KNOW, calculations based on…..

KNOW, calculations based on…..
CHAPTER 3 Mole Molar Mass Avogadro’s Number Calculations - mass, mole, particle - mass % - determining formulas Emperical/Molecular - mass/mass - Molarity (mols/L) - Dilution Stoichiometry Limiting Reagent KNOW, calculations based on….. Formulas Balance Eqn Subst Mass TERMS: formula/molar/molecular weight or mass - sum total of all atoms in chem formula - CaCl2; 1 Ca + 2 Cl = 1(40.1) + 2(35.5) = g/mol states: (s) - (l) - (g) - (aq) rxns: combination, decomposition, combustion

FORMULA--MOLECULAR WEIGHT
The sum of the Atomic Weights of all atoms present in one formula unit or molecule Formula Wt: ionic cmpd Molecular Wt: covalent molecule

What is the formula weight of sodium carbonate,
PROBLEM What is the formula weight of sodium carbonate, Na2CO3. This is an industrial chemical used in making glass. Also, equivalent to 1mole of a substance SOLUTION 2 * = 2 Na 1 C 3 O 1 * = 3 * = + g 3

What is the % composition of chloroform,
CHCl3, anesthetic? PROBLEM SOLUTION 1 C 1 H 3 Cl 1 * = 12.0 1 * 1.0 = 3 * = 106.5 1. Identify # each element 2. Calculate formula weight + 119.5 g 3. Calculate % Composition by each element

% Composition + 100.0 %

BALANCE EQUATIONS “STOICHIOMETRY”
Satisfies “Law of Conservation of Mass”; contains equal number of atoms of each element on both sides of “----” STEPS Write correct chemical equation & fomulas Do not change subscript #’s to balance, but place coefficient in front Coefficient * Subscript = # of Atoms Balance “H” last

Sodium Phosphate (aq) + Barium Nitrate (aq) ------
Barium Phosphate (s) + Sodium Nitrate (aq) 2 Na3PO4 (aq) Ba(NO3)2 (aq) ----- Ba3(PO4)2 (s) + NaNO3 (aq) 3 6 Aluminum (s) + Hydrochloric Acid (aq) > Aluminum Chloride (aq) + Hydrogen (g) 2 Al HCl ---- AlCl H2 6 2 3

BALANCE CHEMICAL EQUATIONS – write and balance each reaction
1. Aluminum metal reacts with oxygen to produce a solid oxide compound. 2. Two solutions are mixed together, sodium phosphate and barium nitrate. Upon mixing they form two new products, solid barium phosphate and aqueous sodium nitrate. 3. Ammonia gas (NH3) will break down into its two atmospheric components of nitrogen and hydrogen.

Mole Ratio The coefficient in front of each component in a balanced equation 2 Al (s) + 6 HCl (aq) ----- 2 AlCl 3 (aq) + 3 H2 (g) This states: 2 moles of Al reacts w/ 6 moles HCl to produce 2 moles AlCl3 and 3 moles H2 gas

COMBINATION REACTIONS
FORM: A B  A+B- 2 reactants form 1 product DECOMPOSITION REACTIONS FORM: A+B  A0 + B0 1 reactants breaks apart into 2 or more product COMBUSTION REACTIONS FORM: A + O  CO2 + H2O hydrocarbon reacts w/ air to produce CO2 & WATER

Hydrogen (g) + Oxygen (g) ----- Water (l)
Combination 2 H2 (g) + O2 (g) ---- H2O (l) 2 Lithuim Chlorate (s) > Chloride (s) + Oxygen (g) Decomposition 2 LiClO3 (s) ----- 2 LiCl (s) + 3 O2 (g) Ethene Oxygen  Carbon Dioxide + water Combustion C2H4 (g) O2 (g) --- 2 CO2 (g) + 2 H20 (l)

MOLES Mole: A collection of objects
A collection of Avogadro’s number of objects Avogadro’s number 6.022 * 1023 Atoms Particles Ions Cations Anions Molecules Formula Units MOLES mass = 1 mole Multiply by Divide by MASS

