Mole. The SI base unit used to measure the amount of a substance.

Size: px

Start display at page:

Download "Mole. The SI base unit used to measure the amount of a substance."

Lucy Greer
5 years ago
Views:

1 Stoichiometry

2 Stoichiometry The study of quantitative relationships between the amounts of reactants used and products formed by a chemical reactions; it is based on the law of conservation of mass.

3 Mole The SI base unit used to measure the amount of a substance.

4 Avogadro s Number The number of representative particles in a mole, and can be rounded to three significant digits: X molecules/mol

5 Molecule Forms when two or more atoms covalently bond.

6 A number smaller than 1.0 x or greater than 1.0 x should not be found in front of molecule or atom.

7 To find the number of molecules it takes to equal a certain number of moles. Take the number of moles given and multiply by Avogadro s number.

8 How many molecules of Sucrose is in 3.50 moles of Sucrose mol Sucrose X 6.02x10 23 molecules/mol =2.11x10 24 molecules of Sucrose

9 How many molecules of Sucrose is in 4.80 moles of Sodium Hydroxide mol Sodium Hydroxide X 6.02x10 23 molecules/mol =2.89x10 24 molecules of Sodium Hydroxide

10 To find the number of moles it takes to equal a certain number of molecules. Take the number of molecules given and divide by Avogadro s number.

11 How many moles of Sucrose is in 3.54x10 24 molecules of Sucrose. 3.54x10 24 molecules of Sucrose 6.02x10 23 molecules/mol =5.88 moles of Sucrose

12 How many moles of Sodium Hydroxide is in 5.63x10 24 molecules of Sodium Hydroxide. 5.63x10 24 molecules NaOH 6.02x10 23 molecules/mol =9.35 moles of NaOH

13 A number smaller than or greater than 10,000 should not be found in front of grams.

14 Atomic Mass The weighted average mass of the isotopes of that element.

15 Mass Number The number of protons plus the number of neutrons found in that element

16 Molar Mass The mass in grams of one mole of any pure substance. g/mol

17 To find the number of grams of substance. Take the number of moles given and multiply by the substance s molar mass.

18 Find the mass of 3.2 moles of Butane needed to complete the reaction. 3.2 mol of Butane X g/mol = g of Butane

19 Find the mass of 4.5 moles of Pentanol needed to complete the reaction. 4.5 mol Pentanol X g/mol = g of Pentanol

20 To find the number of moles of substance. Take the number of grams given and divide by the substance s molar mass.

21 Find the moles of 23 g of water needed to complete the reaction. 23 g water g/mol =1.28 mol of water

22 Find the moles of 112 g of Hydrochloric Acid needed to complete the reaction. 112 g of Hydrochloric Acid g/mol =3.07 mol of Hydrochloric Acid

23 To find the number of moles of an element in a compound, multiply the moles of the compound with the ratio of number of elements to 1 mol of compound.

24 How many moles of Fluorine is found in 5.50 moles of Freon (CCl 2 F 2 ) mol CCl 2 F 2 X 2 mol F atoms/1 mol CCl 2 F 2 = 11.0 mol F atoms

25 How many moles of Oxygen is found in 4.75 moles of Glucose (C 6 H 12 O 6 ) mol C 6 H 12 O 6 X 6 mol O atoms/1 mol C 6 H 12 O 6 = 28.5 mol O atoms

26 Molecular Mass The mass of the molecule found by adding the mass of each atom in the molecule.

27 To find the molar mass of compound, for each element multiply the number of element with the ratio of the molar mass of the element to 1 mol of the element. Add all the grams up.

28 What is the molar mass of Potassium Chromate K 2 CrO 4. 2molK X 39.1gK/1molK=78.2g 1molCr X 52gCr/1molCr=52g 4molO X 16gO/1mol=64g 78.2g + 52g + 64g=194.2g K 2 CrO 4

29 What is the molar mass of Sodium Hydroxide NaOH. 1molNaX22.99gNa/1molNa =22.9g 1molH X 1.01gH/1molH=1.01g 1molO X 16gO/1mol=16g 22.9g+1.01g+16g=139.9gNaOH

