Chemical Calculations: The Mole Concept and Chemical Formulas Chapter 9 – Chem 160 Chemical Calculations: The Mole Concept and Chemical Formulas

The Law of Definite Proportions Compounds are pure substances and They are a chemical combination They can be broken down They have a definite, constant elemental composition

Problem 9.5 – which two of three experiments produced the same compound? X & Q react to produce two different compounds, depending on conditions. Which two of the three experiments produced the same compound?

Problem 9.5 cont’d Experiments 1 & 3

Calculation of Formula Masses Definition Calculation

Example 9.3 a) C7H6O2

Significant Figures and Atomic Mass Uncertainty in data vs. uncertainty in formula mass We’ll use atomic masses rounded to hundredths Consider formula mass calculations as a pure addition

Percent composition Definition Calculation - % composition of Au(NO3)3 % composition by weight

The Mole Avogadro’s number # of molecules in 1.20 moles of CO 1.20 moles CO x 6.022x1023 CO molecules 1 mole CO Cancel “moles CO” Answer 7.2264 x 1023 – what units? Significant figures?

Mass of a Mole Molar mass of an element is … Molar mass of a compound is …

Problem 9.37 (d) Mass of 1.357 moles of Na3PO4 3 x Na = 3 x 22.99 = 68.97 g P = 30.97 g 4 x O = 4 x 16.00 = 64.00g 1 mole 68.97+30.97+64.00=163.94 g 1.357 mole x 163.94 g = 222.466 g Note significant figures

Significant Figures and Avogadro’s Number The mole is the amount of substance … Avogadro’s number should never be the limiting factor in s.f. considerations

A.M.U. and Gram Units 6.022 x 1023 amu = 1.000 g Proof 6.022x1023 atoms N x 1 mole N x 14.01 amu 1 mole N 14.01 g N 1 atom N 6.022 x 1023 amu/g Note that we can change “N” with another atom and corresponding atomic mass; the cancellations will still occur and we will get the same answer.

A.M.U. and Gram Units (cont’d) What is the mass, in grams, of a molecule whose mass on the amu scale is 104.00 amu? (Example 9.8) 104.00 amu x 1.000 g 6.022 x 1023 amu 1.7270009 x 10-22 g 1.727 x 10-22 g accounting for s.f.

Counting Particles by Weighing Atomic ratio to mass ratio Table 9.2 Cl (35.45) / Na (22.99) Extension to molecules

Counting Particles by Weighing – Problem 9.58 b) Grams of Cu that will contain twice as many atoms as 20.00 g of Zn 20.00 g Zn x 6.022 x 1023 atoms x 63.55 g Cu 65.41 g Zn 6.02 x 1023 atoms = 19.43128 g Cu contains ….. as many atoms as 20.00 g Zn Answer is 19.43 x 2 = 38.86 g Cu

Mole and Chemical Formulas Microscopic level interpretation Macro level

Mole and Chemical Formulas – Problem 9.60 6 mole to mole conversion factors from the formula K2SO4 2 moles of K / 1 mole of K2SO4 1 mole of S / 1 mole of K2SO4 4 moles of O / 1 mole of K2SO4 2 moles of K / 1 mole of S 2 moles of K / 4 moles of O 1 mole of S / 4 moles of O

Mole and Chemical Calculations Two general types of problems that this pathway will help in solving. Calculations for which information (moles, grams, particles) is given about a particular substance, and information (moles, grams, particles) is needed concerning the same substance. Calculation for which information (moles, grams, particles) is given about a particular substance, and information is needed concerning a component of that substance. For the first type of problem, only the left side of the figure (A boxes) is needed. For the second type of problems, both sides of the diagram are used.

Mole & Chemical Calculations Problem 9.70 d) Mass of 989 molecules of H2O {(2x1.01)+16.00} g x 989 molecules 6.022 x 1023 molecules = 2.95945 x 10-20 g Text answer is 2.96; must use “3” significant figures for 989. I took it to be an exact number. Can you go from the mass of an element to the mass of a compound and vice versa?

Purity of Samples Definition Problem 9.86 a) calculate the mass in grams of Cu2S present in a 25.4 g sample of 88.7% pure Cu2S 25.4 g of sample Cu2S x 88.7 g Cu2S 100 g of sample Cu2S = 22.5298 g

Empirical and Molecular Formulas Empirical formula – smallest whole number ratio of atoms Molecular formula – actual number of atoms in a formula unit Table 9.4

Empirical and Molecular Formulas – cont’d Problem 9.96 a) write empirical formula for P4H10 P2H5 9.96 d) C5H12? No change Table 9.4

Determination of Empirical Formulas Elemental composition data Empirical formula + molecular mass

Empirical and Molecular Formulas – cont’d Problem 9.98 b) determine the empirical formula if 40.27% K, 26.78% Cr and 32.96% O Table 9.4

Empirical and Molecular Formulas – cont’d We would use the same approach to solve this type of problem if weights rather than weight percentages are given. Empirical formula for 9.98 b) is K2CrO4

Determination of Molecular Formulas Molecular mass information needed molecular formula = (empirical formula)x; where x is a whole number x= molecular formula (experimental) empirical formula(calculat’d from atomic masses) For example (CH2)5 = C5H10

Determination of Molecular Formulas – 9.116 b) P2O3, empirical formula; molecular mass is 220 amu P: 2 x 30.97 = 61.94 amu O: 3 x 16.00 = 48.00 amu Empirical formula = 109.94 amu Molecular/Empirical = 2.00 Molecular formula is P4O6

Determination of Molecular Formulas – 9.120 b) Citric acid, molecular mass 192 amu; 37.50% C, 4.21% H and 58.29% O Mole Citric acid molar mass Mass % of comp. Atomic mass of comp. Moles of comp. units g/mole g/100 g mole/g moles C 1 192 0.3750 12.01 5.995004 H 0.0421 1.01 8.003168 O 0.5829 16.00 6.9948 divide by