Given mass, find moles How many moles are in g NaNO3 ? Step 1: find formula wt. of NaNO3 1 Na = g 1 N = g 3 O = 3 * 16.0 = 48.0 g 1 mole of NaNO3 = g Step 2 : Find # of moles Use factor-label method 4.07 moles NaNO3

How many grams are in 0.6 moles N2O ?
Given moles, find mass How many grams are in 0.6 moles N2O ? Step 1: find molecular wt. of N2O 2 N = 2 * 14.0 = 28.0 g 1 O = g 1 mole of N2O = g Step 2 : Find mass Use factor-label method 26.4 g N2O 15

Given moles, find # molecules
NUMBER of PARTICLES Divide by by Multiply Avogadro’s Number 6.02*1023 Given moles, find # molecules How many molecules are in 1.6 moles oxygen? Step 1: Recognize that mass does not apply here. But Avogadro’s # is used Step 2 : Find # molecules Use factor-label method 9.63 * 1023 molecules O2 16

Given # atoms, find moles
How many moles are in 3.01 *1012 atoms of Chromium? Step 1: Recognize that mass does not apply here. But Avogadro’s # is used Step 2 : Find # moles Use factor-label method 5.00 * mols Cr 17

There is not a direct relationship between MASS and PARTICLES
MOLES 6.02*1023 mass = 1 mole Multiply by Divide by Avogadro’s Number Multiply by Divide by MASS NUMBER of PARTICLES There is not a direct relationship between MASS and PARTICLES 18

2 Step Conversion Problem
Will need to use MASS at some point in the problem Given # formula units, find grams How many grams are in * 1024 formula units AgCl ? Step 1: find formula wt. of AgCl 1 Ag = g 1 Cl = g 1 mole of AgCl = g Step 2 : Find moles Step 3 : Find mass Use factor-label method 288 g AgCl Converts formula units to MOLES Converts moles to MASS 19

2 Step Conversion Problem
Will need to use MASS at some point in the problem Given grams, find # atoms How many atoms are in grams of Mercury ? Step 1: find molar mass of Hg 1 mole of Hg = g Step 2 : Find moles Step 3 : Find atoms Use factor-label method 3.84*1023 atoms Hg Converts mass to MOLES Converts moles to ATOMS 20

EMPERICAL FORMULA Problem: Determine the formula for a cmpd that is
26.6% potassium, 35.4% chromium, and 38.1% oxygen. Solution: step1) assume, using 100 g sample, thus the percentages can be used as your masses for each diff. element 26.6% -- 26.6 g K % --- 35.4 g Cr % --- 38.1 g O step 2) convert your masses to moles of each diff. element Determine Formulas Convert mols of parts into whole # subscripts STEPS 1. Determinemolefraction for each element 2. Divide each subscript by the smallest value 3. If not all # are whole #, find smallest factor to multiple all subscripts by to obtain a whole # 21

step 3) Divide each value by the smallest “mole” number
from step 2 step 4) calculate whole # subscripts Potassium Dichromate 22

MOLECULAR FORMULA Problem:
An unknown substance containing 82.7% carbon and 17.4% hydrogen has a molar mass of 58.2 grams. Find the molecular formula. Solution: step1) assume, using 100 g sample, thus the percentages can be used as your masses for each diff. element 82.7% -- 82.7 g C % --- 17.4 g H step 2) convert your masses to moles of each diff. element 23

step 3) Divide each value by the smallest “mole” number
from step 2 step 4) calculate whole # subscripts step 5) Find formula wt. of C2H5 and the ratio to the molar mass C2H5: C = 2 * 12.0 = 24.0 g H = 5 * = g 29.0 g step 6) Determine the molecular formula There are 2 empirical formula units 2(C2H5) = C4H10 C4H10 = 4(12.0) + 10(1.0) = 58.0 g 24

subst A subst B MASS MOLES MOLES MASS formula wt molecular wt
2 Al (s) + 6 HCl (aq) ----- 2 AlCl 3 (aq) + 3 H2 (g) subst A subst B 5.026 g Al what quantity, g? MASS subst A MOLES subst A MOLES subst B MASS subst B formula wt molecular wt molar mass coefficents in balanced eqn formula wt molecular wt molar mass Al = 27.0 g/mol 5.026 g*(1 mol/27.0 g) = 2 & 3 3 mol H2 /2 mol Al = H2 = 2.0 g/mol H2 mol*(2.0 g/1 mol) =

