl (Na = 23, Cl = 35)
2. KBr (K = 39, Br = 80)
3. CaCO3 (Ca = 40, C = 12, O = 16)
4. ZnCrO4 (Zn = 65, Cr = 52, O = 16)
5. Al(ClO3)3 (Al = 27, Cl = 35, O = 16)
6. Li3PO4 (Li = 7, P = 31, O = 16)
7. Fe2(SO4)3 (Fe = 56, S = 32, O = 16)
8. (NH4)2S (N = 14, H = 1, S = 32)
9. Sb2(Cr2O7)3 (Sb = 122, Cr = 52, O = 16)
10. Pb(HCO3)4 (Pb = 207, H = 1, C = 12, O = 16)
Molecular Weight (MW) or Molecular Mass (MM) – The sum of the atomic weights of each atom in a molecular formula
1. C6H12O6 (C = 12, H = 1, O = 16)
2. N2O5 (N = 14, O = 16)
3. P4O10 (P = 31, O = 16)
4. CCl4 (C = 12, Cl = 35)
5. NH3 (N = 14, H = 1)
6. C2H6O (C = 12, H = 1, O =16)
7. Si2Br6 (Si = 28, Br = 80)
8. (NH4)2CO (N = 14, H = 1, C = 12, O = 16)
9. C8H10N4O2 (C = 12, H = 1, N = 14, O = 16)
10. BF3 (B = 11, F = 19)
9.3 Mole and Mass Composition of Compounds Interconverting Masses, Moles and Number of Particles
mass massmol = ---------------- OR n = -------------
Molar Mass MM
1. Calculate the number of moles of glucose, C6H12O6, in 5.380 g of this substance? (C = 12, H = 1, O = 16)
2. How many moles of sodium bicarbonate, NaHCO3, are present in 508 g of this substance? (Na = 23, H = 1, C = 12, O = 16)
3. Calculate the mass, in grams, of 0.433 mol of calcium nitrate, Ca(NO3)2.(Ca = 40, N = 14, O = 16)
4. What is the mass in grams of 3.00 x 10 –2 mol of sulfuric acid, H2SO4?(H = 1, S = 32, O = 16)
5. A. How many glucose molecules are in 5.23 g of glucose, C6H12O6? (C = 12, H = 1, O = 16)
B. How many hydrogen atoms are present in the same sample of glucose?
6. A. How many nitric acid molecules, HNO3, are in 4.20 g of this compound?(H = 1, N = 14, O = 16)
B. How many O atoms are present in this acid?
9.4 Percent Composition of Compounds
Percentage Composition by Mass
Total Atomic Weight of Element
% Element = ------------------------------------------------------- X 100Total Formula Weight or Molecular Weight
1. Calculate the percentage by mass of nitrogen in calcium nitrate, Ca(NO3)2.(Ca = 40, N = 14, O = 16)
2. What is the percentage of oxygen in sucrose, C12H22O11? (C = 12, H = 1, O = 16)
3. Determine the percentage of chromium in chromic sulfate, Cr2(SO4)3.(Cr = 52, S = 32, O =16)
Percentage Composition by Mass
Total Atomic Weight of Element
% Element = ------------------------------------------------------- X 100Total Formula Weight or Molecular Weight
4. What is the percentage by mass of water in cupric sulfate pentahydrate, CuSO4l5H2O?
(Cu = 64, S = 32, O = 16, H = 1)
5. What are the percentages of all the elements in the compound ferric dichromate, Fe2(Cr2O7)3? (Fe = 56, Cr = 52, O = 16)
6. Compute for the percentages by mass of all the elements in the compound barium acetate, Ba(C2H3O2)2? (Ba = 137, C = 12, H = 1, O =16)
9.5 Empirical and Molecular Formulas
Empirical Formula (simplest formula) – A chemical formula that shows the kinds of atoms and their relative numbers in a substance.
Ex. H2O, NaCl, HgCl2, CH4, CH, etc.
Steps in determining the empirical formula of a compound:
1. If the given elements are given in mass percent, assume 100 g of the sample. Convert the mass percent into grams simply by replacing the percent sign with the unit grams (g).
2. Compute for the number of moles of each element.3. Establish the mole ratio of the elements.4. Divide the mole ratio with the smallest value.5. Round off to the nearest whole number ratio if applicable. If the decimal place is
too ambiguous, use the following multipliers to remove the uncertainties.
0.1 – Round off 0.6 x 50.2 x 5 0.7 x 30.3 x 3 0.8 x 50.4 x 5 0.9 – Round off0.5 x 2 0.0 – Round off
6. Use the whole number ratio as subscripts of the empirical formula.
Examples:
1. Ascorbic acid (Vitamin C) contains 40.92% C, 4.58% H, and 54.50% O by mass. What is the empirical formula ascorbic acid? (C = 12, H = 1, O = 16)
2. A compound called methyl benzoate has the following percentage composition by mass: 70.57% C, 5.93% H and 23.50% O. Determine the empirical formula of methyl benzoate. (C = 12, H = 1, O = 16)
3. An unknown compound contains 10.5% C, 27.8% S and 61.7% Cl. What is the empirical formula of the compound? (C = 12, S = 32, Cl = 35)
4. A compound has 21.7% C, 9.6% O and 68.7% F. What is the empirical formula of the compound? (C = 12, O = 16, F = 19)
Molecular Formula – A chemical formula that indicates the actual number of atoms of each element in one molecule of a substance.
