How to Calculate Molecular Weights of Compounds The Molecular Weight (also referred to as the Formula Weight) of a chemical compound is calculated by adding the atomic masses (weights) of the atoms (elements) that constitute the compound. Example: Calculate the molecular weight of sodium carbonate [Na2CO3] given atomic weights: C = 12 ; O = 16 ; Na = 23.

How to Calculate Molecular Weights of Compounds Continued Assumption: All sodium carbonate molecules are made up of elements in fixed combining proportions. Molecular weight of Na2CO3 = (2 x atomic weight of Na) + atomic weight of C + (3 x atomic weight of O) = (2 x 23) + 12 + (3 x 16) = 106

The Reference Atomic Weight In using a scale (chemical balance), it was possible to obtain the weight of one compound relative to another compound by examining chemical change. Should we take the weight of the hydrogen atom as 1? Should we take the weight of 16O as 16? Finally, it was decided to take the weight of 12C as exactly 12.

Determination of Atomic Weights by Mass Spectrometry Concept of positive and negative ions Concept of multiply charged ions Concept of gas-phase versus solution-phase ions

Ions Atoms (and indeed molecules) can lose or gain electrons to form electrically charged ions. The reasons they would want to do this will become clearer later. Elements from the left hand side of the periodic table (metallic elements) have a tendency to lose electrons to form positive ions or cations: e.g., sodium atoms can lose electrons to form sodium ions Na → Na+ + e- Magnesium atoms form doubly charged ions Mg → Mg2+ + 2e- Aluminum atoms form triply charged cations, Al → Al3+ + 3e-

Ions (cont.) Elements toward the right hand side of the periodic table (the non-metals) conversely tend to gain electrons forming negative ions or anions: e.g., chlorine atoms can gain one electron to form chloride ions Cl + e- → Cl- Oxygen atoms can gain two electrons, O + 2e- → O2-.

Ions (concluded) Some elements can exist in several different ionic states: e.g., iron can exist as Fe2+ (the ferrous ion) or as Fe3+ (the ferric ion) likewise chromium can take the form Cr3+ or Cr6+, etc. As we will see, polyatomic or molecular ions, e.g., NH4+, SO42-, H3O+, NO3-, OH-, are particularly important in the chemistry of solutions.

Relative versus Actual Atomic Masses Distinguish between: Relative atomic masses referenced to 12C set exactly equal to 12 and hence dimensionless Actual atomic masses measured in mass units (grams, kilograms, etc.) The missing link is Avogadro’s number: No = 6.0221420 x 1023 It is the number of 12C atoms in 12 g of pure 12C.

The Mole Concept Definition: One mole of something equals the amount that contains Avogadro’s number of that something. The atomic mass unit, u: By definition 1 u equals exactly one twelfth of the mass of a single atom of 12C. Hence, 1 g = No u = 6.0221420 x 1023 u

Use of the mole: It follows that: 1 mole of He contains N0 atoms. 1 mole of water (H2O) contains N0 molecules (actually 2N0 H atoms and N0 O atoms) The mass of a mole of substance is called its Molar Mass with symbol M . Hence MC = 12.00 g mol-1 (if pure 12C) The same holds for molecules and M (H2O) = 18 g mol-1.

Calculating Amounts It follows trivially that, if we have N atoms of an element we have an amount, n, given by More usually though we need to calculate how many moles of a substance we have based on the mass of a substance we have:

Example: How many moles are there in 9 g of 4He gas? MHe = 4.00 g mol-1 (because 1 He atom has mass 4.00 u) Hence 9 g is 9/4 = 2.25 mol which is 2.25 x 6.022 x 1023 = 1.355 x 1024 atoms What is the mass of 5 moles of Kr gas (the relative atomic mass of natural Kr is 83.80)? Since RAMKr = 83.80, the molar mass is MKr 83.80 g mol-1. Hence 5 moles has mass 5 mol x 83.80 g mol-1 = 419 g

Radioactivity: Not All Isotopes Are Stable  In 1899, Ernest Rutherford discovered that uranium compounds produce three different kinds of “radiation.” He separated the rays according to their penetrating abilities and named them alpha, beta, and gamma radiation, after the first three letters of the Greek alphabet. The  particles can be stopped by a sheet of paper. Rutherford later showed that an alpha particle is the nucleus of a He atom, 4He. The  particles were later identified as high-speed electrons. Six millimeters of aluminum are needed to stop most  particles. Several millimeters of lead are needed to stop  rays , which proved to be high-energy photons.

Balanced Nuclear Reactions When it is undergoing nuclear decay, radium emits alpha particles and actually becomes another element. We represent this nuclear reaction by the following equation: 22688Ra → 42He + ? What is the unknown element? Subtraction tells us that it must be the element whose atomic number is 86 and whose mass number is 222. We look at the periodic table and find that element 86 is radon (Rn). So we can write the reaction in this manner: 22688 Ra→ 42He + 22286Rn In a balanced nuclear equation, the sum of the superscripts, and also the sum of the subscripts, are the same on both sides of the arrow.

Nuclear Fusion and Nuclear Fission Fusion is a nuclear process in which two light nuclei combine to form a single heavier nucleus. Fission is a nuclear process in which a heavy nucleus splits into two smaller nuclei.

Nuclear Fusion and Nuclear Fission Can Provide the Release of Energy

Example: Binding Energy All stable nuclei (and thus atoms) have a mass less than the mass of their constituents. e.g., By definition, the mass of a 12C atom is exactly 12 u. Consider the constituent particles: 6 protons: mass = 6 x 1.00728 u = 6.04368 u + 6 neutrons: mass = 6 x 1.00867 u = 6.05202 u + 6 electrons: mass = 6 x 0.00055 u = 0.00329 u total = 12.09899 u In other words we have a “mass defect” = 0.09899 u

So Why the Difference? The 0.09899 u discrepancy (or mass defect) arises because: Nuclear binding energies are enormous (they have to bind all the positively charged protons in a minute volume) Mass and energy are equivalent (E = mc2 Einstein, 1905) (where c is the speed of light in vacuo c = 2.998 x 108 ms-1) The nuclear binding within the 12C nucleus is so large as to result in an a loss of mass. Binding energy in 12C = (0.09899 u) x c2 = 1.477 x 10-11 J (or 8.897 x 109 kJ mol-1 – enormous compared to chemical energies)