Atomic Weight: fomula wt; molecular wt; molar mass
TERMS Atomic Weight: fomula wt; molecular wt; molar mass sum total of amu (g) of all atoms in cmpd Stoichiometry: coeff used to balance chem rxn; uphold Law Conserv. of Mass (Matter) Mole: quantity of subst present; based on Avogadro’s Number; 6.022*1023 “label” RECAP: 1 mol of “anything” contains 6.022*1023 “parts” Elements on p.table = 1 mol RECAP: -- Balanced chem. equation -- then, GIVEN USE FIND grams Molar Mass mols mols coeff mols mols Molar Mass grams limiting reagent: the reactant that produces the least amt of moles or mass of a specific pdt

Theoretical Yield: based on calculations from balanced rxn
max amt (g) pdt formed when LR totally consumed Actual Yield: amt (g) of pdt acutal formed in rxn % Yield: ratio of (actual)/(theoretical)*100 Limiting Reagent: reactant totally consumed in rxn; produces smallest amt of pdt; present in smallest stoichiometric amt Excess Reactant: amt of reactant not totally consumed in rxn

Ca(s) + O2(g) -------> CaO(s)
MASS - MASS CALCULATIONS & LIMITING REAGENT (amts of 2 reactants given: limiting reactants) Solid calcium metal burns in air (oxygen) to form a calcium oxide cmpd. Using 4.20 g Ca & 2.80 g O2 how much pdt is made? Find 1) grams of pdt made from g of Ca 2) grams of pdt made from g of O2 3) limiting reagent 4) how much pdt can be made 5) % yield if 3.85 g produced PLAN: Need > balanced chem. rxn STEPS: g Ca -----> mols Ca -----> mols pdt > g pdt M g Ca coeff X pdt/Y react M g pdt (same steps converting oxygen) Ca(s) + O2(g) > CaO(s) M g Ca M g O coeff X pdt/Y react M g pdt 40.1 g CaO:Ca 2: g 32.0 g CaO:O2 2:1

5.89 g CaO 2 Ca(s) + O2(g) -------> 2 CaO(s) 4.20 g 2.80 g X g
M g : g g Coeff : mols : 0.105 0.0875 0.105 0.175 3) limiting reagent: Which produced the least amt-- Ca or O2? Ca ---> CaO = 5.89 g O2 ---> CaO = 9.82 g Ca, limiting reagent 4) how much pdt can be made: 5.89 g CaO 5) % yield=

Sample Exercise 3.10 Converting Grams to Moles
Calculate the number of moles of glucose (C6H12O6) in g of C6H12O6. Solution Analyze We are given the number of grams of a substance and its chemical formula and asked to calculate the number of moles. Plan The molar mass of a substance provides the factor for converting grams to moles. The molar mass of C6H12O6 is g/mol (Sample Exercise 3.9). Solve Using 1 mol C6H12O6=180.0 g C6H12O6 to write the appropriate conversion factor, we have Check Because g is less than the molar mass, a reasonable answer is less than one mole. The units of our answer (mol) are appropriate. The original data had four significant figures, so our answer has four significant figures. How many moles of sodium bicarbonate (NaHCO3) are in 508 g of NaHCO3? Answer: 6.05 mol NaHCO3 Practice Exercise

