Chapter 3: stoichimetry of formulas and equations, Cheat Sheet of Chemistry

Download Chapter 3: stoichimetry of formulas and equations and more Cheat Sheet Chemistry in PDF only on Docsity! 3-1 CHEM 1A: GENERAL CHEMISTRY Chapter 3: Stoichiometry of Formulas and Equations Instructor: Dr. Orlando E. Raola Santa Rosa Junior College 3-2 Amount - Mass Relationships in Chemical Systems 3.5 Fundamentals of Solution Stoichiometry 3.1 The Mole 3.2 Determining the Formula of an Unknown Compound 3.3 Writing and Balancing Chemical Equations 3.4 Calculating Quantities of Reactant and Product 3-5 The molar mass (M) of a substance is the mass per mole of its entites (atoms, molecules or formula units). For monatomic elements, the molar mass is the same as the atomic mass in grams per mole. The atomic mass is simply read from the Periodic Table. The molar mass of Ne = 20.18 g/mol. Molar Mass 3-6 For molecular elements and for compounds, the formula is needed to determine the molar mass. The molar mass of O2 = 2 x M of O = 2 x 16.00 = 32.00 g/mol The molar mass of SO2 = 1 x M of S + 2 x M of O = 32.00 + 2(16.00) = 64.00 g/mol 3-7 Table 3.1 Information Contained in the Chemical Formula of Glucose C6H12O6 ( M = 180.16 g/mol) Carbon (C) Hydrogen (H) Oxygen (O) Atoms/molecule of compound 6 atoms 12 atoms 6 atoms Moles of atoms/mole of compound 6 mol of atoms 12 mol of atoms 6 mol of atoms Atoms/mole of compound 6(6.022x1023) atoms 12(6.022x1023) atoms 6(6.022x1023) atoms Mass/molecule of compound 6(12.01 u) = 72.06 u 12(1.008 u) = 12.10 u 6(16.00 u) = 96.00 u Mass/mole of compound 72.06 g 12.10 g 96.00 g 3-10 Sample Problem 3.1 Calculating the Mass of a Given Amount of an Element PROBLEM: SOLUTION: amount (mol) of Ag mass (g) of Ag Silver (Ag) is used in jewelry and tableware but no longer in U.S. coins. How many grams of Ag are in 0.0342 mol of Ag? PLAN: To convert mol of Ag to mass of Ag in g we need the molar mass of Ag. multiply by M of Ag (107.9 g/mol) 0.0342 mol Ag x 1 mol Ag 107.9 g Ag = 3.69 g Ag 3-11 Sample Problem 3.2 Calculating the Number of Entities in a Given Amount of an Element PROBLEM: Gallium (Ga) is a key element in solar panels, calculators and other light-sensitive electronic devices. How many Ga atoms are in 2.85 x 10-3 mol of gallium? To convert mol of Ga to number of Ga atoms we need to use Avogadro’s number. PLAN: mol of Ga atoms of Ga multiply by 6.022x1023 atoms/mol 3-12 SOLUTION: 2.85 x 10-3 mol Ga atoms x 6.022x1023 Ga atoms 1 mol Ga atoms = 1.72 x 1021 Ga atoms Sample Problem 3.2 Figure 3.3. Amount-mass-number relationships for compounds. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. MASS (g) of compound chemical formula Avogadro’s number (molecules/mol) 3-15 3-16 Sample Problem 3.4 Calculating the Number of Chemical Entities in a Given Mass of a Compound I PROBLEM: Nitrogen dioxide is a component of urban smog that forms from the gases in car exhausts. How many molecules are in 8.92 g of nitrogen dioxide? number of NO2 molecules PLAN: Write the formula for the compound and calculate its molar mass. Use the given mass to calculate first the number of moles and then the number of molecules. divide by M multiply by 6.022 x 1023 formula units/mol mass (g) of NO2 amount (mol) of NO2 3-17 SOLUTION: NO2 is the formula for nitrogen dioxide. M = (1 x M of N) + (2 x M of O) = 14.01 g/mol + 2(16.00 g/mol) = 46.01 g/mol 8.92 g NO2 x = 1.17 x 1023 molecules NO2 1 mol NO2 46.01 g NO2 6.022x1023 molecules NO2 1 mol NO2 = 0.194 mol NO2 0.194 mol NO2 x Sample Problem 3.4 