30 To find the moles of a compound, take the mass of the compound and divide by the molar mass of the compound.

31 How many moles are there of 47g of water? 47g of water g/mol of water = 2.61 mol of water

32 How many moles are there of 21g of Benzene? 21g of Benzene g/mol of Benzene = 0.27 mol of Benzene

33 Percent Composition The percent by mass of each element in a compound

34 To find percent composition, take molar mass of element divided by molar mass of compound and multiply by 100.

35 What is the percent composition of H in water. (2.02 g H g water) X 100 = 11.2% H

36 What is the percent composition of O in Glucose (C 6 H 12 O 6 ). (96 g O g Glucose) X 100 = 53.28% O

37 Empirical Formula The formula of a compound with the smallest whole number mole ratio of the elements.

38 To find empirical formula, take the grams of each element and convert to moles by dividing each element by their molar mass. Then take each mole and divided by the smallest mole. This gives you the subscript for each empirical formula.

39 What is the empirical formula with 40.05g S and 59.95g O g S 32.07g/mol S = mol S 59.95gO 16g/mol O=3.747mol O mol S/1.249 = 1 mol S mol O/1.249 = 3 mol O SO 3

40 What is the empirical formula with 35.98g Al and 64.02g S g S 32.07g/molS=1.99molS 35.98gAl 26.98g/molAl=1.33mol Al 1.33 mol Al/1.33 = 1 mol Al 1.99 mol S/1.33 = 1.5 mol S Al 2 S 3

41 Molecular Formula It specifies the actual number of atoms of each element in one molecule or formula unit of the substance.

42 To find the molecular formula, take the experimental mass and divided by the mass of the empirical formula. Take this ratio and multiply by empirical formula to find molecular formula.

43 What is the molecular formula of a compound with empirical formula of CH and exp. mass of g/mol g/mol g/mol = 2 C 2 H 2

44 What is the molecular formula of a compound with empirical formula of CH and exp. mass of g/mol g/mol g/mol = 6 C 6 H 6

45 Mole Ratio In a balanced equation, the ratio between the numbers of moles of any two substances.

46 To find mole ratio, put moles of one substance over the moles of another substance in chemical equation. 2KClO 3 2KCl + 3O 2 2mol KClO 3 /2mol KCl 2 mol KClO 3 /3mol O 2

47 Limiting Reactant It limits the extent of the reaction and determines the amount of product.

48 Excess Reactant It is the reactant in which there is more than what is needed for the reaction to take place.

49 To determine the limiting reactant. Take mass used and divide by molar mass of substance. Take the moles of each substance and develop a mole ratio. Compare to chemical equation.

50 S 8 (l) + 4Cl 2 (g) 4S 2 Cl 2 (l) 200g of S 8 with 100g of Cl 2 100g 70.91g/mol=1.41mol 200g 256.5g/mol=0.7797mol 1.41molCl 2 /0.7797molS 8 = 1.808molCl 2 /1molS 8

51 To find the grams of product, take the moles of limiting reactant and multiply the mole ratio of product to limiting reactant. This give you moles of product, take this and multiply by its molar mass.

52 How many grams of S 2 Cl 2 is produced? 1.41molCl 2 X 4molS 2 Cl 2 /4molCl 2 =1.41molS 2 Cl molS 2 Cl 2 X 135g/molS 2 Cl 2 =190.4g S 2 Cl 2

53 To find the actual amount of excess needed, take moles of limiting reactant and multiply mole ratio of excess to limiting reactant. Take moles of excess and multiply by molar mass.

54 How much of the excess is actually reacted? 1.41molCl 2 X 1molS 8 /4molCl 2 =0.3525molS molS 8 X 256.5g/mol S 8 =90.42gS 8

55 Theoretical Yield The maximum amount of product that can be produced from a given amount of reactants.

56 Actual Yield The amount of product actually produced when the chemical reaction is carried out in an experiment.

57 Percent Yield It is the ratio of the actual yield to the theoretical yield expressed as a percent.

58 Percent Yield = (actual yield/theoretical yield) X 100

Chapter 3. Mass Relationships in Chemical Reactions

Chapter 3. Mass Relationships in Chemical Reactions Chapter 3 Mass Relationships in Chemical Reactions In this chapter, Chemical structure and formulas in studying the mass relationships of atoms and molecules. To explain the composition of compounds and