Sample Exercise 3.11 Converting Moles to Grams
Calculate the mass, in grams, of mol of calcium nitrate. Solution Analyze We are given the number of moles and the name of a substance and asked to calculate the number of grams in the sample. Plan To convert moles to grams, we need the molar mass, which we can calculate using the chemical formula and atomic weights. Solve Because the calcium ion is Ca2+ and the nitrate ion is NO3–, calcium nitrate is Ca(NO3)2. Adding the atomic weights of the elements in the compound gives a formula weight of amu. Using 1 mol Ca(NO3)2 = g Ca(NO3)2 to write the appropriate conversion factor, we have Check The number of moles is less than 1, so the number of grams must be less than the molar mass, g. Using rounded numbers to estimate, we have 0.5  150 = 75 g. The magnitude of our answer is reasonable. Both the units (g) and the number of significant figures (3) are correct. What is the mass, in grams, of (a) 6.33 mol of NaHCO3 and (b) 3.0  10–5 mol of sulfuric acid? Answer: (a) 532 g, (b) 2.9  10–3 g Practice Exercise

Sample Exercise 3.16 Calculating Amounts of Reactants and Products
How many grams of water are produced in the oxidation of 1.00 g of glucose, C6H12O6? C6H12O6(s) + 6 O2(g)→6 CO2(g) + 6 H2O(l) Solution Analyze We are given the mass of a reactant and are asked to determine the mass of a product in the given equation. Plan The general strategy, as outlined in Figure 3.13, requires three steps. First, the amount of C6H12O6 must be converted from grams to moles. Second, we can use the balanced equation, which relates the moles of C6H12O6 to the moles of H2O: 1 mol C6H12O mol H2O. Third, we must convert the moles of H2O to grams. Solve First, use the molar mass of C6H12O6 to convert from grams C6H12O6 to moles C6H12O6: Second, use the balanced equation to convert moles of C6H12O6 to moles of H2O: Third, use the molar mass of H2O to convert from moles of H2O to grams of H2O: The steps can be summarized in a diagram like that in Figure 3.13:

Sample Exercise 3.18 Calculating the Amount of Product Formed from
a Limiting Reactant The most important commercial process for converting N2 from the air into nitrogen-containing compounds is based on the reaction of N2 and H2 to form ammonia (NH3): N2(g) + 3 H2(g)→2 NH3(g) How many moles of NH3 can be formed from 3.0 mol of N2 and 6.0 mol of H2? Solution Analyze We are asked to calculate the number of moles of product, NH3, given the quantities of each reactant, N2 and H2, available in a reaction. Thus, this is a limiting reactant problem. Plan If we assume that one reactant is completely consumed, we can calculate how much of the second reactant is needed in the reaction. By comparing the calculated quantity with the available amount, we can determine which reactant is limiting. We then proceed with the calculation, using the quantity of the limiting reactant. Solve The number of moles of H2 needed for complete consumption of 3.0 mol of N2 is: Because only 6.0 mol H2 is available, we will run out of H2 before the N2 is gone, and H2 will be the limiting reactant. We use the quantity of the limiting reactant, H2, to calculate the quantity of NH3 produced: Comment The table on the right summarizes this example:

a Limiting Reactant Solution (continued) Notice that we can calculate not only the number of moles of NH3 formed but also the number of moles of each of the reactants remaining after the reaction. Notice also that although the number of moles of H2 present at the beginning of the reaction is greater than the number of moles of N2 present, the H2 is nevertheless the limiting reactant because of its larger coefficient in the balanced equation. Check The summarizing table shows that the mole ratio of reactants used and product formed conforms to the coefficients in the balanced equation, 1:3:2. Also, because H2 is the limiting reactant, it is completely consumed in the reaction, leaving 0 mol at the end. Because 6.0 mol H2 has two significant figures, our answer has two significant figures. Consider the reaction 2 Al(s) + 3 Cl2(g) → 2 AlCl3(s). A mixture of 1.50 mol of Al and 3.00 mol of Cl2 is allowed to react. (a) Which is the limiting reactant? (b) How many moles of AlCl3 are formed? (c) How many moles of the excess reactant remain at the end of the reaction? Answers: (a) Al, (b) 1.50 mol, (c) 0.75 mol Cl2 Practice Exercise