3-20 41.6 g (NH4)2CO3 x = 2.61x1023 formula units (NH4)2CO3 1 mol (NH4)2CO3 96.09 g (NH4)2CO3 6.022x1023 formula units (NH4)2CO3 1 mol (NH4)2CO3 = 0.433 mol (NH4)2CO3 0.433 mol (NH4)2CO3 x 2.61x1023 formula units (NH4)2CO3 x 3 O atoms 1 formula unit of (NH4)2CO3 = 7.83 x 1023 O atoms Sample Problem 3.5 3-21 Mass % of element X = atoms of X in formula × atomic mass of X (u) molecular (or formula) mass of compound (u) x 100 Mass % of element X = amount of X in formula (mol) × molar mass of X (g/mol) mass (g) of 1 mol of compound x 100 Mass Percent from the Chemical Formula 3-22 Sample Problem 3.6 Calculating the Mass Percent of Each Element in a Compound from the Formula PROBLEM: Glucose (C6H12O6) is a key nutrient for generating chemical potential energy in biological systems. What is the mass percent of each element in glucose? PLAN: Find the molar mass of glucose, which is the mass of 1 mole of glucose. Find the mass of each element in 1 mole of glucose, using the molecular formula. The mass % for each element is calculated by dividing the mass of that element in 1 mole of glucose by the total mass of 1 mole of glucose, multiplied by 100. 3-25 Mass Percent and the Mass of an Element Mass of element X present in sample = mass of element in 1 mol of compound mass of 1 mol of compound mass of compound x Mass percent can also be used to calculate the mass of a particular element in any mass of a compound. 3-26 Sample Problem 3.7 Calculating the Mass of an Element in a Compound PLAN: PROBLEM: Use the information from Sample Problem 3.6 to determine the mass (g) of carbon in 16.55 g of glucose. The mass percent of carbon in glucose gives us the relative mass of carbon in 1 mole of glucose. We can use this information to find the mass of carbon in any sample of glucose. mass of glucose sample mass of C in sample multiply by mass percent of C in glucose 3-27 Mass (g) of C = mass (g) of glucose x 6 mol x M of C (g/mol) mass (g) of 1 mol of glucose SOLUTION: Each mol of glucose contains 6 mol of C, or 72.06 g of C. Sample Problem 3.7 = 16.55 g glucose x 72.06 g C 180.16 g glucose = 6.620 g C 3-30 Sample Problem 3.8 SOLUTION: Using the numbers of moles of each element given, we write the preliminary formula Zn0.21P0.14O0.56 Next we divide each fraction by the smallest one; in this case 0.14: 0.21 0.14 = 1.5 0.14 0.14 = 1.0 0.56 0.14 = 4.0 We convert to whole numbers by multiplying by the smallest integer that gives whole numbers; in this case 2: 1.5 x 2 = 3 1.0 x 2 = 2 4.0 x 2 = 8 This gives us the empirical formula Zn3P2O8 This gives Zn1.5P1.0O4.0 3-31 Sample Problem 3.9 Determining an Empirical Formula from Masses of Elements mass (g) of each element amount (mol) of each element PROBLEM: Analysis of a sample of an ionic compound yields 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. What is the empirical formula and the name of the compound? preliminary formula empirical formula change to integer subscripts use # of moles as subscripts divide by M (g/mol) PLAN: Find the relative number of moles of each element. Divide by the lowest mol amount to find the relative mol ratios (empirical formula). 3-32 Sample Problem 3.9 SOLUTION: 2.82 g Na x 1 mol Na 22.99 g Na = 0.123 mol Na 4.35 g Cl x 1 mol Cl 35.45 g Cl = 0.123 mol Cl 7.83 g O x 1 mol O 16.00 g O = 0.489 mol O The empirical formula is Na1Cl1O3.98 or NaClO4; this compound is named sodium perchlorate. 