Ex. H2O2, C2H6, C6H12O6, etc
Steps in determining the molecular formula of a compound:
1. Determine the empirical formula of the compound.2. Compute for the empirical weight of the compound.3. Use the formula below to get the factor used to multiply the subscripts of the
4. Multiply the factor with the subscripts of the empirical formula to get the molecular formula.
Examples:
1. Ethylene glycol, the substance used in automobile antifreeze, is composed of 38.7% C, 9.7% H and 51.6% O by mass. The molar mass of the compound is 62.1 g / mol. Determine (a) the empirical formula and (b) the molecular formula of ethylene glycol. (C = 12, H = 1, O = 16)
2. Determine the empirical and molecular formula of ibuprofen, a headache remedy. The compound contains 75.69% C, 8.80% H and 15.51% O by mass and the molar mass of the compound is about 206 g / mol. (C = 12, H = 1, O =16)
3. What is the empirical and molecular formula of epinephrine (adrenaline), a hormone secreted into the bloodstream in times of danger or stress? The compound contains 59.0% C, 7.1% H, 26.2% O and 7.7% N by mass and the molar mass is about 180 g / mol. (C = 12, H = 1, O = 16, N = 14)
9.6 Stoichiometry
Quantitative Information from Balanced Equations
The coefficients in a balanced chemical equation can be interpreted both as the relative numbers of molecules (or formula units) involved in the reaction and as the relative number of moles.
2 H2 + O2 ----------> 2 H2O
2 molecules 1 molecule 2 molecules
2(6.022 x 10 23 molecules) 6.022 x 10 23 molecules 2(6.022 x 10 23 molecules)
2 mol 1 mol 2 mol
Stoichiometrically Equivalent Quantities
2 mol H2 1 mol O2 2 mol H2O
– “stoichiometrically equivalent to” = These quantities can be used to give conversion factors for relating reactants and products in a chemical reaction.
Examples:
1. How many moles of water will be produced from 1.57 mol O2?
2. How many moles of O2 will react with 0.672 mol H2?
3. How many moles of water will be produced from 3.83 mol H2?
Given the equation below and their respective molar masses, answer the following questions below.
8 HNO3 + 6 KI ----------> 6 KNO3 + 3 I2 + 2 NO + 4 H2O
Molar mass: 63 166 101 254 30 18(g / mol)
Mole to Mole Conversion:1. How many moles of KI will react with 5.76 mol HNO3?
2. How many moles of HNO3 will react with 0.374 mol KI?
3. What is the number of moles of NO that will be produced from 4.08 mol HNO3?
4. Compute for the number of moles of I2 that will be produced from 0.882 mol KI?
5. How many moles of KNO3 will be produced from 0.311 mol HNO3?
6. How many moles of KNO3 will be produced from 7.39 mol KI?
Given the equation below and their respective molar masses, answer the following questions below.
8 HNO3 + 6 KI ----------> 6 KNO3 + 3 I2 + 2 NO + 4 H2O
Molar mass: 63 166 101 254 30 18(g / mol)
Mass to Mass Conversion:1. How many grams of KI will react with 40.7 g HNO3?
2. How many grams of HNO3 will react with 67.4 g KI?
3. What mass of KNO3 will be produced from 85.3 g HNO3?
4. Compute for the mass of KNO3 that will be produced from 113 g KI?
5. Determine the mass of NO produced from 25.6 g HNO3?
6. What mass of I2 will be produced from 90.4 g KI?
9.7 Limiting and Excess Reactants
Limiting Reactant or Limiting Reagent – The reactant that is completely consumed in a reaction because it determines, or limits, the amount of product formed.
Excess Reactant or Excess Reagent – The reactant that is left after the reaction has completely stopped.
Examples:
8 HNO3 + 6 KI ----------> 6 KNO3 + 3 I2 + 2 NO + 4 H2O
Molar mass: 63 166 101 254 30 18(g / mol)
1. If 50.0 g of HNO3 and 100 g KI will react to form KNO3, which of the two will be the limiting reactant? Excess reactant?
2. How much of the excess reactant was consumed?
3. How much of the excess reactant did not react?
2 LiOH + CO2 ---------> Li2CO3 + H2O
Molar mass (g / mol): 24 44 74 18
1. Determine the limiting reagent if 87.5 g LiOH will react with 94.2 g CO2 to form Li2CO3.
1. If 67.8 g of Na3PO4 will react with 73.7 g Ba(NO3)2 to form Ba3(PO4)2, what is the limiting reagent?
2. How much of the excess reactant was consumed?
3. How much of the excess reactant did not react?
9.8 Percent Yield
Theoretical Yield (TY) – The quantity of product that is calculated to form when the entire limiting reagent reacts.
NOTE: When the limiting reagent is identified, the amount of product that is computed becomes automatically the theoretical yield.
Actual Yield (AY) – The amount of product obtained from laboratory experiments. The actual yield of a substance cannot be computed. They are determined by actual measurements done in the laboratory. Actual yield is normally less than theoretical yield.
Percent Yield (PY) – The ratio of the actual yield to its theoretical yield multiplied by 100. (PY = AY / TY x 100) The percent yield can never be more than 100%.
8 HNO3 + 6 KI ----------> 6 KNO3 + 3 I2 + 2 NO + 4 H2O
Molar mass: 63 166 101 254 30 18(g / mol)
1. If 86.2 g of HNO3 and 74.3 g KI will react to form KNO3, which of the two will be the limiting reactant? Excess reactant?
2. How much of the excess reactant was consumed?
3. How much of the excess reactant did not react?
4. In the laboratory, it was observed that 40.0g of KNO3 was obtained. What is the percent yield of KNO3?
2 LiOH + CO2 ---------> Li2CO3 + H2O
Molar mass (g / mol): 24 44 74 18
1. Determine the limiting reagent if 71.5 g LiOH will react with 56.2 g CO2 to form Li2CO3.
2. How much of the excess reactant was consumed?
3. How much of the excess reactant did not react?
4. If 90.0 g of Li2CO3 was obtained in the laboratory, what is the percent yield of the compound?