a Limiting Reactant Consider the following reaction that occurs in a fuel cell: 2 H2(g) + O2 (g) → 2 H2O (g) This reaction, properly done, produces energy in the form of electricity and water. Suppose a fuel cell is set up with 150 g of hydrogen gas and 1500 grams of oxygen gas (each measurement is given with two significant figures). How many grams of water can be formed? Solution Analyze We are asked to calculate the amount of a product, given the amounts of two reactants, so this is a limiting reactant problem. Plan We must first identify the limiting reagent. To do so, we can calculate the number of moles of each reactant and compare their ratio with that required by the balanced equation. We then use the quantity of the limiting reagent to calculate the mass of water that forms. Solve From the balanced equation, we have the following stoichiometric relations: Using the molar mass of each substance, we can calculate the number of moles of each reactant:

a Limiting Reactant Solution (continued) Thus, there are more moles of H2 than O2. The coefficients in the balanced equation indicate, however, that the reaction requires 2 moles of H2 for every 1 mole of O2. Therefore, to completely react all the O2, we would need 2  47 = 94 moles of H2. Since there are only 75 moles of H2, H2 is the limiting reagent. We therefore use the quantity of H2 to calculate the quantity of product formed. We can begin this calculation with the grams of H2, but we can save a step by starting with the moles of H2 that were calculated previously in the exercise: Check The magnitude of the answer seems reasonable. The units are correct, and the number of significant figures (two) corresponds to those in the numbers of grams of the starting materials. Comment The quantity of the limiting reagent, H2, can also be used to determine the quantity of O2 used (37.5 mol = 1200 g). The number of grams of the excess oxygen remaining at the end of the reaction equals the starting amount minus the amount consumed in the reaction, 1500 g – 1200 g = 300 g. A strip of zinc metal with a mass of 2.00 g is placed in an aqueous solution containing 2.50 g of silver nitrate, causing the following reaction to occur: Zn(s) + 2 AgNO3(aq) → 2 Ag(s) + Zn(NO3)2(aq) Practice Exercise

a Limiting Reactant (a) Which reactant is limiting? (b) How many grams of Ag will form? (c) How many grams of Zn(NO3)2 will form? (d) How many grams of the excess reactant will be left at the end of the reaction? Answers: (a) AgNO3, (b) 1.59 g, (c) 1.39 g, (d) 1.52 g Zn Practice Exercise (continued)

Sample Exercise 3.20 Calculating the Theoretical Yield and the Percent
Yield for a Reaction Adipic acid, H2C6H8O4, is used to produce nylon. The acid is made commercially by a controlled reaction between cyclohexane (C6H12) and O2: 2 C6H12(l) + 5 O2(g) → 2 H2C6H8O4(l) + 2 H2O(g) (a) Assume that you carry out this reaction starting with 25.0 g of cyclohexane and that cyclohexane is the limiting reactant. What is the theoretical yield of adipic acid? (b) If you obtain 33.5 g of adipic acid from your reaction, what is the percent yield of adipic acid? Solution Analyze We are given a chemical equation and the quantity of the limiting reactant (25.0 g of C6H12). We are asked first to calculate the theoretical yield of a product (H2C6H8O4) and then to calculate its percent yield if only 33.5 g of the substance is actually obtained. Plan (a) The theoretical yield, which is the calculated quantity of adipic acid formed in the reaction, can be calculated using the following sequence of conversions: g C6H12 → mol C6H12 → mol H2C6H8O4 → g H2C6H8O4

Sample Exercise 3.20 Calculating the Theoretical Yield and the Percent
Yield for a Reaction Solution (continued) (b) The percent yield is calculated by comparing the actual yield (33.5 g) to the theoretical yield using Equation 3.14. Solve Check Our answer in (a) has the appropriate magnitude, units, and significant figures. In (b) the answer is less than 100% as necessary. Imagine that you are working on ways to improve the process by which iron ore containing Fe2O3 is converted into iron. In your tests you carry out the following reaction on a small scale: Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) (a) If you start with 150 g of Fe2O3 as the limiting reagent, what is the theoretical yield of Fe? (b) If the actual yield of Fe in your test was 87.9 g, what was the percent yield? Answers: (a) 105 g Fe, (b) 83.7% Practice Exercise

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