0.123 0.123 0.489 0.123 Na and Cl = = 1 and O = = 3.98 3-35 SOLUTION: Assuming there are 100. g of lactic acid; 40.0 g C x 1 mol C 12.01 g C = 3.33 mol C CH2O empirical formula mass of CH2O molar mass of lactate 90.08 g/mol 30.03 g/mol C3H6O3 is the molecular formula O3.33 3.33 H6.66 3.33 C3.33 3.33 6.71 g H x 1 mol H 1.008 g H = 6.66 mol H 1 mol O 16.00 g O 53.3 g O x = 3.33 mol O Sample Problem 3.10 = 3 3-36 Combustion apparatus for determining formulas of organic compounds. Figure 3.4 CnHm + (n+ ) O2 = n CO(g) + H2O(g) m 2 m 2 3-37 Sample Problem 3.11 Determining a Molecular Formula from Combustion Analysis PROBLEM: When a 1.000 g sample of vitamin C (M = 176.12 g/mol) is placed in a combustion chamber and burned, the following data are obtained: mass of CO2 absorber after combustion = 85.35 g mass of CO2 absorber before combustion = 83.85 g mass of H2O absorber after combustion = 37.96 g mass of H2O absorber before combustion = 37.55 g What is the molecular formula of vitamin C? PLAN: The masses of CO2 and H2O produced will give us the masses of C and H present in the original sample. From this we can determine the mass of O. 3-40 176.12 g/mol 88.06 g = 2.000 mol = 0.0341 mol O 0.545 g O 16.00 g/mol O 0.409 g C 12.01 g/mol C = 0.0341 mol C 0.046 g H 1.008 g/mol H = 0.0456 mol H Convert mass to moles: Sample Problem 3.11 = 1 0.0341 0.0341 C 0.0456 0.0341 = 1.34 H 0.0341 0.0341 = 1 O C1H1.34O1 = C3H4.01O3 C3H4O3 C6H8O6 Divide by smallest to get the preliminary formula: Divide molar mass by mass of empirical formula: 3-41 Table 3.2 Some Compounds with Empirical Formula CH2O (Composition by Mass: 40.0% C, 6.71% H, 53.3% O) Name Molecular Formula Whole-Number Multiple M (g/mol) Use or Function formaldehyde acetic acid lactic acid erythrose ribose glucose CH2O C2H4O2 C3H6O3 C4H8O4 C5H10O5 C6H12O6 1 2 3 4 5 6 30.03 60.05 90.09 120.10 150.13 180.16 disinfectant; biological preservative acetate polymers; vinegar (5% soln) part of sugar metabolism sour milk; forms in exercising muscle component of nucleic acids and B2 major energy source of the cell CH2O C2H4O2 C3H6O3 C4H8O4 C5H10O5 C6H12O6 3-42 Table 3.3 Two Pairs of Constitutional Isomers Property Ethanol Dimethyl Ether M (g/mol) Boiling Point Density at 200C Structural formula 46.07 78.50C 0.789 g/mL (liquid) 46.07 -250C 0.00195 g/mL (gas) Butane 2-Methylpropane C4H10 C2H6O 58.12 58.12 Space-filling model -0.50C 0.579 g/mL (gas) -11.060C 0.549 g/mL (gas) C C C C H H H H H H H H H H C C C H H H H C H H H H H H C C H H H H H OH C O H H H C H H H 3-45 Features of Chemical Equations Mg + O2 MgO Reactants are written on the left. A yield arrow points from reactants to products. Products are written on the right. The equation must be balanced; the same number and type of each atom must appear on both sides. 3-46 translate the statement balance the atoms using coefficients; formulas cannot be changed specify states of matter adjust coefficients if necessary check that all atoms balance Balancing a Chemical Equation magnesium and oxygen gas react to give magnesium oxide: Mg + O2 → MgO 2Mg + O2 → 2MgO 2Mg (s) + O2 (g) → 2MgO (s) 3-47 translate the statement Sample Problem 3.12 Balancing Chemical Equations PROBLEM: PLAN: SOLUTION: balance the atoms specify states of matter Within the cylinders of a car’s engine, the hydrocarbon octane (C8H18), one of many components of gasoline, mixes with oxygen from the air and burns to form carbon dioxide and water vapor. Write a balanced equation for this reaction. adjust the coefficients check the atoms balance C8H18 + O2 CO2 + H2O 2C8H18 + 25O2 16CO2 + 18H2O 2C8H18 + 25O2 16CO2 + 18H2O 2C8H18(l) + 25O2 (g) 16CO2 (g) + 18H2O (g) C8H18 + O2 8 CO2 + 9 H2O 25 2 3-50 SOLUTION: The reactant circle shows only one type of molecule, composed of 2 N and 5 O atoms. The formula is thus N2O5. There are 4 N2O5 molecules depicted. The product circle shows two types of molecule; one has 1 N and 2 O atoms while the other has 2 O atoms. The products are NO2 and O2. There are 8 NO2 molecules and 2 O2 molecules shown. Sample Problem 3.13 The reaction depicted is 4 N2O5 → 8 NO2 + 2 O2. Writing the equation with the smallest whole-number coefficients and states of matter included; 2 N2O5 (g) → 4 NO2 (g) + O2 (g) 3-51 Stoichiometric Calculations • The coefficients in a balanced chemical equation – represent the relative number of reactant and product particles – and the relative number of moles of each. • Since moles are related to mass – the equation can be used to calculate masses of reactants and/or products for a given reaction. • The mole ratios from the balanced equation are used as conversion factors. 3-52 Table 3.4 Information Contained in a Balanced Equation Viewed in Terms of Products 3 CO2(g) + 4 H2O(g) Amount (mol) Mass (amu) 3 molecules CO2 + 4 molecules H2O 3 mol CO2 + 4 mol H2O 132.03 amu CO2 + 72.06 amu H2O Reactants C3H8(g) + 5 O2(g) Molecules 1 molecule C3H8 + 5 molecules O2 1 mol C3H8 + 5 mol O2 44.09 amu C3H8 + 160.00 amu O2 Mass (g) 132.03 g CO2 + 72.06 g H2O 44.09 g C3H8 + 160.00 g O2 Total Mass (g) 204.09 g 204.09 g 3-55 Sample Problem 3.15 Calculating Quantities of Reactants and Products: Amount (mol) to Mass (g) PROBLEM: During the process of roasting copper(I) sulfide, how many grams of sulfur dioxide form when 10.0 mol of copper(I) sulfide reacts? PLAN: Using the balanced equation from the previous problem, we again use the mole ratio as a conversion factor. use the mole ratio as a conversion factor mol of copper(I) sulfide mol of sulfur dioxide mass of sulfur dioxide multiply by M of sulfur dioxide 3-56 = 641 g SO2 10.0 mol Cu2S x 2 mol SO2 2 mol Cu2S SOLUTION: 2 Cu2S (s) + 3 O2 (g) → 2 Cu2O (s) + 2 SO2 (g) Sample Problem 3.15 64.07 g SO2 1 mol SO2 x 3-57 Sample Problem 3.16 Calculating Quantities of Reactants and Products: Mass to Mass PROBLEM: During the roasting of copper(I) sulfide, how many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide? PLAN: divide by M of oxygen mass of oxygen mol of oxygen mass of copper(I) oxide multiply by M of copper(I) oxide mol of copper(I) oxide use mole ratio as conversion factor 3-60 Sample Problem 3.17 Writing an Overall Equation for a Reaction Sequence PROBLEM: Roasting is the first step in extracting copper from chalcocite, the ore used in the previous problem. In the next step, copper(I) oxide reacts with powdered carbon to yield copper metal and carbon monoxide gas. Write a balanced overall equation for the two-step process. PLAN: Write individual balanced equations for each step. Adjust the coefficients so that any common substances can be canceled. Add the adjusted equations together to obtain the overall equation. 3-61 SOLUTION: Cu2O (s) + C (s) → 2Cu (s) + CO (g) 2Cu2S (s) + 3O2 (g) → 2Cu2O (s) + 2SO2 (g) Write individual balanced equations for each step: Adjust the coefficients so that the 2 moles of Cu2O formed in reaction 1 are used up in reaction 2: 2Cu2S (s) + 3O2 (g) → 2Cu2O (s) + 2SO2 (g) 2Cu2O (s) + 2C (s) → 4Cu (s) + 2CO (g) Sample Problem 3.17 Add the equations together: 2Cu2S (s) + 3O2 (g) + 2C (s) → 2SO2 (g) + 4Cu (s) + 2CO (g) 3-62 Limiting Reactants • So far we have assumed that reactants are present in the correct amounts to react completely. • In reality, one reactant may limit the amount of product that can form. • The limiting reactant will be completely used up in the reaction. • The reactant that is not limiting is in excess – some of this reactant will be left over. 3-65 PLAN: Write a balanced chemical equation. To determine the limiting reactant, find the number of molecules of product that would form from the given numbers of molecules of each reactant. Use these numbers to write a reaction table and use the reaction table to draw the final reaction scene. Sample Problem 3.18 SOLUTION: The balanced equation is Cl2 (g) + 3F2 (g) → 2ClF3 (g) There are 3 molecules of Cl2 and 6 molecules of F2 depicted: = 6 molecules ClF3 2 molecules ClF3 1 molecule Cl2 3 molecules Cl2 x = 4 molecules ClF3 2 molecules ClF3 3 molecule Cl2 6 molecules F2 x Since the given amount of F2 can form less product, it is the limiting reactant. 3-66 Sample Problem 3.18 Molecules Cl2 (g) + 3F2 (g) → 2ClF3 (g) Initial Change 3 -2 6 -6 0 +4 Final 1 0 4 We use the amount of F2 to determine the “change” in the reaction table, since F2 is the limiting reactant: The final reaction scene shows that all the F2 has reacted and that there is Cl2 left over. 4 molecules of ClF2 have formed: 3-67 Sample Problem 3.19 Calculating Quantities in a Limiting- Reactant Problem: Amount to Amount PROBLEM: In another preparation of ClF3, 0.750 mol of Cl2 reacts with 3.00 mol of F2. (a) Find the limiting reactant. (b) Write a reaction table. PLAN: Find the limiting reactant by calculating the amount (mol) of ClF3 that can be formed from each given amount of reactant. Use this information to construct a reaction table. SOLUTION: The balanced equation is Cl2 (g) + 3F2 (g) → 2ClF3 (g) = 1.50 mol ClF3 2 mol ClF3 3 mol F2 3.00 mol F2 x 2 mol ClF3 1 mol Cl2 0.750 mol Cl2 x = 2.00 mol ClF3 Cl2 is limiting, because it yields less ClF3. 3-70 Sample Problem 3.20 divide by M mass (g) of N2H4 mol of N2H4 mol of N2 mole ratio divide by M mass (g) of N2O4 mol of N2O4 mol of N2 mole ratio mass of N2 select lower number of moles of N2 multiply by M 3-71 3 mol N2 2 mol N2H4 3.12 mol N2H4 x 1.00x 102 g N2H4 x 1 mol N2H4 32.05 g N2H4 Sample Problem 3.20 For N2H4: = 3.12 mol N2H4 For N2O4: SOLUTION: 2N2H4 (l) + N2O (l) → 3N2 (g) + 4H2O (g) 2.17 mol N2O4 x 3 mol N2 1 mol N2O4 2.00x 102 g N2O4 x 1 mol N2O4 92.02 g N2O4 = 6.51 mol N2 4.68 mol N2 x 28.02 g N2 1 mol N2 = 131 g N2 = 4.68 mol N2 = 2.17 mol N2 N2H4 is limiting and only 4.68 mol of N2 can be produced: 3-72 Sample Problem 3.20 Moles 2N2H4 (l) + N2O4 (l) → 3N2 (g) + 4H2O (g) Initial Change 3.12 -3.12 2.17 - 1.56 0 +4.68 0 +6.24 Final 0 0.61 4.68 6.24 All the N2H4 reacts since it is the limiting reactant. For every 2 moles of N2H4 that react 1 mol of N2O4 reacts and 3 mol of N2 form: 3.12 mol N2H4 x 1 mol N2O4 2 mol N2H4 = 1.56 mol N2O4 reacts 3-75 Sample Problem 3.21 Calculating Percent Yield PROBLEM: Silicon carbide (SiC) is made by reacting sand(silicon dioxide, SiO2) with powdered carbon at high temperature. Carbon monoxide is also formed. What is the percent yield if 51.4 kg of SiC is recovered from processing 100.0 kg of sand? PLAN: write balanced equation find mol reactant percent yield find mol product find g product predicted 3-76 x 100 = 77.0% 51.4 kg 66.73 kg = 66.73 kg 1664 mol SiC x 40.10 g SiC 1 mol SiC 1 kg 103g x mol SiO2 = mol SiC = 1664 mol SiC = 1664 mol SiO2 103 g 1 kg 100.0 kg SiO2 x 1 mol SiO2 60.09 g SiO2 x SOLUTION: SiO2(s) + 3C(s) → SiC(s) + 2CO(g) Sample Problem 3.21 3-77 Solution Stoichiometry • Many reactions occur in solution. • A solution consists of one or more solutes dissolved in a solvent. • The amount concentration of a solution is given by the amount (mol) of solute present in a given volume of solution (L). 3-80 Sample Problem 3.23 Calculating Mass of Solute in a Given Volume of Solution PROBLEM: How many grams of solute are in 1.75 L of 0.460 M sodium monohydrogen phosphate buffer solution? volume of solution moles of solute grams of solute multiply by M multiply by M SOLUTION: PLAN: Calculate the moles of solute using the given molarity and volume. Convert moles to mass using the molar mass of the solute. 1.75 L x 0.460 moles 1 L = 0.805 mol Na2HPO4 0.805 mol Na2HPO4 x 141.96 g Na2HPO4 1 mol Na2HPO4 = 114 g Na2HPO4 3-81 Figure 3.14 A •Weigh the solid needed. •Transfer the solid to a volumetric flask that contains about half the final volume of solvent. B Dissolve the solid thoroughly by swirling. C Add solvent until the solution reaches its final volume. 3-82 Converting a concentrated solution to a dilute solution. Figure 3.15 3-85 Sample Problem 3.25 Visualizing Changes in Concentration PROBLEM: The beaker and circle represents a unit volume of solution. Draw the solution after each of these changes: PLAN: Only the volume of the solution changes; the total number of moles of solute remains the same. Find the new volume and calculate the number of moles of solute per unit volume. (a) For every 1 mL of solution, 1 mL of solvent is added. (b) One third of the volume of the solution is boiled off. 3-86 SOLUTION: Ndil x Vdil = Nconc x Vconc where N is the number of particles. (a) Ndil = Nconc x Vconc Vdil = 8 particles x = 4 particles 1 mL 2 mL Sample Problem 3.25 (b) Nconc = Ndil x Vdil Vconc = 8 particles x = 12 particles 1 mL mL 2 3 (a) (b) 3-87 Sample Problem 3.26 Calculating Quantities of Reactants and Products for a Reaction in Solution PROBLEM: A 0.10 mol/L HCl solution is used to simulate the acid concentration of the stomach. What is the volume (L) of “stomach acid” react with a tablet containing 0.10 g of magnesium hydroxide? PLAN: Write a balanced equation and convert the mass of Mg(OH)2 to moles. Use the mole ratio to determine the moles of HCl, then convert to volume using molarity. divide by M mass Mg(OH)2 mol Mg(OH)2 use mole ratio divide by M mol HCl L HCl 3-90 = 5.0x10-4 mol HgS = 2.0x10-3 mol HgS Hg(NO3)2 is the limiting reactant because it yields less HgS. 5.0 x 10-4 mol HgS x 232.7 g HgS 1 mol HgS = 0.12 g HgS Sample Problem 3.27 SOLUTION: Hg(NO3)2 (aq) + Na2S (aq) → HgS (s) + 2NaNO3 (aq) 0.050 L Hg(NO3)2 x 1 mol HgS 1 mol Hg(NO3)2 0.010 mol Hg(NO3)2 1 L Hg(NO3)2 x 0.020 L Na2S x 1 mol HgS 1 mol Na2S 0. 10 mol Na2S 1 L Na2S x 3-91 Amount Hg(NO3)2 (aq) + Na2S (aq) → HgS (s) + 2NaNO3 (aq) Initial Change 5.0 x 10-4 -5.0 x 10-4 2.0 x 10-3 -5.0 x 10-4 0 +5.0 x 10-4 0 +1.0 x 10-3 Final 0 1.5 x 10-3 5.0 x 10-4 +1.0 x 10-3 Sample Problem 3.27 The reaction table is constructed using the amount of Hg(NO3)2 to determine the changes, since it is the limiting reactant. Figure 3.16 An overview of amount-mass-number stoichiometric relationships. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. MASS (g) MASS (g) MASS (g) of element of substance A MASS (g) of substance B of element ft M (g/mol) ih (g/mol) M vn M wn chemical formula chemical i molar ratio formula Avogadro's M (mol/L) of number Avogadro's solution of A Avogadro's M (mol/L) of Avogadro's solution of B number VOLUME (L) VOLUME (L) of solution of A of solution of B 3-92